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\tcTfW ^6l ,0l.^
HARVARD COLLEGE
SCIENCE CENTER
LIBRARY
i
COLLEGE ALGEBEA
BY
JAMES HARRINGTON BOYD, Ph.D.
Hie University of Chicago
^ . '"^^'^^iy narrow ^
PAN^
./ c ^ L !. u : I 5
\^.. -y
COPTRIOHT 1901
By
SCOTT, FORESMAN AND COMPANY.
To
Chaelbs Myeon Yooum
Axn
Bellb Boss Yogum
D. D. D.
PREFACE
The selection and arrangement of topics in this book and the
method of treatment represent that which the author's experience
has led him to believe is best adapted to the requirements of a good
college course. Of the topics usually treated in college algebras,
continued fractions, choice and chance, and probabilities have been
omitted, since few applications of these subjects are made until
the student reaches more advanced courses. Their omission also
gives time for more extended study of such topics as irrational
and complex numbers, series, the elementary properties of deter-
minants, and the properties and solutions of numerical equations of
higher order.
Throughout the book, great care has been taken to secure that
rigor and logical sequence which are being demanded by the best
teachers. Special attention is called to the mode of presenting the
number concepts in the earlier part of the book, to the use of
geometrical illustrations, and to the extensive collection of exer-
cises, in lai^e part hitherto unused in American colleges.
In developing the fundamental laws and theorems of the num-
ber system of Common Algebra, the notation used by H. B. Fine
in the first articles of his book The Number System of Algebra
has been adopted for the sake of uniformity. Since the discovery
of Quaternions by Hamilton, of Linear Associative Algebra by
Benjamin Peirce, and the Ausdehnungslehre by Grassman, mathe-
maticians have generally accepted the doctrine that Algebra is
completely defined formally by the laws of combination which the
fundamental operations are required to obey.
In the development of the principles of the number system of
Common Algebra, emphasis is placed upon the principle of the
▼* PREFACE
permanence of form, the discussion of the irrational, the theory
of fractional exponents, -and complex numbers. The principle of
the permanence of form arises in extending the fundamental laws
and operations which are applicable to positive integers to the new
numbers which arise in Algebra— zero, the negative, the fraction,
irrational and complex numbers. This point of view was first
suggested by Peacock in his Arithmetic and Symbolic Algebra in
1842, and fully developed for the negative, the fraction, and the
imaginary by Hankel in his Complexe Zahlengystemey in 1867, and
was completed by Cantor's theory of the irrational in 1871.
A careful distinction is made between an equation and an identity.
It is pointed out that the solution of an equation or a system of
equations depends upon one's ability to construct equivalent equa-
tions or equivalent systems of equations ; that the same is true in
solving inequalities or systems of inequalities. The great central
problem of Algebra is the solution of the equation, and with the
irrational and complex numbers the system of algebraic numbers is
complete. Full discussions are given of equivalent systems of
simultaneous quadratic equations, of the graphs of their solutions
and of the equations themselves, and of problems in maximum
and minimum values of fractions which can be solved by means
of quadratic equations.
Following the principle of the permanence of form and the equa-
tion, are infinite series and their properties, the tests for their con-
vergence and divergence, the expansion of fractions into infinite
series and their summation. Because of their value in discussing
the properties of infinite series and other problems, emphasis is
placed upon the method of mathematical induction, upon the prop-
erties of a variable and its limit, and upon the rigorous proof of the
theorem of undetermined coefiadents.
Attention is directed to the fact that the sum of an infinite
series is the limit of a variable sum; to the geometric illustrations
of the derivation and meaning of each theorem ; to the distinction
between absolutely and conditionally convergent series; to the fact
PREFACE ^»
that an absolutely convergent series may be treated like any other
number in algebraic calculation, — a property which a conditionally
convergent series does not have; and finally to the value of infinite
convergent series in numerical calculation.
The author herewith expresses his obligations to the many others
who have preceded him in this field, some of whose works he has
used in the classroom for many years. He would mention espe-
cially Dr. E. Bardey, who courteously granted permission to use
exercises from his Au/gabensammlung, He also desires to express
his indebtedness to Dr. E. R. Moulton for his special care in read-
ing proofs of the book, to Dr. Henry Gale for his assistance in
reading the manuscript and proofs of Book IV, and to Mr. A. W.
Smith for critically reading a part of the manuscript.
James Harrington Botd.
The University of Chicago,
May, 1901.
TABLE OF CONTENTS
Preface
Book I
The Fundamental Operations op Common Algebra —
Their Laws and Applications
INTRODUCTION
The Nature of Nambers and the Fundamental Postulate of Arithmetic 11
The Equality of Two Groups . . 12
Symbolic Representation of Numbers 13
The Equation and Inequalities 14
Counting 15
CHAPTER I
Addition and Multiplication
The Sum of Two Groups— Parenthesis 17
Addition of Po<^itive Integers and its Fundamental Laws . . .18
Multiplication of Positive Integers and its Fundamental Laws . . 19
The Exponent and Index Law of Multiplication 21
Addition and Multiplication of Monomials and Polynomials . 24
CHAPTER II
Positive and Negative Numbers — AoDmoN and Subtraction
Definition 27
Series of Natural Numbers 28
Positive and Negative Numbers 29
The Absolute Value of Numbers — Unlike Signs 30
Addition of Algebraic Numbers, Rule 31
Addition of Monomials 32
Subtraction of Algebraic Numbers, Rule ...... 35
Subtraction of Monomials 36
ix
z COLLEGE ALGEBRA
CHAPTER III
Subtraction and tbs Nbqativib Integer — Generauzed DiscuasioN
Numerical Subtraction 39
Determinatenessof Numerical Sabtraction— Formal Rules of Subtraction 40
Zero and the Negative .' • 42
Limitation of Subtraction— Symbolic Equations .... 45
Principle of Permanence 46
Subtraction of Polynomials 48
Use of Parentheses 50
CHAPTER IV
Positive and Negattvb Nuicbers — Multiplication
Multiplication of Monomials 63 .
Multiplication of Polynomials 54
Polynomials in Ascending and Descending Powers of x . .55
CHAPTER V
POSFTIVB AND NbQATIVE NuMBERS — DfVIBION
Numerical Division 57
Determinateness of Division 58
Formal Rule of Division— The Index Law 59
Exponent Zero — Division of Monomials 60
Divisionof Polynomials— First and Second Rules . . .61
Third and Fourth Rules of Division 67
Indeterminateness of Division by 0 — ^Determinateness of Symbolic Division 69
The Vanishing of a Product 70
CHAPTER VI
Applications of the Fundamental Operations — Simple Equations
An Identity— An Equation of Condition— The Unknown Quantity 71
Axioms 72
Transposition 73
CHAPTER VII
Application of Addition and Multiplication— Powers of MoNOMiAiii —
Binomials
Definitions: Rational Integral Polynomials, etc. . .79
Power Formulae — Tlie Double Sign 80
Binomial Theorem 82
CHAPTER VIII
Factoring and Solution of Equations by Factoring
Case I: To Factor a Polynomial 86
Case II: To Factor a Trinomial 87
Case III: To Factor the Difference of Two Squares ... . .88
TABLE OP CONTENTS xi
OaaelV: To Factor the Sum and Difference of Two Cubes . . 89
GaaeV: To Factor the Trinomial a;* + par +9 .... 90
C&se VI: To Factor Polynomials 92
Remainder Theorem 93
Fbcior Theorem 94
Case YII: To Factor Polynomials which have a Binomial Factor . 96
CaaeVIII: To Factor a Polynomial of Four Terms ... 98
Solution of Equations — Imaginary Numbers 99
CHAPTER IX
Greatest Common Divisor and Least Common Multiple
G. C. D. of Quantities Readily Factored 101
G. C. D. of Two or more Algebraic Expressions .... 102
Least Common Multiple of Quantities readily Factored and those not
readily Factored 109
CHAPTER X
Fractions
Definitions: Rational Fractions, etc. 113
Rule of Signs 114
Reduction of Fractions to Lowest Terms 115
Reduction of Fractions to a Lowest Common Denominator . 121
Addition and Subtraction of Fractions 124
Multiplication of Fractions .... ... 129
Powers of Fractions 131
Division of Fractions 133
Complex Fractions 134
Continued Tractions 136
Special Theorems in Fractions ........ 139
Book II
Solution op Equations of the First Degree
CHAPTER I
Equations of the Firot Dbgreb
•
Identity— The Equation 145
Root of an Equation — Degree of an Equation .... 146
Equivalent Equations 147
Theorem on Transformation of Equations 147
Removal of Denominator 151
The Solution of an Equation of the First Degree .... 154
Formulae for the Solution of an Equation of the First Degree in One
Unknown Quantity 158
xli COLLEGE ALGEBRA
CHAPTER II
Problems which lead to Equations of the First Degree
Problems Solved for Illustration 164
CHAPTER III
Literal Equations in One Unknown Number
Numerical and Literal Equations 175
Exercise for Illustration 176
CHAPTER IV
Problems Involving Literal Equations
Formulae and Rules 180
Problems Solved for Illustration 182
CHAPTER V
Interpretation of the Solution of Problems
Positive and Negative Solutions 187
^iero Solutions 190
Indeterminate Solutions 192
Infinite Solutions 193
CHAPTER VI
Simultaneous Linear Equations in Two Unknown Quantities
Indeterminate Equations 198
Independent Equations 199
Incompatible Equations 200
Equivalent Systems of Equations 201
The Solution of two Equations of the First Degree — Elimination by
Addition and Subtraction 204
Elimination by Substitution 207
Elimination by Comparison 212
Elimination by Undetermined Multiplier 218
CHAPTER VII
General Solution of a System of Two Equations
General Solution 230
Composition of Formulae 231
Symmetry of Equations 232
Discussion of Cases a6^ — 6a^=^ 0, etc 233
Homogeneous Equations 236
Two Equations of the First Degree which have a Common Root . 237
TABLE OF CONTENTS 3dii
CHAPTER VIII
General Solution ok a System of Three Equations
Examples of Three Equations of the First Degree .... 238
Number of Solutions of a System of n Linear Equations . . 242
Problems involving Three or More Unknown Numbers . - 251
CHAPTER IX
Graphical Reprebentation of Points and Lines
Graphical Representation of a Point 256
Graph of the Solution of a Conditional Equation .... 259
Graph of the Line y = mx+b 262
Intercepts 264
Intersections of Pairs of Lines . . . 265
CHAPTER X
Diophantian Equations of the First Degree
Simple Indeterminate Equations 268
Rule for Solution of aa:+6y=c 270
Given One Solution of ax — by =c in Positive Integers, to Find the
General Solution 272
General Solution of Two Equations 274
Book III
Involution— BvoLUTiON — Fractional and Negative Exponents
CHAPTER I
Involution
Index Law of Involution 277
Powers of Monomials, Fractions, Binomials, and Polynomials • 278
CHAPTER II
Evolution
Definition of a Root 283
Law of Signs of Roots of Quantities 284
Principal Root .286
Theorems in Evolution 287
Square Root of Compound Quantities 290
Square Root of Arithmetical Kumhers 295
Cube Root of a Polynomial 299
Oabe Root of Arithmetical Numbers 303
xiv COLLEGE ALGEBRA
CHAPTER III
Fractional and Negative Exponents
Distributive and Associative Formulae' 909
Principles 1, 2, 3, 4 312
Theorems: I, a'"Xa*=a'"+"; II, a"'+a"=rt'"-"; III; IV; V; etc. . . 313
Multiplication and Division of Fractional and Negative Powers . 321
CHAPTER IV
Relative Magnitude of Positive and Negative Quantities
Conventions concerning Relative Magnitudes 323
Inequalities between Two Algebraic Expressions — ^Theorem I, II . 324
Solutions of Inequalities of the First Degree in One Unknown Quantity 330
Special Theorem, III 331
Powers and Roots 334
CHAPTER V
Irrational Numbers and Limits
System of Rational Numbers Insufficient 337
Introduction __ 338
Properties of the Series which define r V ^ 343
Limit 343
Irrational Numbers 344
Zero, Positive, and Negative 345
Addition, Subtraction, Multiplication, and Division of Irrationals . 345
Principles of Limits 348
CHAPTER VI
Surds
Definitions: a Radical, a Surd, etc 349
Orders of Surds — Reduction of Surds 350
Addition and Subtraction of Surds 355
Multiplication of Surds 359
Division of Surds S63
Rationalization of Surds 366
Reduction of Special Irrational Expressions 368
CHAPTER VII
Imaginary and Complex Numbers
Pure Imaginary Numbers and their Properties . . . 373
Addition, Multiplication, Division by i and Powers of t . . 874
Addition, Subtraction, Multiplication, and Division of Pure Imaginaries 376
Complex Numbers 379
TABLE OF CONTENTS xv
Book IV
Theory of Equations of the Second Degree
CHATTER I
Introduction
Theorems I, II. Ill, IV, V 385
^CHAPTER II
Solution of Equations of the Second Degree
Solution of ox =6 and aa:»= 6 388
Solution of ox* + &a:+c = 0 392
CHAPTER III
Equal and Imaginary Roots
Imaginary Roots 398
Real, Equal, and Imap^nary Roots, Condition for . . . 400
Solution of the Equation j^+px+q^O and Condition for Real,
Equal, and Imaginary Roots 401
CHAPTER IV
Factoring of a Trinomial
Yaclonoi j^+px+q 404
FactOTSof ax*+&i4-c and ax* + 2 6j:y + cy"+2<ir+2«y+/ . . 406
CHAPTER V
Problems in Equations op the Second Degree in One Unknown Quantity
General Problems 409
Problems connected with the Theorem of Pythagoras . . . 412
Problems concerning the Areas of Plane Figures . . .418
CHAPTER VI
Relations between the Coefficients and Roots of a Quadratic Equation
An Equation of the Second Degree has but Two Roots . • . 422
Rules concerning the Sum and Product of the Roots . . . 423
Properties of the Roots 424
CHAPTER VII
Equations which are Reducible to the Second Degree
TheBiquadratic Equation or* + />jc*+c = 0 . . c . . 427
Irrational Equations . . 428
8olutionofax* + 26x + 2IV^aa;« + 26a;+c = p and ox*" + ftx" + c = 430
Beciprocal Equations 43 L
xvi COLLEGE ALGEBRA
CHAPTER VIII
Factorization
F9ctorEo(x* + p3*+g 438
Cases I, II, III 439
CHAPTER IX
Roots of Surds ^
Transformation of Ka± 1/6 442
Lemma on a+v^6 = a'' + 1/6'', etc 443
Va + Vb+Vc+V^d and Va+V6 44<5
The Variation in Sign of or*-}- 6.r+c 448
CHAPTER X
Systems ok Simultaneous Equations Involving Quadratics
Type I 4oI
Type II 453
Type III 456
Irrational Simultaneous Quadratics 457
CHAPTER XI
Systems op Three or More Equations Involving Quadratic Equations
Special Methods 466
Ftoblems in Simultaneous Quadratic Equations .... 480
CHAPTER XII
Graphical Representation of Solutions op Systems of Simultaneous
Quadratic Equations
Graph of aa:*+6j: + c 488
Type I, Gr2Lph8otAj^ + 2Bxy-^Cy* + 2Dj' + 2Ey'\-F=0 . 480
Type II, Graphs of or* + bx-y +ci/=d and Ai^ + Ihy + (y = D . 491
Graphs of Miscellaneous Forms and Exercises .... 493
CHAPTER XIII
Graphs and Problems in Maxima and Minima
Definition of Maxima and Minima 405
Graph of u = ^((LV — ji^) 498
Graph of y = ^^^ 505
Graph of y = {cu^+bx+c) + (biX'\-Ci) 508
Graph of y = ((u^+bx+c)-+-{ai3^'\-bix+ci) . , , . 509
TABLE OF CONTENTS acvli
Book V
Peoportiomt — Variation — Progression — Loo arithhs —
Arrangements and Combinations
CHAPTER I
Ratio and Proportion
Definitions 517
Properties of Proportions 518
Incommensurable Numbers 522
Application of Quadratic Equations and Ratio and Proportion to
Geometry 528
CHAPTER II
Variation
De6nition 533
Formulae (i), (ii), (iii), (iv) 534
Problems 536
CHAPTER III
Progression
Arithmetic Progression 538
Formulae (i), (ii), (iii), (iv) 539
Geometric Progression 543
Formulae (i), (ii), (iii), (iv) 544
Infinite Geometric Progression (iv) 549
Value of Recurring Decimal 550
Geometric Means .551
Harmonical Progression and Definitions 552
CHAPTER IV
Logarithms
Definitions 557
Properties of Logarithms 558
(Comparison of Systems of Logarithms . . . . v 560
Common System of Logarithms and Rules 563
Use of Table of Logarithms 565
Solution of Numerical Problems by Logarithms, Powers, Roots, etc. 571
Exponential Equations . . 575
Compound Interest— First Convention — Second Convention 578
Annuities: Contingent, and Forborne 585
Refunding of a Debt by Annuities 588
xviu COLLEGE ALGEBRA
Book VI
Induction — Arrangement — Combinations
CHAPTER I
Mathematical Induction
Sum of the First n Integers 591
Sum of the First n Odd Integers 592
Proof of the Binomial Theorem for Positive Integral Exponents . 593
Steps in the Mode of Mathematical Induction .... 595
CHAPTER II
Arrangements and Combinations
The Problem of the Number of Arrangements of n Things taken r at
a time and 7i at a time 59S
Permutations 599
Circular Permutation 600
Combinations and Special Theorems 601
Problems 606
CHAPTER III
Formula for the Expansion of a Binomial
Product of n Different Binomial Factors 610
Characteristics of the Binomial Expansion 612
Maximum Coefficient of Binomial Expansion 615
Summation of the Same Powers of Numbers which are in A. P. . 617
Barbier' 8 Theorem 619
Book VII
Limits and Series
CHAPTER I
Limits
Constants — Variables — Limits 623
Infinitesimals and Infinites— Theorems concerning the same , 625
CHAPTER II
Convergence
Definition of an Infinite Series 630
Convergence of Series whose Terms are Positive — Graphical Repre-
sentations 631
Fundamental Theorems 632
Standard Series for C<>mpari8on Tests of Convergence, 1 +^+ Aj+i+* • • 634
TABLE OF CONTENTS xix
Harmonic Series 636
Ratio Test for Convei^nce ^ . 637
Alternating Series 639
General Theorem 642
Absolutely and Conditionally Convergent Series; Caution 646
CHAPTER III
Undetermined CoEFFiaENTs
Theorems I, H, HI, IV 650
Development or Expansion of Functions 653
Development of Irrational Expressions 655
Reversion of Series 657
CHAPTER IV
The Binomial Theorem for any Exponent
Convergence of the Binomial Expansion 662
Extraction of Roots by the Binomial Theorem .... 665
CHAPTER V
Partial Fractions
Case I, II, HI 667
CHAPTER VI
Exponential Series, Logarithmic Series, and De Moivre's Theorem
The Table of Napierian Logarithms 6.76
The Number called e in Mathematics is Incommensurable 677
Remainderof the 6 Series— Calculation of c 678
Error in the t Series — Special Exponential Forms . . . 679
De Moivre's Theorem and Certain Series which may be deduced from it 681
Value of » Calculated by means of Gregory's Series — Enter's Series 686
CHAPTER VII
The Summation of Series
Recurring Series, Definition— Scale of Relation .... 689
Order of a Recurring Series — To Find the Scale of Relation . 690
Summation by Undetermined Coefficient-s .... 695
Series which may be given the Form of the Alternating Series . 696
Summation of Arithmetical Series of Higher Order .... 698
Miscellaneous Series 701
CHAPTER VIII
Limiting Values of Indeterminate Form
Definition of the Value ^ . . 705
Indeterminate Form ^ 707
Table showing the Region of Convergence of Series . * . .710
zx COLLEGE ALGEBRA
Book VHI
chapter i
Introductory Chaptkr on the Theory of Detehinantb
Determinants of Two Rows — Determinants of Four Elements — Prin-
ciple of Development of r 711
TheEliminant 712
Solution, of Two Equations of First Degree when Determinant of Un-
known Numbers is not 0 713
Homogeneous Equations 714
Determinants of Three Rows — Determinants of Nine Elements . 715
Relation between Determinants of the Second and Third Orders 717
Propertiesof Determinants — First Property .... 718
Second and Third Properties . 719
Fourth Property .... 720
Properties of Minor Determinants 721
First Property .... 722
Second Property 723
Solution of Equations of the First Degree in Three Unknown Quantities 723
Principle of Addition of Rows or Columns 727
Product of Two Determinants of the Third Order . . - . 730
Determinants of the Fourth Order 731
Elimination 733
CHAPTER II
The Cubic Equation
Definition of the General Equation of the n^ Degree-— A Root . . 738
Definition of Cubic Equation —Cube Roots of Unity . . 737
The Equation 3^=a — Symmetrical Cubic Equation — Cubic Equation
with One Rational Root 738
Cardan's Solution 739
Trigonometric Solution 742
Trigonometric Solution of Cubic Equations with Two Imaginary Roots 744
CHAPTER III
The Equation of the Fourth Degreb
Resolvent Cubic 746
CHAPTER IV
The n Roots of Unfty
Solution of Equation x^=l 752
Cube and Biquadratic Roots of 1 — Quinary Roots of 1 — Geometrical
Representation of Complex Numbers by Points . . . 753
Modulus— Argument — Sine 754
Cosine— Geometrical Representation of then Roots of 1 . « • 755
TABLE OF CONTENTS
xzi
CHAPTER V
Thboby of Equations
Properties of Eqaations
Fractional Eoots— Transformation of Equations .
Synthetic Division •
Standard Form of Reciprocal Equations
Deecartes's Rule of Signs
Derived Functions
Equal Roots
Continuity of a Rational Integral Function of :r .
Sturm's Function and Theorem
Calculation of Incommensurable Roots by Horner's Method
767
761
764
767
768
770
771
772
775
777
Ikdkz
783
INTRODUCTION
1. In the study of Common Algebra the notion number is funda-
mental, and it is therefore necessary first of all to define it. What
are the properties or characteristics of number?
Given a group of objects, as marbles, a party of boys, a herd of
horses, a village, or the like, distinctness or separateness of the things
in any one of these groups is an intuitive property of these objects
which enables one to realize that there is a marble, boy, horse, or
house which is different and distinct from another marble, boy,
horse, or house of the same group of marbles, boys, horses, or
houses. If each marble of the group of marbles were replaced by
an apple, then each apple by a nail, and so on; or if the marbles
were painted different colors, arranged differently; or finally if any
change were made in the things of the group which would not
destroy their distinctness, the group of objects would Contain as
many individuals after any such change as it did before the change
was made.
The notion of number is based upon this property of the separate-
ness of the things in a group, and is defined as that property of a
group of different things which is unchanged no matter what change
is made in the things of the group without destroying the distinct-
ness of the individual things.
Such changes affect only the character or arrangement of the
things and do not cause any individual thing to be divided into two
or more, or two or more to be merged into one. These characteristics
of number expressed in the form of a theorem constitute the funda-
mental postulate of Arithmetic:
The number of individual things in a group of things does not depend
upon the order of their arrangement in the groupy their character-
istics, or the toay they may be related to one another in smaller
groups.
12
COLLEGE ALGEBRA
[«2
2. The Equality of Two Groups. — Consider two groups of letters,
1 and 2.
On comparing the individual objects of the groups 1 and 2, we
may assign ^ to a, B to h, C to c, I) to d, and E to e, and recipro-
cally, ato A, b to B, cto Q dto Dy and 6 to ^, i. e. , there are just
as many things in group 1 as in group 2.
The number of things in two groups of things is the same if to every
thing in the first there may be assigned one in the second, and, rexdpro^
cally, to every thing in the second there can be assigned one in the first.
Such a relation is called a one-to-one correspondence.
In case of the groups of letters 3 and 4:
a b
d
« 3 4
to every letter in 4 may be assigned a letter in 3, thus, a to ii, 6 to
B, c to C, but, reciprocally, a letter of 4 can not be assigned to each
letter of 3, thus: A to a, B to b, C to c, since the letters of group 4
are exhausted and there remain no letters of this group which can
be assigned to the remaining letters d and e of group 3.
The number of things in group 3 is greater than the number of
things in group 4, or the number of things in group 4 is less than
that in group 3, when there is one thing in the first group for each
thing in the second, but not reciprocally one thing in the second for
each thing in the first
Reciprocally, the number of things in group 5 is less than the
number of things in group 6:
A B C Z) etc.
13]
INTRODUCTION
13
for, while it is possible to assign a letter of group 6 to each letter of
groap 5, thus, A to a, B to h, C to c, it is not possible, reciprocally,
to assign to each letter in groap 6 a letter of group 5, because after
a has been assigned to ^, 6 to ^, c to 6^, there remain no letters in
group 5 to assign to the remaining letters, Z>, etc., of group 6.
3. Representation of Numbers by Symbols. — When it is desired
to compare the number of things in several groups of objects (sheep,
cattle, horses, potatoes, bricks, etc.)) the convenience of practical
affairs demands that symbols be used to represent numbers — the
totalities of things in groups of objects.
The number of things in a group can be represented by another
group, e. g., by the fingers or any set of simple marks, thus:
The number of things in group I can be represented by any one of
the groups 1, 2, 3, since there is a one-to-one correspondence between
the objects of groups 1, 2, 3 and group I.
The difference between the primitive and modem methods of rep-
resenting groups of things is this: that the symbols in the middle row
below or the numeral words of the third row are respectively used,
14
COLLEGE ALGEBRA
General groups
^ \ /"A B
[34
// .
Groups of marks
/// ////
Modem symbols
3 4
one
two
Numeral words
three
four
/////
5
five
instead of the groups of marks in the first row, for the groups of
things in the circles to which the groups of marks respectively
correspond.
The positive integer is the symbol used to represent the number of
things in a group. For example, the symbols 1, 2, 3, 4, 5, in the
illustration given above are positive integers. The primary use of
the word number is that of the positive integer, a numerical symbol
denoting the totality of the things in a group.
In the paragraphs which immediately follow, 1, 2, 3, 4, etc., in
particular discussions, and the letters a, 2», c, etc. , in general dis-
cussions, are used as positive integers.
4. The Equation. — If a and h are the numerical symbols which
represent the number of things in two groups, and there is a one-to-
one correspondence between the objects of the groups, this relation of
a group 6 group
the two groups to each other is indicated by the symbolical relation,
az=h\
for example, 5 = 5,
85]
INTRODUCTION
15
which is called an equation or equality. The symbol = is read
equals, and the equation a = 2) is read a is equal to h. If the first
group is greater than the second, the relation of the groups is
a group
represented by the inequality,
b group
a>6j
7>5.
for example
If the first group is less than the second, the relation of the
b group
a group
groups is represented by the inequality,
for example 5 < 8.
A numerical equation simply declares, in term^ of the symbols
which represent tlie groups, the numei-ical relation which exists between
these groups; and the symbol ==, that these groups are in a one-to-one
correspondence.
6. Counting. — To count the things in a given group of things is
to seek a one-to-one correspondence between the things of this group
and the individual things of some group (or groups) which is known.
The fundamental operation in Arithmetic is counting.
Counting the things in a group leads to a numerical expression in
terms of the representative groups; if the representative group is a
group of marks, in terms of this group of marks ; if it is fingers, in
terms of the group of fingers ; if it is one of the numeral words or
symbols in common use, to one of these words or symbols.
16
COLLEGE ALGEBRA
[35
For example, since there is a one-to-one correspondence between
the letters of the group to the left and the marks of the group
to the right, counting the letters in the first groap leads to the
group of marks (or 9) which may be taken to represent the number
of objects in the group of letters.
BOOK I
CHAPTER I
ADDITION AND MULTIPLICATION
Addition
6. Two groups a and b
ABC
D E F G
H
a b
i^ed into a single group c
A B C D E F a
H
by removing the vertical bar which separates them.
The nmnerical symbol 8 of group c is the number of letters
in the groups a and 6, or the sum of their numerical symbols,
3 and 5. This result written in the form of an equation is
8 = 3 + 5
and is read eight is equal to the sum of three and five^ or is eqwil
to three plus Jive,
The sum-group 8 is formed by uniting group h, to which 5
belongs, to group a, to which 3 belongs.
In general, if a, 6, c, etc., are the numerical symbols for the
number of things in the groups 1, 2, 3, etc., then the number
6,
17
18 COLLEGE ALGEBRA. [W
corresponding to the sum-group including the a's, b'&y c's, etc., is
8 = (a + ^) + ^ + ^'^•
The sign of continuation, , is read and so on, thus,
1, 2, 3, 4, 5, 6,
is read one, two, three, four, Jive, six, and so on.
The symbol ( ) is read parentheses, and when it encloses the
sum or sums of two or more numbers it indicates that all within it
is to be treated as a single number.
s will be the numerical sum of the groups 1, 2, 3, etc., that is,
the number corresponding to the group d.
The sum-group is found by joining the group, 2, of b's to
the group, 1, of as, giving a group with the numerical symbol
{a-\-h); then the group, 3, of c's to the resulting group of the
a's and i's, giving the numerical symbol (a + 6) + c; and so on
(figure d).
Addition is the operation of finding 8 when 3 and 5 are given,
or of finding s when a, b, c, etc. , are given. That is, a group 8 has
been found as the result of bringing together the groups 3 and 5,
and a group s as the result of uniting the groups a, b, c, etc
Thus it follows that addition is abbreviated counting.
Addition, in consequence of its definition (addition of groups a
and b and of groups 1, 2, 3, etc.), is subject to the following laws,
called the Commutative and Associative Laws respectively, viz. :
In particular In general
I. Commutative
3+5=5+3 a+b=b+a
To add 5 to 3 is the same as To add 6 to a is the same as
to add 3 to 5, to add a to 6.
J7] ADDITION AND MULTIPLICATION 19
In particular In general
II. Associative
34. (5 -|_ 7) = 3 + 5 + 7 a + (6 + c) = a + 6 + c
To add the sum of 5 and 7 To add a to the sum of b and
to 3 is the same as to add 7 c is the same as to add c to
to the sum of 3 and 5. the sum of a and b.
These laws are the immediate consequences of the fact that the
sum-group (e. g., groups c and d, 26) contains the same things as
the individual groups combined, and therefore the number of things
in it will be the same, whatever be the order of combination in
which the different groups are brought together. (See Law of Arith-
metic, il.)
Multiplication
7. How is it possible to find the number of things there are in
a group which is formed by bringing together 7 groups, each con-
taining 5 things?
By addition the result would be the sum
5+5+5+5+5+5+5
which is written 5 • 7 or 5 X 7 = 35
where 7 indicates the number of 5's taken. The result 5 • 7 is
read the product of five by seven or seven times five.
Similarly, the sum of b numbers each Qqual to a is the product
of a by 6, and is written a - b or a x ^ or simply ab,
Holtiplication is the operation by which the product of a by fe is
found when a and b are known.
Thus multiplication is abbreviated addition.
5 and 7 are called the factors of the product 5 • 7 ; a and b the
factors of the product ab. The factor a is called the multiplicand^
and b the multiplier.
There are three laws, called respectively the Commutative, Asso-
ciative, and Distributive Laws for Multiplication, which are the con-
sequences of its definition, viz. :
In particular In general
III. Commutative
5 • 7 = 7 • 5 ab = ba
The product of 5 by 7 is the The product of a by 6 is the
same as the product of 7 by 5. same as the product of b by a.
20
COLLEGE ALGEBRA
[«7
lY. Associative
3(5 • 7) = (3 • 5) • 7 a(bc) = (ab)c
The product of 3 by 5 -7 is the The product of a by 6c is the
same as the product of 3 * 5 same as the product of ab
by 7. by c.
V. Distributive
3(5 + 7)=3-5 + 3-7 «(6+c) z^ab + ac
The product of 3 by the sum The product of a by the sum
of 5 and 7 is the same as the of b and c is the same as the
. sum of the products of 3 by sum of the products of a by
5 and 3 by 7. b and a by c.
These laws are the consequences of the commutative and associ-
ative laws of addition.
The Commu&tive Law (III). — The units of group 1, which cor-
respond to the sum of 7 numbers, each equal to 5, may be arranged
in 7 rows of 5 marks each. But in such an arrangement there are
Columns Columns
Rows
Rows
1 2
5 columns containing 7 marks each; so that if the same gronp of
marks be arranged by columns instead of rows, the sum becomes
that of 5 numbers of 7 marks each, or 7 '5.
In general, the things of a group which correspond to the sum
of b numbers each equal to a may be arranged in J) rows of a things
in each. But in such an arrangement there will be a columns of b
units each; so that if the same group of things be arranged by col-
umns instead of rows, the sum is that of a numbers each equal to
b or ba.
The Associative Law (IV). —
3-5-7 = 7 sums (3 + 3 + etc., to 5 terms)
= 3 + 3 + 3 + etc. , to 5 • 7 terms (by the associative
law for addition)
^ 3 (5 • 7) (by the definition of a product, {7).
81 8, 9] ADDITION AND MULTIPLICATION 21
The Distributive Law (V).—
3 (5 + 7) = 3 + 3 + 3 + etc. , to (5 + 7) terms (by the definition
of a product)
= (3 + 3 + 3 + etc. , to 5 terms) -f (3 + 3 + 3 + etc. ,
to 7 terms) (by the associative law for addition)
= 3 • 5 + 3 • 7 (by the definition of a product).
The commutative and associative laws for the product of any
number of factors, and the distributive law for the sums of any
number of terms, follow immediately from the laws I-V, ??6,
7. Thus, the product of the factors 3 • 5 • 6 • 7 • 9 taken in any
two orders is the same, since any one order can be transformed into
the other by successively interchanging the consecutive figures.
For example,
3 • 5 • 6 ■ 7 • 9 is the same as 3 ■ 7 • 9 • 5 • 6
for,
3-7-9-5-6 = 3-7-5-9-6 = 3-5-7-6-9
= 3 -5 -6 -7 -9 (by III, «7).
Or more generally, ahcd is the same as adcb : for adch = acdh = achd
= abed {hj III, 27).
8. The Exponent. — Suppose that the factors of a product are
all equal, for example, 4 • 4 • 4. This product is written
4 • 4 • 4 = 43
whert the figure ^ indicates the number of times 4 is used as a fac-
tor, and is called the exponent of 4 for this product.
Similarly, the product a • a • a • a • a is written a*, where the
exponent * shows that a has been used five times as a factor.
The exponent of a number is a number written to the right and
above it, to indicate the number of times it is used as a factor.
Thus:
a is written a\ and read a first, or a to the first jtovcer;
aa is written «', and read a square, or a to the second power;
ana is written a*, and read a cube or a third, or a to the third power;
aaaa is written a*, and read a fourth, or a to the fourth power;
aaa to p factors is written aP, and read a p^^ power, or a to the p^^
poiver,
9. The Index Law of Multiplication.
1. To find the product of 3* by 3*.
3« • 3* = (3 • 3) (3 • 3 • 3 ■ 3) = 3 • 3 • 3 • 3 • 3 • 3 = 3«.
The final exponent * is the sum of the two exponents * and *.
22 CJOLLEGE ALGEBRA W 10, 11
2. To find the product of a' by a\
a* ' a^ z={a ' a) {a ' a ' a) =z a • a ' a ' a ' a =a'.
The exponent ' is the sum of the exponents • and '.
3. To find the product of a"* by a" where m and n are poeitive
integers.
d^=:a' a ' a- , . . torn factors (by {8).
a"* =a- a- a- • • 'tow factors (by {8).
Hence,
^m . ^ii_- (a • a • a to wi factors) . (a • a • a to n factors)
=:a' a- a* ' • • to (m + n) factors (by IV, §7)
= a"'+» (by ?8).
The final exponent ("• "^ ") is the sum of the exponents *" and * of
the factors of the product.
Hence, the exponent of a letter in a product is equal to its exponent
in ike multiplicand plus its exponent in the multiplier,
10. The Index Law for multiplication may be extended to the
product of several powers of a number a, thus,
a^' a^' a^^a^"^^ • a^^ia^'^^'^^zzz a*« (by the index law for
the product of two powers of a).
Similarly,
a' • a*** • a" = a' ■•■ "* • a^ -=. a^ +m+ii^
'By means of the laws of indices and of multiplication we can
simplify products as follows:
3 • 4abcabbdadhc = 3 • 4: anabbbbccdd = 12a'6*c*eP;
13a6VaV75 = 7 • 13 a • a»fe«Wc« = 91a*W+«;
8 a'"c»(f''a'oP = 8 a'^a^c^c^d'' = 8rt'+"'c"+'»(f'';
2 a^Ab^acb^= 2 Aa^ab^b^c = 2 Aa^b^c.
In the product 12 a^b*c^(Pj 12 is called a numerical factor and
a', 6*, c*, d*y and a^b*c^d^ are called literal factors.
DEFINITIONS
11. A product may consist of a numerical factor and a literal
factor; in this case the number represented by the numerical factor
is called the coefficient of the latter. Thus in the product 7 abc the
factor 7 is called the coefficient of the factor abc. Where there is
no numerical factor, we may supply unity; thus we may say that,
in the product abcy the coefficient is unity.
In case the product consists entirely of literal factors, any one
factor may be called the coefficient of the product of the remaining
factors. Thus, in the product abc, we may call a the coefficient of
« 12-15] ADDITION AND MULTIPLICATION 23
be, or h the coefficient of oc, or c the coefficient of ab. If it is
necessary to distinguish these two kinds of coefficients, the latter
may be called literal coefficients, and the former numerical
coefficients.
Of the results at the end of 310 it may be said:
12, 91, 8, and 2 are respectively the numerical coefficients of
a^b*€*<P, a*6V+«, a'+'»c«+''d'', and Aa*b^c.
12. Monomial is the name given to a single factor or the product
of two or more factors; for example,
5 a, lahc, 9a^b^€^, tt'"6«+^c'*, etc.
If two or more monomials be connected by one or more of the
operations of Algebra — multiplication, etc. — the result is called an
a Jgehraic expression; thus :
5a + 7a6c, ea^+COa't'c+a^t^+^cO, etc.
The monomials 5 a, lahc, 6 a', 9a'6*c, a"»Z>"+^c'' are called the
terms of the algebraic expressions,
5 a + 7 afcc and 6 a«+ (9 a%^c + a"»6»+''c'').
Positive Terms are those which have the plus sign prefixed, e. g.,
+ 7 abc, + a"'6"+^c'*. If no sign precedes a term the plus sign is
understood; thus 5 a, 6 a*, 9 a^h^c are respectively the same as + 5a,
+ 6a», +9a»«^«c.
13. Similar or Like Terms are those which do not differ at all,
or differ only in their numerical coefficients; otherwise terms are
said to be unlike. Thus 3 a, 5 ab, 7 a*, and 6 a^bc are respectively
similar to 15 a, 9 ab, 11 a*, and 13 a%c. And ab, a^b, ab^, and abc are
all unlike.
14. Each letter which occurs in an algebraic product is called
a dimension of the product, and the number of the letters is the
degree of the product Thus ab^c^ or a - b * b * b ' c ' c is said
to be of six dimensions or of the sixth degree. A numerical coeffi-
cient is not counted ; thus 9 a'6* and a'&* are of the same dimensions,
namely seven dimensions. Hence the degree of a term or the
number of dimensions of a term is the sum of the exponents. It
should be remembered that if no exponent is expressed the exponent
^ is understood as indicated in 28.
16. An algebraic expression is said to be homogeneous when all
its terms are of the same degree or dimensions. Thus 5 a^-\- 3 a*6'4-
9a*6c* is homogeneous, for each term is of five dimensions.
24 COLLEGE ALGEBRA [1116-18
16. Addition of Similar Monomials (US.)
1. The sum of 7 a and 9 a is required.
By ?7, Law V, 7a + 9a = (7 + 9)a
= 16 a.
Hence^ to add two positive similar termsy find the sum of their
coefficients (ill) and affix to the result the common letters,
2. Find the sum of 9a6*c and 16 a6*c.
By the rule above,
9 ah*c + 16 al^c = (9 + 16) ah*c
= 25at»c.
17. Addition of Polynomials of Plus Terms. — A polynomial is
an algebraic expression of two or more terms.
The addition of polynomials is accomplished by means of the
second law of addition (II, 3 26 and 16| Rule).
Example. — Find the sum of
6a + 9x«, 3x«+5a + 6y», and 2x«+a + mn.
It is convenient in practice to write the expressions one underneath
the other, with similar terms arranged in the same column.
Find the sum of the terms in each column (H6), and write the
results connected with the plus sign.
Thus:
+ 6a+ 9x«
+ a+ 2x« +mn
12 a+ 14x* + 6y» + mM.
18. Multiplication of Monomials. — Find the product of
ha^c and 11 ah\^,
5a'cX 11 w^^c* = (5 aaac) {Wahhbcc) = 5 • 11 • aaaahbbccc = 55a*6V
(by index, commutative, and associative laws of multiplication.)
Similarly, the product of dal^c'^d'' by ISa^h^cPd^ is
9a¥c'^d''XlSa%^cPd''= 9 • 13 • CTa*&'6»c'"c''rf»rf'-= 1 1 7 a^+^t'+'c^-^d"-^.
The coefficient 55 of the resulting product 55a*6'c' is the product of
the two coefficients 5 and 11 of the multiplicand 5a'c and the multi-
plier liable] the literal part a*b^(^ is the result of forming a product
of all the different letters occurring in both multiplicand and multi-
plier, each with an exponent equal to the sum of the exponents
of their letters in both multiplicand and multiplier. The product
U9] ADDITION AND MULTIPLICATION 25
117a*+"6'+'c*"'^d"+'' is formed in a similar mamier. The coefficient
117 is the product of the two coefficients 9 and 13, and the expo-
nents ^'^\ •'+', ^"^^j "■^'' are respectively the sums of the exponents of
a, b, c, d in both the multiplicand and multiplier. From 27, laws
III, IV for multiplication, and ? 18, the following rule for the pro-
duct of two monomials is derived:
To the product of the two monomial coefficient^ (?^1) <^^''^^ ^^«
letters, etich with an exponent equal to its exponent in the multiplicand
plus its exponent in the multiplier.
Example. ^-Multiply 7 x'^yz by 3xy ; m being a positive integer.
By rule 7 x'^yz x 3 xV= 7 • 3 x«+y +»z= 21 x'^+yz.
19. The Multiplication of a Polynomial of Plus Terms by a
Plus Monomial. — From {7, Law V,
a{h -\- c) ■= ah -\- ac
the following rule is derived:
Multiply each term of the multiplicand by the multiplier, and add
the partial products.
Example. — Multiply 2x' + 5x + 7 by 7x*.
By the rule above,
(2x«+5x + 7)x (7x«)=(2x»+5x)7x«+7(7x«)
= (2 x») (7 x») + (5 x) (7 x«) + 7(7 x«)
= 14x»+35x»+49x«.
The following exercises will serve as illustrations of the pre-
ceding definitions and rules of addition and multiplication.
EZEBCISE I
If a = 1, 6 = 3, c = 4, d= 6, e = 2, /= 0, X = 3, y = — 3, find the
numerical values of the eight following algebraic expressions:
1. Find the value of a + 2 6 + 4 c; here, a = 1, 6 = 3, c = 4,
and a + 2H-4c = l + 2-3 + 4 • 4 = 1 + 6-f 16 = 23.
2. ab + 2bc+Sed, 3. ac + 4cd+3c6.
"• 2c+8o '■ <P + dc+<*
8. (9+y)(x+l)+(x+6)(y+7) + 112.
26 COLLEGE ALGEBRA [119
Add the following (see JU6, 17):
9. 2a^+bz^+z+7, 3x« + 2 + (xr» + 8u-, a: + 3a:« + 4, and l + 2z«+5Lr.
10. 2a + 36+4<f, 26 + 3d+4c, 2c?+3c + 4a + 46, and 2c+3ff.
11. a^+Sxy + y'+x + y + l,2a^+^xy + 3y^+2x+2y+3,3a^+5xy+
4y« + 3a: + 4y + 2, and Qafi + lOxy + by^ + x+y.
12. 2a^ + ajc«, x^ + Scu^, a^+2ax^+a*x,
13. 4a:'+10a» + M5jr+6a), 3(2a«+:r3) + 2rtj(2a:+a), a^l7ar+19a) +
16a*-r, and 60,-8 (-j;4.3a) + o« {7x+5a).
(Remove parentheses by Law V, J7.)
14. ^ab + 3^,Sx^ + 2aby2x{a + b)&nd6a(x + 7b) + llx(x+l3b + 7a),
Find the following products by ??18, 19:
15. 3a^y and 5ai/*.
16. 2m, 36, 4c, a6, be, and a6c.
17. Simplify (Sxy^) ' {5xys^) ' (4y*zw) * (xyzw).
18. Find the product of;
a: + y + 2 and 2^:2/0
3 a;« + 6yS + 72^and9V^
3j: + 2 a (ar + 2 ay + 628 } and 6 a^jcySz.
CHAPTER II
POSITIVB AND IVEGATIVE NUKBERS
Addition and Subtraction
20. In the conducting of business, capital is increased by gains
and decreased by losses. Suppose that for a week's business the
account stands:
CREDITS
$25.50
60.00
75.50
21.50
39.00
100.00 .
1321.50 total credits.
LOSSES OR DEBITS
$75.50
35.00
50.50
49.50
40.00
$250.50 total debits.
The total losses of $250.50 neutralize $250.50 of the gains or
credits, leaving a net gain of $71.00. Suppose that for another
week's business the above losses and gains were reversed:
CREDITS
LOSSES
$75.50
$25.50
35.00
60.00
50.50
75.50
49.50
21.50
40.00
39.00
100.00
$250.50 total credits.
$321 . 50 total debits.
The credits, $250.50, are less than the losses, $321.50. Arith-
metically the net loss is found by subtracting the credits, $250.50,
from the total loss, $321 50, which gives $71.00.
This result can be calculated algebraically as follows: Designate
28 COLLEGE ALGEBRA [t21
the credits pita ( -|- ) or positive dollars and the losses minus ( — ) or
negative dollars, then the account can be represented :
+ $75.50 — $25.50
+ 35.00 — 60.00
4- 50.50 — 75.50
4- 49.50 — 21.50
+ 40.00 — 39.00
— 100.00
+$250.50 total + dollars.
—$321 . 50 total — dollars
where the relation between + dollars and — dollars is that, in bal-
ancing accounts, a given number of — dollars cancels or neutralizes
the same number of + dollars; i. e., a loss of $75.00 neutralizes or
cancels gains of $25.00 and $50.00. This result is expressed
algebraically:
+ $25. 00 + $50. 00 — $75. 00 = + $75. 00 — $75. 00 = $0. 00.
The gains and losses are said to balance, or there is a balance of
^ero dollars, $0.00,
Or if there should be again of $75.00 and two losses, one of
$25.00 and one of $50.00, the result of the transactions would be
expressed algebraically:
+ $75,00 — $25.00 — $50.00 = + $75.00 — $75.00 = $0.00.
Zero is defined as the difference between two equal numbers.
Suppose a loss of $321 .50. It may be separated into two losses,,
one of $250.50 and the other of $71.00, and the operation of
balancing the -|- dollars and — dollars in the problem above would
be indicated algebraically as follows:
+ $250.50 — $321.50 = + $250.50 — $250.50 - $71.00
= $0.00 — $71.00 = — $71.00.
That is, there remains a debt of $71.00, or the balance is a
negative number.
21. The Series of Natural Numbers. — If from a fixed point 0,.
in a line A B, units of length are laid off to the right, the successive-
points so found can be designated by the series of natural numbers,,
1, 2, 3, 4, 5, etc.
1 2 3 4 5 6 7 8 9 10 11 12
0 I I I I I I I I I I I I
522] POSITIVE AND NEGATTV^ NUMBERS 29
If the measurement proceeds from 0 toward B it is positive
measurement; if the direction is reversed, say at the 11th division,
and proceeds from 11 toward 0, this is another kind of measure-
ment, namely negative.
If the measurement starts from the 5th and ends at the 9th
division, 4 units are added to the original 5. The final distance
from 0 is
+ 5 + 4 = +9.
This illustrates positive measurement.
If the measurement is from the 11th division to the 8th, the act
of moving from the 8th to the 11th division is neutralized. This
operation which neutralizes the operation of addition, is indicated by
the sign—, read minus. In moving from 0 to the right to the 11th
division, then reversing the direction and moving to the left to the
8th division, the total result or distance from 0 is represented alge-
braically:
+11—3 = +8+3 — 3 = + 8 +0 = +8 (see ?21, definition of zero).
This may briefly be translated thus; move to the right 11 units
from 0 and then move 3 units to the left from 11, stopping at a
distance of 8 units from 0 to the right.
If the motion takes place from 0 to the right to the 6th division,
and is then reversed over 6 divisions to the left, the final position is
0 and the distance from 0 is zero units. The result of the motion
is indicated algebraically:
+ 6 — 6 = 0.
The operation of moving from zero to the right or from any
point of division to the right can be expressed by addition of units,
and of moving from the right to the left by the subtraction of units.
22. Positive Numbers. — Suppose that it is desired to subtract 8
from 5. Move from the 5th division to the left over 5 unit spaces.
The resulting position is 0, after but 5 of the 8 units have been sub-
tracted. The act of moving over 8 unit spaces from the right to
the left can be separated into one motion of 5 and another of 3
units. Therefore the operation of moving 8 units to the left from
the 5th point of division may be indicated algebraically:
+ 5-8 = + 5 -5- 3= 0-3 = - 3.
0 — 3 is represented by the simpler symbol — 3, called a negative
number. This result and the corresponding result —$71.00, derived
30 COLLEGE ALGEBRA [«l 23-26
when the losses in a basiness transaction were $71.00 greater than
the gains (220), necessitate the introduction of a system of nega-
tive numbers into Algebra.
23. Negative Numbers. — Begin at zero and lay off unit lengths
to the left; by the repetition of the unit, a series of negative numbers
is formed. These two series of numbers, the series of positive or
natural and the series of negative numbers, are called series of
algebraic numbers, and are represented on the line as follows :
-1S-12-1M0 -! -« -7 -4 -5 -4 -S -2 -1 0 +1 +2 +S +4 +5 H +7 +8 +J +10+11+12+11
I I I I I I I I I I I I I I I I I I I I I I
ml k j i h g f e d c b a ABCDEFQHIJKLM
The subtraction of 8 from 5 will be expressed by a motion from
E to the left over 8 unit divisions to c, and the result is — 3; i. e.,
the final position is a place 3 units to the left of zero.
The result obtained by subtracting a greater number frorn a lea
token both are positive^ is always a negative number.
In general, in case a and b are any two positive integers, the
expression a — b is a positive integer when a > i, is zero when
a = 6, and is a negative integer when a < 6.
In a series of algebraic numbers, in counting from left to right,
numbers are said to increase, in counting from right to left they are
said to decrease in magnitude. Thus — 4, — 2, — 1, 0, +2, -|-4
are arranged in ascending order of magnitude.
24. The Absolute Value of a Number.— The absolute value of a
number is its value without its sign. Thus the absolute values of
— 4, —2, —1, -|-3, +5» are respectively 4, 2, 1, 3, 5.
25. Every algebraic number -f- 5 or — 5 consists of a sign -|- or —
and the absolute value of the number. The sign shows whether the
number belongs to the positive series or the negative series of num-
bers; the absolute value of the number shows the place which the
number has in the positive or negative series.
When no sign is written be/ore a number, the sign -|- is understood.
The sign — is always written.
26. Unlike Signs. — Two algebraic numbers which have respect-
ively the signs + ai^d — are said to have unlike signs. Thus -|- 7
and — 9 have unlike signs.
li 27-29] POSITIVE AND NEGATIVE NUMBERS 31
Addition of Algebraic Numbebs
27. Since algebraic numbers may be positive or negative, four
different problems arise in the addition of them :
^ In particular
I. Addition of two positive numbers, as . . . . -j- 3 + (+ 4)
II. Addition of a positive and a negative number, as -f- 3 + ( — 4)
III. Addition of a negative and a positive number, as — 3 -|- (+ 4)
IV. Addition of two negative numbers, as . . . — 3 -|- (— 4).
-IW2-11-1I -5 -8 -7 -I -5 -4 -S -2 -1 • +1 +2 +S +4 +5 +« +7 +8 +f +l(Kt1+12+IS
I I I I I I I I I I I I I I I I I I I I I I I I I I I
m I k j % h g f e d c b a ABODE FGHIJKLM
I. The sum of 4*3 and +4 is found by counting from (7,
whose distance from 0 is + 3, 4 units to the right, or in the positive
direction, and is therefore -\- 7, the number of units from 0 to 6r.
II. The sum of -|- 3 and — 4 is found by counting 4 units to the
left, or in the negative direction, from C (or + 3), and is therefore
— 1, the distance of a from 0.
III. The sum of — 3 and -{-^ \b found by counting from c
(or — 3) 4 units to the right, or in the positive direction, and is
therefore + 1> the distance of A from 0.
IV. The sum of — 3 and — 4 is found by counting from c
(or — 3) 4 units to the left, or in the negative direction, and is
therefore — 7, the distance of g from 0.
28. If a and h represent any two integers, the results in 227
are therefore expressed as follows:
In particular In general
I. +3+(+4)= +7 ^a+{+l)= +a + h
n. +3+(— 4)= ~1 +a-|.(_6)= +a-h
III. -.3+(+4)= +1 _a+(+Z.)= -a + h
IV. -3+(— 4)= —7 —a+{-h)=: -a-h.
These four cases give rise to the following rules.
29. RULES FOR THE ADDITION OF ALGEBRAIC NUMBERS
I. If ttco numbers have like signs, find the sum of their absolute
values, and prefix the sign common to both numbers to the result.
II. Jf two numbers have unlike signs, take the difference between their
absolute values, and prefix the sigh of the number with the greater
absolute vaJue to the result.
32 COLLEGE ALGEBRA [*30
The results in the several cases, I, II, III, IV, are called the
algebraic sums in distinction from the arithmetical sum, which ia
simply the sum of the absolute values of the numbers.
III. If there are more than two numbers to he added, add two of the
numbers, then this sum, to the third, and so on; when the numbers to he
added are positive and negative, take the difference between the absolute
values of the sum of the positive numbers and the sum of the negaiivt
numbers and prefix the sign of the greater suin to the result, which icill
be the algebraic sum of the numbers,
EXBBOISE n
Find mentally the results of the indicated additions:
1.
S.
S.
4.
J.
6.
+ 9
-8
-6
-12
+ 12
— 13
-5
+ 3
-5
+ 7
- 7
— 7
7.
8.
9.
10.
11,
if.
+ 5
-7
-6
-30
+ 64
-29
+ 7
-6
-4
+ 16
-35
+ 50
-9
+ 8
-3
15.
+ 44
IG.
-19
n.
-33
13,
18.
+ 17
-19
-27
-60
-19
+29
+ 12
— 11
+ 49
-20
+ 64
-27
-27
+ 37
go.
-38
21.
+ 50
22.
-48
2S,
-23
19,
H-
17
-21
-91
29
-11
60
-10
-16
+ 05
- 8
-21
— 30
+ 15
-25
- 7
23
-64
- 40
-29
34
+ 75
-55
-17
- 60
+ 30
5
16
100
80. The Addition of Similar Monomials. — 1. Find the sum of
3 a, a, 4 a, 7 a.
By I, 229, and by 816,
3a + a + 4a + 7a = (3 + l + 4 + 7)a [Law V, 17]
= 15a.
Hence, the sum of the monomials is 15 a.
€30] POSITIVE AND NEGATIVE NUMBERS 33
2. Find the sum of — 3 6, — 5 ft, — 7 6, — 11 b.
6 J I, 229, find the sum of the coefficients, which is
_3-5-7-ll = -26
Hence, the sum of the monomials is — 26 6.
The same result would be obtained by assuming Law V, J 7 to
bold for negative numbers, thus:
— 36-56 — 76-116 = (-3-5 — 7 — 11)6
= -26 6, by I, {29.
3. Find the sum of 6 ax', —5 ax', — 2 ax", -|-13ax*, —19 ox',
By II and III, 229, the sum of the coefficients of the positive
terms is
6 + 13 + 1 = +20,
and the sum of the coefficients of the negative terms is
_- 5 — 2 ^ 19 = — 26.
The difference between 26 and 20 is 6, and the sign of the greater
is — . Hence, the sum is — 6 ax*.
Remabk. — ^The divisions AB, BCy etc., in the figure in J27 would, in
case of example 1, be a; in case of example 2, be 6; and in example 3, be ox*.
Here, as in example 2, the same result would be obtained by assum-
ing Law V, 27 to hold for positive and negative numbers, thus:
6aa;* — 5ax« — 2ax* + 13ax«— 19ax« + ax« =
(6 — 5 - 2 + 13 — 19 + l)ax« = - 6ax«, by II, 229.
Therefore,
To find the 9um of similar monomiah, find the algebraic sum
of the coefficients and prefix this sum to the letters common to the sev-
eral terms,
EXBBOISE HI
Find mentally the sum of the indicated additions:
1, t, S. 4. S. 6,
3a
4aT
4xy
2 6c
+ 66cd
- 9«
5a
— box
-lOxy
- 6c
- Sbcd
- Sz
— 2a
+ 6ax
+ 16:i:y
- 66c
+ IS bed
+ z
— 4a
+ 1100?
-17a;y
+ 13 6c
+ bed
+ 4z
- 66c
+ 2bcd
-lOz
34 COLLEGE ALGEBRA C230
7.
8.
P.
10.
ii.
/^.
18j:«
6n«
-IIA
-19x«y>
P*
5*»
- 3a^
-llii«
+ ;r*z
- :r*i/«
-lip*
15 *»
+ 7:c«
-14n«
+ Z.x*z
+ 13^^«
+ 6p*
— ^
-Uj^
+ 10n«
+ nj*z
+ 9.r«/
-31p*
— 29«»
- 5A
- 7x«y>
+ 29p*
-|-16«»
BZEBOISE IV
Add the following monomials:
1. 3a, —6a, +2a, —7a, +12a.
2. lOon, — 6an, 4an, 7an, — 9an, an.
3. 4a^, — 3j:«y» — 5x«y.
4. 3:cy, 6a:y, — 7j^, +xy.
5. -3di^», 4dy», -8<fy», -13(/y», 2(iy«, 18(fy».
6. 16c, —lie, —2c, 3c.
7. 13c, 12c, —24c, 2c.
8. -62«, +22«, -52«, 42*, -32«, «•.
9. 13c, 12c, —26c, 4c.
10. \Zj^, — 10x«y, — 6jr«y, 5a:V» — 4a;«y.
11. — 3a2, 7a2, — 2aa;, — az.
12. ojrz, — 7ax2, +8axz, ^axz, —Sazz, *^9aa»i.
13. -7a», -4a», +a», +13a», -9a».
14. 5a6x, — 2a6jr, — 3a6x, lOo^x, — 4a6x.
15. 3i/2», y2«, -7i/2«, 102/2«, -yz*.
16. 2(a + 6), -3(a4-6), -7(a + 6), 7(a + 6), 4(a+ft).
17. 9(x + y), 6(x + y), -ll(a; + t/), 9(x + y), - 19(a; + y).
Simplify the expressions:
18. 9.r«-6j:« + 6^-3j;« + 3^-6:c«.
19. 5a«-19a« + a« — 5a« + 6a« — a« + 8a«.
20. 6a«a: + 8a«j:-lla«x-27a«x4-6a«^.
21. -3.i:«/+7j:«y«-6xV+13j*y-4jrV-9:rV-
22. — 11 a6cjr + 4 after — 5 a6cur-|- 29 a6«; + a6cr.
23. 3a»-7a»-8a» + 2a»-lla».
24. 2«jr — 3«x + &r — ax — 5«Li: + 5«ar.
25. ix-ix + x+ix-lx.
26. -56 + i6-i6 + 26-i6 + }6.
27. — |x« — }x«-}x«-}^-ar«.
i31J POSITIVE AND NEGATIVE NUMBERS 35
Subtraction op Algebraic Numbers
81. Subtraction is defined as the inverse operation of addition.
Thus, to subtract 5 units from a number is to undo or neutralize the
result of adding 5 to that number.
Therefore^ to subtract one algebraic number from another, toe begin
at the place in the series occupied by the number and count in a direc-
tion opposite to that indicated by the sign of the subtrahend, as m^ny
units CM there are in the absolute value of the quantity to be subtracted,
-8 -7 -6 -6 -4 -3 -2 -1 0 +1 +2 +3+4+6+6 +7 +8
I I I I I I I I I I I I I I I I I
hgfedcha ABCDEFQH
Since the numbers may be positive or negative numbers, there will
be four distinct problems to solve :
In particular
I. Subtraction of a positive number
from a positive number, as . . . + 5 — (+ 3)
II. Subtraction of a negative number
from a positive number, as . . . + 5 — (— 3)
III. Subtraction of a positive number
from a negative number, as . . . — 5 — (+ 3)
IV. Subtraction of a negative number
from a negative number, as . . . — 5 — (— 3).
I. The result of subtracting + 3 from + 5 is found by count-
ing from ^3 units in the negative direction, to the left; that is,
in a direction opposite to that indicated by the sign + before 3, and
is therefore + 2, the distance of B from 0.
II. The result of subtracting — 3 from + 5 is found by count-
ing from E 3 units in the positive direction, to the right; that is,
in a direction the opposite to that indicated by the sign — before 3,
and is therefore + 8, the distance of H from 0.
ni. The result of subtracting + 3 from — 5 is found by count-
ing from e 3 units in the negative direction, to the left; that is, in a
• direction opposite to that indicated by the sign + before 3, and is
therefore — 8, the distance of /* from 0.
IV. The result of subtracting — 3 from — 5 is found by count-
ing from e 3 units in the positive direction, to the right; that is, in
a direction opposite to that indicated by the sign — before 3, and
is therefore — 2, the distance of b from 0.
36 COLLEGE ALGEBRA [SI 32-34
Hence, if a and h represent any two integers, the results of I-IV
are represented as follows:
In particular In general
L -|-5-(+3) = + 5-3 = +2 4.a_(4-fe) = +fl_6
IL +5-(-3)=: + 5 + 3 = +8 -|.a-(-fc) = + a+6
IIL _5-(+3)=:-5-3 = -8 —a — {+h).= — a—h
IV. _5-(-3) = ~5 + 3=-2 _a_(-6) = -a + [».
32. It follows from the four cases /, //, ///, IV, that to subtract a
positive numher is equivalent to adding an equal negative number; and
that to subtract a negative number is equivalent to adding an equal
positive number,
33. To subtract one algebraic number from another^ change the
sign of the subtrahend and add the result to the minuend,
EXEBOISE V
Perform mentally the indicated subtractions:
1. g, S. 4- 5, 6. 7. 8.
13
5
13
- 5
-13
5
-13
- 5
7
13
7
-13
- 7
13
— 7
-13
9.
10,
11.
le.
13.
14.
15,
16.
9
21
13
-25
-19
-47
65
-13
-55
-29
-46
— 46
-33
17
75
-26
34. The Subtraction of Similar Monomials.
1. Subtract — 5nV from 13 w*x*.
The number 5n*x* can be represented by the five intervals from
0 to ^ in the scale for positive numbers in §31, in case each
space Oi4, etc., contains as many unit lengths as there are units in
the product n^x^. The number — 5 n^x^ then would be represented
by the distance 0^, where O^i, etc., are each equal to 7i*x*. SimUar-
ly, any monomial may be regarded as a number in the positive or
negative series of numbers. Therefore,
By rule in 232, 13 n^x^ — (— 5/i V) = 13 n»x« + 5 nV
By addition, = 18»'x*.
2. Subtract -\-2a — 3 a from + 4a,
Bynile, 833, +4n — (— 3rt) — (2flr) = +4a + 3a — 2a
By addition, 830, =: + 7rt — 2a = 5a,
Hence,
JW] POSITIVE AND NEGATIVE NUMBERS 37
To subtract a monomial /mm a similar monomial y change the sign
of the coefficient of the subtrahend; add the coefficients^ and prefix the
result to the common letters.
BXBBCISB VI
Perform mentally the indicated subtractions:
i. 2. s. 4> 6.
-S6xy^ — 6a» -7m«
+ 9xy* -16a» +57n«
8. 9. 10,
— 75j: 4a6c — ah^x
— 26j: — ahc -\-bal^x
19 j:
29 r^
14 a:
-IZxz
6.
7.
29y
-61c
51 y
19 c
11. From X take y\ from x take — y,
12. From 3 j: take 6; from 4 x take — 6.
13. From 13x"y take 9 j-"«/; and from — 21 ?/"+« take 5t/"+«.
14. From — 19 a«»-i take — 26 a^-\
If a = 5, 6 = — 3, c = — 4, find the values of:
16. a+6 + (-c). 18. +(-a)-(~6)-(«c).
16. a-64-(-c). 19. -(-a).+ (-6)-(-c).
17. -o + (-6) + c. 20. -(-a)-(-6)-(-c).
21. From 4f ar« take 2} ax^\ from 13i ar» take — 9J a-*.
22. From — 17 axi^ take ax^* and 4 cwy*.
23. From — 13i xj? take — 3J x^ and 9 J xi^.
Perfonn the indicated subtractions:
24. a-36 25. —2a
4-46 fl-26
2vS. 5a 29. 4a — 1
1 -4a +7a
32.
n-1
n + 1
36.
bx + Z
1 +x
40.
7 -r
2x-10
44.
n-3
m + 5
48.
a + x
ar-1
33.
71-7
n + 5
37.
a-1
1-a
41.
a-5
2-3a
45.
n + 1
a-3
49.
a-6
1+a
26. -56
27.
3a — 56
a+46
+ 36
30. 3a +1
31.
8 a
-4a
35.
1 -Sa
34. n-1
n-8
3-2n
7-n
38. a + x
39.
a — X
x — a
43.
x + a
42. 7a — X
2a-36
2.r — 5a
36 + a
46. a-1
47.
3a — n
S-x
n -6
50. 7n — n
51.
m-1
3n — 771
n+1
38 COLLEGE ALGEBRA [?35
REMARKS
35. I.— The definitions and rules established in connection with the
introduction of positive and negative integers and their addition and
subtraction in the two illustrations used — gains and losses, and i>06itive
and negative distances — are equally applicable in case of fractional unit&
For example, in case of a gain of three-fourths of a dollar, or +$0.75, and a
loss of three-fifths of a dollar, or— 10.60, the result of the transaction would
be indicated thus:
-f $0.75 - $0.60 = + $0. 15;
and + (2 ft. and 8 in.) = + 2 J ft. would be represented by a point between
B and C in the scale of positive numbers, found by dividing the unit of
space B C into three equal parts and taking two of them. ( J83.)
II.— Besides the two cases discussed in ii22, 28, and the paragraph
following, there are many other instances in which it is desirable to repre-
sent not only the magnitude but also what may be called the quality or
affvciion of the things under consideration. For example, in questions of
chronology it may be desired to distinguish a date before a given time from
a date after that time. In case of the readings of the temperature indi-
cated by a thermometer, the reading 25° above zero is written -\- 25°, and
19® below zero is written — 19°, and so on. The definitions of positive and
negative numbers and the rules derived for their addition and subtraction
in JJ22, 28, etc., apply in these cases as well, and in many others.
In the following it will be shown that the family of algebraic numbers
contains still other kinds besides those already described.
III.— The student should be careful to keep in mind the fact that the
symbols + and — , when used to represent an indicated addition and sub-
traction, are symbols of two specified operations common to both Arithmetic
and Algebra, but that they are also used to show that a given number
is in the positive ornegativeseriesof numbers, and in this sense are used in
Algebra alone. In Arithmetic a set of symbols is used in calculating the
results of the addition, subtraction, multiplication, and division of w^hole or
mixed numbers, in all problems in which whole or mixed numbers are
involved.
In the preceding chapters it has been found necessary to introdnoe
negative numbers, and rules for the addition and subtraction of negative
numbers and also of positive and negative numbers have been derived.
It is now proposed to establish rules for calculating the results of per-
forming a finite number of the fundamental operations, addition, subtrac-
tion, multiplication, and division, upon the symbols
t i\ . a a
+ a, 0,-a, + ^,-^,
TWt only when ihey represent numbers, hut also when they are regarded as mere
symbols.
CHAPTER III
SUBTRACTION AND THE NEGATIVE INTEGER
Generalized Discussion
36. Numerical Subtractions. — To every mathematical operation
there corresponds another, usually called the inverse, which exactly
undoes what the operation itself does. Subtraction stands in this
relation to addition, and division to multiplication.
To subtract 5 from 9 is to find a number such that if it is added
to 5, the sum will be 9. The result is written 9 — 5; by definition
it satisfies the equation
(9 — 5) + 5 = 9.
(9 — 5) group 5 group 9 group
a
c
+
E F
a
c
E F
b
d
GUI
h
d
GUI
That is to say, 9 — 5 is the number belonging to the group which
with the 5 group makes up the 9 group.
In general, to subtract h from « is to find a number to which if
h is added, the sum will be a. The result is written a — ^; by defi-
nition, it identically satisfies the equation
VI. {a — h)-\-h = a',
that is to say, a — 6 is the number belonging to the group which
added to the h group makes up the a group.
It is evident that subtraction, as thus far defined, is always pos-
sible when h is less than a. In addition, the relative sizes of the
numbers added is unessential, but in the operation of subtraction,
this makes a difference.
The symbol — is read minus, and if placed between two numbers
a and h, then (a — h) is read a minus h; a is called the minuend and
h the subtrahend.
Note.— 7%^ tigri'^Xs used to denote the difference of the numbers between which It is
nlAced; thus, a^^ b signifies a— 6 if a Is greater than b, or b^a if b is greater than a.
39
40 COLLEGE ALGEBRA [SI 37, 38
37. Numerical Subtraction is a Determinate Operation. — Sab-
traction, when possible, is a determinate operation. That is, there
is but one number which added to h will produce a; or there is but
one number which will satisfy the equation x-\-h z= a.
For if c and d satisfy this equation, they may be put in place of
cc, then c-\-h = a and d-{-b = a, and therefore c + 6 = d -{-b,
since things which are equal to the same thing are equal to each
other. Hence, a one-to-one correspondence may be set up between
the individuals of the (c -|- h) and {d -j- h) groups, (24). But tbere
is a one-to-one correspondence between 6 individuals of the (c -|- ^)
group, and h things of the {d-^-h) group; hence there must be a
one-to-one correspondence of the c other individuals of the first
group and the d other individuals of the second group; i. e., cz=d.
This characteristic of subtraction is of the same importance as the
Laws I- V, ?26, 7 of addition and multiplication, hence we add to
the group of Laws I - V and the definition of subtraction VI, which
like them is a fundamental principle in Common Algebra, the theorem
VII. if a + c = h + c
then a = i;
which may be stated in the fonn: if one term of a sum changes
while the other remains constant, the sum changes.
38. Formal Rules of Subtraction. — All the rules of subtraction
are derived by means of the fundamental laws I, II (J 6), III, IV,
V (27), VII (837), and definition VI (3 36). The assumption that
these rules must follow whatever the meaning of the symbols «, b,
c, -|-> — J = ^^y ^®) ^8 * ^^^^ which has an important bearing on
the discussion which immediately follows. The rules of subtraction
are first derived under the restriction that the minuend is greater
than the subtrahend, and then generalized. .
In any computation involving subtraction it is sufficient to con-
sider the following five equations ; because, if they are constructed
in the proper way, they will determine the result of any series of
subtractions or any compound operation composed of addition, sub-
traction, and multiplication.
1. a — (b-\-c) = a — b — c = a — c — b,
2. a — (b — c) = a^b-{-c.
3. ^f -|- ^ — ^ = r/.
4. a -{-(b — c) = a-{-b — c = a — c-\-b,
6. a ( b — c) = ab — ac.
838] SUBTRACTION AND THE NEGATIVE INTEGER 41
PROOFS
1. a — h — c is the form to which if first c and then h is added;
or what is the same thing (by I, ^6), first b and then c; or what
again is the same thing (by II, ?6), & + c at once; the sum produced
is a (by VI, 236). Therefore, a — h^c is the same as a — c— 6,
which is the form to which if 2>, then c, is added, the sum is a ;
butr a — (^ -j- c) is the form to which if h -{-c'ls added the sum is a
(by VI, J 36). Therefore, a — b—c or a — c-^b is the same as
a — (/> 4" ^'). Equation 1 shows that two consecutive subtractions
may be interchanged, that they are commutative.
2. a — (6-c) = a — (/> - c) - c + c [Def. VI, JS6]
= a — {b — c + c)-\-c [Eq. 1]
= a-b + c. [Def. VI, J36J
The sign of deduction, . *. , is read therefore or hence.
To subtract b — c from a is the same as to subtract b and to add
c to the remainder.
3. a-\-b—b-\-b z=: a-\-(b — b)-\-b
= a + b [Def. VI, J36]
a-i^b-b = a. [Law VII, {37]
That is, the operation of subtracting b undoes the effect of the oper-
ation of adding b to a. The equation defines the subtraction as the
inverse operation of addition.
4. a + b — c=:a + {b — c + c)^c [Def. VI]
=^ a-{-(b — c) + c — c [Law II, {6]
= a + {b-c), [Eq. 3J
To add the quantity {b — c) to a is the same as first to add b and
then subtract c from the sum. Equations 1,2, 4, together constilute
an associative law for subtraction, and with law II, constitute a com-
plete associative law for addition and subtraction.
5. ab — ac = a(b — c-\- c) — ac [Def. VI]
= a{b — c)-\- ac — ac [Law V, 27J
= a(b-c). [Eq. 3]
The product of a by the difference of b and c is the same as the
difference of the products of a ])y b and a by c.
Equation 5, ?38, supplements Law V, constituting with it a com-
plete law of multiplication of a number by a sum or difference.
For the present the assumption will be made that the equations
1 — 5 hold for all integral values of o, b, and r, without regard to
their relative magnitudes.
42 CX)LLEGE ALGEBRA LJ39
39. Zero. — If b is made to equal a in the general equation for
subtraction (VI),
(a — h) -{-b =z a,
then this equation takes the forms
(1) {n — a)-\-a=za if b = a,
(2) {h^b) + b = b if a = 6.
It can be proved that a — a = b — b
For {a — a) + {a + b) = (a — a) + a+b [Law 11, 86]
= a + b,
since by (1) {a — a) -\- a = a.
Also (b-b) + {a + b) = (b-b) + b + a [Lawsl, II, {6]
= b + a=a + b.
[«39, (2), and Law I, 86]
Therefore (a — cr) + (a + ?>) r={b — b)-{- (a + 6)
and a-a = b~b. [Law VII, 887]
Similarly b — b = c — c.
or in general a — a = b — h =ic — c^ etc.
That is, a — a is independent of the value of cr, and may be
represented by any symbol unrelated to a. The symbol which
mathematicians have given this number is 0, called zero. Thus
(3) « — a = 0.
For example,
3-3=4-4 = 5-5 = 7-7 = 11-11= • • • - = 0.
That is, the result of subtracting any integer from its equal can
not be expressed in terms of integers, and therefore a new symbol,
0, must be introduced to represent this result, namely zero.
For this symbol, 0, and for negative numbers, rules of calculation
must now be established.
Addition is defined for 0 by the equations:
1. 0 + a = a [{39,(1)]
a + 0 = a. [Law I, 86]
Subtraction is defined (partially) by the equation.
2. a — 0 = a.
Since (a — 0) + 0 = a, [Def . VI, 836]
a-0 = a. [1]
Multiplication is defined (partially) by the equation
«40, 41] SUBTRACTION AND THE NEGATIVE INTEGER 43
3. axO = Oxa = 0.
Since a x 0 = a (5 — i) [Def. of OJ
=:ab — ab [{88, 5]
ax 0 = 0. [Def. of 0]
40. The Negative. — When b is greater than a, say, equal to
ci + </, then
6 = a + J
b — a =z a-\-d — a
= a — a-fj [J88, 4]
= 0 + d [Def. of 0]
1. .-. b^a = d, [J89, (1)]
Also a — b=za — (a + rf)
=:a — a^d [J88, 1]
= 0 - cf [Def. of 0]
2. .-. a — b = --d.
Substitute the shorter symbol — d for 0 — d, in view of the
lack of significance of 0 in relation to addition and subtraction. The
equation 0 — d = — d supplies the missing rule of subtraction for
0 (589, 2).' The symbol —J is called the negative of rf, and in
opposition to it the number d is called positive,
Thougli the sign — in its origin is the sign of an operation,
(subtraction from 0), it is here to be regarded as a part of the sym-
bol d.
— d is as practical a substitute for a — 6 where a < 6 as is a
simple numerical symbol -f- d when a^b,
RULES OF CALCULATION FOR THE SYMBOLS, 0 AND —d
41. The rules of calculation for negative numbers together with
definitions for their aefc/iV ton, subtraction^ 9,ndL multiplication^ are easily
deduced from the laws I-V, «{6, 7; VII, ?87; definition VI, {86;
and equations 1-5, 288.
In particular In general
1. 6+(-6) = -6+6=0. b+{—b) = -bJrh=(S,
For _6+6=(0-6)+6 _[»+6=(0-^>)+6 [Def. -6]
=0. = 0. [Def. VI, {86]
Therefore, — 6 can be defined as a symbol, such that the sum of -\-b
and — 6 if 0.
44 COLLEGE ALGEBRA [HI
In particular In general
2. 7+(— 6) = — 6+7 = 7-6. a+{-h)=—h+a=a-b.
For 7+(— 6)=7 + (0-6) a+(—h)=a+{0—h) [Def. —b]
=7 + 0-6 =a + 0-h [S38, 4]
= 7-6=— 6 + 7. =«— 6=— fe+a. [J39, 1,4]
Therefore, the sum of a positive and a negative number is equal to the
first minus the second.
3. _6+(-5) = -(6+5). ^a+(-6) = -(a+6).
For— 6+(— 5)=0-6+(~5) — a+(— />)=0-a+(-6) [Def.— a]
= 0-(6 + 5) =0-(a+i) [838, 1]
= _^(6+5). =— (a+6). [Def.of neg.J
Therefore, the sum of two negative quantities is minus their sum.
4. 6-(— 5) = 6+5. a-{—h)=a+h.
For 6-(-5)=6-(0-5) a_(_6)=a-(0-6) [Def.— 6]
= 6-0+5 =a-0+6 [J38, 2J
=6+5. =a+6. [S39, 2]
Therefore, to subtract a negative number from a positive number ^
change the sign of the negative number and add.
5. (_6)-(-5) = 5-6. (-_a)__(_6) = 6-a.
For -6-{-5) = -6+5 _a-(-6) = -a+ft [4]
=5-6. =fe-a. [2]
Therefore, to subtract a negative number from a negative number ^
change the sign of the second and add.
Corolla rg,
__6— (— 6) = — 6 + 6=0. —«_(_«) = — a+a = 0. [Def. of 0]
6. 5(-6) = (-6)5 = -5 • 6. a{-b) = {-b)a = --ab.
For 0 = 5(6-6) Q=a{b-b) [{89,3]
= 5 • 6+5(— 6). =zab+a(—b). [Eq. 5, 138]
.-. 5{-6)=-5-6. a(-b) = -ab. [1 ; Law VII, J37]
Therefore, the product of a positive number by a negative number is
minus ihe product of the numbers.
7. (-7)x0 = 0x(-7) = 0. (-a)xO = Ox(-«)=0.
For {-7)x0 = (-7) (6-6) (_^)xO=-a(6-fe) [Def . of a]
= (_7) 6-(-7)6 =^^a)b-(-a)b [{38,5]
= _7 . 6— (-7 • 6) = 0. = — a • b-{—a • b)= 0. [6, 5, Cor.]
Therefore, the product of a negative quantity by 0, m Oi. .
«42, 43] SUBTRACTION AND THE NEGATIVE INTEGER 45
In particular In general
8. (—5) (-6) = 5 • 6. (-a) {-h)=ah.
For (-5) (6-6)=0 (—a) (6_/>)=0 [7]
(_5)6+(-5) (-6)=0 {.^a)h+(-a) (-h) = 0 [LawVJ7J
__5 . 6+(-5) (— 6) = 0 -ab+i-a) (-6) = 0 [6]
... (_5) (-6)= +5 • 6. {^a){-h)=aL [l;Law VII,J37]
Therefore, the product of two negative quantities is equal to plus the
product of the quantities; t. c, i» multiplication minus times minus
gives plus,
42. Limitation of Numerical Subtraction. — Equations 1-5,
i88| show that subtraction conforms to the same general laws as ad-
dition, and therefore it might seem perfectly possible to interchange
the role of direct and inverse operations.
This interchange, however, is seen to be impossible upon examina-
tion of these equations. The requirement that the minuend must be
greater than the subtrahend sets a comparatively narrow limit to the
field of subtraction, making its range much narrower than that of
addition. This limitation restricts the use of equations 1^5 of 238
to particular classes of values. For example, such a simple inference
as2a — (2rt + 3fc) + 56 = 21 does not hold since 2r/ + 3^» > 2a.
The use of subtraction as so far defined in any reckoning with sym-
bols must be regarded as unwarranted unless the relative values of
the symbols are known.
Accordingly the question arises, how is this limitation upon sub-
traction to l)e removed? This question is answered in 2H3y 44, 45.
43. Symbolic Equations. — Definition VI, S38, that is, the equa-
tion (a — ft) -f- i = a, as has been seen, is suflftcient to define subtrac-
tion when a >• 6.
Moreover, {a — ft) -j- ft = a, according to definition, only when
a — ft is a numl^er as defined ({36, VI, and 242).
However, an equation can be defined in a broader sense.
An equation is any declaration of the equivalence of a definite com-
bination of symbols; t. e. one of the combinations may be substituted
for the other, — and accordingly {a — ft) + ft = «, may be an equation
whatever the values of a and ft.
Now if no other meaning is attached to a — ft, except that it is a
symbol such that associated with ft in the expression (a — ft) -|- ft it
is equal to a, then the equation
(a _ ft) + ft = a
46 COLLEGE ALGEBRA [85 44-46
is a definition of the symbol (a — h). This symbol is not numerical,
but purely symbolical. The sign -|- ^^^ indicate numerical ad-
dition only in case the symbols which it connects represent numbers.
44. Principle of Permanence. — The assumption of the perma-
nance of form of the equation
{a-h) + b = a,
which is the result of the definition of subtraction, gives at once a
symlK)lic definition of subtraction which is to hold for all values of
a and b.
The symbolic definition is more general than the definition of
numerical subtraction, which is the particular case of the symbolic
definition when a and h are numbers and ay>h.
From the point of view of symbolic subtraction, it is irrelevant
whether {a-^h) is a number or not; only such properties can be
attributed to {a — h), considered by itself, as follow directly from
the generalized equation
(a—h) + h=:a.
Similarly, each of the fundamental laws, I-V, 326, 7, VII,
J 87, as soon as it fails to be interpreted numerically, becomes, on the
assumption of the permanence of its fonn, a mere declaration of the
equivalence of certain particular combinations of symbols. Thus,
equations 1-5, 238, become definitions of symbolic addition, subtrac-
tion, multiplication, and their combinations. — The symbols a, h, etc. ,
are purely symbolic and are unrestricted as to meaning.
46. Some illustrations (2227-34) of the increased power gained
by considering a, b, etc. , as symbols merely have been met with al-
ready and many will occur later.
In 2239 and 40, the introduction of zero and the negative number
are the immediate consequences of symbolic definition of subtrac-
tion. They greatly increase the simplicity, scope, and power of the
operations of Algebra.
48. Review. — It is profitable at this point to review the nature
of the argument which has been developed in this and the preceding
chapter.
1. The associative and commutative laws (Laws I, II, 26) of
addition and subtraction, and the determinateness of subtraction
(Law VII, 237) followed directly from the definitions of the
positive integer, and the operations of addition and subtraction.
{?47, 48] SUBTRACTION AND THE NEGATIVE INTEGER 47
2. The result of sabtracting h from a, namely a — 6, is uniquely
defined by the equation (a — b)-{'b = a for all values of a and b.
This assumption led to the definitions of the two symbols 0 and —d,
zero and the negative number (3239, 40).
3. From the assumption of the permanence of the Laws I-V
and VII, were derived the definitions of addition, subtraction, and
multiplication of the symbols 0 and — d (see Ml), and as has been
shown in i41, these assumptions were sufiScient to determine the
meaning of these operations without ambiguity.
4. The Laws I-V, VII, and Definition VI were derived from the
properties of numbers and the definitions of their fundamental oper-
ations; on the contrary, in the case of the symbols 0 and — c?, their
characteristics and the definitions of their operations were derived
from Laws I-V, VII, and Definition VI.
5. With the introduction of the negative, the character of Arith-
metic undergoes a decided change, which gives rise to a symbolic
Arithmetic or Algebra.
Arithmetic is already in a sense symbolic, since equations and
inequalities involving letters as symbols for numbers are used in
ilrithmetical investigations. But its equations, symbols, and opera-
tions can be interpreted in terms of the realities which give rise to
them, namely, the numbers of things in actually existing groups of
things.
The introduction of the negative cuts off this connection with
reality. The negative, (— c/), is purely symbolic, because it is a
symbol which stands for an operation that can not be effected with
groups of things which actually exist.
47. Not only do the symbols and the fundamental operations per-
formed on them lose all reality, but the equation, which is the fun-
damental instrument in all mathematical calculations, also loses its
reality. In its primary definition, the equation is a declaration (J4)
of the existence of a one-to-one correspondence between two groups
of things. With the introduction of the negative, it loses this
interpretation and becomes a mere statement regarding two combi-
nations of symbols, that, in any reckoning, one of them may be sub-
stituted for the other.
48. Subtraction of Polynomials.
Example 1. Subtract Ix^y — bab-^ 2m« from Axhj — 3 afe -f 5 n.
48 COLLEGE ALGEBRA CW9
Changing the sign of each term of the subtrahend (238, 1 and 2,)
and adding the result to the minuend, we have ^
— 3 x^y -|-2aZ»-|-5n — 2 m*.
49. It is customary in subtraction, to perform mentally the oper-
ation of changing the signs of the subtrahend.
Example 2. Subtract l — m — 4(p — q) from 3(/ — m)— 2p+2 q.
3(l-m)-2{p-q)
2(Z-m) + 2(p-9).
The expressions (l — m) and (^ — g) are to be treated as simple
numbers in the subtraction.
Add the following:
1. m — 2 2.
n-1
5.
+ 7a
-3a
+ 4a
-6a
9.
+ I2y2
— 9yz
+ Syz
— ISyw
10.
EXEBCISB Vn
wi — 3
3. m — 2n
4.
Sx-'2y
3-n
n — 2m
8.
Sy-2x
-lSb(x-y)
7. +13
-9x
+ 56(x-y)
-17.r
+ ^
+ I0h(x-y)
- 8.r
-9:r
- 96 (X- 3')
+ 2x
-7t
+ 7p
11. +a •
12.
+ ^
-97
-6
+ y
-2/,
+ c
— z
+ Sq
+ b
-2x
13. 7a-36 + 2c — 3cf 14. 9j* + 3t/-42+ 8
5a-46-5c+7d -7.r-3y-22-17
15. 8m— ri + 7u + 3t; 16. a + 6 — c+ d
— 9m + 4n — 7u — 5t> a — 6 — r + 3cf
17. 3a-46 + 5c+3d+7«-8/+ g-h-Sk-t
2a+ 6-3c — 7d-7^-9/-2^ + ;i+ it
18. — a+ 56 + 8c— 9d-10c+12/— 7^
+ 8a-10fe + 5r-10</-12tf-13/+7^— ;i + 2ib
19. 75a — 55 6 + 199c-28cZ-23€-45/-25<7~78^
21a + 436-271c + 87d+14e— 9/-25<7+78;i
H9] SUBTRACTION AND THE NEGATIVE INTEGER 49
20. li:r-3i3/ + 2j2-7jM + Ut;-5ip *
21. ia-ib + ic-id + ie--^,f+ig
22. 0.8a — 3.47ft— 1.73r + 0.05rf — 38.7 <j-41i a: + 63i y
1.9a-3.a5fe + 5.7 r-8.1 (/+ 9.87f + 37.8 r- 61.06 y
23. 5.3 a + 0.5fe — 9J c+ 3f rf+7.75e — 17} p + 2.1 q
L86a-91 /> + 7.8r+14.4d-8xV ^- 2.25 /> -1.729
24. 5a — 364-3C— d 25. 7t— y+ u— t>
— 3a+ b +7d -5.r + 4t/-8u4-4t>
+ 2a-56-8c+ d -2.c + 52/ + 3u — 7t>
— 3a + 4/> + 7r-9/i + or - 8 ?/ -f4u-4t
Subtract the second quantity from the first:
26. 9«-86 + 7c — 3d 27. a-26 + 3c-4d
5a-66-3c + 2(i 7a + 3/>-5c+8d
28. 4x — 3y + 9u — 8t 29. wi — 3n+p-7
5jr + 4y — 3u — 8v m — 4n— p + 8
30. a- 6 + c-J + 5^-7/+3/i-7it+ 1 + 5
— 5a + 36-c-d+4<' + 8/-7/t + 9ib-31 — 7
31. a + 6-c — d+ c+ f-^g^h + k — m-S
a — b+c — d—2e+Sf—g + h + l— n — 9
32. 15a-76 + 3c-7d-8<r + m-7j:-2y- 2 + 4
10a + 7/>-3r + 4rf + 4«-p- a+ i/ + 5z-2
33. 73a-52ft-71c + 21rf-52:r + 17y + 59z+lU
54a-606 + 81c + 37d+18j;-33^ + 992+ 7
34. 8.37 a — 9. 49 6 + 8.5 c + 57.6 rf - 5.37 c- 9.07 j: + 0.09 y
3.97a-9.8 6+ 83c- 3.46rf+2.63f-0.57j--8.91y
35. lJa-li + 63ic-55(Z-4ic+}/+};i
-3|a + Ufe + 4}c-3;rf-3lc-^/+i/i
36. |a+J6- |c-V<i+V<?- I/+ i
+ |a-|ft-Vc+Vd- jg-iV-A
37. 5.66a+ 7§6- 27}c-5.73cf + 0.76j'- lij/-27.5«
4ta + 9.386 + 2.65c— 13|rf — 53.7 j: — 0.375 y— 19|z
• Tbe student Is expected to handle the fractions In Nos. 20-28 from bis knowledge
of tbem in Arltbmetic
50 COLLEGE ALGEBRA [2850-52
38. 7}a-4.456+ 19ic + a85d-1.75j~ 83;/-9.5
0.25a- 4|fe-0.625c + 47.5d- 2T"g3: + 1.125 // - OJ
no. 2(a-6)-c+d 40. -{a + b)x + {b + c)y
a — 6 — 2(r~6) (a — 6)jr — (ft — r)y
SIGNS OP AGGREGATION
60. It is often convenient to introduce in Algebra symbols of
abbreviation.
The parentheses { ), J6, the brackets [ ], the braces { |,
and the vinculum , indicate that the numbers enclosed by them
are to be taken collectively. Thus,
{a + b)Xc, [a-^-h^Xc, {a + h\Xc, and a-^-bxc
all indicate that the number obtained by adding 6 to a is to be
multiplied by c.
51. The use of parentheses is very frequent, and it is necessary
to have rules for their removal or introduction.
The rules governing such removals or introductions are the inune*
diate consequences of equations 1 — 5 of ?88. Thus,
3a_4!»+(2c— rf+5c) = 3a— 4fc+2c— ^+5e, [Eq. 4, !38 and
and Law II, (6]
2a—3b^{3c--d+2e) = 2a—3b-^3c+d^2e. [Eqs. 1 and 2, «S8]
In the first case the signs of the terms within the parenthesis are
not changed when the parenthesis is removed; but in the second
case, the sign of each term in the parenthesis is changed from
4- to — and — to + .
The following are the'rules for removing parentheses:
A parenthesis preceded by a -\- sign may be removed tcithout
changing the signs of the terms enclosed.
A parenthesis preceded by a — sign may be removed if the sign
of each term enclosed is changed front + ^o — > or from — ^o -|-.
52. In the removal of the brackets, braces, or ^^nculum, this rule
applies equally well, since the use of each has the same object in
view.
It should be noticed in the case of the last, that the sign which
is apparently prefixed to the term underneath is in reality prefixed
to the vinculum.
Thus, + a — b and — a~— Tare respectively equivalent to + (a — b)
and —{a — b).
{«53, 54] SUBTRACTION AND THE NEGATIVE INTEGER 51
53. Parentheses may enclose others; in this case they may be re-
moved in succession by the rule of 251, the innermost parenthesis
being removed first.
Example. — Simplify 6 .r — j 2 x + (— 3 .r _ 4.r — a)j .
First remove the vinculum and then the others in succession, thus:
5x— j2x+(— 3x — 4x — a) j = 5x— j2x + (— 3x — 4x + a)|
= 5x— j2x — 3x — 4x + aj
= 5x — 2x + 3x+4x--a
= lOx — a.
54. To enclose any number of terms in a parenthesis, take the
converse of the rules in 3238 and 51.
Any number of terms may he enclosed in a parentJiesis preceded by
a -f- sign, without changing their signs.
Any number of terms may be enclosed in a parenthesis preceded by
a — sign, if the sign of ea^h term w changed, from -f- to — , or from
— to +.
Example 1. Enclose the last three terms of
3x» — 8x*+3x' — 5x« — 2x
in a parenthesis preceded by a — sign.
Result, 3 X*— 8x* — (— 3x» + 5x« + 2x).
Example 2. Simplify 3 [jx — (x — 3) } + (2 x —'3 — 2x)].
3 [ j X — (x-3) } + (2x - 3 - 2x)] = 3 [(x - x + 3) + ( 2x - 3 + 2x) ]
= 3 [3 + 4x - 3] = 3 [4 x] = 12 X.
BZBB0I8E Vm
Simplify :
1. 7a-96 + (a + 6).
2. 15a-76— (7a-56).
3. 6a + (3a — 26) + (a + 26).
4. (a + 6~c) + (a-6 + c).
6. (a+h — c) — {a — h + c),
6. (7a-36)-(5a + 36)-(a-56).
7. (8.r-5) + (3x-7)-(9a:-ll).
B. 12-(5a:-6)+(3x+l)-(3r+10).
52 COLLEGE ALGEBRA [454
9. {6a-36 + 7r)-(a-6 + c) + (2a + 6-6(r).
10. (3 m — 7n-5p) + (2m + 4?i-3p)-(4m— 3n— 6p).
11. (6:r + 52/-32)-(5x-3»/ + 22)-(j:+7y-42).
12. 56 X + (934 y- 307) - (1000 1/- 44 j:- 207) + 100.
13. (738a -967 6) -(69a -8036) + (766 -643a).
14. 6a«-(3a6 + 2ac)-(2a€-3a6) + (5ac + 7a«).
15. (6 ax + 2pq) - (7 + 4 ax) - (4 pq - 7) +pq,
16. 9x»-(17 + 3.r«) + (17-x)-(8x'3-2:j:«-a;).
17. 2y-(i«x+ly) + (n^-iy)-(iy~M:p).
18. (4iar — 76) — (2}ax-8i6) — (liar + 5|6).
19. 8.3a-(3.7a-2.376) + (0.7a-L76) — (3.2a + 4.76).
20. (2.7.r + 0.07n) - (9.15p - 0.62n) - (0.69n - 1.46p + l.Tor).
21. rn+[(a^b) + (b + d)l
22. m+[(6 + c)-(m+rf)].
23. w-[(rt-6)-(c-m)].
24. m-[(x-2/)-(a-m)].
25. (7a -2 6) -[ (3a -c)- (26 -3c)].
26. (9a-4c)-[((36-4c) + 5a)-3 6].
27. (8a + 36)-[36-(4o + (a:-7a))].
28. (3a: + 52/)-[(7:r-3T/)-(5x-72/)]+(a;-y).
29. ((3a-46)-2x)-((3.r + 36)-(4jr-2a + 6)).
30. (8m-l) + 5p-((3? + 4p-l) + 7m-(29-p)).
31. ((8x-3i/)-5t/ + 6)-((5a:-72/)-(3r-6))-(6x-y).
32. 8}n- (3ip -(p-5.5n) ) - (5ip + (2n-0.5/)) ).
33. (2i.r-(3}y + O)-((0.75.r-O.5y) + (iar+iy-0).
34. (7.01p-(2.5r- 1.74) )-((4ir-0.79p)- 3.26) -Itp.
36. 8.08ar-(0.55y-(p-7|2-) + 7iy)-(0.33ar-}y).
36. (6.45a6-(0.8a;-3.7))-((3ia6-7.3ar) + 4.2)-6lx.
CHAPTER IV
POSITIVE AND NEGATIVE NUMBERS
Multiplication
65. Multiplication of Monomials.
1. Multiply 7 a by —36.
-36 = (-3)6. [841]
Whence 7a x (—36) = 7a x (—3)6
= 7x(-3)Xax6 [Law III, {7, 6]
= -21a6. [Ml, 6J
2. Multiply — Ox^ya by 6ajy; m being a positive integer.
(— 9x"'^«)X6xy= — 9 • 6x"» • x*yy»-z [Law IV, J7]
= — 54 x^'+Vz- [Ml, 6 ; 89, rule]
Hence the following rule for the product of two monomials:
To the product of the numerical coefficients (841, 6, 8,) annex
the letters^ giving to each an exponent equal to the sum of its exponents
in b*}th the multiplicand and multiplier,
56. From preceding sections
(-a) (-6) (-c) = a6(-c) [841,8]
= -a6c [841,6]
and (— a) (— 6) (— c) (— cO= (— a6c) (— d) [Preceding eq.]
= ahcd, etc. [841, 8J
The product of three negative terms is negative; of four negative
terms is positive ; and so on.
In general, the product of any number of factors is positive or
negative according as the number of negative factors is even or odd.
Example. — Find the product of — 3aV^', 56V, and — 9cff.
Since there are two negative factors, the product is positive.
Whence (- 3 a«6») (5 6V) (- 9 c(P) = 135 a«6Vrf«.
54 (X)LLEGE x\LGEBRA H § 57, 58
57. Multiplication of Polynomials by Monomials. — The third
law of multiplication gives
a(h + c) = ah + a€ [87, V]
a(b^c)z=ab^ac [«7,V; ?38, 5]
a (- 6 - c) = - a6 - a<!. CJ7,V ; HI, 8J
Ilence the following rule:
Multiply each term of the multiplicand by the multiplier, and add
their products, observing the law of multiplication that like sigtis give
plus and unlike signs give minus.
Example. — Multiply 2x'--4x + 5 by — 9x'.
By the rule,
(2x'— 4ar+5)x(-~9x') = (2.r«) (— 9x') + (— 4x) (_9x») + (5) (— 9x')
= — 18x«+36u:*— 45x».
68. Multiplication of Polynomials by Polynomials.
From the preceding sections,
1. {a+b) {c + d) = (a+b)c + (a + b) d [Law V, 27]
= rtc + ^c + ad + bd.
2. (a + 6) (c — d) = (a + b)c + (a + 6) ( — d) [Law V, 87]
= ac +bc— (a + b)d [Law V, and J41, 6]
= ac + 6c — {ad + bd) [Law V, {7]
= ac +bc — ad — bd. [{88, 1]
3. (a — 6) {c + d) = (a ^ b) c + (a -^ b)d [LawV, §7]
= ac + (— b) c + ad + (— b) d [Law V, J7]
= ac-^bc+ad — bd. [HI, 6]
4. (a — 6) (c — 1/)= (a — 6)c+(a — />)(— f/) - [Law V]
= ac+(—6)c-fa(— </) + (— 6)(—cO [Law V]
:=^ac-^bc — ad + 6(/. [§41, 6, 8]
Whence follows from equations 1, 2, 3, 4, the rule:
Multiply each term of the multiplicand by each term of the multi-
plier; if the terms have the same sign, prefix the sign -{- to their product ;
if they leave different signs, prefix the sign — ; then add these partial
products to form the complete product.
1. Multiply 2a — 36 by 3a — 7 6.
According to the rule, multiply 2 a — 3 6 by 3 a and then by — 7 6
and add the partial products.
2i59, 60] POSITIVE AND NEGATIVE NTJMBERS 55
In practice the work is usually simplijied by arranging similar
terms in the same column. Thus,
2a— 3ft
3a— 7b
6a«— dab
— 14a6 + 21fe«
6a«— 23a6 + 216*
2. Multiply 6 x*+ 6 x — 2 — 3 x»— x« by 2x» + 2 + x.
It is convenient to arrange the multiplicand and multiplier in the
same order of powers of some common letter and to write the partial
products in the same order.
A polynomial is arranged in descending powers of some letter,
X, if the highest power of x comes in the first term, the next highest
in the second term, and so on; in ascending powers of x, if the
powers of x are arranged in the reverse order.
Arrange the multiplicand and multiplier with respect to the
descending powers of x. For example,
6 X*— 3 x' — x« + 6 X — 2
2 x'+ X + 2
12x'— 6x«— 2x*+ 12x*— 4x'
+ 6x'*— 3x*— x'+6x«— 2x
+ 12 X*— 6 x'— 2 x'+ 12 X — 4
12x^—6x«+4x'^+ 21x^—11x3+ 4x2+ lOx — 4
69, The rule in 154 has an application which is very useful in
consequence of the different forms in which the product of several
binomial or polynomial factors may be written. Thus,
(a— ft) (c — cZ) = (ft — a) {d-c)
= - (6 _ a) (c — d)
= ^(a-b) (d — c).
60. In like manner it can be shown that in the indicated product
of more than two expressions, the signs of any even number of them may
he changed without altering the product^ but if the signs of any odd
number of them are changed^ the sign of the product is changed (?66).
Thus, (a— ft) (c — cf) (c— /) may be written in any of the forms
(a-ft) W-c)(/-e)
(b-a)(d-c){e-f)
(b--a) {c — d) (/— c), etc.
56 COLLEGE ALGEBRA [*5«
EZSBOISE IZ
Find the prodact of the monomials:
I. 5a, 9ab, 46a«6. 2. —3, 7m«, — 21m«.
3. (-3a«)(-4a»)(-12a*). 4. {-7xy) {-6yh) {^2xyh).
5. (5 a^x) (— 3 ««x") = — 15 a"«+« j:"+>. 6. (9 xy) (— 13 ojr) (15 ab) =
7. (6a6«j»)(3a»6«)(-9ay«)= 8. {Sp^) (Sp^^ (\9j^f^ =
9. (8 jry«2») (- 9 x»yh^) (- 5 jr»y2) =
10. (7 am^n) (- 3 6«/i«) (- 4 a6) (a6«n«) (- 2 6«n«m) (— m«n) =
Multiply:
II. 2.r-ybya: + 3y. 12. 3j« + 4j:* + 8x+ 12 by 2jr — 7.
13. (5x + l)(7y--2). 14. (2a-36-5x) (5m — n).
15. {x + y){x + y), 16. (7x + 5)«.
17. {u-v)(u — v), 18. (3a-4)«.
19. (6x-by)K 20. (l-x)*.
21. {m^n){m + n). 22. (a + l)(a-l).
23. {7x + Sy){7x'-Zy). 24. (3j:-2) (2x + 3).
25. {Sx-7y)(7x+Qy). 26. (3.2 a - 5 6) (5 a - 2.8 6).
27. (2.6.r + 0.3y) (5r + 0.7.v). 28. (3.5 a: + 0.2) (8.4 jf — 0.3).
29. (7.25 + 4. r) (2.8- 3.6 j:). 30. (3y+2}) (4.8y — 1.5).
31. (7i a -0.3) (2.8 a + 5}). 32. (a + 6 + c) (a+6-c).
33. {a + b-c)(a-b + c). 34. (a + 6 + c)«.
36. (3a + 6-uf. 36. (2a-36 + a:)«.
37. (3jr-5y-2)«. 38. (oa - 2 aft + 66) (a + 6).
39. (oa + 2 a6 + 66) (a — 6). 40. (xx + xy + yy) (x — y).
41. (j:ar — ary + yy) (.r + y). 42. (ra: + xy + yy){xx — xy + yy).
43. {aaa — aa6 + a66 — 666) (a + 6).
44. (8 aoa + 4 aa6 + 2 a66 + 666) (2 a — 6).
45. ((a + 6) + (-r + y)) ((a + 6)-(.r + y)).
46. (a + 6 + c + rf)(a-6 + c — d).
47. (a-b + c-d) (a + 6-c-d).
48. (3a + 26 + 5.r-y)(3a + 26-5a; + y).
49. (6ac-3a(i + 26c-6rf)(6ac-3ad + 26c + 6d).
50. (4a6 — 6aj- + 26y — 3a-y) (4a6 + 6aar — 26y — 3ary).
61. (a + 6)(a + 6)(a-f 6). 52. (a-6) (a-6) (a-6).
53. (x + l)«. 54. (l-y)«.
55. (2a-6)». 56. (3a:-4y)«.
57. (j:-l)(j--2)(jr — 3). 58. (x — a) {x — b) {x — c).
59. (2ar-3)(3j:+7)(6jc-5). 60. {Sx + 6) {7x + b) {2x — l).
61. (x-3) (x + 4) (2--5) (J- + 6).
62. (a« + a6 + 6«) (a» - a«6 + 6») (a - 6).
CHAPTER V
positive and itegativs numbers
Division
61. Numerical Division. — Division is the operation inverse to
multiplication. To divide a by 6, is to find the number which mul-
tiplied by b produces a. The result is called the quotient of a by bj
and is written ^. Hence, in accordance with the definition of division,
if follows formally;
IX.
©'="•
As in the case of subtraction, division can not always be accom-
plished. It is only in a special case that a group of a things can
be regrouped into subgroups each containing b individuals. For
consider the following a and b groups .
The a group can be subdi-
vided into the 3 groups on
the right of 4 things each;
a
a
a
a
a
a
a
a
a
a
a
a
The a group or 12 group. The b group or 4 group.
Here the 4 group can be counted out of the 12 group three distinct
times; that is, the quotient of 12 by 4 is 3.
But if the a group contained 15 things and the b group 4 things,
then after counting out 3 groups of 4 things each from the a group,
there would not remain enough things to form a fourth group of 4
things; and accordingly 4 is not contained an exact number of
times in 15.
67
58 COLLEGE ALGEBRA CW2
The number a to be divided is called the dividend^ the number
divided by is the dtvtsor^ and the result of the division is the quo-
tit nt. Formula IX translated, is:
The product of the quotient hy the divisor is equal to the dividend.
The number of times a group of 4 individuals can be counted out
of another group of 4 things is 1, or a group of a things out of
another group of a things is 1. Thus:
X. 7=1 and - = 1,
4 a
which satisfy the definition of division:
(quotient y 1) X {divisor ^ 4) = divideudy 4 and
(quotient y 1) X (divisor y a) z:z dividend y a,
<|a. Kuin^tcal Divisi«a Gives a Single Result When division
^n be effected at all, it leads to but a single result; it is determinate.
There is but one number whose product by h is a. Suppose the
quotient ^ has two values, c and d\ then by IX, i61|
ch=i a
and dhzzz a
and then ch = dh
c=Ldy,
because h groups of c individuals each can not be equal to 5 groups
of d individuals each unless c = ^ (?4).
I^oTi.— The case & — 0 is excluded, sJoceOL&uot % niunber in the sense in which tfaAt
word Is used in this discussion.
Then formally
XL if ch = db,
c = d„
This theorem is of vital rm,poi:tance. It declarea that if a product
and one of its factors are determinate thie other is also determinate;,
or should one of the factors, of the prodii*ct change while the other
remains unchanged the product changes. The possibility of divisions
in the arithmetical sense depend^, upon the truth of this theoremi
alone. The fact tha^t Laiv XI does, not hold for 0, that we can not-
divide by 0, is clear; fo^ if oofi oi the factors of a pr<K&ict & 0, the
product is 0, however t^hje o>tber factor may changOi Thus let the^
quotient of - be g ; tlpt^n by definition of division: q X: 0 should equall
5; but g X 0 = 0 so long as q is any known fixed: quantity whatever;.
hej^f^ \)^. assumption that j- can have a definite fixed, value is false:.
« 63, 64] POSITIVE AND NEGATIVE NUMBERS 59
68. The First Formal Rule of Division.— The theorems of
division are the formal consequences of the fundamental laws of
multiplication, 37, namely,
III.
ab = ba,
IV.
V.
of the definition
a {be) = a6c,
a (6 -f c) = a6 -f oc,
(1)'=-
of the theorem
XI. if ac= be
a=zb unless c = 0,
and the corresponding laws of addition and subtraction.
The rules of division can be deduced in the same way as the
rules of subtraction and multiplication (338, 1-5).
Equation 1.
a c ac
bd"^ bd'
Because
= ac
..)
[L<
»w III, J7]
[Def. IX]
and
ac , -
[Th. XI]
Hence
/a c\ - , ac , ,
(-X-,)xW = -,;^x6d
.'.
a c ac
[Th. XI]
TTie ^oduct of two quotients is eqttal to the quotient of the product
of the dividends by the product of the divisors.
64. The Index Law.
1. Required the quotient of a^ by a\
[«8]
a
X«
X
a
X«
X
a
a
Xa
X
a
X 1
X
1
a
a
a
a
a
1 , [«63, 1]
a a a I \
= 111yJ [LawX, J61]
8
= a*, • [Index Law]
a
since ^ -=. p\ for by definition of division, i> X 1 >« equal to p.
60 COLLEGE ALGEBRA H 2 65, 66
2. Required the quotient of a"* by a", where m and n are posi-
tive integers 8uch that m is greater than n.
a^ a • a ' a to wi factors
a" a ' a ' a to n factors
[«8]
a a
a a
=z a- a ' a to *'"" factors [Law X]
to n factors a • a • • • to m — n factors,
[863, 1]
That iSy the exponent of a letter in the quotient is equal to the exponent
in the dii^idend minus the exponent in the divisor,
65. Exponent Zero.
According to the index law, the quotient of a"* by a* is a*"**,
a^ What numerical value has a®? According to rule, J64,
«*„. = ?.£. «.^ = l .1.1.1 = 1. [LawX, «61]
a* a a a a
^ a^ ^ a a a .-. ^ ^
and — =a° = -•-•- torn factors =1-1 torn
a"* a a a
factors = 1.
An integer raised to the zero power is unity ^ t. c, a® = 1.
66. Division of Monomials.
— 14a6
1. Find the quotient of
2rt
Uab (—7) -2 a- 6
[Ml, 6]
[«73, 1]
2 a 1 • 2 • a • 1
_(-7) 2 a h
"~ 1 " 2 a \
= (_ 7)^, [Law X «61; Note, 862]
= -76. [841, 6]
By the rule of signs in 841, 6 and 8, for the products of positive
and negative factors, plus times plus or minus times minus produces
plus, and plus times minu^ or minus times plus produces minus.
For the present, the analagous laws are assumed for the quotients
of positive or negative numbers by positive or negative numbers,
-|- divided by -f-, an(l — divided by — , produce +, -f~ divided by
— , and — divided by -f-, produce — .
This law will be established in Chapter X.
J67]
POSITIVE AND NEGATIVE NUMBERS
61
2. Divide 105 x»y V by - 15 7^fz\
105ccy2'
105 x» / z'
— 15
x*y>z*
~~"
— 15
x«
y' z'
=
— 7 •
x»-
..^-s
•:zz.
-7x»z».
[1, J68]
[2264, 65]
Hence the following rule:
The quotient of one monomial hy another is the product of the quo-
tient of the numerical coefficient hy each letter with an exponent equal
to its exponent in the dividend minus its exponent in the divisor^ and
omittinq any letter having the same exponent in the dividend and
divisor.
1.
4.
10.
13.
16,
19.
20.
22.
BXEBOISB Z
?=+»•
2.
— a
3.
a
^'=+'-
6.
4m
6.
M=-^"-
12u6c
2ac ""
8.
^
—x*"
9.
^=
A2ahcd
11.
aby
lay
12.
27 a»
lac "
—9 a*""
~54rt*6»c«
— 2a6« ""
14.
64a*xy
— Say "~
15.
2x^ ~
llx«
= 17.
4aWX lOrWz _ . _ 21^*/*22« .
^M,2 - 1®- YxyH (-2jryz)=
5a/>V
(24a»6»x + 8a«6«) + (36a«&*;r« + — 6a»6«x)=:
84a*"+« ^, 81 a«-*
7a*»-» "~
106a*"+*»-*-
21.
3a»
• 5 a*'*+''~*.
For
67. Division of Polynomials by Monomials.
Equation 2. — Second formal theorem of division.
a±& a h
c c c
(a h\ a h
c cj c c
z=i a :±zb.
[261, IX]
[Law V]
[261, IX]
62 COLLEGE ALGEBRA [168
Hence (?^^) , = (?± ^),
and .-. iM'^l^^i. [W2,XI]
c c c
Accordingly the following rule holds:
Divide each tenn of tlw dividend hy the divisor and add the partial
quotients.
Example.— Dmde 12 a«6»— 20 a»c — 16 ay>c3 by — 4a«.
12 g'fc^ — 20 a'^c^ 16 a'hc' _ 1 2 aV>» _ 20 a^c _ 16 a^h^
— 4 a* ~~ — 4 a* — 4 a* — 4 a*
= _3/>3^5ar+4a*6c».
BXEB0I8E ZJ
Divide:
1. a:* — 4x2/ by :r*. 2. a:^ — 7 :r* + 6 2** by a^.
3. 10a:*— 8a:^ + 3j^bya:S, 4. 27.i'»— 36j* by 9a:*.
5. —2^xy*—mx^y by — 12aj/.
6. 35mV— 2Im«//« + 28my* by —7 my.
7. 12 a'^^* — 8 a»6« + 10 a%^ — 14 aM by 2 oft*.
8. 36 Jt^yz^ — 8 .r*i/23 — 16 x^y^r^ + 28 a^y^^ by — 3^z,
9. - 9 a^U^d' + 12 a*//^c* — 18 a^b^(? by — 3 a*6*c«.
10. 14a:P+V^^— 21J^^-V-^+49a'*Py*« by —1j(^-^\^''\
11. Divide 36 mV+ 28 m*y*-- 4 rwt/8 by 4m*y.
12. Divide 6 a^Jt^ - 14 a^a^ + 12 a*a-» — a^a^ by - 2 a«a:».
13. Divide2aV"*'* — 3a:*y*-«» — 4a-»+"»y»-» by ar*-*^"*"-
H. Divide a^"j/" + . I* V" + y*'* hy a:«y».
16. Divide wV-"W*^~'i/^'*""+^'^"*'y" hy m'"*^"-**.
68. Division of Polynomials by Polynomials. —
In case both dividend and divisor contain more than one term,
the operation of division in Algebra must be performed in the same
way as Long Division in Arithmetic. . The following rule can be
given:
Arrange both dividend and divisor ax^cording to the powers of some
common letter^ either both according to ascending or both according to
descending powers.
Divide the first term of the dividend by the first term of the divisor,
and write the result as the first term of the quotient. Multiply the
whole divisor by this term, and subtract the product from the dividend.
568J
POSITIVE AND NEGATIVE NUMBERS
63
arranging the remainder in the same order of poioers as the dividend
and divisor.
Regard this remainder as a new dividend and repeat the operation
till all the terms are brought down.
Thus in Long Division in Arithmetic:
Example 1.
256800
282159 Ml^ Divisor 282159
256800 879 Quotient ^'' 321
25359
22470
"" 321
+ 22470
2889
2889
+
321
2889
321
=
800 •>
■ +
70
+
9
Partial
quotients.
879 Quotient
Example 2. Divide a' — 2 ocy + y* by a; — y.
05* — 2 xy + y*
05* — xy
y
Divisor.
Quotient.
— ar.y+y»
The reason for the rule is that the whole dividend may be divided
into as many parts as may be convenient, and the complete quotient
is found by taking the sum of all the partial quotients.
Thus,
sc* — 2 xy -(- y' X* — xy — xy -|- y*
X — y ~" ^ — y
_ x(x — y) — y(x — y) .
~ X — y
_ x-y _ X— y [?67, Eq. 2]
""^*x-y ^'x-y [?68, Ex. 1]
= X — y. [Law X]
Divide x* '\-x^ — 4iX^-\-bx-
[By addition]
[LawV, ?7]
Example 3. Divide x* + x' — 4 x« + 5 x — 3 by x* + 2 x — 3.
Arrange the dividend, divisor, and quotient according to descend-
ing powers of x,
x*+ it' — 4x« + 5x~3
x« + 2 X — 3 Divisor.
a;4 4.2x' — 3x«
.c* — X + 1 Quotient.
_ x'— x«+5x — 3
- .x'-2x«-|-3x
x«+2x — 3
x«4-2.r — 3
64 COLLEGE ALGEBRA LU69, 70
The same result will be obtained by arranging the dividend and
divisor according to the ascending powers of x. Thus,
— 3+2a;+ x«
3 + 2x + g*
l_x + a«
3x — 5x« + x»
3x — 2x« — X*
— 3x« + 2x» + aJ*
— 3x« + 2x»+aj*
69. The operation of division can often be shortened in certain
cases by the use of parentheses.
Divide (a — 6) x» + (6» — a»)x + ab{a* — h*) by (a — 6)x + a« — ^►«.
(a^b)x + a* — Ij^ Divisor.
jc* ■— (a -|- 6) X + a6 Quotient
— (a«~6«)x« + ih^ — a») X + a6 (a» — i»«)
a6(a — &)x + a6(a«— Z»*)
ah((i — h)x + ab(a^ — h*)
70. It may happen as in Arithmetic, that the division can not be
exactly performed. Thus, for example, if a' — 2 a6 -f 36* is divided
by a — 6:
a»-2a6 + 36«
ab
a-b
— ab + 3b^
- ab+ b*
2 6' Remainder.
The result is expressed in a manner similar to that in Arithmetic: '
r =z a — b-A -: that is, there is a complete quo-
a — 6 a — b «--»
2 6*
tient, a — 6, and a fractional part, . In ifl and Chapter X,
a — 0
algebraic fractions will be further considered.
If we multiply both members of the equation above by a — 6,
we have
^ X(a-6) = {a-b){a-b) + -- rCa-^)
a — o a — o
which gives the identity
a» — 2a6 + 36« = a« — 2a6 + 6« + 26».
170] POSITIVE AND NEGATIVE NUMBERS 65
Hence, if the division of one polynomial by another is not
exact) and if the dividend, quotient, divisor, and remainder are re-
Bpectively represented by 2>, q, d^ R, we have the formula:
I) = qd + R.
That is, the dividend is equal to product of the quotient, at any sta^e,
by the divisor , plus the remainder at this stage. Thus,
49^5 = 9 + 1 = 91; .-.49 = 9x5+4
3
and a5« + « — 9-5- (x + 3) = x — 2 — — -r
x + o
x« + x-9 = (x— 2)(x + 3) — 3.
NoTB.— It Is very Imporiiint to arrange the terms of the dividend and divisor in the
ascending powers or descending powers of some letter, and to keep this order through
out the operation.
BZBB0X8B ZH
Find the quotient of:
1. a« + 2a6 + 6«by a + 6
a+b Divisor
o + 6 Quotient
a« + 2a6 + 6«
o«+o6
<r6 + 6«
a& + 6«
4. (ac — a<J + 6c — 6d)-H (c— ef). 6. (mm — mar — m+a?)-i- (m — 1).
6. (6am — 9an — 46m + 66n)-i-(3a — 26).
7. (6ac-2arf + 4a/-96c + 36d-66/) + (2o-36).
8. (2ax — 66ar + 8cx — oy + 36y — 4cy) -i- (2a: — y).
9. (a'+a6-2ft») + (o-6). 10. (3a« + a6-26«) +(3a-2 6).
11. (a»-o*6 + 2&»)H-(a + 6). 12. (63:» + a;»-29a; + 21)+ (2j:-3).
13. (2:r»-2-p«-61x + 71) +(2\i:-3).
14. (a»-6^ •♦- (a- 6). 15. (a»+6») + (a + 6).
16. (81a*-166*) + (3a-26). 17. (a» + 6*) +(a + 6).
18. (9a«fc«-4aV + 4a6c»-6«c«) + (3a6-2ac + 6c).
19. (a«-t» + 26c-c^ + (a+b-c),
20. (3a« — 4a6+8ac — 46« + 86c-3c«) -i- (a — 26 + 3c).
21. (j:« — 2jr« — 4y«+8y2 — 3z«) + (a: — 2y + 2).
22. (16a:«-4a«+9o«6«-366«a:«) + (3a6- 2a + 66a:-4a-).
23. (32962r-208(ur + 87a6-153a^- 1566*+ 153a«) + (17a- 136 + 9X).
24. (0.4a^+1.47ar-8.5) + (0.8a;-2.5).
26. (2.21 n« - 1.8 np - 1.61 p«) + (0.7i> +- 1.3 n).
66 COLLEGE ALGEBRA [{70
26. {3.9x«-4.1a^-llfy«) + (liar-3.5y).
27. (2a«- Aaar-lli.r«) + (3.5a?+1.5a).
28. (6a;»-21.38:ri/-6a:2+18.5y»+L64y2-362»)-f-(2.5ar-3.7y + 6z).
29. (0.06rn«+0.01mn — aiSwip — 18.2n«+13.67ni> — 2.4p«) -i-
(1.5p-5.27i + 0.3m).
30. (26«+c«-6a«+Jjf oft + VflMJ- W6c) + {2a-fc+lW.
31. (2ix»-16y»-62^-^-xy^S2xz+^^yz) + (j2-Yy + 3x).
32. (ia»-ffia6 + 2i}ac+i6«-2J&c-2ic«) -h (Ja-fft-Jc),
«o /r «" I 16 , 9,g\ /4« , r 3ft\.
34.
/lO , 343 3 ., , 407. .\ /6a 36 , 2x>
36. 6a;«i( + y» + a:»+6xi/«by4:ry + y«+a:«.
36. 15+2a-3a«+a»+2a*-a6by5 + 4a-a».
37. a;«-2x« + lby^-2a:+l.
38. a;« — 6a?* + 9A-«~4bya;« — 1.
39. Divide the product of ar» — 12 a: + 16 and a;» — 12 a; ~ 16 by a:» — Id.
40. Divide the product of f* — 2a:+l and a;* — 3a;+2
bya;'-3a* + 3ar-l.
41. Divide the product of .r« — :i: — l, 2j;«+3, .r* + ar+l andar — 4
by.T^-3.c8+l.
42. Divide the product of a* + or + x^ and a* + z* by a* + ah^ + a:*.
43. o^ + a*6 + a*c — a6c — l^c — bc^ by a* — 6c.
44. .vi^ + 2 i/z — x}/z + xyz^ — j(^ — 2yjfl-\-a^z — xz^hyy+Z'-'X.
45. a» + 6»— .c3 + 3a6cbya+6 — c.
46. a;»+?/» + 3ri/— Ibyjr+j/— 1.
47. a^ + x^y + 3^ — a^i^-- 2x1^ + 1/ hy j^ + xy — "/.
48. {x + yf-2(x + y)z + z^hyx+y-z.
49. ai-« — a6» + 62jr-j^by(j;+6)(a:-a).
50. (6-c)a' + (c-a)6» + (a-6)c'bya« — a(6 + c) + 6c.
51. 3!^ + (a — b -\- c) J^ -\- {ac —ab — bc)x — abc by ar + c.
52. a^+(3-6).r»+(c-36-2):Ea+(26 + 3c)a: — 2cbya:« + 3x-2.
63. (a2-3a6)a:«+(2a2 + 4a6 + 362)ar — (2a6 + 56*) by aar-6.
54. :r3+6j: + 16 by ar« + 2a: + l.
55. x^ + ax* + b.ir^+cx^-\-dx+e by x^+ax'+bx+c
56. a:* — 6* by x^+ax+b.
57. 3a-s_3 by i^ + Ja:+i.
58. iaH» + ^a:-iV by i-r-f
69. ia:»-Jy» by Ja^^+ary + iy*.
271] POSITIVE AND NEGATIVE NUMBERS 67
60. J:r«-fx«y + V-:n/-27i/» by Ja^-3:ry+9y'.
61. 6:r*»+V"'"^— a^"!/"*"*"*+i-^""*y"'"^*+i^""*y-A'**'"*
by (3:f»"+^^"'+* — Jz*"^"').
62. 6a»»-lla*» + 23a»« + 13a*» — 3o»+2 by 3a»+2.
a3. a:«/»+2j:i/"»+"2+2j:y'»r + 2/^2 + 2y«2r+r« by xy'^+zy'^+r.
(A. 3*^-6** by 3*- 5*.
65. 32jc*» — 9j;»'«(/« + 12a:«"^ — 18arV"* — 523/*" by «" — y«».
Carry the division to five terms in the following:
1 1 1 — x 1— 3r
66.
l-x' 1 + x ' 1 + x' 1+2^
67. -4— and — ^- 68. ^ ''
a-^-x a—x 1— 2a; — a:*
1— ar — -c« 1+x + a:*
Find the remainder in each of the following indicated divisions, and
verify the work by applying the principle in {70.
71. a:» + 3a^+3a:+l by a;«+ar+l.
72. 2 — 3ar-2a«x« by l + 2aar.
73. 18ar»-5x+l by 6a*+2a:+l.
74. (x-y)*-2(a:-y)«+l by (:r-y)«+2(a;-y)-l.
75. 4a^2/»» — 13a*»y*»+14a;*»y» — 2 a;*'* by a;V"— 2a;*'y»+a:*».
71. The Third and Fourth Rules of Division.
Equation 3.
\h) ^ad
G)
For 7^ • 3=r [I>ef. IX, 861]
G)
•^ i7X5=rS »«».i;i."iv,!7]
a
6*
dc - , , . [Def. IX, m; Law
Smce __Xcrf=</c = lx</c xi, ?62; 118]
68 COLLEGE ALGEBRA [272
dc
|a)=:4a)
and .-. V&/^nrf. [LawXI, «68]
DC
0
Equation 4.
b'^ d
if ad=: he
For ^^) 6 • e/=(rf)^=a^
and (^\ d • ^ = (c)fe = 6c.
If ad=.hc
and
a c
l^d'
72. Equation 1, §63, has a numerical interpretation only when
^ and ^ are numbers. The numerical definition of a quotient gives
it such narrow limits as to make division an unimportant operation
as compared with addition, multiplication, and subtraction as dis-
cussed in Chapter III.
This restriction on division can be removed in the same way as
subtraction was generalized.
We accept as the quotient of a by any number 6, which is not 0.
(i. e., b=^Q) the symbol ^ defined by the equation
(1)'="
which simply declares, that the symbol (^\ b is equivalent to the
symbol «, and that either may be substituted for the other in any
reckoning.
li 73, 74] POSITIVE AND NEGATIVE NUMBERS 69
By this definition it does not matter whether 7 is a number or
not. When - is used as a symbol it is called a fraction^ and on
the contrary a symbol a is called an integral symbol.
Definitions of the addition, subtraction, multiplication, and divi-
sion of this symbol have been given. Moreover they are definitions
which are consistent with the corresponding numerical definitions
and with one another, a3 soon as 1, 263, 2, 267, 3 and 4, 271
are assumed to hold as symbolic statements as well as numerical
statements.
The purely symbolical character of ^ and its operations detracts
nothing from the right to use them, and they establish division on a
footing of at least formal equality with the addition, subtraction,
and multiplication of Aiithmetic.
The complete discussion of the fraction and its properties will be
given in Chapter X.
78. The Indeterminateness of Division by Zero. — Division by 0
does not conform to the law of determiuateness (262, XI) ; Equation
1, 263| and 1, 2, of index law of division, 264, are therefore not
valid when 0 is one of the divisors.
The symbols j: , ? , of which but little use is made in mathe-
matics, are indeterminate.
1. - is indeterminate. Because - is completely defined by the
equation (-\ 0 = 0; but n x 0 = 0, whatever the value of n; there-
fore - may be any number whatever.
2. ^ is indeterminate. Because, by definition, (^\ 0 = a. If
^ were determinate, since then by 263, 1, (^)O would be equal
to -^, or to -, therefore, the number a would be equal to ^, an
indeterminate expression. Therefore, division by zero is not an
admissible operation,
74. Determinateness of Symbolic Division. — The exception to
the determinateness of division just pointed out might seem to raise
an objection to the right to assume that symbolic division is deter-
minate, as is done when the demonstrations 1, 263; 2, 267; 3 and
4, 271, are made to apply to symbolic quotients.
It must be observed that - , ^ are indeterminate in the numeri-
70 COLLEGE ALGEBRA [J75
cal sense, whereas by the determinateness of symbolic division is^
indeed, not meant actaal numerical determinateness, but < symbolical
determinateness, " that is conformity to Law XI, 263, regarded
merely as a symbolic statement For from the present standpoint
the fraction t is a mere symbol which does not have numerical
meaning apart from the equation (^\ b=ay with which, therefore,
the property of numerical determinateness has no possible connec-
tion. The same is true of the sum or difference, product of two
fractions, and of the quotient of one fraction by another.
As for symbolic determinateness, it needs no justification when
assumed, as in the case of the fraction and the demonstration 1, 2,
3, 4 cited above, of symbols whose definitions do not exclude it
The deduction, for example, that because
(M)'^=(S)'^
b d^bd
which depends on this principle of symbolic determinateness, \b of
exactly the same nature as the inference that
(fi)"=f'i^
which depends on the associative and commutative laws. Both are
pure assumptions made of the undefined symbol for the purpose of
giving it a definition identical in form with that of the product of
two numerical quotients.
75. The Vanishing of a Product. — ^It has already been shown
that the suificient condition for the vanishing of a product is the
vanishing of one of its factors (2239, 3; 41, 7; 63, 1). It follows
from the determinateness of division that this is also a necessary
condition. 1/ a product vanishes, one of its factors must vanish. Thus:
Let ab = 0, where a and b may represent numbers or any of the
symbols which have been considered.
Since a6 = 0
ab-\-ac = ae [289, 1]
or a (6 + c) = ae [Law V, 27]
and if a is not 0 b -{- c =z c [Law XI, 268]
or 6=0.
CHAPTER VI
APPUCATIONS OF THE FUNDAMENTAL OPERATIONS
Simple Equations
76. The results learned in addition, subtraction, multiplication,
and division, can now be applied to the solution of some simple
examples and problems.
77. When two algebraic expressions are connected by the sign of
equality, the whole expression thus formed is called an equation
(24). The expression on the left of the sign of equality is called
the first, and that on the right, the second member of the equation.
78. An Identity is an abbreviated term for an identical equation.
An identical equation is one in which the first and second members
are equal for all numbers which the letters may represent; for
example,
(x -\-a) {x — a) ^ x* — a*.
The symbol ^ is read identical with.
70. An Equation of Condition is one which is true, not for all
values of the letters, but only for a certain definite number of
values of the letters; thus,
x + 3 = 9
can not be true unless x = 6.
On account of its frequent use, an equation of condition involving
only the first power of the unknown quantity is called a simple equa-
tion. Here the question always is: <*What number must x be in
order that," say,
x + a= b ?
80. The Unknown Quantity in an equation is a letter to which
a particular value or values must be given in order that the equation
inay be true. Such a particular value of the unknown quantity is
said to tatitfy the equation, and is called a root of the equation; thus,
71
72 COLLEGE ALGEBRA [*81
7 is the value which must be given to x in order that the eqiiAtion
X — 2 = 5
may be satisfied. To solve an equation is to determine the particu-
lar value of the unknown quantity for which the equation is satisfied
or is an identity. Up to this point four operations have been dealt
with — namely, addition, subtraction, multiplication, and division.
81. In the discussions which follow, certain propositions are
needed which are obvious axioms in Arithmetic and which are still
true when the extended meanings in Algebra are given to their terms
and symbols; thus,
1. If equal quantities are added to equal quantities^ their sums will
be equal (Law VII, J38)
2. 1/ equal quantities are taken from equal quantities, the remainders
will he equal Thus, if x + 3 = 8, then taking 3 from each of these
equal quantities will leave x = 5. (Law VII, ?38)
3. Jf equal quantities are multiplied by the same or equal quantities^
the products will be equal Thus, if 5 = 2 -(- 3, then 5x9 =
(2 4- 3) 9 = 18 + 27 = 45; and if a = ft, then a" = 6», |/a = y K7
(Law XI, ?62)
4. If equal quantities are divided by the same or equal quantities^
their quotients will be equal (Law XI, {62)
5. If the same quantity is added to and then subtracted fron^ an-
other, the value of the latter will not be altered. (See 238, 3)
6. If a quantity is both multiplied and divided by the same quan-
tity, its value will not he altered, (J63, Definition IX)
7. Quantities which are equal to the same quantity are equal to each
other.
8. General Axiom, — If the same operation is performed on two
equal quantities, the results will be equal
REMARK.— The student should note that these axioms are true whether the quan-
tities are positive or negative, and when the four fundamental operations have thelr
extended meanings. For example. If < = m, n=p, then <ii = mp, which Is evident
If the quantities are all positive quantities. Suppose that n is a ne^tive quantity,
say —a\ then p Is a negative quantity, since fl =p; and we shall represent p by — ^.
We have:
I (-a) =: m (r-a) = -^ma l|41, Q
and m (— 6) = —mb = (—m)b. [|41^ 6]
But sinoe ~~a = —b, then a = b.
Hence, mi—b) = (—m)a = ^ma. [fAi, ft]
lint ma ZI rna
and —ma 13— ma
/(— a)=m(— ft) [Axioml]
and In = mp.
ii 82, 83] APPLICATION OF FUNDA^IENTAL OPERATIONS 73
88. The axioms can be used to establish some simple rules for
solving the simple equations of the first degree.
1. Any quantity may he transferred from one member of an equation
to the other by changing it$ sign. Thus, suppose
(1) x-3 = 9.
Add 3 to each side by 281, 1 ; then
a; — 3 + 3 = 9 + 3
that is, (2) X =9 + 3.
Here, — 3 has been removed from the first member of equation (1)
and in its stead + 3 appears in the second member of (2).
Again, suppose that
(1) x — a = 6+y.
Subtract h from each side, thus,
(2) X — a — b = b — ?> + 2/ = y.
In equation (1), +6 has been removed from the second member
and in its stead — 6 appears in the first member of (2).
2. If the tign in every term of an equation is changed, the equality
still holds.
This rule can be proved by the preceding section.
Thus, suppose
(1) X — I z:z m — y.
Transposing y — «* = Z — a;,
or I — X = y — *^-
and (2) — x + ? = — m +y.
It will be noticed that the signs of the corresponding terms in
equations (1) and (2) are opposite,
8S. The unknown quantities of an equation are usually repre-
sented by the last letters of the alphabet x, y, z^ u, etc. , and the
known, by the digits or the first letters of the alphabet, thus, 1, 3,
5, 6, 7, a, b, c, dj e, etc.
The following is a rule for the solution of a simple equation:
Transpose all the term^ which involve the unknown quantities to the
first and the known quantities to the second member of the equation;
divide both sides by the coefficient, or by the sum of the coefficients of the
unknown quantity.
The principles just established justify this rule.
Examples. (1) Solve for the value of x,
8x — 29 = 26 — 3x,
74 COLLEGE ALGEBRA [183
Transposing 8x + 3x= 26 + 29
or llx = 55;
by division x = || = 5.
To verify the result substitute the value of x in the original
equation.
Thus, , 8x5- 29 = 26-3-5
40 — 29 = 26 — 15
11 = 11.
(2) Solve for x,
{b + 2)x + ah z= h{a + x) + 4a.
Simplifying hx-\-2x-\-ah = ah + hx-\-4a
4a „
x=— = 2a.
Verification. — Substitute 2a for x in the original equation, thus,
(h + 2)2a + ab=h(a + 2a)-{-4a
2a64-4a + ai = 3a6 + 4a
or 3 aft -|- 4 a = 3 afc -|- 4 a.
BZEBOISE XIII
L 8 + 6a:=20. 2. 4t-8 = 16.
3. 24-7ar = 3. 4. 4a;+5-ar = 8.
5. 9 + 3a-— 2x = 10. 6. Ill— ar-7x = 3L
7. 3I-7x = 41-8x. 8. 19-2x = 5x-16.
9. a:-3 + 6jr-9 + 12x-15 = jr.
10. x=3x+2 + 5x + S+7x+9.
n. a: = 7-5ar+10+8x-7 + 3x.
12. 0 = 6 + 12x-9-8a:+10 + x.
13. 100 + 2j:-9j:+15 = I0-7j: + 6 — liar.
14. 10x-ll-12j;-13=13 + 12x+ll-10a:.
15. 7x-9-9a: + 7 = 9i: + 9-7.r-7.
10 3j+(7-jr)=ll. 17. a:-(8-x) = 10.
18. x-9 = b(x-b), 19. 6x-(3 + 2ar) = 9.
20. 3(ar-2)-7 = 8. 2L 10(x4-l) = llar+7.
22. 4(10-2jr)-3(j:-5) = 0. 23. 3<9-2x)-6(2ar-9) = 0.
24. 7(4ar — 3) + 3(7-8i:) = L
25. 8(3x-2)-7.c-5(12-3x) = 13x.
28. a(x — a^)=h(x-b*). . .
27. (a-l)«(a-J^) + (2a+l)(a-l) = 3ax.
28. f{x-a)+^(x-h) = x,
29. cfl{a-x)-^l?{b'-'x) + ab{a — b)x = 0.
184] APPLICATION OF FUNDAMENTAL OPERATIONS 75
30. a(3h+2x)-2c^=b(b + x).
31. 6a(jr-a)-76(a;-6) + 2a!>=0.
32. (x+b) {x-7) = (x-3) (x- 15).
(Sabtract x* from each ^ide, i. e., cancel it.)
33. 2(16-2r) + 3(5ar-4) = 12(3+x)-2(12-ar).
34. 3(a;-3)«+5(x+5)»=10+8(x-8)«.
35. 10(jr-2)»+5(a;-3)« = (5:c+ll) (3x-21) + 300.
36. (j:-2)(5-a:) + (x-3)(a:-7)-2(x-l) + 15 = 0.
37. (2ar-7) (x+4) = (7-2x) (4-x) + 44.
38. {a+b)x+(a-b)x = icfi.
39. (x+a)(x+6) = (x-a)(j;-6)+(a+6)«.
40. ax(x+a)+bx{x+b) = {x+a)(x+b){a+b).
PBOBZiEXS
84. 1. Divide the number 91 into three numbers so that the
second shall be double the first and the third double the second.
If we know the first number the others are readily found.
Hence, let x = the first number,
then 2 X = the second number,
and 4 a; = the third number,
x-]-2x-\-4x = the sum of the three numbers;
but (1) x + 2x + 4aj = 91,
7x = 91,
X = 13, the first number,
2 X = 26, the second number,
4x = 52, the third number.
Verification, — Substituting the values of x, 2x, and 3x in (1)
13 + 26 + 52 = 91 as required.
2. Find two numbers differing by 9 whose sum shall equal
twice their difference.
Let X = the greater of the numbers,
then X — 9 = the smaller of the two numbers,
hence 2 x — 9 = their sum,
and 18 = twice their difference,
.-. (1) 2x-9 = 18,
2x = 27,-
27
X = — , the greater number,
X — 9 = '-* the smaller number.
76 COLLEGE ALGEBRA [284
Verification. — Substitute the value of x in equation (1).
2(f)-9=18,
27 — 9 = 18, an identity.
3. How many rods of fencing will be required to enclose a
plantation containing 8000 acres in the form of a rectangle if it is
twice as long as it is wide?
Let X = the width of the plantation in rods,
then 2 X = the length of it in rods,
and (2 x) • (x) = 2 x* the area of the plantation in square
rods, or 8000 x 160 square rods.
Hence (1) 2x« = 8000 -160,
x« = 4000 • 160 = 640000 sq. rods,
(2) X = 800 rods, the width,
and 2 X = 1600 rods, the length.
Hence the length of the fence required to enclose the plantation
will be 2 • 800 + 2 • 1600 rods = 4800 rods.
4. The respective ages of a man and his wife are. now in the
ratio of 6:5, but 22 years hence their ages will be in the ratio of
8 : 7. How old are they now?
Let 6 X = the age of the man,
and 5 X = the age of the woman,
then their ages, 6 x and 5 x have the ratio
r^ = - = 6 : 5 as required,
then 6x-|-22 = the man's age 22 years hence,
5 X + 22 = the woman's age 22 years hence,
6x + 22:5x + 22 = 8:7,
or 7(6x + 22) = 8(5x + 22),
42x4-154 = 40X + 176,
2x = 22,
and X = 11,
6x = 66, the man's age,
5x = 55, the w^oman's age.
5. A person who possesses $15, 000 employs a part of the money
in building a house. He invests one-third of the money which remains
at 6 per cent and the other two-thirds at 9 per cent, and from these
investments he obtains an annual income of $450. What was the cost
of the house?
f84] APPLICATION OF FUNDAMENTAL OPERATIONS 77
Let X = the cost of the house in dollars,
then $15000 — x = the money to be invested,
hence = the money invested at 6 per cent)
, 2(15000^0;) ^, . ^ ^ * a
and = the money mvested at 9 per cent,
o
( "" J .06 = the interest in dollars on the first invest-
ment for 1 year at 6 per cent
- (15000 — x).09 = the interest in dollars on the second invest-
ment for 1 year at 9 per cent
That is, (15000 — x). 02 + (15000 — x). 06 = 450, the money earned
by both investments.
300 — .02a; + 900— .06aj = 450
• .08a; = $750
750
xz=—-=i $9375, the cost of the house.
Oo
6. Divide the number 181 into two parts so that 4 times the
greater may exceed 5 times the less by 67.
7. The property of two persons amounts to $3870, and one of
them is twice as rich as the other; find the property of each.
8. A company of 266 persons consists of men, women, and
children; there are 4 times as many men as children and twice as
many women as children. How many of each are there?
9. Divide the number 148 into four such parts that the first
exceeds the second by 10, the third by 18, and the fourth by 24.
10. The sum of two numbers is 5760, and the greater is three
times their difference. Find the numbers.
11. The difference between two nimibers is 21, and the greater is
to the less as 11 : 4. What are the numbers?
12. Two shepherds own a fiock of sheep and agree to divide its
value equally; A takes 81 sheep, and B takes 105 sheep and pays
A $108. Find the value of a sheep.
13. The combined ages of a father and a son are 94 years; twice
the son's age is 11 years greater than the father's age. What are
the ages of father and son?
14. Two casks contain equal quantities of vinegar; from the first
36 quarts are drawn, and from the second 80 ; the quantity of vine-
gar remaining in one cask is now twice that remaining in the other.
How much did each originally contain?
78 COLLEGE ALGEBRA [«84
15. The difference of the squares of two oonsecutive numbers is
25. Find the numbers.
16. A vessel containing oil was filled by pouring in 54 gallons,
and there was then in the vessel seven times as much as at first.
How much did the vessel hold?
17. Forty yards of cloth and fifty yards of silk togetker cost
$1500; and the silk cost twice as much per yard as the cloth. How
much did each cost per yard?
18. A grocer has two kinds of sugar, one worth 7 cents, and the
other 12 cents per pound. How many pounds of each must be taken
to make a mixture of a hundred pounds worth 9 cents per pound?
19. A farm contains 100 acres. Three times A's part is 8 less
than 4 times B's part. How many acres had each?
20. The length of a room exceeds its breadth by 5 feet; and if
the length had been increased by 3 feet, and the breadth diminished
by 2 feet, the area would not have been altered. Find the dimen-
sions of the room.
21. The head of a certain fish is 9 inches long; the tail is as
long as the head and half the body; and the body is as long as the
head and the tail together. What is the length of the fish?
22. A father has six sons, each of whom is four years older than
his next younger brother; and the eldest is three times as old as the
youngest. Find their respective ages.
23. A gentleman gave some beggars 10 cents each, and had a
dollar left. He found that he would have required a dollar more
to enable him to give them 15 cents each. How many beggars were
there?
24. A man has three times as many quarters as half dollars,
four times as many dimes as quarters, and twice as many half dimes
as dimes; the whole sum is $7.30. How many coins had he
altogether?
25. The width of a room is two thirds of its length; if the width
had been 3 feet more, and the length 3 feet less, the room would
have been square. Find the dimensions.
26. What is the number whose third, fourth, sixth, and eighth
parts together are three less than the number itself?
27. What is the number whose m^^ and n^ parts are together
equal tojpf
28. What is the number which increased by m times the number
itself gives a f
CHAPTER. VII
APPLICATION OF ADDITION AND MULTIPLICATION —POWERS OF
MONOMIALS, BINOMIALS, POLYNOMIALS, AND IM-
PORTANT PRODUCT FORMULAE
86. It is desired to derive a rule for raising a rational integral
monomial to any power whose exponent is a positive integer.
Definitions. — A monomial is said to be rational and integral if it
is expressed in terms either of common numerals, or of a single
letter with unity for its exponent, or of the product of two or more
sQch numbers or letters; thus 5a'&', which is equivalent to 5 * a * a
b ' b ' b, IS rational and integral.
A polynomial is rational and integral if each of its terms is
rational and integral ; as 3 x* — J ab^ — c\
An expression which can be reduced to either of these forms is
said to be rational and integral.
1. Find the third power of 5xV-
(5 aV)« = (5 x«y«) - (ba^y^ - (5 x«y») = 125 x«y». [Law III, {7]
2. Find the fourth power of — a.
(— ay = (— a) {— a) (- a) (~ a) = a* [{66 Rule]
3. Find the third power of (— 5 am^.
(-_5aTO»)»==(— Sam^X— 5aiii»)(— 5am») = — 125a«wi». [566,Rule;and 1]
4. Required the value of (a^)" when m and n are positive integers.
(a"')»= a'^'a^'a^ to n factors [Def. of exponent, 18]
^m-Hn-Hn-^- • • • • to n terms __ Qum _— ^rnn
5. Required the value of (ai»)" where n is any positive integer,
(afe)* = (ab) (ab) (ab) .... to » factors [Def. of exponent]
= (a "a'a.... ton factors) (6 •& •&•... to n factors) [Law III, J7]
= a"6«. [Def. of exponent]
The following rule is inferred from these examples:
Raise the absolute value of the numerical coefficient to the required
poweTy and multiply the exponent of each letter by the exponent of the
required power.
80 COLLEGE ALGEBRA [«86, 87
Give to every power of a positive term, and to every even power
of a negative term the plus sign, and to every odd power of a nega-
tive term tlie minus sign ({66).
86. The following examples deserve special notice. The square,
the cube, the higher powers of binomials, and the products of fac-
tors of special forms lead to a series of formulae very useful in
practice.
The most simple of these are:
I
a-f b
a+ b
II
a- ft
o- b
III
a+ b
a- b
a«+ ab
+ ab + b^
a* — ab
- ab + b^
a^—2ab + 6«
a^+ab
-^ab^b*
a«+2a6 + b*
a« — 1«
The first example gives the value of (a + 6) (a + 6), that is, of
(a +6)*; we thus find
L (a + 6)« = o« + 2a6+6«.
Hence, the square of the sum of two numbers is equal to the sum of
the squares of the two numbers plus twice the product of the first times
the second.
Again we have
II. {a — b)^=za* — 2ab + b\
Hence, tfie square of the difference of two numbers is equal to the
sum of the squares of the numbers diminished by twice their product.
Finally
III. (« + b) (a — b)=a^ — b*.
Thus, the product of the sum and the difference of two numbers is
equal to the difference of their squares.
87. The Double Sign. — The sign =h is sometimes used, and is
called the double sign.
Since {a + by= a«+ 2ab + b\
and (a — 6)»= ««— 2 ab + 6«,
it follows {a±:by=a^±:2ab + b\
Thus d= indicates that either the sign + or the sign — may be-'
taken, adtib is read a plus or minus b.
?S88. 89] APPLICATION OF ADDITION AND MULTIPLICATION 81
88. The results in 285 furnish simple illustrations of the use of
Algebra; Algebra makes it possible to prove general theorems con-
cerning numbers and to express these theorems simply. Thus,
{a-{-b) (« — 6) = «.*— 6', is a result expressed in symbols more
compactly than in words.
89. Besides formulae I, II, III, {86, there are many others
which are not so important as these. Among those which occur
with frequency sufficient to warrant their being given are the
following:
X -fa
x + b
ac"-|- ax
-j- 6x -j- ah
TV. 3i^+{a + h)x + ah or (x-f a) (ic+6)=a;«+(a+6)x+afc.
And X* -f (a -|- 6) X + aft
X -|- c
x»-f (a + 6)x« + a6x
cx' + (ac -f- ft c) X 4- oftc
35*+ (a -|- ft -f c) x*-\- {ah -|- ac + ftc) x + ahc^ hence we have
V. (x+a) (x+ft) (x+c) = x'-f (a+ft+c)x*+(ttft-|-ac+ftc)x-|-aftc.
Again
a»-f aft + ft«
a — ft
a» -f a«ft + rtft«
— a«ft_aft«_/,s
a' — ft' hence the formula,
VI. (a — ft) (a* + aft + ft«) = a« - ft».
Example. — (3 X — 2 y) (9x« + 6xy + 4y«) = (3x)' — (2y)»
= 27x»— 8y».
Similarly by multiplication,
VII. (a 4- ft) (a« -ah + ft«) = a» + ft».
Example.— (2 x -f y) (4 x« — 2 xy + y«) = (2 x)« + y» = 8 x» 4- y».
82 COLLEGE ALGEBRA L589
Particular cases of the Binomial Theorem can be found by mul-
tiplication. Thus:
{a + hy = a + b.
(a + 6)«= a* + 2ah+b*. [J86, H
(a -f 6)« = (a + 6) (a + 6)» [Def. of exponent]
f a«+2a6 + 6«
a» + 2a«6+ ai*
a«6 + 2a&« + 6»
= (o + 6) (a» + 3a«6 + 3a&«+fe»)
The expressions for the first, second, third, and fourth powers
of (a -|- 6) all have the following properties:
VIIL 1. The number of terms in the expansion is one greater than
the exponent of the binomial,
2. The first term of the binomial appears in the first term of the
expansion with the exponent of the binomial. It appears in the second
term ioith an exponent diminished by unity; and so on in the other
terms,
3. 7%« second term of the binomial appears to the first power in the
second term of the expansion^ and its exponent is increased by unity in
each succeeding term,
4. The coefficient of the second term of the expansion is equal to the
exponent of the binomial^ and if the coefficient of any term is mul-
tiplied by the exponent of a in this term and divided by the number of
the term the quotient is the coefficient of the term following.
5. 7%€ degree of any term is etiual to the exponent of the binomial.
The results arrived at in 1-5 constitute as a whole the Binomial
Theorem,
Example. — Find the cube of 3x — 2y'.
By the theorem,
(3x-2y«)» = (3x)»-3(3x)M2i/«) + 3(3a;)(2y«)«-(2yV
= 27x'-3(9x«)(2^«) + 3(3x)(4y*)-8y«
= 27 x» — 54 xY + 36 xy* — 8 y\
S90] APPLICATION OF ADDITION AND MULTIPLICATION 83
The sqaare of a polynomial of three or more terms is useful.
(a + b + c)*= l{a + b) + cy
= ia + by + 2(a+b)c+c* [J86, I]
[Here (a -|- b) is regarded as a single quantity.]
= aF + 2ab + b* + 2ac + 2bc+c*
= a*+b*+c* + 2ab + 2ac + 2bc.
The square of a trinamtal is the sum of the squares of its terms plus
twice the partial products of each term by those which follow it.
Again
(a+6+c+eD»= [(a + 6)+(c + rf)]«
= (a+6)« + 2(a+6)(c + <f) + (c + rf)« [{86, I]
= a*-\-2ab+b*-\-2{ac-\-ad+bc-\-bd)-\-<^'\-2cd-\-<P
= a*+6«+ c«+ cP+ 2a6+2ac+2cw/+ 26c+ 26<]?+ 2cc?.
In case any term, as c alone, should be negative, then all the
terms in which c occurs, excepting c\ would be minus.
IX. In general^ the square of a polynomial of any number of terms
is equal to the sum of the squares of the terms plus twice the partial
products of each term by all of the terms that follow it.
Example. — Find the square of a — 2 x + 6*.
(a — 2x+fe«)« = a«+(-2x)«+ (ft«)« + 2a(-2x) + 2a6«+2(-2x)6«
X. Below are some formulae which will often be found useful and
which the student can easily verify by multiplication.
(6 +c)(c-{' a) (a + 6) = a*(6 + c) + 6«(c + a) + c«(a + b) -\-2abc.
(h — c) (c — a) (a — 5) = a*{c — 6) + 6«(a— c) + c*(6 — a).
(a+6+c)«= aJ+ 3a«(fe + c) + 3a(6 + c)«+ (6 + c)»
= a»+3a*(6+c)+3a&«+6a6c+3ac*+/>«+36«c+36c«+c»
= a«+6«+c»+3a«(6 + c) + 36«(a+c)+3c«(a+6)+6a6c.
90. In case of the examples already solved and those which are
left to be solved, the following laws may be noted with respect to
the result of multiplying algebraic expressions.
The number of terms in the product of two expressions is never
greater than the product of the number of terms in the two expres-
sions, but may be less, owing to the arrangement of partial products
in columns of similar terms.
In case the terms of both the multiplicand and the multiplier are
arranged in the same way with respect to some letter, the first and
84 COLLEGE ALGEBRA [890
the last terms of the product are unlike any other terms. To
illustrate: consider example 2, 258. The multiplicand and the
multiplier are arranged with respect to x\ the first term is 12x^ and
the last is — 4 , and there are no other terms which are similar to
these. The other terms all contain x to some power less. than x^
and are different from 12x^-, the last term does not contain x at all
and is therefore different from the other terms.
If the multiplicand and multiplier both are hoviogeneott* of the
degree 3 and 4, then the product will be homogeneous of the degree 7.
In example in IV, 189, the multiplier is homogeneous of the degree
1 in X and hj and the multiplicand of the degree 2 in x, a, and h,
and the product is homogeneous of the degree 3 in x, a, and b.
Example 1. Simplify (3x*a + bbi/)\
(3 x^a + 5 byy= (3 xhi)^ + 2 (3 x*a) (5 by) + (5 by)* [?86, I]
=9 xV + 30 abx^y + 25 bY-
Example 2. Simplify (2 x' — 7 ay) (2 x« + 7 ay)
(2 x»- 7 ay) (2 a^+ 7 ay) = (2 .t»)« — (7 hy)*
=4x«— 49av. [«86, nr]
Example 3. Multiply together a — x, a + x, and a* + x*.
Thus, (a - x) (a + x) (««+ x«) = (a« - x«) (a« + x«) [«86, lU]
= a>-x*. [?86, III]
Example 4. Required the product of a+fc+c+rf by a+fc — c — d.
Thus,
(a + b+c+d) {a+b--c-d) = [{a+b) + {c+d)-] [(a+6)-(c+^]
= (a+6)«-(c+rf)« [186, III]
= a«+6«+2««^-c«-d!»-2c^ [J86, IJ
= a«+6«-c«— <^+2a6— 2c<f.
BZEBCISE XIV
Find the powers (881):
1. (3a2j^)3. 2. {^7xy^z^)\ 3. (-3:r*|/»2)».
4. (a*fn«)«. 5. (2.i;'yzS)P. 6. (-ary»«»)"*.
Expand the following by inspection:
7. (38)«= (40 - 2)« = 1600 - 160 + 4 = 1444.
8. (63)«=(50+3)2=2500+300 + 9 = 2809.
9. (2jc+ll2/)2 = 4a,^ + 44.ry + 12l2/«.
10. (5a«j:-7a:«i/)«=25a*i:«-70a2a^/+49a^j/».
11. • (3 al^c + 5 a«d) (3 ab^c — 5 cfid) = 9 a«6*c« — 25 a*d«.
190] APPLICATION OF ADDITION AND MULTIPLICATION 85
12. (x-y)«= 13. (Z + m)«= 14. (3ar+2)« =
15. (2a-56)«= 16. (y«-l)«= 17. (2ar-3a:«)« =
18. {bxy+7f= 19. (3-ar)(3 + a:) =
20. (2a6 + 36«)(2a6-36«) =
2L (6xy+7!/«)(6j:y-7»/«) =
22. (a«x»+6y)(a«-c«-6«y«) =
23. {2^ + xy+f){j* + xy^f) =
24. (x« + :ry + 3/^(.r«-2^ + y«) =
25. (tr + wiy)(te-my)(^2:«+my) =
26. (x«~ar+l)(x«+a:+l)(:c*-a;*+l) =
Square the polynomials (J 89, IX):
27. (x + y+2)«= 28. (x-y-z)« =
29. (/ + m-n-/))«= 30. (3j:« — 5ar+7)« =
31. (x8-3/«+2«)«= 32. (a» + 6»+c»)« =
33. (:c*-y»+2«)«= 34. (:c«- 2ar + 2)« =
35. (2x*-5x-7)«= 36. (2:c+3y + 42)« =
Using rule IV, {89, write out by inspection:
37. (x + 3)(z + 5)= 38. (:c-2)(a:-4) =
39. (x — 8)(z-2)= 40. (a:+ll)(jr+l) =
4L {x'-2a){x + 2]a)= 42. (a; + y) (x-4y) =
43. (r-3a)(2r+2a)= 44. {x-7b) (x + 6b) =
45. (ar-9)(<«: + 5)= 46. (a: + 12) (:r - 9) =
Write by inspection the results (J89, VIII):
47. (x+a)»= 48. (a:-a)8= 49. (y+l)»=
50. (y-l)»= 51. (a:+y)*= 62. (a:-y)*=
53. (x+y)»= 54. (ar~y)*= 55. (i/ + l)* =
56. (y-l)*= 57. (y+l)»= 58. (y-l)» =
Simplify:
59. (a+6)(6 + c)-(c + d)(d + a)-(a + c)(fe-rf).
60. (a + b+c + rf)«+ (-a-M-c+c/)2+ (a -6 -c + d)«+ (a+b-c-d)^
61. (a» + 6»+c«)«-(a + 6+:c)(a + 6-c)(a+.c-6)(6 + c-a).
62. (a + 6)« (a-6)».
63. (a-6)» (a+6)«.
CHAPTER VIII
FACTORING AND SOLUTION OF EQUATIONS BY FACTORING
Factoring
91. If two factors are given, their product can be found by
multiplication.
If one of two factors and their product are given the second
factor can be found by division.
It is often important to determine the factors of a given product.
There are many useful examples wherein factors can easily be found
by means of the theorems proved in the previous chapters I-V; thus:
92. Case I. — To Factor a Polynomial, Every Term of which
has a Common Factor.
For example,
1. x^+xi/ = x(x + y).
2. 3a^-6ab + 9a%^ = 3a{a — 2b + 3ah^).
3. 72.r2y — 84x^2 +60 xV= 12xi/{Qx — 7 ij + bxy).
The rule for factoring this case is an immediate consequence of
Law V of multiplication, namely, that
a (6 + c) = ah -\-ac
a{b -{■ c -\- (I) — a {b -{- c) -\- ad = ah + ac -\- ad^ etc.
Rule. — Divide each term of the jmlynomial by the product of the
factors common to all of the tertns of the polynomial, the divisor and
the quotient will be the factors required.
EXEliCISE XV
Factor the following expressions:
1. j^-x». 2. a^-ab.
3. 9j^-7jry. 4. 5ai«-15a'.if
6. 7x' — 3o,r^y. G. lGy^ + G4x/A
7. 49-84X. 8. 42.(**/3-84jV + 203^V-
9. 8 aVr^j^ + 2 a^bu^ + 6 abx\ 1 0. 90 .t V - 1 ^0 aV + 270 ai/^.
11. 21 a;"- V"»+* — 28 x^^-Y'"' ^^ + 63 a'3«-'y ^+5 _j_ 49 jAuyim^
12. ox'y 6j:«- V*"^ + ex**- V*"' + do-"" V^' — car"" V*"*-
193] FACTORING AND SOLUTION OF EQUATIONS 87
93. Case 11. — When the Trinomial is a Perfect Square.
1. Find the factors of a;^ _|_ I4x + 49.
Since this is a trinomial with plus terms and is to be a perfect square,
it must have the form
(x + a)« = (x-\-a)(x + a) = x« + 2ax + a« [{86, I]
and
x* -|- 2 rtx + a* must be the same as x' + 14 x + 49
Hence the second term a of the binomial whose square is x^+ 14x4-49
must be a number such that
two times a is 14
and a squared is 49.
The only number whose double is 14 and whose square is 49, is 7.
x«+14x + 49 = (x + 7)(x + 7) = (x + 7)«.
2. Factor x«— 18 x + 81.
Since the sign of the middle term of the trinomial is — and the
trinomial is to be a perfect square, it must have the form
x« — 2ax + a« = (x — a)« = (x — a)(x— a) [?86, II]
and
X* — 2 ox + a' must be the same as x* — 18 x + 81.
Hence the second term of the binomial whose square is x* — 18 x + 81,
must be a number (—a) such that
two times (— a) is — 18
and (—a) squared is 81.
The only number which multiplied by 2 is — 18 and whose square
is 81, is — 9.
x«— 18x + 81 = (x— 9)(x — 9) = (x -9)«.
loh
^e into factors:
EZEBCISE XVI
1.
3,
5.
T
f •
9.
a*+22j:+121.
y*+16t/« + 64.
a;«+12ax + 36a«.
4x*i/«-20.r«y'2 + 252/*2«.
25 m«n« - 30 mn + 9.
2. re* -38 ^ + 361.
4. s/«-262r*+169.
6. a:«— 8ai + 10a«.
8. l-6a36+9aW
10. 16^^fe«-406-i"«?/ + 25xy.
11. Six*i^ — lSOx*i/z + lO0x^z^. 12. 64a26V-48a6c2/ + 9y«.
13. 49X-2 — 42a:-iy-i + 9y-«. 14. 64.r! -16a;i i/! +j/l.
15. 25arV* — ^ + 36aV- 1^- 10a:-V* — 24x-iy-»+9jr«3^.
88 COLLEGE ALGEBRA [?J 94-96
94. Case III.— The Factoring of the Difference of Two Squares.
Definition, — If a number can be resolved into the product of two
equal factors, one of them is the square root of the number.
Thus: 16 = 4 X 4 = 4»
By definition, 4 is the square root of 16, and is indicated
f/16=4 or *i/(4)* = 4; and in general i/(x)*= x.
That is, the operation indicated by the sign y/ is the inverse of
the operation indicated by the sign ( )*, and undoes the work of
squaring a number. Further, the
l/4 a'x^ = l/(2 ax«)« = 2aa^.
An expression in the form of two squares which have a negative sign
between them, is the product of txco factors which can he determined as
follows:
Take the square root of the first number, and the square root of the
second number. The sum of these square roots will be the fir9t factor
and their difference will form the second factor.
Thus,
1. x« -y^ = {x + y){x- y). [ J86, III]
2. x^^{y-zY=\x + (y-z)\ \x-{y-z)\
= (^ + 2/ — 2) {x — y + z).
3. {x-yy-{z-wY = \(x-y) + {z-w)] j(x-y)- U -ie)|
= {x — y-{'Z—w)(x — y-^z-\-u)).
96. The terms of an expression can often be arranged so that it
is equivalent to two squares with the sign — between them, and the
expression can then be resolved into factors, thus
x^ + y^ -^ z^ ^ d^ — 2xy — 2dzz=z x^ — 2 xy + y^ ^ {z* + 2dz + iU)
^{x-yy^{z + dY
= \{^-y)+{^+d)\ \{x-y)^{z+d)\
^{^ — y+z+d) (x^y — z—d).
96. The difference of two squares may often be resolved into
several factors,
1. x"-2/"=(x«+y«)(x«-2/«)
= {x'+y'){x*+y*)(x*-y^)
= (x«+,v') U*+y') i^'+y') {x^-y^
= (x«+ y') (.r*+ y') ix'+ y') (^c + y){x^ y).
* Owing to the frequent use of the operation of square root the symbol ^/ Is used
instead of V-
897] FACTORING AND SOLUTION OF EQUATIONS 89
2. 4 {xy + zw)*— (a;«+ y«— s«— w?«)«
= ! (x+y)+(2-M?) } 1 (x+y)-(2-w) ! i (2+tc)+(x-2/) ( ) {z+w)-{x-y) f
= (^+y+«—«') (x+y— 2 + tc) (x— y+2+w?) (— x+y+2+«?). ^
EXEBCISE XVii
Resolve into factors:
1. a«-9. 2. 9a«-25. 3. a* — 1.
4. a*-6*. 0. a«-l. 0. a»-6».
7. 25j:« — 49!/«. 8. 144a268- lOO^V- 9- 1-64j:«.
10. z*-8l2/«. 11. (.r-j/)«-2«. 12. 2^-(y-2)«.
13. (x+y)«-(r + M;)«.
14. {x + yf-{z-u)\
15. a« + 6«-2a6-4. '
16. a;«-2/«--23__2y2 and (a8-62 — 9)2-366«.
17. a:« + ^— 2j7/ — z«.
18. a»-&«-c2+<i2-2(ad-6c).
19. 4a«i>«-(a«+t«-c«)«. 20. 4(ad + 6c)«-(a«- 6«-c« + d«)».
21. :r«"«-(y-2)«« 22. (« + 6+c + d)«-c>.
23. (3j^ — 4.T-2)« — (3.t« — 4jr + 2)«.
24. (4 ah — cdf - (a« + 6^ - c* - d^)*.
25. a* + x^-^(y^ + z^)-2(yz-(u:).
97. Case IV.— The Factoring of the Sum and the Difference
of Two Cubes.
Definition, — The cube root of a quantity is one of the three equal
factors into which it may be resolved. Thus,
the cube root of 64 is 4, and is written f ^64 = 4.
The operation of finding the cube root is the inverse of the opera-
tion of finding the cube. Thus,
(5)» = 5 • 5 • 5 = 125.
^"125 = f 5 • 5 • 5 = f (5)» = 5; and in general f'Xxy = x.
Further
^64 tt'xy = # (4f/.ry/ = 4axY-
To find the cube root of a quantity A^ is to undo the work accom-
plished by finding the cube of ^1.
90 CX)LLEGE ALGEBRA [298
= x« + a:y + y*, .'. a:' — y' = (x — 3/) (a5« + xy -f y«)
= ar« — xy + y*, .'. x' + y' = (x + y) (x* — xy + y«).
By
1.
and
2.
division,
as — y
x» + .v'
a^ + y
It follows from 1 that:
The difference between the cubes of two quantities is equal to tlie product
of two factors; the first is the difference between the quantities and thf
second is the square of the first teinn plus the product (f the first by the
second term,, plus the square of the second tenn of the first factor.
It follows from 2 that:
The sum of two cubes is equal to the product of tico factors; the fimf
is the sum of the quantities^ and the second is the square of the first
term minus the product of the first term bi/ the second term, plus the
square of the second term of the first factor.
Example 1. — Factor x* —y'.
x'-y^ = {u'^y _ y' = (x« - y) {x* + x'y + y«). [{97, 1 ]
Example 2. — Factor Sa^ — 27b^.
Sa^ — 27 b^ = (2ay - (3hy= {2a ^3b) i^a"^ + Gab + 9b^-).
Example 3. —Factor m'+ 8 w^
wi« + 8 n^ = im^y + (2 n)' = (m« + 2 n) (m* — 2 mhi + 4 w«).
BXEBOISE XVm
1. ^ — 216. 2. 34Sx^ — t/. 3. SI ji* — 64 y^zK
4. a^^ — c^. 5. mhi^—p^ 6. 1000 o^ — 1331 6».
7. l-729a»6«. 8. 12^i^"i/"* — 64 z^p. 9. 1+j^.
10. x^ + 27i/. 11. a^^ + h^. 12. 8n86«c5+l.
13. 64.r3+1252/'. 14. Sl:i^"* + 2\Gfz^P.
15. (2a + Shy+Six-7ijy. 10. 64 (j: - 4 »/)» - (2 a + 4 6)«.
98. Case V. — When the Trinomial has the Form 3c^ + px + q.
It frequently happens that the product of two binomials is a tri-
nomial (§86). Conversely, some trinomials can be separated into
the product of two factors, thus :
1. Find the factors of x'' + 9 x + 20.
i^8J FACTORING AND SOLITION OF EQUATIONS 91
The first term of each of the bioomial. factoi*s must be x\ then, if
a and h are the other terms of the factors,
{x + a){x -f- h) must bQ the same as x' -f- 9 as + 20
or X* -[- (« + ^)^ + «^ must be the same as x* -)- 9 x + 20.
Hence, the second terms a and 6 of the two binomial factors must
be two numbers
whose sum (a-\-h) \s 9
and whose product ah is 20.
The only two numbers whose sum is 9 and whose product is 20
are 4 and 5.
.-. cr*+9x+20 = (x + 4)(a^ + 5).
2. Find the factors of a:« + 7. ry + 12/.
Here, as in example 1,
X* -|- (a -f- '>)»c + oh must be the same as x* + (7y)x -|- 12 y' .
The second terms a and h of the two binomial factors must be
two numl)ers
whose sum is 1 y
and YfhosG product is 12/.
The only two numbers whose sum is 7 y and whose product is
12/ are 3y and 4y.
.-. x« + 7x^/ + 12/=(x + 3y)(x + 4y).
3. Factor the trinomial x^ + 3 x — 54.
Here
X* + (a -|- h)x 4- ah must be the same as x^ + 3 x — 54.
Hence, the second terms a and h must be two numbers
whose »Mm is + 3
and whose product is — 54.
If the product of ah is — 54, the factors a and h must have
opposite signs and the greater must be positive, since their sum is
+ 3. The only two numbers whose sum is -f- 3 and whose product
is ~ 54 are + 9 and — 6.
.-. x= + 3x — 54 = (.r+9)(x — 6).
4. Find the factors of x^ — 7 x — 44.
Here the sum of a and ^ is — 7,
and the product of a and ^ is — 44.
The only two numbers, a and />, whose sum is — 7 and whose
product is — 44 are — 11 and + 4.
,., a^_7x— 44 = (x — ll)(x + 4).
Note.— The student. In examples of this kind, should always verify the results, by
forming mentally the product of the facto- b he has chosen ( g89, IV).
92 COLLEGE ALGEBRA [899
EXEBOISB XIX
Resolve into factors:
1. a^ + 9x + lS. 2. a^»-7ar+12. 3. a^+13:r + 42.
4. a;»_a:_2. 5. a^ + 2ar — 3. 6. :r« — 2a: — 3.
7. a;»-3a: — 40. 8. a^ — 4 a«a:« + 3 a*.
9. 3^ — 27 xy + 2e. 10. j/«2» — 28 aftyz + 187 a«6«.
11. Factor 90 + 9a: — a;*.
90+9a: — x« = — (a:«-9x — 90)=— (a: + 6)(ar-15) = (15 — ar)(ar + 6).
12. Factor 85 + 12 a: — :r«. 13. Factor 110 — a: — a;*.
14. Factor 22 — 9i7/z-a:«i/8z«. 15. Factor 84 + 5 y*2« — 3/«z*.
16. a«c« — a»c — 2. 17. a'j* — 15 oa: — 54.
18. c«d«-llcd— 180. 19. lS7a*h^'-2Sd)yz + y^z'.
20. c«cP-4a6cd-221a«62.
99. Case VI. — A polynomial of more than three terms, if it has
factors, can often be factored readily by the principles already ex-
plained. Thus, it is seen at a glance that the expression
aj'— 3xV + 3a:^« — y»
fulfills, in respect both to exponents and to coefficients, the laws for
expanding a power of a binomial, stated in ?89, VIII ; and it follows
at once that
x^^Sx^y + 3xt/* - y^ z= (x- yY
In the discussion of the solution of quadratic equations, it will
be shown how to find the factors of expressions of the following
forms:
(a) ax* + hx + c,
(b) Aa^+2Bxy + hy* + 2Bx + 2Ey + F,
(c) Lx'+Ma^y^ + AY]
which in general can not be factored by inspection.
ZiXEHCISZi XX
Resolve into factors:
1. 1— 3a + 3a«-a». 2. «' + 3a« + 3a+L
3. (^ + y?+S(^ + y)*a + S{x + i,)a^ + a\
4. a»" + 3 a*"6» + 3;a"62« + b^".
EXEBOISE XXI
MISCELLANEOUS EXERCISES IN FACTORING
1. a;*+4a:2a« + 4a*. 2. 49^ — 81m«.
3. 9a26« + 24rt26c2+ir)^V*. 4. aV + 512.
5. j^-.a^^2(!h-h\ 0. (2a-+3y)8 — (7a- — 4y)«.
J 100] FACTORING AND SOLUTION OF EQUATIONS 93
7. (a — 6)* — c*. 8. xy — xz--t^-\'yz.
9. c^—j^ — ah — hx, 10. a* — 6« — a — 6.
11. 2«-2ary + 3/«-a« + 2a6-fe«.
12. a;« + 2a:p + a8-5(-p + a).
13. 7(jfi-f)-^x + ^y. 14. x"y + y«
15. a»6-6«. 16. a:» + llx-2ia
17. 6jcV + 4^i/— 2j:*. 18. l — x-\'X^ — x^.
19. 3TO«-21win + 30n«. 20. (a - 6) (a« - c«) - (a - c)(a« - 6«).
21. {z^ + ^ — x^f — ^y^z\ 22. 7j:8 — a;« + 7a:-l.
23. l-18a»x+81a«j«. 24. (:c3 - y») - a: (a^ - 1/«) + y(a: - y)«.
ore 1 y + c^-g' 26c-(y4-c«-a')
^- ^"" 26c - 26c
26. l + ^"^^^^"""> and (u;» + 8aV-4 6«^-32 6V.
27. 81j* — a: and (a^- 4)8 — 8.
28. (x + yY-z*. 29. ^ + 26a:+133. 30. a*-3^i/.
31. 3:« + y' + 3j-y(2r + y). 32. a;* — j/* — z^ + «;« — 2 (ii/; ~ yz).
33. (x + y? + (x-yf. 34. (x+y)'- (x-t,)».
35. 6^c«Pr«i'— 6»c»a«a:". • 36. 81 a*» — 99 a8»6*« + 25 6»"».
37. a»j:«it64(r»y«. 38. x"-i/".
2a6 )' ^^' 8i^-36:r«i/ + 54a:j/«-9y».
41. a«;— 2«d + d2 — 4a;«+12jy-9t/».
42. (x~^yf — 3^-.f, 43. (a + 6)7 — a^ — 6^
44. 2* + 3/« — 2« — u« — 2 2u + 2 xy.
45. 256 j^!/* — (x* + 64 1^— 2*)«.
100. Remainder Theorem* — If a rational integral expression in
X, X* + j3jX""* + PgX""' + ' ' ' H"-Pn-i^"l"i^n> ^^ divided by X — r, <Ae
rtmainder will he
Proof. Divide the polynomial by x — r and continue the process
of division until x no longer appears in the partial dividend. The
quotient will he an integral polynomial of the degree n — 1 in x, and
the remainder independent of x, and therefore a constant. Hence,
if Q is the quotient and i? the remainder, it follows from the defi-
nition of division (^70, formula) that
(1) ^" +i>i^"-^+ • • • +Pn^i^ +p^=Q{x-r)+R
a relation which holds, whatever x may be; hence, in equation (1)
give X the value r. Then
94 COLLEGE ALGEBRA [1101
(2) r-+p,r«-*+p,r"-« + • • • +p^_^r + />„ = {Q)^^^{r~^r) + R:
here r^r = 0, and (0,_^ is the value the expression Q has when
X is replaced by r. Hence it follows from (2) that
r«+;9jr"-i+p/"-2+ • • • +Pn-t^+P„= ^^•
which was to be proved.
Example. — "What will the remainder be on dividing
x3 + 2x2 — 4x+9 by x — 3?
By rule, remainder = (3)»+ 2(3)« — 4(3) + 9 = 42.
101. The Factor Theorem* — If a rational integral expression in x
vanishes^ that is, becomes equal to 0, when r is substituted for x, then
X — r is an exact divisor of the expression.
Given (1) x"+j9jX"-* + j)gX"-2+ • • • -fj^^^^x+p^.
By supposition (2) r'* + p^r""* + 7?/"~^ + * * * +i'„-i^ + Pn = ^•
By (1) in the discussion in ?100,
(3) X- + Pj.T"-» + p^x-'^ + • • • + P„^,^ + P„ = Q(^ -t) + R.
Put in (3) X = r.
(4) r-+Pr"+Pr^^+ ' • • +Pn-/ + Pn= «?).»r(''- »-) = ^•
By (2), the first member of (4) is 0 ; and also r — r = 0, therefore
from (4)
0- <?,_,. 0 + 7?
.-. R = 0,
Hence, when we divide the expression, (1), by x — r there is no
remainder, that is, expression (1) is exactly divisible by x — r.
Note. If expression (1) Is exactly divisible by x—r^ then Pn Is exactly divisible by
r, because the product of the last term in the divisor by the last term In the quotlCQl is
equal to the last terra In the dividend. Hence, in sonrchinj? for a numerical \'alue of x
which will make the given expression vanish, only exact divisors of the last term of the
ex]>ression need bo tried.
Example 1. Find the factors of re' + 5 x^ -f- 7 ar + 2.
The factors of 2 are + 1, +2, — 1, — 2; of these four numliers
— 2 only will make the expression 0, thus,
(_2)»+5(-2)^-14 + 2 = —8 + 20 — 14 + 2 = 0.
.r — (— 2) = x + 2 is a factor of x^+bx^ + 7.r + 2.
Vrrijication:
,r.3+5.T2 + 7.T + 2
x + 2
x^ + 2j''
,r^+3x+l
3x«+7.r + 2
3.r2+6rc
x + 2
x + 2
2101] FACTORING AND SOLUTION OF EQUATIONS 05
Therefore the factors ofx'4-5d^ + 7x + 2 are
x + 2 and a:* + 3x-f-l.
NoTK.— An expression can sometimes be resolved into three or more factors.
Example 2. Factor x'+ 5x« — 2x — 24.
The factors of 24 are 1, 2, 4, 3, 6, 8, 12, — 1, — 2, — 4, — 3, — 6,
-8, -12.
On trial, 2 will make x' + 5.r* — 2 x — 24 vanish, thus
23 + 5 (2)^ - 2 • 2 - 24 = 0.
Hence, sc' + ^ac' — 2x — 24 is divisible by x — 2; the quotient is
j^ + 7 X + 12, whose factors are {x + 3) (x + 4).
.-. a^+5x« — 2x — 24 = (x — 2) (.r + 3) (.r + 4).
£X£BCISB XXII
Show by means of the factor theorem, that x + 7 is a factor of the
following expressions:
1. 2^ + 12j- + 35. 2. :i^ + 15.r + r>G,
3. a^ + Ux+eS. , 4. a-2 + 8x+7.
5. Jt« + 10x+21. 6. a«+13x + 42.
Without actual division show that:
7. J*+i/^ -J^y—ry^ is divisible by (1) :r— y, (2) x + y,
8. .»-» — 12j« + 27x + 40 is divisible by x — 5.
0. fr* — i «* — I a + 0 is divisible by a + J.
10. 0^ + V«^— V a+ V is divisible by a— J.
11. a» — 5a*6 + llr{3/>2 — I4o««r» + 9rt6* — 2/^5 is divisible by a — b,
also by a — 2 6.
Factor:
12. j:« + 3.««— 5x— 15. 13. a-3+5.t^+3j:+^
Resolve into factors:
14. T*-22^+7.r-14. 15. j-^ + 5 .<•* + j« + 5 r.
1(>. j-» + 8j^+17j + 10=0. 17. j-* + 2^V4-5r«-V.r — 1 = 0.
Find the remainder:
18. \\Tien3j:*— 5./.^+10j^ + lI.r— G is divided by x — 3.
19. AVhena:3 + 5j4!+3j.+ 2isdividedby.f — 1.
20. AVhen j* — 4 a-* + 7.1-'- 11. r— 13 is divided by .r —5.
21. AVhenj* + 3j-7-15a:2 + 2is divided hyj-2.
22. What must be the value of the coefficient of .i^ in .r' + (lc^ — lOr +113
in order that the expression may be divisible by j+4?
23. For what values of a in.i-'+7(/.r2+16a*?v-4a' will the exiH*wsion
be divisible by r — 1?
^6 COLLEGE ALGEBRA [»02
102. Case VII. —Compound Expressions which haye a Linear
Binomial Factor.
A compound expression involving x and y is divisible, according
to the factor theorem (SlOl) by a* — y if the expression vanishes when
+ y is substituted for uc; and is divisible by x + y if the expression
vanishes when — y \s substituted for x.
1. x" — 3/" is divisible by x — y, whether n is odd or even.
Put y for X in x'* — 3/", then x" — y" = y* — y" = 0 whether « is odd
or even.
Since y^ — y" = 0, x" — y" is divisible by x — y, wliether n is odd or
even,
2. x" — y" is divisible by x + y, if n is even.
Put — y for X in x" — y", then x" — y" = ( — y)" — y".
If n is even, ( — y)" = y", and ( — y)" — y" = y" — y" = 0.
Since y" — y" = 0, x" -|- y" ^s divisible by x -f- y, «/ »* *« ^i''^'^.
3. x" + y" 18 divisible by x -[- y, (/* w »« or/r/.
Put — y for X in x" -|- 2/") ^^^^^ •^'* + y" = ( — 2/)" + y"-
If M is odd, (— y)" = — y** and (--y)" + i/" = — 3/" + ^** = ^•
Since — y" -|- y" = 0, x** -f-^" ^^ divisible by x + y, i/n i« (x/<i.
4. x" -\- y" is in no case divisible by x — y.
Put y for X in x" -[- y"; ^li^^^i •'^" + y" = i/" + .V" = 2 y".
Since 2y" is not 0, x" + y" is not divisible by x — y.
It follows from these four cases, that:
i. For all positive integral values of n
x» - y"= (x - y)(x"-» + x"-2y + x'-y + • • • + xy"-« + y"-»).
ii. For all positive evm integral values of v,
X"— y^^Cx + yXx"-*— x'^-Sy + x^-y— ,+,• • • +xy"-«— y""^).
iii. For all positive odd integral values of w,
x''+y" = (x+y)(x"-»-x"-2y + x"-y-,+,- • • -.ry^-'+y-').
iv. x"4-y" is never divisible by x — y, and is not divisible by
x-f-y if w is even.
Note.— If the terms of the expression havea commoD monomial factor It should first
be removed before applying the preceding rules.
The proofs of the formulae in 1, 2, 3, 4 can be established by
division, thus:
1102] FACTORING AND SOLUTION OF EQUATIONS 97
Divide x"« — «"• by a; — a!
X — a
ax"*"* — a"* .*. Ist rem. ax"*"* — a*"
a'x"*"' — «"• .*. 2d rem. fi'x'""' — a*
-f- rt'x"'"'^- «"• . *. 3d rem. flr'x"'"' — a"
a*" *x — a"^
rt"» — fx»» -^ last rem. a"* — </'"= 0.
The law of formation of tlie terms of the quotient is readily
observed. All the terms are positive; the exponents of x decrease
continually by unity, and those of a increase continually by unity.
Since the remainder from the division is exact, whatever the integral
exponent of m is, it follows, therefore,
X — a ' ' ' ' '
This result has many applications in Algebra. When the difference
of two squares is involved in an example, the principle of {94 should
always be used.
Examples. — 1 . Factor x* -f- a*.
By {102, 3,
X* -f- «* = (-c + «) (^* — ^'^ + ^^* — -^^^ + «*)•
2. Factor 32 6*+ 243 c^
326'^+243c*= (26)'^+(3c)'^
= (2b+3c)l{2by-i2by (3c)+(2/>)«(3c)2-(2/>) (3c)»+(3c)*]
= (26 + 3c) (16 6*-24 6»c + 36 6«c«-54^c»+31c*).
EXBBOISE yxii I
Resolve into factors:
1. ^-1. 2. 0^+1. 3. aT + b\
4. a«+6«. 5. x^ + y^K 0. aio+6W>.
7. 2* -64/. 8. l+2« 9. 64j^+;A
10. 3^ — 729 (fi. 11. j*r/* — a*fc-\ 12. x^^-1024yK
13. (^b^+m\ 14. 720 + a«. 15. 1024 j^ + y^^
16. 729a:»-1728y». 17. a«» + 6»«, 18. ar^^-y*".
98 COLLEGE ALGEBRA [11103, 104
103. Case VIII.— To Factor £spressioiis containing Four
Terms.
When a polynomial contains four terms which can be arranged
in pairs so that each pair of terms contains the same binomial fac-
tor, the pol}^omial ma}' be factored as follows:
Divide the polynomial hy the common binomial factor; then the
divisor will be one factor and the quotient the other. Thus:
(1) Resolve into factors x* — ax + bx — ab.
Notice that the first and the second terms contain the common
factor X, and the third and the fourth terms have the common fac-
tor b ; therefore, the first two terms and the last two can he factored
by Case I, and the result is x{x — a) + fc(x — a) ; divide by x — a, and
the quotient x -|- 6, the other factor, is obtained. Thus,
x^ — ax -f- bx — nb = x{x — n) + b{x — a) = (x + b){x — a),
(2) ac-^ad — be + bd = a(c — d) — (6c — bd)
= a{c — d) — b(c — d)
= (a^b){c^d),
EXEBCISE XXIV
Resolve into factors:
1. x^-\-ax-^bx-\-ab.
3. am — bm— .an -f- bn.
5. aj^ — ab^ + b^x — r^.
7. a* + a3+a«-fa.
9. mn -\-np — rnp — n*.
11. j^t/ — j^i/-ji^f/+\.
Solution of Equations by Factoring
104. The ability the student has acquired in factoring ranonal
integral expressions enables him to solve an extensive class of
equations which are of a degree higher than the first.
Suppose that it is desired to solve the equation
x«_9x + 20 = 0.
Factoring the first member,
(1) x« — 9x + 20 -. (x — 5)(x — 4) = 0
Any value of x which makes the first member of equation (1)
vanish is called a root of the equation.
2.
ab+ay — by — '!^.
4.
(yax — 3bx — 6ay+Sby,
G.
a«6-6jc8 + «*-^-^.
8.
a^c^ -f acd — 2 abc — 2 bd.
10.
7ar^ + iajij — 7bxy — 4b^,
12.
2j^— a-»+4j: — 2.
8104] FACTORING AND SOLUTION OF EQUATIONS 99
To solve equation (1) is to answer the question, what values of x
will make the first member of this equation equal to zero? The
values of x which will make either of the factors x — 5 or x — 4
equal to zero will make the first meml)er (x — 5) (.c — 4) equal to
zero. Therefore the roots of (1) are the values of x obtained from
the equations
X — 5 = 0 X — 4 = 0;
Xj = 5 and x^ = 4
are the roots of equation (1).
Examples — 1. Solve the equation .r' — 2 x* — 9 x -f- 18 = 0.
Factoring by UOl or by 2103,
x» — 2x« — 9x+18. (x — 2)(x — 3)(x + 3) = 0.
The equation is satisfied by putting either
X— 2 = 0 or X — 3 = 0 or x + 3 = 0;
Xj = 2, Xj = 3, and Xj = — 3
are the roots of the given equation.
2. Solve the equation a^-^-^ax-^Aa^— b^ = 0.
Here .
x«-f-4ax + 4ri«_?>2 . (x + 2fi)2 — 6*
I (x + 2a — i»)(x + 2^ + 6) = 0.
The equation is satisfied by putting either
X + 2 a — /> = 0 or x + 2 « + 6 = 0 ;
the roots are x^ = 6 — 2 a and x^ = — 6 — 2 a.
3. Solve the equation x* + 2 x* + 5 x' + 10 x« -- 6 x — 12 = 0.
Here
x^+2x*+5x^+10x«-Gx— 12 . (x+2) (x^+Sx^-G) [8101]
. . (x+2) (x^-l) Cr^+6) [298, 3]
.- (x + 2) (x+ l)(x-l) (x^+6) = 0.
The ecjuation is satisfied by placing
x + 2 = 0, x+l=0, .r — 1=0, or .r- + ()zz:0;
hence,
.fj = — 2, Xg = — 1, X3 = 1, and x^, x^ = ± I — i>
are the roots of the given equation.
V — 6 can not be found, because there is no number which multi-
plied by itself will give —6. j/— 6 is called an imaginary number.
100 COLLEGE ALGEBRA [^104
EXEBOISE XXV
Solve the following equations by factoring:
1. a«+10x+25 = 0. 2. a:«+2z-63 = 0.
3. x«-6x+8=0. 4. 2a:« — 3jr — 2 = 0.
6. 4:r«~lla: + 6!=0. 6. 5C^ + 9.r-2 = 0.
7. 35^+ 17 X +.2 = 0. 8. a:«-2^-^ = 0.
9. (x+6)«-c8 = 0. 10. (22r-a)«-(i:-6)«=0.
11. 2i Jt^b -723^1^ -\-b4xb^ = 0, 12. (j^- II x+'30f-{i^-x -30)^=0.
13. :c*~5ar+6a«+a6-6«=0. 14. :c« — (2a — 1) j-6-a + a« = 0.
15. j:-«+14j:-i + 45 = 0. 16. x-»+3a:-» -40 = 0.
17. a:-«-2x-i-99=0. 18. x»+3^-4x = 0.
19. ic»-a:« — x-l = 0. 20. :c»-2a*+3x-6 = 0.
21. 5:r«+2^-15jtr-6 = 0. 22. 8r»-6j:»-28x+21=0.
23. 6x»-2:c«-21a:+7 = 0. 24. gj*- 12^ + 6:r- 1 = 0.
25- ^ + S-2 = 0.
a* ' or
26. a:5-(a + 26-3c)a:* + (2a6-3ac — 66c)2r-6a6c = 0.
27. ar»-8 = 0. 28. a:»-l = 0.
29. a:* — 7:c« — 8 = 0. 30. 16a:*-l = 0.
31. :r*- 625=0. 32. 1-^=0.
33. 128 — a;«=0. 34. ^r^i* — 3a;«»' — 70=0.
35. (2ar — 3) (4jr — 5) (6a:- 20) (a;« — a: — 20) (2a;«+7r-4) = 0.
36. (ar+6)(cx-rf)(a^-3cLr-28a*)(mjc«— w) = 0.
CHAPTER IX
6RSATEST COMMON DIVISOR AND LEAST COMMON MULTIPLE
105. In Arithmetic the Greatest Common Divisor of two or more
positive integers i8 the greatest number which will divide each of
them without a remainder. This term is used much in the same
sense in Algebra. Its meaning in this subject will be understood
from the following definition of the Greatest Common Divisor of two
or more algebraic expressions:
If ttoo or more integral algebraic expressions he arranged accord-
ing to the descending powers of some common letter ^ the factor of highest
dimensions in that letter which divides each of these expressions without
a remainder will be their greatest common divisor.
RsHABK.— For the sake of brevity the letters G. C. D. will often be used for this term.
Highest Common Measure or Highest Common Factor is sometimes
used for G. C. D.
106. An expression which will exactly divide two or more
expressions without a remainder is called a common factor of these
expressions. Thus,
2 ax is a common factor of 6 a*x' and 18 axb\
Two algebraic expressions are said to be prime to each other
if they have no common factor except 1.
107. G. C. D. of Quantities which are Readily Factored.— If the
quantities are readily factored their G. C. D. can be readily found
by the rule:
Resolve the quantities into their prime factors. The product formed
of all the prime factors common to all of the quantities will be their
O. C D. Thus:
101
102 COLLEGE ALGEBRA [JIOS
1. Find the G. C. D. of 17 j>^, 34 p% and 51 j^y.
17^5*= 17 -^ • gr • J
Mp\ = 2 '17 'p'p -q
51 ^?y = 3' 17 'p'p'p-qq-q
G. C. D.= 17 'p' q= 17 pq,
2. Find the G. C. D. of 3x«— 6x+3, 6x«+6x— 12, and 12;r*-12.
3a:«— 6.'c+3 = 3(x« — 2x + l)= 3(x — l)(x — 1)
ex^ + Gx — 12 = 6(x«+ x-2)= 2 •3(x + 2){x-l}
12 x2 — 12 = 12(x»— 1) = 2 • 2 • 3 (x+ 1) (x-1).
.-. G. C. D. = 3(x — 1).
EZEBOISE XXVI
Represent the G. C. D. as the product of all the common prime
factors, of the following:
1. 6a»6, 9a*6, 27a*ry.
2. 12 a«63c and 75 6*c*.
3. 48 aV^', IBr^y^z^, 2^x^z\ 54a*y*2».
4. a«-6«, 03-6', and tt«-7a6 + 66«.
5. 4a;» + 8x8-32x and 8j.5^-128a:.
6. 3a« — 6a + 3, 6a« + 6a— 12, and 12rt* — 12.
7. a;«»+a;~ — 30 and a-*" — ar» — 42.
8. a6c (a* — ft^) (a^ — c*) and ac (6 — a)2 (c — a)\
9. lOxSy — COaV+Sjy* and 5 a V — 5 a^ — 100 y*.
10. 6(x-2/)*, 8(^-1/2)8, and 10 (^i-* - j/*).
n. a«-l, a;s_i^ and (.r- 1)« (.r+1).
12. 12 m^^ + 24 mn\ 8 mV — 56 m^n* — 64 m».
13. 3fi + ^f + j^y*^+,/ and a;3 + 2 :i48y + ^y^ + 2 jr*.
14. a^-2,», a:»-j/», a:»-y3, ori^-yUi.
15. Pr — /2|/ + 2 //wo: — 2 Imy + wi'j; — 77i'«/ and l^x — /*y — rt^x + mV
108. The G. C. D. of Two Algebraic Expressions As in
Arithmetic, the following is the rule for finding the G. C. D. of
two algebraic expressions :
Let A and B he the two expressions; arrange A and B with respect
to the descending powers of some common letter and suppose thai the
exponent of the highest j^owcr of that letter in A is equal to or grcakr
than the exjwnent of tJie highest power of that letter in B. Divide
22109,110]
G. C. D. A^D L. C. M.
103
A by B; make the remainder a new divisor and divide B hy it. Pro-
ceed- in this way until there is no remainder; then the last divisor will
he the G. C. D. required,
109. Example. —Find the G. C. D. of 817 and 1763; also of
a:«_4x + 3 and 4x»— Qa^ — 15x+ 18.
In Arithmetic
817)1763(2
1634
129)817(6
774
43 ) 129 ( 3
129
43 is G . C. D. required.
In Algebra
1st Dividend. Ist Divisor.
4X-''— 9x2— 15x + 18 |x2 — 4x+3
4x» — 16x2 + 12x
7x2 — 27X + 18
7x« — 28X + 21
-4:r-+T'
X — 3 Ist remainder.
2nd dividend. 2nd divisor.
x* — 4x + 3 |x — 3
x^- 3x Ix — 1
— x + 3
- x + 3
. •. X — 3 is G. C. D. required.
The division is continued until the first term of the remainder is of
a lower degree than that of the divisor.
110. The truth of the rule given in {108 depends upon the follow-
ing principles:
1. If P divide A^ then it will divide mA. For suppose that P
is contained in ^1, r times, then A=LrP and mA = mrP. i^ is a
factor of mA and therefore a divisor of mA.
Any divisor of an expression is also a divisor of any multiple of
that expression.
2. If P divide A and B, then it will divide mA ±:n B. For,
since P divides A and B we may assume that A = rP and B = sP
(by 1), then mA^nB = mrP^nsP=: {mrdans) P; .'. P divides
mA ± nB by definition of division.
HencCj any common divisor of tiro expressions is a divisor of the
sum or the difference of any multiples of the expressions.
These two principles make it possible to prove the rule given
in n08.
104 COLLEGE ALGEBRA W HI, 112
111. Let A and B be the two expressions. B) A (a
Suppose that they are arranged according to the aB
descending powers of some common letter, and ^ ^ /^
that the exponent of the highest power of that ^ q
letter in A is equal to or greater than the highest
power of that letter in B. Divide A by B] let ^) ^ ^^
a be the quotient and C the remainder. Divide ^ .
B by C; let h be the quotient and Z> the remainder. Divide G by
D; let c be the quotient with no remainder. The laws of subtraction
give the following results:
(1) A = aB+C; (2) B = hC+D) (3) C=cD,
To prove that D is a common divisor of A and B. /> is a divisor
of C since by (3) C = cD; i. e. , Z> is one of the factors of C; hence
by 1, UIO, Z> is a divisor of bO and .-. of bC+D, by 2, {110; or
by (2), B = b{cD) + I) = {bc + l)D .*. D divides B. Since D
divides B it divides o2?, and divides aB-\- C by 2, {110, that is,
divides A. Hence /> is a divisor of A and B,
To show that D is the. greatest common divisor of A cuid B.
We have from (1), (2), and (3),
(!') A — aB = C; (2') B-bC=D.
By {110, 2, any expression which divides A and B divides A — a 5,
that is, divides 6', (1'); thus an expression which divides A and B
is a divisor of B and C. Similarly, from (2'), every expression which
divides B and C is a divisor of C and Z>. That is, every expression
which divides A and B is a divisor of D, But no expression of a
higher degree than D can divide D. Therefore D is the G. C. D.
required.
112. In order to avoid fractional coefficients in the quotients in
the operations of finding the G. C. D., (1) certain factors may be
rejected which do not form a part of the G. C D. required; and
(2) a factor of a certain kind m/iy also be introduced at any stage
of the process.
Example— Find the G. C. D. of a^ — 2x* — Gar' + 4jc^ + l3x-\-6
and 3 u-* — 10 .r» + 15 X + 8. Thus,
3x* — 10x« +15x+ 8 |x^— 2x^—6. r»+4y»+13x-f6
6x*+ 8.c'—12x« — 24^ — 10
«113] G. C. D. AND L. C. M. 105
Before proceeding farther divide the new divisor by 2 and multiply
the new dividend by 3. Then continue the operation thus :
3x*— 6x*— 18ir»+12x2 + 39x+18 |3x* + 4a^--6x«- 12x- 5
3x^+ 4a;*— ex^ — 12a^— 5x Tx~
— lOx* — 12x3+24x«+44x + 18
Divide the last remainder by 2 and multiply the quotient by 3. Then
we have:
— 15x*— 18x3 + 36x« + 66x + 27
Continuing the division :
— 15x* — 18x»-f 36x« + 66x+27 |3x* + 4x8— 6x« — 12x — 5
— 15x* — 20x» + 30x' + 60x + 25 [Z^5
2x^+ 6x«+ 6x+ 2
Dividing the last remainder by 2 and continuing the operation, thus:
3x* + 4x»— 6x«— 12x — 5 |x3 + 3x« + 3x+l
3x* + 9x»+ 9xg+ 3x [3x-5 ^
— 5x»— 15x« — 15x-5
— 5x^~15xg'— 15x-5
Hence x»4-3x« + 3x + l is the G. C. D. required.
113. The factor 2 was omitted from the first remainder in accord-
ance with rule 1, of ?112. The justification of this rule will be now
given.
Suppose that it is desired to find the G. C. D. of A and B] and
that at any time in the process P and Q are respectively dividend
and divisor. Let Q =z mS where m does not have a factor which
jP has: m may be rejected; that is the process may be continued
with P and JS instead of P and Q,
For it has been shown (Jill) that A and B have the same com-
mon divisor that P and Q have.
But any common divisor of P and S is a common divisor of P
and Q and is therefore a common divisor of A and B.
Any common divisor of P and Q is a common divisor of P and
mS, But w has no factor which P has. Hence any common
divisor of P and (? is a common divisor of P and S. Therefore any
common measure of A and jB is a common measure of P and JS.
Thus A and B have just the same common divisors which P and S
have ; the fact which it was desired to prove.
106 COLLEGE ALGEBRA [?? 114-116
114. By J112, rule 2, a factor of a certain kind may be introduced
at any stage of the process. Thus, in the example in ill2| after the
second division we divided the remainder — 10 x* — 12x^ + 24ur*
-^ ^4c X -\- IS hy 2 and then multiplied the quotient by 3, which is
not a factor of the divisor 3.f*+4ar'— 6x*— 12x— 5. This rule is
formed as follows: Given the problem,' to find the G. C. D. of -4 and
B, At any time in the process suppose that the expressions L and M
are to be respectively the dividend and the divisor. Let P = nL,
where n has no factor which ^has; then n may be introduced with-
out aflfecting the G. C. D. ; that is, instead of continuing the division
with L and M we maj' continue it with P antl M.
By Jill, A and B have exactly the same common divisor as L
and M have. Moreover, any common divisor of L and JIf is a
common divisor of P and jV; so that any common divisor of A and
j8 is a common divisor of P and M,
Any common divisor of P and J/ is a common divisor of nL and
iT (since P = nL)] but n does not contain a factor of M, There-
fore the common divisor of P and J/ is a common divisor of L and
M and is therefore a common divisor of A and B.
Thus it is evident that A and B have exactly the same common
divisors as P and M have.
115. As has been shown, certain factors may be removed from
either dividend or di\dsor, or introduced into either-, in practice fac-
tors are usually removed from divisors and introduced into dividends.
These factors are, as a rule, numerical factors. The results of
SH13, 114 show that these operations may be made at any time in
the process ; for example, at the beginning.
116. Suppose that A and B have a common factor L, which is
readily seen. Let A = aL and B = hL, Then L will be a com-
mon factor of A and B and a part of the G. C. D. Find the G. C.
D. of a and h and multiply it by L and the product will be the
G. C. D. of A and B. For example,
Find the G. C. D. of
2a*-|-3a3.c — 9(/2x2 and 6«*x— 17aV+ 14aV— 3ax*.
Arrange the quantities with respect to the descending powers of a
and factor them, thus,
fi« (2 a2 _|_ 3 ax — 9 x^) and a.r (6 a» — 17 a^x + 14 aa;» — 3 ic«).
Set aside the factor a common to both expressions as a part of
»n] G. C. D. AKD L. C. M. 107
G. C. D. and omit the remaining factors, a from the first and x
from the second, because they will not affect the G. C. D. (1112)
and proceed as follows:
6a'+ ^a'x — 21ax^ ' |3a — 13x
— 26ci«x+41f/x2— 3a^
— 26riV — 39 (i:r« + 117 x»
80«u^— 120a^' = 40x2(2a — 3j:)
Omit 40x*, because it is not a factor of the last divisor, and continue
the process:
2rT — 3j-
2a»
+
^nx
-Ox«
2a«
3r/.r
Qax
-9x2
6<iar
~9x«
c/ + 3 X
Therefore, the G. C. D. is a (2a — 3 x).
117. Similarly, if at any stage of the operation a divisor and a
dividend have a common divisor, it may be set aside and the opera-
tion continued with the remaining factors. The G. C. D. will be
found by multiplying the final divisor by the factor set aside (1107).
Suppose that it is desired to find the G. C. D. of three expres-
sions, Ay By and C. Find the G. C. D. of A and B; let D be this
G. C. D. Then the G. C. D. of I) and 6' will be the G. C. D. of
Ay By and C.
For any divisor of D and C is a divisor of Ay By and C, since
D or each of its factors is a divisor of A and B; and likewise any
divisor of Ay By and C is a divisor of D and Cy since the G. C. D.,
D, of A and B must be divisible by any factor common to A and By
and the divisors of. Ay /?, and Cy are by hypothesis divisors of C.
Therefore, the G. C. D. of D and (7 is the G. C. D. of J, By and C.
Similarly, the G. C. D. of four algebraic expressions. A, By Cy Z>,
will be found by finding the G. C. D. of any two, A and B for
example, then of the other two, C and />, and finally the G. C. D.
of these two G. C. D's just found will be the G. C. D. of Ay By Cy
and D.
Example. — Find the G. C. D. of ^' — 9 j^ + 2G x — 24,
x»—llx« + 38^-40, and x^- lO.r^ + 31 x — 30. Thus,
108
COLLEGE ALGEBRA
[«118
x»-. 9x«+26x— 24
x3— 10x«+31ir— 30
a:»— 10.7:«+31ar— 30
1
x»— 10x*+31x— 30
7? — h^-\- 6 X
— 5x«+25x— 30
— 5x»+25x— 30
-^x»-_5x+6
x»— llx«+38x— 40
x* — 5 x* 4" ^^
— 6x«+32x— 40
— 6x«+30x— 36
x«— 5 x+6
2x— 4
x-2
x* — 5x + 6-T-x — 2 = x — 3.
.-. the G. C. D. required is x — 2.
118. In accordance with the rule derived in Jill and the results
deduced in the paragraphs immediately following it we have the
more general rule for finding the G. C. D. of two or more integral
polynomials.
1. Remove the simple factors of the given expressions and reserve
the factors common to all of the expressions as a part of the G. C. D.
sought.
2. Arrange the remaining factors of each quantity in the order of
the descending powers of a common letter. That expression which is of
the lowest degree is to be taken for the divisor; or if both are of the
same degree, that whose first term has the smallest coefficient,
3. Continue each division until the degree of the remainder is less
than that of the divisor,
Jf., If the last remainder of any division is found to contain a
/alitor which is not a common factor of the divisor , this factor is to he
removed; the resulting expression to be used as the next divisor, and the
Inst divisor a« the next dividend,
6, If at any stage of the division a factor is discovered which is
common to a dividend and a divisor it may be removed and the division
continued; but it will be a part of the G, C. D,
6, A dividend whose first term is not exactly divisible by the first
term of the divisor may be multiplied by such an expression as will
make it divisible.
8119] G. C. D. AND L. C. M. 109
EXEBOISE XXVn
The choice of method is left to the student. Find the G. C. D. of:
1. :r*-62^ + 5a;« + 5a:-6 and ar»-32»-6a:+8.
2. j*-20^+30a;* + 19x-30 and x^+b^+bx^-bx-Q.
3. x*+4r^-'9x^-lGx + 20 and x^ — 2ji^—23x+(jO.
4. x^-lQx^ + mx^-llOx+lOb and x^— Id jfi + 128 x'^-Sb6x + 336.
5. j*-nar* + 39^-41j:«-32ar+60 and rc»+22^-7a*-10.^+10ar+12.
G. x^-l0x^+Sbx^-b0x + 2'i&nd x^-Sx^+Ux-lO.
7. ;i4_5^^5^_|_5j._e and j;* + 2x5-13:c«- 14^ + 24.
8. x* — b(u^+ba*ji^ + b(^x-'6a* and ar' + 4ar«+a«r — 60^.
9. z* + 2ar» — 3a8j;« — 8a»j: — 4a* and x*—a*.
10. 2*-3 ar*+2 aV+2 a»j^— 3 o^x+o^ and a:*— 4 (u^+3 a«j;2+4 a»jr— 4 a*.
11. j* — Sx*—bj^ + lbr^ + 4x-'12 and a;* — 7a:3+6a^ + 28a:-40.
12. 2*-14-t* + 122« + 49:r;« — 84ar + 36 and a*-14jr* + 49a*-36.
13. a^ — 14j:« + 49i^ — 36 and a* — 2x*— ll.r« + 40a:3--44jr+ 16.
14. j:* + 3-t*-5a:5-27^-32a: — 12 &nd x'^ — 6x^+dx^ + 4x^-12x.
15. a:* + 3:c*-17a:* — 39.r«+88j:«+108ar — 144 and
a^+3a;* — 23:r» — 51a«-f94x+120.
16. x^-yx^-7y^x^+y^x + 6i/^ and 2x^ — dyx^+7ii'x+(yi/^.
17. 2j* — :E«y — 14:cV-5a:i/8 + 6y* and 2ar*-a^«/-14^V^+19.cy»— 6y*.
18. ^-13^+67:r*-175r»+244jc«-.172ar+48 and
:i*-2:c»-202:*+10j«+79a:«-8a:-60 and
a^ + 2a?»-23.r»-64.^^ + 4ar + 80.
19. 24j^-502* + 69x*-60.t^ + 36jt;«--10ar+l and
24j^+14jr*+lla;*+12a:3 — 12a;* — 2x + l and
24^— 14a^+ll:r* — 12:r»__i2j:« + 2a. + l.
20. :r*— 12jr« + 48;r*— 46a:« — 153a:« + 378ar — 216 and
24j_i2iJ* + 38a;* + 28a^ — 243x2+80a: + 300 and
a* - 39.«« - 34 a:« + 252ar + 360.
21. a*-46j<-36:c«+609ar* + 828.x«-1516a:« — 1440a:+1600 and
a:'-5j:« — 42a:6+182a:* + 497xs— 1533a:«-1800a: + 2700 and
;c« + 5:c* — 2oa^- 145a:»-16x« + 500ar + 400.
119. Least Common Multiple. — In Arithmetic the Least Common
Multiple of two or more whole numbers is the smallest number which
each of the numbers will exactly divide. This term is also used
in Algebra, and the sense in which it is used in this subject will be
understood by the following definition :
The least common multiple of two or more expressions which are
arranged in the order of the descending powers of some common letter
is the expression in the lowest degree of tltat letter which is exactly
divisible by ea^h of them.
110 COLLEGE ALfiEBRA Hi 120-123
120. Any expression which another will exactly divide is a
multiple of it,
121. If two expressions have no common factor their least
common multiple will l)e their product, since it will contain each of
them.
For brevity, let L. C. M. be used for Least Common Multiple.
122. When the Expressions can be Readily Factored. — If the
expressions can be readily factored their L. C. M. will be an expres-
sion which is the pro<luct of every factor of each expression taken
the. greatest number of times it occurs in any one of the given
expressions.
1. Find the L. C. M. of lSax\ 90«y«, 12 axi/,
ISax^ = 2 '3 -3 ' a x X
90a2/* =5-2-3-3ayy
I2axi/ z=z 2 ' 3 ' 2 ' a ' X • y
the L. C. M. = 3« • 2« • 5 • a • a:« • y« = ISOaxV-
2. Find the L. C. M. of x^j/ — xf, 3 xix — yf, 4y{x — yY-
xhj — .r^' = xy{x — y)
3 .r(.c — yf = 3 X {x — yy
the L. CM. = 3 • 4 • y • x{x-yy = Uxy{x - y)\
3. Find the L. C. M. of f/« — x^, a* — 2 ax + x^, a« + 2 ax f x«.
ffi — .r* = {a -j- ar)(rt — :r)
a^ — 2 ax + :r2 = {a — x)^
«« + 2ax + X- = (« + xY
the L. C. M. = {a- x^ {a + x)»
= (««_.r.«)(a«_a;«) = (a« — x')'.
The rule can be more simply stated: The L. C. M. of tvoo or
more, expressions which can he readily factored is the product of each
factor taken the greatest number of times it occurs in any one of the
expressions.
123. When the Quantities are not Readily Factored. — The
L. C. M. of two or more quantities can be found by finding their
G. C. D.
«124] G. C. D. AND L. C. M. HI
Suppose that D is the Gr. C. D. of A and B; then
A =aD and B = ID,
Since Z> is G. C. D. of A and By a and ?> have no common factor,
.'. L. C. M. of a and b = ah.
Hence the L. C. M. of A and B or of aD and 6/> is a&Z>.
But A = aD and B=bD:
.-. AB = ahIJ^={ahD)D.
.-. a&/> = L. C M. of ^ and ^='^ ^"^ = ^ ' 2? or ^ • | .
That is,
The L, C. M. of two expressions can he found hy dividing their
product hy their G. C. D.; or by a method which is usually more
simple; hy dividing one of the quantities hy their G, C D. and multi-
plying this quotient hy tJie other.
Example. Find the L. C. M. of 20.x* + x«— 1,
and 25 X* + 5 .t' — x — 1.
100x* + 20a;» — 4a;— 4 | 20x*+x»— 1
100 x*+ 5:r«
20 X*
20 X*
+ x« — 1 pOa;»— bx^—4x + l
5a;' — 4a;« + x |x+l
LI
5x»
+ 5x«
— X
— 1
4
20x» +
20 x» —
20x*-
5x«-
-4x
-4x
— 4
+ 1
M
25 x«
— 5
5x«
-1
20 x«
20 x«
4x+l
4x
5x« — 1
4x -1
-5x«
- 5x«
+ 1
+ 1
5 X* — 1 is G. C. D. required.
20x* + x«— 1 -^5x« — 1 =4x*
Hence, L. C. M. required is (4 x«+ l)(25x*+ 5x3 — x
"^ + 1.
1).
124. To find the L. C. M. of three quantities, A, B, and C; find
the L. C. M., say if, of A and B; then the L. C. M. of M and C is
the L. C. M. required, say M\
For M will contain each factor of A and B tlie greatest number of
times it occurs in either of them; and M' will contain each factor of
112 COLLEGE ALGEBRA [2125
M and C the greatest number of times it occurs in either M or C.
Therefore, M' will contain each factor of -4, B, and C the greatest
number of times it occurs in A or B or C; and that is what is
required to be proved.
126. For an understanding of the remaining chapters in this
book, complete theories' of the greatest common divisor and the
least common multiple are not necessary. Such discussions will be
found in works on the theory of equations. However, the solution
of the examples illustrating the principles discussed in the preced-
ing sections is a valuable exercise in the fundamental operation of
algebra.
EXEBOISE XXVni
Find the U C. M. of:
L ba%c\ (Sabc\ 2. 6aV, 8c6«, bhcK
3. Za^by 4 6«c, 2cV, 13ad«. 4. o^ and ar+a:*.
5. al^c^Jt^, M^c'j^ and (^Ir^cx, 6. ^, x^—2x,
7. 27u-», Ojc^Ijt+I). 8. x+y, x — y, x^—f.
9. 2x{x-y), 6xi/(x>-y«), Sx^y{x + y),
10. x^ + xy, xy+t/. IL 4a«6 + 4a6«, 6a— 66.
12. a{x — b){x — c)y b{c-nx) (x — a)y c{a^x)(b — x).
13. l+y + ^, l-y+j/«, l + 2/« + y*.
14. (l-:r), (l-.r)«, (l-.r)3.
15. (a + cf — 6«, {a + 6)« - c*, (6 + c)« - a*.
10. a* + 2.H + a,^+8-*^ + 16:r + 8 and x^-ix^ + x* — 4.
17. x*-4:r3 + a^«~4 and ar»+2x« — Cj-— 12.
18. 2x«+2x-l, 3x'— 4x+l, and 2a:«-3a: + l.
19. a^—JT — 2, r»- 4j«+3, and a;^ — 3^:2^2.
20. a;2-3:r+2, x^—Qx^+Ux — d, and a:«— 5jr + 6.
CHAPTER X
FRACTIONS
126. If a and h are two integral numbers such that the group of
things represented by h can not be counted out of the group of
things represented by a, then the symbol ? is a fraction. For
example, -. a is called the numerator and h the denominator of the
o
fraction -.
o
The numerator and the denominator of the fraction are called
the terms of the fraction.
The sjTxnbol (?) is always subject to the relation
(1) (f)^ = «- t?72]
An integral expression may be regarded as a fraction whose
denominator is unity: thus, a-\-h 19 the same as ^^-j— > by the defini-
tion -^ *'^--^ ^"^^^^
of quotient, (^^y^) -1 = a + b.
7 is a rational fraction when a and b are integers.
6 ^
c
The denominator of the fraction - declares that the number of
8
things in a certain group is 8, and the numerator 5 declares that
of this group of 8 things 5 are taken ] and in the use of the
fraction - the unit group is a group of 8 things. Thus, if one has
o
a quantity of vinegar to measure, and finds that a gallon measure,
which contains 8 pints, can be filled 13 times and that besides a
pint measure can be filled 5 times, the measure of the quantity of
vinegar is 13*- gallons. A gallon is the unit, and the denominator
8 declares that the unit gallon is divided into 8 parts, and that 5 of
these parts, or pints, are taken in the fraction '-• Here - gal. means
o 8
one of the eight equal parts of a gallon (a pint), or briefly, 1 eighth of
a gallon ; and - gal. means 5 of the eight equal parts of 11 gallon (or
o
113
114 COLLEGE ALGEBRA [« 1:^7, 128
5 pints) or briefly, 5 eighths of a gallon; and similarly for any other
fraction.
127. The rules of division are purely the formal consequences of
the fundamental laws of the multiplication of numbers. III, IV, V,
definition IX (?61), theorem XI (J63),and the corresponding laws of
addition and subtraction.
The rules of division, or, what is the same thing, the rules of
the operation of fractions, can be deduced in the same way (?72) as
the rules of subtraction (188, 1-5). They follow without regard to
the meaning of the symbols a, 6, c, = , + , — , ah^ ^ (172).
128. The rule governing the dependence of signs of a fraction
upon the signs of its terms is deduced from the rules of the signs of
products when the factors have different signs (S41, 6 and 8), thus:
^-\ = -^. for +15 = (±^)(-5)=(-3)(-5) [m,8]
-;z^= _ 3, for - 21 = (^) (+7) =(-3) (+ 7) [J41,6]
^ = + 7, for -35= (^) (-5) =(+7) (-5). [S41,6]
Hence, + divkhd hy — ghea —
— divided hy -\- gives —
— divided hy — gives -^-.
In general, if g = ^^ we can, as above, show that
-\- (I — a -\- a
Proofs: According to the rules used in establishing the preceding
rule, we have:
{—h) = {—q){—h)r, (1= ^^, which is true since J- = y
^^^ and «= bqi
and a= bq }
«12J», 130 j FRACTIONS 115
3. — — J = — (^— q) = q = jy [i and ii above]
— i^ = — {— q) = a = -• [i and ii above]
— 6 ^6
The sign written before the fraction is called the sign of the
fraction.
Thus, if the sign of both numerator and denominator are changed^
the sign of the fraction is not changed; but if the sign of either one is
changed, the sign before the fraction is changed.
In case the numerator or denominator is a polynomial, we must be
careful, in changing the signs, to change the sign of each of its
terms (J 41, 3, 4, 5). Thus, the fraction - can be written, by
chauging the signs of both numerator and denominator, in the
. b — a
form-
d — c
129. It follows from HI, 6, 8, that if the terms of a fraction are
the indicated products of tu:o or more i^arcnthcses^ the sign of the fraction
will remain the same, if the signs of an even number of the paren-
theses be changed, but the sign of the fraction will be changed if the
signs of an odd number of parentheses be changed.
Thus the fraction - — — ■ can ])e written in any one of the
(n— i>)(7 — r)
forms,
m — / w^ — I m — I
( « — p) iq — f) (i> — w) (</ — /•) (/> — w) {r — q)
If the integer in the numerator of a fraction is less than its
denominator the fraction is said to be a proper, or pure fraction,
and if greater, an improper fraction.
Reduction of Fractions
130. The Reduction of Fractions to their Lowest Terms.
Let the line ^^ be divided into seven equal parts, at />, E, F, G, If, I.
I 'I I > I I I 1 I I > I I I I I i I I I I
A D E F G H I B .
Then (1) AG\^^ of AB. [a26J
116 (X)LLEGE ALGEBRA t?131
Now let each of these parts be subdivided into 3 equal parts.
Then ^5 contains 21 of these subdivisions and AG contains 12
of them.
(2) ^6^ is ^|of -45.
Comparing <1) and (2) it follows that
4 _ 12
7 "~ 21'
That is, the value of the fraction - is not altered by multiplying
12
both its terms by 3, and the value of the fraction — is not
altered by dividing both of its terms by 3.
131. The result of the previous section is a particular case of the
following:
TJieorem I. It does not alter the value of a fraction to multiply or
divide both of its terms by the same quantity.
Let q be the value of the fraction 7?
b
then ? — ?
and a = bq. [§126, (1)]
On multiplying both members of the equation by c, it becomes
ac = bqc = bcq [Ax. 3, J81; and Law XI]
whence Z^ ^ [281»4]
(1) - = -.
^^^ be b
If we multiply both terms of the equation
a z= bq
by _ we have
f = ? = (;)'• [«««• ^- iJ
(2) _^^„ = 5 (Inverse of U26, (D)
o o
Hence, it follows from (1) and (2), to reduce a fraction to lower
terms, divide both numerator and denominator by any factor common
to both.
28 132, 133]
FRACTIONS
117
132. A fraction is expressed in its lowest terms if its numerator
and denominator haye no common factor; and therefore any fraction
can be reduced to its lowest terms by dividing both numerator and
denominator by their G. C. D., because it contains all the factors
common to both terms of the fraction.
Eeduce the following fractions to their lowest terms.
1.
y'
{x — y) (x«-f-a;y + y«) a^-\-xy-\-y^
3.
x«_ y« (x _ y) (x + y)
x« — 9x + 20 __ (x — 4)(x — 5)
x«_7x+ 12 " (x — 4)(x — 3)
6 x* — 11 x*y + 3 xy*
6x2y— 5x/— 6y'*
X — 5
Since in example 3 no common factor can be determined by
inspection, it is necessary to determine the G. C. D. of the numera-
tor and the denominator by the method of division.
Omit the factor y from the denominator and divide.
Bx'-llxV + Sa'y'
6x« — 5xy — 6y*
6x'— Sx'y — 6ary«
X — y
- 6x«yH
- &x*y-\
h9a;y»
-5xy» +
6y»
2y'
4 xy* — 6 y' Remainder.
6.T' _5a-y — 6y»
2x — 3y Divisor.
6x»— 9ry
3x +2y
4a:y — 6y'
4xy-6y»
.-. G. C. D. =2z-3y.
Now (6x« — 5x^ — 6y«)yH-(2x-3y) = y(3.x + 2y)
and 6 x»— 11 xV + 3a-y' -h (2x — 3 y) = X (3x — y).
6x»-llxV + 3
•• 6x*y— 5xy»— 6
xy* x(;
y' ~y(5
5x — y) (2x — 3.y) x(3x— y)
tx + 2y)(2x-3y)~y(3x + 2y)
133. IFAcn <Ac terms of the fraction can not he readily factored^
then the G. C. D. mmt be found by division and the terms of the fraction
divided by it
118
COLLEGE ALGEBRA
[J134
ff.r4-T«
3 b.r — ex
17.
20.
20.
21).
:n.
33.
BXEBCISE XXIX
ar^ — he'* — r"^ „
21 aVj^c—^ a f?c^
'Sbc^ + c* 15 a*fe«c +3 frV/*<^— 12 a6»c
14 «2- 7 aft.
10 a^ — 5 fee
45 o'ftV 4-^27 rtWr d! J
12a-V4-2a«J:^.
18a62a; + 36«c«'
6ar + 9ftr — 5r«
12ad/+18M/— lOrd/
8.
10.
13.
30a«6V»(i<+18a76»c3c/*
5rf*-|- 5rrr o^ — ^^
a* — .t- ' (a — .r)*
(T'4-(l-ha)a.//4-v^
a* — ?/*
6c2 + 9crf-2c-3(^
9.
12.
30 (^'-^b^c""^* -6fl'»-*6\;'-d^'-
20 a»6'-ic«(P —4 a-56«cr+»
n«-2n+l.
?i«— 1
14.
16.
ac + 6d 4- <T<f + '"■_
af+2bx-\-2ax-\-hj'
n*
-2n«
n'
3^4-2^-3
jf^ + ^x+6'
a* — ar — a + •'^
.j*-3r-f-J
:r«-l
23. ^-^-
18.
21.
24.
1
^4j^4_3^
2a>— ff+10.r— 5
rt — 2rw:— lO.r + 6*
gg -}-/,«- ^24- 2 gfe
/>24-c«4-2ao'
r»2 —
.t^-.r-20
a^ + x-30*
, 2.x7/4-3g>/4-4.r4-6z
' if + y(z + 2) + 2z
4n4-4
m J- — ?» — jT-h I
(m-l)«
a6 4" ^^' + ^ ~ ^
2rt« — aft — 3A*
2 a«- 5 aft 4- 3//
19.
22.
25. :f ,
28.
x^-^
x^-/
;?;* — :r« — 2^4-2
"2^^ — x—l
.,-3_6j-«-37 .r 4- 210
.>4-4.r»-47j, — 210'
8 r>3 4- 36 ggft 4- 54 aft« 4- 27 ft»
4rt2 + 8aft4-3ft2
30.
32.
34.
6 ac 4- 10 ftc 4- 9 iid 4-15 bd
12 rtc 4- 20 ftc 4- 24 cu/ 4- 40 ftd'
{x-\-yV~ix'^-^y^)
{x + yf-^-1^
{•r + yY-{x-y)^
134. To Reduce a Fractioii to an Integral or Mixed Quantity.
Change
3.r«4-2x + l
.r + 4
- L
0 a m
3.r» +
2.r
+ 1
3x' +
12. r
10. r
+ 1
—
10. r
— 40
to a mixed quantity. Thus:
3.r-10
Therefore,
If the degree of the numerator br equal to or greater than the degree of
the denominator^ the fraction may be changed to the fonn of a mixed
or integral expression by dividing the numerator by the denom,{nator:
n35] FRACTIONS 119
The quotient will be the integral part, and the remainder, if any,
will be the numerator, and the divisor the denominator of the frac-
tional part of the mixed quantity.
EXBBCISB
Reduce to integral or mixed quantities the following:
23a .T^-h4
5 ' x-i
9 «'+2afe „ «'+6»
^' a+b ^- ^^nfTiF+l?'
^ .%fiV + 3c. ^ 3* — ^
9 X +y
Si^4-f}h* ,^ 8j:* — 5j:^— 3
4^ ' 4j:«+3
'^" 2x+3 * ar« + 6j;+c
135. To Reduce a Mixed Expression to the Form of a Fraction.
We have learned in Arithmetic that
4 4 4 4
In Arithmetic, the sign connecting the fraction and the integral
part of a mixed number is always +, but in Algebra, it may be +
or — ; so that a mixed expression may have either one of the follow-
ing forms:
Reduce i ^ it r to a fractional form.
0
U+'^\ Xh = Ah + ^ 'b = Ab + a [Law V, ?7; Def. IX]
But (Ab + a)-^b = dA±_^ [Def . of division]
A + 'I^=^-1:l^- [Ax. 6, 181]
Similarly: A — - =
^ b-
"- b
Ah + a
b
Ah— a
h '
— Ab +a
b '
-Ab-a
120 COLLEGE ALGEBRA [W36
To reduce a mixed quantity to a fraction, multiply the integral pari by
the denominator, to the product annex the numerator^ and under tk
result write the denominator.
136. The sign before the fraction shows that the number of
things of the group b indicated by the numerator must be added oi
subtracted according as the sign is + or — from the number of
things in the integral part of the kind in the b group, i. e. , from Ab.
If the sign — precedes the fraction, when the numerator is
annexed, the sign of every term in the numerator must be changed
Thus: ^_^^^Ab_-p-y)^Ab-x + !,,
b b b
1. Change to fractional form 1 — -• Thus,
a -\- b
a—b_{a + b) — {a — b) a+b^a + b 26
a-{- b a-{- b a -{-b a -^ b
2. Change to fractional form 3 x — — ' ~ •
za
bax — 3 Qax — (5 ax — 3)
dx — — ^
2a 2a
_ 6 ax — 5 ^.r + 3
2a
ax -|- 3
"^ 2a
BXEBCISB
Change to fractional form:
1. 13r\>9Jf' 3||. 9. <^-(a:+2^-
a+ X
2- ''■+-a- 10. l-^°^-
ix 11. l+x+.c»+j;»-
3. 3 + ^^- „ ,^,_^.^._ 2(^+1).
4. 2.-3,+ ^ ^ ^_^_3fc£).
a—b ^--2
x~l
5. 1-f . ,
a+6
6. 1-
13. ,-a+y^^~+J+^'
8. a+6-- 7—^^ 15. x-\-Z — -z — ^r-
a-\-b ' a;* — 9
« 137, 138] FRACTIONS 121
137. Reduction of Fractions to a Lowest Common Denomi-
nator.— Some propositions concerning fractions in Arithmetic will
now be recalled, and be proved to hold universally in Algebra. In
the following paragraphs the letters represent positive integers,
unless it is otherwise stated.
138. 1. Rule for multiplying a fraction by an integer. Either
multiply the numerator hy that integer^ or divide the denominator hy it.
«,, nr a a c a y, c ac ^,^« ^
Thus, I. -xc=-Xi--=,-^,=y [J63,Eq.l]
Or, in the fractions ^ and ^ the unit is divided into h equal parts,
and c times as many parts are taken in y as in ^ ; hence ^ '^^ c times
-• This proves the first part of the rule.
^«^°- rc'"' = lM = l- [2«3.Eq.i;n3i]
Or, in each of the fractions ^ and ? the same number of parts is taken
in each case, but each part in - is c times as large as the parts in — »
since, — X c = ,-— — = r- [U31]
he b X c 0 *-
Hence, II. 7 is c times 7-?
6 be
which proves the second part of the rule.
2. Rule for dividing a fraction by an integer. Either multiply the
denominator hy that integer^ or divide the numerator by it.
Let ^ be any fraction, and c any integer; then will - -^ c = ^'-.
For, ^xc = ^- [1; ?131,Th.I]
Therefore, ? is c times —j that is, ^ is -th of 7- This proves the
'6 be be c b
first part of the theorem.
Let now ^ ^^ *^y fraction and c any integer; then prove that
-^ -1- c = 7 • By the first part of the theorem
6 b
f -=-c = f; but^=f, [Uai.Th. IJ
b be be b
ae , a
This demonstrates the second part of the rule.
122 COLLEGE ALGEBRA Hi 139, 140
3. We have from {131 a third rule of frequent use in reducing
fractions to common denominators.
ac a
5^ ""6*
139. Rule for Reducing Fractions to a Common Denominator.
Multiply the numerator of each fraction by all the denominaton
except its own for a new numerator of that fraction^ and multiply all
the denominators together for the comrnon denominaior.
Thus, let ^» % and y be the given fractions; then, by S138, 3,
a aSf c cbf e ehd
h~' bdf d" dhf f~ Jbd
But ^» j-^j and ^— are fractions which have respectively the
same values as 7* % and 71 and have the common denominator
u a f
hdf; and, further, each numerator is found according to the rule
above.
140. If the denominators have one or more common factors, the
rule for reducing them to equivalent fractions, with their lowest
common denominator, will be:
Find the Z. C. M, of tlieir denominators; then for a new numerator
corresponding to each of the given fractions divide the L. C. M. hy the
denominator of that fraction and multiply its numerator by the quotient,
1. Suppose, for example, that the given fractions are -—, - -^■
The L. C. M. of the denominators is 10 abc.
And lOabc-T- bbc=2a
10 abc -r-10 ab = c
10 abc -^ bb = 2 ac.
Hence, ax2a=2a\ the new numerator of first fraction;
b X c = be, the new numerator of second fraction ;
c X 2ac = 2 rtfr*, the new numerator of third fraction.
o 2 a' 6 be c 2 ac^
Vbc ^ lOabc' 10 ab ~ 10a6c' bb ~ lOabc'
The second members are the equivalent fractions of lowest common
denominator, which are respectively equivalent to the given fractions.
2. Reduce ^ . — r— . and , 3 to equivalent fractions hav-
ar — lar+l 1 — x^ ^
ing the least common denominator.
«140] FRACTIONS 123
The L. C. M. of the denominators a — 1, x-\-l, and 1 — x* is
(- 1) {x + l)(x-l) = -{x'-l) = l- x«;
and (l_a:«)-4-(x — 1) =—(x+l)
(l_x2)^(x + l). = _(x-l)
(l_x*)^(l— a:«)=:l.
Hence, x x [ — (ic + l)] = — ic(u:+l), new num. of first fraction;
1 X [ — (x — 1)] = — {x — 1), new num. of second fraction;
1x1 = 1, new num. of third fraction.
.'. the equivalent fractions are respectively -— — ~- — , — ^ — ■ ,
I — X* 1 — x"
1
1 — x«
3. Reduce —-— — : » ^n :» w/ 2 "^{ ^ equivalent frac-
2(1 — «) 6(a' — a) 12(cr — 1)
tions having the least common denominator. •
The L. C. M. of the denominators
2(1 — «), 6(fi« — a), 12(a«-l)
is 12 «(1 — a)(l + o) = 12 a(l - a«).
And 12a(l — a^)^ 2(1 ~ «) = 6a(l +a);
12 <i(l — a«) -f- 6(«« _ a) == — 2(1 + a) ;
12 «(1 - ««) -f. 12(a« _ 1) = _ «.
Hence a X [6/i(l +a)] = (}a\l + a), the first new numerator;
1 X [— 2(l + «)] = — 2(l + «), *^ 2d **
(3a+l)[— a] = — a(3a+l), ** 3d '*
.-. the equivalent fractions are respectively
6 a«(l + a) - 2(1 +a) — ri(3 a +_1) ^
12a(l— a^) ' 12rKl — «') ' 12a(l'— a^) '
EXERCISE XXXII
Reduce to equivalent fractions with the lowest common denominator:
^- 6c' oc' ab ^' bx 10 r*' 15:1:*
23: — 5 33: — 7 a — b a+b b^
'^- 6 12 *• 3 a 4(1^ 6a8
2x_^3jii^ 3r — 6y. 2(z— 5c^ 3a — 4r.
^- 5a:* ' 15^ " ^- Sac* ' 20a2c
a: + 5 x+G x + S ^ 7 8
7 1 , , 1 . Q ,
ab a^b al/^ °- l + x l-./-^
1 1 1 ,^ '"J
9. zzTx' ::F-r' rii* lo.
ar-l x'-l x+\ ^"* (j:-y)(^-2) (x - y) (j: - 2)
124 COLLEGE ALGEBRA [^141
1 1 f J , 1 1 a y a .
^^' a + b a«-6« a-b ^^' 3(x + y) 6(jt*-y«)
2a ^ 3a« ^ 5q» §£±2, 2.r-l^ 3.r-f9.
^^- a: + 2 i+3' j:« + 5x+6 ^^- or— 3 ' 5x- 15* 7x— u'
a 6 c ^ x—my z — ny
A^- a»-b^ a«+6« a* -6* ^*^- ' mV n*x
17. '^ . 6 - c .
(a+b)(a+c) (b + c)(b + a) (c + a)(c + b)
a+b b-\-c c + a
1Q ' » ' 1 ' •
^^' (b — c)(c-a) (c-^a)(a — b) (a— 6)(c — 6)
1 2 3 5
a — b b^ c c — a (a — 6) (6 — c) (c — a)
^- 12-c3_2^-20x-6 ^^ 4a^-6a:»-4ar+6
1 _ 1 , 1 ^ L_..l_.
oo 'l±£_lzii-__ JL, 1-2j^ 2-3j^ 3 .
23. ^+^-^+^n-i + ^+~+/
24. ^_„ -Tpx-n +pa-r "T j^i + '•
y* . :r 1_
97 -4- 1 '—*!— — . 2-.
'• a* + 6P ' o*— 6P x"* — y" a:"* + 3^
Addition and Subtraction op Fractions
141. The sum or the difference of two fractions having a comimm
denominator is a fraction whose numerator is the sum or the difference
of the numerators of the given fraction^ and whose denominator it the
common denominator.
It follows from equation 2, ?67, that,
ah a-X-h _ a h a — h
- + - = — ■ — } and - — - =
c c c c c c
3« — 5Z» , 26— « 3ri — 56 + (26 — a) _ 2a — 3fe
and,
3a 3a 3 a 3a
6_6 — ^a _ 46 — 3a _ 66 — 5 a — (4 6 — 3a) _ 26 — 2j«
3a 3a~~ 3a """Sa
J142] FRACTIONS 125
142. If the fractions to he added or subtracted do not have a com-
mon denominator^ they shoidd be reduced to equivalent fractions having
a least common denominator; then proceed according to the rule in 8141.
Thus we shall prove
a c a^4- cb
l + d = -Td- *'•'*
For {l + ^)M=ll., + la.l>
z= ad -\-bc
a
c ad — be
b
"d" bd
[Lawsiri-V;{38, 5]
[Def. IX, J61]
[Def. IX, ?61]
a , c ad -V- cb ^^ ^^^ .«^-,
b + d = TTd- [Law XI, 562]
A similar proof holds for the difference of the two fractions.
Compare U39 and 2140. Thus,
X— l"^x+l "" (x— 1) (x+1) "■ x«— 1 '
a a' a^-\-ab — a* ah
1 a-\-b a-\-b a-\-b
In case the denominators b and d have common factors, for example,
1. Simplify ^ + -,-
Reduce the fractions to equivalent ones with a least common
denominator,
y*z yz* y^z* y*z^ y^z^
2. Simplify -^ + -f-.
X— y x+y
Since the denominators do not have common factors use the
principle just proved above.
^_ y X (x + y) -\-y (x --jy)
x—y x+y {x—y) {x+y)
X* + ?/x + xy — ?/* X- -|- 2 xy •
x^ — y^ x^ — y^
126 COLLEGE ALGEBRA [2142
3. Simplify — -— — -•
n n _ »(x" — 1) —11 (a;"+l)
X»-f 1 :€" — 1 (x»+ 1) (iC»— 1)
wx" — n — fix" — n — 2/1 2»
3
— >
4. Simplify - + «---;;——«- -,r-
*^ -^ x^3x+6 x8+2x
2 x^_G^ _ _ 1 ^ 6(.r4.2) x(x-6) _
x"^3x+6 x« + 2x 3x(x+2)"^3x(x+2) 3x(x + 2)
- 6a; + 12 + x^— 6.r — 3 _ _«« +_9
"" 3x(x + 2) ~~3x(x+2)'
5. Simplify ^-_l-^-___L_
1 _ j;« ^ ;>. _L 1 _L r8
X* + x«+ 1 u:— 1— x«'x+l+x'
X* + x«+ 1 ^ X*+ 1 — X ^ x« + 1 4- X
^1 _ (*^'+_l_+^^) + C^'+ 1 Ti5l «
^*+ X* + 1 ^ tlt^+l)- .r] [V+ 1) + x]
1 2(x'+l) ^ 1±2U«+1)
j.4-1- x« + 1 "^ (X« + 1 )»— X« X*+ X2+ 1
_ 2x«+3
Note.— In case the denominators are binomials, it will always simplify the reduction
to arrange the fractions so that ail the denominators are arranged in the order of the
descending or the ascending powers of some letter.
EXERCISE XXXm
Combine and simplify the following expressions:
564a b ^ a J h a b c
^i ^1 _JL, . JiL _ ^^''^ I 2. fl _ h 2cd,
^- 4b^ Sl ly ^' 4bg 12 bh'^ 'S ^' 4 bed 2bcg'^bbg
2a . 6df ^ deq . _ ; _ A _ _? .
'• Zbc'^^t^c 66V* ef eg
_-__j7i ,_£L. fM _ ^nd _6«.
^- * -^ 2f/"*"3ri7 ^"- 36vr» 2 6*c« C(i'
W42] FRACTIONS 127
13. c + 2ab — Sac — r^ t
(r — bc
a-^b a ~ b a-\-b _a — b^
^** 2 "'' 2 ^•^- 2 2"
13« — 5fe 7a-2/> 3 a
^^- " 4 ~ 6 ~ 5 *
3a — 46 2a — b — c , 15a~4c
17. --y— -— 3 — + -j^
9j--h2_7j:-h5_8-7T 5-3j:_3-73:.
^^•3 4 0 "^ 8 12
7fi — 36 4 (? — 56 _ 3ff— 86 5a — Oj: _ 36 — 8a:
19- 4 "^ ' 0 1) "*" 18 24
^ 5(2 -r- 3) 2(7 r- 5) . 4(3 r-M)
'^' ~ i - 3 + 5
3(2rt-36) 2(3 g- 5 6) , 5(« — 6)
21.
8 3 +
6
22.
a — 3 6 4a -6 5a-|-3.r
6a "•" 26 "^ 9^'
a« — 6.F 2 a
2ajc 6
23.
3a— 56 a — 7c 56 — 4r
15 a6 12 ac 206c
^4a^55^3c
24.
6a-|-c 5a — 46 36-5c
6 6c 4 ac 5 a6
. J._i . A,
2 _ 1 _2a-h3 1 3a — 26.
^'^* 3a 26 6a« "^2^^"*" Oa6
1 j_ a — 2b _ _3^ _3a — 4j^ _l A 1 '"^-^ — _5^.
^' Ga"*" 3a6 46 8at^ "'"8^*"*' 126j^
5 a^^x __ 36— 4.r 4a« — 5 6 _ a^-jr __ a-6 ^.
^- " lOfur 126jr "^ 20a«6 4a«r 5rt6'^3 6'
a (3 6^ 2 c} _ 6j4 a - 5 c) Sa^-f 3 6^ _ 5a-46.
^- 66c " lOac "^ r)a6 10 c
a(3j^— 26)_6(5a'g-3a) 5a-66 a 2 6.
^- 12 6j« "l5a^ "•" 30a^ "^12 6 "'"3 a
01 <L+fe ■ «-^
'^^' a-b^a+b
^ ^ ^
30.
a + b^ a-b
32.
7 5
a + 6 a
34.
6 5.
x4-3 3
33.
a + 1 (( - 1
3r-l 2.r-7
^* l-3.r 7
128 COLLEGE ALGEBRA IU42
36. ^7:3T;-.-7r^' 37. ' '
2(j:-l) 3(j:-1) "^^^ 3^-9 6x-15
(Kf+fcc , ad- be . T-1 _3j--4 2x-l.
^- 2cd(c-rf)'^2c(i(c + d) ^•'- 2J-+2 3j:+3"*"6x+6
40.
5^r-f 4 _ 3j:-2 _ j«— 2jr — 17
a: — 2 a:-3 a« — 5x + 6
4, ^+y"'__a-*" — y**. .r— 1 a- — 3 _ a- 4-1.
43. i-r^-,4 '^
x-S x+3'^(j: + 3)»
Sm _ /8?n* — 15mj 3: \
^' 6a' Vl2wa: — 6a:*"*"a?--2m/
*^- x + y Va:« + y« x-y^
46.
r-4 _3t-5 5.r«4-9j--H4
2x-3 ar+2 "*" 2a,'« + 3j — 2 '
47 1 _ 4 8 3T-h7
ar-l 1-x l + x^ j^-l
48.
8 . 6 3t — 4
2a: — 3"^3 — 2jr 2j:2-a; — 3
49 -A- . _1.__5_. ^ 2 9 4 .
x+l^a:-l a; + 2 ^'' 2a--l 3 ar - 1 "^ 2^^::^
.. .J_. _^__3__. __2 ,1,21
^'- x + a"'"ar+6 ar + c ''^* (j:- i)8-r (x- l)t"rar - 1 x
.« ^___J_. L_ . 2 4 4
54.
65.
a--l a:+l^(x- l)«^(ar+l)« x^ + I (a*- 1)«
a--|-l _ x-h2 _ a:+3
{x + 2) (a-+3) (a: + l) (ar + 3) (x+1) (ar+2)'
(a--l) (a:-2) (x - 2Ma- - 3) "^ (a: - 3) (x — 4)*
^* a-1 a«+l a+l"*" a«-l
57.
2x . 3x 5x
x«-x-20"^a^-8x+15 x«-9x + 20
g« -^'-f « __ r+b . x+c
x« — (6 + c)x+6c a* — (a + c)x + ac"^x2-(a+6)x + a6
50 ^ -y^ . J^^--*: I ?!=j£^
(-r + y) (a: + ^)^(y+2) (y + ^)^(2 + x) (2 + y)
"^^ 2x-2y"^2x + 2y x*-y*
8113J FRACTIONS 129
Multiplication op Fractions
c
143. The rule for the multiplication of two fractions is:
multiply the numerators together for a new numerator and the denomi-
nators for a new denominator.
It has already been proved in {63, equation 1, that if ^ and
are two quotients, which are integers, then
a c ac
b^Tl^bd'
This formula is also satisfactory when one of the quotients t or .^
is an integer (U38, 1, 2).
In case, however, both ^ and ^ are fractions (see {126), we can
not speak of multiplying them together without defining what we
mean by the term multiplication, for, according to the usual mean-
ing of this term, the multiplier must be a whole number. As
already explained in {72, the so-called rule of multiplication of
fractions is really a definition of what we find it convenient to
understand by the multiplication of fractions. And this definition
is so fashioned that in case one of the fractions which we may desire
to multiply together is an integer in a fractional form, or. when both
are integers, the result of the definition coincides with the conse-
quences deduced from the common use of the word multiplication.
The symbolic definition of the quotient a by 6, e. g. , ^, and its
operation as described in {72, formula, justify the rule for the pro-
duct of two fractions ^ and ^ as expressed by the equation
a ^ __ ^^
h^ d^bd'
NoTB.— The following verbal definition also will show dearly the connection between
tbe meaning of the word multiplication when applied to Integers, and when applied to
fractions. When the product of an Integer a by 6 is formed, the operation may be
described as follows: What is done with unity to form b, one must do with a to find 6
times a. To obtain 6 from unity the unit is repeated b times; hence to find b times a
the number a is repeated 6 times. Therefore, if one desires to find the product of
- by - , by adopting the same definition as above, it follows that what is done with
<7 a c a
unity to produce - one must do with - to find - times -•
To obtain ^ from unity one divides the unit into d equal parts,
and takes c of them; hence, to find the product of ^ by ^ the
fraction ^ is divided into d equal parts, and c of such parts are
130 COLLEGE ALGEBRA [^44
taken. If one divides ^ into d equal parts, one of these parts is
,-. ({138. 2), and if c sucii parts are taken the result will be ^•
bd tMi
Q. E. D.
1. Simplify - -— r X
x + 1 x* — 27
x-3 x' + 2.T+l X— 3 (x4- 1)* x+l
x + l ^ x^ — 27 x4- 1 "^ (x— 3)(x2+3x+9) .r«+3.r+9
The factor (x — 3) of the firet numerator will cancel the factor
(x — 3) of the denominator of the second fraction; and x-|-l, the
denominator of the first fraction, will divide the numerator of the
second fraction (x+ 1) times.
2. Simplify -— — —rrr- X o,. , .o>7 X
24 6x+ 15/>^ 20 6 — 12 f^ n
10a/>-6m/ 72^^ -1-^5 (7f/ 2 />_ 2a{bh — '^d) 9^(8r+5v) 2 6
24 6j: + 15 6i/^ 20fe-12d ^ a"~36(8z + 5y) ^ 4(56-3(yV^ a
The factor (5 h — 3(1) in the numerator of the first fraction will
cancel the factor (5 6 — 3 d) in the denominator of the second frac-
tion, and the factor (8 x + 5 y) in the denominator of the first fraction
will cancel (8 x + 5 y) in the numerator of the second fraction.
Similarly the factor a is common to the numerator of the first frac-
tion and the denominator of the third fraction, and b is common to
the denominator of the first fraction and the numerator of the third
fraction; 3 is common to the denominator of the first and 9 d in the
numerator of the second fraction; and the two 2's in the numerator
of the first and the third will cancel 4 in the denominator of the
second fraction.
nrx 4- x' 2 hx — ex' a + x
3. Simplify jrj--^ — X -7—,—.^ X -\-
2 6 — ex {a -f x)* X*
ax+x' 2hx — ex* a-\-x _x(a-\-x) x(2h — ex) a-\-x
2h — ex ^ ~~{a+x)^~ ^ "^ ~ 26 -7!^ ^ {a+x)^ ^ ~x*~ ~
4 ^J^4^ .. ^^-?/' i2g4^\>gTr>?'-n«)_(x«— yg)(a^^^^^
.(m-h^3-^ J^ ^ m-n ^ .^j^-^-TT 2(.w-=«)(j»-r^
rn+n 2 t t
_ ^- — .V
~ 2
144. The principle proved in ?64, 2, namely,
«*" -f- a" = a"*""
can now be extended to the case in which m is a smaller integer
than n.
«45] FRACTIONS 131
For example,
rf7 . «"-7
W^ 1
In general — = in <rn,
^ a* a ' a to w factors
For — = —
a" a * a to 7i factors
a ' a to m factors
to m f actora • a • a • • • to h — wi factors
1 •
a ' a ' a
1
to w — 711 factors
..... 15r/'6« 14 xy
For example, simplify — -, x ^5^ •
15a»/>« 14 xy Sa^-^^-^-l
X -
22 x^y' ^ 25 a«6 "" 11 x^-y-' -5
_ 21 nh
~ "55 :r^* *
When m and n are positive integers the quotient of a"* by a",
when TO <:^?i, is unity divided by a with an exponent which is equal
to the exponent of the divisor minus the exponent of the dividend.
145. The Powers of a Fraction.
The power of a given fraction is a fraction whose numerator and
denominator are the respective results obtained by raising the numerator
and the denominator of the given fraction to the required power,
^^'"' fe) =u^
where n is a positive integer. For example,
5 a%\^ (5 aV>')3 5» • (««)»• (b^Y 125a«6»
(3 xV)' "■ 3»- (.r/*)3 ■ {iff 27 A"
The formula above is, when n is a positive integer, the immediate
consequence of the definition of an exponent, equation 1 of ?63,
and again the definition of exponent, thus,
(a\ /a\ /a\ , ^ , a - a - a - • • • U) n factors
b) ' [bj " Uj "*^ ^ ^^^^=^.6.6....to.rfa^tors"
_a^
" b^'
13^ COLLEGE ALGEBRA [?145
EXEBGISE XXXrV
Multiply:
5xy Say chir 4 clt* 4jr^ - 5ff*or
6. —J— by \ ' ,• 7. 77 TTTr by (3a — 2 6).
a+a: "^ a' — jc* 15 a — 10 6 -^ ^
Simplify the following expressions:
g^feV 5 a*ft»c« 21 cV • 3 c»// 4^ 8 ^"^^j^**
cV ^ 7 J,V 10 a»6«' 4 a:«-i ^ \bb^^ 9 r^y*
"• 15 63x«-i ^ a»+i '^ X "»-i 3 a + 36 ^ (x - y)«
(^-y)' (r+y) (.t-^-j/^) '^- a^-7a: ^ Jt« + 2x
3:«-11j: + 30 jg-Sj; .i^-O a«+2a6 q6 — 2fe«
>-6.r + 9 ^ 0^-5 ^^-36' a« + 4 6« ^ a« — 4 6*'
62^7^, + C ^,8_|,8 ft + 12 /,»-8/;«
62+26-8 ^ />«-146 + 48 ^ 6« + (>6
a« — 6« a6 -26f a« - a6
^^- a2-3«6 + 26«^'a«+a6 ^(a-6)«'
a»-3aV>4-3a6«-6» 3 a« + a6
^®- a«-6* ^2rt6-26«^"a-6 '
(x4-.v)2-rV z«-(.r-.v)« (:r-a)«-6«^-(6-a)«
^^- :c«-(i/-2j«^22-(r-y)3 ^- (i;-6)»-a«^:r»-(a-6)"
jr + V r«^i/8 4(m + n)» 6(m«-n«)
'^'^- (m + n)«"'" 16 "*■ (m-w) "^ (^+y)«
a»-6» a + 6 (a«-a6 + 6»)«
^■*- a»+63'^a-6^(tt«+a6 + 6*)2
Multiply:
25. 5-f + l by ^ + ^ + 1. 26. :r.-x+l by ^ + ;+l.
Simplify:
a^4.(a-f?>)x-f-a6 a:«-a«
H146-148] FRACTIONS 133
Simplify the following:
35.
87. (^r^yx
(^»_j^,«
(a + 6)«
Division op Fractions
146. The Reciprocal of a Fraction.— 7%e reciprocal of a frac-
tion i$ a fraction whose numerator is the denominator, and whose
denominator is the numerator of the given fraction. Thus, the recip-
rocal of
a , h
b a •
Conversely ^ is the reciprocal of —
147. Two fractions which are reciprocal have the following
property:
The product of a fraction and its reciprocal is unity.
„ a b ah ah
For - X - = J- = -r = 1.
o a ba au
148. Suppose it is required to divide j- by y then the rule estab-
lished in 3, 871, holds, and we have
\~h) _ad
134 CX)LLEGE ALGEBRA [21149, 150
In case either of the fractions is not an integer the quotient of
7 by ^ will also be
o "" a
ad
where the general definition of a quotient of A by B is expressed
by the equation
A.
Q-
See 171, 3. Rule for finding the quotient of two fractions: form ike
product of the fraction used as the dividend hy the reciprocal of the
fraction used as the divisor.
Proof — Suppose it is required to divide t by - • If an integer A be
divided by another integer -8, one must find a quantity whose product
by B is equal to A. One gives precisely the same definition to the
quotient of i hy ^- A quantity must first be found such that its
product by ^ is 7-
a 0
ft. f* (14* /*or
Let y -7- - = x: then by definition - = x x , = -r (?)•
b d h da
Therefore (^-) d = (Jj\ d (?) and y = ex-
. ad a c aA
hence — z^ x\ ,\ T'^l,'= T"
be b d be
149. Hence, from the definition of the quotient of two fractions,
we have from the equation in ?147
a _ 1
0
3 ^ 8 24
(s)
160. Complex Fractions. — In all the discussions hitherto it has
been assumed that the letters a, b, c, d represented tohole numbers;
and thus only rules which are familiar to the student in Arithmetic
have been recalled. But by reason of the extended definition it can
be proved that all the rules and formulae already given are true
when the letters denote any whole or fractional number. Take, for
n50] FRACTIONS 135
example, the formula ~ = ^, and suppose that it is desired to show
that it is true when
w r mr
q s qs
ac __ mr pr __ mr ^ qs
he ns ' qs ns pr
mrqs
nspr
mq
~ np
a = — » o = _» ana c =: — ,
n q s
here 2=1=^X^(2148) ='5«
o p n p np *
also
and
hence
oc
V = 7~> which was to be proved.
o be
Moreover, these rules and formulae hold when the letters stand
for negative quantities (1 272, 81, 8, Remark).
16a»6« ^ Aa% _ 16a»6« 9 x'y _ 4g6
27 xV "^ 9 xV " 27 xV '^ 4 a«6 " 3xV'
o Tx. .^ 8(a« — 6«)« 4(a + 6)
2. Divide ' , ,/ by ' ^ -
7(x«— 1) -^ 1 — X
2 a+6
8(a«— by ^ 4(o + 6) _ ^(a — b)g0^^friry' ix-^^
7(a;»— 1) • 1-x ""7(ji>--rrj(x«+x+l)^4XiH'^
—1
2 (a~6)»(a + 6)
- -7(x« + x+l)
[(^/+x)'— 4ax] [(a — x)*+4ax] a*x — ^rx*
a* — X* ' [(a+x)*— ttx] [(a— x)*+fixj
a — X
_ (g — a;)* - (« + x)« (q« + ctx + x») (a' — ax + x')
(a* — x') (a' + x*) ax (a — x)
a — a; a+x
a+x
_ (g— x)(a + x)' _ g + a;
ax(a — x) (tt+ x) ax
136 COLLEGE ALGEBRA [1151
4. Simplify 1±1.
1 + 1 = 1 and4 + | = V;
hence 1±|= |-^ V = | X ,', = H •
5. Simplify
1_I
n m
• — m '
n
1 _ i- m — n
n w. inn "» — w n m — n
-74'
- — m
n
m* + ?<' — 7n?i wm ?»• — m»-|-»' m(»»* — mn+n*)
1-i
n m + n
151. Continued Fractions. — A continued fraction is a fraction
whose numerator is an integer, and whose denominator is an integer
plus (or minus) another fraction whose numerator is an int^er and
whose denominator is an integer plus (or minus) a third fraction, etc.
1 1 1 1 36
6. 2+g— ^ 24-- 2+-- ^^
7 7
^"*"ar+2 x+2
= x — \
""^^ (-r+l)«
= jr-l
(r-f-2) (J+I)' - {x+\){T-{-%
(x+\f
^"^ U + l)(x-f-2)[x+l-l]
_ _ 1 _ J- ^.r -4- 1 ) a^+3:-2-jr-l
"•^ (x + 2)jr"" a;+2
rr^-3
x + 2"
SlolJ
]
FRACTIONS
EXEBCISE :
KXXV
Simplify the following:
1.
2.
4a«6"*'
(-lo-
3.
/ 15&« 27c«x
V ICc ^ lOod)
abc
2SiP'
4.
27 7i2j/ /^
7/1 n
5.
U* ^ a6^ or/
m^
■^V
6.
z+1 _j_:r«-l
X a:
7.
a:* — w* J* + V
(1 + 6 "^ 1
8.
6-c 1
b+c'^ c + b
137
9. (l+x)^^(x + l).
10.
x^ + y^ x* + y
3:*— 7x+12 x — 4 J-^ + T/'— 2.n/-~2« , r— ?/-i-g
^^- 2«+4jr+4 "*'x + 2' ^^- a«-9 + 46« + 4a6 "^0+26-3
^^' 15ar+21ay"*' 5z+7y ' ^^- 35 6«-426d "*■ 20a6-24a<i*
/ 3t \ , xix—2) /T« 1\ . /T , 1 . 1\
1^- V W V x) ^^' a»-6« a* -6*
19.
24.
/gg — 3a«fe + 3ay-6» 2q6-26«\ a«+«6
21. V a2-ft« "^ 3 / a-6 '
-^^^ l^y ^x + x^ x^ r^«+» + 0x»+3 + 9x3^a:"+« + 3i^
,- (■r + ?/)«-(z + 7r)' ^ (j--2)g -(//;-?/)«
z« — 3!* — 2j7/ — y* ^ ^4i?/ -tJ
^- j:« + 2ry + 3/«-2« ' :i + t/-2
27. a«--6«-c« + 26c-^-^?^^.
a + o+c
g8+3«tj-f3qjg + j4» ^ _ (« + J)*
ic* — y» .i-^* + ^»/+?/5
a+6+c
a + 6 — c
28.
29. n«_6«_c«-26c-*-
z* ^x ***• 2«'« ^ x^ 2 a 3:
138 COLLEGE ALGEBRA [W51
Simplify the complex and continued fractions:
5— c a4-h ax-^h
32. 9~- 33. ^- 34. ^•
a b h
X + 1 3^2. n^
36. — ±- 36. -75-. 37. ^•
* »+i «arj/ 356*
« o + x J"— I
38. — • 39. ±r- 40. —
y + K? „ ^ 1+ •'•
X
1 —X
1
4L . 1_ 42. , _ 1 ' , !_
\-\-r. 1 + x 1-x
a:* 4-?/* __ « I C_
y a-*- v* 6"^d
1 1 ^^ + ^' ^' TTs!
y X f h
45. ^-i-f. 46, *"=' "^«
A: m
a . h c* c*
a_£|Vf6^ £_a±6.
*^- _a 6_ ***• (!« cy
a- 6 a+6 a+6 rf/t«
^^' 24x+16ax+l2xy + Saxy (2ax + Zx 8a+12i
50. ^-^^H^-^-]n?7^i[-
51.
1+ ^-IT-^ 1-
1+. + --'^ . 1- ^
l—x l—x
53. :3 54. ar+l
a+-^^ :, + 2--^-+--l
a: y a6
2152] FRACTIONS 139
('^in'-ir
58.
57. frr: mr y +
y
bjf^ — b . 2*" — 2
(,+l)-(.-l)' ' »+•
59.
X*-r — 2'^ X"4-l
Special Theorems in Fractions
152. Theorem I. — If several fractions are equaly a fraction whose
numerator is the sum of the numerators and whose denominator is the
sum of the denominators of these fractions ^ is equal to each of these
fractions.
Thus, if the fractions 7 > 7, > 777 *re equal, each of them is equal
b b b"
to the fraction
b~+h'+b"'
h
„ a a' a"
For, put -= - = ~ = g
and it follows that
az=: hq
a' z= b'q
rt" = b"q
and, in adding these equations member to member,
a + a' 4. a" = (& 4- 6' + h'')q;
since a -+- a' 4- a" is the product of 6 + Z»' + ^" ^Y ?» ? ^^ the quo-
tient of a + a' + a" divided hy b + h' + b". Thus it follows:
^-6 + 6' + 6"
and consequently
a _ a' _ a^ _ a + a' + a"
^ ' b^V^V'" b~+~b^+b'''
More generally, let the fractions he
fi_a[_a^
b^b'" b"
140 COLLEGE ALGEBRA [«52
and let w, m', m", be any positive or negative numbers whatever; it
follows that
a a' a" ma -|- m'a' -f- tn^^a''
b^ b'~V'^ mb+ m'b' + m"b" '
since each term of a fraction can be multiplied by the same num-
ber without altering its value (1181), one has
b'" b'" b"'~ mb ~" m'b' ~ m"b''
and, on applying the preceding theorem to the equal fractions
ma mJa
mb m%' m''b"
it follows that
. ^ _^' __ ^" __ wia _ m'a' _ m"a" __ ma -|- m a' + w"a"
^ ^ b~b'~Tr^mi)'^ ^' "" m"«^" " mb + w'Z^' + m"6"'
Theorem II. — Tf several fractions are eqiuil, each of them is equal
to a fraction which has for its numerator the square root of the sum
of the squares of the numerators, and for its denominator the square
root of the sum of the squares of the denominators of these fractions.
Thus, if one is given
b~ b' ■"^"'
it also follows that
b~ b''~ b"" i/^.|. h'^^h^J
For, since the fractions t ' ?7 » and ^ are equal, the squares of
these fractions are equal fractions, and we have
__ __ " " i~" -t- ^ . r3152 Th II
/>*""Zi'«~ 6"*"" 6«+ 6'«+ ^"* L*****, -J
Since these last fractions are equal, their square roots are equal,
and we have
b " b' ~ b" ~ |/^2_|.^'>i _^ t'/«
For, let _ = _^_-:=^.
then a^=bY, a'^=ib'Y, «"« = i,'/*^.
a=«»9; a' = ?>'<7; a^rr^''^ [S81, 8]
and _ = ^_=_^_ = j. Q.E.D.
J153J FRACTIONS 141
It may be demonstrated, as in the preceding theorem, that what-
ever the numbers wi, w', wi" are, one has
a __ a' _ a^ __ l/ ma* + m' a'* + m" a''^
153. Application of Theorems I and 11.
Example 1. Given that (1) l^?SK±^ and (2) - = ^
T m 1^ f r(r + k) r U
Lrove that - = — . = —•
t h' 'T
R II
Since — = -r' hence, by Theorem I, \ 152,
r h
R H R+II
r h r + A
J? \ JT
Therefore, on substituting these values for — ^7^ in (1) we obtain
'/■ -\- h.
T_R R+n R :?_1^__^ E-^?l
^^ t " r^ r + h " r^ r" 7^~ h^ h" h*'
Example 2. Show that {r'+r'* + rr') {h — A') and {r*h — r'W)
T T*
are equivalent when \—\-'
By Theorem II, J 152, it follows that 7^ = 77^* and since \ is equal
to — » the square of 7 is equal to the product of - by - : one has
therefore, -=- = —,
and by the preceding theorems
A« '^ h'^'^hh! ■" A« + A'« + /iA' "" ¥ — h'^
• • ;,2 ^ ^/2 _|_ /,/,. ~ (A - A') (A« + A'2 + AA')
Hence, on omitting the common factor in the denominator, it follows
that
r«A — r'*A'
{'^-\-T'*—rr') (A — A') = rVi — r'^h'\
142 COLLEGE ALGEBRA [il53
BZBBOI8E XXZVI
Simplify:
<B a
f ^ I o \ /B q 6_\
^- (a-6)(a-c)^(6-c)(6 — a)"^(c — a)(c-6)
"■ '«'™ ('+!)'+(;+')'+ (f +5) =•+(!+!) (;+l)(f+!).
6. Multiply 2"'^3 + 2 ^^ l + i'
7. Multiply -+6-^^>' a-6*
8. Divide(i + y'-lby(i + i)-l.
9. Divide 8a;3+ p by 2:r+^.
Simplify:
in 1 4- 1 -^ ^ - (^ + 6-fc)«
^"- a+6'^6 + c"^c+a (a + 6)(6+cKc+a)'
a+6 flg+6« a— fc rt»-6»
12.
a-6 T ^i_j^ . ^^^ ^j3+63
c + 6 c»+^ "*" c — b'^(?-l^
A'>- a« — 6» a«+6« ^ a + 6 __ g-fe
a — h a+6
ar+a r — a
14.
T
a: J' — a x-\-a
x—a ^ x + a^ x-\-a.x — a
X — a~^ x-\-a
a 6+c
^"' 2» — 1 a" + 1 x« — 1 "^ a" + 1
J153] FRACTIONS 143
17. Given
18- Verify {(^+1^) {c*+d^ = {ac + hd)*+{bc—adf
Leonard dx Pisk.
Simplify: *
«^ (4i+°)(.-ii-')-(4i+')(j^--)-
^/(aix2_ X _2^ 1
, \ 4aa: /' _1_ b{abc + a+c)
21. -^
9t-9 ..gjr^iea* x+l y+- z-^\
23. ./■8(x4-2)v^ 10^-10 24. ^x fx— j.
26. f-^±if4-f-^l±2?. 26. !lI£_ + JLJ_-.
1 _ a+ X \__ a — x i 4- i 1—1
07 o*— 1 ^ Ti 1 1 ^ 1 -h n — f)' — n*
n'4-M L 1— -J 1 — a'
q — JT q — y^(q — a:)« (fl — y)
(a-y)(a-r)« (a-x)(a-3/)»
29.
l+r+;r*+...+a5-'+ll^
H-2:r+^-|^.x*
30.
144 COLLEGE ALGEBRA [3153
31.
32.
-1-
7-^
34-:r«^
COLLEGE ALGEBRA
4
ar*
x«+«-f 3a:*
6x*"
-24
., 2x
a.«»+8+6.
r"+s^_i)^
1
1 . 1 . ]
(a — 6)(a— c)U+a)^(6 — a)(6 — r)(jr+6)^(c — a)(c — 6)(t-f-c)
'*^- (p-7)(P-»-)^ (9-^J(9-P)^ (r-p)(r-g)
35
(a»6 + c + rf)+ (i + i-^ + y (a+6-c + d)
+ /'l + i + l-n (a + 6+c-rf) = 16.
' \a ' 6 ' c d J
r r'
36. Given " = ;:; ; prove that (r+»*^) (c— c^) and (tc—t'c') are equivalent.
37. Given that \=\f \ prove that {f'\-T^ — rK)(/i+/iO and (r«^ + r^h^
are equivalent.
38. Given that — , = — = ^ ;
prove that (i4 + ^'+i/XP"+ ^ + 5^+ ^5^4-C+(r + i/C(7) and
^ + B+ C+^'+-B'+ Cy + i/(^ + -B+ 0(^' + J?'+ C) are equivalent.
^- r 4(xy+zu;)« | | 4(a:y + 2t£;) J
g 6« + q^ . « ^ + «*.
1 6+a "^1 6-a'
BOOK II
CHAPTER I
SQUATIONS-SQUIVALENT EQUATIONS— TRANSFORMATION OF AN
EQUATION INTO AN EQUIVALENT EQUATION— THE SO-
LUTION OF AN EQUATION OF THE FIRST DE-
GREE IN ONE UNKNOWN QUANTITY
154. In Chapter YI, Book I, some simple equations and probr
lems involving equations of one unknown quantity have already
been solved and some of the properties of such equations discussed.
It is now proposed to take up a more complete study of equations
in one unknown quantity and of their properties.
166. Identity. — Identity is an abbreviated term for an identical
equation, 278. In an identity any numerical values whatever can
be assigned to the letters which enter in the two members of the
equality, and if the indicated operations are performed the two
members have the same numerical value.
For example 5 x 4 = 20
(a+6)« = a« + 2a6+6«
(a+6 + c)« = a« + fe« + c* + 2a6 + 2ac + 26c.
Replace now n, 5, c by any numbers whatever, and perform the
calculations indicated in each of these equalities. Both members
in each case will be equal numbers, and, for this reason, these equa-
tions are called identities.
166. The Equation. — When an equality can not be verified,
except by assigning to one letter or to several letters particular
values, the equality is called an equation of condition^ 179, or sim-
ply an equation. The letters to which it is necessary to assign the
particular values in order to render both members of the equality
equal are called the unknown quantities,
1. Consider, for example, the equality,
5x + 4 = 19.
146 COLLEGE ALGEBRA 118157, 158
Assign any value whatever to x (say x= 7; then 5x4-4 =39,
dilferent from 19); then 5x-|-4take8 in general a value diirerent
from 19; hence this equality is an equation in one unknown quantity.
When the equation is written in the form
5x + 4 = 19
it is supposed that only such a value is assigned to x for which 5j--|-4
is equal to 19. The number 3 satisfies this condition. It remains
to be proved that no other number will satisfy the same condition.
2. Let a second ei^uality be
x« + 40 = 13x;
when any value whatever is assigned to x, the two members of the
equality will have in general different values. Accordingly, this
equality is an equation in one unknown quantity. Give x the values
5 and 8 ; then both members of the equation have the same valoesi,
respectively 65 and 104. It will be proved later that these two
values are the only values of x for which the equality holds.
3. Consider the equation,
3x — 2^ + 4 = 6y — 4x + 7.
If to X and y are assigned any values whatever, the two members
of the equality will in general have different values. This equality
is an equation in two unknown quantities.
157. Root of an Equation. — To solve an equation is to find all
the values, which, substituted for the unknown quantity, satisfy the
equation. Example :
The equation 5x -[- 4 = 19
has the root 3, and no other, as will be shown later.
The equation x? + 40 = 13x
has the two roots 5 and 8, and these only.
158. The Degree of an Equation. — When both members of an
equation are rational and integral in each of the unknown quantities,
and the sum of the exponents of the unknown quantities in every
term is found, that sum which is the greatest is the degree of the
equation.
Consider, for example, the following equations:
(1) 3x-4 = 9-5x,
(2) 4x-7y + 3 = 6y-5x-7,
(3) x« + 21=:10:c,
(4) 4x — 5j-y — 9 = 4y — llx + 3
8H59, 160] EQUATIONS 147
(1) is an equation of the first degree in one unknown quantity,
(2) is an equation of the first degree in two unknown quantities,
(3) is an equation of the second degree in one unknown quantity,
(4) is an equation of the second degree in two unknown quantities.
159. Equivalent Equations.— Two equations which have the
same roots are called equivaltnt equations.
The following two theorems relate to the transformation of an
equation into another which is equivalent to it
Theorems Concerning the Transformation of an Equation
INTO AN Equivalent Equation
160. Theorem I. — If the same finite quantity is added to or sub-
tracted from both members of an equation j the result is a new equation
equivalent to the first.
Let A and B represent the two members of the equation in one
or more unknown quantities:
(1) A = B.
Let C be any expression which may involve the unknown quantities*
but which remains finite for any finite values assigned to these
unknown quantities. By adding C to both members of equation (1),
equation (2) results.
(2) A+C=B+a
It is necessary to prove that equations (1) and (2) are equivalent; that
13 to say, that every solution of equation (1) is a solution of equation
(2.) and, conversely, every solution of equation (2) is a solution of
equation(l).
Let X = n, y = 6, 2 = c, be a solution of equation (1).
Sul)stitute X = a, y = 6, « = c, for the unknown quantities
in A and B, then A and B will take equal values, (1).
For these same values of the unknown quantities, O will have
some finite value, and, consequently, A-^ C and B -\- C will have
ecjual values (?81, 1), and equation (2) is satisfied. Therefore any
solution of equation (1) is a solution of equation (2).
Conversely, let x = a', y = 6', 2 = c', be a solution of
equation (2). If x, y, 2, .... in the expressions A-\- (7 and B -\- ('
are replaced by a', 6', </ these txpressions will take equal
148 COLLEGE ALGEBRA [§1161, 162
values. But for the same values oi x, y^ z, the expression
C takes a certain finite value; and therefore A and B will be equal
if ^ + C' is equal to B-}- C) since, if the same finite quantity C is
subtracted from equals (§81, 2), the remainders are equal. Therefore
any solution of equation (2) is a solution of equation (1), and equa-
tions (1) and (2) are equivalent.
By adding— C to both members of equation (1), a new equation,
(3), is formed:
(3) ^-C'=5-(7,
which is equivalent to equation (1).
161. Application.— This theorem makes it possible to transfers
term in one member of an equation to the other member.
Rule. — In order to transfer a term from one member of an equation
to the other, it suffices to omit this term in the member in tchich it u
found and write it in the other member with its sign changed.
For example, consider the equation
7x — 5 = 10 + 4a-.
In order to suppress —5 in the first member, add 5 to both members
and find
7.c_5 + 5 = 10 + 4.r + 5. [Th. I, H60]
The terms - - 5 and + 5, whose sum is 0, cancel. Hence
7x = 10+4.r + 5.
In order to remove 4 x from the second member, add — 4 x to
both members.
7.r — 4x = 10+4.X — 4x + 5.
But 4x — 4x =z 0, and therefore
7.1 — 4x = 10 + 5.
All the terms which involve unknown quantities have been writ-
ten on one side, and those which involve known quantities only,
on the other; simplifying,
3x = 15.
Then equations 7x — 4x = 10-|-5 and 3x = 15 are equivalent to
the original equation 7x — 5 = 10 + 4x (Th. I, §160).
162. Remark.— If the signs of all the terms of both members of an equation are
changed an equation remains which is equivnient to the first (see {82. 2) ; for by theorem
I. il60. all of the terms may be transferred from the first member to the second, and tbo«e
of the second to the first by changing the signs of the terms. Thus, the equation,
11.C — 7 = 5x— 9
is equivalent to the equation,
0 — 5x = 7 — llx
or 7 — llx = 9 — 5a;.
8163] EQUATIONS 149
163. Theorem II. — Ifhoth members of an equation are multiplied
or divided by the same quantity y which has a finite and determinate
value different from zero, a new equation, equivalent to the first, is
formed.
Let A and B be the two members of an equation and C a finite,
determinate quantity different from zero. It is necessary to show
that the equations,
A = B
AC'= BC,
are equivalent.
According to the theorem in the preceding section, these equa-
tions are respectively equivalent to the equations,
(1) A-B = 0,
(2) C{A—B) = 0.
It is sufficient therefore to show that these last equations are
equivalent. They are equivalent because an}- system of values of the
unknown quantities which satisfies equation(l) reduces the expression
A^B
to zero.
Since C is a finite quantity, this same system of values substituted
for the unknown quantities will reduce the product,
CU-B\
to zero and will therefore satisfy equation (2). Conversely, any
system of values which, when substituted for the unknown quantities,
satisfies equation (2) will reduce the product
C{A-B)
to zero; and since Cis a finite quantity different from zero, this can
happen only when A—B = 0 (§76); that is, the equation,
A^B = 0,
is satisfied.
This same system of values, therefore, when substituted for the
unknown quantities in equation (1), will satisfy it.
BxMABX.— The preceding discussion practically assumes that the multiplier C has
a determinate value, and that this value is neither zero nor infinity. If the multiplier
is an expression which inyolves the unknown quantities it can become zero or infinity
for a system of values of the unknown quantities, and, consequently, the reasoning in
the preceding section no longer holds.
150 COLLEGE ALGEBRA [81164, 165
164. * In case C involves the unknown quantities, the equation
(2) C{A-B) = 0
contains all the solutions of the equation
(1) A-^£ = 0;
and, besides, all the solutions of the equation
C= 0;
because, for any system of values assigned to the unknown quanti-
ties, the expressions C and A—B take finite values. If one of these
factors is zero the product
CU-B)
is zero.
Equation (2) is more general than equation (1). The solutions of
equation 6^ = 0 are called foreign solutions which are introduced by
multiplication. Thus, let the equation be
4x — 7 = 53 — 2x.
By multiplying both its members by x — 5, the equation
(4x-7) (x-5) = (53 — 2x) (x — 5)
is formed, which is more general than the given equation, sc = 10
is a root of the first equation and also a root of the second, as is
seen by substituting, x = 5 is a root of the second equation, but not
of the first equation. Hence x = 5 is called a foreign root which
belongs only to the second equation.
166.* If, however, the expression A^B is not integral in the
unknown quantities and the multiplier C is integral in these unknown
quantities, then equation (2),
(2) C(^ — 5) = 0,
possesses all of the solutions which equation (1) has, namely,
(1) ^ — 5 = 0,
because the factor C has a finite value for any system of finite values
of the unknown quantities which make A—B zero. Thus, consider
(2) (a.-8x+15)(^-^-i^)=0.
and
(1) — -^— = 0.
^^ x-5 x + 4
• A thorough grasp of the Ideas discussed in {164 and |160 need not be Insisted upon
In the first reading, though they are essential to a full understanding of the tbeoroK
in 1168.
il66] EQUATIONS 151
z = — — satisfies eqaation (2) and also equation (1), but C, or
x« — 8a;+ 15 = ~+ 2(43) + 15, which is finite and different
from zero.
It can not further be affirmed that equation (2) possesses also all
the solutions of the equation,
67= 0,
because, in this case, for certain values of the unknown quantities
the factor C might be zero and the factor A — B infinite. Then the
value of the product C (A — B) is not determinate and can not be
said to be zero.
Return to the specific example,
C-=x«-8x + 15 = (a;-3) (x — 5).
Hence x = 3, x = 5 satisfy the equation,
C=:(x-3) (x-5) =0. [a04]
But ^^^^(-I^)-.(_^) forx=5
7 3 _7 3
^^ 5-5 5+4~0 5 + 4
which is indeterminate (273, 2), and therefore,
C(A-^B) = [(x-3)(x- 5)] (^^ - ^), for x = 5, is
(5-3)(5-5)(I-?)
which is also indeterminate and can not be said to be zero.
186. Application. — Removal of denominators, — Theorem 2, 2 163|
makes it possible to replace an equation containing terms which are
fractions by an equation which is integral; this process is called the
clearing of fraction$, or the removal of denominators.
If the denominators of the given equation do not involve the
unknown quantities, the new equation is equivalent to the first.
Consider the equation,
-—2 =? + ? — !.
2 4^5
All the terms can be reduced to the same denominator, 20, and
the equation written as follows:
lOx 40 _ 5x 4x 20
20 20 "" 20 "*" 20 20*
152 COLLEGE ALGEBRA [J 167
Hence, by multiplying both members of the equation by 20 the
equation will be transformed into
lOx — 40 = 5x + 4x — 20,
which contains only integral terms, and is equivalent to the given
equation (2163).
Therefore^ to remove the denominators of an equation, reduce all the
terms to a common denominator and omit this common denominator.
Or, multiply both members of the equation by the L, C, M, of the
denominators.
In practice the process is abbreviated. Write immediately the
equation obtained by removing the common denominator. Thus,
consider the equation,
o 5 , 3x 5 4x
3^4 12 3
12 is a common denominator. Proceed as though to reduce all
the terms to fractions having a common denominator 12, but instead
of writing these fractions and then finally omitting the common
denominator, write only the numerators, and obtain at once the
equation,
24x — 20 + 9x = 5 — 16x.
167.* In case the denominators involve unknown quantities,
reduce all the terms to the same common denominator, the simplest
possible; e. g., the least common denominator (?140)* Then mul-
tiply the two members by this common denominator, thus suppressing
it. A new equation is thus formed which contains all the roots of the
given equation, but which can contain, besides the roots of the
given equation, foreign roots also (J 164). These foreign roots can be
introduced only through the multiplier and are obtained by equating
to zero the common denominator by which the two members of the
given equation were multiplied.
Example 1. Consider the equation,
(1) _^ + _J_ = l^±l.
^^^ x« — 9 x(x — 3) x(x + 3)
Reduce all the terms to the same denominator,
a:(x-3)(x + 3),
and put the equation in an integral form by multiplying both mem-
bers by x(x — 3) (x + 3). Equation (2) follows :
• See Note to J$ 164, 166.
8167] EQUATIONS 153
(2) oc'^x+d = (x-3) (5x + 3).
This new equation contains necessarily all the roots of equa-
tion (1), (2163); but it might have as roots, values of x which reduce
the multiplier,
x(x-3)(x + 3),
to zero ; e. g. , the values,
X = 0, X = 3, X = — 3.
But equation (2) is satisfied by x = 4, or — J, and wo^ by x = 0,
X = 3, or X = — 3, hence the multiplication does not introduce any
foreign roots, and the two equations (1) and (2) are equivalent
Example 2. — Let the equation be
(1) -J ? = 1.
^^ x«— 4 x(x — 2) X
Reduce all the terms to the common denominator,
x(x-2)(x + 2),
and the equation to an integral form by multiplying both members
by this common denominator. Thus equation (2) is obtained :
(2) 6x-3(x + 2) = x«-4.
This new equation has necessarily all the roots which equation
(1) has, and, besides, may have as roots the values of x which reduce
the multiplier,
x(x-2)(x + 2),
to zero; that is to say the values (U66)
X = 0, X = 2, X = — 2.
Equation (2) is not satisfied either for x = 0 or for x = — 2 ;
but it is satisfied for x = 2. The number 2 may, therefore, l>e a
foreign root introduced by multiplication. Put x = 2 in the first
member of equation (1). Then the value of the first member is not
determinate, because it is the difiference between two indeterminate
expressions (273, 2), namely,
6 3 ^6 3
22—4 2(2—2) 0 0*
The first member of equation (1) has a determinate value for
a value of x which is different from 2, but which may be as nearly
equal to 2 as one would like.
It is desired to find the value which
6 3
x«-4 x(x — 2)
154 COLLEGE ALGEBRA [U68
approaches when x approaches 2. Instead of the difference,
take its eqaal,
or better,
x(x — 2) (x + 2)'
which for all values of x different from 2 has the same yalae as the
fraction,
x»
— 4 X
(x-2)
6x— 3(a
5 + 2)
x{a?-
-4) '
Z{.x-
-2)
x(x + 2)
obtained by dividing both terms by x — 2. For x = 2 the last
q
fraction becomes -• Therefore when x tends toward 2, the valne of
o
the first member of equation (1) tends toward^- But the second mem-
1 1
ber of equation (1), namely -) tends toward ^ as x approaches
2; and therefore 2 is not a root of equation (1). Hence equa-
tion (2) has as roots, not only the roots of equation (1), but also a
foreign root equal to 2.
Solution of an Equation op the First Degree in One
Unknown Quantity
168. It is a simple matter now to solve by the aid of these theo-
rems an equation of the first degree in one unknown quantity. To
solve the equation,
/1\ vi 2 X X
(1) 4*_- + - = 5_-,
reduce all the terms of both members of the equation to equivalent
fractions with the least common denominator 12. Then suppress
this denominator, or multiply both members by 12.
Equation (2) is the result, and is equivalent to equation (1),
(2) 48x-.8+4x = 60— 3x.
Transpose all the terms involving x to the first member (il61)
21169,170] EQUATIONS 155
and all the known terms to the second member, leaving the equation,
48x + 4x + 3x= 60 + 8,
or, after simplifying,
(3) 55x = 68.
Finally, divide both members of the equation by the coefficient of
X, that is to say, by 55, and find equation (4), which is equivalent
to equation (1),
(4) X = il
Therefore equation (1) has but one root, which is ^|.
169. Rule. — Therefore, it follows that, to solve an equation of
the first degree in one unknown quantity x, it is necessary: (1), to
remove the denominators; (2), to transfer all the terms involving x into
one member and all the known terms into the other and collect the terms;
(3), to dimde the known term by the coefficient of x» The quotient thus
obtained U the root of the given equation,
170. In practice, the steps in solving an equation of the fir^
degree in one unknown quantity need not always be the same.
Thus, consider the equation,
After simplifying, it becomes,
4 . 3aj 2 . ^ ,_ 4 2.r
*^+T + 5+3 = " + 5-T-
Collecting all the terms in x in the first member and all the
known terms in the second member; hence
^ , 3.x . 2x ^^ 4 2 „
4^ + T + T = ^^+5-5-^'
or, after further reducing,
3x . 2x ,, 2
^x + -+~ = U + -.
Remove the denominators and find
80x + 15x + 8x = 280 + 8
or 103x = 288.
Finally, x = |f f = 2. 796126
166 COLLEGE ALGEBRA * [J8171-173
171. BxxABK.— It may happen In applying the rule in $169 that the coefficient of x
in the first member and every known term in the second member are negative. In this
case, change the signs of all the terms In both members, as has been explained in SlfiS.
Example. Solve the equation :
bx — 9— Y =7a! — 1».
Remove the denominators and obtain
15a;— 27-4 x = 21a;— 57.
Transfer all the terms which involve x to the first member and the known terms to
the second; then
15a? — 4a;— 21a; = 27—57;
or, after reduction,
— lOx = —St).
Finally, change tbe signs in the two members; then
10x = 30
or X = 3.
172. When the coefficient of x in the first member and the final
known term in the second member have contrary signs, the root of the
equation will be a negative number. Thus, consider the equatioo,
7x-2x 5 3^_5x^
3 ^ G^ 2 6
It is equivalent to 42x — 4x + 48 = 54-9x — 5 a,
or 42x — 4x— 9x+5x = 5 — 48,
or, by simplifying to 34 x = — 43
and finally to x = — J|.
The equation has therefore the negative number — 1| for its root
173. The Solution of More General Equations.
Example 1. Solve the equation,
.r_3 2x-5_41 3x-8 5x-f6
^ ^ 4 ' 6 ""60'^ 5 15 ' .
After the denominators have been removed by multiplying all
the terms of the equation by 60, it follows that
15(x - 3) - 10(2 X - 5) = 41 + 12(3 x — 8) — 4(5'x + 6),
or by multiplying out,
15x^45 — 20x + 50 = 41 + 36x — 96 — 20x — 24.
Collect similar terms, then
15 X — 20 X -- 36 X + 20 X = 41 — 96 — 24 — 50 + 45
or — 21 X = — 84
and finally x = 4.
{173] EQUATIONS 157
Example 2. Solve for x:
3 — X — 2(a; — l)(x + 2)= (x-3)(5 — 2x).
Multiply out the parentheses:
3 — x-2(x» + x — 2) = llx— 2x«— 15
or 3 — X — 2x» — 2x + 4 = llx — 2x« — 15.
Transpose the terms involving the unknown quantities :
— X — 2x»-2x— llx + 2x«= —3 — 4—15
— 14x= —22
— 22 11
-14"" 7
Example 3. Solve the equation
X — 1 X — 2 X — 5
X — 2 X— 3""x— 6 X — 7
Combine first the terms in the first member of the equation, then
X— 1 _ x-J _ (x — l)(x-3)-(x-2)«
X — 2 X — 3~ (x — 2)(x — 3)
x» — 4x+3— (x» — 4x + 4) X* — 4x+3 — x« + 4x — 4
~ x« — 5x+6 "" x« — 5x + 6
—J
x' — 5 X -|- 6
Similarly, the terms in the second member combined give,
X — 5 X — 6 __ (x^~JO (x- 7) — (:c--6)«
x^==r^""x— 7^ ' (x— 6)(x"— 7)
x« — 12X + 35— x«+12x-36 -1
x« — 13x 4- 42 ~ x' — 13x + 42
1 -1
Therefore, ^.^5^.^ g- ,.^ 13, ^ 4,
and —1 (x«— 13x + 42)=: — 1 (x* — 5x+6)
— x«+13x — 42= _x« + 5x — 6.
Transpose, and get
— x« + 1 3x + x« — 5x = 42 — 6'
8x=36,
and finally x = 4 J = 4|.
158 COLLEGE ALGEBRA [21174-176
Formulae for thb Solution of an Equation of the First
Degree in one Unknown Quantity
174. Every equation of the first degree in one unknown quantity
can, as has been seen, be reduced by addition, subtraction, and mul-
tiplication to the form,
ax=. 6,
where a and 6 are known numbers. To arrive at this result, remove
fractions, and render the equation integral throughout by multiplying
both members by the least common denominator. Transpose all the
terms in x to the first member, and all the known terms to the second.
Then combine all the terms in x into one term, and, similarly, all
the known terms into a single term.
The equation having been reduced to the form,
(1) Yix = h,
two principal cases can arise : either a, the coeflicient of x, is difiTer-
ent from zero or it is equal to zero.
176. When a is Different from Zero. — If a is different from zero,
divide both members of the equation by a and form equation (2),
equivalent to equation (1):
(2) x = -^
a
Since a is different from zero, equation (1) has a determinate root,
and this root is given by formula (2).
176. When a is Equal to Zero. — In case a is zero, it is no longer
possible to divide both members of equation (1) by a. It is accord-
ingly necessary to study this equation more minutely.
Two cases can arise: at the same time that a is zero, b can be dif-
ferent from zero or equal to zero.
1. When a = 0, but 6 =^ 0.
In this case, no number substituted for x can satisfy equation (1),
because the product of any number whatever by a, that is to say,
by zero, is equal to zero, and consequently is different from 6. The
equation is therefore impossible.
Suppose that instead of a's being zero, a is very small ; then the
equation,
ax = by
J177] EQUATIONS 159
will still be eqniyalent to the equation,
=0
and accordingly will have a determinate root. If b remains fixed and
the number a decreases indefinitely, and approaches zero, the root
- will increase indefinitely, and in case a is equal to zero, the equa-
tion is said to have an infinite * indeterminate root (ITS, 2).
2. When a = 0, and 6 = 0.
Then any number put in place of x will satisfy the equation,
because the product of any number whatever by 0 is equal to zero.
The value of x, -, is then indeterminate (273, 1).
177. Numerical Applications.— ^Consider the equations:
(1) '-3='> + T-4'
<3, |_»|+a = , + ^'_|_4.
In case these equations are reduced to the form
ax =zb
it follows for the first equation:
or
and
or finally
x=i 12.59259. . .
Consequently equation (1) has a determinate root 12.59259
• An infinite number is one which Is larger than any number one can choose.
»-T + i =
-+!■
»
('
-t-iy-
17
ay-
17
X =
17_^ 9 _17
3 "^20 3
x?^ =
^ 9
340
27'
160 COLLEGE ALGEBRA [J177
In the case of the second equation:
2x 13x bx 3g_
3 30 6 "^ 5 "" '
/2 13 5 , 3\
(3-30-6+5)"=^^
/20— 13 — 25 + 18\
( 30 j^ = ^'
30''~*'
Hence a = 0 = .— and 6=4.
Hence this equation has an indeterminate root,
4
0
and
x=t' ({73,2)
For the third equation,
f2 5.3 13
or
/^ 5 . 3 13\ ^ . o
(3-6 + 5"3o)^ = ^-^"^>
/20-25 + 18-13\
( 30 ) " = ^^
Ox = 0.
That is, a = 0, and 6 = 0. Therefore, equation (3) is satisfied
when X is replaced by any quantity whatever. The value of x is
wholly indetemiinate (J78, 1).
EXBBOISE XZXVn
Solve the following equations:
,. 3^-2£±5_^g__Lrj-i9_2r±l.
7 2 3
« 2r — 1 , 3j-2 , 6x— 4 , 7x-6
^- 2 ^ 4 ^ 8 ^ 8
^ 13.r + 5 l(U-f 6 llj + 4 5x--l ^
Q 1 —. 1 n^ 1 — Xm
^'2 3 3 2
S-j-Hr _ 43--7 ^ 16 J- -27 _ a-+3
^•2 3 21 6 '
3jr + 4 9jH-44 ^5.r-4-12 9j:-f30
^•7 6 3 4 *
6jr — 2 3.r-2^j— ft , 7j: — 4 . 2(2a:4-3)
6-12 40 18 "^ 30 "•" 45 *
7. ^+'V"+'t^+^=2('^+^)-
J 177] EQUATIONS 161
3 — jr _ ^7— r _ jr-l-3 >^ _,_ (1 —X ___ 9_+3£>^
10 ,4^9 ,1 7 . 1^23-jr .7 1 .
x''"9 ar''"2' ^"- a? "^3 3a; "^12 4r*
7 , 13^13t-24_37 , 10
^^- 3"^5jr 3x 20"^ a;*
10-T , 134--r ^7j: + 26 17 4-4ar
^^' 3 "^ 7 x+21 21
^^* 8ar — 15 15 3 "^ 5 '
14. -f^4
33.
34.
1
ar+3 2(x + 3) 2 2(ar+3)
15. 5^:^ = 6:5. 16. i^«:^=2:l.
,7 Ji£l=4)_^l. L(4_£zil) = 2.
^'' }(3ar+5) 6 ^*- }(6a;+l) 3
19. (x— 3) (x — 4) = (a: - 6) (x - 2).
20. (2a:+7)(ar+3) = 2(x+5)(a + 2).
21. (x-8):(x-9) = (a:-5):(ar-7).
22. (x+1):(j: + 3) = (x-5):(x-7).
^Lui^T^^, 2r~l ^gja^zij)^
"^^ ar-5 a:-3 ^^' 2{x — Z) 3x-l
„ 5 2j- — 5_2 y 5x— 2 ^ 3 ^ 4.r-5_5 ^ 7a: — 3
25. 7^3a:-7""3 ^7x-3' ^0. 4 ^ g^. 7- 7 ^ 5^_4-
27. -^5|^(2a:-ll)=6(a:-6). 28. ^+"1 (3 a;- 11) = 3 (a: -3).
29. 2(0.6— 0.04 r)- 0.2 (0.5 a:— 2) = 0.02 a:.
30. 3(2ar-0.3) = 0.6 + 5(ar-0.1).
31. ll.la:-3(2ar-5) = 7(1.8a:— 3)-3.9.
5a:-0.4 . 1.3-3x^1.8-83-
^- 0.3 "•" 2 1.2
4(13j-0.6) , 3(1.2-x)^9x4-0.2 . 5-4-7x ,
5 ''"10 20 "^ 4 "^^^
9X-0.7 _ 7x-l.l ^ 5X-1.5 _ 5 (0.4 -2a:)
4 3 7 6 *
35. (1 + 6x)« + (2 + 8 x)« = (1 + 10x)«.
36. 9(2x-7)«+(4x-27)« = 13(4x+15)(x + 6).
37. (3 - 4x)« + (4 - 4 x)« = (5 + 4x)«.
38. (2-x)(3-x) + (l-8x)(l-3x) = (l-5x)».
39. (9-4j)(9-5x) + 4(5-x)(5-4x) = 36(2-a?)».
40. 3[3(3(3x— 2)-2)-2]-2=l.
41. 9[7(6(3ar-2)-4)-6]-8 = l.
162 COLLEGE ALGEBRA [«77
42. 4[Hi(J(^+2) + 4) + 6) + 8] = L
43. J[i(i(ix-li)-li)-li]-li = 0.
45. m(i(J(l^+2) + 2)+2) + 2] = l.
46. ?[A«(fa;+5)-10) + 3]-8 = 0.
47. (7ix-2i)-(4|-l(3J-6i:)) = 18i = 5(liir-10).
48. 4.709- <(5.7a: - 31) -0.3(2 J -5.3 a:) = 0.
49. 5J-2J(4.6-3}x) = 4.7x-0.8(3iar-}).
50. 5.7x-2i(7.8— 9.3a:) = 5.38— 4J(0.28+36x).
51. 738a:- 73.8(0.738 — 7.38a:) = 73.8-0.738(7.38-73.8^).
52. 5.05a: — 505(505 — 5.05a:) = 50.5j: — 50.5(50.5a: — 5.05).
53. 3.37ar — 337(337 — 3.37 a:) = 33.7 a: — 2(337 a: -33. 7)3. 37.
i-1 i-1 ,_i a-1 x-i
54.
56.
60.
11^1 ., 2a^-3a:-i-5_2
2^=-l£. 7:r«-4x-2~7-
1 + 1 a + 1 aj,t^hx + c__a
ax X 62.
ma:*- na:+l> *»
^ 13a^J0^ + ^^55^ + 3qxl
-a:*
16 ' '10
4 -a- 1 , 5.5 67 (\ 3a:«\
^- a:* a:«"^3a:""15a:« \x« 5aH»A
66. 8a:«— }a:»+i = 7a:»» + Ja:"+». Divide by a:",
67.
2a:*4-7x"-^ , 7.r"— 44j:"-^ __ 4x"-f 27a:"-^
9 *^ 5a:- 14 18
68. j^ j^ = -^-ar"-..
^^ 6a^— a: 3.r 4.^+2x ^^ a: — 9 , a: — 5 «
a:-17 ' a:-9 ' ''• a:-7 ' a:-12 ' a:-7
1177] EQUATIONS 163
'^- 2x+1^3(x-3) 6 '**• i(x + 2ybx+l3 20
5(2^-f3)_7x--5^ 7x + 55^3^^^_3^+8.
^- 2x+l 2x-5 ^^ '^^ ^^- 2x + b 2 2ar-4
^ 2x — 3 . 3x-2 5r«-29x-4
79. -— r+ ~
80.
x-4 ' X — 8 *«— 12jr:+32
3x — 7 3(x+1)^ llj-f3
2jr — 9 2(x+3) 2j:»-3a:-27*
7£^ . 8_x:^ , _J0ri7_^
^** 3x-2^3jr-1^9a:«-9j:+2 ^'
3£:-j Tx^;^ x+100 ^
^-^- x + 3 ^ x+2 ^a;»+5x+G
6T-17 . 7(x-4) . 12(10x-h73) ^
^ 4x-3 ^4x-5^16x«-32x+15
7X-13 . 13X-28 . 28x-f43 ^
^- 2x-l "^ 2x-3 "^4x«-8x+3
..3,1 4 ^^ 61 , 37 98
" 86. ~
87.
z — 7 ' X — 9 X — 8 °"- x-38 ' x — 62 r — 50
_9 5_^_9 5__
X — 7 X — 8 r — 2 x + l'
**• x-6^x-3 x-2^x-5
X — 8 , X — 3 . X — 9_.x— 1 I X— 13 , X — 6
®®* x-3'^x-5"^x-7 x-S"*" x-5 "^x-7*
90.
x+2 . x4-7 . x+l_x + 9 . X — 3 . x-f4
x+7"^x+5"^x+3 T + 7'^x + 5"^x + 3*
3x — 5 . 5x — l_8x~17 6x-6 . 7x-8_4(3x— 1)
^^- x-2 • x-3 x-6 ' ^' x-3"^x-4 x-1
3x-5 . 2x-5^35(x — 2)
^' X — 3 "^ X — 4 7x — 24 '
^ 2(x~l) . xj^^3(5x+16),
*^- X — 7 ^x — 4 6x-28
95. o""^"f'Q~^~'"'fi" Show that the equation is impossible.
W>. 2^3^5 6^3 5
Show that this equation is satisfied for any values that may be assigned
tox.
CHAPTER II
PROBLEMS WHICH LEAD TO SIMPLE EQUATIONS OF ONE
UNKNOWN QUANTITY
178. In the present chapter the methods already given will be
apphed to the solution of problems, in order that the student may
understand their practical application. In a problem certain quanti-
ties are given, and others, which have certain assigned relations to
the first quantities, are to be determined. The method of solving
the problem may be thus described in general terms:
Represent the unknown quantities hy letters, and express in algebraic
symbols the relations which hold between the unknown quantities and the
given quantities; then equations will be obtained from which the values
of the unknown quantities may be determined.
In the present chapter only problems which may be solved by
using one unknown quantity will be discussed.
179. Problem 1.— Divide 53 quarters between two persons so
that the first shall have one-third more than the second, plus four
quarters.
Let X = the number of quarters belonging to the second person.
Then x -(- J ;e + 4 = the number of quarters belonging to the first
person. The sum of the quarters belonging to the first and second
persons will be equal to 53 quarters.
Hence x -|-x + - x+4 = 53.
or 2x + ^ = 49;
and, removing the denominator 3 by multiplying both members of
the equation by 3,
6 X + X = 147,
or, 7x=147
. *. finally x = 21 , number of quarters belonging to second person.
Hence x+| + 4 = 21 + 7 + 4
^ 32, number of quarters belonging to first person.
164
«180, 181] PROBLEMS 165
180. Problem 2. — How much money is there in a purse the
sum of whose fifth and fourth parts is $225?
Let X = the number of dollars in the purse. Then | = one-
fifth of dollars in the purse, and \ = one-fourth of dollars in the
purse. But a fifth and fourth part of the number of dollars in the
purse is $225; hence,
Remove the denominators by multiplying the equation by 20 and get
5x + 4x=: $4,500;
or, 9 x=: $4,500,
and .-. a; =$500.
181. Problem 3. — Two persons have the same capital; the first
lends his at five per cent, the second lends his at three per cent
The revenue of the first exceeds that of the second by $400. Find
the capital.
Recall the principle established in Arithmetic : To find the interest
on a given capital, multiply the capital by the rate and divide the
result by 100. Let x be the capital desired. The revenue of the first
person will be -—that of the second —-' Since the revenue of the
lUU lUU
first person exceeds that of the second by $400, the equation follows,
5a5 3 X ^^^
iTo = Too + ^««-
To solve this equation transpose tlie terms involving x to the
first member,
Too - 100 = ^'*'*'
'o'o=400;
hence, 2 x = 40000
and, X = $20000.
Then the capital sought is $20000.
VerificcUion, — The interest on the capital, $20000, at 5 per cent
is $1000; and at 3 per cent is $600. The first income of $1000
exceeds the second income, $600, by $400.
166 COLLEGE ALGEBRA [J« 82-184
182. Problem 4. — The denominator of a fraction exceeds its
numerator by 2 ; and, if 1 is added to both numerator and denom-
inator, the resulting fraction will be equal to |. What is the
fraction?
Let X = the numerator of the fraction.
Then x-\-2 the denominator of the fraction.
If 1 is added to both the numerator and the denominator, the
fraction will be ,, —
X+2 +
2
r ^^3
Hence,
x + l 2
x + 3~3'
clear fractions and
get
3(a:+l)=2(x + 3),
or
3;r + 3=2x + 6;
transpose and get
3x-2x=6 — 3,
and .-.
X = 3, the numerator,
05 + 2 = 5, the denominator;
.-. the fraction is
X 3
x + 2""5
Verification. —The numerator 3-|-l =4; the denominator 5 -J- 1 =6,
and .*. the new fraction will be -^- = ^, — - as required.
5-|- 1 0 6
183. Problem 5. — If A can do a piece of work in 8 days and B
in 10 days, in what time will they perform it together?
Let X = the number of days required. If A can perform the
work in 8 days he can perform J of the work in one day, and B can
perform -j^ of the work in one day; and consequently A and B
working together can perform J + ^V ^^ ^^® work in one day. But
if they both can perform the work together in x days, they can
together perform I of the work in one day. Thus,
8^10 X
Clear fractions, and get 10 jr + 8 x = 80,
or 18x = 80;
X = 4J days.
184. Problem 6. — Find the number of passengers who were in
a train leaving New York^ under these conditions: the train lost \
of the passengers at the first station, 12 passengers at the second
station, \ of the remaining passengers at the third station, and \ of
the remaining passengers at the fourth; 42 passengers continued
their journey from the fourth station.
1185] PROBLEMS 167
Let X = the number of passengers. Then the statement is:
Passengers leaving Passengers remaining
First station -— x — -r- = -— i
5 5 5
3x
Second station 12 ——12,
3x ,^\ X , 2x
Third station ^(t "" ^^) ^l "" ^
Fourth station J (-^ — g] =:|^ —
5
2x r.\^^ 3x
and, since the number of passengers who continued their journey on
the train was 42, the equation follows:
^=-
and 3 X = 480
x=160.
The train therefore contained on its departure 160 persons.
186. Problem 7. —At what time l)etween one o'clock and two
o'clock 18 the long hand of a clock exactly one minute in advance of
the short hand?
In accordance with the construction of a clock the long hand
passes over 60 spaces on the dial while the short hand passes over 5
such spaces; therefore the long hand moves 12 times as fast as the
short hand.
Let X = the time past one o'clock
required when the long hand points to D
and the short hand to C, Now, when
the long hand points to A the short
hand points to B; and since the long
hand moves 12 times as fast as the
short hand, the minute spaces passed
over by the short hand (space B (J) will
be .jJy of X or A D,
But AD^AB-\rBC^CD.
168 COLLEGE ALGEBRA [81186, 187
Hence follows the equation in a?, since ^ ^ = 5 and CI)=1
x = 5 + ^ + l;
clear fractions and get
12x = 60 + x + 12;
hence 11 « = 72
x= 6-^ minutes past one.
186. Problem 8. — A tank can be filled by one pipe in 15
minutes, by another in 12 minutes, and by a third in 10 minutes.
In what time can it be filled if all are left open?
Let X be the time required to fill the tank when all the pipes are
open. If the first pipe alone can fill the tank in 15 minutes, it will
fill ^^ of the tank in 1 minute.
The second pipe will fill ^j of the tank in 1 minute.
The third pipe will fill ^^ of the tank in 1 minute.
In X minutes, the time required for the pipes all running at the
same time, the first pipe will fill j^ of the tank; the second fz of
15 12
the tank, and the third -- of the tank ; but in x minutes the tank is
filled. Therefore
— 4----f — - 1
15^12^10"" '
or 8x+10x+12x= 120
and 30 X = 120
X = 4 min., time required.
187. The student may give himself practice in the solution of
the following problems. It may be remarked that in such problems
the main difficulty lies in translating ordinary verbal statements into
algebraic language, and the student should not be discouraged if at
first he becomes involved in difficulties which seem at the moment
unexplainable. Nothing but practice can give quickness and accu-
racy in these processes.
9. By what number must 702 be diminished in order to obtain
twice that number?
10. What number must be subtracted from 875 in order that the
remainder may be equal to 787 plus the number?
11. What number is as much smaller than 7^ as it is greater
thanSp
12. What is the number which, increased by three times itself, is
equal to its double plus 7^?
1187] PROBLEMS 169
13. The sum of the products of a certain number by m and ti is a ;
what is the number?
14. A young man was asked how old he was. He answered:
After 10 years I will be J more than 3 times as old as my brother,
who is 7^ years old. How old was the young man?
15. A farmer estimated that he would harvest 10 bushels of wheat
for each bushel which he sowed. He harvested 637^ bushels, which
was 45 bushels less than he had estimated. How much wheat did
he sow?
16. A father is now 40 years of age and his daughter 13. How
many years ago was the age of the father 10 times the age of his
daughter?
17. A man's age and his wife's age have to each other the ratio of
6:5. But 15 years hence they will have to each other the ratio 9:8.
How old are they now?
If X = the man's age, then the wife's age would be represented
by the fraction — , since x-4- — = -= 6:5; but fractions can
be avoided by proceeding as follows:
Let 6 X = the man's age ;
then 5 flc = the wife's age,
and 6 X -|- 15 = the man's age 15 years later,
5 X -[- 15 = the wife's age 15 years later.
Hence, by the conditions of the problem,
6x+15: 5x+ 15 = 9:8,
6x+ 15 9
or =: ~ *
5a:+15 8
Clear fractions, and get
48 X + 120 = 45 X + 135.
Transpose and get 3 x = 15
x = 5.
6x = 30
= 30 I
= 25 )
Ans.
18. A man is now 25 years old and his oldest sister is 15. How
many years must elapse before their ages will be as 5:4?
19. The sum pf $5,301 is to be divided between A and B in the
following manner: as often as A receives $7|, B shall receive $8^.
How much will each receive?
170 COLLEGE ALGEBRA [?187
20. The prices of two books are to each other as 2^ : 3^. What
are the prices of the books, if the first book cost $0.75 less than
the second?
21. Divide the number m into two parts which are in the ratio
a : h.
22. If Cleveland had 10,000 inhabitants more than it has, then
the ratio of the population of Cleveland would be to that of Milwaukee
as 9 : 4. What is the population of each city if Cleveland has 39,000
more inhabitants than Milwaukee?
23. A certain sum of money was invested at 4^ per oent^ and
the amount paid in 3 years at simple interest was $6765. What was
the sum of money originally invested?
24. There are two numbers, x and a; + 1, such that one-seventh
of the smaller is equal to five times one-ninth of the larger, plus
five. What are the numbers?
25. The difference of the squares of two consecutive numbers is
15. Find the numbers.
26. A can do a piece of work in 7 days, and B can do the same
work in 11 days. How long will it take A and B working tc^ther
to do the work?
27. A can do a piece of work in 2^ days, B in 3^ days, and C in
4 J days. How long will it take them to do it working tt^ether?
28. The area of a rectangle is 5 square yards greater than that
of a square. How long is the side of a square, if it is 7 yards
shorter than the longer and 3 yards greater than the shorter side of
the rectangle?
29. The longer side of a rectangle is 20 yards longer than the
shorter. How much larger will the rectangle be if the longer side
is made 10 yards shorter and the shoii:er side 10 yards longer, and
what are the dimensions of the rectangle?
30. A wrote daily 14 pages of manuscript. When he had worked
6 days B began to work and wrote 18 pages daily till the woA was
finished. How many pages did each write, if B wrote in all as many
pages as A, and how many days did B work?
1187] PROBLEMS 171
31. In going from Boston to Portland, a passenger train, at 36
miles an hour, occupies one hour less time than a freight train at 27
miles an hour. Find the distance from Boston to Portland.
32. A number has three digits which increase by 1 from left to
right The quotient of the number divided by the sum of the
digits is 26. What is the number?
33. Find three consecutive numbers, such tbat, if they are
divided by 9, 10, 13 respectively, the sum of their quotients will
be 49.
34. The width of a room is two- thirds of its length. If the
width were 5 feet more and the length 8 feet less, the room would
be square. Find its dimensions.
35. The sum of a third, a fourth, and a fifth part of a number
exceeds the half of the number by 17. Find the number.
36. There were engaged in building a wall, 3 master-masons,
17 masons and 5 helpers, who together received daily $92. 50. Each
master-mason received $0. 50 more than a mason ; each mason received
$0.40 more than a helper. What was the wages of a master-mason?
37. A woman desires to have f^ piece of linen spun from a certain
number of pounds of flax. The first servant said that she would
have the linen finished in 36 days; the second said^that she could do
the work in 48 days. Because the woman desired to have the linen
spun as soon as possible, she joined the servants in the work and
spun each day ^ pound more the second servant; on account of
which they together finished the work in 8 days. How many pounds
of flax were there?
38. A peasant brought eggs to market to sell at 50 cents for 25
eggs. In bringing the eggs to market he broke 15 of them. Then
he found it necessary to ask 50 cents for 22 eggs in order that he
might receive as much for the eggs which were not broken as he had
first asked when none were broken. How many eggs did the peasant
bring to market?
39. A man has $2000 invested in a mill from which he receives
a certain per cent, and $1000 in real estate from which he derives
only J of the previous rate ; and he has an income from both of $330.
What rate per cent does he receive?
4/4
3r
^"^2 COLLEGE ALGEBRA [il87
40. A merchant adds yearly to his capital one third of it, but
takes from it at the end of each year $4000 for expenses. At the
end of the third year, after deducting the last $4000 he has tirioe
his original capital. How much had he at first?
Let X = number of dollars he had at first
Then ^-4000 or 1^11^2000
3
represents the number of dollars he had at the end of the first year,
and
^4x-12q00x_^^^^ or 16:Lzi84000^
3 / 9
represents the number of dollars he had at the end of the second
year, and
|(^-^^"^)_4000 or e^^-444000^
represents the number of dollars he had at the end of third year.
But 2 X represents also the number of dollars he had at the end of
the third year.
64.^ — 444000
27- = 2x;
whence x = $44,400.
41. A merchdiht adds yearly to his capital one fourth of it^ but
takes from it at the end of each year $1000 for expenses. At the
end of the third year, after .deducting the last $1000, he has 1^^ of
his original. How much did he have at first?
42. There are two places 154 miles apart, from which two per-
sons start at the same time with the design to meet; one travels at
the rate of 3 miles in two hours, and the other at the rate of 5
miles in four hours. How far from A will they meet?
43. A desires to pay $2007 in 5 months, $3395 in 7 months,
$6740 in 13 months. In how many months will the whole sum be
due in one payment?
44. A general, on attempting to draw up his army in the form
of a solid square, finds that he has 60 men left over, and that he
would require 41 men more in his army in order to increase the side
of the square by one man. How many men were there in the army?
4187]
PROBLEMS
173
45. A merchant maintained hhnself for three years at an expense
of $1000 a year, and each year increased that part of his stock which
was not BO expended by one third of it At the end of the third year
his original stock was doubled. What was his original stock?
46. A person bought a certain number of eggs, half of them at
2 for 1 cent, and half of them at 3 for 1 cent. He sold them at the
rate of 5 for 2 cents, and lost 1 cent by the bargain. What was
the number of eggs?
47. A number of men have $72 to divide. If $144 were divided
among three more men, each one would receive $4 more. How many
men are there?
48. A person hired a laborer to do a certain work on the agree-
ment that for every day he worked he should receive $2, but that for
every day he was absent he should lose $0.75; he worked twice as
many days as he was absent, and on the whole received $39. How
many days did he work?
49. At what time between 8 and 9 o'clock are the hands of a
clock pointing in opposite directions?
It will always facilitate the solution of such problems to make a
diagram of the dial of a clock.
We mark the minute hand by m and the hour hand by h,
hour hand points to 8 when the minute hand points to 12.
X = the number of minutes or min-
ute spaces pa8t eight when the hands
are opposite. Since the minute hand
moves over 12 minute spaces while the
hour hand moves over one minute space,
the hour hand will move over ^ minute
spaces while the minute hand moves
from 12 to P, when the hour and minute
hands are opposite. If the hands are
opposite, 2 P = S Q. We have just
found that 8 © = f^^ and 2 P = x — 10
minute spaces.
Hence, x — 10 = /^ and x = 10 |^ minutes.
The time requiped is 10 ^^ minutes past eight o'clock.
The
Let
174 COLLEGE ALGEBRA [J187
50. The hands of a clock are at right angles to each other at
three o'clock. When are they next at right angles to each other?
Ans. 32^ min. past 3.
61. How many minutes does it lack of four o'clock, if three
quarters of an hour ago it was twice as many minutes past two
o'clock?
52. At what time between one and two o'clock is the minute
hand of a clock exactly one minute in advance of the hour hand?
53. At what time between three and four o'clock are the hands
of a watch pointing in opposite directions?
54. A watch gains as much as a clock loses; and 1799 hours by
the clock are equivalent to 1801 hours by the watch. Find how
much the watch gains and the clock loses per hour.
55. An express train that travels 40 miles an hour starts from
a certain place 50 minutes after a freight train, and overtakes the
freight train in two hours and five minutes. Find the rate per hour
of the freight train.
56. A cistern has two supply pipes which singly will fill it in
4^ hours and 6 hours respectively; and it has also a leak by which
it would be emptied in 5 hours. In how many hours will it be filled
when all are open? Ans. 5^^.
57. The national debt of a country was increased by one fourth
in time of war. During a long peace which followed $250,000,000
was paid off, and at the end of that time the rate of interest was
reduced from 4^- to 4 per cent. It was then found that the amount
of annual interest was the same as before the war. What was the
amount of the debt before the war? Ans. $2,000,000,000.
58. A ship sails with a supply of biscuit sufficient for 60 days
at a daily allowance of a pound a head. After being at sea twenty
days she encounters a storm in which 5 men are washed overboard,
and damage is done that will cause a delay of 24 days; and it is
found that each man's daily allowance must be reduced to five
sevenths of a pound. Find the original number of the crew.
Ans. 40.
CHAPTER III
UTESAL EQUATIONS IN ONS UNKNOWN NUMBER
188. It is of great importance to determine the unknown numbers
of an equation in terms of general numbers, i. e., in terms of
numbers expressed by letters. In the preceding chapter the known
numbers have been represented by numerals. Known quantities when
represented by letters are usually represented by the first letters of
the alphabet, a, 6, c, /, m, n, a, )8, 7. etc. ; the unknown quantities
by the letters re, y, «, . . . . Thus, in the equation,
05 -|- a = 6
a and b are the known numbers and x is the unknotcn number.
From this equation x has the value,
x=zb — a.
It is purely an arbitrary agreement to represent the known
numbers by a, 5, c, . . . and the unknown numbers by x, y, 2, . . .
The number a might be supposed to be unknown and x and b to be
known numbers. Then the solution of the equation would give
a =: b — X.
If b were the unknown number and a and x were known we
would have as our solution
5 = x-j-a.
189. A Numerical Equation is one in which all the known
numbers are numerals.
E. g., 7x-9=26; ?_e^ = 4x-17.
A Literal Equation is one in which some or all of the numbers
are represented by letters.
E. g., 3ax + a« = h^ — 2bx; ax + by = c; 3x + 56 = 19.
190. The principles of equivalent equations hold when the
eqiiations are literal.
175
176 COLLEGE ALGEBRA [J190
Example 1. Solve 'the equation ax + 6 = ex -j- e?.
By transposition, ax — cx — d — h
that is (a — c)x = <£ — 6;
by division, x = ■ •
^ ' a— c
Example 2. Solve (a + x) (6 + x) = (w + x) (» -|- x).
By multiplication, afc + (a + 6) x + x* = mn -|- (m + n) x -f a:^
by transposition, (a -j- 6) x — {m-\-n)xz= mn — ah
or [a + t — (»» + «)] X = mn — ah,
mn — ab
a -|- 6 — (m + n)
Example 3. Solve the equation,
By multiplying by 12,
or 6x — 2o — (4x-~a)+3x— ^ = 0,
5
6x — 2a — 4x+a4-3x — — =0.
^ ^ 5
Simplifying,
5x
— a-
3a
5
= 0,
5x
8a
X
8a
~25
« + />
X — c X — a X — h
a -f- h a h
Example 4
Transpose and get _
X — c X — (I X — b
By reducing terms in the first member to common denominator,
{a -\-h)(x — a) — a(x — c) b
(x — a) (x — c) ~ X — b
ax + bx — fi* — (lb — ax -\- ac b
(x — a) {x — c) ^ x—h
Clear fractions and get (6x — a*— ab 4- ac) (x— fc)=(x — a) (x — e)h
or 6x^— a*x— a?>x+«cx— 6*x-|-a (a-j-ft— c) 6 = 7jx*— (a+ c) 6x-|- ahe.
{190] EQUATIONS IN ONE UNKNOWN NUMBER 177
Transpose and get
abx-^hcx — a*x — ahx-\-acx — fe'x= — ab (a+6 — c)-}-abc]
(— a*— 6« + ac+bc)x= — a6(a-|-ft — 2c)
ab{a + b — 2c)
xz= - ——- — -> Ans.
a* + 6' — ac — 6 c
Or, by clearing fractions at first, the result is obtained,
(a-|-5) (x — a) (x — 6)=a(x — b) (x — c) + 6(x — a) (x — c)
(a-|-Z>) as*— (a+5)'x-|-a6 (a+i) =ax'— a (6 + c) x-|-a6c
+ 6x* — b {a-\-c) x-{-abc,
[6(a+c)— (a-|-6)*+a(6 + c)]x=a6(2c— a— 6)j
^ ^ a6(a + 6-2c) _ ab{a + b^2c) ^^^^
a'-|-2rt6 + 6* — ab — be — ab — ac a*+ 6* — be — ac
BZBBOisB xzzmn
Solve the following equations:
1.
a — bx^cx — d.
2.
ax + x = m.
3.
a — bx = cx^z.
4.
a{X'-'l) — b=zx — a,
6.
{a-\' b) X = m — ex.
6.
(a — 6) X — c = d — (6 — c) a?.
7.
ab-{X'-c)d=(cd+x).
8.
ax-\-cx = ab-}-c.
9.
ajc-\-bx=m-^-x.
10.
ax~^bx — m (jr — 1) = w.
11.
or = 6 (c — jt).
12.
{a-b)x = 2a-'(a+b)x.
13. {a — b)(c - x) + (b — c) (a — x) + {c — a) (b ^x) = a-x.
14. (a— j:)6 + (a— c— 2r) (j:--6)=x(a— ^).
15. m{a-\-b — x)=n{a+b — x).
16. (a- 6) (a- c + ar) + {a + b) (a+ c -a:) = 2a«.
17. {m + x)(a+b'-x) + (a-x){b-x)=:a(m+b),
18. (ar--l)(6x — l)(cjr — l)+l = ar + 6a:+cj:.
19. (a + x)(b + x) {c+x)-(a-x) (b-x) {c-x)=2(x^+abc).
20. (a — 6)(a— c)(a+x)+(a + 6)(a + c)(a-a:) = 0.
21. 5a»cjr+ ac«j: — 5 a6c» — 3 a»c' = 6 a»fecx+ fec'o: --3 a«fec' — 5a«c».
22. 2a«6«c + a6«ar — 2a6»c-a6c«d-3a»x = (6»-3a«6)jr-6«c«d.
23. «5 = 6c + d+l. 24. 3^^ir_5_6.
XX XX
26. T^^ + <fc = 6x-ac. 26. c=a + VL(l=A.
b — c Sa-\-x
178 COLLEGE ALGEBRA [^190
27.
a(fP^a^)^ or 3.r« + 7^-25^3
29.
30.
X
2— 5x 7-f3:^ 148-5j^ o
5x + l 3-2x 3+13jr-10.r«
31. f(^_5)_^^=?x. 41
6' Zr-O 6 2x — d
32 _1 J+«-- ^ 33 cj-"* _. /j"* ,
jr+rt X — a a* — X* ' a+6j: d + cr
a4. lUL^^-J'.^l+^'-A^^l-f^. Divide by X-.
or 4 6,2^
35. H ' = 0.
X — 4 X — 6 X — 2
36. ^i£zL5£+2a-^_a+/_^.
a— c a a— c
37- f + r + f+f=*-
OX ox /x hx
Sabc . a«fe« . {2a + h)b^x _o^^ , 6x
"^^ a+6"*"(a + 6)»"*" a(a + 6)« "^ a '
^ 2b-a 2ab{a+b) 3c-d 2a6(a-6) (i«-6« '
40. (a + x)(6 + x)-.a(6+c)=^+x«.
o
.- a— 6x,6 — rx.c — ax ^
41. — r h 1 7 — =U.
oc ac ab
Ao q(^— J") I h{c —a) __a-\-h (^\^\,
bx ex X \c b)
Ao g f ^ — ^) I bix — h) __ X ,<4 5 _ ffx _. £ _ ex
a + 26"*'2a + 6 2 c ex - 1 a ax— 1
-fc. 6 — x.e — X a(c — 2x)
a+x a — X a* — x*
46.
a+x
qx+6 __ 6x __ ax __ (ox* — 2 6) 6
ax — b ax-\- b ax — b a'-r* — 6*
.^ ar . ex a t c
47. 1 = 1
mx — p nx — q m n
48.
ax-
b , ex — d ^^ (bn + dm)x — {bg -|- dp) __ ^ i c^ ,
mx—p nx — q ^ (mx—p){nx — q)
^^^ + ^:^^=2. 50. ^^±^ + ^^-+^ = a+c.
49.
X — m X — n
a 90] EQUATIONS IN ONE UNKNOWN NUMBER 179
51 (a-h6k [ ^ (b-c)x a^d^{a-^r)x b-d
c* a — b c a — 6 c
52, ^'^-^-^ _L P^i^ = -£ 4- ^ .
mx — n px — q m p
X — a X — 6 X — c
54. -— +-^+ ''— = . . .
X — a X— 0 X— c X— c X —a x — 6
55 o+g (3a— 6r)x ■ (3fl — 26)(x — 1) __ (6c ~-2fe)x a— c
a — b 2a — 36 a— 6 2a — 36 a — 6'
^- (a + 6)» ^ (a-6)« ^ a«+6«
x + 6 x+a
58. 1 ±^ 1.
RQ /x — <in'__x — 2a — 6
• \x+6>' :r+a+26
m+x ^
ea «+3? — 1 — 2ax
a + ^ (« + -«^)'
61. (x — a)» + (x — 6)»+(x- c)»=3(x -a) (x- 6) (x- c).
CHAPTER IV
PROBLEMS INVOLVING LITERAL EQUATIONS
191. Formulae and Rules. — Represent the given namberina
problem by letters, thea its solution will be an expression invohing
these letters and will include all problems of its particular form.
Such an expression is called a formula^ and the translation of ibis
formula into words is called a rule.
Thus for example:
I. Find two numbers whose sum is s and whose difference is d.
Let X = the smaller number;
then X + (7 = the larger number.
But the sum of the two numbers is «.
05 -f- ^ + ^ = '»
or 2 X = « — <f ,
x = — jr— J the smaller number;
, , s — «?., s — d-{-2d « + ^^, , ,
and X + c? = — r— + rf = ^ = — Jr— ) the larger number.
T-... . m, . * d , 9-\' d 2«
Verification. Their sum — 1 — = — = «,
... __ « + rf % — d ,-(-f/ — , + .Z ^d ,
and their difference — — = - — — = — - = a.
ii A Ji Z
Since these formulae hold true for all values of the numbers t and
d^ the following rule for finding two numbers when their sum and
difference are given, can be formulated.
Rule : Tlie greater number is found hy adding the difference to thi
sum and taking one half tlie result.
The smaller number is found by subtracting the difference from t^
sum and taking one half the result,
IT. A can do a piece of work in a days, and B can do the same
work in b days. In how many days can bofh together do the woA?
ISO
2191] PROBLEMS INVOLVING LITERAL EQUATIONS 181
Let X = the numl>er of days required. If A can do the work
in a days, he can do - of it in one day; and similarly B can do - of
the same work in one day. Therefore working together they can
do in one day
i + iofit;
a h
but if both together can do the work in x days they can do - of it in
one day.
— rr = -'
a b X
or hx -\- ax =. ah,
ah
and X = r-
a + 6
The translation of this formula will give a rule for finding the
time required by any two agents working together to produce a given
result, if the time that it takes each separately to produce it, is given.
Rule. The time required hy any tico agents to produce a given,
rtJtfdt is the quotient found hy dividing the product of the numhers
which express the tin^e in units required by each to produce the result^
hy the sum of the numhers,
R»tARK.--€oTDpare problem 26. |187.
III. A person has just a hours at his disposal. How far may
he ride in a coach which travels h miles an hour, so as to return
home in time, walking back at the rate of c miles an hour.
Let X = the distance AB in miles which he is to ride in the coach
and to return by foot:
A X miles B
X
then - = the time required to travel from .4 to ^ in the coach,
X
and - = the time required to walk from 7i to ^.
Therefore the total time required to ride out to B and walk back
to A is: X X
or
and
b + -c
or a.
5 . ?
b^ c
= a,
cx -{- hx
= abcy
X
abc
6 + c
182
COLLEGE ALGEBRA
[5191
Rule. Divide the product of the time at disposal expressed in
any unit (a), the number of miles the coach can travel in that unit (b)
and the number of miles he can walk in the sam-e unit (c) by the swn of
the rates of travel {b + c).
IV, One man asked another what time it was and received the
answer that it was between n and (w -|- 1) o'clock and the hour hand
and the minute hand pointed in opposite directions. What was the
time?
At n o'clock the minute hand points to 12 and the hour hand to
fi. The hour hand is therefore 5n minute divisions in advance of the
minute hand.
Let X = the number of minute divi-
sions passed over by the minute hand
from n o'clock until it is directly oppo-
site the hour hand or the minute divi-
sions in the arc AnPF^.
Then ^ = minute space in arc nP,
since the minute hand passes over 60
minute spaces while the hour hand pas-
ses over 5 minute spaces.
Now from the figure,
arc AnPF' = arc An + arc nP+ arc PP\
X
or
or
5»+To+30
12
X — -^ = 5 » + 30,
Ux
12
= 5 « + 30,
.T = ^(5n + 30).
If n = 1, it was |f (35) - 38 j\ min. past 1 o'clock.
If n = 3, it was |f (45) = 49y^y min. past 3 o'clock.
If 71 = 5, it was \\ (55) = 60 min. past 5 o'clock or 6 o'clock.
In case n is 7 or any integer between 6 and 12, then a little
care will show that x will have the value,
x= }f (5 71-30).
If n = 8 it was Y^o = 10 |f minutes past 8 o'clock. In this
connection compare problem 49, ?187. For the case n = 3, com-
pare problem 52, of the same set of examples.
1191] PROBLEMS INVOLVING LITERAL EQUATIONS 183
V. A train, starting from a point Ay travels m miles per hour;
a second train, starting from a point B, p miles behind A^ travels in
the same direction n miles daily. After how many hours will the
second train overtake the first, and at what distance from B will the
meeting take place?
It is assumed that n^m.
Let X = number of hours after which the trains meet;
then mx = number of miles travelled by the first train = AC,
and nx = number of miles travelled by the second train, = BC.
But BA = BC- AC,
or/) = nx — mx.
n — m
the number of hours after which the trains meet.
The distance travelled by the first train is
mx= miles;
n — m
and the distance travelled by the second train is
nx = — - — miles.
They meet, therefore, — "-^ miles from jB.
7B0BLEKS
Solve the following problems:
1. Find a number which added to m gives a sum equal to n times
the number.
Let w = 10, 71 = 11.
2. Divide a into two parts so that -J of the first plus j^ of the
second shall be equal to 6.
3. What number is that whose i,, i and J parts are together
equal to 1 ?
4. A man's age and his wife's age now have to each other the
ratio n : w; but r years from now they will have the ratio Qi p :q.
How old are they now? (Compare problem 17, J187.)
184 COLLEGE ALGEBRA [U91
5. The sum of two numbers is m, and the quotient formed by
dividing the less by the greater in ^ . What are the numbers?
6. A and R can do a piece of work in m days, A and C in n
days, and B and C in jy days. In what time can they do the work
all working together?
7. A passenger train, going from Boston to Portland at the rate
of m miles an hour, occupies h hours less time than a freight train
at 27 miles an hour. Find the distance from Boston to Portland.
(Compare problem 31, §187.)
8. Two towns, A and B, are a miles apart. One person sets
out from A and travels toward B at the rate of b miles an hour; at
the same time another person sets out from B and travels toward
A at the rate of c miles an hour. How many miles from A will
they meet? (Compare problem 42, J 187.)
9. A merchant adds yearly ^ of his capital to it, but takes from
it at the end of each year d dollars. At the end of the third year
after deducting the last d dollars he has ^ of his capital left. Find
his original capital. (Compare problem 41, ?187.)
10. A asked B what time it was, and received the answer that
it was between n and n -f- 1 o'clock, and the hour hand was directly
under the minute hand. What time was it?
11. A was employed a days on these conditions: for each day
he worked he was to receive h dollars, and for each day he was idle
he was to forfeit c dollars. At the end of a days he received d
dollars. How many days did he work? (Compare 48, 1187.)
12. A has m dollars and B has n dollars. A gives to -S a cer-
tain number of dollars and has left q times as many dollars as B.
How much money did B receive from A?
13. A can do a piece of work in 2ni days, B and A together in
n days, and A and C together in wt + » days. In what time will
they do it working together?
14. A broker invests J of his capital in a% bonds, and the
remainder in i!> % bonds ; his annual income is c dollars. Find the
amount in each kind of bond, and the sum invested.
15. A banker has two kinds of coin: it takes m pieces of the
first kind to make one dollar, and n pieces of the second kind to
make a dollar. A person wishes to obtain r pieces for a dollar.
How many pieces of each kind must the banker give him?
J191] PROBLEMS INVOLVING LITERAL EQUATIONS 185
16. A grocer wishes to receive a certain sum for his eggs and
intends to sell them at m cents a dozen. But he broke n eggs, and,
in order to receive the desired sum he then sells the unbroken ones
atp cents a dozen. How many eggs had he originally?
17. The annual dues of a certain club were at first a dollars.
Subsequently the yearly expenses increased by d dollars, while the
number of members decreased by n. In consequence the annual
dues were increased b dollars. How many members were originally
in the club?
18. Two couriers start from the same place and travel in the
"same direction, one m hours after the other. The first travels at
the rate of r^ miles an hour, and the second at the rate of r^ miles
an hour. After how many hours will the second courier overtake
the first?
19. At what tame between n and w -f- 1 o'clock will the minute
hand be 20 minute spaces in advance of the hour hand?
20. Find four numbers such that the sum of the first and j the
second equals a, the sum of second and j^ the third equals b, the
sum of third and ^ the fourth equals c, the sum of fourth and ^ the
first equals d,
Sugge9tion: Let x = first number. Then I (a — x) =seeond number,
m [6 — I {a — x) ] =third number, n \c — m[6 — I {a — x)] J -f-
P
21. A person after doing ^ of a piece of work in p days calls for
an assistant and together they finish it in q days. In what time
could each do it separately?
22. A ship, having on board m persons, is provisioned for n
days. After sailing d days, m^ persons are lost overboard in a storm
and in consequence the allowance of food for each person is in-
creased by J of a pound. After sailing d^ days longer, m^ per-
sons were landed. It was then found that the journey could be
completed d^ days sooner than was supposed, and accordingly the
allowance for each person was again increased by ^ of a pound.
What was the original allowance of food for each person?
23. A particle has a uniform motion in a straight line with a
given velocity; it passes the point 0 at a certain instant Betermine
the position at any instant •
186 CX)LLEGE ALGEBRA [tl91
24. At what time between n and /i + 1 o'clock will th3 hour and
minute hands be m minute spaces apart?
25. Two particles, which have a uniform motion in the same
straight line, pass at the same instant, the first the point A and the
second the point B. Given that the velocities of the particles are
respectively a and b and that they move in opposite directions find
the point at which they meet.
26. A man hired a servant for I months, and agreed to allow
him ip and a suit of clothes if he staid I months; but at the end of
m months the servant went away, and received $q and the livery as
a proportionate part of his wages. What was the value of the snit
of clothes?
Solution. Let x = value of clothes in dollars.
Then p -\- x = number of dollars he would have received
at the end of I months.
q -\-x = number of dollars he received at end of m
months;
m . .
q+X=:- (p + x)]
Iq -\- lx=: mp -|- mx ;
Ig — mp
m — I
LetZ = 12; m = 8; p = 20; ^ = 12.
27. A composition of two metals, tin and copper, A and B,
containing n cubic inches weighs m ounces. Supposing the weight
of a cubic inch of tin to be I ounces, and that of a cubic inch of
copper to be m ounces, find the number of cubic inches of tin.
28. Find the number such that after it has been divided by p,
the sum of the quotient, dividend, and divisor shall be equal to q.
29. A and C can do a piece of work in in days ; B and C can
do the same work in n days; A and B can do the same work in /
days. How many days will it take each person alone to do the
same work? How many days will it take A, B, and C, together
to do the same work?
30. A broker has two kinds of tea; one worth a cents a pound
and the other b cents a pound. How many pounds of each must
be taken to form a mixture of 7l pounds which shall be worth n cents?
CHAPTER V
INTBSPRSTATION OF THE SOLUTION OF PROBLEMS
192. Usaally in solving integral algebraic equations the meaning
of the results is not emphasized. But when an equation has arisen
in connection with some practical problem, the question of the
interpretation of the result is an important matter.
In this chapter is illustrated the interpretation of the solution of
equations of the first degree in connection with the problems from
which they arise, — problems much like those solved in the preceding
chapters.
Positive and Neoativb Solutions
103. Problem I. — A has a company of 151 soldiers and B has
a company of 40 soldiers. How many soldiers must A give to i5 in
onler that A shall have three times as many soldiers left as B has ?
Let X = the number of soldiers A is to give to 5. Then, as A
gives X soldiers to J5, he will have 151 — x left, and B will have
40 + 05. According to the conditions of tiie problem,
151 — X = 3 (40 + x)
or 151— x = 120 + 3x
X = —- = 7 J men.
4 *
The result 7 J men shows that the problem is impossible, for the
assumption is that A gives so many whole men to B. But the
value X satisfies the equation.
Suppose that the problem was stated more generally: A has a
company of m soldiers and B a company of n soldiers. How many
soldiers must A give U> B in order that A's company may contain q
times as many men as B's company?
188 CX)LLEGE ALGEBRA CJ194
Let X = the number of soldiers which A gives to B,
Then m — x = q {n -]- x) =: qn -\- qx
and qx-{- xz=m — qn
m — an
«= r^ •
• Suppose that m = 185, n = 25, j = 5,
.u 185-25x5 60 ,^
then x= ^^-j- =y=10;
which is a possible solution.
(1.) Any values of m, n, and 5 which give x a positive integral
value will give a solution.
(2.) If values are assigned to m, n, and q such that x will have a
positive fractional value, the problem is impossible because the
assumption is that a certain integral number of men is transferred.
Thus, m = 151, w = 40, ^ = 3, and x = 7J is an impossible solutiou.
(3.) If positive values are given to m, n^ and q such that m is
less than nq, m— nq will be negative, and therefore x will have a
negative value, which has no meaning for the problem in question.
It is assumed in the statement of the problem that m, «, and q
are positive numbers.
194. Problem II. — What number must be added to a number a
in order that the sum may be twice a given number bf
Let X = the number required.
Then a + x=2/>;
therefore, x = 2 6 — a.
This formula gives the value of x corresponding to any assigned
values of a and b. Thus, if «= 12 and b = 10, it follows that
X = 20 — 12 = 8. But suppose that a = 35, and i = 15; then
X = 30 — 35 = — 5, and the question of the meaning of this nega-
tive result arises. The problem now reads: What number must be
added to 35 in order that the sum may be 30? It is evident that if
the word added and the word sum are to keep their arithmetical
meanings, the given problem is not possible. However, it is clear
that the following problems can be solved: What number must be
subtracted from 35 in order that the difference may be twice 15, or
30? 5 is the answer to the question. The difference between the
first and the second statements of the problem is this: the words
added to are replaced by taken from, and the word sum is replaced
by difference.
« 195-197] PROBLEMS ' 189
Rbmabk.— Accordingly, in this example, the negative result indicates that the
problem in a purely arithmetical sense is impossible; but the problem may be so
enunciated that the absolute value of the negative result will bo the correct answer.
195. This problem illustrates the convenience of using the word
add in Algebra in a broader sense than it is used in Arithmetic.
Suppose that it is desired to add x algebraically to a; this algebraic
sum is a + x, whether x itself be positive or negative. Thus the
equation a -\-x = h will be possible algebraically whether a is
greater or less than h.
196. Consider now another problem.
Problem III. — As age is 35 years and B's is 20 years. When
will A be twice as old as Bf
Let x = the number of years hence when A will be twice as
old as B,
Hence 35 + x = 2 (20 + x),
A's age X years hence.
This equation may be written
(1) 35 + x = 40+2x;
and it is evident that if only a purely arithmetical solution is allowed^
this equation is impossible, because 40 is greater than 35 and 2 x
is greater than x, so that the two members can not be equal. The
solution of this equation gives a negative result, x = — 5.
The following would have been the solution if the problem had
been worded thus: A's age is 35 years and B's 20; when was As
age twice B's age?
Let X = the number of years from the present time. Then accord-
ing to the problem,
(2) 35 — X == 2(20 — x);
thus, 35 — X = 40 — 2x,
and X = 5.
Here again observe that the negative result obtained in solving
equation (1) indicates that the problem is impossible in a strictly
arithmetical sense, but that a new problem can be formulated so
that the absolute value of the negative result will be the correct
answer for the newly formulated problem.
197. Problem IV. — Suppose that this problem is formulated as
follows: As age is a years, and B^s age is h years; find the time
when As age is twice B's age. This statement of the problem does
190 COLLEGE ALGEBRA [{{198, 199
not imply whether the time is before or after the present date.
Suppose that it is before ; then the equation is
a— « = 2 (6 — x) = 2 ft — 2 X,
and (1) X = 2 5 — a.
If the time is in the future then
a + x =^(6 + 0-) = 2h + 2x,
and (2) x = a— 26.
If 2 6 is greater than a, the first supposition is correct and leads
to an arithmetical value of x, since, by equation (1) the value of
2 6 — a is -^ ; but the second supposition is incorrect and leads to a
negative value for x, since by equation (2) the value of a — 2i
is — . If, however, 2 6 is less than a, the second supposition is cor-
rect and leads to an arithmetical value for x, and the first supposition
is incorrect and leads to a negative value for x. It happens some-
times that a negative result indicates that the wrong choice has been
made out of two possible suppositions which the problem allowed.
When such a wrong choice has been made, it is not necessary to go
through the whole investigation again, for the result obtained from
the wrong supposition can be used. It is necessary only to take the
absolute value of the negative result and place the time before the
present date if it was supposed to be after, or after the present date
if it was supposed to be before.
198. The equation a + x = 2 (6 -f- x) may be regarded as repre-
senting symbolicall}'^ what is enunciated in the following: Let a and
b be two quantities. What quantity must be added to each in order
that the first sum may be twice the second? Here the words *«i»,
quantity^ and added may be used in the algebraic senses, so that x,
a, 6, may be + or — . One of the admissible senses of this alge-
braic statement is found in the arithmetical question concerning the
ages of A and B, More may be included in the algebraic statement
than in the statement of the problem. It appears then, that when
a problem is translated into an equation, the same equation may he
the syml>olical expression of a more comprehensive problem than
that from which it was obtained.
Zero Solutions
199. A zero result may, in some cases, be the answer to a qaes-
tion. In other cases it proves the impossibility of the equation.
«199] PROBLEMS 191
Problem I. — A man is 50 years old and his son is 10 3^ears old.
Aft4?r how many years will the father be 5 times as old as the son?
Let X = the required number of years.
Then 50 + x = 5(10 + a;)
or 50 + x = 50 + 4x
whence 3x = 0
This result is the correct answer to the problem. At the present
time the father is five times as old as his son.
Problem II. — ^The denominator of a fraction is four times its
numerator; if 9 is added to the numerator and 15 to the denomina-
tor, the fraction is |. What is the f racoon?
Let X = the numerator of the fraction.
Then 4 x = the denominator of the fraction.
From the conditions of the problem the equation is
_x + 9 3
4.c + 15~^5
or 5x + 45 = 12x + 45
x = ^ = 0 and 4x = 0;
X 0
and the fraction is -— = - [J73, 1]
4x 0 0
which is indeterminate; that is, no determinate fraction will fulfill
the required conditions.
Problem III. — One kind of flour can be bought at $9.50 per
barrel, and another at $6.25 per barrel. How many barrels of each
kind of flour must be purchased in order to make a mixture of 100
barrels worth $625?
Let X = the number of barrels of the first kind of flour.
Then 100 — x = the number of barrels of tiie second kind:
and 9Jx + 6i(100 — x) = the number of dollars the mixture is
worth.
Hence 9^x4-6^ (100 — x) = 625,
145 +625-?^ =625.
13x ^ ^
— ;— = 0 and X = 0.
4
Henoe, no mixture which contains the first kind of fiour can be
made to satisfy the conditions.
192 COLLEGE ALGEBRA [8200
Indeterminate Solutions
200. Problem I. — A man is 50 years old and his son 15 yeais
old. After how many years will the father be 35 years older than
his son?
Let X = the number of years required. Then, liy the conditions
of the problem,
50 + X = 15 + X + 35 or 50 + x = 50 + x.
The two members are identical, and the equation is satisfied for any
finite value of x whatever (2156), i. e., the problem is indeterminate.
From solving the equation in the usual way, the equation follows:
X — X = 50 — 50
or (l^l)x = 50 — 50
50 — 50 0
1-1 "^0
K78,l]
The symbol - means therefore that the condition of the problem
is satisfied when x = any finite number. It is evident from the
problem that the father will be at any time 35 years older than his
son.
Problem II. — The solution of the equation
(1) ax + ft = CX+ J
is (2) ' x=^-=^.
a — c
Examine the case when a — c is equal to 0.
1. If a — c is 0, but d — ft is not zero, formula (2) gives
which has no meaning. It is easy to show that the equation is im-
possible, because if a — c = 0, then a = c an J equation (1) becomts
ax -f- ft = ax -[- d^
which can not be true, since ft is not equal to d.
2. Let a =:i c and ft = d; then formula (2) becomes
X = = - [173, 1]
a — a 0
which has no meaning. It is easy to see that in this case, equation
(1) is satisfied for any finite value of x, because it becomes
ax + ft =3 ax + ft,
the members of which are identical.
J201] PROBLEMS 193
Infinite Solutions
201. Problem I. — What number must be added to the numera-
tor and the denominator of the fraction f to make the fraction equal
to 1?
Let X = the required number;
then ^r-^ — = 1,
8 + x '
or (1) 5 + X = 8 + a:.
This equation is impossible in an arithmetical sense, as there is
no number which added to 5, is equal to 8 plus the same number.
The greater the value of x, the nearer the value of the fraction
—5— approaches 1, or the more nearly is equation (1) satisfied.
The impossibility of satisfying equation (1) by a finite value of x
means that the problem from whose conditions this equation was de-
rived was impossible.
Problem II. — One pump can fill a reservoir in 15 days; another
can fill it in 25 days; and a third can empty it in 9| days. If all
the pumps are set working at the same time, how long will it take to
fill the reservoir?
Let X = the number of days required to fill the reservoir when
all the pumps are working.
Then 1 = the part filled in one day when all the pumps are working;
and r}^ = the part filled in 1 day by the first pump,
^ly = the part filled in one day by the second pump,
j: == the part emptied in 1 day by the third pump.
118
Hence, ^ + 5^~^^ = ^® P*^* ^^^ in 1 day when all the pumps
are working.
^^ 16^25 75 X
Simplify the first member of equation (1) ; it then becomes
(2) 1=0.
There is no finite quantity x such that 1 divided by x is equal to
zero. But x can be taken so large that the value of ^ differs from
0 by less than any assignable quantity. Such a value of x is called
numerical infinity and is represented by Ihe symbol 00.
194 COLLEGE ALGEBRA [{{202-204
This result shows that the reservoir will never be filled, since the
part filled in one day is just the same as the part emptied in one
day; i. e.,
(') 15 + 25 = 91'''^ 75 = 75-
202. In accordance with what is given in 2203, when the formula
for the solution of an equation in one unknown quantity gives for
the value of this unknown quantity an expression of the form J, we
say that the equation is impossible ; but it does not follow that the
problem is impossible — we can affirm, and only affirm, that the
quantity taken for the unknown ceases to exist.
203. When the denominator of a fraction decreases, the fraction
increases and will increase indefinitely if the denominator diminishes
indefinitely. However, it is customary to say that, in case the de-
nominator becomes zero, the fraction becomes infinite. This relation
is expressed, x = 00. This is an incorrect conclusion, because the
fraction whose denominator is zero does not represent anything. If
the given quantities of a problem vary in such a way that the de-
nominator of the unknown quantity approaches zero, the unknown
quantity itself increases indefinitely ; but, when the denominator is
actually zero, the solution does not exist and the equation is
impossible.
204. Problem III.— Two particles, with given velocities, have
uniform motions along an indefinite straight line, LL' \ at the same
time that one of them passes -4, the other passes a second point A'.
Find the point on the straight line where the two particles meet
There are several cases to consider in this problem according to
the position of the points A and A'^ the magnitudes of the velocities
of the two particles, and the direction of these velocities.
0 A A' R
1 I I I
—77 \ 1 1 1 z—
Suppose, first, that the two points A and A' are situated on the
same side of 0, and respectively at the distances AO -= a and
A'O z=^ a\ Suppose that both of the particles move in the same
direction, from left to right, the first with the velocity t?, the second
with the velocity, v', which is less than v. Evidently, the two par-
ticles will meet at some point R to the right of the point A\ Let
this distance OR be x.
J205] PROBLEMS 195
In order to express this problem in an equation, it is suflScient to
set up the condition that the two particles have reqaired the same
time in travelling till they meet, the first travelling the distance A R
and the second the distance A*R, Let t be the time required; since
the distance AR\% x-^a^ it follows that
X — a = vt^
whence
(1) t =
V
Similarly, since the distance A*R\s x — a',
(2) . = '-^:
whence finally = r— ,
t? t/
or (3) (t? — t/)x = va' — av\
The following convention concerning the signs of the magnitudes,
distance, velocity, and time is adopted:
1. The distances OA^ 0A\ OR are reckoned positively from left
to right, and negatively from right to left
2. The velocity of a particle is regarded as positive or negative
according as this particle moves from left to right or from right to
left
3. Finally, the time which has elapsed between the moment that
the particles passed A and A' and the moment that they meet at R is
reckoned as positive or negative according as the moment of passage
of the point R took place after or before the moment the particles
passed A and A' ; the formulae (1) and (2) are applicable in every case.
Therefore, the equation (3), which was established for a particular
case, is true for all cases.
205. Discussion. Consider equation (3), {204,
(t? — v') X = va' — av'y
and examine the various cases which can arise:
1. If r — t/ =5^ 0, the equation has a root
va' — av'
X = ^ )
V — V
then the two particles meet at a point to the right of the point 0 if
V V
is positive, and to the left of the point 0 if this fraction is negative.
196 COLLEGE ALGEBRA C«205
2. If V — v' = 0, two cases must be considered, according as
va' — av' is different from zero or equal to zero.
(a) If va' —av' is different from zero, the equation does not
have a finite root; the two particles will not meet.
It is easy to explain what this result means a priori. Since one
has by hypothesis
t? = t; ' and va ' — av' =^ 0,
it follows that va' — av' =^ C,
and, therefore, that a is different from a ' . Since the two particles
are always at some distance apart, a' — a, and move at the same
rate, they will never meet.
{h) If, however, va ' — ay ' is zero, the equation is satisfied for
any finite value assigned to x,
since, 0 • » = 0.
In this case the two movable particles do not separate ; since, if
V = v' and va' = at/
it follows that
a z= a'
and the points A and A' coincide. Therefore, the two particles are
at A at the same time, and as they travel to the right, with the same
velocity, they do not separate.
The results of the preceding discussion can be arranged in the
following table:
V — v' z:^ 0 ; the particles meet,
f « f va* — av' =^ 0; the jHirtides do not meet,
1 va' — av' ■=!=. 0 ; the particles do not separate.
Numerical Application op the Same Problem
Certain applications are now given in order better to fix the
meaning of the formula in the previous discussion:
1. The points 0, -4, A' are arranged as indicated in the figure
annexed; the two particles travel from left to right The distance
OA is 12 ft. and the distance OA' is 14 ft. ; the velocity of the par-
ticle which passed A, when the second particle passed A'y is 2. 5 ft. ;
the velocity of the second particle is 1 ft.
-12 14
A .-- ~.^ O ,.--' ---... A' R
— \^ ^-^^^z: ^H j:.4—
'"-^ ^ ^-'''
'■----^ 1 ---"'
2.5
31.333
«205] PROBLEMS 197
The values of the quantities a, a' v, v' are
a = — 12, a' = 14, v = 2.5, v' = 1.
^ ^ m^-^.- ^ 2.5X14 + 12X1 ^ 31 333 ^^
t,_v' 2.5 — 1
The particles meet at a point which is 31.33' ft. to the 'right of
the point 0, and after the particles passed the points A and A\
2. The points 0, A, A' are arranged as above, but the particles
travel in opposite directions ; the particle which passed A, when the
other passed A\ moves from left to right, with a velocity of 1.5 ft. j
the other travels from right to left with a velocity of 1.4 ft.
-12
14
-K >/i^ =H-
1.448 ^IJ-
1.5
One has in this case,
a = —12, a' = 14, v = 1.5, v' = 1.4.
^ __ va' _ gyf __ 1.5x 14 — 12x1.4 _ ^ ^^g ^^
v^v' 1.5 + 1.4
The particles meet at a point which is 1.448 ft. to the right of
the point 0, and after the particles have passed the points A and A\
3. The points 0, Ay A* are arranged as above, and the particle
which passed A when the other passed A\ travels from right to left
with a velocity of 2.5 ft; the other travels from left to right with a
velocity of 4 ft.
-12 14
k: ^\^ H
< _9 ^
-1.5 2 4
In this case, it follows that,
a = — 12, a' = 14, t; = — 2.5, t?' = 4.
^ _ yg/ _ at/ _ _ 2.5 X 14 + 12 X 4 _ _^
r — v' __2.5 — 4 "
The particles meet at a point which is 2 ft. to the left of 0, and
before the particles passed A and A!,
CHAPTER VI
SIMULTANEOUS LINEAR EQUATIONS
Systems op Equations in Two Unknown Quantities. A Single
Equation in Two Unknown Quantities has an
Indefinite Number of Solutions
206. Indeterminate Equations. — Often two unknown quantities
satisfy an equation of the first degree. Assign arbitrarily any finite
value to one of the unknown quantities; then the other will take a
finite and determinate value. Consider the equation
(1) 7x-2y= 18.
Give any value whatever to y; then for determining oc, the above
is an equation of the first degree, whose root is
._18 + 2y.
For example, assign to y the values:
1
then X will take respectively
the values
6
18 + 2_«6
___18 + 4 _
= 3
_ 18 + 6 _
= 3
x = l-«+^==3^
7 7
18 + 10_
x = li+l? = 42
7 7
On the other hand, x might be given any arbitrary value and the
corresponding value of y would be determined by the equation,
7x_18
y = — ^i —
198
8207] SIMULTANEOUS LINEAR EQUATIONS 199
Any such set of corresponding values of x and y satisfies the
given equation, and therefore gives a solution. •
An equation which, like the above, has an infinite number of
solutions is called an indeterminate equation.
207. Independent Equations.
The equation,
(2) 4y + 5aj = 40
also has an infinite number of solutions.
Solve this equation for y. Then
40— 5x
When X =
y =
0
1
2
I the corresponding values of y will be
5
6
7
10
H
5
^
li
Observe that equations (1) J2P6 and (2) have one common solution,
namely, sc = 4 and y = 5. Later it will be shown that these equa-
tions have only this solution in common. One might suspect that
this is the case by comparing the two systems of values of x and y
for equations (1) and (2).
"With the single exception, x = 4 and y = 5, equations (1) and
(2) have no solution in common, and are for this reason called inde-
pendent equations. If every set of values of x and y which satisfies (1)
were also a solution of (2) the equations would no longer be inde-
pendent
The three equations
(1) 7x-2y = 18,
(2) 5x + 4y = 40,
(3) 2x— y= 5,
are not satisfied by any common set of values of x and y. For, by
{206 and 2207, equations (1) and (2) are satisfied by the values
X = 4, y = 5. But equation (3) is evidently not satisfied by these
values because 2 -4 — 5 = 8 — 5 is not equal to 5. These equations
are said to express three independent linear relations between x and y.
200 COLLEGE ALGEBRA [82208-210
208. However, the equation
(1) . 7a;-2.v = 18
and the independent equation
(4) 21a; — 6y = 54
furnish the same values for y in terms of a?, namely,
21rr — 54 7x-18
,, = ..____ - = — — •
Consequently these equations are satisfied by the same unlimited
number of common sets of values of x and ^, and accordingly are
said to be not independent.
209. Incompatible Equations.
The equations
(1) 7x-2y = 18
and (5) 14x — 4y==25
are not satisfied by any common set of values of x and y. For any
set of values of x and y which will make 7jc— 2y= 18 will make
14 a — 4 y = 2 (7 x — 2 ^) = 36 and not 25. These two equations
express incompatn)le relations between x and y, and accordingly are
called mcompatihle v.quations.
210. Systems of Simultaneous Equations. — A group of equa-
tions which are satisfied by the same set, or sets, of values of the
unknown quantities, is a system of simultaneous equations.
And any set, or sets, of values of the unknown quantities, which
convert a system of simultaneous equations into identities, that is,
which satisfy all of the equations, is a solution of the system.
The examples in { 2 206-209 are illustrations of the following
principles which will be proved later.
A system of equations will have a definite number of solutions,
I. When the number of equations is the same as the number of
unknown quantities,
IT. And when the equations are all compatible and independent,
BXEBOISB ZZXIX
^ Of the following equations, which are not independent? Which
are incompatible ? Which arei iti'd^pendent -and. consistent?
2x— y=16 ,o .r3x+4;r=r8
|2x— y=16 ^2 P^
lx + 3y = 36. ' l6x
x + 3^ = 36. (6x+ 82 =36.
x + 4y = 21
2x + 3y=-22.
3 (4x— 127/ = 5 ^ ( x + 4y = 21
I 5x- 15.1/ = 41. ' 1
5S211, 212] SIMULTANEOUS LINEAR EQUATIONS 201
■ t7x-lly = 70. ' llOx — 15y = 20.
^ f x + 3^ = 50 g ( 7x — 3y = 15
|3x-2z= 7. ' ll4x-6y = 40.
9 (4x— 5y= — 9 ^^ j 4x— y = 5
• Isx— 10y= — 13. * i^x + 3y=z27. ]
^j f9x-3y = 9 ^2 | x + 2y=20
ISx— y = 21. ' l2x-3y = 5.
j3 r7x-2y=12 ( 7x-23^=18
I8x — 5y = 30. ' (28x^8^ = 72.
-• fax+Z»y = m -^ (" mx — ny = c
1 ^ (x 4" y) = «wi. I awix — anj/ =; c.
Equivalent Systems of Equationb
811. 7W systetiu of simitltaneoiis equations are said to be equiv-
alent when they are satisfied by the same solutions.
For example, the systema I and 11^
j|2x + y = 42 jj( 2x + y = 42
ISx — y = 33 I6x — 2y = 66,
3ie equivalent, for both are satisfied by the same solution, x = 15,
y = 12, as is seen by substituting these values in the equations.
212. The principles of equivalence of equations discussed io
S160 and following, were there proved for equations which contain
one or more unknown quantities. They can be applied, therefore,
to any equation in a system of equations. The solution of a system
of equations depends upon the following theorems concerning the
equivalence of systems.
1. If any equation of a system of equations be replaced by an
eqwivalent equation^ the resulting system will be equivalent to the given
one. Thus, the system
(2x + y = 42
(3x-y = 33
is equivalent to the system
( 2x + y = 42 .
(6x — 2y =. 66
^here the equation 3 x — y = 33 of the given system is replaced by
the equivalent equation 6x — 2y = 66.
202 COLLEGE ALGEBRA [J212
2. If any equation of a system be replaced by an equation obtained
by adding to or subtracting from it the same multiple of corresponding
members of two or more of the equations of the system, the resulting
system will be equivalent to the given system.
Thus, the system,
j|2x + y = 42
XSx — y = 33
is equivalent to the system
nil 2x + y = 42
t(6x-f-3y)-(6x-2y) = 126 — 66,
or ■ f2x + y = 42
I by = 60.
3. J[f one equation of a system be solved for one of the unknown
quantities, and the value found be substituted in each of the other equa-
tions of the system, the derived system will be equivalent to the given one.
Thus, the system,
j(3x — y = 33
(2x + y = 42
is equivalent to the system,
jy( y = 3x-33
l2a;-(3x-33) = 42.
The proofs of theorems 1, 2, 3 follow.
Proof 1. Let iH--^
be two equations involving two unknown quantities, x and y. Sup-
pose that C" = 2/ is an equation which is equivalent to C = Z>. Then
the system
is equivalent to system I. For, since the equation C = 1/ ia equiv-
alent to the equation C = D, every set of values of x and y which
satisfies the equation C = D, satisfies the equation C = I/, and
conversely. Therefore every set of values of x and y which satisfies
the equations A = B and C= D will satisfy the equations A=z£
and 6" = ly, and conversely.
In like manner the theorem can be proved for a system of any
number of equations involving the same number of unknown
quantities.
Prouf ^. Let 1 1 ^ = ^
1 C=D
be two equations in two unknown quantities, x and y.
1212] SIMULTANEOUS LINEAR EQUATIONS 203
Then the systems,
id ^=^ and IIl{ ^=^
(,A + kC=zB+kD \ A-^kC=B — kD
are equivalent to system I, where k is an}' finite real number. For
every set of values of x and 1/ which makes A = B and C=D will
make A + kC= B + kD and A — kC=z B — kD; since if equals are
added to or subtracted from equals the results are equal. Therefore
any solution of I is a solution of II and III, and every set of values
of X and y which makes A = B, A-}- kC = B-{- kCy or A = B, and
A — kC= B — kD makes C = D. Because, if
A^kC^B-\-kD
and A=iB
by subtracting, kC—kD\ . • . C= D, etc.
Similarly the theorem can be proved for a system of any number
of equations involving the same number of unknown quantities.
Proofs, Let I
(Az=B
\ C=D
be two equations in two unknown quantities, x and y; and suppose
that x=: B' \s the value of x found by solving ^ = ^ f or x. B'
contains y and known numbers. Let C" = Z)' be the equation
resulting from the equation (7=2> when a; =-5' is substituted in it.
Then it is required to prove that the system,
III - = ^'
is equivalent to system I.
Since the equation x = 5' is equivalent to the equation Az=. B,
every solution of this equation must satisfy the equation x = B']
i. e. , must also furnish the same values for x and B\ Now, observe
that the equation C —U differs from C = Z> in this, that where x
occurs in the latter, B' takes its place in the former. Since x and B*
have the same value, any value of x with the corresponding value of
y which makes / ?~ -2 must make C=:iy.
Conversely, every solution of system II is a solution of system I.
Since the equation x=B' is equivalent to the equation A=B^ every
solution of system II must satisfy equation A = B, And also,
since every solution of system II makes x = B\ and since equation
G= D differs from C = ly in this only, that where B' occurs in the
latter x occurs in C= D; therefore the values of x and y which
satisfy system II and which make (7'= />', make C=D, Hence
204 COLLEGE ALGEBRA [JJ213, 214
every solution of system II is a solution of system I. Therefore,
the two systems are equivalent.
This theorem can be proved in a similar manner for a system of
any number of equations involving the same number of variables.
The Solution op Two Equations op the Fiest Degree in Two
Unknown Quantities
213. Elimination.— X is said to be eliminated from a system of
equatiofas when this system is replaced by another equivalent sjstem
in which none of the equations except ojie involves x. For exam-
ple, at the end of 3212| system II is substituted for system I.
Elimination by Addition and Subteaotion
214. Example 1. Solve the system,
J f(l) 7x + 4y=94
1(2) 5x — 2y = 38.
To eliminate y, multiply both members of equation (1) by 5, and
both members of equation (2) by 7, thus rendering the coelficients
of both terms in x the same. Then
(3) 35x + 20 3^=:470
(4) 35aj-14.y=i266.
System II is equivalent to system I by 2212, 1. Further, sys-
tem II is, by {212| 1 and 2, equivalent to the system,
IIIJ(I) 7x+4y = 94
"l:
I
(5) 35 X + 20 y — (35x — 14 y) =470 — 266
or, through simplifying, equivalent to
„,U1) 7x + 4y = 94
1(6) 34y = 204
orto Vi^ 7x + 43. = 94
I (7) 3/ = 6.
The required solution can now be found from system V by using
the theorem 3 in 2212. Substitute, therefore, ^ = 6 in equation (1)
and obtain
(8) 7x + 24 = 94
(9) .-. x=:10.
The required solution is, therefore,
yjj(9) x=10
I
(7) y= 6
expressed by system VI, which is equivalent to system I.
{214] SIMULTANEOUS LINEAR EQUATIONS 205
The legitimacy of the successive steps made in arriving at the
solution a; = 10, y = 6, is shown by exhibiting the successive sys-
tems of equivalent equations. In practice this work may be abbre-
viated as follows:
(3) Multiply (1) by 5 35 x + 20 y = 470.
(4) Multiply (2) by 7 35x-14y = 266.
(5) Subtract (4) from (3) 34 y = 204.
(6) Divide by 34 y = 6.
(7) Substitute 6 for y in (1) 7 x + 24 = 94.
(8) Whence x— 10.
To eliminate y, multiply both members of equation (2) by 2.
Then the coefficients of y in the resulting equation and in equation
(1) are the same with different signs, and the work may be arranged
as follows:
(1)
7x+4y = 94.
(3)
Multiply equation (2) by 2
lOx — 4y = 76.
(4)
Add (1) and (3).
17x--=170.
(5)
, ',
x = 10.
(6)
Substitute 10 for a; in (1)
70 + 4y = 94.
(7)
.'.
y = 6.
Example 2. Solve the equations
(1)
x-4 y-3_
4 ' 3
:3.
(2)
ar_3 y-4_
3 4
2i
(3)
Clear (1) of fractions; 3x
- 12 + 4y— 12 = 36.
(4)
Clear (2) of fractions ; 4 a:
_12_3y + 12 = 30.
(5)
Transpose and unite terms
3x + 4y = 60.
(6)
Transpose and unite terms
4x — 3y = 30.
(7)
Multiply (5) by 4
12x + 16y = 240.
(8)
Multiply (6) by 3
12x — 9y =90.
(9)
Subtract (8) from (7)
25y = 150.
(10)
, ',
y = 6.
(11)
Substitute 6 for y in (5)
3x+24 = 60.
(12)
, * ,
x=12.
Observe that (1) and (2), (3) and (4), (5) and (6), (7) and (8),
(10) and (11) form systems of equivalent equations. With (10) is
associated the simplest of the equations, number (5), which is equiv-
alent to (11).
206 COLLEGE ALGEBRA [221 5
215. The examples discussed in the preceding section illustrate
the following method of elimination by addition and subtraction for
two equations in two unknown quantities.
Simplify the equations hy removing parentheses or clearing fractions^
and transpose the x and y terms to the first members and the terms free
from X and y to the second members^ multiply both members of each
equation by such a number as will make the absolute values^/ the coef-
ficients of one of the unknovm quantities the same in the two resulting
equations.
The unknown quantity whose coefficients are now equal, with the same
or opposite signs, can be eliminated in the first case by subtracting and
in the second case by adding corresponding members, and equating the
results.
The solution of the given system of equations can now be found by
solving this derived equation, and substituting the value of the un-
known quantity thus found in the simpler of the preceding equations,
EXEBOIBE Zli
Solve the following systems of equations by the method of addi-
tion and subtraction:
|lla:+12y=100 |3x + 7y = 7
1 9aj+ 8y= 80. •l5x + 3y = — 36.
f5x + 3y+2=0 f2ix = 3Jy + 4
• l3x + 2y + l=0. • l2jy = 3ix-47.
rl.5x — 2y = l I 7x — 10y = 0.1
t2.5cB-3i/=6. • lllx— 16y=0.1.
(0.16x — 0.043^ = 1 f 3.9 x-0.08y= 2.77
I0.19x — 0.1l3/=l. I 26x+ 0.4y = 18.
f25.9x-60.1y=l fix-2y==l
1 24.1 X — 55.9 y= 1. liaJ— y = 0.
11,
f^x = it/ + l f
Ux = |y-10. 'l
.ix + fy=19.
|7x~5y = 3.042 f5x-4.9y=l
l3x-2y = 1.323. •l3x-2.9y = l.
( 5x-4y + l = 0 r2.7x + 2.6y = 8.8
ll.7x-2.2y + 7.9 = 0. lo.9x + 2.2 3/ = 4.4.
"•{
27.4x — 31.5y = ll
21. 4x— 26.53/= 1.
«216, 217] SI^rcLTANEOUS LINEAR EQUATIONS 207
( 2.60x — 0.41 y — 2.222 + 2J a; = 0
" 10.51a;- 3.60y + 3.333— ^y= 0.308.
(3.5x + 2iy = 13 + 4^x-3.5y
• I2^x + 0.8y = 22i + 0.7aj-3i3
X
0.8 , 3.6 .
1 = 5.
X y
f5(x + 2)-3(y+l)=23
l3(y-2) + 5(y-l)=19.
Elimination by Substitution
216. Particular Case. — Consider first the case in which one of
the equations inyolves but one of the unknown quantities.
Solve the system, | 4x — 9 = 19
l3x-4y = 9.
The first equation is equivalent to
4x = 28 or X = 7,
and the system of two equations is equivalent to the following,
r X = 7
|3.7__4y = 9
or further to the system,
fx = 7
\y = 3.
217. General Case. — Reduce the solution of the general to this
particular case, by replacing the proposed system by another system
equivalent to it, in which one of the equations involves but one
unknown quantity.
Let, for example, the system of two equations be
((1) x-12y = 3
^1(2) x + 4y = 19.
Derive the value of x from equation (1) as though y were known,
thus, (3) X = 3 + 12y;
substitute ^-\-12y instead of x in equation (2) ; thus,
(4) 3 + 12i/ + 4y = 19;
208 COLLEGE ALGEBRA [??218, 219
"!:
and, according to the theorem 3, i2li, the given system is equiv-
alent to the system,
(3) x = 3+12y
(4) 3+12y + 4y = 19.
But the equation (4) in system II involves y only, and by solving
fory,
19-3 ,
y = = 1.
^ 16
y = l.
Then substitute 1 for y in equation (3), thus,
X = 3 + 12 = 15.
System II is therefore equivalent to
X = 15, and y = 1.
Consequently system II has the solution x = 15 and y = 1, and
no other solutions.
218. Rule. To solve a system of two equations of the first
jdegree in two unknown quantities, the following rule can be stated:
Derive from one of the equations the value of one of the unknoum
quantities^ as if the other were known, and substitute this value in the
other equation; and thus obtain an equation of the first degree in one
unknown quantity.
Solve this equation. Substitute the value found in the first equation^
and solve the resulting equation for the first unknoum qvtantity,
219. Applications. — In practice circumstances arise which tend
to simplify calculation.
Example 1. Solve the two equations,
|(1) 4x + 3y = 61
1(2) Ix-y = 38.
Deduce the value of y from equation (2) ; then,
(3) y = 7x— 38.
Substitute 7 x — 38 for y in (1) ; then,
(4) 4x + 3(7x— 38) = 61,
an equation all of whose coefficients are integers. From simplifying^
it follows, that
4x + 21x = 61-^-114
25 X = 175
(5) X = 7.
J219] SIMULTANEOUS LINEAR EQUATIONS 209
Substitute 7 for x in (3). Then
y = 7 -7-38 = 11.
The given equations have, therefore, the solution,
X = 7, y = 11,
and no others.
Example 2. Solve the equations,
f(l) 4x-7y = 19
1(2) 4x + 9y = 67.
Since the coefficients of x are equal, solve (1) for 4x; thus,
(3) 4x = 7y + 19.
Substitute 7y-f 19 for 4 x in (2); then
(4) 7y + 19 + 9y = 67
or 16y = 48
y = 3.
By substituting 3 for y in (3) the equation follows:
4x = 7 • 3 + 19 = 40
a; = 10.
The given equations have the solution,
x= 10, y = 3
and no others.
ExA3fPLE 3. Solve the equations,
|(1) 3x + 10y = 12
((2) 12x — 5y=3.
Since the coefficient 5 of y in (2) is a divisor of the coefficient 10 of
y in (1), solve (2) for y; thus
(3) y = ^^-
By substituting ^~ for y in (1) the equation is derived,
(4) 3x+10i?-Y^=12
or, by simplifying successively,
3x + 2(12x — 3) = 12
3x-f24x— 6 = 12
27 X = 18
x = ?.
* * 2 3
Substitute - for x in (1), then
3.|+10y = 12
10y = 10
210 COLLEGE ALGEBRA [J220
2
The solution is ac = -. y = 1.
o
Example 4. Solve the equations,
f(l) 7x-f.l8y = 110
1(2) lla;-12y = 52.
Since the coefficients of y in the two equations have a common
factor, deduce y from one of these equations, from the second, for
example. Then
(3) y='^^-
Substitute "^7^^ for y in (1); then
7a; + 18^j^^= 110,
or, by simplifying,
7aj + 3^i^^^ = 110,
or Ux + 3(11 ic— 52) = 220,
47x = 220 + 156 = 376;
x=8.
Substitute 8 J or x in (3) ; then
88 - 52 o
The solution is x = 8, y = 3.
220. The process used for solving two equations in two unknown
quantities hinges on the fact that this system can be replaced by
an equivalent system in which one of the two equations involves
but one unknown quantity. In general, this equation in one
unknown quantity has a unique solution, and the conclusion is
drawn that the given system of equations has a umque solution.
But it can happen that this system does not have a solution (2209),
or that it has an infinity of solutions. In the first case the equations
have been called incompatible (2209). In the second case, the
system is equivalent to a single equation of the first degree in two
unknown quantities. The system is indeterminate (2206).
To illustrate an indeterminate system try to solve the equations
Solve (1) for
(3)
x;
f(l)
1(2)
then
5
35 X
X =
x + Sy =
+ 56y =
9-8.V
" 6
:9
:63,
Substitute —^
^^ for X in (2);
then
35irzli+56y = 63,
J220]
SIMULTANEOUS LINEAR EQUATIONS
211
5
and 63 — 56 y + 56y = 63,
(4) 0 • y = 0.
The given system is equivalent to the system,
((3)
1(4) 0-y = 0.
Equation (4) is satisfied for any finite value of y. The system
of two given equations is equivalent to the single equation,
""- 5
This system shows the solution to be indeterminate^ since for any
value of X there is a corresponding value of y, and conversely.
BXBBOISBXLI
Solve by substitution:
y = 18
(x-y = 18
1 X = 4y.
(5x — 8^ = 7y — 44
l2x = y + |.
( x + y = a
1 X — y — 6.
3x + 2y = 118
x + 5y = 191.
( 7y = 2x-3y
• I 19x = 60y + 621f
I 7x+ |y = 411i
' (39x— 14y = -935^^
|5Jy-llx= 4y + 117i
8x+ 175 = 2y.
8 \ ^+^
l0.56x+13.
= 18.73
421 y = 763.4.
3(2x-y) + 4(x-2y) = 87
2(3x-y)-3(x— y) = 82.
ix-i(y + l) = l
'"• l^(x + l) + J(y-l) =
11.
x + 2y
7
2x + y
5
12.^
^x + 3y __ g
X — y
7x— 13
L3x — 2 6 — y
14.
3y-5
X — 3
y + 2
U-2
= 4.
2
3
3
2'
212
COLLEGE ALGEBRA
[1221
15.
2x-y + l
3x^y + l ^ 5
x-y + 3
16.
3a; + 2.y + 12.3 _g
4x + 3y — 44
4x+ lOy — 6.7 _^
Scc + y — 10
17 U^-iO^ + i) = H
• U(x-l)~Jy = 4f
19.
15x + l^
45 -y
12iMli? = 25
ic- 10
23.
|^2x — 0.3
(9 X -0.7y + 7.3_
13x-15y+17 ■"
1.2x — 0.2 j^ + 8.9 __
13;r. ""
22.
0.2
0.3.
ar + 3y + 13 ^3
4x + 5y — 25
Sx + y + 6 ^^
5x + 3y — 23
15y + 17
Elimination by Comparison
221. Example 1. Solve the equations,
jj(l) 4x+ 9y = 51
(2) 8a: — 13y = 9.
To eliminate y proceed as follows:
Solve both (1) and (2) for y; then II
(3) y =
(4) y =
51 — 4x
9
8x — 9
13
System 11 is equivalent to system I (by S212, 1).
Substitute the value of y in (3) for y in (4) ; then
51 — 4x 8x — 9
(5) ^— _ = -^^,
an equation which involves but one unknown quantity and which,
together with equation (3), forms a system equivalent to systeiQ
II, and therefore equivalent to I.
Substitute in (3) ; then III | |
J221] SIMULTANEOUS LINEAR EQUATIONS 213
Solve (5) for x; then
13(51 — 4x) = 9(8x-9),
663 — 52x = 72x — 81,
124 X = 744.
(6) X = 6,
(7) y = 3.
The ^stem formed by (6) and (7) is equivalent to the system
formed by (5) and (3), and therefore to the given system.
Hence the given system has the solution x = 6, and ^ = 3, and
this solution only.
To eliminate x proceed as follows :
Solve (1) and (2) for x; then
(8) -^^.
(9) . = l±^.
Equate the values of x,
Mm 51-9y_9 + 13y
(10) -^ — g
whence y = 3.
Substitute 3 for y in (8) ;
then, X = 6.
Example 2. Solve the equations,
.^^ 2x 3y a; + 2y__o 5x — 6y
^^^ T"T — 4—-^ r-
(9\ ^+2y 3x — .y _ K , ac
/2) ^ 5""^" ^15'
Clear (1) and (2) of fractions; then
(3) 40x — 36y — 15x — 30y = 180 — 75x + 90y
(4) 15x + 30y — 18x + 6y = — 150 + 2x.
Simplify (3) and (4); then
(5) 25x — 39y = 45
(6) __5x + 36y = —150.
Solve (5) and (6) for x; then
(7) ' = -tr^
214
COLLEGE ALGEBRA
L2222
Equate the values of x,
45 + 39.v_36y+150
25
(9)
Hence (10) 3/ = — 5
Substitute — 5 for y in (8) ; then
X := — 6.
Hence, the given system of equations has the solution x = — 6,
y = — 5, and this solution only.
Observe that (1) and (2), (3) and (4), (5) and (6), (7) and (8),
and (10), with any preceding equation excepting (9), form equivalent
systems. With (10) is combined the simplest of the preceding equa-
tions, in this case (8).
222. The examples of the preceding section will illustrate the
following rule of elimination by comparison.
Solve eacli equation for the vnknown qiLantity to he eliminated^ and
equate the two results thus obtained. The resulting equation will involve
hut one unknown quantity.
Solve the equation thus obtained for the unknown quantity and sub-
stitute the value in the simplest of the preceding equations. The solution
of the resulting equation and the value of the unknown quantity already
found will be the solution of the given system.
EXEBCISE XLII
Solve the following systems of equations by comparison:
2x — 5y = — 16
1.
3.
(2x-5y =
{3x+ 7y =
{
2.
-5x-|-3y = 51
7x + 2y = 3.
3x — 19
(y = 3x — 19
I X = 3 y — 23.
j 3x — 5y = 19
( 7x+4^ = 2.
|Gx+ 15y = — 6
ISx — 21y = 74.
( 7y — 3x = 139
l2x + 5y= 91.
7.
^ + ^=42
9^8
43.
(^8^9
fBx
— +5i/=13
19 ^ ^
2x.
<.'y
= 33.
10.
3
x — y
+ X = 15
( x = 3y-
\ y z= 3x —
19
23.
5223]
11.
13.
14.
SIMULTANEOUS LINEAR EQUATIONS
X
12.
215
fx + l y + 2_2(x-y)
! 3 4 5
^z:^^y-3^
4 3^
3x — 2v,5x — 3y ,-
2x~3y 4g-3.v^ , ^
3 "^ 2 ^"^ •
2x — y + 3 X — 2y + 3__.
3 4
3g — 4y + 3 . 4x — 2y — 9_,^
4 3
+ 32^ = 7
= 3y-4.
3
4x-2
5
Literal Simultaneous Equations
Fractional Equations
223. Example 1. Solve the equations,
Solve (1) for y,
12x — a
(1) 12a; — 6y = a
(2) 13x + lly = 4a.
12« — a
Substitute
6
(3)
for y in (2),
y =
6
(4)
Clear fractions,
(5)
(6)
13X+11 .l?-^-=^ = 4a.
78x+ 132x — 11a = 24a
210 x = 35 a
_ 35a a
^"2l0""6'
122 — a a
Substitute - forx in (3); then y = — ^^ — = -•
6 6 6
Therefore the solution of the given system is
a a
6' "^ 6
Example 2. Solve the equations,
a -\- o a — o
216 COLLEGE ALGEBRA [2223
Solve (1) and (2) for y,
(4) .V = a? — ^ah.
Put the values in (3) and (4) equal to each other,
(5) x-4a6 = ^^<^'-^^')-^^^^^^^^
a-\- b
or (6) x(a+h)—Aah{ii + h)=2a(a^-^h^)^(a—h)x.
Transpose, (7) xia-\-h +a-h'] - 2aW—h*-\-2ah +2&«J
(8) 2ax= 2a[a« + 2a6 + ^«] =2a(a + «^)«
(9) x = (« + W
Substitute in (4), y = (a + i)« — 4a6 = a« + 2a6 + ^« — 4a6
or y = rt* — 2 at + 6*
y = (a-6)«.
Example 3. Solve the equations,
(1) - + -=1
a; y
(2) - + - = 1 .
x y
Multiply (1) by iw, and (2) by w,
(3) ^'+— = ^
X y
X y
The numerator of y in both equations is the same. Then, by
subtracting, (5) — ~ — = m — n
X
or (6) m* — n' = (m — n) x
(7) x = = m + w.
Substitute (m + n) for x in (1),
(8) -=1-^=1= "^
Simplify, (9)
Divide by «, (10)
y X m-f- w
» m + 71 — wi. w
1 1
y m-^n
y =z m + n.
Hence, the solution of the given system is
x = y =z m -\-n.
J223] SIMULTANEOUS LINEAR EQUATIONS
BZEBOISB XTiTIT
Solve the following systems of fractional and literal equations:
1.
217
ax = by
{ax =
x + y =
(ax+by = i
1A +9!/ = ^
4.
axy — by •=. c
dxy + ey=f.
C 5^9
}. J 4 + y 20-.1
(9y — 5x = 26.
8. ^
a 6
-2^ = 0.
[6 a
10.
12.
14.
16.
18.
20.
10
20.
^ y
:2
-2 + ^ =
:3.
6 , 1G_
= 79
. ^ y
= 44.
Zy^^jc
= 7
7 1
.6y 10
- =3.
X
' ^_y-
a h"
m
ft.
2.
5.
7.
9.
11.
13.
15.
17.
19.
■r — 3_y — 6
4 3
4(x-3)+5(i^-5)=62.
6 + j/ 3a+y
aX'{-2by = d.
wix 4- =1
2/
+ 111 t
- = 1.
y
^ + 2^-5
n_7_3
X x~ 2
{ 3
X
1
2y
= 16
^ +- = -15.
2x y
1
+
2
or-l y + 1
3,4
+
_5
~6
= 2.
Lx-l y+l
_5 ?_ = _1
x-1 y-1 6
_3 1_ = 1
a: - 1 y - 1 30*
{(a -f- 6) a: — (a — 6) y = 4 at
(a = i*) X — (rt + ^) y = 0.
218 CX)LLEGE ALGEBRA [J224
224. Elimination by Undetermined Multiplier.
The process of elimination by undetermined multiplier is not
much used. The discovery of the method is attributed to Bezout.
Example. — Solve the equations,
(1) x + ayzzzb
(2) ax + 6y = c.
Multiply (1) by m; then
(3) mx + may = mb
(2) ax -\- by = c.
Add (2) and (3),
(4) (a + m) X -|- (6 + ma) y = c -\- mb.
To eliminate x, put
(5) a + m = 0.
That this may be the case, choose m so that
w = — a.
Substitute — a for m in (4) ; then
(6) Ox + (6— a')y = c — afc.
c — ob
^ 6 - a«
To eliminate y^ put
fc +fwa = 0;
i.e., choose m so that
111= •
a
Substitute — - for m in (4), then
a
(5)
(a )x — (b — a-jy = c — -b
a) \ aj a
or (6)
a
-X
+ 0
ZIZ
a
Hence
in'
—
b)x
= ac -
-b*
.*.
X
a«
-b*
— b
Therefore the solution
is
x =
nr
-fc'
__/.«.
— fIC
6 — a*
J224] SIMULTANEOUS LINEAR EQUATIONS 219
EXBBOIS£ XI<rV
Solve the following examples by any of the preceding methods :
x+ y = 6912
1.
{ X — y = 4444.
|x4-13^ = 176
\x+ 7i/ = 98.
( x-\-ay = b
( ex 4- ^ = <^-
X + 17y = 300
104.
( 1.543689 x-y = 1.543689
lx-0.^
|x + 17y =
• \ nx-!,=
2.
4.
6.
8.
(x+y= 8
\x — y = d.
(x+lfy = 26,V
l4|y-x = 443.
1
mx -f- y = />
wx + y = 7>.
2?x — Jy = 116
40.
10.
12.
13.
16.
17.
19.
21.
22.
23.
. 8392867 i/
X — 5143
3i/ + ll
3262 — X
2 y - 11
= 37
0.8392867.
11.
= 43.
4x + 81
10,y-17
12x + 97
I5y-17
= 6
( (^ +.3) {y +.5) = (x +.1) {y +.8)
((2x-.3)(.5y+.7)=.2(5x- 6) (//+.]).
fii54.i2? = 6 14 |-^ + iVy=71
I 0.7 ^ y • U--,Vx = 61
I l«-^. 1^-31 15 M3x + ll3/ =
[ .7 +y-^'- ^^- ll3x-ll3/ =
+ 3.14159^ = 3.141593 + 1
3.14159
3.14159 X-
^- - = 3.141592 — 1.
3.14159
!ax -\' hy z= c
18.
{
ax -\- by =1 c
20.
\i
( mx + wy = c
( X :y = a :b.
(a + />)x — (a—b)y = 4«fc
t + i,)a; + (a — ^)2/ = 2 (a^ + /,2).
(a + /.)x+(<i-^>)i/ = 2{a'+b')
.h)x+{fi + b)y = 2(/r + />2),
ax + by = 2a
i*x — Z**^ = (r + />-.
a (x + y) = m
({a.
((a-
U«-
220
COLLEGE ALGEBRA
LS224
24.
25.
27. ^
28.
29.
31.
32.
33.
34.
35.
36. -I
j ax + hy =z a^ + 2a*b + 6»
( bx + ay = a^ -\- 2ab* + b\
2 {a« + 6«)
a* — 6*
4a6
26.
X — y =
a« — i»«
ax -\- by = 2 a
g' + fe'
» + y =
ab
a' — 6*
a-|-6 tt — b a — b
X y _ 1
la+ b
a + b
30.
a
X y
a, , b.
_i + -1 = Cj.
a b
(a — b) X -\- {a -^ b) y = a -|- i
a? y__ 1
a-f-6 a — b a -{- b
/ i\ . a + b 4-1
{a — b)x + y = —T-L^
a-\- b
I x + (a + b)y = ^-^=±±l.
I a — 6
{{a -\- b — c)x — (a^b -\- c)y = 4 a (6 — c)
X : y = (a + 6 — c) :{a — b -{- c).
( (ic + y) : (a; — y) = a : (6 — c)
1 (x + c) : (y + i) = (a + i) : (a + c).
f (x — a):{y — a) = (a — 6) : (a + 6)
1 x:y = (a» - 6') : (a» + «^').
x+1
— 2 — = a
y
y-±l = b.
37.
x _^a
y b
X 4- 1 c
J224]
38.
SIMULTANEOUS LINEAR EQUATIONS
39
221
40.
42.
44.
45.
46.
47.
48.
50.
51.
ras + l _«
+ h + c
y + 1 o
-h + c
X — 1 _ a
+ h-e
Ly— 1 a
— b-c
a; + y + i
_a + l
x-y+l
a — I
as + y + l
_! + «•
LX — y — 1
1-6
' X — c a
Lx + y —
y _ c 6
X — y = a — 6.
^ 1 y -
41.
43.
aj + y —
= a
= b.
=ib.
Lx — y —
X — a -|- c __ 6
y — a -\- b c
x-f- c a -f- 6
.y + 6 a+ c
a 4- i» a — b
a + b
5 + | = 2«.
La 6
i» a — c
^- = b^
b
X + c
+ 2lzii? = i + e.
»+i^2
a-\- b a -\- c
x-b _|_ y-^ ^ 2.
-•a) — c a — 6
(a -{- c) * — (a — c) y = 2 afr
(a+fc)y — (o — b)x = 2ac.
a-\-x
b
+
b-y
a
a+x b — y
(mx + »y) : (px — jy) = a : 6
. (rx + «y) : (to ^uy) zn c : d.
a
«= + 3'-l_a
b
49. -
x-y+1
b
y--+l-«6
a
x-y+l
OX = 6y -f-
2
52. -
(a — 6) X := (a -f~ ^)y'
71 + y m — X
P ■_ q
Iq — x p + y
222
CX)LLEGE ALGEBRA
[J224
53. ^
a a — c a
rt -f- c b a-f- c
c a ac
55.
54.
56.
57.
58.
X
m2_ 1 a«— 1
X y
-y—-a^^n^
I -" _j ^
\ m — a m —
S = '
y _
n — a n — b
= 1.
. V . . , . 4-2 = tt« + n«.
[ a'^^ 1 • w«+ 1 ^ ^
( (rt + 2i*)x — (tt — 26)^ = 6ac
t (tt -|- 3c)y — (a — 3c)x = 4a6.
— a z= 0—
X X
1 1^ __ ci»+/>«
(a — 6)x {a-\-b)y abxy
(a - by - 2M
1 + X = y - 1 + 2-
a" — b^
ab
^y-ttx=— ^— -y^-(a + «^)4-
tt* — 6* tt 4- o
f 306tt»+324a«i— 1015ttt«-810«>» 1
59.
60.
1
120tt6{3tt+26) (7a+6b)xy {3tt+26)y (7tt+66)x
1026rt* — 393tt«i*« — 430M _ 7tt«_-6i« __ 3q*— 2b*
X y
(x+l) (y-2) = (3-x) (4-.y)-l
61. < 2x— 3 3x— 4 5
62. ^
12i}abxy
j ^'-/=« 61. j 2x-3 _ 3x-4 ^ ^5
{x-y=b, U^-5 6i/-7 2(4j/-5)
i2x-3',+ 17 + ^— ^^ + ^^-^
(7-6y)
!.>:
|^2x — 3^y + 17
1 1
+ 16i/ = 10x + 88J.
l—x+y x+y—l 3
1 ^ 3 64.
"4
1
1
1 — i«+i/ 1-x— ^
1
X + --
.1
X —
y —
K'-i)
b
y — -
X X
1.
J225] SIMULTANEOUS LINEAR EQUATIONS 223
Problems Which Lead to Two Equations of the First Degree
IN Two Unknown Numbers
225. Problem I. The sum of two numbers is 100 and their
diflference is 22. Find the numbers.
Problems like this have been solved by using one unknown quan-
tity. This can be solved by using two unknown quantities.
Let X = the first number,
and y = the second number.
Then, since the sum of the two numbers (or x -{- y) is 100 and
the difference of the numbers (or x — y) is 22, it follows that
(1) x + y=100
(2) x-y= 22.
By adding the equations,
2x =122
x= 61.
By subtracting (2) from (1),
2y = 78
3/ = 39.
Hence x = 61 and y = 39 are the numbers required.
Problem II. A certain fraction becomes 1 when 3 is added to
its numerator, and ^ when 2 is added to its denominator. What is
the fraction?
Let X = the numerator of the fraction,
and y = the denominator of the fraction.
Then - = the fraction.
y
Since the fraction becomes 1 after 3 is added to its numerator,
x+3 .
(1) -X- = l.
y
And since the fraction becomes ^ after 2 has been added to its
denominator,
(2) _^ = -^
^^ y + 2 2
Clear (1) and (2) of fractions and transpose; then
(3) X — y = — 3
(4; 2x-y = 2.
By subtracting (3) from (4)
(5) X = 5.
224 COLLEGE ALGEBRA [?225
By substituting 5 for x in (3),
(6) y = x + 3 = 8.
X 5
Therefore the fraction is - = ;;•
Problem III. If a rectangular field were 8 feet wider and 2 feet
longer, it would contain 960 square feet more; if it were 2 feet
narrower and 8 feet shorter, it would contain 760 square feet less.
What is its area?
Let X = the width of the field,
and y = the length of the field ;
then xy = the original area of the field.
If the field were 8 feet wider, the width would be a? + 8 feet; and
if 2 feet longer, its length would be y + 2 feet; and, therefore, its
area would be
{x +8) (ij + 2) or xy + 960 feet; i.e.
ix + S)(y+2) = xy + 960
(1) 2x+Sy = d4L
If the field were 2 feet narrower, its width would be x — 2 ; and
if it were 8 feet shorter its length would be y — 8, and, therefore,
the area would be
{x — 2) (y — 8) or xy — 760,
hence (ac — 2) (y — 8) = xy — 760
(2) _2y_8x = -776.
By simplifying (1) and (2),
(3) x + 4y = 472
(4) y + 4x=:388
(5) Multiply (3) by 4, 4x+16y = 1888
(4) and subtract (4) y + 4x = 388
(6) hence 15 y = 1500
(7) .-. y = 100.
By substituting 100 for y in (4)
(8) 100 + 4x = 388
(9) or 4 a; =288
(10) .-. x = 72.
Problem IV. After working 2 days on a certain job with B,
A said to him: **I can finish this job alone in ten days." B re-
plied: **If we work together one more day, I can finish it alone in
5 days. ** If what they said was true, how long would it take each
alone to finish the job?
J225J SIMULTANEOUS LINEAR EQUATIONS 225
Let X = the namber of days that it would take ^ to do the work,
and y = the number of days that it would take ^ to do the work.
Then - = the part of the work which A would do in one day,
and - = the part of the work which B would do in one day;
2,2
then — h - = the part of the work which A and B^ working to-
gether, would do in two days.
— = the part of the work A would do in 10 days;
5
- = the part of the work B would do in 5 days.
Then, according to the first conditions of the problem,
(1) ?+2_i-l«
X y X
3 3
Moreover, - + - = the part of the work that A and B, working
• ^ y
together, would do in three days.
Hence, in accordance with the second condition of the problem,
(2) ? + ?=!- ^
X y y
By transposing in (1) and (2),
12 2
(3) - + - = 1
X y
(4) ? + ^ = l
X y
AQ Q
(5) Multiply (3) by 4; _ + ° = 4
X y
(4) and subtract (4); ^ + ? = 1
X y
AK
(6) tlien =3
X
(7) .-. X =15 days.
Substitute 15 for x in (4); then
(8) 4 + ^ = 1-
15 y
u 8,14
Hence, ~ = 1 _ - = _
y 5 5
(9) .-. y = 10 days.
226 COLLEGE ALGEBRA [?225
Problem Y. A man bays 570 pulleys, some at 16 for a dollar
and the remainder at 18 for a dollar. He sells them all at 15 for a
dollar and gains three dollars. How many of each sort does he buy?
Let X = number of pulleys of the first kind ;
y = number of pulleys of the second kind.
Since x-\-yj the sum of both kinds of pulleys, is 570,
(1) x+y = b70.
The cost of X pulleys, bought at 16 for a dollar, is :p» and of y
pulleys at 18 for a dollar, ^> and the cost of both kinds
16^ 18
But the 570 pulleys are sold at 15 for a dollar, or for $38; and,
if he gains three dollars by so doing, then
(2) — ,
^^ 16 ' 18 15
+ Z. = 570_3or38-3
(3) or 1+9 = '^^-
By combining (1) and (3), and clearing (3) of fractions,
(1) x + yz=b70
(4) 9ic+8y = 5040
(5) Multiply (1) by 8 g .c + 8^/ = 4560 '
and subtract (4) 9 x + 8^^ = 5040
(6) then __ ir = — 480
(7) or x = 480.
Substitute 480 for x in (1),
(8) 1/ = 570 — 480 = 90.
Hence, the number of pulleys of the first kind is 480 and of the
second is 90.
Solve the following problems involving two unknown numbers.
6. Find two numbers whose sum is 857142, and whose differ-
ence is 571428.
7. If A gave B $100, A would then have one-half as much
as B would have then; but if B gave A $100, B would have one-
third as much as A, How much has each?
8. A boy was asked his age and that of his sister, and replied:
* *If I were three years older, I should be three times as old as my
sister; but if she were two years older, she would be ^ as old as I
am." How old was each?
J225] SIMULTANEOUS LINEAR EQUATIONS 227
9. Find two numbers whose diflference is ^^ of their sum, and
3 less than | of the larger number.
10. A boy says to another: **Give me 5 (a) of your nuts; then I
shall have 3 times (n times) as many as 3'ou will have." The second
boy says: <'Give me 2 of your nuts; then I shall have 5 times
(/> times) as many as you." How many has each? (Solve in terms
of a, n, p.)
11. In Mecklenburg the longest day is 10 hours and 2 minutes
longer than the shortest night. How long are the longest day and
the shortest night?
12. The sum of two numbers is 350. If the first is divided by the
second, the quotient will be 8 and the remainder 8. What are the
numbers?
13. If the first of two numbers is divided by 8, the second by 3,
the sum of their quotients will be 310; if the first is divided by 3 and
the second by 8, the sum of their quotients will be 150. What are
the numbers?
14. What numbers satisfy the following conditions: add a to the
first, and the sum is m times as great as the second ; add h to the
second, and the sum is n times as great as the first?
15. A receives each year $2160 interest. If he had lent the
same capital at ^% higher, then he would have received each year
$240 more interest. How much capital did he have, and at what
interest was it invested?
16. A capital earns each year a dollars interest. At p % higher
it would earn only b dollars more interest. How large are the cap-
ital and the interest?
17. Find two numbers, such that ^ of the first and f of the second
together shall be equal to the excess of three times the first over the
second, and this excess equal to 11.
18. A receives each year from his capital $1172.50 interest.
He would receive $1200 interest, if his capital were $550 greater.
Find the capital and the rate of interest.
19. A has $7000 to lend at a certain rate per cent, and B has
$8500 to lend at another rate. Under these circumstances B's
yearly income would be $135 greater than A's, Both would have
the same annual income at the present rates if A should lend $11000
and B $9500. At what rate per cent has each lent his money?
228 CX)LLEGE ALGEBRA [S225
20. A certain capital earns each year a dollars interest. If this
capital were diminished by m dollars, it would earn only h dollars
interest Find the capital and the percentage.
21. A has two kinds of silver. $1.25 of the first kind and $2.00
of the second kind melted together give 13^ parts of pure silver;
$ . 75 of the first and $2. 50 of the second kind melted together give
13f parts of pure silver. How fine was each kind of silver?
22. A man has two quantities of gold of different purity. 71
pounds of the first kind and 190 pounds of the second kind give 800
parts of pure gold; 171 pounds of the first kind and 95 pounds of
the second kind give 900 parts of pure gold. How pure is each kind
of gold?
23. A miller expended $375 for wheat and rye. He paid $1,875
for a bushel of wheat, and $1.25 for a bushel of rye. Had he sold
4 weeks earlier, he would have gained $15.*25, because, at that time,
wheat was 2^ cents and rye 10 cents higher per bushel than now.
How many bushels were there of wheat and rye respectively?
24. What value do a dollar and a rouble have in marks, and in
what ratio does the dollar stand to the rouble, if 48 roubles plus
three marks are equal to 36 dollars, and 1 dollar and 1 rouble are
exactly 7^ marks?
25. The sum of two numbers is 15390, the first of which con-
tains one figure, the second, five figures. If the first is placed to the
left before the second, the number thus formed is 4 times as large
as the number which is formed by placing the first to the right,
behind the second. What are the numbers?
26. Two numbers have a given product. If the first were 8 less
and the second 25 larger, their product would be increased by 5000.
If the first were 12 greater and the second 25 less, their product
would be 4000 less. What are the two numbers?
27. A farmer brought eggs to market, and hoped to sell them at
a certain price. Had he sold the eggs for -^ cent each more than he
had hoped, then he would have realized his total price, if 12 of the
eggs had been broken on the way. But if he had been obliged to
sell the eggs at ^ cent each cheaper than he had thought, then he
would have needed 12 eggs more, in order to receive 3 cents more
than he had at first hoped. How many eggs did he have, and what
should each egg sell for?
J225] SIMULTANEOUS LINEAR EQUATIONS 229
28. The difference of the squares of two numbers is 840. If
each number were 3 larger, the difference of their squares would be
900. What are the numbers?
29. Two amounts of money, one of which is $1000 more than
the other, are lent at different rates — the second at ^% higher
than the first — and both amounts earn the same amount of interest.
If the first amount were lent at the rate at which the second was
lent, and the second at the rate of the first, then the first would
earn $95 more than the second. How large were the two capitals,
and at what rate was each lent?
30. A composition of lead and zinc, which weighs 149 pounds,
loses 18 pounds in water. How many pounds are there in each
metal, if a quantity of 11^ pounds of lead and one of 6f pounds of
zinc each loses 1 pound in water?
31. A composition of two metals loses p pounds in water. How
many pounds are there of each metal, if a pounds of the first loses
m pounds in water, h pounds of the second loses n pounds in water,
and the entire mass weighs q pounds?
32. There are two numbers, one of which contains two figures
and the other four. If the second is divided by the first, the quotient
is 204, with a remainder of 1 ; if a number is formed by writing the
first before, and to the left of, the second, this number is half as
large as the number which is found by writing the second number
before, and to the left of, the first. What are the numbers?
33. A boy made a cork belt in order that he might swim with
greater ease. The boy and the belt weigh 139 pounds, and the boy
is of exactly such weight that he can keep his head, which weighs
12 pounds, out of water, without its being raised above or lowered
into the water more than is necessary for the movement of his arms
and feet in swimming. How much did the boy and the cork belt
each weigh, if 120 pounds of the boy*s body immersed in the water
weighed 3 pounds and the specific weight of the cork belt is 0.24?
34. The fore wheel of a carriage makes six (a) revolutions more
than the hind wheel in going 120 (h) yards; if the circumference of
the fore wheel be increased by \{\) of its present size, and the cir-
cumference of the hind wheel by ^ (i,) of its present size, the six (a)
revolutions will be changed to four (c). Find the circumference of
each wheel in both cases.
CHAPTER VII
GENERAL SOLUTION OF A SYSTEM OF TWO EQUATIONS IN TWO
UNKNOWN QUANTITIES— SYMMETRY OF EQUATIONS— DIS-
CUSSION OF THE EQUATION— HOMOGENEOUS EQUA-
TIONS—THE CONDITION THAT TWO EQUA-
TIONS OF THE FIRST DEGREE IN ONE
UNKNOWN QUANTITY HAVE
A COMMON ROOT
In {§203, 205 (^), reference was made to the forms
- and ^ which may occur in the solution of an equation of the first
degree. The meaning of the forms when they occur in the solution
of simultaneous equations of the firet degree is here treated. First
review the results already obtained.
227. Any equation of the first degree in one unknown quantity
can be reduced to the form ax = b, from which x = -' When a ap-
' a ^
proaches 0, x becomes a quantity which may be as large as is desired.
In this case we have Ox = h, an equation which can not be satisfied by
any finite value of x because, as long as x is finite, xO = 0. But the
equation requires that Ox shall be finite and equal to 6. Therefore
a solution is impossible. Again, if a = 6 = 0, then ax = 6 takes
the form Ox = 0, which is indeterminate, since for any finite value of
X, Ox = 0. Therefore an infinite number of values of x would sat-
isfy the equation. In case h =0, and a =^ 0, then .x = 0 is a pos-
sible solution of the equation.
General Solution of a System op Two Equations in Two
Unknown Quantities
228. Two equations in two unknown quantities can always be put
In the form,
jr(l) ax + by = c
1(2) a'x + b'y = (f.
230
J229] SOLUTION OF LINEAR EQUATIONS 231
Solve (1) for x; then
(3) x=5---i^.
a
Sabstitate in (2); then
(4) a'£_ZL^ + 6'y = c'.
a
By reducing, a'c — a'hy -\- ah'y :=ac'
or {aJ/ — a*b).y = ac' — a'c.
System I is, therefore, equivalent to system II,
„!<„ x = £^
1(4) (a6'— a'6)y = ac'— a'c,
and the second equation involves but one unknown quantity.
Therefore the solution and the discussion of a system of two
equations of the first degree in two unknown quantities resolves
itself into the solution and the discussion of an equation of the
first degree in one unknovm quantity.
Suppose aU—a'h=^^. Equation (4) has the root
rtc' — a'c
y
ab' — a'b
ac —— a c
and. on putting for y in equation (3), it follows
ab* — a'b
_{ac^ — a^c)r
ab^—a^b __acb' — a'cb — abc'-{-a'cb _^a{b'c — c'6).
a a(ab' — a'b) a(ab' — a'b)
and system I is equivalent to system III,
cb'-^bc'
lll\
(5)
(6) y =
ab'— a'b
ac' — ca'
ab'—a'b
The Composition op the Formulae
229. The composition of these formulae is easy to exhibit.
The values of x and y have the same denominator, ab' — a'b.
The denominator is formed by taking the product of the coefl3cients
of the unknown quantities crosswise, first from left to right and then
from right to left, and then taking the diflference of the products.
The numerator of x is formed by substituting, in ab'—a'b, c and
c' for a and a', the coefl9cients of x in the two given equations.
The numerator of y is formed by substituting in a^'— a'b, c and
c' for 6 and 6', the coefficients of y in the two given equations.
232 COLLEGE ALGEBRA [8J230, 231
Example. — Solve the equations,
3x+5y=7
4x— 8y=12;
then ^ _ 7 X (-8) — 5 X 12 ^ -JIG ^ 29
3(— 8) — 5x4 —44 11
3x12-7x4 ^ J-8^ ^ _ ^
3 (-8) — 5x4 -44 11
Symmetry op the Equations
230. If in the given system of equations,
( (1) ax -^hy = c
1(2) a'x+h'y^c\
a is changed into or', and «' into a, 6 into ft', and h 'into fc, c into r',
and c' into c, the first equation will be transformed into the second
and the second into the first, and the system will not be changed.
Hence, if the same changes are made in formulae III, the values
found for x and y should be the same. This is exactly what hap-
pens. The expressions,
ah' — 6a', ch' — hc\ ac' — ca\
are transformed respectively into
a'b — ft'a, c'6 — 6'c, a'c — c^a,
and, consequently,
ch' — hc' c'b — h'c
X = Yf Tf becomes -^7 — -.
ah' — a'6 a'h — b'a
ac* — ca* , a'c — c'a
y = becomes — - :
a6' — a'b a b — ab*
Hence the values of x and y are not changed.
231. Suppose that in the given system
J ( (1) ax + by =1 c
1(2) a'x+b'y = c'
X is changed into y and y into Xy a into b and h into a, a' into b' and
b' into a'; there is formed a second S3'stem of equations,
jY ( (3) by-\-ax=c
1 (4) b'y + a'x = c',
which does not differ from the first except in the order of the terms
in the first members of the equations. The value of x found from
system I was
b'c — bc'
X =
b'c-
-he'
1h'-
-a'h
nc' -
ah' -
-ca'
-a'h
ab' — a'b
J232] SOLUTION OF LINEAR EQUATIONS 233
By operating in the same way on equations IV the value of y can
be found, and this value of y will be the result found by changing,
in the value of x, a into h and h into a, a' into h' and 6' into a'. This
determination gives
ca' — ac' ac' — a'c
^ = 6a' — h'a = a6' — a^'
which is the value of y found above.
Discussion
By a discussion in Algebra is meant the examination of the vari-
ous ' cases about which any question can be raised. The discussion
DOW arises of the principal circumstances presented by the solution of
equations (1) and (2) of the first degree in two unknown quantities.
232. Case when ab' — ba' is different from 0.— It is clear that
every system of values of x and y which satisfies the two equations
(I) and (2) of system I satisfies also the two equations (3) and (4) of
system II, which were derived as shown in J 228; but, these equations
(3) and (4) are satisfied in accordance with J212, 2, only by the values
of X and y given by the formulae (5) and (6) of system III; there-
fore the proposed equations have but one solution. To show that
these values satisfy the two given equations, substitute for x and y
the values given by formulae (5) and (6) of system III in the first
member ax + by of (1) and arrange the result with respect to c and
c'f thus:
ch' — c'6 , ac' — ca' ab^c — abc' + abc' — a'bc
^aV—ab''^ a'b — a'b ~~^ ab' — a'b
_c{ab' — a'b) 4^ (/(aft — oft) _
ab' — ab
Since the coefficient of c' is zero, and that of c is equal to the
denominator, the above quantity is equal to the second member c
and the equation is satisfied. The first member a'x-\-b'y of the
second equation becomes, in like manner,
, c7/ — c'b _,T,(ic' — ca'_ a*b' c — a'bc' -\- ab' c' — a'b' c
" ab'—a'b^ ab' —a'b ^' ab'—a'b
cja'b' — a'b') 4- c'(nb' - a'b) _ ,
~" ab' — a'b~ ~^'
The coefficient of c is zero and the coefficient of c' is equal to the
denominator, and this quantity is equal to the second member c' of
equation (2) ; and the equation is therefore satisfied.
234 COLLEGE ALGEBRA [2233
Theorem. When the denominator ah' — a'6 i» different from zero,
the two given equations have one solution and only one.
Example.
f 5ac— 3y = 9
(7a; + lly = 43.
Here a = 5, 6 = —3, c = 9, a' = 7, 6' = 11, c'= 43.
Since a6' — a'6z3 5 • 11—7 • (— 3) =76=^=0, the two equations
have the solutions x = 3 and y = 2 given by the formulae (5) and
(6).
Case when ab' — a'b = 0. Suppose now that the denom-
inator ah' — a'h =0 and that one of the numerators, for example,
ch* ^hc' —/- 0. If the two given equations are satisfied bj' the values
of X and y, then the equation,
x{ah' — a'h) = ch' — hc',
is satisfied, but, since ah' — a'h = 0, a; = 00 ; therefore the given
system has no solution and the equations are said to be incom-
patihle.
Example. Apply the theory to equations (1) and (5) of 2209:
(1) 7x — 2y = lS
(5) 14x — 4y = 25.
Here a = 7, 6 = —2, a' = 14, h' = -4;
hence ah' — a'6 = 7 (— 4) — 14 (— 2) = — 2S + 28 = 0.
Hence the equations are incompatible.
But it is easy to show from another point of view that the equa-
tions are impossible. Since ch' — hc'z^=0, then one at least of the
coefl3cients, h and 6', is different from 0. Suppose that h is different
from 0; hence from the condition ah' —a'h = 0, it follows that
a' = ^ ; this value being substituted in equation (2), the equation
becomes
ah'
X 4- h'y = c' or h'(ax 4- hy) = c'h.
0
If the coefficient h' is zero, the first member is zero and the
second member c'h is diffei-ent from zero, since ch' — c'h =^0 by
hypothesis. This is impossible. If 6' is not zero it is possible to
write
h'{ax + hy) = c'h
J234] SOLUTION OF LINEAR EQUATIONS 235
in the form
ax + hy = - ~ ,
but the first member of eqaation (1) is also ux-\-hyj which would
lead to the conclusion, c = -^ or
0
h'c-hc' = 0,
which is not allowed. Hence the equations are incompatible.
Put ab' — a'6 = 0 in formulae (5) and (6) of III; then
cf/ - c'h
(5) ^= i7 ,7'
(6) y= 1/ TT-
ab — a'b
It follows from (5), when ch' ^ c'b =/= 0 and ab' — a'b = 0, that
X = CO]
and from (6), since a' = ^> that
0
, arf/
' y^ab'—a'b b{ab'—a'b) ^
in case a =?^ 0. But, if a = 0, since the binomial ab' — a'b = 0,
it follows that
0
^=0
which is indeterminate.
234. The case when ab' — a'b = 0, cb' — be' = 0, ac' — ca'=0.
It has been seen that the second equation becomes
b'{ax + by) z= be'
if ab' — a'6 = 0.
Suppose that at least one of the four coefficients a, 6, a', 6', is
not zero, say 6 ; consider the equation in the form
ax+by = — ,
be/
then, from the relation cb' — 6 c' = 0, c = ~r; a^^^l equation (2)
becomes identical with equation (1). If the coefficient 6' = 0, then
also c' = 0, since cb' — be' = 0; and 6 =^ 0, and equation
6' (ax + 6^) = be'
becomes an identity
0 = 0.
236
COLLEGE ALGEBRA
[JI235, 236
The system is then reduced to the first equation,
ax -\- hy =: c,
which can be satisfied by an infinite number of values of x and y,
K the four coefficients, a, fc, a', 6', were zero, the given equa-
tions would become 0 = c, 0 = c', which would be impossible unless
c and c' were zero; and in that case, the equations would become
0 = 0, an equation which is absolutely indeterminate. Any values
whatever might be assigned to x and y and the equations would be
satisfied.
236. Resume of the Discussion. — For the sake of brevity put
D = ah' —a% Nj, = cl' -^c'l, Ny = ac' —a'c.
The following table gives a resume of the discussion of the
system.
D
not both zero
D :=f=. 0. One unique solution \ x~. -~y y =
Z>=:0
J incompatible equations
both zero
1
a, a', 6, 6' ) C indeterminateness:
j I one arbitrary
not all zero
a, a', 6, 6', -
all zero
b^thzero{^^^^°^P*^^^^^^3^
c and c'
both zero
indeterminateness :
two arbitrary
Homogeneous Equations
236. When the second members of system I become zero, i. e.,
c and c' equal zero, the system becomes
ax-\-hy = 0
a'x + b'y = 0.
Since all the terms in both equations are of the first degree in x
and y, they are called homogeneous equations. If ah' — a*b is not
zero, then they have the solution, x = 0, y = 0, but if ah' — a'6 = 0,
then they have an infinity of solutions, since by (5) and (6), III,
ch' — c'h 0
X =
and
ah'
-a'b
0
y =
ac' .
^a'c _
_0
ah'
-a'b
0
Ox
= 0,
yO =
0.
since c = c' = 0,
S237] SOLUTION OF LINEAR EQUATIONS 237
Hence, the two equations are satisfied by a single infinite set of
values of x and y.
If a and a* are zero, there is still an infinity of solutions, since
the values of x and y are indeterminate. From the equations,
hy = 0, h*y = 0, i =5^= 0, h' :=f=. 0, y can tjike only the value zero,
but X is arbitrary ; if 6 and V are zero, and a and a' are not both
zero, then x can take only the value zero, but y is arbitrary.
The Condition that Two Equations op the First Degree
HAVE A Common Root
237. Let the two equations in one unknown quantity be
ax+6 = 0
a'x + 6' = 0.
Suppose that one at least, say a, of the coefficients is different
from zero. Then the first equation has the unique root,
h
a
In order that the two equations may have a common root it is
necessary and sufiScient that the second equation have the root »
that is,
- a'^ + ^' = 0,
a
or since a =^ 0,
aV — a'6 = 0.
If both coefficients a and a* are zero at the same time, then
ah' — a'h = 0; but, if h or 6' is not zero, then one of the equations
could not have a finite root; and therefore, in this case, the given
equations could not have a finite root in common.
CHAPTER VIII
THB GENERAL SOLUTION OF A SYSTEM OF THREE EQUATIONS IN
THREE UNKNOWN QUANTITIES— THE SOLUTION OF THREE
OR MORE EQUATIONS IN AS MANY UNKNOWN
QUANTITIES
238. The first step in solving a system of two equations in two
unknown quantities, is the elimination of one of the unknown quan-
tities; this elimination results in an equation in one unknown quan-
tity which can at once be solved and which, with one of the given
equations, forms a system equivalent to the given S3'stem of equations.
The solution of a system of three equations in three unknown
quantities is an extension of the principle stated above.
First combine any two, say the first and the second, equations of
the given system and eliminate one of the unknown quantities; then
combine say, the second and the third equations, to eliminate the
same unknown quantity. This gives two equations in two unknown
quantities, which can be solved in the usual way, and the third
unknown quantity can be found by substituting the two values
found in any one of the three given equations.
Example 1.
Multiply (2) by 2,
Add (4) and (1),
Subtract (3) from (4),
Solve the equations,
(1) hx + 2y-4z=lb
(2) bx-3t/+2z = 2S
(3) 3y-f-4 2— iB = 24.
(1) 3x + 2y-4z =lb,
(4) lOx— 6y+42 = 56.
(5) 13x — 4y = 71.
(4) 10 a; — 6^^ + 42 =56
(3) 3y-f-42-x = 24
(6) llx— 9j/=32.
{2381 SOLUTION OF LINEAR EQUATIONS 239
To eliminate y from (5) and (6), multiply (5) by 9 and (6) by 4 and
obtain
(7) 117 a: - 36 y = 639
(8) 44a:-36y = 128
and after snbtracting (8) from (7),
(9) 73x = 511
(10) jr = 7.
Substitate 7 for x in (6), then
(11) 77-9y = 32
hence 9 y = 45
(12) y = 5.
Substitute x = 7, y = 5 in (1), then
(13) 21 + 10-4^ = 15.
After collecting, 42 = 16
(14) ^ 2 = 4.
Hence the given system has the solution,
(15) X = 7, y = 5, 2 = 4,
and this solution only. The system of equations (15) is equivalent
to the given system I.
Example 2. Solve the equations,
(1) -+-=1
X y
1 1
(2) i + A=2
X z
(3) '+'=1.
y z 2
To eliminate x from (l)and (2), subtract (2) from (1),
(4) l-i=-l.
y 2
This equation contains y and z ; so also does (3).
By adding (4) to (3), ~ = \
y ^
(5) y = 4.
Substitute 4 for y in (1),
(6) \ + \ = ^
hence - = .
X 4
240
(7)
COLLEGE ALGEBRA
4
LS238
Substitute y = 4 in (3),
hence
(8)
1,1^3
4 "^ a 2
1^31 ^5
2 2 4 4
4
Z =z - '
Therefore the only solution of the given system is
3
y = 4
4
Example 3. Solve the equations,
f(l) ax — fey = 0
(2) * y^z = ^C
(3) az + cx=:a*+ c(a+ b).
To eliminate y from (1),' solve (2) for y, i. e., y = « — c and sub-
stitute z — c for y in (1); thus,
ax — b {z — c) = 0
or (4) ax — bz z= — be.
Multiply (3) by 6, (4) by a,
(5) abz -\-bcx=i a^h ~\~ abc -{- b*c
(6) a^x — abz := — abc.
By adding (6) and (5),
(7) bcx -\-a^x=b (a« + be)
or {be -\- a^) X = b(a^ + be)
(8) . = 'S^±^l=h.
By substituting b for x in (1),
By dividing by 6,
(9)
be-\-a'^
ab — by = 0.
a — y = 0
y = a.
Substitute a for y in (2) ; then
(10)
a — 2 = — c
z = a-\- c
(X = b
y = a
z =1 a -\- e.
8239] SOLUTION OF LINEAR EQUATIONS 241
Rule. — Hence, to solve three equations in three unknoum numbers,
eliminate one of the unknoion numbers from any two of the equations,
and eliminate the same number from any one of these and the third
equation, The^e two steps give two equations in two unknown numbers.
Solve these two equations by the usual method; then substitute their
values in the simplest of the three given equations; the third unknown
number will be found by solving the resulting equation.
239. To solve a system of four equations in four unknown quan-
tities, solve one of the equations for one of the unknown numbers
and substitute this value in each of the other three equations; we
then have, instead of the given system, an equivalent system of four
equations, three of which contain three unknown quantities. These
three equations can be solved by the method already explained
in the preceding paragraph.
Example 4. Solve the equations,
'(1) 2x^3y + 2z = 13
(2) u + 4y + 2z = 23
(3) 3x + 4u — 2z = 35
(4) 4x — 5y + 3M = 34.
Solve (1) for x, (5) x = ^^-^^y-^^ .
The substitution of this value of x in each of the given equations,
gives the equivalent system II:
f (6) M + 4y + 22 = 23
(7) 3(^^ + 'Y~^^)+4u — 2g =:35or8w + 9y — 10a =31
II X (g) 4^i3±^L=22^_5y + 3,,=:34or2y + 6M-8a = 16
13 + 3y — 22
and (5) x = — -'— ^
Equations (6), (7), and (8) may now be solved by the method
of 2239.
Multiply (6) by 5, (9) 5 u + 20 y + 10 a = 115
(7) 8M+9y— 102 = 31.
Add (7) and (9), (10) 13 w + 29^^ = 146.
Multiply (6) by 4, (11) 4 u + 16y + 8 2 = 92
(8) 6u + 23^ — 82 = 16.
Add (8) and (11), (12) 10 u + 18y = 108
or (13) 9i^ + 5w = 54.
Multiply (10) by 5, (14) 65i* + 145y = 730
and (13) by 13, (15) 117y + 65i* = 702.
242
COLLEGE ALGEBRA
[S240
Subtract (15) from (14), 28y = 28
(16) y = 1.
Substitute y = 1 in (13), 9 + 5 « = 54
(17) « = 9.
Substitute 9 for u and 1 for y in (6) and get 9 + 4 + 2 a? = 23
(18) 2 = 5.
Substitute 1 for y and 5 for ;: in (1) and get 2 a? — 3 -|- 10 = 13
(19) X = 3.
The work can often be shortened by introducing some simple device.
Example 5. Solve the system,
(1) x + y + z =a
(2) x + y + u = b
(3) x + z -\-u = c
(4) y+ z + u = d,
(5) z — M = a — b,
(6) y-z = b^c. '
(7) y + 2z=a—b + d.
3z = a^2b + c + d
(8) ^^"-^^+c + d
o
Subtract (2) from (1),
Subtract (3) from (2),
Add (4) and (5),
Subtract (6) from (7),
Substitute from (8) in (6), (9)
a + b — 2c + d
(10) x=:?L±M-^jr^.
3
This example can be readily solved also by finding the sum of
(1), (2), (3), and (4), then dividing by 3 and subtracting from the
resulting equation each of the equations (1), (2), (3), and (4).
NuMBEE OF Solutions op a Svstbm of n Linear Equations
240. The examples which have been solved in the preceding
section illustrate the following principles:
{!,) A system of n independent and compatible linear eqttations in
n unlcnown numbers has onCy and only one^ determinate solution.
Deduce from the first equation the value of the first unknown
number x, as though the others were known, and substitute this
value for x in each of the n — 1 other equations ; thus a system
equivalent to the first is obtained, composed, first, of one equation in
n unknown numbers, second, of m — 1 equations in w — 1 unknown
numbers.
1240] SOLUTION OF LINEAR EQUATIONS 243
From the first of the equations in n— 1 unknown quantities
deduce the value of the second unknown number y^ as if the others
were known, and substitute this value in the place of y in each of
the other n — 2 equations; thus a third system is obtained equiva-
lent to the first, and composed, first, of one equation in n unknown
quantities, second, of one equation in n — 1 unknown numbers, third,
of n — 2 equations in n — 2 unknown numbers.
By continuing in this manner, a system equivalent to the first is
obtained, in which the last equation contains but one unknown, the
last but one contains two unknown, and the third preceding the last
contains three unknown quantities, and finally the first contains n
unknown numbers. From the last equation is obtained the value of
one unknown ; substitute this value in place of this unknown in the
equation preceding the last; the solution of this equation gives the
value of a second unknown. By working backwards in this way,
step by step, the values of all the unknown numbers are successively
calculated. Thus but one unique system of values of the unknown
numbers is derived.
(2.) A system, of n independent equations^ in more than n unknown
numbers, Jias an infinite number of solutions.
As is shown above, by each elimination of an unknown number
a set of equations is obtained containing one less unknown number
than before. By n — 1 successive eliminations, » — 1 unknown
numbers will be eliminated, and the last elimination will result
in a single equation with at least two unknown numbers, since
there were originally at least n-]-l unknown numbers involved.
But an equation in two unknown numbers can have an infinite
number of solutions; therefore the given system can have an infinite
number of solutions.
{S.) A system of n independent linear equations in less than n
unknown numbers can not have a determinate solution.
Since there are more equations than unknown numbers, a set
of equations can be selected consisting of the same number of equa-
tions as there are unknown numbers involved. By (7) this system
will have one unique solution.
But since the other equations are independent of the equations
selected, i. e., express different relations between the unknown
numbers, they will not in general be satisfied by this solution.
Hence the given system can not be satisfied by any one definite
set of values of the unknown numbers
244 COLLEGE ALGEBRA [?240
EXBBOISE XliV
Solve the following systems of equations:
x + y = 37 /2x+3y = 12
5.
.)x+« = 25 2. ]3x+22 = ll
(y + z =22. L3y + 4z=lQ
rx + y + z=36 (1*35+1^^ = 10
.} Ax = 3y 4. J 2f a; + 2f 2 =20
( 2x = 3z, (3i:y + 3|2 = 30.
.2ix + 3iy + ^z=U rx = 2iy-^6
) Hx = 2iy 6. ]y = 3i2-l
( 3Jy = 2ij5. (.z = Hx-^S,
rx + y—z = n ry+z—x=a
7. Jx+s-~y = 13 S.}z + x — y=b
Ly + z—x=z 7. (x+y — 3=c.
j^(a5 + y + «=99 ^^ (x + y + z=m
' \ X : y : z = b : 3 : 1. \ x : y : z = a : b : c ,
/-x-|-y-f'S=26 /-ax + 6y-f'Ca = r
11. ^ X : 2 = 11 : 7 12. ) x : y = m: n
(^ y:2 = 14:9. C y:2;=2>:5r.
( ac + y + 2 =9 r
) x + 2y + 4z = lb 14.]l
( a; + 3^ + 92 = 23. (.1
f7x+Qy + 7z= 100 /■ 3 X -|- ::y -f iJ2 = iiu
) X— 2y+2= 0 16. ]5x+.v — 4^= 0
( 3x + y — 2z = 0. (2x — 3i^+ 2= 0.
/-X+ y+ a= 9 r x+2y+ 35; = 32
17. ]x+ 2y + 32 =14 18. J2x+3y+ 2 = 42
(a; + 3y+ 6z =20. ( 3x + y+ 2 z = 40,
rx+y+2z = 34 r3x+3y+2 = 17
19.) x+2y+ 2 = 33 20.J3x+ y + 32 = 15
(2x+ y+ 2 = 32. ( x + 33^ + 32 = 13.
/-5x— y+32=a /'7x+lly+2=-
21. J5y— 2 + 3x=6 22. )7y+ll2 + x =
lbz—x+3y=c, (72 + llx + y =
: + y + 2= 9 r x + y + z=: 3
13. -I x + 2t/ + 42 = 15 14.j2c;;+4y+ 8^=13
a; + 3y + 92 = 23. (3x + 9y + 27 2 = 34.
7x+6y + 7 2 = 100 f3x + 2y + 32 = 110
15 "
c.
r x + 2y+ 32 = 15.4 r x+2y — 2= 4.6
23. ]3x + 5y+ 72=37.4 24. j .y+22— x = 10.1
(5x+8y+ II2 =59.4. ( z + 2x — y= 5.7.
J240]
SOLUTION OF LINEAR EQUATIONS
245
26.
25.
r 0.2a;+ 0.3y-
) 0.3x+ 0.4y-
(o.4x + 0.5y
0,2x+ 0.3y + 0.42 =29
r+ 0.5 2 = 38
+ 0.7 2 =51.
rx + 2y — 0.7
]3x + 0.2y-
(0.9a; + 7y— 2
-x + 2y — 0.72 = 21^
2 = 24
/— 22 = 27.
27. }3jy = aj + 2 + 12
(4j2 = aj + y + 15.
28.
30.
32.
36.
i^ + Jy + 1« =
1^ + iy + i^ =
i^ + y + \^ =
^+i = 2
36^
27
18.
y + 2
2 + 1
2+3
Lx + 1
= 4
1
2*
y 2
X z
X 1/
2b
34. ^
x?/
X2
X + 2
yg
.y + ^
29.
31.
33.
35.
(U + 2)(
]U-2)<
Uy + D'
(x + 2)(2y + l)
•(32 + 1)
l)(^+2) =
2Jx+3Jy+4j2 =140
3Jx+4Jy+5i2 = 175
2fx+3|3/+4|2 = 157.
1J — z
^+i = 9
X — i/
y_+i__
x + 5
1.
^-? = i
X y
X 2
•r.'/
45^ — 3;e
= 20
Sa
= 15
»2
4y — 52
= 12.
r(2a;_l)(y+l) :
37. ]u + 4){z+l) =
({y-2)(z+3) =
f(x+l)(14y-3)
38. ](4u;-l)(z + l) =
(9(y + 3)(«+2) =
= (2x+7)5^
= (x + 3){3«-l)
(y + 3)(2 + l).
= 2(x+l)(y— 1)
(x + 2)(2 + 2)
(y -!)(« + 1).
= (7x + l)(2y— 3)
:2(a! + l)(2.. -1)
(Zy-6){Zz-l).
246
39.
40.
41.
42.
43.
44.
45.
46.
47
COLLEGE ALGEBRA
\(2x + y):{3x+z):(i/ + z) = 1:2:3
t21a; + 31y + 422; = 115.
[ {x^2y):(2x — 3z):(2i/ + 3z) = 1 :3:5
l21x + 31y + 4l2 = 135.
^^(y + z):y{x-j-z) :z{x + y) = a:b:c
Ix y z
r ax -\- by — cz = 2 ab
^. by -\- cz — ax = 2bc
icz -\- ax — by = 2ac.
r (a — b)(x -\- c) — ay -\-bz = 0
I {c — a){y + b)-cz + ax = 0
LX+y + 2 = 2(a + 6 + c).
- (a + 6) X + (a — ft) 2 = 2bc
\(h+c)y + {b-c)x = 2ac
^ (c -f- a) 2 + (c — a) 3^ = 2ab.
' X -\- y -{- z = a-|-fe-|- c
^ hx+ cy + az = a^+b^+ c*
^ ex -^ ay -{- bz = a' + 6^ -f" ^'•
•a;-f'y + ^ =a + ft-|-c
ax -{- by -\- cz = ab -\- ac -\- be
. (6 — c) X + (c — a)y + (a — ft) a =0.
' (a -\- b) X -{- (b -{- c) y -\- (a -{- c) z = aft -f- «c + ftc
(a -f- <^) a; + (a + ft) y 4" (^ 4" ^) ^ = o,b -\- ac -{- be
^(b+e)x+(a + c)y+(a+b)z = a«+ft«-f-c«.
[{240
48.
50.
x+f--=a
b c
y H =ft
c a
2 H f = C.
a ft
49.
ft + c
y
c -\- a
= a+ft
-— = ft+c 51.
— ft
3 iC
c4-«.
b -{- c c -\- a
b-a
+
ft + c a+ft
+
c -f- a
x
ft + c
x
a+ ft
= c-ft.
X
.y
a — ft
2
= 0
c — a a -\- b
= 0
ft+c c — a a-j-ft
= 2a.
1240]
52.
54.
56.
58.
60.
61.
62.
64.
!
SOLUTION OF LINEAR EQUATIONS
X : y : z : u =z a : b : c : d
mx -\- ny -^ pz -^ qu = r.
247
50 {x:y:z:u=l:2:3:^
(9x + 7y + 32 + 2i* =
x:y = 2:1
x:z = 3:1
y : t« = 3 : 1
u + y + z = a
z + w + ^ = ^
u+ x + y = c
^+ y + ^ = «^-
200.
55. \
57. i
X'\-y'\-z-\-u = 144
x-\-2y+2z + 2u = 267
"j x+2y+3z + 3u = 359
j^x+2y+3z + 4M = 410.
59.
x+2y = 5
y+2z =S
s + 2m = 11
14+ 2x = 6.
x+ 3y — 2 = 1
y _|_ 32 — 14 = 4
z + 3i« — X = 11
M+3x — y = 2.
X — 2y+32 — u = 5
y — 2z + 3w — X = 0
z — 2w+ 3x — y = 0
u — 2x + 3y — z =5.
rx+y+2 + u = 60
\x+2y+3z + 4u = 100
] X + 3y + 6 z + 10 It = 150
! x+ 4y + lOz + 20m = 210.
J^ + !/ + z + u = 1
2x + 4y +Sz + 16m = 5
3x+9y + 27 z + 81m = 15
4x + 16y + 64« + 256m = 35.
2lx^iy+2z =4
l|x — IJy + 3m = 1
2x — 3j2 + M = 2
1^ + iy + i^ + i^ = 53
i^ + U + 4^ + l*^ = 5^
i^+iy+i«+i'* = 56
63. ^
11x4" 9y+ 2 — M = a
lly + dz + u — x = b
llz -{- 9u + X — y= c
11m+ 9x+y— z = d.
248
COLLEGE ALGEBRA
[S240
r ax -\-by -{- cz = d
65. ) a^t + bj/ + Cj2 = d^
07.
a a — r a — 8
- + - = a
*+ y + * =1
68. ,
X 2
?+ » + ^ -1.
K. c c — r c —8
'2X + 3.V _„,
70. -
.T -1- «
5(x— z) *
9^12^15^
lOx-32 „.
L 4X-22 -^''*
71. J
H h - = wi'
X y 3
1 1 = m"
X y 2
- - H 1 = 7/l'".
X 1J Z
{ X : y : z =. a : b :c
7.). ] ^
( ?iix '\' -ny '\' pz :=z 8.
73. J
f
77.
I ^f -|~ /> ^ — c c -\- a
j^ .V_ , J
— b b — c <- — ft
X 3y+z-^"
4x
74. -^
6x y 2 *•
X +3y = 19
y+ 32= 8
2+ 3u= 7
u+ 3t? = 11
t? + 3.r = 15.
f x+ 2y — 2 =12
^ + 2 2 — « = 10
76. ■] 2 + 2 tt — t' = 8
a-\- 2v — x= 1
y+2x — y= 9.
= 2c
= 2a
a — b
= 2a-
c c -\- a
2 c.
J24UJ
SOLUTION OF LINEAR EQUATIONS
249
78. J
» + y = «
y -\. z = h
z -\- u = c
n -\- V = d
V -\- X = e.
79.
X -\- y — z =z a
y -\- z — i£ = h
2-1-1/ — V = c
u -\- V — X •=(!
V -\- X — y = e.
X — y -\- z z=z a
y — z -\- u :=ih
80. . z —u -\- V = c
n — V -{- X :=z d
V — x + y r=e.
2^ 8 ^5 ^^
y + z + w = 248.
84. J Jc + y + 2 + < + M? = c
•«^ + y + w + < + M7 = rZ
X + Z + M + « + «? = €
2/ + Z + M + <+«? = /.
81.
83.
85.
.X + y + z — M. = a
y + 24-7/. — V =z h
Z -\- U -{- V — X =: C
u -{- V -\- X — y = d
V+X+y— Z=:z€.
xy
cz + fx
^ + y + * — ^* — V =. a
y -\- z -\- u — V — X =. h
z -\- ti '\- V — X — y ^= c
a '\- V -{- X — y — z z=z d
v+ X -{- y — z — M =e.
I x + y+z+it=:a
X -\- y — z — u =^ e
x — y—z-\-ii— I.
2^4^ 5
87.
88.
4^2^5
3^5^2^6
5^3^6^2
8;). -'
^1+ 3'j = "i
*, + y, = 3
^, + Xj = e
y, + «^, = io
''3 + y» = ^'^
•'"1+ ^a+ *3 =^1 + ^8+ Vz-
250
COLLEGE ALGEBRA
[mo
90. ^
01.
93.
94.
92. J
^+ y+ z + u = 1
16x+ Sy+ 4z + 2u = 9
81x + 27y+ 92 + 3ji*= 36
256x+64y + 16z + 4u = 100.
2x — y — z-\- 2u — t7 = 3a
2y — z—u+2v — x = 3h
2z — u — t?+2ic — y = 3c
2w — r — a;+2y — z = 3rf
2i? — a; — y -^ 2z — « = 3c.
' V — 2x4-31* — 2y+ a? = a
X — 2y + 3r— 2« + M=6
y — 22+3x— 2M-ft;=: c
a_2M+3y— 2u-fx= <i
M~2i?+32J — 2x+y=: c.
x+y+^ + ^+^= 15
x+ 2y + 4« + 8M-f 16v = 57
a; + 3y + 9 5; + 27m+ 81 v = 179
fl: + 4y + 16 2J + 64 m +256 17 =453
X + 5 y + 25 2 + 125 1* + 625 r = 975.
' X + ay + a*z + ahi + o* = 0
X + fey + [,«2 + fe3,j + fe* = 0
X -{- cy -{- c^z -\- chi + c* = 0
X + <fy + c?«2+ c?»« + c/* = 0.
95. Show that the four eqaations
a c \ hj
a;- __ « __ 1 /| _ 3/\
a c c/'\ I)
can not be satisfied by a system of values assigned to x, y, and z, if </ is
different from d\
96. Demonstrate that the equations
X .z
a c
X
a
a c
c u\ h)
are satisfied by the same values of Xy y, and z, and determine those valoes.
«241, 242] SOLUnON OF LINEAR EQUATIONS 251
Problems Involving Three or More Unknown Quantities
241. In solving problems which involve two or more unknown
qaantities, the pupil should first discover from the statement of the
problem the number of quantities which are to be determined. When
he has learned this fact, he must then obtain from the conditions of
the problem, as many equations connecting the unknown quantities
as there are unknown quantities to be determined. The unknown
quantities can now be found by solving the equations (2238).
242. Problem I. Find the numbers which are in the ratio
2 : 3:4, and whose sum is 999.
Let ac, y, 2 be respectively the first, second, and third of the
required numbers. Hence, according to the first condition of the
problem
X :y : 2 = 2 : 3 : 4
i. e., (1) 5 = ^
y 3
and (2) ^- = ?.
z 4
According to the second condition of the problem
(3) x + y + z = 999.
From (1) X = -^ I and from (2) 2 = 7^; substituting these values
o o
of X and 2 in (3),
(4) ?l+y + ^ = 999
or 33^ = 999;
y = 333, the second number,
X = ^ = 222, the first number,
3
and z = S. = 444, the third number.
3
Verificatian: 333 + 222 + 444 = 999.
Problem II. A lent his money at 4 %, 5 at 5%, C at 6 %.
How much money had each, if A and B received together $1592 in-
terest, B and C together $1766 interest, A and C together $1638
interest?
252 COLLEGE ALGEBRA [1242
Let «, y, and z be respectively the capital of Ay B^ and (7. Then
the interest which each received will be respectively the following :
4
A' 8 interest —- . u; = $. 04 x
5
B"s interest --- • v = $.05y
Cs interest -— - • z = $. 06 2
Hence, according to the conditions of the problem,
(1) .04x + .05y = 1592 '
(2) .05 y + .06 z = 1766
(3) .04x + .062 = 1638.
Subtracting (2) from (1)
(4) .04 .r-. 06 2 = —174.
Adding (3) and (4)
(5) .08^ = 1464
x=z $18300, .4'* capital
Substituting the value of j in (1)
.05y = 1592 — 732 = 860
y = $17200, 5'« capital.
Substituting the same in (3)
z = $15100, C's capital.
Problem III. A miner has three ingots composed of gold, silver,
and copper, which contain respectively:
the first, 50 gr. of gold, 60 gr. of silver, 80 gr. of copper;
the second, 30 gr. of gold, 50 gr. of silver, 70 gr. of copper;
the third, 35 gr. of gold, 65 gr. of silver, 90 gr. of copper.
What weight should be taken from each ingot in order to form an
ingot containing 79 gr. of gold, 118 gr. of silver, and 162 gr. of
copper?
Let X, y, z be respectively the weights taken from each of the
three ingots.
The first ingot contains 50 + 60 + 80, or 190 gr., of which 50 gr.
is gold, 60 gr. silver, and 80 gr. copper; therefore in the composi-
2242] SOLUTION OF LIXEAR EQUATIONS 253
iton of the ingot -^W or ^^ is gold, j\% or -^\ is silver and ^^\ or ^^
is copper. Therefore, if x gr. be taken from this ingot one takes
Y§ ^' ^^ g^^^' 1^ S^- ^^ ^^^v^^' Ysi ^' ^^ ^^PP^^-
Similarly, the second ingot contains 30 + 50 + 70 or 150 gr., of
which 30 is gold, 50 silver, and 70 copper; on taking y gr. from
this ingot, one takes
15 ^^' ^^ ^^^^' 15 ^^' ^^ ^^*^^^' 15 ^^' ^^ ^^PP^^-
Finally, the third ingot contains 190 gr., of which 35 gr. is gold, 65
silver, and 90 copper; in taking z gr. of this ingot, one takes
y gr. of gold, -^ gr. of silver, '^^^ gr. of copper.
And, since the ingot should contain 79 gr. of gold, 118 gr. of silver,
and 162 gr. of copper, one has to determine the x, y, z of the three
equations
(3) ^ + !^ + L^ ^ 162
/ ^ 19 ^ 15 ^ 19
which after clearing of fractions become
50x + 38y + 35z = 15010
36.r + 38i/ + 39^ = 13452
120j- + I33f/ + 1352 = 46170.
On eliminating y from the first two equations one has
(4) 7x-^2z= 779.
On elimmating y from the last two equations one has
(5) 228 UJ + 57 z =34656.
On solving equations (4) and (5) for x and z, it follows that
X z= 133 and z = 76.
Now substituting these values for x and z in (I)
iV^^^^+^^+^-^^ = ^^
y = 150.
4. Three cities have together 532,000 inhabitants. The first
and second have together 206,000, the second and third together
200,000 inhabitants. How many inhabitants has each?
254 CX)LLEGE ALGEBRA ti242
5. The different sums that can be formed from three given
numbers by taking two at a time in all possible ways are respectiyely
a, 6, c. Find the nambers.
6. From the sum of every two of three given numbers the
remidning number is subtracted, and the numbers so obtained are
a, h, c. Find the given numbers.
7. Three numbers whose sum is 1332 are to each other as
3:4:5. Find the numbers.
8. A father said: <'My age now is twice the sum of the ages of
my sons Otto and Max. Two years ago I was four times as old as
Otto, and four years ago I was six times as old as Max." What
was the age of the father and his two sons?
9. A grocer pays $2. 50 for 7 lbs. of coffee and 5 lbs. of sugar,
$1.50 for 3 lbs. of coffee and 10 lbs. of rice, and $1.50 for 7 lbs. of
sugar and 6 lbs. of rice. Find the price per pound paid for each
article.
10. A has lent money at 2 per cent interest, ^ at 2 per cent,
and (7 at 3 per cent A and B together receive $1592 interest, B
and C receive $1766, Cand A receive $1638. Find the number of
dollars each has lent out
11. Three towns form a triangle ABC. The distance from A
travelling through B to C along the triangle is 82 miles; the distance
similarly from B, travelling through (7 to J. is 97 miles ; and from
Gy travelling through A to B, 89 miles. How far are A, J?, and C
from each other?
12. Divide the number 96 into three parts, such that the first
divided by the second gives 2 with a remainder of 3, and the second
divided by the third gives 4 with 5 as a remainder.
13. Find five numbers such that the sum of each and four times
the sum of the remaining numbers gives respectively, 49, 43, 55, 61,
64 as a result
14. A man has seven baskets of apples. From the first basket
he puts into each of the other baskets as many apples as are con-
tained in them; then trom the second he puts into each of the
other baskets as many as they then contain, and so on, to the last
basket, when he finds that each basket contains 128 apples. Find
the number of apples in each basket before the distribution.
I242J SOLUTION OF LINEAR EQUATIONS 255
15. A and B can build a wall in 12 days, B and Ccan do the
same work in 20 days, A and C can do it in 15 days. How long
will it take (1) each one alone, (2) all three together, to build the
wall?
16. A miner, has three ingots composed of gold, silver, and
copper; the first ingot contains 2 kg. of gold, 3 kg. of silver, and
4 kg. of copper; the second contains 3. kg. of gold, 4 kg. of silver,
and 5 kg. of copper; the third contains 4 kg. of gold, 3 kg. of silver,
and 5 kg. of copper. How many kilogrammes is it necessary to
take from each ingot in order to make a fourth ingot which contains
9 kg. of gold, 10 kg. of silver, and 14 kg. of copper?
17. A number is composed of four figures whose sum is 21 ; the
figure in thousands' place is one-half the sum of the other three
figures; the figure in units' place is one-half the figure in tens' place;
finally, if 3906 be subtracted from the number, the remainder is the
required number reversed. What is the number?
18. A merchant bought wheat at the rate of $2.40 ioT 4 bushels,
com at the rate of $1.60 for 7 bushels, and barley at the rate of
$1.10 for 3 bushels. He spent $546.90; the cost of the wheat
exceeded that of the com by $80; the cost of the com, that of the
cost of the barley by $85.10. How many bushels of wheat, com,
and barley did he buy?
19. In one hour 150 persons enter a theater at the first door,
250 at the second door, and 400 at the third ; and the receipts were
$1625. During the next hour 120 persons enter the first door, 210
the second, and 324 the third; and the receipts were $1329. During
the third hour 135 pei-sons entered the first door, 280 the second,
and 366 the third; and the receipts were $1606. What is the price
of seats at the first, second, and third doors?
CHAPTER IX
GRAPHICAL REPRESENTATION OF POINTS AND LINES
The solution of an equation of one unknown quantity, and solu-
tions of systems of equations of two and three unknown quantities,
discussed in the preceding chapters, have very beautiful geometrical
interpretations.
The Graphical Representation of a Point
243. The first problem in this geometrical discussion is to give
a geometrical representation of a point in a plane.
Consider two fixed lines, OX and OF, which are drawn at right
angles to each other; these lines are called respectively the x-axis
and the y-axis, — they are lines of reference.
To represent a point P in the plane of the paper proceea as
/ollows : Draw PM perpendicular to OX. P will be determined in
position by the perpendicular Pilf and the distance OM oi M from
0, Fig. 1. The line QM\% called the abscissa, and PM the ordinate^
of the point P, They are, for the sake of brevity, represented by
the letters x and y, respectively, which are called the coordinates of
the point P, Thus, by definition,
OM=XP=+x
^^ PM= NO = +y.
As in the case of positive and negative numbers we may lay off
positive oc's from 0 to the right toward JT, and negative x's from
0 to the left toward X\ Similarly, +y'« from the a^-axis along
the y-axis or parallel to it upward, and — y'« from the x-axis
downward.
S243J REPRESENTATION OF POINTS AND LINES 257
+e
+ 5
+4
N
• P
+ 3
+ 2
-fr
y
+1
4
X
M
— 0
— B
—4
— 3
— 2
— 1
O 4
1 -i
2 4
3 4
4 4
5 4
6
—1
-2
— 3
-4
-6
-e
Figure 1
o
P.(-*
l*,+ 3
%)
P,(2.
3)
P.(-3
.2)
y-
+3
P.
<— 6
M.
X
--3
X
-42
M,
*"*
(-•.c
)
M.
M,
X
«-2
O
y-
-4
y-
-2
-2)
Pa^a
-4)
Figure 2
258 COLLEGE ALGEBRA [2243
This convention of signs for plus and minus xs and plus and
minus y's is sufficient to determine a point in any of the four parts of
the figure JTOr, YOX\ X'OY', and VOX.
Thus the points,
ix = 2 (x=-3
* U = 3 « U = +2
(x=-2 j.= +4
' U = - 4 M ,/ = _ 2,
in Fig. 2, are constructed as follows:
For 1\. Lay off on OX to the right OM^ = + 2 units, and on
the line through M^ vertical to OX, upward, I\M^ = + 3, locating
the point P^.
For P^. Lay off on OX' to the left 0 J/, = — 3, and on the line
through Jfg, vertical to OX' upward, P^^f^ = + 2, locating the
point P^.
For Py Lay off on OX' to the left OM^ = — 2, and on the line
through Jfj vertical to OX' , downward, P^M^ = — 4, locating the
point P^.
For P^. Lay off on OX to the right OM^ = + 4, and on the
line through M^ vertical to OX, downward, M^P^ = —2, locating
the point P^.
It is observed that in the angle XO Y the abscissa and the ord-
inate are both + , in the second angle YOX' tlie abscissa is — and
the ordinate is + ; in the third angle X' OY' the abscissa and the
ordinate are both — ; in the fourth angle Y' OX the abscissa is +
and the ordinate is — .
For brevit}^ any point x = a, y = b is written (a, b). For example,
/*3 { ^\J ~ ^ is written />^ (_ 2, - 4).
The point A { ^' " o" ^
is situated on the rr-axis six units spaces to the left of O.
Construct the point
For X = — 4 J lay off four units + ^ of a unit from 0 to the
left to 3/g and for y = -f 3J lay off 3 units + J of a unit on
the vertical line through if^, upward, locating the point P^ (Fig. 2).
«244]
REPRESENTATION OF POINTS AND LINES
259
Construct the points:
1.
a; = — 2, y = — 3.
2.
3.
X = 0, y = + 4.
4.
5.
X = - 1 J, y = 3J.
6.
7.
X = - 3i, y = 0.
8.
9.
x= -3, y= +6i.
10.
11.
x = +i, y = -7|.
12.
13.
X = — bm, y = — 6 n.
14.
X = +3, y = -4.
X = 0, y = — 5.
X = -h 4, y = 0.
x= -3i, y = -If
» = -5f, y = — f.
X = 0, y = - 3t.
x = v^. y = -V^-
The Graph of the Solution of a Conditional Equation
844. I. The Conditional Equation.— A conditional equation of
the first degree can be reduced to the form
ax = h,
which, solved, gives
X = - or A,
where the A can be plus or minus.
Thus, for example, the solution of
.,v 2x — 1 3x — 2_5x — 4 7x+6
\*/
3
4 6
12
is
and of
x= + 4,
(2)
6x+l
15
2x-4__2x-l
7x-16 5
is
x = -2.
II. The Graphs of the Solutions of (1) jr = + 4 and (2) jr = — 2.
Lay off OM = + 4, from 0 to the right. Fig. 3. Here y may be
anything, since equation (1) determines only the value of x. Hence,
giving y all possible values :
x = + 4,
and y = 0 determines the
point M
y = l
11 11
<* J/,
y = 2
11 a
** iV,
etc.,
iC ti
*« etc.
y = -i
a ( (
*^ 3/'
y = ~2
(( iC
ic M"
y=-.3
u u
i: M'"
etc..
i, n
^' etc.
260
COLLEGE ALGEBRA
(2) y (O
[2244
H-6
+B
+4.
(-2
+ 3
.+ 2)
Na
+3
(-f-4.
f2)
+2
N.
+2
Oa
+ 2
W.
+1
N
+1
o,
+1
M
-
3 -
S -
4. -
3 -
2 -
1
o t
1 4-
2 43+4-45-f|6 4
7
-1
N«
-1
O'
-1
M>
(-a
—2
.-3)
N"
-2
O"
-2
-3)
-3
Niii
-3
O'x
-3
M"«
-4.
-5
-G
(3)
(4)
Figure 3
Moreover, there can be as many values as are desired fory
between 0 and 1, for the same »= + 4, i. e., as many points as ar.»
desired between M and i/^, M^ and M^ etc. Similarly there can bv?
as many points as are desired between M and M' , corresponding to
X = + 4, and y equal to any value between 0 and — 1. In like
manner there may be an infinite number of points between M' Jf ',
M"M"\ and so on. All these poiats lie in the line through M
vertical to OX, and are ail the same distance from OY^ and the
vertical line M'MM^ is said to be parallel to 0 K
Similarly, in the case of equation (2), laj- off ON'=z — 2, then,
as above,
X = — 2 and
etc.
+ 3
+ 2
+ 1
0
— 1
3
etc.
J244J REPRESENTATION OF POINTS AND LINES 261
i. e., if X = — 2 and y = any + or — number, the line through N
vertical to L X' and parallel to YO Y' is a solution of equation (2).
Again, the solution of
(3) 3y-l ^ 16-y^T.y ^ 4(.v+3) "
5 2 3 6
is y=+2,
and of
... 5y + l 9y+5^9y + 7
^ 7 "^ 11 5
is y = — 3.
Here, as above, for y = -f 2 and x any number desired, + or
— , lay off on OY^ 00^ = + 2, and the line through 0, vertical to
O !F will contain ail the points whose ordinate is -|- 2, and x any de-
sired number, + or — .
Similarly, the points whose ordinate y = — 3 and x any desired
number, -f or — , will lie in the line through &" vertical to OY,
and parallel to OX,
On inspecting Fig. 3, the following valuable results are reached;
(x = 4
The point M^ \ _ o > ^^ ^^ intersection of lines (1) and (3).
(x = —2
The point JV, j __ _i_ o » ^^ ^^® intersection of lines (2) and (3).
{ o
^ Q > is *^® intersection of lines (2) and (4).'
y — — *>
{X = +4
_ , is the intersection of lines (1) and (4).
y — — «i
Draw the graphs of the solution of the examples:
(1) ^ = 1^. (2) i±i + §i^^ = i^+J.-
(4) 19y + l(7y-2)=4y+V-
(5) Find the coordinates of the intersection of lines (1) and (3),
(2) and (3), (1) and (4), (2) and (4).
(6) _i5_+_i! 0.
m 8-6y 38-2y^^^
^*' l-2y 12-y
262 COLLEGE ALGEBRA LJ245
(8) Find the coordinates of the four corners of the figure formed
by the lines (1), (2), (6), and (7).
(9) Find the coordinates of the four comers of the figure formed
by the lines (2), (4), (6), and (7).
The Graph of the Line t/ = mx-{-b
245. In answering the question, what is the geometrical figure
represented by y =. mx + hy begin with the simplest case first
1. y = X. Here, on giving different values to a-, one gets for
I X = - 1 (x=-2 ^„, i ^ = - 3 etc.
A little o])servation and reflection will convince one that the
points 0, P^, Pj, P^, etc., and P' , /"', P'", etc., will have the loca-
tion given in Fig. 4. All other sets of values x and y which satisfy
the equation y =: x will represent points situated on the straight
line P' OP^,
For example, .x = 1|, y =1| is the point Q.
2. y = mar, where m is any number.
The points corresponding to the values .r = 0, 1, 2, etc, . . . and
— 1^ ._2, —3, etc., substituted in y = mx are:
^ fa-=0 r f^ = l r fa: = 2 ,. ( x=3 .
(y=0 ' ty=7/i ''(y=27H My=3m
^,|a:=-l. /y.{^=-2 ;X-]-^=-3 . etc.
On comparing the corresponding values of x and y in 1 with
those in 2, it is seen that
J/j L^ = m ' M^ P^— m^ since M^ ^\—^
3/j L^ = m • 3fj Pg = 2 m, since 3/,P, = 2.
__ iV'// =z _ m • N'P' = — w, since N'P' = 1
^N"L"=— in • A^"7"' = — 2 m, since A""P" = 2.
So that, by division of these equations,
2245] REPRESENTATION OF POINTS AND LINES 263
/
/
/
H-B
J
r/
rv
/
/
+#
/
h
r
+3
vk
©y
•■,
*/
AQk
%
/•
j
N"
N'
A
^\
M,
—
e — .
<a —
■• —
3 -
'i
fy^
0 /
1 4
2 -t
3 4
4 4
B -t
a
pii
/
P'/l
y
1
7- a
pii.
/
^«"y
u /
o
/
/
/
/
/
/
/ *
/
/
Ai
/
/ ,
[■■'7
-8
/
-©V
1
/
i
"1
/
/
1
^
/
1
/
1
f
Figure 4
Hence,
MJ^ M^_ m _1
^' M^L^ M^F^ 2m. 2
(2) ^
N'JJ _
{-m)N'P' _ N'r
— m
_1
I 2'
^"L"
{-m)N"P" N"F" -
-2»
On comparing (1) and (2), it is found that
.v,/
If.
'J
-i =
J
t
etc.
) —
N
N"
-
But, according to a geometric principle, if 0, P^, P^, etc., lie
on a straight line, then 0, 2/^ , i/, , etc. , lie on a straight line.
Nora— In case a pupil has not had this geometric principle, it will do no harm to
assume the fact for the present, as this is the only assumption that will have to be made
In connection with this subject, and he will soon have enough geometry to understand
this proof
Therefore, y = mx also represents a straight line through the
origin O, the line L' OL^, Fig. 4.
264 COLLEGE ALGEBRA [JJ246-248
3. y = mx + 6. For the same values of x used in 2, the follow-
ing points are determined:
{x=0 ( x=l
t y= -2m+b= -N"L"+ 0R=: -N''L''+R''L''=i — N^'R".
All the points /?, R^ , /?", etc. , are located at the same distance,
b = OR = L^R^ = 2/"/?" .... measured along the vertical lines
through 0, L^y N'\ etc., and lie in the line R'^RR^^ parallel to the
line L'OL^,
An important property of the line y = mx -|- 6 is to be noted,
namely, that it cuts off the intercept OR = ft on the positive portion
of the y-axis. If b were negative, the line would have the posi-
tion Ay By Cy cutting off the negative intercept — t = OB on O Y'.
246. The Solutions of /=imjc + b. — It follows from the pre-
vious section that the infinite sets of solutions which x and y can
have in the equation y — mx + 6 represent an infinite number of
points distributed along the straight line R''RR^ ; for example,
(y=—2m+b Ly = b ^(y = m+b
247. The Intercepts. — It has already been noted that the inter-
cept of the line y = nix+ 6 on the y-axis is + 6 = OR, It is
found by making the variable point on the line y = mx + b move
along the line till it falls on OF, and this will happen when x = 0,
and the corresponding value of y, namely, y = + ft, is the intercept,
ORy of the line on OY.
Now make this variable point move along the line y = mx + ft,
i. e., along i?"/?/?^, until it falls on OX' at JV, where y = 0; then
the corresponding value of x will be found from the equation
0 = mx + ft, i. e. , x = ^^ , which is the intercept of the line
WRR, on the axis XOX', If ^"^— is called — a, the equation
y = mx + ft may be written
(1) _.^ + i^=i or -^^y = \,
^^ (*)^ft -a+ft
248. The rule for the signs of the intercepts on the axes will be:
Intercepts on OX' and OY' are negative and those on OX and OY
are positive.
1249]
EEPRESEXTATION OF POINTS AND LINES
265
Hence, equation (1) and the equations
(2) ? + ? = !,
a b
(3) ^+-i4- = l.
— a — o
(4) ?+-i^ = l
a — 6
represent lines crossing the second, first, third, and fourth angles
respectively, as shown in Fig. 5.
>
Vs.
'y
X
^
^
N
sx
y
•4
b
X
\y
y
y
^
\
—a
o
-!•»
y
^A^
\
iiN
•v
b
^
^
For example, to draw the lines,
(1) x+2y= 5
(2) 2a;-y = -4
(3) 4x+3y + 12 = 0
(4) 3x-Sy= 24
(5) 2x — 3y= 12
FIGURES
e lines,
(1)
?+? = l
the
(2)
-2^4
equations
may be
written
respectively
(3)
(4)
-3 ^-4
8^-3
(5)
6^-4
Therefore, by the preceding rule the
intercepts of the lines
1
2 ' 3
1
4
5
on the X-axis are
5
— 2
^3
8
6
and on the y-axis are
s
If
4
-4
-3
— 4.
Intersections op Pairs of Lines
249. 1. The solution of equations (1) and (2), ?248, is x = -^| ,
and y = 2 J. Since these values of x and y satisfy equations (1) and
(2), they are the coordinates of the point of intersection of (1) and
(2), namely, F^, which is shown in Fig. 6.
266
COLLEGE ALGEBRA
[2250
2. The solution of (2) and (3) is x = — 2f , y = — |. This is
point Pj , which is the intersection of lines (2) and (3), Fig. 6.
3. The solution of (3) and (4) is « = — ff , y = —3^^. This
is point P^ the intersection of (3) and (4), shown in Fig. 6.
4. The solution of (4) and (1) is x = 6f , y = — ^^ , whidi are
the coordinates of the point P^, the point common to lines (1) and
(4), Fig. 6.
\
^^
^
/
*
>
\
n
%
(»K
\
fey/
^
<
/'
y
\
v^
v\
•"
-3
V
-2
o
V
k
^
+•
/
\
<A
^
"•
^
^
/
\
\.~'
^
■ /^
?^
«
A
-
^^
y
y
(*)>
■
^y
y
\
(^
y
\
\
/
/
y
^
s
^
y
\
a
k
FIGURB 6 "
250. RhfsuMR Thus is reached the beautiful geometrical repre-
sentation of an equation in one and two unknown quantities' a
conditional equation in one unknown quantity is in every case repre-
sented by a line parallel to the x- or y-axis, either on the podtive
or the negative side of the axis, 1244, II; an equation in two unknown
quantities (y = 7?ia; + 6) is represented by a straight line, 2245, 3;
and, when it is written in the form of '^ + ^ = 1, a and h are respec-
tively the intercepts of the line on the x- and y-axes, 2247. If
both intercepts are positive the line crosses the first angle XOY; if
a is — and 2> is +, the line crosses the second angle; if both a and
?251] REPRESENTATION OF POINTS AND LINES 267
6 are — , the line crosses the third angle; and, finally, if a is -f and
6 is — , the line crosses the fourth angle, {248, Fig. 5.
If ajj and y^ be the values of x and y which satisfy two equations
of the first degree in x and y, then x^ and y^ are the coordinates of
the point of intersection of the straight lines represented by these
equations.
BXEBOISE XLVI
Construct the lines represented by the equations, 3 §248, 249:
1. x = 2. 2. y = 3. 3. y = l^x. 4. 3x + 4y = 0.
5. y = 2x — 1. 6. aj=5 — 2y. 7. 3x — 2y = l.
8. 3x-2y = 6. 9. ^+| = 1. 10. | - ^ = 1.
Draw the lines represented by the following equations and locate
their points of intersection :
11. 3x + 4y = 10, 4x + y= 9.
12. x-f9y = 13, 3x+y = 14.
13. 8a5 — y=34, x+8y = 53.
14. 14x — 3y = 39, 6.r+17y = 35.
15. X = 5, 4x+5y = — 37.
16. y= — 3, 4x — 3 3/ +11=0.
17. Find tlie coordinates of the comers of the rectangle formed by the
lines
x = 3, y=-2, x=-4, y = -6.
18. Find the coordinates of the points at the vertices of the triangle
formed by the lines
1+1=1, 2y = -3x-6, -4x+5.v = 20.
251. It may be stated without proof what the student will learn
later in Analytical Geometry of three dimensions, that an equation
of the first degree in three unknown quantities, x, y, and z, repre-
sents a plane where x, y, z are the three perpendiculars from any
point of the plane to three fixed planes, perpendicular to each other;
for example, the fioor, a side wall, and the adjacent end wall of a
room. The plane represented by the equation - + ^ + - = 1, would
be the plane cutting across the three straight line intersections of the
planes (or walls); it would cut off intercepts r/, />, and c on the lines
in which the three perpendicular planes running out from the corner
of the room intersect.
The values of x, y, 2, which satisfy any three equations of the
first degree in these quantities are the co-ordinates of the point of
intersection of the three planes represented b}' the three equations.
CHAPTER X
DIOPHANTIAN EQUATIONS AND PROBLEMS*
Indeterminate Equations op the First Degree
252. It has already been learned that, in case the number of
unknown quantities is greater than the number of independent equa-
tions, there will be an unlimited number of solutions, and the equation
will be indeterminate. However, it is possible to limit the number
of solutions by introducing conditions which the unknown quantities
must satisfy. When it is required that the unknown quantities shall
be positive integers, the equations are called* simple indeterminate
equations,
253. In the present chapter the solution of indeterminate equa-
tions of the first degree, containing two and three unknown quantities,
will be considered, in which the unknown quantities are restricted to
positive integers.
Every equation in two unknown quantities can be reduced to the
form,
(1) ax ztf>!/ = zt c,
where a, fe, c, are positive integers which do not have a common
divisor.
The equation
ax -f fey = — c,
included in form (1) can not be satisfied by positive integral values of
X and y; because if «, 6, x, y are positive integers, ax + by must be
a positive integer which can not be equal to a negative integer, — c
Furthermore, the equations ax zhhy = c and ox — by = — c can
not be solved in positive integers if a and b have a common divisor.
For, if X and y are positive integers, the common divisor of a
and b must also be a divisor of ax dc by, and consequently of c ;
which is contrary to the hypothesis that a, 6, and c have no common
divisor.
• Diophanii Arithmeticorum, libri VI. Diophantus lived, according to Abulforag,
about 840 A. D., in Alexandria.
2254] DIOPHANTIAN EQUATIONS AND PROBLEMS 269
254. The placing of this restriction on the variables enables one
to express the solution in a very simple form.
Example 1. Solve in positive integers 2 x + 11 y = 49.
After transposing, 2 a; = 49 — 11 y
x=24-5y+ 1^
where the quotient is written as a mixed expression. Transposing
(1) x + 5y~24 = ^^.
Since the values of x and y are restricted to having positive inte-
gral values, then x -j- 5 y — 24 will be an integer, and, therefore,
1-y
2
will be an integer, although written in a fractional form.
Let — — ^ = n, an integer;
ij
then 1 — y = 2 »
.-. (2) y = l-2n.
Equation (2) shows that n^ with respect to y, can be zero ; or can
have any negative value, but can not have a positive integral value.
Put 1 — 2n for y in (1), then
(3) a; = 19-flln.
This equation shows that w, with respect to x, can be — 1, 0, + 1»
etc., but can not have a negative integral value greater than — 1.
Hence it follows from (2) and (3) that, for
n = -l, 0,
X = 8,19,
and y = 3, 1,
which are the only positive integral solutions of the given equation.
Example 2. Solve in positive integers,
(1) 8x — 21y = 33
by means of the process used in Example 1,
l + 5y
a: = 4+2y +
8
tr- 2y- 4 = i-±^ = + integer,
o
On multiplying the numerator by 5,
5 (ii^)='-±|^=+ integer.
270
That is,
Let
COLLEGE ALGEBRA
^ y + T = + integer.
o
[2254
(2)
.V+5
//, a+ integer;
/ (2) y = 8 n - 5
I ^ and from (1)
((3) x = 21n — 9.
The artifice of multiplying — - — '- by 5 saves calculation in the
o
above example, as will be seen in what immediately follows.
The rule is, in any case multiply the numerator of the fraction by
such a number that the coefficient of the unknown quantity shall exceed
some multiple of the denominator by unify.
In case this had not been done the work in the last part of the
solution of Example 2 would stand as follows:
Let (2) ^J~^ = n or 5y + 1 = Sn.
8
On dividing by 5, y -{- ^ z= n 4-
o
o *j 1
Then must be an integer,
o
3n
Let
(8) •'^"- ^ = i^ or 3h - 1 = bp.
0 "•
On dividing by 3, n — }^=p + ^^ ;
hence, ^~^^-~ must be an integer.
Let
(4)
^JP+A-
= q OT 2p = 3q—l]
on dividing by 2, i> = ^ + ^ - •
Let (5) ^~^ =rorq = 2r+l.
Substitute in (4) ; then 2p = 6r+3 — 1;
',p = 3r+l.
Substitute in (3); then 3r — l=15r+5;
J .•.n = 5r+2.
Substitute in (2); then 5^+ 1= 40 r+ 16;
\y = Sr+3,
Substitute in (1) ; then (G)8 x — 168 r — 63 = 33
L .-. a' = 21r+12.
(7)
W55] DIOPHANTIAN EQUATIONS AND PROBLEMS 271
The values of x and y differ in form from those found above, but
the same system of values for x and y is obtained, since here for
r = 0, 1, 2, 3, . . .
it is possible to substitute
X = 12, 33, 54, . . .
y= 3, 11,19, . . .
and by giving to n in system I any positive integral value, an
unlimited number of values for x and y is obtained ; thus
r n= 1, 2, 3, . . .
(8)] a; = 12, 33, 54, . . .
( y= 3, 11, 19, . . .
Solution (7), or (8), is called the general integral solution of
equation (1).
The student will see, in the solutions of Examples 1 and 2, that
there is a further limitation to the number of solutions introduced
according as the terms in x and y are connected by the plus or minus
sign.
265. Should there be two equations involving three unknown
quantities, they could be combined so as to eliminate one of them
and have an equation containing two unknown quantities; then the
process would be that of the previous examples.
Example 3. In how many diflferent ways can the sum of $5.10
be paid with half-dollars, quarter-dollars and dimes, so that the
whole number of coins used shall be 20?
Consider that the sum $5.10 is reduced to dimes.
Let X = the number of half-dollars,
y = the number of quarter-dollars,
z = the number of dimes.
Then bx+^y+ z = bl,
or (1) 10x+5y-f 2^ = 102;
and according to the last condition of the problem,
(2) x + y+z = 20.
From (2) z = 20 — (x + y\
and by substituting in (1) 10 x + 5 y -f 40 — 2 (x + y) = 102
(3) Sx+ 3y = 62]
. 62— 8x „^ „ , 2 — 2.C
hence y = = 20 — 2x -| — j
o o
.-. (4) 5 f — ^ — j = ^ =3 — 3xH — = + integer.
272 COLLEGE ALGEBRA [1256
1 X
Let (5) = n, a + integer, or a; = 1 — 3it.
o
Substitute 1 — 3 n for x in (3), then
8 — 24»+3y = 62
54-4- 24» io , o
or (6) y— — --- - = 18 + 8n.
Therefore from (2)
(7) z - 20 — l + 3n — 18 — 8n = 1 - 5».
Hence it follows from (5), (6), and (7), that for
it is possible to substitute ^* / J
^ X = 1, 4, 7,
y = 18, 10, 2,
z = I, 6, 11,
and there are no other positive values of ar, y, z which will satisfy
equations (1) and (2); and therefore there are three ways in which
the given sum can be paid: one half-dollar, 18 quarter-dollars, and
one dime; 4 half-dollars, 10 quarter-dollars, G dimes; and finally,
7 half-dollars, 2 quarter-dollars, 11 dimes.
256. Theorem L — Given one solution of ax ^hy = c in pantive
integers, to find the general solution.
Suppose that I and m is one solution of ax — hy = c, so that
al — hm = c. By subtraction,
(1) a{x^l) — h{y-^vi) = 0.
Since a is a divisor of a (x — Z), it must be a divisor of h(y — m);
a must therefore be a factor of y — m, since a is prime to 6.
Let y — m z= at, where t is any integer; then from (1)
a{x — l) = b(y^m) = aht^ and therefore x— I = ht.
Hence if a; = /, y = m be one solution in integers of the equation
ax — hy = Cy all other solutions are given by the equations,
(2) x—l=ht and y — m = at,
where / is a positive integer; therefore
(3) X = I -\- h(, and y z=z m-{- at.
Hence, if one solution is known, it is possible to obtain as many
solutions as may be desired by assigning to t different positive inte-
gral values. It is possible also to give t such negative integral
values as make ht and at less than I and m respectively.
Example. — The smallest positive integral numbers which satisfj
9x — 5y = 1, are a = 4, and y = 7; what are therefore the next
five solutions?
2257] DIOPHANTIAN EQUATIONS AND PROBLEMS 273
On comparing 9j; — 5y = 1 with ax — 6y = c ({ 266), a = 9, ft
= 5, ? = 4, m = 7, and with equations (3),
aj = 4 + 5<
y = 7 + 9e
will be the general solution. We have
for < = 0, 1, 2, 3, 4, 5,
X = 4, 9, 14, 19, 24, 29,
y = 7, 16, 25, 34, 43, 52.
267. Theorem II. — Having given one tet of integral values xchick
satisfy the equation ax -\-by = c, to find all other possible integral
solutions.
Let x = ly y z=m, be one integral solution of the equation
(1) ax -\- hy = c
then al -\- hm =1 c.
Hence, by subtraction, a(x — I) + b(y — m) = 0.
Since a is a divisor of a {x — l), it is also a divisor of 6 (y — m),
because a{x — l) = —b(y — m)\ but a is not a divisor of ft, 2263|
therefore a is a divisor of y — wi.
Let then y — m = ta, where t is any integer.
Hence, a(x — 1)= — b {y — m) =z — tab]
and, therefore, x= l — tb.
Hence, if a; = Z, y = w, be one solution in integers of the equation
ax + fty = c, all other integral solutions are given by
(2) x = l — tb and y =m + ta
where t is any integer.
It follows from this discussion and 2266 that there are an
infinite number of sets of integral values which satisfy the equation
ax -\- by =zc. The number of positive integral solutions of the
equation is, however, limited in number.
Example. — One solution of the equation 31x+12y = 1350
is X = 42 and y = 4; what are the other positive integral solutions?
On comparing this equation with ax+ by = c, of Theorem II, J267,
one has a = 31, ft = 12, Z=42, m=4) hence from equation (2), J267,
X = 42 — 12 i and y = 4 + 31 ^
From these equations it follows
for < = 0, 1, 2, 3,
X = 42, 30, 18, 6,
y = 4, 35, 66, 97.
274 COLLEGE ALGEBRA L??258, 259
If / = 4, etc., or — 1, — 2, etc., the values of x or of »/ will be
negative.
268. To solve the equation ax-\-hy-\- cz = d in positive integers
one may proceed thus: write it in the form ax+ hy ^=id — rz, then
give to z the values 1, 2, 3, . . in succession and determine in
each case the values of x and y by the preceding articles.
259. General solution of two simultaneous equations in three
unknown quantities. Solve in integral values the equations,
(1) ax-\-hy-^cz^:zd
(2) aU + h'y + c'z = d\
Eliminate, «, for example, we obtain an equation connecting the other
two variables, Ax-\- By = (\ suppose. Now if ^l and B have only
such common factors as are contained in C, then proceeding as in
§2256, 257 we may find
(3) x= 1+ nt, y =m-^ At .
Substitute now these values in either equation (1) or (2) and obtain
an equation in t and z, which can be written AU -\- h'z = (?. If A'
and B* do not contain any common factors except such as are also
common to C", we may write
(4) tz=VJ^ B't', z-w! — A't' .
Substitute the value of t in equation (3), then
x=zl+ B(l' + B' t% y = m-A{r + B^)
or (5) x=l-^ BV-^ BB't, y=zm — Ar — An't\
Hence we obtain for each of the variables x, y, an expression of the
same form as that already obtained for z.
EXSBOISE XLVn
Solve in positive integers:
x+ y+ z = e.
5x+7y + 4 = 56.
123 x+ 567 y __^
5028 "
3875 x+ 2973 y = 122362.
5x+ 8^ = 29.
12. 17x4-53y— 123 = 441 — 19x+ 15 y.
1.
x + y= 10.
2.
3.
2x+ 3y = 25.
4.
5.
y= 13 + ^V(15-x).
6.
7
2373 __ J
13x+24y
8.
9.
3x+ 5y = 10.
10.
1.
16x+ 4y = 1830.
J259] DIOPHANTIAN EQUATIONS AND PROBLEMS 275
13. 3x+by+7zz=G7, p+3y+5s = 44
• (3a: + 5y+7« = G8.
15 (^+2y+3;5=50 ( x + y - 4z = - 19
' |4x — 5y — 6z= — 66. (3u:+73^— 8.- = 3.
x + y + 22 = 17 ,^ (x + y+s = 20
• "[x+3y+4z = 28. 1 3x+7y+ llz = 100.
3/+ 2 = 50
13 y+ 17;i = 500.
19 1^ + ^+^ = 30 (x+3/
* l4x+ lly+162= 300. • I2x +
.x + y + z=100 .-^ = 17
\7x+23y + 59^ = 1000. ^
23. 8;r = lly.
25. 5 X = 7 ^ = 9 2.
27. 391x = 493^ = 6672.
29. 17x = 11^+86.
31. 11 X — 13i/ = 36 j/ — 3 X
73 X +_17 _ 58j/ -56
19 - 21
32
34.
r 2x
|3x
2x + 5y — 72 = 22
+ 4y — 8 s =0.
24.
91x = 221i/.
26.
12x = 15y =
202.
28.
3x = 5?/+ 1.
30.
Sdx — lUy =
= 1.
- 133.
33.
j Sx + 3t/-2z
[7x+2!,^z-.
= 8
= 8.
rx+22/ + 3z
= 14
35.
^2x+32,+ 4/
= 24
( 3x+ 42 + 5/ = 35.
36. What are the next five solutions of the equations 4 x — 13^
= 10, if the first is x = 9 and y = 2?
37. A solution of the equation 15 x + 11 y = 1000 is x =41
and y =35; what are the other solutions?
38. Find a particular solution of the etjuation 13 x + 5// = 444
and by means of it determine the general solution and finally all
the positive integral solutions of this equation.
39. Find a particular solution of the equation lOx — Sy = 11,
then determine the general solution and finally the next five positive
integral solutions.
40. Find two numbers which, when multiplied respectively by
14 and 18, have for the sum of their products 200.
41. What is the least number which, when divided by 3 and 5,
leaves remainders of 2 and 4?
42. Find two fractions whose denominators are 5 and 7, and
whose sum is — •
oo
43. A farmer spends $752 in buying horses and cows. If each
ho'^e costs $37, and each cow $23, how many of each does he buy?
276 COLLEGE ALGEBRA [i259
44. A farmer bought 125 animals — sheep, P^gs, and hens. He
paid $225 for the whole number bought; the sheep cost $5 each, the
pigs $2. 50 each, and the hens 25 cents each. How many of each
did he buy? How many solutions? How solve with two unknown
numbers?
45. In how many ways can $100 be paid in dollars and half-
dollars, including zero solutions?
46. A owes B $5.15. A has only 50-cent pieces and B only
3-cent pieces. How may they settle the account?
47. Find a number which, being divided by 39, gives a remain-
der 16, and by 56, a remainder 27. How many such numbers are
there?
48. In how many ways can 100 be divided into two parts, one of
which shall be a multiple of 7 and the other of 9?
49. Solve 39x — 6y = 12 in positive integers, so that y may
be a multiple of x,
50. Solve 20.x — 31 y = 7, so that x and y may be positive, and
their sum an integer.
51. Solve 1 x-\-lby = 145 in positive integers, so that x may
be a multiple of y,
52. Some men earning each $2. 50 a day, and some women earn-
ing each $1.75 a day, receive altogether for their daily wages $44.75.
Determine the number of men and the number of women.
53. Show that 323x — 527y = 1 000 can not be satisfied by inte-
gral values.
54. Find all the positive integral solutions of the simultaneous
equations, 5x + 4i/ + « = 272 and 8a;-f9y + 32 = 658.
55. Divide 70 into three parts which shall give integral quo-
tients when divided by 5, 7, 11 respectively, and the sum of whose
quotients shall be 20.
56. A number consisting of three digits, of which the middle
one is 4, has the digits in units' place and hundreds' place inter-
changed by adding 792. Find the number.
57. A number of lengths, 3 feet, 5 feet, and 8 feet, are cut.
How may 48 of them be taken so as to measure 175 feet altogether?
58. Two wheels are to be made so that the circumference of
one is to be a multiple of the circumference of the other. What
circumferences may be taken so that, when the first has gone round
three times and the other five, the difference in the lengths of rope
coiled on them may be 17 feet?
BOOK III
CHAPTER I
INVOLUTION
If a quantity is repeatedly multiplied by itself, it is said to
be raised to a power, or involved, and tbe power to wbicb it is raised
is expressed by tlie number of times the quantity has been used as a
factor in the multiplication. The operation is called Involution.
Thus, as has been stated,
a X « or a* is called the second power of a ;
a X a X a or a' is called the third power of a ;
and so on.
In J J 85, 86, 89, VIII, some examples in involution have been
given, but it is now desired to give additional rules more concisely
stated and of more general character.
The theory of involution, however, involving fractional and neg-
ative exponents, will not be discussed now, but later in a chapter
on the theory of exponents.
261. Index Law for Involution. In case nis sl positive integer
it has been proved, in 285, 4, that
Hence, any required power of a given power of a number is found
by multiplying the exponent of the given power by the exponent of the
required power,
262. The Law of Signs. 1/ the quantity to be raised to a given
power has a negative sign, the sign of the even powers will be positive^
and the sign of the odd powers will be negative.
Thus, (— a)« = (—a) ( — a) = + a*
(_ a)» = (- o) (-a) ( - a) = (+ ay{-^a) = -a'
(-a/ = (-«)«(- «)« = ( + a«) (+ a«)= + a*
(— a)»- = [( - ")']" = ( aY = + a^" [2261]
(_ a)««+* = (— a) (— a)»» = (— a) (+ a«») = — a«'»+'.
Here n is any positive integer. These results show that, when the
277
278 COLLE(iE ALGEBRA [«?263-265
exponent is even the result of the involution has the + sign, and
when the exponent is odd, the result has the — sign.
263. The Positive Integral Power of a Positive Quantity..
It has been proved that
1. («")"• = o•"^ [2261]
In case n is a positive integer, it also follows from §86, 5, that
2. (a/>)" = a"-fe".
3. (abcy = (abc) (abc) . . . to the product of n factors {abc)
=: (a ' a ' , . to n factors) • (6 • 6 • . . to n factors)
• (c • c • . . to » factors).
(abc)"" = a" • 6" • c\
4. For example,
(a^b'^c)^ = (a^)* • (b^)^ • (c)» = a'n'^cK
5. Also by the preceding section,
( — a'^y = db"*"", where the positive or negative sign is to be
prefixed according as n is even or odd. Or, since
— a^ = (— 1)^*", then
(—a*")" = [(—1) a"*]" = (—1)" • (a"*)" [2]
3= (_l)nf,»"»,
which is -f or — according as n is even or odd. These five observa-
tions give the following rule:
A quantity is raised to any power by multiplying the exponent of
every factor in the quantity by the exponent of that power, and prefix-
ing the proper sign, determined by the preceding rule.
264. The Positive Integral Power of a Fraction.— By defini-
tion, when ? is a rational fraction,
• • to n factors.
B"* ll; = l^ = 'i;^ [«68,Eq.l]
_ a a a a^ a a^ , _
a'l^ IT 'I = 71*^ = 7:?' and so on;
265. Powers of Binomials.— It has already been proved in 289
that
1. (a + by = a^ -\- 2 ab + ^*, the second power of (a + h)
2. (a + by = a^+3a^b+ 3ab^ + b^, third power of (a + h)
/a'
i" a a a
U,
1 ~ b' b' b' '
a
b
'b~'b'l~b^
a
f 1 « _ a' , <» __
b '
b'b~~h^'b'
©"=?•
«266, 267] INVOLUTION 279
3. (a + by = «*+ 4a»6+ 6a*6*+ 4^6'+ 6*, fourth power of (a+ 6) :
Similarly, the second, third, and fourth powers of (a — 6) are:
4. (a — by = (r' — 2ab+ b^
5. (a — by = a''—3a^b + 3ab^ — b^
6. (a — t)* = «* — 4a'6+6aW — 4a6»+fe*
That is, wherever the odd power of b occurs, the negative sign is
prefixed.
Later the theorem called the Binomial Theorem will be proved
which provides a method for finding any positive integral power of
Iha binomials a-^ b or a — b without multiplication. This theorem
has been stated in J89, VIII. It may be expressed for the exponent
H in a formula as follows:
♦ (a-f 6)« = a» + na"->6+^^ . a"-»6« + ^ <^ ~ ^^ ^^ ~ ^> » a»-»&»+ • . .
I '1-2 1 • 2 • H
'1-2 1-2-3
These formulae have n + 1 terms in case n is a positive integer,
but have an infinite number if n be negative or fractional.
266. These rules for the formation of a power of a binomial
hold in case the terms of the binomial have coefficients or exponents,
1. Find the third power of 2 x^ — 3 y\
Since (a — by = a» — 3 a*6 + 3 a6« — 6»,
by putting 2 x^ for a and 3^' for b, it follows that
(2 x« — 3 yy = (2 x«)» -3(2 x*)* (3 y^) + 3(2 x«) (3 y»)* — (3 y»)»
= 8 x« — 36 xV' + 54 xY — 27 y*.
2. Find the fifth power of x' — J yh.
Since (a -^ by =: a^ - b a*b + 10 a^b^ — 10 a*Z^» + 5 oft* — 6^
by putting x' for a and ^ ^/'^^ for b, the result is
(j:')'^-5(a^)Mi2/'2) + 10(^)»(i2/«2)«-10(x2)«(i2/«2)' + 6(x«)(lt/«2)*-(
= x» - 1 oV^ + } :iV^* - i-^/^' + j% ^ V^* - Ay^^'.
267. It is evident that the m*** power of a" is the same thing as
the n*** power of a"", namely, a*"" ; that is, the same result is arrived
at by different processes of involution. For example, the 6^ power
of a -f- 6 may be found by repeated multiplication by (a + ^) ; or
* It wUl be shown later that the law of formation of these formulae holds wheA n Is
a negative integer or a positiye or negative fraction when — 1< r < + 1.
280 COLLEGE ALGEBRA [«268, 269
the cube of a -|- ^ may first be found and then the square of the
result, since the square of (a -f ly is (a + 5)*; or the square of
(a -f- b) may first be found and then the cube of (a + 6)*, which
268. Powers of Expressions of more than Two Terms. — It
has already been shown (J 89, VIII) that
(a+ h-{- c)«= ««+ 6«+ c^+2ab + 2ac + 2hc,
{a+h+c-\-dy=a^'^h^+c^+d^-^-2ah+2nc+2ad+2hc+2hd+2cd;
and hence is obtained the following rule, which holds good in the
preceding examples and others similar to them: Tlie square of any
polynomial consists of the square of each term^ together with twice the
product of every pair of terms.
These results may be written in another form:
(a + i* + c)2 = a« + 2 a (?> + c) + 6* + 2 Z>c + c«
(a + 6 + c +t/)« =rt*+2 a {h^ c+d) +6«+ 26 (c+rf) + c«+ 2ctZ+(l*.
The following rule holds good in these and similar examples:
The square of any polynomial consists of the square of each term plus
ttvice the product of each term by the sum of all the terms which follow it.
A general proof of these formulae for general cases can be de-
duced by the process of mathematical induction, which will be
explained later. Thus, it may be proved that:
+ a/+2a,(«3+a^+ . . . . +aj
+ a,«+2«3(f.^+a^+ .... +aj
+ ««-,'+ 2 a„.,(a^_^+aj
+ a\
•^ n
269. The following additional examples illustrate the first of the
rules in the preceding article.
(o + Z> — c)« = a» + 62 + c^ -\-2ab —2ac — 2bc.
(1 — 2x+ 3x2)2 = 1 + 4a:2^ 9 :r* — 4 x + Gx*— 12x»
= 1 _4x+ 10.c«— 12x3+ 9jj4
(l-2x + 3x«—4x»)'= 1 + 4x2+ 9x*+ 16 x«— 4a+6a:« — Bit'
—12 x»+16 X* — 24 x'* = 1— 4x + 10 x2 —20 x3+25 x*— 24x»+16x«.
J270] IN\^OLUTION 281
270. The results given in 289| X, for the cube of a + 6, a — h,
and of rt -f /> + c should be carefully noticed. The following may
also be verified.
(a-{-h + c + dYzzz a'+ 5»+ c^-\-d}-\- 3 aHJj+c+ d)-\-Z 6«(a+ c+d)
-|-3c*(«+^+fO + 3(/«(a+6+c)+6Z>cJ4-6acc?+6aM+6a6c.
+ 3a^«(a,+ «3+ . . . +aj-f 3a/m-a,+ a^ . . +a^)+ • '
BZBBOISE XLVm
Raise to the required power and simplify:
1. (If^Q'- 2. (-a6)«.(ai)»-«.(-a)».
3. (— hah^c^y, 4. (— ax)* • (— hyY • (aZ^x^)"-'.
5. (_ 7 xV^')'. 6. (a&)« . (y * • (^^y.
^' V 2x*i^W * U y ' Uaj ' (3a)«'
(3 a-y)' • (4 xzy ■ (5 ../z)« /a+?A' fa-b\» /x'-y'y
^'*- (25 xy«)» • (6 xyz)» ■ " Va! + W 'U -^ ' U"^-W '
- ©■• (')■•©■• -(g)'©'©'
(6 a6a;)» • (10 abyY /^_«Y
(4at)«- (3ax)»- (25ty)«' V 3 ij"
27. (— 3xy«2»)«". 28. ( — 2 ««Z*V)**n
29. (2x — 3 6)». 30. (3a*h^-2cdy.
31. (a — 2fe+c)». 32. (1 + x — x'+x*)*.
33. (1 — 2x+x«)».
282 COLLEGE ALGEBRA [2270
34. Simplify (1 -f 3 x + 3 x« + ;r')«— (1 _ 3 x + 3 x« — x»)«.
35. {2a—\-c^)\ 36. {a + 2h+c-2d)\
37. Show that (27a^-18.W-M)^, (o_.^- ^)3(^-«3)
38. Find the middle term of {x + I)***.
39. Find the two middle terms of (x — y)".
Find the expansion of the following to six terms:
40. (l_x + x« — x'+a-* — cr/»+ . . )«.
41. (x + 2x2+3.r'+4x* + 5x'*+ . . )\
42. (« + hx + cT^ + </x5 + ej-.* + fx^' + <7.T* +..)«.
43. (a — a^x + o^^* — ^j*"' + «^c* — a^x^ -f- a^x*^ — • . )'
44. (x+^x'+ix'+i.c'+ijx^+ia;«+^x'+ . . )'•
45. (l + x + x'+ cr»+.-e«+ar»+ . . )'.
46. (l+Jx + Jx* + i.T» + ^x«+ . . )»
47. Find the term independent of x in ( ~- ) .
48. Show that {ax^ + 2 Z/xy + ry^) {aX^ + 2 6.1^+ cF*)
= {axX+cyY+h{xY+yX)\^+{ac-h^{xY-yX)\
49. Show that (x« + pxy + (^i/«) ( A'« + pXY+ q Y^)
= (xX+ pyX+ qyYY-itp(xX+pyX+ qyY) •
50. Show that (x2+ y»+ s« + w?«) (p»+ ^»+ r*+ ««)
= (xp — ^5' + 2;r — icsY + (x^ + yp — zs — wry
^ {xr—ys — zp + xcqY J^ {xs + yr -\- zq-{- wpY.
Find the value of the following to the fifth decimal place:
51. l + .r + x2+x3+ • ■ .forx = (l) 0.2; (2) 0.3; (3) \\ (4)^^.
52. I_x+x2-x^+ ■ • .forx = (l) J; (2) Vg*, (3) ^V
53. x+ 2x2+ 3;^s_,_4^4 . . forx = (l) ^V; (2) ^V-
54. i + f+r-2+i^5+r^'V4 + --^-^=^
56. 1 - •^^ + p/yr-^ - rYT^i^^ + . . . for x= -,4^.
57. x-Jx3 + lx^-lx^+ . .forx=(l) |; (2)^; (3)^;
CHAPTER II
EVOLUTION
Definition and Principles
271. Evolution, or the extraction of any root of a given quantity,
is the inverse of the operation of raising a certain quantity to a power
which will produce the given quantity. The extraction of the r***
root of a quantity undoes the act of raising the r**» root of that quan-
tity to the r*** power. Thus, l)y definition,
V^ = a and (V«)" = «.
272. Definition of a Root. — 1. The definition of the square root
of a number has already been learned (§ 94). Thus it is known that
l/a* = a and (k «)' = a.
2. The meaning of the cube root of a number has also been
learned (? 97). Thus it is known that
'l/a' = a and ('i/a)'=a.
Since {a + by = a' + 3a«t + 3a6« + h^, then
V(a3 + 3 a«6 +3a6« + b^) = V'(« + by = a + b,
3. It follows from the definition of evolution, §271, that the n^^
root of a number is one of the n equal factors of the number. Thus
-f- 3 or — 3 is one of the two equal factors of 9, and {a + b) is one
of the three equal factors of
a' + 3 a«^^ + 3 ab^ + b\
273. The radical sign, |/, is used to denote the square root,
and is placed before the number whose root is desired (894).
The radicand is the number or expression whose root is desired.
The index of a root is a number which indicates what root of the
radicand is to be found, and is written above the radical sign.
284 COLLEGE ALGEBRA [{1274-277
Thus,
the square root of 9 is written *i/9 or 1^9= 3;
the fourth root of 16 is written *v^i6 = 2;
and the n^ root of a is written "l/a . Here the indices of
the roots are respectively 2, 4, and n.
274. A parenthesis, or vinculum, is often used to express the
root of a quantity consisting of more than one term. Thus V 16+25
means the sum of l/l6 and 25, while l/l6 -|- 25 means the square root
of the sum of 16 and 25. Moreover, 'i/ar* • y^ means the product of
y^ and the cube root of a^, while 'v/x'x J/^ means the cube root of the
product x'j/'.
Parentheses are sometimes used instead of the vinculum in con-
nection with the radical sign. Thus, the same result may be ex-
pressed by i/16 + 25 or i/(16 + 25).
275. Like and Unlike Roots. — Two roots are said to be like or
unlikf according as the indices of the roots are equal or unequal^
whether the quantities under the radical sign are equal or not.
Thus,
'|/x and 'i/y are like roots; i/x and 'i/y are unlike.
276. In this chapter will be considered the roots of numbers
which are powers whose exponents are multiples of the indices of the
roots.
An even root of a number is one whose index is an even number;
thus*
\/a*i Vl6^ *V^
An odd root of a number is one whose index is an odd number;
thus,
•|/^, V32, *""^V^*»^, when n = + integer.
277. The Law of Signs of Roots of Quantities.— From the law
of signs in involution, 2262, it is evident that:
1. Any even root of a positive number will have the double sign Jb;
because either a positive or a negative number raised to an even
power is positive, 2262. Thus,
Vl6 = d=2, for (di2)* = 16; Va* = zta, for Cdza)* = a*.
J277] EVOLUTION 285
2. Among the odd roots of a number there w at least one root of
the same sign as the number itself. Thus,
since (- 2)» = — 8, .-. V^ = — 2;
since 3» = 27, .-. '1/27 = 3;
since (— a)* = — a*, /. *l/— a* = — a.
In general, since (— a)«"+i= (— l)«»+V+»= — a«"+»;[||263,2;262]
«"+V(_o)««+i = (__ l)(a) = - a.
The principle stated in 2, when the radical is negative, may also
be stated as follows:
3. An odd root of a negative number is minus the sam^e root of a
number which has the same absolute value. Thus,
since V— 27 = — 3 and — V27 = — 3,
V^^ = -V27.
Since """"V— «««+» = — a and — '"■*"Va«»+* = — «,
Hence, to find an odd root of a negative number j find the same root
of the positive number which has the same absolute valve, and prefix the
negative sign to this root.
4. Since 0* = 0, therefore i/O = 0. In general, since 0" = 0,
.-. *l/0 = 0.
5. 7%c even root of a negative number can not be taken; because no
real number raised to an even power can produce a negative number.
Such roots are called impossible. Thus,
l/ZTg can not be + 3 or — 3, since (+ 3)« = 9 and(— 3)« = 9.
l/ZT^ can not be + » or — x, since (+ x)* = a^ and (— x)* = x*.
■"|/— a*" can not be + a or — a, since (+ «)*" = a*" and
(_ a)*" = a*".
Even roots of negative numbers can not be expressed in terms of
numbers hitherto used, i. e. , in terms of positive or negative inte-
gers, positive or negative fractions, or of positive or negative roots
that can be found.
The roots of numbers which are not powers with exponents which
•are multiples of the indices of the required roots and even roots of
negative numbers will be discussed later.
L— It has been shown above that a poeitive number which is the nth power of
a number has at least one nth root and, when n Is even, at least two; also that any negsr
live number which is an odd power of a. negative number has at least one odd joot.
286 COLLEGE ALGEBRA [?278
It will be shown that any number has two square roots, three
cube roots, four fourth roots, and five fifth roots ; and in general it
may be proved that any number has n, n^ roots.
278. Principal Root. — 1. The principal root of a positive num-
ber is its one positive root. Thus,
3 is the principal square root of 9, and 6 is the principal cube root
of 216.
2. The principal odd root of a negative number is its one neg-
ative root.
Thus, — 3 is the principal cube root of — 27, and -- a the prin-
cipal (2 n + ly^ root of — a*"^\
3. It should be noticed at this point that the relation,
holds for the principal n'** root only. For, by the preceding article,
the "k a" has n values, the principal value being a. But, by the
definition of a root, ("i/n)" = a for every ?i'** root of a. Thus,
v/5' = it 5, if the negative root — 5, as well as the principal root
-|- 5, is admitted; but
l/5« = (v/5)'
in case of the principal square root only.
In the work which follows, the radical sign will be used to repre-
sent the pnncipal root only, unless the contrary is expressly stated.
Thus, 1/25 = 5, -1/36 = — 6, V-27 = — 3,
2«+l /-
EXEBOISB XIjIX
Write:
1. Two square roots of 125. 2. Two fourth roots of 81.
3. Two sixth roots of 64 and 729.
4. Two square roots of 5^" and a^".
Find one cube root of:
5. 125. 6. —216. 7. —1000. 8. — a*».
Find the values of the principal roots indicated in the following
examples:
9. 1/225. 10. V-216. 11. n -512. 12. 1^625.
13. V - :m 14. *v 625. 15. '] —243. 16. ViOOG^
«279, 280] EVOLUTION 287
Using the definition of a root, express x as the root of the second
member in each of the following equations:
17. x«=6. 18. x^=h\ 19. x*=t» 20. x'^-W, 21. x»=6'".
22-26. Express h as a root of the first member of each of the
equations in 18-21.
Theorems in Evolution
In any case evolution is merely a special case of factoring, in
which all the factors are equal. That is, the square root, the cube
root, the fourth root, etc. , are found by taking one of two, of three,
of four, etc. , equal factors, respectively of the given expression.
Since even roots of negative numbers are not considered in
this chapter and since the odd root of a negative number can be
found by taking the like root of the same positive number (1277, 3),
methods and rules for finding principal roots of positive numbers
and expressions only will now be given.
It is to be assumed in what follows that the radicand is a positive
number or quantity and that the roots taken are principal roots.
279. Theorem I. — The n*** root of the product of several positive
factors is equal to the product of the n^^ roots of each factor. Thus,
(1) %/J^ = "l a . "l /7 . "v c.
For, by the definition of the n^^ root,
("v abcY = «6c, and
(Va . V6 . V^" = (V «)" • Cl/Zy ' Cl/'cy-.= ahc, [2263, 3]
Hence equation (1) is true.
For example, |/9 .36 = 1/9 X l/36 = 3 • 6 = 18.
V— 27 a«6»x» = •]/— 27 . Va« • V^* . Vaj» = — 3 • a* • ^» • x.
280. Theorem II. — The n^ root of the quotient of two quantities
is equal to the quotient of the n*** roots of the dividend and divisor.
Thus,
n |^_ Va
For, by definition of the ?t'** root,
{■S'=i-
and
V", t/ ("r
= \ 1L"JL = 1
; Def. of root!
288 COLLEGE ALGEBRA [11281-283
Thus, for example,
/25 /25 5 ^^^^ J27xY V27 ar^j/ 3xy«
\64 "" VU ~8*°^ \ — 125a« ~ V -^1 25ir» "" ITs^*
281. Theorem III. — To raise a radical to the n^ power it is 9ufi-
<cient to raise the quantity under the radical to the w*** power.
Thus,
For ('"i/a)"=("l a) C^Va) (""la) ... to n factors, and, ac-
'cordiug to Theorem I, §279, the product of the n radicals, each
«equal to "*1 Vi, is equal to "*V^a • a - a . . . . to n factors = "i a".
Thus, _
(Vl0)2 ^ *yH^2 = *|./(4«)8 3= H 4* = 4
4rtV
A l8a»xn'^3 l/8a«x»\'_s |64aV».
\ \27 6« / \ V27 ^« y \ 729 b^
9 6*
3. Theorem IV. — Li order to extract the n^^ root of a radical,
the index of the radical is multiplied by n.
Thus,
For, in order that a number may be raised to the mn^ power, it
can first be raised to the n^^ power and the result to the m"* power.
This would give a for the mn^^ power of the first member, and a for
the mn^^ power of the second member, by definition of a root
Or, the n*^ power of V "*y/a is ""v a; the m^^ i)ower of "l a is a:
therefore, the mn^^ power of 1 ' *"i'' a is r? ; and consequently.
Thus, for example.
\ a= 1/ a.
I 1/729 tt"x« ='l ' 3«(a')*^*= 3 a«x.
283. Theorem V. — The arithmetical value of a radical u not
changed by multiplying^ or dividing the index of the radical and the
exponent under the radical by the same positive integer,
Letp be any positive number; then
For, «i'i/^^"v/''v («-^^"j/^ [Th IV;«S823
«284] EVOLUTION 289
Thus, for example,
Voi = V[J?ry = V^8« = 8
8 1^1^=^ //(3ay y _, /(3ay ^ 3a«^
>|8x»y» >(V(2xy«W . \(2V)* 2xy''
284. In case the root of a number or quantity is not readily
detected, it may be found by resolving the number into its prime
factors. Thus, to find the square root of 3111696 and the square
root of the result:
2»
3111696
2*
777924
3»
194481
3»
21609
7
2401
7«
343
•*• V^|/^111696 = V2*3*7* = 2- 3 -7 = 42.
BXBBOISB I4
Simplify the following examples by means of the preceding theorems
and principles:
V-32x«
216 xV'
343 2"
\25c«' \64y«' \ 125 x»« * \
3. V-1728xV2", V3375^V^, '\ ^»'
4. V729^V^ "l/a«"6«", "'i/i«V"'^^» V— a^t'^c'^.
6. Vr00032 x'*'-»y«<*»+*^>.
7. »^
243 x"y''
6^z«^
\ 512x»"'y»P ' \a"6«"''* \3«'6-
38»^S(na:-x)
290 COLLEGE ALGEBRA [2285
9. l/'l/(64» X 27'), 'i/i/(5«xV"), vVH 29x^1/^).
10. 1^ V'^(25 or^"-*), h V(8"x V*)*, V^l/( — 32«x»y«^).
11. V''V'^(x"'>y"''''-*-*i>*-"P), F(100x*V)', '\|(|f|)'-
1 3. ''l/(3'^Pa;P*M'^2''^)*" .
14. V 25 a«^<c« — '] '64j-y^ + "f 2«^«V^7*^.
15. 1 25~ff«6*7« — '] -SSa^b^^ + *v/81 a*6*c* — V— 32a'*6>V.
16. ]/ (V64^^ X *V^(4 VV') X "1/3 na^*n^2n2-7n ^
285. The Square Root of Compound Quantities. — The extraction
of square roots of numbers in Arithmetic is based upon the method
for finding the square root of a compound algebraic expression.
This method will now be explained.
Since it is known that the square root of a* + 2 aft + 6' is a + 6,
a general rule may be deduced for finding the square root of an alge-
braic expression by observing in what manner a + ^ is derived from
a« + 2afe + i>«; thus,
a' + 2ab + h^ \a + hz= root
2a + b
2 ab + b*
2ab + b^
Arrange the expression according to the powers of a; then the
first term is a*, and its square root is a, which is the first term of the
required root. Subtract the square of the term of the root just
found, namely a^, from the expression and bring down the remainder,
2 ab + b^. Take twice the part of the root already found, namely 2 a,
for the first term of the new trial divisor, and divide the first term
of the remainder, namely, 2 ab, by 2 a, obtaining a quotient 6, the
second term of the root; annex this to the first term of the trial
di\i8or 2 a, obtaining 2 a -\- b sls a complete divisor and multiply it
by by the second term of the root; this gives 2a6 + ^*, which sub-
tracted from the remainder leaves zero. This completes the opera-
tion in this case. II there were more than three terms in the ex-
pression, then the process with a + 6 would be like that with a.
{286J
EVOLUTION
291
Thus, find the square root of a* + 2 afe + [*• — 4 ac — 4 />c + 4 c*.
1st step; a* ~
2d step; 2a +6
2ab + />«
I 2ah + b^
3d step; 2a + 2 6— "2 c
— 4ac — 4^c+4c'
— 4ac — 4/>c-(-4c*
Up to the third step the process is the same as above. At the be-
ginning of the third step the root already found, namely, a + 6, is
doubled, 2 a + 2 6 being obtained for the first part of the trial divisor.
Divide the first two terms of the remainder by it, obtaining — 2c; this
is the third term in the root: annex this term to the first two term9 in
the trial divisor for a complete trial divisor and multiply the sum by
the third term of the root, i. e. , (2 a + 2 6 — 2 c) (— 2 c), and sub-
tract the result from the remainder. In this case the operation is
now complete. In case more terms are left in the remainder after
the third step, the process must be continued till the square root is
found.
286.' Examples. The method just explained may be extended
to expressions of more terms, if care is taken to obtain the trial
divisor at each step of the process, by doubling the part of the root
already found, and to obtain the complete divisor by annexing the
new term of the root to the trial divisor.
I. Find the square root of x*-f 25x«+10x*— 4^*^— 20 x'-f 16— 24x.
Arrange the terms in the ascending powers of x; thus:
16— 24X+25 x2— 20 x^-f 10 x*— 4 x'^+x* | 4—3 x+2 x«— x'
16
8-3x
-24 x+25 x«— 20 x'-f 10 x*— 4 x'^+ x«
-24 x+ 9x3
8 _ 6 X -f 2 x«
16 x«— 20 x^+lO x<— 4 x^+x«
16x«-12x'+ 4x*
8 — 6x + 4x«
— 8x5+ 6x*— 4x'*-fx«
— Sx^+ 6x*— 4x'^+x«
Here the square root of 16 is 4; this is the first term of the root.
Subtract 16, the square of 4, from the whole expression ; and the
remainder is — 24 x + 25 x« — 20 x' -f 10 x* - 4 x* + x«. Divide
— 24 X by tvnce 4, or 8, obtaining — 3x, the second term of the root.
292 COLLEGE ALGEBRA CJ287
and annex it to 8, the first term in the trial divisor ; multiply the
result by — 3 .X and subtract the product from the remainder, leaving
a second remainder, 16 x ' — - 20 x' + 1 0 x* — 4 x* + a;*. Double the
root already found, obtaining 8 — 6 x, the first part of the second
trial divisor; divide 16x' by 8, obtaining + 2 x* for the third term
in the root; annex this term to the trial divisor 8 — 6x, multiply the
sum 8 — 6x + 2x^by2x* and subtract the product from the second
remainder; this operation gives a third remainder, — 8x' + 6x* —
4x* + ^* Double the root already found, obtaining 8 — 6 x + 4x*
for the first part of the third trial divisor; then divide — 8 x^ by 8,
obtaining — x' as the fourth term of the root. Now from the above
process there is no remainder on the completion of the last step
Tbfe operation is completed in this case.
This problem could have been solved with the same ease by ar-
ranging the expression with respect to the descending powers of x.
II. Find the square root of
Arrange the terms with respect to x thus:
9x^-6jcb + SOxc+(xi'd+b»-10bc-2bd+2bc^+l0cd+d»[
9x»
ar-M-^r-H
6ar— 6
-exb + d0xc+6xd+b»
— 6.r6 +^«
6ar — 26-
+5c
+ 30 xc +6xd -10bc-2bd+25c^
+ dOxc -106c +25c«
6a;— 26+lOf
+ d
+i)xd -26d +I0cd+d»
+exd -2M +10cd+d«
It will be noticed that each trial divisor is equal to the preceding
with the last term doubled.
287. The fourth root of an expression can be found by extracting
the square root of the square root. Similarly, the eighth root may
be found by extracting the square root three successive times; and
the sixteenth root by four successive extractions of the square root
and so on (see f282).
For example, find the fourth root of
81 X* — 432 x' + 864 x« — 768 x + 256.
By proceeding as in the foregoing example the square root of the
proposed expression will be found to be 9 x' — 24 x -|- 16; and the
square root of this is 3 x — 4, which is the fourth root of the given
expression.
2287] EVOLUTION 293
EXEBOISB U
Find the square root of the following polynomials :
1. x« + 2 X + 1. 2. a« + 2 + ^-
2r
3. X* — 2x»+cc«+2a; — 2 + ^- 4. /• + 6/»x* +9 x».
5. y* — 2 y»x + SyV - 2yx» + x*.
6. y* + 2yV — y«x« — 2yx» + x*. '
7. i;* — 2tjV+ t7V + 2i?V — 2rV4- !*•.
8. xV — 2x«y« + x« — 2 xV + 2 xy + y«.
9. ?|? + 2 a*n» + 1|?. 10. -«/a«6« - f a6c« + |c^
11. X* — ax» + JaV. 12. a«"» + 2a"'x» + x«».
13. a«- — 4a"«+»+4a«". 14. ^'-|^ + |^.
y*y« y 'a:a:*x*
16. ^+2a«+ a«6«+ 2 + 26«+ V
17. 2^_y«(2_a;«)-2y(l-x«) + x«.
18. ^-x«(2-y«)-2 + 2y« + 2j.
19. a*+ 2ab+2ac + h*+ 2bc + c\
20. 1 + 2x+3x«+ 2x'+ X*.
21. l + 4x+ 10x«+ 20x'+ 25x*+ 24x» + 16x«.
22. 1 — 22 + 2z«-2» + |*•
23. «**+ 2e«' + c**+ 2c*+ 2 + €-«'.
24. 1 + 2 gr + 3 j«r» + 4 q^r^ 4. 3 j V + 2 gV + 5V.
25. l + 2c"+3e*^+4c*'+5c*^+4e*'+3e«+2e''*+c«.
26. 9x« — 30ax — 3a«x+ 25a«+ 5a»+ ^•
27. 1-2 L+2^_2^+£;.
aVj^ 2 a* . a«c« , 2 6V h^c^ 2 c« . «^«r^« 2 &« ^ 2 /y^
a«_2a6 + 6« a V + 2a6V+g>^r*
x*+4ax« + 4a"' a«"» + 2 a^'x" + x*" "
31. a*"x^ + 10 ca*"«"*x*«+* — 6 a^'+^x"""* + 25 c'a*"~*x*"+»
-30ca"'-»x» + ^.
«.« s >i I 1 ^12 2a , 8& . 4 , 4 , 1
0 a or ab 6"
294 COLLEGE ALGEBRA L2287
33. n«x«" — n(n - l)x«"-» + ^^ ~ ^^' x«"-« .
4
34. n«* — 4 (n — l)'n* + 4 (n —!)«' + 2 (w — 2)*»«
_4(» __ l)'(n — 2)' + (n — 2)^.
35. «^-'-4+ ^
X* X'
JIn+S
2» 2"/j«*
3 '9
37. r I - I r-— I h ---— H- — •
a* ba 144 a 12 4
38. j^z* — 2/>»2^ + />*«• + 2 y (l — 2^) 2« - 2 y(p — y)«»
p* J>* p*
39.
-2 (y»-^)+yV*»
»»»+2»i+l , 2(m«+3n»+2) , 3 »»• 4. 12 m + 10
40. r i r 1
y* y' »*
2 (ct' + 5 m + 6) m« 4- 6 wt + 9
y' y«" ■
41. Prove by extracting the root that:
V/1 + X = 1 + Jx-ia5«+7'jx'-,|,x*+,J,a!»-TH^»+ . .
l/T3^ = l-ix_ix'-,>,x'— rf,x*-,J,x»-TH,x»- . .
How can one of these roots be derived from the other?
42. Calculate to seven decimals the values of j/ 1 0 by means of one
of the series in 41» Here v 10 = Vl^l — fV» ^^^ P^^ ^ = tV ^^ ^^'
43. Calculate in a similar manner to five decimal places l/2.
•Here l/2 = j] U^.
44. Calculate to six decimal places y^ll.
Here v U = Vl^^-ji^j
45. Calculate likewise 1/ 3 , V^5 , and 1/ 6 to four decimal places.
Here
l/3 = V ^^1+^^; 1/5 = Vv'1 + ii'ir; a^^ Ve = iVl-^j,
H288, 289] EVOLUTION 295
46. Calculate to five decimal places l/7 and i/l3.
Find the fourth root of:
47. x«+ 4x« + 6x* + 4x« + 1.
48. (. + l)--4(. + !-l).
50. x*" + 8 x*"-» + 24 x*»-« + 32 x*»- + 16 x*»-* — 12x»»
— 72 x»"-' — 144 x»«-» — 96x»»-» + 54x«" + 216 x**"*
+ 216x«»-« — 108x» — 216x«-' + 81.
The Square Root of Arithmetical Numbers
288. The rule for finding the square root of an algebraic expres-
sion makes it possible to derive a rule for finding the-square root of
numbers in Arithmetic.
The square root of 100 is 10, of 10000 is 100, of 1000000 is
1000, and so on; therefore the square root of a number less than
100 consists of one figure, of a number between 100 and 10000
consists of two figures, of a number between 10000 and 1000000
consists of three figures, and so on. If, therefore, a dot be placed
over the figure in units* place of a number equal to or greater than 1,
and over every alternate figure, the number of dots will be equal to
the number of figures in the root of the number. Thus, the square
root of 4096 consists of two figures, the square root of 611524 of
three figures, and so on.
289. Find the square root of 4489.
Point the number according to the rule; hence the root will consist
of two figures.
Let a + 6 denote the root; then a may
4489 I 60 -I- 7 ^^ taken as the value of the figure in tens'
3600 place, and b the figure in units* place.
Then a is the greatest multiple of 10 whose
120 4-7
^^^ square is less than 4400; this is found to
QQQ ^ '
^°^ be 60. Subtract «« or 3600 from the given
number and the remainder is 889. Divide the remainder by 2 a, that
is, by 120, and the quotient is 7, which is the value of h. Hence
(2 a+6)6, which is (120+7)7, or 889, is the number to be subtracted.
Therefore, since there will be no remainder, the conclusion is that
the required root is 67. The ciphers may be omitted for the sake
of brevity and the following rule be derived from the process:
296
COLLEGE ALGEBRA
[8J290, 291
4489
36
167
127
889
889
Point off the number into 2}eriods of two figurcM
each, beginning with units' place.
Find the greatest number whose square is contained
in the first period; this will be the first figure of the
root; subtract its square from the first period and to
the remainder bring down the next period. Drop
the right hand figure of the remainder and divide the number so found
by twice that part of the root already found. Annex this quotient
to the part of the root already found and also to the trial divisor.
Then multijyJy the divisor as it now stands by the figure of the root last
found and subtract the product from the last remainder.
Jf there are more periods to be brought down the operation must be
repeated,
290. Extract the square roots of 481636 and 11566801.
129
481636
36
1216
1161
694
11566801
9
3401
64
1384
256
256
5536
5536
6801
6801
6801
NoTE.~The student should note the occurrence of the cipher In the root.
291. If the square root of a number has decimal places, the
number itself will have twice as many.
Thus, if . 23 is the square root of some number, this number
will be (.23)* = .0529; and if .113 is the square root of some num-
ber, the number will be (.113)'= .012769.
Therefore, there is an even number of places in a decimal which
is a perfect square, and the number of decimal places in the root
will be half as many as in the given number itself. Hence this
rule for extracting the square root of a decimal may be deduced:
Place a dot over the figure in units' pla^, and over every alternate
figure, continuing to the lift and to the right of it; now proceed as in
the extraction of the square root of whole numbers, and mark off as
many decimal places in the result as there are periods in the decimal
part of the given number.
22292, 293]
EVOLUTION
297
Example. Find the square roots of 556.0164 and 0.667489.
= (2)« =
556.0164
4
23.58
0.667489
64
[_817
2 0=2(20)= 40
6 = 3_
(2 a + 6) 6=43 -3=
156
129
2a =2(230)= 460
b = b
{2a+b)b =4r>5-5=
2701
2325
161
1627
274
161
11389
11389
2a=2(2350)=4700
5=8
I 37664
(2 a+b)b = 4708-8= | 37664
It follows from the dotting that the root of the first example will
have two integral and two decimal places, and that the root of the
second example will have no integral but three decimal places.
292. The student will readily see that many integers haye,
strictly speaking, no square root. Take, for example, the integer 7.
It is clear that 7 can not have a square root; for the square of 2 is
4 and of 3 is 9, therefore the square root of 7 lies between 2 and 3,
and consequently the square root of 7 can not be an integer. The
square root of 7 can not be a fraction, for if any fraction, which is
strictly a fraction, be multiplied by itself, its square will be a
fraction.
If the square root of a number consists of 2/1+1 figures,
when the first n + 1 of these figures have been found by the usual
method, the remaining n may be obtained by division.
Let JV represent the given number; a the part of the square root
already found, that is, the part in the first n-\-l figures found by the
rule, with n zeros annexed; and x the part of the root which
remains to be found. Then
(1)
\/N= a-\-x
N= a'^+2ax + x^
N—a* . x«
= x+ — .
2 a 2a
Now iV— a' is the remainder after n+1 figures of the root, repre-
sented by a, have been found ; and 2 a is the corresponding trial
divisor. Equation (1) shows thatiV— «* divided by 2 a gives x,
the rest of the square root desired, increased by — -.
298
COLLEGE ALGEBRA
[J293
It will now be shown that — is a proper fraction, so that, by
neglecting the remainder arising from the division, ar, the rest of the
root, is obtained. By hypothesis, x contains n figures, and,
therefore, x' contains 2 n figures at most; but a contains 2ii-f-l
(the last n of which are zeros), and hence 2 a contains 2 n -f-1 figures
at least; and therefore — is a proper fraction.
According to the preceding discussion, on putting n = 1, n + 1
= 2, it is seen that two at least of the figures of the root must
have been found in order that the method of division used to obtain
the next figure of the square root may give that figure correctly.
Now let this method be applied to finding the square root of 129
to five decimal places. First it is necessary to find the square root
to four figures by the usual method ; and then the remaining three
may be found by division ; thus.
129
1
11.35
2l[
29
21
223
800
669
2265
13100
11325
1775
Now divide the remainder, 1775, which corresponds to J^— a*, by
twice the square root already found, namely, 2270, which corre-
sponds to 2 a, and obtain the next three figures ; thus,
2270) 17750 (781
15890
18600
18160
4400
2270
130
Therefore to five decimal places v'112l) = 11.35781.
2294]
EVOLUTION
299
EXERCISE JaU
Extract the square root of:
1. 4225
7. 1555009
13. 0.464270527876
2. 7056
8. 46335249
14. 40967526745744
3. 14161
9. 537729721
15. 6785644.675329
4. 95481
10. 6402720289
16. 15747849342736
5. 119025
11. 1420913025
17. 306402103296
6. 877969
12. 285970396644
18. 2091478843170721
19. 1656371402322849 20.
8528.91037441
Find the square root of the following, when not perfect sqqares, to
seven decimal places:
21. 5
29. 15.2379 37. ff
45. i
22. 13
30. 0.056 38. Ill
46. V,
23. 22
31. 0.5 39. 1
47. tV
24. 96
32. 0.00789 40. V
48. lO^j
25. 153
33. 0.003 41. im
49. 133412.218990...
26. 101
34. 0.014 42. 7JI
50. 3.14159265358979...
27. 7.65
35. { 43. 8if
51. 2.718281828459...
28. 9.6
36. A 44. f
52. t\v^; nn
53. Reduce to decimals and extract the square root of the
following to within .00001, i. e., to five decimal places:
i; h f; *; 4; ih\ if; 5|; tlr
54. Use the method of 1293 to find the square roots of the num-
bers in 21-31 correct to seven decimal places.
55. Prove that, if a number contains n digits, the square root
contains ^
2n + l — ( — 1)«} digits.
The Cube Root op a Polynomial
294. Since the cube of a + 6 is a^ + 3 a* 6 _|- 3 aft' -f- ^'> then, by
definition, the cube root of a* + 3a^b + 3ab^ -|- ft' is a + 6.
It is desired to find a method for extracting the cube root
a -|- ft when a^ -{- 3 a* b -{- 3 ab^ -\- b^ is given.
Example I. Find the cube root of a' -f 3 a' ft + 3 aft* -f ft*.
a> + 3 a«ft + 3 aft« + ft' [g-fft
a'
3a«
+ 3 aft + ft2
3 a* + 3 aft + ft*
3a«ft 4-3aft« + ft»
3a«ft + 3aft«+ft»
300 COLLEGE ALGEBRA [J295
The first term, a, of the root is evidently the cube root of the
first term, a', of the given expression.
Subtract its cube, namely, a', from the whole expression, and
bring down the remainder 3 a'fe -|- 3 ab^ + b\ The second term b of
the root is found by dividing 3 a*6, the first term in the remainder,
by 3 a*; i. e. , by three times the square of the first term of the root, a.
Since the remainder 3 a^t + 3 ab^ -\- b^ = b (3a^ + Sab -{- fc'),
the complete trial divisor is 3 a^ -f Sab + b\ and it is found by add-
ing 3afe + 6* to the trial divisor, 3 a'; i. e., by adding three times
the product of the first and second terms of the root plus the square
of the second term.
Example II. Find the cube root of 8 P+ 36 Z*m -f 54 Zi»*+ 27 m\
The work analyzed as above is:
8 /8 :}- 36 Pm + 54 Im* + 27 m» I 2 /+ 3»
8P
12 ?»
+ 18/m+ 9m«
12P + lSlm + 9m^
36 I^m -f 54 Im* + 27 m'
36 ?«m + 54 Im* + 27 m»
The cube root of 8 Z^ is 2 Z, which is, therefore, the first term of the
root.
The second term of the root, 3 m, is found by dividing the first
term of the remainder, 36 /*m, by 3(2/)' = 12/*, which corresponds
to 3 a* in example I ; the divisor is completed by annexing to the
trial divisor 12/«, 3(2/) (3 m) + (3m)« = 18/m+ 9 m*, which corre-
sponds to 3 «6 + />* in example I,
295. The method of the preceding paragraph may be extended
to finding the cube roots of polynomials of any number of terms
by regarding a in the model example I, as (he part of (he root already
found, and at the same time regarding 3 a' + 3 aft -|- 6* as the com-
plete divisor.
Thus, if the part of the root already found be x + y, then a*
of example I will be represented by (x + y)*, and if the third term
of the root be + z, the complete divisor 3a* + 3aft + ft* will be
found by adding 3 (.r + y) 2 + «*, which corresponds toSab-^-b^ to
the partial divisor 3 (x + y)*, which corresponds to 3 a*, hence the
complete dix'isor will be
«295] EVOLUTION 301
Example III. Find the cube root of «• — 6x*^+ 15a;* — 20ar» +
^ I x« — 2 a;+ 1
x«_6x«+ 15a;*— 20x8+ 15x«— 6x+ 1
3(x«)*= 3x*
3 (x*)(—2x)+(—2x)«==— 6x8+4x2
3x*_6x8+4x«
— 6x5+ 15x*— 20x3
— 6x*+12x*— 8x»
3 (x«— 2x)«=3x*— 12x3+12x»
3(x«— 2x) (+l)+(l)«=+3x«-6x+l
3x*— 12x3+15x«— 6x+l
3x*— 12x8+15x«— 6x+l
3x*— 12x8+15x»— 6x+l
KoTS.— For economy of space tbe root is placed above the expression.
The first term of the root, x*, is found by taking the cube root of
the first term of the expression ; the first trial divisor, 3 x*, is found
by taking three times the square of the first term of the root.
The second term of the root, — 2 x, is found by dividing the
first term of the remainder, — 6 x^, by the first trial divisor, 3 x*.
The complete divisor is found by adding to the trial divisor, 3 x*,
3 (x*) ( — 2 x) + (— 2 x)* = — 6 x' + 4 X*, which corresponds to
3ab + b* in example I, 8294.
The part of the root already found, a, is now represented by
X* — 2 x; therefore 3 (x* — 2 x)* = 3 x* — 12 x' + 12 x* corresponds
to 3 a*, the second trial divisor.
The third term of the root is found by dividing the first term of
the second remainder, 3 x*, by the first term of the second trial divi-
sor, 3x*, which gives + 1.
The second complete divisor is found by adding 3(x* — 2x)(-f- 1)
+ (+ 1)*, which corresponds to 3 afe + i', J 294, to the second trial
divisor, 3 x* — 12 x'+ 12 x\ which gives 3 x* — 12 x»+15 x* __ 6 x + 1.
EXERCISE Lm
Simplify :
1. V'54x — 36 x« + 8 A» — 27.
-2. VS — 60 X + 150 x2 — 125 x\ .
3. V27 x' — 189 x^^/ + 441 xy^ — 343 y^
4. V300 ab^ — 240 a^b — 125 6» + 64 a\
5. % 'a» — 3 a'^6 + 6 a*b^ — 7 a^b^ + 6 a^b"^ _ 3 aft"^ + b\
6. V'8x» — 36x'^+ 66.x* — 63x=» + 33x« — 9x+ 1.
7. V8 x3 — 12 x2 + 30 X — 25 + 30 x"^ — 12 x"* + 8 x"'.*
•ABSume that negative integral indices observe the same laws as positive integral indices.
302 COLLEGE ALGEBRA [i295
Find the cube root of:
8. ^x*-^W3(i^'(? + |)x*-i)5(pV+6)x> + 3(p(/^+|)ar
P P'
9. 8 x« + 48 cx« + 60 c*x* — 80 c V — 90 cV + 108 c»x — 27 c\
10. (a 4- 6)«'»x' + 6 caP (a + 6) *'"x« + 12 c V (a + fc) «"-x + 8 c*a*.
11. z^ — 3 xi/z"" + 3 xy (1 + xy) z* — xV (6 + xy) z^
+ 3 x«y« (1 + xy) a* — 3 x'y'z + xy.
13. A' + 3x»y«(.v*+l)+3xV(y« + »» + l)
+ x»(y«-3[y« + y»] + l)-3x»y(y« + y«+l)
14. x»— 3xV + 6*V' — 4xV+6a:y — 2x»y*+ Sxy + y*.
15. x»y»+3xV.+ 3x(y«+l) + ?(y+i)+l-
16. xV+3xy+3xV+ ic'y+3y) + 6xV+3xy«
X 2/ x^ys
17. »;+4+3jf(l^_l) + 3,/l n^A/l^l.,1)
+ _L + 3 uv (m« + v* - 1) — uV - i- - ?-!i'.
tt'y' It r
18. /^u + 2?Jt;*+|«tr»+yi;.*
19. at — 3 a (6i — cl) + 3 al (t * — 2 ft ^d) — 6 + 3 6lcl +
3(rti_ii)c* + c«.
20. x'+3xMy+l) + 6xi/ + 3y«(x + l) + y»
+ (x + y)l{\ + nx + y-]),
21. x»y«+ 3 xV + 3x«y + xt — 3 x^yi — 6 xty* — 3xyt — 3xy*
+ 3xi2/'-3/.
•Assume in examples 18—22 that fractional exponents obey the same laws that into-
^TdA exix>nents do.
Ii296, 297] EVOLUTION 303
22. a'x^— 3 a*ba^y^ — b^y^ + 3 ah^o^y*^ + 3 a« cx«»«»
— 3 hc^y^'^z**,
23. Verify the correctness of the following expressions by ex-
tracting the cube roots:
How can we derive one of these formulas from the other?
24. With the assistance of one of the series in 23, calculate the
value of ']/ 37 to seven decimal places.
Here V37 = V Vl - ttfVit* etc. Put x=^^^ in 23.
25. Similarly find the value of 'i/28 to six decimal places.
The Cube Root of Arithmetical Numbers
296. The first step in finding the cube root of numbers expressed
by figures is to point the number off into periods.
Since
1»=1 .-. Vl=l,
10« = 1000 .-. Viooo=io,
100» = 1000000 .-. VlOOOOOO = 100, etc.
Therefore, the cube root of any number between 1 and 1000, i. e.,
of any number which has one, two, or three figures, is a number of
one figure; the cube root of any number between 999 and 1000000,
that is, of any number which has four, five, or six figures, is a num-
ber of two figures, and so on.
Hence, if a point is placed over every third figure in any number,
beginning with the units, the number of points will show the num-
ber of figures in the cube root.
297. If the cube root of a number contains decimal places, the
number itself will have three times as many.
Thus, if .5 is the cube root of a number, the number will be
(.5) (.5) (.5) = .125; and if 2.3 is the cube root of some number,
the number will be 12.167.
Hence, if the given cube number hat decimal places, it mill have
three times as many decimals as its cube root. Therefore, if the given
number has decimal figures, and a point is placed over the units* figure.
304 COLLEGE ALGEBRA tJ298
and over every third figure to the right and left of it, then the number
of points in the decimal part of the number will indicate the number of
decimal places in its cube root.
If the given number is not a perfect cube, ciphers may be annexed,
and a value of the root may be found as near to the true value as one
chooses by repeating the process for finding the cube root,
298. Some examples in the extraction of the cube root of arith-
metical numbers are now given, the rule being derived from the rule
for finding the cube root of a polynomial.
Example I. Find the cube root of 2628072.
2628072
1
138
3(100)«= 30000
3(100x30) = 9000
(30)«= 900
1628072
1197000
39900 X 30 =
3(130)«= 50700
3(130x8)= 3120
(8)^^= 64
431072
431072
53884 X 8 =
The pointing shows that the root will consist of three figures.
The largest cube root in the first period, 2, is 1 ; by subtracting
the cube of 1 the remainder 1628072 is obtained. The figure 1 is
in the hundreds* place, so the trial divisor is 3(1 00)* =30000,
which corresponds to 3a', example I, ?285. The trial divisor 30000
may apparently be contained three, four, or five times in 162807,
but it will be found on trial that 4 and 5 are too large ; therefore 3
will be the second figure of the root. The complete trial divisor
3a2+3a5+6«will in this example be 3(100)«+3(100x30)+(30)«=
39900, which, multiplied by 30 gives 1197000, and subtracted from
1628072 leaves the second remainder 431072. The new trial divisor
(2295) 3{x+yy is in this case 3(100+ 30)« = 3 (130)«= 50700.
431072 divided by 50700 gives 8 in the third figure of the root.
The complete divisor 3(x -f y)«+ 3(x + y)z + z* (§295) will, in this
case, be 3(100+30)«+3(100+30)8+82=53884, which multiplied
by 8 and subtracted from the last remainder leaves zero.
22299, 300]
EVOLUTION
305
299. I. In practice the preceding operation is abbreviated as fol-
lows:
a3=(l)»=
2628072
1
138
3a« = 3(10)«=300
1628
3a6 = 3(10-3)= 90
6«=(3)«= 9
(3a«+3«6+6«)6 =399X3 =
1197
3a« = 3(130)« = 50700
3a6 = 3(130)(8)= 3120
ft«=(8)«= 64
(3a« + 3a6+6«)6 =53884X8=
431072
431072
II. Find the cube root of 60236.288.
60236.288 I 39.2
a»=(3)» =
i>
!7
3a« = 3(30)» = 2700
33236
*3a6 = 3-30-9= 810
ft« = 9«= 81
(3a« + 3a6 + fe«)6= 3591X9 =
32319
3a« = 3(390)« = 456300
917288
3a6 = 3(390)2= 2340
6«=2«= 4
(3a«+ 3 a6 + 6«)6 = 458644 X 2 =
=
917288
800. By repeating the steps for finding the cube root of a per-
fect cube, the cube root of a number which is not a perfect cube
can be found to any desired degree of approximation. Thus, find
the cube root of 6.21, correct to the third decimal place.
as=(l)» =
6.210000000
1
1.838
3a« = 3(10)« =300
3fl6=3(10)8 = 240
ft« = 8«= 64
(3a« + 3a6 + 6«)6=604X 8 =
5210
4832
3a« = 3(180)« = 97200
3a6 = 3(180)3= 1620
&«=3«= 9
(3a«6-|-3a6-f /^)6= 98829 X 3 =
378000
296487
3 a« = 3 (1830)' = 10046700
306=3(1830)8= 43920
6«=8«= 64
(3a« + 3a6 + 6«)6= 10090684X8 =
81513000
80725472
787528 etc
306 C()LLE(iE ALGEBRA [1301
801. If the cube root of a number consists of 2 n + 2 figures,
in case the first n-{- 2o{ these have been found by the usual method,
the remaining n can be found by division.
Let ^be the number; a the part of the cube root already found,
that is, the first w + 2 figures found by rule, with n zeros annexed;
X the remaining part of the root.
Then V iV" = a + x
N = a^+3a^x + 3ax*+x^
N — «' , ^' , a:'
3 a* a 3 a'
Now I^ — rt' is the remainder after n -|- 2 figures of the root
represented by a have been found ; and 3 a* is the corresponding
trial divisor. Equation (1) shows that N—a^ divided by 3 a' gives
Xy the remaining part of the cube root required, increased by — \- — .
It can now be shown that
a 3 (V
so that, by neglecting the remainder arising from the division, oj,
the rest of the root required, is obtained. Thus, by hypothesis,
X contains n figures, and a, n -f 2 figures,
X < 10" and a= 10«"+*
a "^lO^^+i' ^^ a ^10*
X* 10^" 1
^^ 37«<3^10^-' ^^ <3-^0(Fi'
a ^3a«^10^3 X 10"+« ^
for any positive integral value of n,
EXEBCISB LTV
Evaluate to five decimal places:
1. V12167. 2. V 79507. 3. 'v 373248.
VO.054872. 5. V0.000343. 6. Vl56590819.
7. 'v 480.048687: 8. V6.331625. 9. Vo. 007762392.
10. Vo. 000050653. 11. % 105890949891.
12. V829789013773. 13. Vo. 005240822553.
<302]
EVOLUTION
307
14.
V'2,
V},
V6|,
Vsif
15.
V4,
Vi,
V'l3i,
Vsoo.
16.
V6,
v;r.
Vj,
'l/ll-
17.
V9,
1
V2|,
V21V
18.
'|/10,
1
Vioo
V80,
vh-
19.
'v/i2,
Vii.
Vi,
'v^-
20.
V8,
VO.8,
VO.08,
VO.008.
21.
't/73,
V9.28,
V7.3,
Vo.928,
VO.73,
V730.
22.
VO.0928,
V9280:
23.
VO.3786,
V37860,
V3.786,
V 37786.
24.
vii.
V'S,
V8i,
V225.
25.
Vw.
V'A,
V424,
V14.7.
26.
vm
V5H3,
, Viooo,
, Vo.ooi
»/0.64, *!/(
Vl8Hf
V23I3I-
27.
V4826809.
28. V308915776.
29.
V1838265625.
'i/io, Vioo,
Vo.i, Vo.oi,
•i/64, Veli, •
30. Vl2810(
Vioooo,
, Vo.oooi,
).064, VO.OC
[)283921.
31.
Viooooo.
32.
VO. 00001.
33.
)64, VO. 00064.
34. V 74300, V7430, V743, V74.3, V7.43, Vo.743.
35. Having found four figures of the cube roots of the numbers
in examples 3, 4, 6, 11, and 13, find three more figures in each of
these roots by 3301.
36. V282429536481. 37. V208827064576.
302. The n^ Root of a Polynomial.— A rule for finding the n^^
root of a polynomial can be obtained by observing the formation
of the n^ power of a polynomial, n being any integral number what-
ever (2265).
Thus (a + 6)" = a" 4- n a"-i6 + • • •
= a + b.
Therefore
•|/a"4.na»-^6 +
308 CJOLLEGE ALGEBRA [1302
One observes that the first term of the root, a, is the n^ root of
a" , the first term of the quantity whoso root is to be found, and that
the second term of the root, 6, is found by dividing the second term
of the given quantity, »a""*6, by na^'^ or by n times the (n— 1)**
power of the first term of the root.
If the root now found be raised to the n^ power, and subtracted
from the quantity whose root is desired, it will be apparent that the
two terms of the required root have been found.
NoTB.— The student sboald obserre that the process Just deacTibtd Is an extension of
that discussed for the square and cube roots of polynomials and numbers.
Rule. — Arrange the terms according to the powers of some letter.
Find the required root of the, first term of the given polynomial,
for the first term of the root, and subtract tlie root noto found raised to
a power equal to the required root, from the given polynomial.
Divide the first term of the remainder by n times the {n — ly^pouxr
of this root for the second term of the root and subtract the n^ power of
the root now found from the given polynomial.
If there is a remainder, use the same trial divisor as before, and
proceed in like manner till the n^ power of the root becomes equal to
the given polynomial,
EZEBOISB LV
1. Find the fourth root of x« — 4 j;^ + 10x» — 16x* + 19 ac*
— 16x»+10ic8 — 4a; + l.
:r«-4a-»+10:r«-16aJ*+19a:*-16r»+10a:«-4x+l |a*-rH-l
4 (x«)» = 4 ofi_\-Ax^
(j:«-a:)*= :r«-4.r7+ 6:r«- 4.r*+a^
4(a-«)3= ^'jfiUj*
(2^—x+lY = a:«-42r7-f.i0a;«_i6:r5+19^-16:r^+10jr«-4x+l
Ans. X* — X + 1.
2. Extract the fifth root of
32 X* — 240 X* + 720 x« — 1080 x» + 810 x — 243.
3. Extract the fourth root of
x«— 4x^+18x'— 40x<^+91x*— 110x»— 54x»4-108x+81.
4. Find the development of Vx* — x« + x*— x + 1 to 4 terms.
5. Extract the fifth root ofx» + x* + x84-x*+x+lto3 terms.
6. Show that the fifth root of 243.1 is 3.00024.
7. Find the fifth root of 1024.68 and 16805.81 correct to 5
decimals.
CHAPTER III
FRACTIONAL AND NEGATIVE EXPONENTS
308. a", when m is a positive integer (J 9, 3) has been defined by
the equation
a^ = a 'a 'a- , . . to m factors.
In other words, a"* is an abbreviated way for writing the product
of m factors, each equal to a.
304. It has been shown that positive integral exponents obey the
five laws expressed by the following equations:
Distributive formulae
with equal base . .
Associative formula
Distributive formulae
with equal expo- -
nents
II,
II,
II..
III.
IV.
a" X a" = <*"•+»• [? 9, 3]
am^a"" = a*"-", w > n [264, 2]
a*" -T- a" = a^ = 1, m = n t?66j
1
a"" -f- a" =
(a . 6)" = a" . 6»
or (tt . 6 . c)" = a** . 6'
V f-Y = "*-
w<n [?84, 2]
[J85, 4]
} [?85, 5]
W 145, 264]
Thus far an exponent has been regarded as a positive integer; but
it is very advantageous to use exponents which are not positive inte-
gers. The meaning of such exponents will now be shown.
305. No definition has been given of fractional and negative
indices, so any definitions may be given to them; since the law
for positive integral exponents has already been obtained, expressed
by the equation,
a*" X a" = rt*""^*" (I)
it would be most natural and most convenient to give fractional in-
dices and negative indices such definitions as will make the important
relations I — V always true whatever m and n may be.
310 COLLEGE ALGEBRA [J8306, 307
For example, the meaning of a I is required. Since relation I is
to be fulfilled, ai • at = «1 + i= a. Thus ai is such a number that in
case it is multiplied by itself the product is a; but the square root of
a is such a number. Therefore al must be equivalent to the square
root of a; that is, , .-
ai = ya.
Similarly, what is the meaning of ai ? If relation I is fulfilled, then
ai ' ai ' a^ z= aJ+i+i = a. Hence, as above, ai is equivalent to the
cube root of a; that is, i 3 -
' ' ai= V a.
Again, the meaning of a* is required. By relation I,
ai ' ai = al"*"! = a'
therefore, ai = i/a'.
These examples will show the student what is meant by frac-
tional exponents.
In like manner the student may show that
ai z= V^
ai = Va'
7 ablci = 7a^\/lFc,
In the next four articles the definition of the law which fractional
and negative exponents obey will be given in general symbols.
1
306. Required the meaning of a" when n is any positive integer.
By hypothesis,
^n.^n _ to n factors = a» "* " "^ * **"* *^""* = a» = a ; hence
t
a" must be equivalent to the n"* root of a ; that is,
i_ _
a" = "|/a.
m
807. Obtain the meaning of a" where m and n are any positive
whole numbers.
By hypothesis,
a»-«». ... to w factors = «H +*» + .. .ton terms -,^^-^m.
m
therefore a" is the n*** root of a"* ; that is,
m
a»*= "l/a*".
m
Hence, when m and n are positive integers, a" means the n^ root of
the m^ power of a ; that is, in the case of a positive fractional expo-
nent, the numerator denotes a power and the denominator a root.
IJ308-311] FRACTIONAL AND NEGATIVE EXPONENTS 311
308. A meaoing has now been given to the tenns, positive inte-
gral exponent and positive fractional exponent; and it remains to
assign a meaning to negative exponents whether they be integral or
fractional.
For example, it is required to find the meaning of a"'. If the
law expressed by 3304, I, always holds, whatever m and n may be,
then by hypothesis,
a' • a"' = a*~' = a^ z=. a\
^_,_ rt _ 1
The definition may now be formulated in general symbols.
309. What is the meaning of a"" when n is any positive number,
whole or fractional?
By hypothesis, whatever m may be,
Suppose that m is positive and that wi>n, then by 2304| I,
QtH-n y^ ^H —. fjm .
and, therefore, a*"""" = — •
By hypothesis a"* " * = a'^'a''* ;
a"'-a- = ^; [J81, Ax. 7;i
a*
Hence, a~* is the reciprocal of a"; or this result may be written
symbolically in any of the following waj^s:
(1) «-" = —, a» = — , a"rt-" = l.
a" a""
310. It will follow from the meaning given to a negative index
that
a"* — a" = a*"""
when m is less than n as well as when m is greater than n. For,
let m be less than n, then
a^^ a»= — = — = -L_, [J309, (1)1
= a-<»— > [i309, (1)]
311. Thus, for example, according to the preceding definitions,
a!='l/^«, a^ = \/a\ aJ = l/a* = a«.
-s 1 -1— 1 _ 1 -f 1 1
312 COLLEGE ALGEBRA [J3312, 313
It may be seen, however, that it is not absolutely necessary to
introduce fractional and negative exponents into Algebra, for they
only give us a new notation for expressing quantities which we already
know how to represent. But the student will soon learn to appre-
ciate that they are a convenient notation in algebraic calculation.
The notation which has been explained will now be used to
establish some propositions concerning roots and powers.
312. Principle 1. — To show that «« x &« ={«i)» .
Let x = a'*x2'";
a;« = («» • i")» = (a«)"(6"~)» =ah ; [2304, IV ; 1271]
i. e. , x" = ab and x = (a6)* which was to be proved.
Likewise i i
As an example of this proposition,
l/a X Vh = Vah.
It has already been seen that a square root admits of a double
sign. Hence, it may be said of this result that the product of one
of the square roots of a by one of the square roots of h is equal to
one of the square roots of ab.
Similar remarks apply to other propositions of this chapter.
Matters of this kind are discussed much more in detail in works on
the theory of equations.
1 m
313. Principle 2.— To show that («>")"» =a\
111 _! 1 1
Now a" X ^" X e" = (a/>)" X c" = (a6c)~-
By continuing in this way,
1 _! 1 1 1
^'i" X a;» X ^a** X X r^nT = (^i ^j «, • • • « J"'
Let now each of the m quantities in the parenthesis be equal to
a; then
(1) {n-r = («")""= a^' [«30e, 807]
Hence, it follows from (1), that the m^^ power of the n^ root of a u
equivalent to the n*^ root of the m*^ poxcer of a.
11314-317] FRACTIONAL AND NEGATIVE EXPONENTS 313
11. _i_
314. Principle 3.— To show that (a»0 *» = a«'\
1 i
Let X = (a*")".
1
Then x" = a"* and x*" ** = a ;
1
x = a'"".
But by hypothesis x = (a"')";
i i J_
(a"*)" = f/""*. [281, Ax. 7]
m mp
316. Principle 4. — To show that a" = a"^.
m
Let X = a".
Then x" = a"* and x"'* = a*"** ;
mp
x = ri"^
m
But by hypothesis x = a".
m mp
a" = a *'^.
The student may infer from what was said in 2311, that the
propositions just established might be proved without using frac-
tional exponents. Take, for example, principle 1, 2312; it is
necessary to show here that
Let x= "i rt X "v ^~;
hence x» = ("i ;;)" x ("l lY = ah, [2304, IV]
X* = "l ah J which was to be proved.
316. The definitions of 22306-310 have been evolved as a conse-
quence of regarding the relations,
I. a*" X a" = a""^^
and IIL («»»)» = a*"",
as true for fractional and negative indices as well as for positive
integral indices for which proofs have been given; conversely it can
be shown that if these definitions are assumed to be true, the follow-
ing theorems hold.
817. Theorem I. — That a"* x «" = a*""^" is universally true what-
ever m and n may he.
1. Show that a'' X «• = <
+'-
314 COLLEGE ALGEBRA [«318
Thus a" X «" = a^' X f^% [4316]
JL J-
= (aP*)«' X (««'*)^*, [J813J
i
= (o^ X a-^)'', [J812]
That is, it has been shown that the relation,
a"* X «" = a"*^"
is true when m and n are positive fractions; so that it is tnie when
m and n are any positive quantities. * It remains to show that it is
true when either m or n is a negative quantity or both are negative
quantities.
2. Suppose that one of the exponents is negative, say n ; and let
n = — V.
Then a"* X a" = (t"*" x «"" = a*". — = -"* = a*-*'= a"'+".
3. Suppose that loth exponents are negative quantities ; let
lit := — p and « = — (?.
Then
a"Xa"=a"^Xa-''=-ix— =-^=— =a~''-''=a'"'^. [1809, (1)J
318. Theorem II. — That a^-^a^rzia"-" « universally true tcJuU-
ever m and n may be,
1. Let m and n be positive rational fractions, ? and -: then it is
required to prove that
It is known that
rt« -^M" =r a« «, when c>-.
P r p* qr
a« -s- a« = a«» -h a««
[«815]
=(«^r^(«'''r
[J813]
= (ai )'"-'''.
5^ are positive integers and ps > rj, i. e.
[WIS]
a? -7- a« = a «*
p_»*
on carrying out the division indicated in the exponent.
« " Positive quantities," as used in Theorems I-V, are restricted to mtional qaanlitles.
J319] FRACTIONAL AND NEGATIVE EXPONENTS 316
2. Suppose that n is a negative number; i. e., n = — v. It is
reqnired to prove that a* -r- a"^ = a^^^"^.
It is known that a"* -r- a"" = — , by properties of fractions,
= a«a*' = d^a-^-'^\ [HI, 4]
= a"'-<-''>; [{317]
a* -5- a~*' = a'"^^~*'\
Of course, — (— r)=u and m— (— t;)=:»i-|-r. The form m— (— i;)
is retained merely to show the subtraction of the negative index.
3. Suppose that m is a negative number; i. e., w = — Z. It is
a"'
known that a"' -r- a" = — , by properties of fractions,
a"
= a-'a-» [J309, (1)]
= a-'-"; [2317]
a"' -~ rt" = a"'"".
4. Suppose that m and n are both negative numbers; and let
m = — r and 71 = — «.
It is known that a''*-T-a'' = — , by properties of fractions,
a~'
= a-'-a-f-">;
a-'* ~ a-' = a-'-^-*'. [1817]
819. Theorem III. — That (a"*)" = a"^ is universally true whatever
m and n may be,
1. It is known that {a^y = a"*** whatever m may be if r is a pos-
itive integer ({313).
2. To show that {a^)'= a^\
Let X = (a^)'; .-. 7f = {c^Y^d^. [{309, (1)]
Hence x''=a^'', and .•, x = a«'', which was to be proved.
3. To show that (a*")" = a*"", when one of the exponents, say n,
is a negative quantity.
Let n =z — V]
then (a*")" = (a«)-^' = 7—— = -— - =a-*»"-^ a*"". [{{309,(1); 310]
316 COLLEGE ALGEBRA [11320-322
4. To show that (a*")" = a*"** when m is a negative quantity.
Let m = — tc.
Then (a"*)" = (a-«^)" = f A V = -J— = — = a'"^ = a"«.
5. To show that (a"*)" = a*"" when m and « are both negatiTe
quantities.
Let m = — w and n = — r.
Then (a*)" = a(-"')-'' = — i— = -?— = a"'^' = a"*". [4809]
320. Hence whether m and n be integral or fractional, positive
or negative, it follows that formulae I, II, III, i304, are universally
true.
321. Theorem IV. — That (a6)" = a^'b" i« universally true^ toAar-
( IT r 7n aw(f w may he.
1. To prove {ah)^ = a«6', i. e., when n is a positive fraction, "-
It is known that (ahfi = [{ab)p] I ■ [8819]
= liaPbP)]^, [8304, IV]
= [(«V • («^^)T^ [H19]
= [(^61)"]^ [1804, IT]
p p #
= «a • ^« > on taking the q^ root of
the q^ power.
,p
(a6)g= ag6 .
2. Suppose that n is any negative number.
Let n = — V.
Then (a • 6)'' = {ah)-^ =1 = 1 [J309; 1804, I^j
{ahy a^b^
= a-'^i-'^ = a"6". [i809J
(a. • 6)" = a"Z;".
822. Theorem V. — That (J)'' = |^ w universally true whatever
n may be.
1322] FRACTIONAL AND NEGATIVP: EXPONENTS 317
1. Suppose that ii is a positive fraction ? •
It is known that (^)^ = [(ij ]*'' t*^^®^
= (^y [2304, V, for positive
integral exponents.]
Am
[2319]
[J804, V, for positive
integral exponents.]
= — ^ taking the q^^ root of the q^^ power.
p
\b) -"6f
which was to be proved.
2. Suppose that n is any negative number; e. g , n = — v.
"^^ Gy-(jr=7h = k »309;J304,V]
\6/ b^
_ 1 a-*' a'
a
which was to be proved.
^r^^v l-v Jn '
[2309]
EXBBOISE LVI
Express with fractional exponents:
1.
2.
3-
l/a+b,
Vb;
Vn,
Vc',
Vn"
Vw""".
4.
V(a-6)',
Va+3,
», Vo^+b',
Vx'+y«.
5.
V(a-6)a;.
6.
'l/a'-b
"j/(rt_6a;)"-S
"V (x-y)'"-".
7.
VW-
a-y + y')',
'r (a.r'-6x« +
c)".
318 COLLEGE ALGEBRA [1322
8. Express with radical signs:
9. Express with positive exponents:
10. In each of the following, transfer all the literal factors from
the denominator to the numerator:
2z 7a« c« a^ . bx
3x»i/i' 3Z;c-«** a«6-«^ sT^' 7 x'^y'i'
11. In each of the following, transfer all the literal factors from
the numerator to the denominator:
2x«j/l. Qc-»w-»- rr-i»y-i- ^^'^y* . 6a-«6-*
-— — , ox 1/ , X sy , ^ ^ — .
2i «' 5c*
Express in roots and powers with integral positive exponents:
12. hi, cJ, dl
13. ni, pi, gi,
14. yi, w"+i, t?"-l.
15. aH, 6H, c'K . e£^l.
16. «-l, x-i, x-H, y-»l.
^w 0.5 1.2 —0.26 — 1.7&
J.I.X, X, X, X*
18. («-^))l, (c+r/)l, (ax-ft)i, (7x-3y)t.
^ 6 n— 1 m—t
19. (/>x — g)9, (a — 2»x)a, (ax — 6) a , (x — y) * .
20. (a'-h^)-h^ (x«-y2)-|^ (,^_n)i'+l, (p— g)"-i.
21. (x« + 3x-5)J, (x« — 2xy + 3y«)-i.
Calculate the value of the following expressions:
22. 36i, 27i, 16l, 32i
23. 48, 8l, 27*, 64l.
24. 32?, 64*, 64*, 81*.
25. (3|)1, (3|)l, (5,Vi, (5iV)*-
26. (0.25)1, (0.027)1, (0.0081)1, (0.00032)*-
27. 32"', 49" ^ • 16"'', 81"-^.
28. 36-i, 27-t, (0.16)-i, (0.0016)-t.
29. {j\)-i, (^j)-t, (V^^)-l, (if)-l.
8322] FRACTIONAL AND NEGATIVE EXPONENTS 319
Apply the theorem and formulas of this chapter to exercises 30 to
54 and express the results with radicals and positive exponents.
30.
aiai, bi-bi,
ci • cA,
di • dA.
31.
mi • m~i, ni • »~A,-
pi p-iV,
q-i ■ q-A
32.
o* • al, a" ■ a~i,
ai ■ a~ij
a-a~i.
33.
ai • y^a, c~i • l/c,
xl-Vi,
yi-Vy.
34.
xiV^, y»-Vir^,
ui-Vu,
vi }/v-\
35.
abi c • a~ibci.
36.
xiyhi'^'ly^iz-i.
37.
ai hi ci </A
ai' 6}' ci' c^i '
38,
mi ni
mi ni
pT2 qi
39.
xA yi iiii vi
xi yT^'i ui t?2v
40.
ai al
-I , ->
ai a*
41.
ai ai aA ai
Va^' *]/a V«' Va
42
X* ' xii
43. ai-6i, ai • 6f , d - di, ci-rfJ.
44. mi • nl, mi • ni, j>l • gA, jji • ji.
ai ai a^ ai xi xi wi wi
45. — » — > — :> — • 4b. — > — » --) — -•
6» 6* 67 6t yt yir t;» t?T5
"■(l)'(#(l)'' ('*)' (^0* ('»)*■
48. (at)'. (6|)«. (i)'», (l,)>. (22l)». (3,)-'.
49. (a«)i • (6»)i • (c»)i • (e?»«)i 50. (x«)» • {y^)\ • (w>»)t • (t;*)f.
51. («*)«. (i»)^ (o*)^ (</*)».
52. («.-i)\ (n*)-, (^-i)^ (,«)-*.
53. (a.6»)i, [ah'^f, [a^d-^)-\ [a^bj.
54. (xV)\ (xV)" (x*/)-^ (.V)".
320 COLLEGE ALGEBRA [8322
Simplify exercises from 55 — 70.
fiZx+9y fj^-iv a^'-v
55.
56.
57.
58.
59.
60.
61.
62.
64.
65.
67.
68.
69.
70.
(^ a^b^ a^
a'b* a^b'^' a'b"^'
(x+l)6»-> (m— n)a"6«» (a + b)ab
(a_l)»(a;— 1)« a^{x-yy nHx+i/y
{a-iya-xf ~a(y-xy' a(x-y) *
2a^x^ ^ay^ by^ 4a^6* Ihbc^ 2cd_
3xV^ • ~c^V*~ a^+^6^V^^
bj. — — . . , . .
^m — nJjn—p^p—m fjH—php—m^m—n x'*7J^Z^
^,^ 2a'b'^c* ^ 4aV,^c^ 4a^x^y _^ Sa^xy*
3xy^ * 5.1^^*' bb'^z* ' 3bc^z^
5^,11 /^n-l^n-2 ^ 3^,n-l/^^.n+l
5^658cn-H ^ _3flV^*C_
^j3x-yJ^2v-Zx fjlx-ZyJyly-tx
J323] FRACTIONAL AND NEGATIVE EXPONENTS 321
323. The laws which apply to the exponents of simple expres-
sions apply also to the exponents of compound expressions.
1. Multiply 2xi—3xi—4+x-iby3xi+x—2xi.
Arrange the work thus:
2xi—3xi—4 +x-i
3 xi+ X —2x1
6x*—dxi—l2xi+3x
2xi^ 3x*— 4x+ xi
— 4x1+6 a:+ar5-2xJ
6x2— 7x3 — 19xi-[-5 x+dxi—2xi
2. Divide x-\-xit/i+yhy xi-{-xii/i+yk
X -\- xiyi + x^y^
x\ + x\y\ + yi
xi — x^y^ +yi
— x\yi + y
— x%y\ — x3 v^ — '^^y^
+ ^r^y'^ + ^h^ + y
+ xiy^ 4" •'**.V^ + y
EXEBCISE LVII
Multiplication
Simplify the following products, leaving the results with positive
exponents:
o. • \' ac ' — : — •
4. (^ v7+'i///-') . (^ ;;^— V6«).
»■ (»v='-.7:.)C"'--w.)-
322 COLLEGE ALGEBRA [J323
9. x'" — x**y* + y*" by a;"* + x"y" + y*".
10. at + 6i + ci-l6 by ab-i — ai+ hi.
11. xi — xyi + xiy — yi by x -(- xiyk + y.
12. xl — x' + xJ — x« + .tJ — ie + xi — 1 by xi + 1.
13. 1 — xy-* + x«y-* by 1 + xy-» + xV*-
1 4. a-'6« - a^ft — 2 a by 2 a«6^* + 2 a»6-« - 4 a*6 -».
ZHvtsion
m p m __£ _m £
15. a" -f- a^. 16. a" -^ a «. 17. o " -t- a«.
18. a"^-^a"«. 19. cai -f- tfa*. 20. aUl ^ a-ib-ic,
21. (a — ?;) -T- (a* - bl). 22. (a + «») -^ (ai + fci).
23. (a» — 2 aM — aUi + 2 6i3) -f- (ai - hi),
24. (a + 6 + c — 2 j/^ — 2 ^/o^ — 2 l^bT)
^ (l/a - v/6 - ]/c + 2 V/^).
25. (5a«-41a6+42fe«)''v^ -^ /"'v ^ - -^ V
26. (*l/ a^ - 'v b^) -^ (V« - *V 6).
Miscellaneous
27. Expand (xi -f yl)'; (2x-« ~ x«)*; (xy"* — yz'')\
28. Extract the square root of:
a^b-i-iaib-i + b^ia'ibi+a-^bl- and ^ + ^ + ^ ^""^ '
29. Extract the cube root of 256 ai — 512 a + 640 ai — 512 ai
+ 304 — 128 a-i + 40 a"! — Sa'^ + a"*.
30. Resolve into prime factors with fractional exponents:
%/l2, *v 72, V96, i/24, V576; and find their product
Simplify:
31. [(—x-")^]-"; (xyP) (x"y~«); (a'*^^6<») (a"^^6"'»).
- (9)'-C-£)'"'' (..-a--)-{(.-)--=^]iT
CHAPTER IV
fiSLATIVB MAGNITUDE OF POSITIVE AND NEGATIVE QUANTITIES-
INEQUALITIES BETWEEN TWO ALGEBRAIC EXPRESSIONS
CONTAINING UNKNOWN QUANTIi;^S
324. The Conventions Concerning the Relative Magnitudes of
Positive Numbers. — Let a and h be two positive numbers; then, in
case the difference a—b is positive, a is said to be greater than 6,
and is written a > 6 (?4) ; and if the difference a — b is negative, a
is less than b, and is written a<6 (34).
This definition is extended to the case when a and b are any
positive or negative numbers, and it is agreed to consider a as
greater than b when the difference a — b is positive, and as less than
b when the difference is negative.
In accordance with this convention,
Every positive number is greater than Oy and than every negative number]
and reciprocally,
Every negative number is less than 0, and than every positive number.
Of two negative numbers, the greater is that number which has
the less absolute value.
For example,
.... -4, -3, -2, -1, 0, 1, 2, 3, 4, ... .
form a series of increasing numbers, and
.... -4<-3<-2<-l<0<l<2<3<4 ....
It may be remarked, moreover, that if the difference a—b is positive,
the difference 6— a is negative; and it follows, therefore, that each
of the inequalities,
a — 6 > 0, a > 6, ?> — a < 0, ^ < a,
expresses the same fact.
324 COLLEGE ALGEBRA [««325-327
The preceding can be illustrated as follows: Let.O be a fixed,
and M a variable point on the indefinite straight line X* OX,
X' 0^^-^ .? ""'if X
Let X be the distance of M from 0 reckoned positively to the right
(JI21, 23), and negatively to the left. If the point M is at first to
the left of 0, at negative infinity, and moves continually from left to
right, to 0 and through it, the M moving on to the right to positive
infinity, the number x, which is the measure, in magnitude and sign,
of the distance 03/, increases from — OO ^ ^» then from 0 to -|-
00 , and it is said that the distance 03f increases from —00 to zero,
then from zero to + 00 .
325. General Definitions of Equality and greater or lesser In-
equality of Negatives.
ThiORIM. — «>,=,<; — 6, according as &>,=,<«.
For f as Z;>, =, <o
^'U^a + a+i,>, =, <-6+6+a. [Hl,l;«39]
The same reasoning, used in proving VII, 837, proves that
(2) I*® «+ c> or <?» + c
• ( a > or < 6.
Hence, it follows from (1), on account of (2) and Law VII, {37, that
-«>, =, <-6. Q. E. D.
Similarly, (3) — a < 0 </>.
Inequalities Between Two Algebraic Expressions which
Contain One, Two, or Three Unknown Quantities
326. Suppose A and B are two algebraic expressions involving
one unknown quantity x ; by the solution of the inequality
A:> B
is meant the values of x for which the numerical values A^ and
B^ of A and B, found by substituting in A and B these values of x,
will satisfy the inequality
A^ > B,
Two inequalities are said to be equivalent when they have the
same solutions.
327. The theorems which were demonstrated for the solutions of
equations (solutions of equations and the transformations resulting
from them) are applicable, with certain slight modifications, to
inequalities. These rules are the following.
then is
(2)
For
or
,* ,
(3)
But from (1),
(4)
or
J328] INEQUALITIES 325
Theorem T. — If the same finite quantity is added to or subtracted
from both members of an inequality^ a new inequality is formed^ equiv'
alent to the first; and reciprocally.
Consider the inequality,
5>3;
w+5>m+3.
(m + 5) - (m + 3) > 0, [Def . J324]
m+ 5 — m — 3> 0
2>0.
5-3>0 [Def. {324]
2>0.
Hence, (1) and (2) are equivalent and (2) can be derived from (1) by
adding m to both members of (1), which is the proof for the first
part of the rule.
Similarly, if (1) 5>3.;
then is (5) 5 — w > 3 — m.
For (5 - wi) - (3 - m) > 0 [ J824]
or 5 — wi — 3 -f- wi > 0
2>0.
Hence, inequality (5) is equivalent to (1), and (5) can be derived
from (1) by subtracting m from both members of (1). ^
328. General Proof of Theorem I. — Consider the inequality
(1) A>B,
in which A and B are algebraic expressions in x; let C be any alge-
braic expression in x which is finite for all finite values of x. By
adding C to both members of (1),
(2) A+OB+ a
It is now necessary to show that (1) and (2) are equivalent inequali-
ties, and the converse. Let A^ and B^ be the values which A and B
take for x = x^; then by definition (J326),
or, what is the same thing written differently,
(3) ^1 - ^, > 0. [8324]
For this same value of x, C takes a finite value C^, and it is necessary
to show that (4) a^+ C^>B^+C^,
That the relation in (4) should hold, it is necessary and sufficient
that the difference A^ + C^^(B^+ C^ is positive (J324); that is,
that (5) a^+C^-{B^+G;)^A^^B;^^,
The result (5) is true on account of the truth of relation (3).
326 COLLEGE ALGEBRA [S329
Reciprocally, let x^ be a solution of inequality (2) ; then by hypo-
thesis,
or, what is the same thing,
or, finally, ^^ — ^^ > 0 and A^ > B^,
Therefore, any solution of inequality (2) is a solution of inequality (1).
Hence, inequalities (1) and (2) are equivalent.
Similarly, by adding — C to both members of inequality (1),
inequality (6) is formed:
(6) A-C> B^ C,
equivalent to inequality (1).
Ajyplication. — This theorem justifies the removal of a term from
one member of the inequality to the other member on changing
the pign of this term.
Example. The inequality,
5x — 2>4x+7,
is equivalent to
5x— 4x> 7 + 2,
or to the inequality,
X > 9.
329. Theorem II. — By multtplying or dividing the two members
of an inequality by the sam^e quantity whose value is always finite
and j)Ositivej a new inequality is formed equivalent to the first.
For example, suppose a + 1 > 5.
If both members of the inequality be multiplied by 4, then it is true
that
4(a+ 1)> 20;
because we have in the first case by definition
f/+l_5>0 or a — 4>0,
and in the second case
4 ri + 4 — 20 > 0 or 4 a — 16 > 0,
i. e., 4(a-4) > 0,
which is true, since « —- 4 > 0.
More generally, consider the inequality
(1) A >A
in which A and B are certain algebraic expressions involving one or
more unknown quantities jc , ... Let C be a positive number
J330] INEQUALITIES 327
or an algebraic expression which may involve x . . . , but such
that, for every finite value assigned to the unknown, the expression
C takes a finite and positive value. If both members of inequality
(1) are multiplied by Ca new inequality (2) is formed:
(2) AC^BC
which is equivalent to (1).
For, inequality (1) is equivalent to
and inequality (2) is equivalent to the inequality
CU-^)>0;
aad, since the factor C is by hypothesis always positive, in order
that the product C{A — B) may be positive, it is necessary that the
factor ^ — ^ be positive. Therefore the second inequality is equiv-
alent to the first.
330. In case the factor V is negative and both members of the
inequality
(1) A^B
are multiplied by — C, the sign of the inequality will be reversed,
that is,
(2) ^AC<-BG,
For example, suppose
(3) aj + 3>7,
and multiply both members of this inequality by — 5, then will we
have — 5 X — 15 < — 35.
The sign of inequality will be reversed, because, by definition
— 35 — (— 5a: — 15) > 0
or —^35+ 5x+ 15> 0
5x — 20>0,
i. e., (4) 5(x-4)>0,
which is true since, by definition, from inequality (3) we have
x+3 — 7>0, orx — 4>0;
and in general, the inequality (2) is equivalent to
or 0< (7(^ — 5).
Since C is positive, and the product C{A—B) is positive, there-
fore A — B\^ positive and A^ B, Therefore the inequality (2) is
equivalent to (1).
328 COLLEGE ALGEBRA C*331
Example 1. Let C= — 5; then the inequality
(1) A>B
is equivalent to the inequality
(2) — 5^<-55;
because the latter is equivalent to
_5^-(-55)<0,
or — 5U — 5)<0;
i. e., — 5(^ — B) is negative, and, therefore,
A—B is positive, and A > B, (1).
In particular, if both members of an inequality are multiplied bj
— 1, or, what amounts to the same thing, if the signs of all the
terms are changed, it is necessary to reverse the sense of the sign
of inequality.
Example 2. Let C = 4 x — 5. The factor 4 a: — 5 is positive
or negative according to the value assigned to x, and consequently,
the inequality,
(1) A> B
is not in general equivalent to the inequality,
(2) ^(4x-5)>^(4x-5).
If it is agreed to give to x only such values as are greater than |,
the factor 4 x — 5 is positive, and, under these conditions, the in-
equality (1) will be equivalent to inequality (2), by i329.
But if it is agreed to give to x only such values as are less than |,
the factor 4 x — 5 will be negative, and consequently, under these
conditions, the inequality (1) will be equivalent to the inequality
(3) .4(4x — 5)<i?(4x — 5).
331. Application of the Preceding Theorem. This theorem
makes it possible to replace an inequality which contains fractional
terms b}' another whose terms are integral. This process is called
the clearing of denominators. If the denominators are positive
numbers, an equivalent inequality is formed by reducing all the
terms to the same denominator, and by multiplying both terms of
the inequality by the common denominator.
Consider, for example, the inequality
If all the terms are reduced to the same denominator, 15, and if both
members are multiplied by 15, the result is the inequality
75x — 9 + lOx > 90 + 6 — 7x,
8331] INEQUALITIES 329
which is equivalent to the first, and in which all the coefficients are
integral.
If the inequality contains an unknown quantity in the denomina-
tors, then, in order to transform it into an equivalent inequality in
integral terms with respect to this unknown, it is necessary that the
expression by which the two members are multiplied, be chosen in
such a way that it can not be negative for any of the admissible
values of the unknown quantity. Consider, for example, the ine-
quality,
4x — 3 5(4x— 3)
This inequality is not always equivalent to the inequality,
(2) 4x5-|-5(7-4x)(4a;-^3) > 3x + 5 (3a; — 2) (4 a; — 3),
obtained by multiplying both members of (1) by 5 (4 a; — 3) because
the multiplier 5 (4 x — 3) is positive or negative according as x is
greater or less than |.
If all values of x greater than \ are considered, inequality (1)
has the same solution as inequality (2), but if values of % less than
\ are considered then inequality (1) has the same solution as the
inequality
(3) 4x 5-f 5(7 — 4x)(4x-3) < 3x + 5 (3x — 2) (4x- 3).
However, if both members of the inequality (1) are multiplied by
5 (4 X — 3)', the result is the inequality,
(4) 4X5(4a;-3)-f 5(7-4a:)(4x-3)«>3.r(42-3) + 5(3j;-2)(4^-3)«,
which is integral in x and which is equivalent to inequality (1) for
all real values of x, Since 5 (4x — 3)* is positive for all real values of
X , and since when both members of an inequality are multiplied by a
positive quantity the inequality subsists in the same sense.
Inequality (4) can, by transposition and by taking out the com-
mon factor 4x — 3, be written in the form,
(5) (4x — 3)(140x«— 282X+115) < 0,
which is also integral in x, and equivalent to inequality (1).
When both members of an inequality are integral algebraic
expressions in the unknown quantity (or quantities), the degree of
the inequality is the degree of that member of the inequality which
is the highest.
330 CJOLLEGE ALGEBRA [1332
Solution op an Inequality of the First Degree in One
Unknown Quantity
332. To solve an inequalify of the first degree in one unknown
quantity cc, is to find all the values of x which satisfy this ine-
quality. Since these solutions are at once evident when the fiist
member of the inequality is x, it is said that an inequality of the
first degree in x is solved when it is replaced by an equivalent
inequality, in which the first member is x.
It will next be shown that every inequality of the first degree in
one unknown quantity x can be written in the form
ax>. h.
If the inequality has the form,
(0 A>B,
transpose all of the terms in x to the first member, and all of the
known terms to the second member; then combine all of the tenns
in X into a single term and all of the known terms into a single
term, thus obtain an inequality of the form
ax > h.
If the inequality has the form
it is replaced by the equivalent inequality
OV) B:>A,
which can be dealt with in the same manner as the inequality (0.
Suppose then the inequality (i) has the form
(1) ax>5.
There will be three cases according as a is positive, negative, or zero.
If a is positive, the inequality (1) is equivalent to the ineqaaiity
(2), by 1329.
(2) X > t
a
If a is negative, the inequality (1) is equivalent to the inequality
(3), by {330.
(8, .<K
If a is zero, the inequality (1) is satisfied by all possible values of
X in case b is zero or negative, and can not be satisfied by any value
of X in case b is greater than zero.
J 333 J INEQUALITIES 331
Example 1. Solve the inequality
(1) 2x + i-7<4-7x-f
Transposing according to Theorem I, 8327,
(2) 2x+7x<4-i + 7-i,
collecting terms, (3) 9 x < 1 0,
dividing according to Theorem II, 2329,
(4) a:<V.
Example 2. Find the limits between which x must vary in order
to satisfy the inequality
(1) M+^-<2-ii-
The inequality (1) is equivalent to the inequality (2), by J327,
Theorem I,
or by uniting the terms in (2),
6 0^+4-30^ + 3
^^^ 2(x-l) <"'
3x4-7
and simplifying, (4) — -^^ < 0.
Multiply both members of inequality (4) by the positive quantity
(x — 1)* and we obtain the equivalent inequality (5),
(5) (3x + 7)(x~.l)<0.
The inequality (5) is satisfied by any value of x between — \ and 1 .
Therefore the given inequality (1) is equivalent to the double in-
equality -J<x<l.
333. Theorem III. —Let 7-^, ^, ^, . ... ^ he fractions of
^l \ ^3 . ^n
which the denominators all have the same sign, then the fraction
«i + «g + ^a +••■•+ ^,.
^+^+^+ • • • • +^
lies in magnitude between the least and the greatest of the fractions
'h 'In "ji 5b.
''.' '',' v ••■■*»
a a a a
Proof. Let 7^, 7*, r*, . . . . 7-" be arranged in ascending order of
(J, f>„ o^ 0
magnitude, and suppose that all the denominators are positive; then
332
COLLEGE ALGEBRA
[«339
^ = 7I, therefore, <i =b x^
^ > -1, therefore, a >b Xj^
? > f N therefore %>h^x ^'
^>JS therefore a^> 6, x^
therefore, by addition,
«i+ ««+«»+ +««>(^ + ^+^+ . .
therefore,
\+^+^+ — +K K
Similarly it may be proved that,
+'-^^
It can be proved in like manner that the theorem holds when the
denominators are negative.
Example. Show that if the denominators of the fractions
b' b'' b"'
are positive, the quantity V\,,'T_^.,, lies between the smallest and
the largest of these fractions.
Suppose the following to be the order of magnitude of the given
fractions,
(1) «<«'<?:.
then it will follow from relation (1) that
(2)
since b b\ 6" are positive.
Hence, it follows from relation (2) that
(3) a + a' + a">(i + 7/ + 6")^-
a a
l~b ■
0
b'^ b
6
b"-^b
.. a" > 6" X r ;
6
2334] INEQUALITIES 333
Therefore, by division, since t -|- fc' + 6" is positive,
(4) ^ ^', ^,, > 7 , the least of the given fractions.
Similarly, it can be shown that
(5) ?-i-^ ~ < ?- > the largest of the fractions.
Hence, from (4) and (5),
a a + a' + a^^ a^
6 ^ 6 + 6' + 6" ^ 6" ^ ■^•
334. Examples.
1. Prove that a« + 6« > a«6 + ab^.
This is the same as to prove that
a» — a«6 - a6« + 6' > 0, [J827]
or that (a« — b^) (a — 6) > 0,
which must be true since both factors are positive or negative accord-
ing as a is greater or less than b,
2. Prove that
It can easily be shown that
(a^ +y" + «') (^" + y'" + «'") - (^^' + yy' + «^')«=
The second member of this identity can never be negative because it
is the sum of three squares and can only be zero when each of the
parentheses is zero, i.e., when
(1) yz' — y'z = 0, (2) zx' — z'x = 0, (3) xy' — x'y = 0.
Divide equation (1) by y'z', equation (2) by z'x% and equation (3) by
x'y' and obtain
(4)^,--, = 0, (5)i-^, = 0, (6)^,-»-=0
y' z' z' x' x' y
- =^ =1
•• x' y' 2''
Hence (x« + y* + ««) (x'« + y'« + «'«) > (xx' + yy' + 22')«,
except when 3 = "^ = "^ » ^^ which case the inequality becomes an
equality.
3. Under what circumstances is
3x — 1 , 2x — 3 ^ ^ -«
-+ -> or <5?
X — 2 X — 5
334 COLLEGE ALGEBRA [J335
1st. Let us suppose that x does not lie between 2 and 5, and
is not equal to either of these values.
Then (x — 2) (x — 5) is positive, and we may.multiply by this factor
without reversing the signs of the inequality (3 {329, 332, Ex« 2).
Hence ^^i:Il=l + 2^> or < 5,
X — 2 X — o
accordingas(3x-l)(j;-5) + (2j;-3)(x-2)> or <o(x-2)(r— 5),
according as 5x* -— 23x + 11 > or < 5x' — 35x+ 50,
according as 12 x > or < 39,
according as x > or < 3J.
Under the present supposition, x can not have the value 3J, but
it follows from the preceding inequalities that if
X > 5, then /' > 5, and if x < 2, then F<ih.
2d. Suppose 2 < X < 5. In this case (x — 2) (x — 5) is negative
and the sign of inequality is reversed (3330) and we must reverse
all the signs in the preceding inequalities after multiplying by
(x_2)(x-5). *
It follows therefore that if
2 < X < 3 J, then F > b, and if 3J < x < 5, then i*^ < 5.
335. Powers and Roots. The following principles are deduced
by means of the lemmas
(i). If both members of an inequality are positive and are raiied to
the same integral power ^ the resulting inequality subsists in the sa$ne
sense; that is, if a > ft, then a"" > 6»,
in which- a and b are positive, and n is a positive integer.
For, if a" > 6», then by definition, 3324,
a« — fe» > 0,
or
(1) (a— ft) (a»-i+a»-«6+a«-«ft«+ t-aft"-«+ft»-')> 0. [3102, i]
Since a > ft then a — ft is positive by definition, and since a and ft
are positive by hypothesis, the second parenthesis in inequality (1) i«
positive, Q. E. D.
E. g., 7 > 5, and 49 > 25.
A similar proof holds for the two following principles:
(ii). If both members of an inequality are negative and are
raised to the same j^ositive odd 2>ower, the resulting inequality subsists
in the same sense; that is, if
— a>—b, then (— «)2«+i>(— ft)«"+i when 7i = l, 2, 3, 4 • • •
E. g., -5>-G, and (_5)'>(-6)», or -125>~216.
1335] INEQUALITIES 335
(iii). If hoik member i of an inequality are negative and are raised
to the same positive even power, the resulting inequality will he of
the opposite specieS; tJiat is, if
_ a > — fc, then (— a)** < (— ft)*".
Kg., _ 3 > - 5, and (- 3)* < (- 5)*, or 81 < 625.
(iv). If the same principal root of both members of an inequality
is taken, the resulting inequality subsists in the same sense; that is,
if
ay>b, then "v^a> Vft.
Proof If Va > V6
then (Va)" > (Vft)**
a>b Q. E. D. tmi, J336, (i)]
E. g., 169 > 49, and 13 > 7; — 64 < — 27, and — 4 < — 3.
(v). If the same negative even root of both members of an inequality
is taken, the resulting inequality subsists in contrary sense; that is, if
ayb, then —*^\/a < — ^\/b.
E. g., 169 > 49, then — 13 <— 2.
The proof is similar to that of (iv).
EZBBCISE LVm
Prove the following inequalities, supposing that all the letters repre-
sent positive quantities :
1. 3 X — 5 > 34 if X > 13.
2. |x+ |x < 19 if X < 15.
3. 6x* + 7x — 3 > 6x« + 17x— 13 if x < 1.
4. (x+2)(x+3) > (x — 4)(x — 5) if x > 2.
Find the limits of x in the following:
5. (6x+l)» +15 > (2x-3)(18x+ 5).
6. (5x - \f - 20 > (3x + 4)« + Ux - 3)«.
7. (x+2)(x-3)(x + 4) < (x-l)(x + 2)(x + 4).
8. For what value of y is ^^ < '^ ~ '' + 2, if a and b are
a b
positive, and a^b?
336 COLLEGE ALGEBRA [?335
Find the limits of x and y in the following:
(4x+6y<100 (15x- 6y > 3
(5x-2y = 13. ^"* I 6x+10y = 58.
11. Find the limits of x when 6a; + 8 < 9x — 13 and
16x— 25 < 12x + 5.
12. When will a^h + a6» > 2a«6«?
Prove that
13. a8 + 66«> 6(2a+ 56) if a > 6.
14. (m« + n«) • (m* + n*) > (m» + n»)« if m > n.
15. a*fc«+a26*>2a»6Mf a>6. 16. (^4^)' > "^•
17. a6(a+4) + ^c(6 + c) -I- ca(c+ a) > 6a6c.
18. a + -Sf 2 if a >0.
19. m — wx>.p — ax if x >• ^~"^.
20. a6+i»c + ca<l if a2 + [^2 + c2 = 1.
21. Which is greater, ^^ or -^?
22. Under what circumstances is
23. Under what circumstances is
x'+5x> <8x«+ 14?
Prove that
24. ^^<-2-T^» if a>xandx>0.
25. (^ + » + ^)(" + ^ + ^)^9.
\a 0 c/ \x y z/ ^
3y + 5:r < 8
27. ax+fey + c2 <1 ifa« + 62 + c« = l and x«+ y« + 2« = 1.
28. 6c(6 + c) + ca(c + a) + afe(a+ 6)<2 (a»+ 6»+c»)-
29. x5+3^+ 25>3xy2 if x+ i/+2>0.
30. If x'=a'+6* andy=c* + t/^, xy>ac+6rf, or x^>ad[+ fee.
31. If a > feja"— 6*"< ?na'""Ha — 6), w being a positive integer.
32. (a+ 6— c) (a+c — 5) (i + c — a) <:ahc,
33. xV + y*« +«*^>3ry*+y2*+2x*.
34. (y_.«)(2;_x)+(2-x) {x-y) + (x-^^) (y- 2) ^ 0.
35. yz + «x + xy ^ x' + y* -f 2'.
26
CHAPTER V
IRRATIONAL NUMBSRS AND LIMITS
886. The System of Rational Numbers Insofflcient.— We have seen in
Books I and II that a system of rational numbers is sufficient for the use
of the four fundamental operations and supplies the means for expressing
the solution of all problems which can be solved by these operations.
However, the system of rational numbers does not fully meet the needs
of Algebra.
A great central problem in Algebra is the equation. A number system
which is algebraically complete should supply the means for expressing
the solution of all possible equations. The system of rational numbers
enables one to express the solution of equations of the first degree in
one, two, or more unknown quantities (Book II); but it does not even
contain symbols for the root« of such elementary equations of higher
d^rees as
(1) ^=2,
and (2) a:«=-l.
To solve equation (1) extract the square root of both members, then
x=±V2= ± 1.4142 .... [J291]
The value of the symbol 1^2 can only be found approximately by the
device of extracting the square root. According to the number of decimal
places, to which this operation may be carried, the V2 is expressed
approximately by a rational number; if to three places, then by
1.414 = lili = lil^.
1000 1000
To solve equation (2), extract the square root of both members, then
The V—1 can not be expressed in terms of rational or irrational numbers,
and belongs to another system of numbers called imaginary or complex
numbers ({104, Ex. 3). But how is the system of rational numbers to be
enlarged into a system of algebraic numbers which will give us the means
to express the roots of all possible equations, and at the same time be
sufficiently simple?
887
338 COLLEGE ALGEBRA r*337
The roots of equations of a degree higher than the first, sach as are rep-
resented by the equation
when n=s 2, 3, 4, etc., are not the results of a simple elementarj' operadon
as are the negative of subtraction and the fraction of division. For
example, the roots of the equation
x^ = a (a positive number)
are found by extracting the square root, x= ± Va; and it will presently
be shown that the roots of the equation of the second degree
are obtained by means of the four fundamental operations and evolntion
(extracting of the square root), and that these operations and evolution often
enough repeated yield the roots of the cubic and biquadratic eqaatioDB.
But the roots of the fifth and higher degrees can not be determined by
means of these five operations.
However, an investigation shows that the forms of numbers necessary
to complete the algebraic system may be reduced to the following: the
symbol V —h called the xraag'tnaTy^ the indicated root of the equation
j;* = — 1 ; and the class of symbols called irrationals, to which the roots of
the equations x*=2, a:'=4, etc., belong.
The remainder of this chapter will be devoted to the discussion of the
irrational, and Chapter VI to the discussion of imaginary numbers.
IXTRODUCTION
337. The first step toward the establishing of the irrational system of
numbers is to prove the following theorem.
Theorem l.^Tliere are numbers, far example the r"* tool of a potititt fro/c-
tion ivhose terms {one or both) are not the r*^ pouter of posUive integers, vhick can
not be expressed as integers or as frcuiions.
As a particular case of this theorem, prove that V2 can not be a fraction
or an inU^ger.
I^t, if possible,
(1) v/2 = 3,
where a and b are integers which do not have a common factor.
Squaring equation (1),
nS
(2) 2 = 'f =
a
6« 6 6
Since - is in the lowest terms, f • ^ = ^ is also in the lowest terms.
6 0 0 0*
Since a* and 6" can not have a common factor, ¥ is not contained 2 times in
a* as equation (2) requires, therefore it is not possible that V2 is a rational
fraction ^.
b
«338-340] IRRATIONAL NUMBERS AND LIMITS 339
PROOF OF GENERAL CASE
Let ^ l^ a fraction in which A and B (one or both) are not the r***
XK)wer of positive integers, then prove that *-^'^ can not be expressed as a
fraction f • Since it does not alter the value of a fraction to divide both of
o
its terms by the same number we may regard ^ and ^ as in the lowest
terms. Let, if possible,
(1)
and hence (2)
r [a a,
B b^'
Since ? is in its lowest terms, then ^==? .^ tor factors = ^
b B b I ^
is in its lowest terms. It follows from (2) that
(3) ^ = ^-lr-
Since A is an integer and a** and b^ do not have a common factor, it fol-
lows that b^ is a divisor of B. Therefore
^^) iand ^^iP,[wh«re*i8 an integer.
But by h3rpothe8is A and B have no common divisor, hence ifc = 1, and
<« i izt.
But since, by hypothesis, A and B (one or both) are not the r**» power of
pKitive integers^ equation (5) (oneor both), and therefore also equation (1),
are not true. Consequently, ^^^ can not be expressed as a fraction.
388. In case b is assumed equal to 1, the preceding proof shows that
'"'SJH ^^^^ ^^^ ^ expressed as a positive integer ; also that if ^ be assumed
equal to 1, that the ''V^A can not be expressed as a common integer, or as a
positive fraction. _
839. It has then been proved thai v^ and ''J-^* in which A and B (one
or both) are not the r^ powers of positive integers, can not be expressed in
terms of integers or rational fractions, the only numbers which have thus
far been discussed. Hence our idea of number mu^t be enlai^cd
We therefore assume thai i/2, and i/i general ''^1> ** « number^ and include
iZinourmwnber tytlem^ caUing UUi tuw member of the family of algebraic nmn-
ben ihe LrratUmaL
The properties of this irrational must be consistent with the definition
of a root; that is, with the relation
84l>* Before irrational numbers can be admitted to the number sys-
tem of Algebra it must be proved that they obey the fundamental laws of
Algebra, which have already been established for integers and fractions
Hi 6, 7, 8e-47, 61, 67, 186-181, 188, 139, 142, 143, 146-150, 152).
340 COLLEGE ALGEBRA [J341
841. Special Cases of Irrational NmnberB Defined.— If the ueual method
for finding the square root of a numher is applied to 2, the following
sequence of rational numbers is obtained by carrying the reckoning ont to
0. 1, 2, 3, 4, . . . . places of decimals, viz.;
1, 1.4, 1.41, 1.414, 1.4142,
/i\ 1 1 4 1 41 T 414 , 4142
egM 1^2=1.4142135623
The numbers of the sequence (1) are rational, and it may be extended
indefinitely by continuing the steps in the process of finding the square
root indefinitely. It is evident from the laws of inequalities that the
following tables of relations hold:
I n
1. 1 <i/2<2 and 2 -1 =1.
2. 1.4 <V2<: 1.5 and 1.5 - L4 = .1 = ?-.
^ ^ 10
3. 1.41 < i/2 < 1.42 and 1.42 - 1.41 = .01 = --
lOF
4. 1.414 <i/2< 1.415 and L415 -1.414 = .001 = ^ .
5. 1.4142 <V2< 1.4143 and 1.4143-1.4142= .0001 = i.
etc., etc.
n. Un <i^2< Vn and Un - Fn == -^.
It follows therefore from tables I and II that there can always be found,
two numbers, one ((/» ) less and the other {Vn ) greater than 1^2, whose
difference is
1
10--1
w hich can be made as small as one chooses by making n sufficiently lai^
These numbers Un and V„ can be made therefore to differ from v^2, which
lies between them, by as little as one pleases.
The sequences in table I may be regarded as a definition of V2. The
sequences are such in the sense that a number can be found in each of
them, Un and Vn, such that Un and Vn or any of the numbers following
them, differ from V2 by less than any assigned number. Either of ^eee
two values is an approximation of V2.
E. g. Thus 1.414 and 1.415 are approximations of i/2, and the error is leas
than .001; 1.4142 and 1.4143 are closer approximations of l/2, because the
error is less than .0001.
{342J IRRATIONAL NUMBERS AND LIMITS 341
842. Theorem U.—Let ^ be a fraction whose terms {one or both) are not
the r** power of positive integers, then numbers can always be founds both greater
and less than »--J4, which differ from r^^ by less than any assigned nuTnbers,
however small.
In the proof of this more general case, it is more convenient to write
the two valoes between which the required root lies at any stage of the
work in terms of common fractions.
For example, 1.41421 < 1^2 < 1.41422
14. 1 . 4 I 2 . 1 ^ /o <- 14. 1 . 4 I 2 ■ 2
10 100 1000 ' 10000 ' 100000 ^ ^2 ^ 10 ' 100 1000 ' 10000 ' 100000
14 , 1_ , _4_ , _2^ , JL^ /- ^14 , JL_ , i_ , _2^ , _2_
10'^10""*'l0»"^10**^10»^^" ^10"^10«"*'l0»"^10*"^10*'
A A
Let ^ be a fraction, in which — - (one or both) are not r^* powers of pod-
tive integers.
The powers of the following series
<•' «-• (r.)'. Q'. (fo)' &'■■■■■
increase without limit as n increases.
One can write {tJ = ^'
for, since n may increase without limit, then the product of r n'«
increases without limit and consequently the fraction ^ increases with-
out limit, the denominator 10^ being constant.
Therefore two consecutive powers of this series can always be found
between which 4 lies, no matter what positive value it may have.
Let /^^y and (^JT ) ^ t^^e powers of series (a) between which — lies.
Here i>i=0, 1, 2, 3, 00 .
(s)'<i<(s^')'
W iQ ^ \B^ 10
E. g. It was found in table I that
141 <- /5<r'l^.
100 ^ ^2^100'
here/>i = 141 andpi + l = 142.
The interval between ^"j" and ^ is — . Divide this interval into ten
eqnal parts, and form the series of powers
<"> (S)' (S+ M' (fo+ M' (fo+ h)'- (^)'-
ffince ^ lies between ^^^ and C^T^ ) » it must lie between two con-
secutive powers of the series (b).
342 COLLEGE ALGEBRA tJ343
be the two powers between which -- lies.
Then (^'-+^y < - < (P' +^^±^V.
Belation (ii) shows that ^\— lies between two numbers whoee difference
1
is -- . Divide this interval into ten equal parts as before, and oontinne
the process indefinitely, and therefore have
where i = 1, 2, 3, 00 .
According to (iii), the two numbers, between which ''a/ — is found to lie
1
at the 1;^^ step of the preceding process, differ by —r, which can be made
' i 4
as small as one chooses by taking k sufficiently laige. Therefore **%,'—
which lies between these two numbers, differs from each of them b
less than any assigned number, however small.
Hence the sequences in the first and second members of inequality (iii)
may be regarded as a definition of ''a/—- The sequences are such, inas-
^ I t
much as a number can be found in each of them, ^ and ^^T. * 8uch_that
10* 10* I J
they or any number following them differ from the number ''\^ by-
less than any assigned number.
Either of the numbers ^ and ^'^ "j^ is an approximation of the i-alue
«"Vi
848. Application. I. Geometrical Uhutraiion. Pythagoras found that
if a triangle ABC be right-angled at C,
(1) aS' = bc+ac. ^^
let BC=AC=1\ hence from (1)
(2) AB = ^1^+1*= 1/2.
The length of the hypotenuse AB oi the right-
angled triangle ABC is just as exact as either of
its sides AC or BC^ yet it can be expressed only
by V2 (1 inch being taken as the unit of length). A
II. Algebraic Example, If r = 2, ^ = 3, B === 1,
'V-
^-•s-
ii344, 345] IRRATIONAL NUMBERS AND LIMITS 343
By definition of a root v^3 X Vs = 3; but the approximate value of 1^3
multiplied by itself will not give 3. Hence the number which multiplied
by^taelf gives 3 must have an exact value. This exact value (in general
rJ^) as we have seen can not be expressed in terms of integers and frac-
tions. Up to the introduction of fractions our number system was restricted
to int^ers and, because we were not able to express the fractions in terms
of integers, fractions were added to the number system. Just so in the case
of irrational numbers.
Properties op the Series which Define ^-^ |1
84i. It was proved in §842, that r^^ lies between the two series of
numbers;
^ ' 10 10^10« 10^10» ^ ^10*
and
(2) ail, a-j_&±l , P14.BL J. .... 4. &L±1,
where ifc = l, 2, 3, 4, 00.
Series (1) and (2), in consequence of the way in which they were con-
structed, {842, have the following properties.
I. T^ numbers
% + ^+ +^' where k = -l, 2, S,... CO,
of series (1) increase as k increases, but are respectively less than the numbers
a + a. 4. + i2fc_±i, ^vhere ifc=l, 2, 3, ... 00 ,
10 10» 10^ ' '
of the second series; and the numbers of series (2) decrease as k increases but
are respectively greater than the numbers of series (1). Thu^ the numbers of
one series continually approach the corresponding numbers of the other series, but
never equal them,
II. The difference between the n^ numbers of the two series, namely^
^^ n^OO* [?»48]
can he made less than any assigned number, however small, by taking n as large as
one chooses.
845. Limit and Definition of the Irrational as a Limit.— The number
which a series of numbers (e. g., series (1) and (2), {344) continually ap-
proaches (or which two sequences, one of increasing and one of decreasing
numbers continually approach) and never reaches, but from which the series
can be made to differ by less than any assignable quantity, is called the limit
of the series.
E. g., the i^2i J841, and r^4 §342, are respectively the common limit
of two series.
•The symbol n~^, is an abbreviation for the pbrase "» taken as large as one chooMa.**
344 COLLEGE ALGEBRA [?346
Two such series, therefore, may be used to define the number which is
their common limit. Thus, the number r^~* which is the common limit
of series (1) and (2) {844, may be defined by the relation
NoTK.— Besides the irrational nambers>/2 and those included In ^V^-, there are
many others. For example, the ratio of the circumference of a circle to its diameter,
namely tt =3.1415926 . . . : the base of the Naperlan system of logarithms, namely
« = 1+ J^+ ^ + Y^ -h . . . . = 2.7182818284 . . . ; logarithms themselves; continued
fractions, etc.
Irrational Numbers
«
846. The properties of the two numbers V2 and ^^~ discussed in
2{887-344 leads to the generalized ^
Definition op Irrational Numbers
|n = l,2,3,
Let (1) 01,03, a» . , , „ „
00
(2) A^Ai, An
be two series of rational numbers, the first an increasing series (on ^ On+i)
and the second decreasing {An"^ ^n+i), and such that the dinerence be-
tween the corresponding n^^ terms,
An — an
approaches zero when n is indefinitely increased ; then, it is said that the
terms of the two series approach a common limit, {846, and that they
define a number which is their common limit.
E. g., in the definition of i/2, {841, the series
1, 1.4, 1.41, 1.414,
is a particular case of Oi, Of, os, a« ;
and 2. 1.5, 1.42, 1.415,
a particular case of Au A2, Ast A^^
In the definition of r^~, §842, the series
Pi Ex i 21 2i ^ J2i . £h
ro 10 ^ 10*' • • • • ' 10 ^ W^ . . • . lOfc, . • . .
is a particular case of Oi, Of, . . . Oit, ;
10 10 ^ 10« 10 ^ iF 10*
is a particular case of Ai, .^^s, ..... , Aki
It will sometimes be convenient to represent irrational numbers by the
single letters a, 6, which have heretofore represented positive
rational numbers only.
• The general discussion of irrational numbers given in |i 846-351 can if desired
be omitted in the first reading.
11347, 348] Il^RATIONAL NUMBERS AND LIMITS 345
847. Zero, Positive, Negative. — ^The common limit of the series
(1) «1, «2 «« »
(2) Au A^ A„,
is either
(1) numerically less than any assignable number,
or (2) greater than some definite positive rational number,
or (3) less than some definite negative rational number.
In the first case the number a which the series (1) and (2) define, is said
to be zero, in the second positivef in the third negative.
The Four Fundamental Operations with Irrational Numbers
848. Let the irrational number a be defined by the two series
T ( (1) fli. a«, a».
I (2) ^1. ^8, .... ^„, ....
and let a second irrational number h be defined by the two series
TT f (3) ^» ^, ftn.
\ (4) Bi,B, Bn,
The numbers of the series (1) and (3) are supposed to be increasing and
those of (2) and (4) decreasing in magnitude. Suppose also that x and y are
two rational numbers which respectively approach a and 6 as limits in
any manner whatever.
AnnrnoN. Consider the two series
(a) fll + &X, Og, + &«, y On+hn
(b) Ai + BuA^, + B^, ,An + Bn.
By definition of (1), (2), (3), (4), the series (a) is increasing and the series (b)
decreasing. The difference between the corresponding n^^ terms of these
series
(5) An+Bn-{an+ 6„) = (An - On) + (Bn - K)
approaches zero when n ~oo , since by the definition of the series I and II
(An — anX^ )
(5„-6„)<| I
when n =!z 00, and (f is as
/ T* . V ^ - ■ small as one chooses,
and
Therefore
•M„-a„) + (5n~6»)<d
where d is as small as you choose. These two series therefore define a
number, J 846. The sum x-\-y has this number as limit, since if the numbers
X and y are comprised, the first between On and An , the second between 6n
and Bnt their sum x-^y lies between
an+ 6n and ^n + ^n.
The limit of the sum ar + y is called the sum of the two given irrational
numbers a + 6, that is
. lim* (x+y) = a + h.
n=oo
•For limit (a„) the more complete symbol ]l^^ (^n) is also used; read "limit
which a^ approaches as n approaches infinity.'*
346 COLLEGE ALGEBRA , [J348
Subtraction. The irrational a is said to be greater than the irrational b
if we always have
an> Bn, n = l,2, oo .
For simplicity suppose that oi > Bi, Consider the two series
(C) Gi—Bi, Og — B^^ . , , . an— Bn\
(d) j4i — 61, idj — 62, . . . . i4ii — 6» .
(c) is an increasing and (d> a decreasing series (J348, definition of I and U).
The difference between the corresponding n**» terms of series (c) and (d) is
An — hn-(an'-Bn)^{An'-an) + (Bn-hn\
which approaches zero as n is indefinitely increased [see discussion of (5)].
These two series, therefore, define a number (§346).
The difference x—y has as limit the number defined, because if the
numbers x and y are comprised, the first between an and An , the second
between &11 and Bnt x — y remains comprised between
an - Bn and An-bn. [J8$0, Ex. 1]
The limit of the difference (x — y) is called the difierenoe of the two num-
bers a and 6, or a — 6, that is
(ii) lim (x — y) = a — b,
» = 00
MuLTiPUCATioN.— Consider the two series
(e) ai6i, Ojfej, (Is&s, .... Onbn
(f) ^B„ AtBt, A^B^, AnBn.
The first is increasing, the second decreasing; the difference between the
corresponding n"» terms,
AnBn — On frn ,
approaches zero as n is indefinitely decreased. For
AnBn — Onbn ^(an + An — an) (bn + Bn — bn) — Onbn ,
or
AnBn — anbn=^bn{An-an) + an{Bn-bn) + (An -an)(Bn — bn ).
By hypothesis each of the differences An — an, Bn — bn , can be made less
than any assignable positive number d however small (^848, I and II), and
since An and bn are less than some fixed number P, then
AnBn-anbn<2Pd + cP
where d is as snmll as one chooses when n is taken as large as one chooses.
Therefore the two series (e) and ( f) define a number, {846, because 2 Pd
-{- cP can be made as small as one chooses.
The product xy has as limit the number thus defined, because if a; re-
mains comprised between an and An, and y between bn and Bn, the product
xy remains comprised between
On bn and An Bn .
The limit of the product xy is called the product of the two given irra-
tional numbers a and &, or aby that is
(iii) \'im(xfj)=ab.
1349]
IRRATIONAL NUMBERS AND LIMITS
347
Division. Consider the two series
(g)
(b)
^1 ^i Bn
Ai A% An
— *•> — *-J •
6i 6j hn
The first is increasing and the second is decreasing (2848); and the differ-
ence of the corresponding n"» terms,
An __ an - - An Bn — fin hn
On Bn bn Bn
differs from zero by less than any assigned number where n is indefinitely
increased. For, the denominator bn Bn is finite and the numerator, as we
have seen above, has the limit zero; hence, the limit of the product is zero.
Therefore the series (g) and (h) define a number (2846).
The quotient - has as limit the number thus defined, because if the
y
numbers x and y are always comprised, x between On and An , y between
hn and Bnf - is always comprised between
-^ and ^
Bn bn
The limit of the quotient - is called the quotient of o by 6 or ?;
y h
that IB
(iv)
lim - = «
n = CO y h
At,
"ik
On,
•An ,
bn ,
Bn ,
848. R6sum^, The results of the preceding discussion are summed up
in the following theorem.
Thborkm.— Cy the numbers defined by
■ it.
and
ht,
1. the mm is the number defined by the series
«i + &i, (h+bt , . .
Ai+ Bi, A^ + Bi , . .
2. the difference is the number defined by the series
ai — Bif Og— Bi f . . . . On — Bn
Ax — 6i, A2 — b% , , , , , An ^bn
3. the product is the number defined by the series
Olblt <hbi , . . . . Onbn f . . •
AiBif A^B^ f .... An Bn , . . . ;
4. the quoHent is the number defined by the series
A ' B^' ' ' ' ' Bn* ' ' '
Ai A2 An
bi ^ * bn *
. an-\-bn ,
An + Bn,
348 CX)LLEGE ALGEBRA LH350, 351
860. Irrational numbers obey the commutative, associative, and dis-
tributive laws of int^ers and rational numbers. This generalization is
readily obtained on considering the approximate values of the given
numbers.
For example, the product of two irrational numbers ab is the limit of
the product xy, x and y being rational numbers which have respectively
the limits a and 6.
Since we always have
xy^yx
and since lim {yx) = &a, it follows that
lim {xy) = lim (yjr)
or a6 = 6a. ■
Hence in the product of two irrational factors we can interchange the foctors,
and thus is established the commutative law for irrationals. Similarly
it may be shown readily that the laws governing the fundamental opera-
tions with irrational numbers are the same as those governing these
operations with rational numbers.
E. g., 1. V'3 ± i/7 = ± 1/7+ 1/3.
2. 1/2 + (-1/5) + VTl = i/2 + i/ii + (-v/5).
3. Vi -f- Wh -*- 1/T3) = 1/7 -*- 1/5 X Vis.
4. (V2-i/5) = (v/5)(V2).
6. (i/7-Vn)*=(i/7)*-(Vn)\
861. Bquality.— Of two numbers a and 6, defined by series I and II, J849,
the first is greater than, equal to, or less than the second according as
a ^ Bn, n=l, 2, 00.
This definition is to be justified exactly as the definitions of the fun-
damental operations on irrational numbers were justified in {848.
Some important principles of limits follow immediately from the results
established in the preceding paragraphs. For by definition, J848,
limar= a
limy = 6;
but by (i), (ii), (iii), (iv), ?848, viz.,
(i) lim(x-fy) = a + 6,
(ii) lim(a: — y) = a— 6,
(iii) lim {xy) = a6,
(iv) '-0 = 1 =
lim (a: -f y) = lini x + lim y,
lim(a: — y) = lima: — limy,
lim {xy) = lim x ' lim y,
y limy
CHAPTER VI
SURDS
In 8311 the student has learned that there are two
notations in use for expressing the root of an expression, one nota-
tion using the radical signs and the other fractional exponents.
Though it is not necessary to have two ways of writing the same
thing, yet, because each notation has special advantages in certain
cases, the two notations are retained.
It has been shown in Chapter V, Book III, that the same
laws I— V of ?{6, 7, which govern the fundamental operations on
integers, zero, the negative number, and the fraction, govern the
operations with roots, in both forms of notations mentioned, 3350.
353. A radical is an indicated root of a number or quantity; as
A radical expression is an expression composed of radicals; as
3>/71, l/a+l/6, (i/a+i/6) + 'i/^.
A surd is a root of a rational quantity which can not be found
exactly; as i/6, i/a, W\
One should notice that i/l + 3 '|/5 is not a surd, since 1+3 ^l/5
is an irrational number.
The distinction between arithmetical and algebraical irrationality
is important Thus, i/x is algebraically irrational; but in case
aj = 16, then l/x = i/16 = 4 is arithmetically rational.
Thus, l/|> *l/4 are surd numbers. Expressions like i/9, '|/8,
etc., are written in the form of surds. Expressions like \/x^
'l/asV, etc., are often called surds, although, of course, they are
such only in case x and y are commensurable quantities whose roots
can not be found exactly.
.349
350 COLLEGE ALGEBRA [{{354-356
In the preceding definitions a distinction is made between the
terms incommensurable j irrational expression^ and surd. According
to the definition just given
'v^3+]/5, 'k |/7, V^i l/e= Napierian base,
are not surds; however, they are irrational and incommensurable.
This limited meaning of the term surd is not only convenient, but
is being used more and more by authors.
Orders of Surds
864, A quadratic surd or a surd of the second order^ is one with
index 2 ; as Vb^ Vx,
A cubic surdj or a surd of the third order, is one with index 3 ;
as V4, Vn, V^Ty.
A biquadratic surd, or a surd of the fourth order, is one with index
4;a8V7, Vx(x + ^).
Similarly, surds are classified according to their indices, as
Quintic, . . . n-tic, as the case may be.
A simple monomial surd number is a single surd number, or a
rational multiple of a single surd number; as 1/2, f ^1/5.
A simple binomial »urd number is the sum of a rational number
and a simple surd number or of two simple surd numbers; as
|+V^5, 1/3 + V7.
355. The rules for operations with surds follow from the prin-
ciples and theorems of Chapter III, 1303, etc. We restate for conven-
ience each principle as occasion for its use in this chapter arises.
As in evolution. Chapter II, the positive values only of the radicals
are considered and likewise the principal roots only.
NoTK.— In operations invoWing surds, arithmetical numbers contained In the surdf
should be resolved into their prime factors.
Reduction op Surds to Their Simplest Form
356. A surd is in its simplest form if the radicand is integral,
and does not contain a factor whose exponent is equal to or a mul-
tiple of the index of the root; as ^ 3, 'Kut^, '"i/x".
«357, 358]
SURDS
351
A surd can be reduced to its simplest form by applying one or
more of the following principles :
Distribntiye
formulas
Associative
formulas
I. Va-"l/6 = Va-6
Prin. I, that is a" -6"= (a-6)»
11. Va^V^»=Va-^6
fill. V^ = (Va)'
since each radical = a".
IV. ''l/V^='^>/a = VVa
since each radical =aw.
[iS21]
[J312]
[2322]
[J307]
[J307]
V. "''l/a"^ = {ayp = {ay = Va« [«307, 316]
kn
or *i/a*''=a" = a*
VL V^ =a^ = a^= WJ [»307, 316]
387. Rule 1. — A rational quantity can he expressed in the form
of a surdf hy raising it to a power whose exponent is equal to the index
of the surd desired.
Thus,
Also
(1)
(2)
(3)
3 = l/3« = l/9; x= l/x^ = Vx»
n
a + X = (a + x)? = (a -|- a^)".
{a-\-x)i= (a+ x)f =(a-fa;)l= . .
[J366,VI]
368. Rule II. — The coefficient of a surd may be introduced under
the radical siyn^ hy first reducing it to the form of the surd (Rule I),
then multiplying OAicording to principle I, S366.
Thus _ _
(1) 2i/3 = i/2« • i/3 = i/4 -3=1/12.
(2)
(3)
(4)
(5)
4 V5 = V4» % 5 = V4» • 5 = V320.
a l/ a* = V «' * V'x = \/a*x.
ajv^2a— X* = Vx^ ' V2a — x* = l/2ax* — x*.
x(x-y)l = (xf)!(x-y)i = [^J(x-y)]} [S367; J366, 1]
= [(xi)»(x-y)»]i
= [x«(x-y)»]i.
352 COLLEGE ALGEBRA [81359-363
359. Rule III. — Conversely ^ any quantity may he made the coeffi-
cient of a surd, if the radicand is divided by the quantity raised to
a power whose exponent is equal to the index of the surd.
Thus, _
(1) V^ = i/(2)n5 = i/2« i/l5 = 2 i/l5. [8856, I, V]
(2) l/x»y«= l/(xy)x = i/xYv'x=x^y f/x. [8366, I, V]
(3) l/l6a2— 16a«x«=Vl6rt«(l-a^) = l/l6aVl— x«=4al/l— x«.
(4) "l/x»+y H= V(x"i/"^")(xy')=VxV" Vx/=xy« Vx/.
)• Rule IV. — If the radicand is a fraction, the surd may be
reduced to its simplest form by multiplying both terms of the fraction
by such a quantity as to make the denominator a perfect power of the
same degree as the surd, and then proceed (w in 8359, using also prin-
ciple II, 8366.
Thus,
la_ In i"-» IT , 1 ,
(3)
361. Rule V. — Since IV, 8356, is ti^e in all cases^ we know thai
the index of a surd can be lowered if the expression under the radical
sign is a perfect power corresponding to some factor of the original
radical index.
Thus,
(1) V9=V^3«j= V3i=i/3. [8356, VI]
or V9 = V^v9 = l/3. [8356,IV]
(2) V64xV* = V(8xy)«=V(8^i=V 8^=2 V'^. [8356, VI]
or V64xy = V^/ 64^/= W^y, [8356, IV]
362. A Burd is in its simplest form: (1) when the radicand is not
itself, or does not contain, a factor which is a perfect power of the
required root; (2) when the radicand is integral; (3) when the
index of the surd is the lowest possible.
363. It is usually supposed in any piece of work, that all the
Burds will finally be left in their simplest form.
1363] SURDS 353
EXAMPIiES T.Tir
Express as pure surds the following:
1. aVb, l\/2, |;70.5, 3a '-J^-
2. ab\/c, (a+6)v/c, (7— a)i/a!, lat/x.
3. a^. ,vre, |,/4. 2«*^||-
4. «'l/6, fVj. |*V^. 2a*i/5.
5. 6*^. 2»»/|, 4VA, |VA.
1^ aft* jxy* a J /6^ a'^ /ft^ aft" jay^
xy^\ab^* h ya^y b \a^y xy^\h^x
8. (a+x)J^, ^±l-^|/«Elr g J«'-2«»+«\
/ \ I 9a+96 , » I g a;* — 6 x»
*• r-*>)^f4a«-8ax+4«'* ^" + ''^9 a»+ 18 a& + 9 6«'
10.
(v^5 - 2) V^9 + 4l/5, (i/lO + 1/6)1/4 — V^15.
11. (2i/2 + v^6)V7-4i/3, (»/3 - ^^2)1/12 + 5|/6.
V */\x'-y» a \(2a«-l)»
Redace each of the following surds to the simplest form:
14.
1/320,
|/405,
|/363,
v/432.
15.
3>/8,
51/80,
81/75,
61/150.
16.
3l/12a»,
4l/20 6»
5»/40 c,
71/48 oo^.
17.
|l/24«»,
|l/27i»,
|v/45c..
|l/80x'y».
18.
ij v/72a»
7^1/966',
3|l/54c',
2\ l/l25aV.
0
19.
V8a6»,
V27 a^x,
Vllx»y»,
Vl2xV.
354 COLLEGE ALGEBRA [8363
20. V5^, Va» + fc', VV— i^ V(a— t)'.
21. l/x^, vx\ i/x«»+S l/x«"-».
22. 'l/^, VV, V^^S V^*^'.
23. "l/x'^+S Vx»+», V5x««+i, «^
ox"
24. l/a6»cS l/4a«6«c», V 7 xV^", l/9x»y«i2^^
27
28
25. Vafc'c*, V8a^6«c», V7 x«y*«*, VBx^'zI*.
26 1^ joTs ITT I0T5
' \2T' \3:6' \0:96' \"5T"
|27^ |3^ |5^ ff^
\36«' \5x»' \67'' \lOfc'
IT" 2 liT~ llT^ 3 (13^
29. 2a^8^,. a\l2«'' ^ax^^^-^ ^\T8^ *
30. lahyj^^,^ 20 6»\^3' ^2 ^tV*' 4^«^-
31 !«' + ^' jam+h^) laHl+b)^ \a+h*
yi xV ' \xM + y*)' \x«(l-3/)«' \(x+y)«'
|a[+7« [7^ l(a«+6*)c« p(6+c«)
33. 1/16T9' v/tV+1. i/1+T» /i-f
34. \/^+j\, Vi-iV, vA-iV i/V + |.
9
35. -^V'l + i^, jV^l-i^iF, 4V^l-i^r, 2>^1-^V-
12, /rr—r 8,/:; r lo,/:; r 10
36. f i/l + ,V §^1 - 1^> y >^1 - tV f v^i-tH.
37. Vax« — 6x« + 1/7 x«, l/a« + 2 a*6 + a«c.
38. i/x» — 2xV + iCi/', i/5x5 — 20x«+20x.
39. v/3 a«c» — 6 a6 c» + 3 6«c», 1/18 x^y — 60 xy» + 50 y».
IgS — 2a«+ g IgS 4. ggfe ^ ab* — 6»
' \ ax«+6x« ' \ 9(a-6)
8364] SURDS 355
\8x — 8x«+2x'' \50y — 20/+2j^'*
42. If v'SO = a, how large are i/8, i/l8, i/32, v/98?
43. If 'l/250 = 6, how large are 'i/l6, V54, Vr28, Vi024?
44. ^l/d^\ 45. **l/a"*^. 46. **i/a"*"6**c'*"'^rf"'"''.
47. -^^^^•48.. V^9a*6«/-V'". 49. ^j^c^x^^^^'\
^^- \(m7)wv^' ^^- \ 86^s;7is
^2- W sL^'-^^ ' 53. »^(2«a-5-X^^)-
54. *'"^'^|/^i^+»ry-45i^^ 55. *^"^ Va^^^ -*»«^.
58. ^^| ^4-H ' • 59. -yj—.,^^'
60. *^l/^ *^l/^, "•^l/C^^.
61. -'■'V^i^^ "^""i/^F^.
62. Find the (9a»— 49fc«)^ root of m^"'^
and the (12 a«+ 61 at + 77 b^^ root of m*«+"*.
Addition and Subtraction op Sitbds
364. Similar or Like Surds. — Surds are said to be similar if
they have the same surd factor when reduced to the simplest
form, as 3>/2, 5l/2, l/l28 = 8i/2. All other surds are unlike,
(a) In case surds are unlike, the result of their addition and
subtraction can only be indicated, thus,
21/ 3 + 4 V5, v'l - 7 V9 + 3 V6.
(b) But if the surds are all similar or can be reduced to similar
surds, they can be united by addition or addition and subtraction
into a single like surd.
356 COLLEGE ALGEBRA [8364
. Example 1. Simplify i/27 + 2i/48 — 3i/75.
l/27 = 1/9 X 3 = 3 v^.
21/48 = 2|/16 X 3 = 8|/3.
— 3|/75 = — 3l/25 X 3 = - 15i/3.
l/27+ 2i/48— 3l/75 = -4i/3.
Example 2. Simplify 5 Vi + 6 V32 — 5 Vi08.
5 V4 + 6'i/32 = 5 V4+ 12 Vi = 17 Vi
-5 Vio8 = - 5 V27xl = -15 Vi.
.-. 5V4+6'l/32-5Vr08 = 17V4-15V4=.f 2V4.
Example 3. 'l/54 + l^i — V 250 — J |/| = ?
'l/54 = V27 X 2 = 3 V2.
1/^ = 1/^X2=1^^2
— V250 = ~ Vl25 X 2 = — 5 V2.
-il/f = - I1/4 X 2 = - Jl/2.
.-. V54+l/^-V250-fl/f=+ii/2-2 V2=j(|/2-8 V2).
Example 4. 6a|/63a&» — 3l/ll2a'6» + 2 a6>/343a6 =
6al/9 6«x7a6 — 3l/l6a«6« x 7a6 + 2 a6l/49 X 7 a6 =
18 a6 \/Tab — 12 abl/ Tab + 14 ab\/Y^z=:20abV7ab.
EXEBOISE liX
1. How do we express the root of a product?
2. How are roots of equal indices multiplied together?
Simplify and perform the indicated additions and subtractions:
3. 1/49:64 + >/l00a«6«c« — V8'6'c«.
4. 1/18+^^28 — 1/75. Ans. : 3i/2 + 2i/7 — 5l/3.
5. 1/20+1/125+ 1/63--1/252— 1/700+ l/567-l/605.
v^+ i/3t + 1/4^ + 1/5^ + v^e^.
6. 5i/48 + 4i/l47 — 2i/3 — 5i/432. Ans. : — 14i/3.
1364] SURDS 357
7. 1/7168 — 2i/l8 - 7|/5 + 2i/'45 - 26|/2 + 4i/363.
8. 3Jl/24-5|v/54+13Jl/99 + 2jJ|/216 -211/44.
9. 2l/2450 — 3|/2048 + 5i/l3122. Ans. : 3791/2^
10. If V'S = 2.2360679, how large is- 1/320?
11. V24; VsT; 5 '1/I6- 2*1/54 + 8*1/2;
12. 2 ViO + 3 VIO8 + V500 — V320 — 2 Vi372.
13. 1/4^ + 1/25 a6» — (a — 5 6) v/^. Ans. : (a+ 10 6) |/^.
14. — |/m»iic« — ~ 1/4 m»'e« + — |/9^W^.
flic ne P9
15. VI6 a*«»*c — V54afe*c* + V250'^*6^.
16. c V^^6V — a V^^ + 2» Va«6«c». Ans. : 06c »|/a6V.
17. "|/a"+«6"+» — "|/o"+»6''+«.
18. Va^—bx^ + Va'6»c' — tf^t'c' + i/4mV — 9mW.
19. Va'""*^"^— a'6-^+i — '"^''i/a«'+«'6'+^ — a*+«»'6*»+»'.
20. 1/ V49' • 64» + VV27-« • 64'.
21. V''2i*a"6»c — V4 • 5*aWc'^ + V4 • 6*a6'^c.
23. 3 6» »/^+-l/^*-c«-E.
c \6*
26. V54a"+«6» — Vl6 a"-'&« + V2 a«"+» + V2 c'o".
27. "l/2'"o"'+'6""'+' + "'i/3"a*'"-""+»6"+' — "l/a't^c*".
28. , (3^^2Vy^. e^/S^. 29 >-, f""^*^""'^"'"
30. |/(a«c + a«rf). 31. V^^*** — «'""/").
358 COLLEGE ALGEBRA [J365
365. Redaction of Surds of Different Orders to Equivalent Surds
of the Same Order.
Surds of different orders can be reduced to equivalent surds of
the same order by the principle
"|/^ = '^i/a^, [{856, V]
Example. — Reduce i/i, V2x, and *V^5y, to equivalent surds
of the same order. According to the preceding formula one has:
V2 ="v^ =V64,
v^ = "i/(5i;)« = "1/257.
We then have the following rule:
JPifid (he L. CM. of the indices of the radicals ^ and multiply the
index of each radical and the exponent of its radicand by the quotient
of the L, C. M. by the index of the radical,
NoTB.— Tbe relative magnitude of radicals may be determined by reducing tbem. if
neoessary, to radicals of the same degree.
ExAMPLB.— Which Is the greater, '1/3 or *|/6?
By rule V 3 =: "i/ 8* = "i/81.
Since 125 >81 .-. "i/l^ > "i/Sl.
BXBHOISB JjXJL
Reduce to equivalent radicals of the same degree :
1.
'1/4
and V6.
7.
Vah, Vfcc, and Vco.
2.
V2
and V5.
8.
V2a, V3 6, and V4c.
3.
Vi
and VlO.
9.
V 8, V2, and 'l^n.
4.
V3
and V12.
10.
Vb, Vl5, and VlO.
5.
y'l
and Vi-
11.
"l/x* and "l/y».
6.
5
and V 10.
12.
"*V^' and""Va»fc.
13.
1/^
-y and Vx + y.
14.
"l/a*xy, "v'axY, "Vax'y.
Which is the greater:
15. t/5 or V9? 16. 2/3 or 8/2? 17. l/3 or Vi2?
18. f vlO or I VlT? 19. V2^' or i/2^, x < 1?
20. Arrange in order of magnitude 'l/l4, 1/6, *|/20.
21. Arrange in order of magnitude 1/}, 'i/f, *l/|.
«366-368] SURDS 359
Multiplication op Surds
366. Multiplication of Monominal Surds. — The product of two or
more monominal surds is found by applying the formula
*l/a • V^ = V^. [J866, I]
Examples:
1. 2i/3 X 3i/5 = 6i/l5.
2. 2 Ve^ X 5 *i/l2^ = 10 V72^ = 20a V9.
3. V4xVX V5xy«x'l/8^ = Vl60xy=2xyV5xy«.
NoTB.— In cftse the sards are not all of the same degrrae, they shoold first be redaoed
to equivalent surds of the same degree.
4. l/2 X X V4 x« = V(2 xy X V(4 x*)« = Vl28 x' = 2x V2 x.
6. Vi2 x'l/ix Vi = V(i2r« X "^^(i)» X VCi? *
=>-^-(i)'a)'i
- Af^ * 2» 4» 6«
= *V3*^ = Vl62.
367. Multiplication of Polynomial Surd Numbers.—The work
is arranged as in the multiplication of rational polynomials.
Example. Multiply (2i/6 + 5i/3 — 7i/2) (i/6 — 2 1/3 + 4 1/2)
Thus 2i/6 + 5 1/3 — 7 i/2
|/6 — 2 1/3 + 4 1/2
12-
f 15l/2 -
-14i/3
-30
-12|/2
+ 141/6
— 56
+
16 v/3 + 20 l/6
— 74 + 3 1/2 + 2 i/3 + 34 1/6
Notice that the terms of the partial products have . been simplified
and that the similar terms are then placed in the same column.
868. Conjugate Surds. — Two binominal quadratic surds which
are the same except in the sign of a surd term are called conjugate
surds.
B. g., 1 + i/3 and 1 — l/3, l/2 -f 3 1/5 and i/2 — 3 1/5.
360 CX3LLEGE ALGEBRA [?369
Either of two conjugate surds is called the conjugate of Hie
other.
The product of any two conjugate surds is a rational number.
Thus {Vx + I Vy) {Vx -lV^)z= (i/S)* - {IV yY = X -r-y.
369. Type Forms. — Many products are more readily found by
using the product and power forms of J89, VIII, etc.
Example. (i/3 - 21/5)" = (i/3)' — 4i/3 • l/^ + (2i/5)".
= 3 — 4i/l5 + 20 = 23-4]/l5.
The Product of Like RooU
Simplify the following:
1. l/3-l/l2, i/28-l/7. 2. i/l0-i/15, i/20-l/30.
3. V2'v/4, V5V50. 4. a}/x'bl/x, l\/x'aVi^
5. (V^0'(V3y. 6. Cv/x)".(Vx)^
7. (v^i^Fvy. 8. Ci/^-)". 9. (^fp)'-
10. (VvsT^y. 11. Vvi6^ 12. (Vyf?)*.
.5. (^.^.H)^ ,e. .^y-:^!^' ./v=^'
17. (2l/6 — i/12 — i/24 + i/i8) V2.
18. (7i/2-5i/6-3l/8+4»/20)-3t/2.
19. (2i/20 — 7i/8 — 3l/5 + 3i/18) • 4|/l0.
20. (i/7— i/3)(i/3-t/2). 21. (3l/2-2i/3)(7l/2+5/3>
22. (8 + 3l/5) (2 — 1/5). 23. (5l/3 + l/6)(5i/2-2).
24. (2o+3l/«)(3a-2l/x). 25. {Wa—V^{Va-\-2V^
26. (2i/6 + 5i/3 - 7i/2)(i/6 - 2l/3 + 4l/2).
27. (2 1/30 -3 1/5 +6 1/3) (1/8 +1/3 -1/5).
J369] SURDS • - 361
28. (5l/ll2 + 1/176 — 1/4375) (31/396 + i/l75 — 2l/539).
29. (•l/3 + »l/2)(2V9-3'l/4)(V24-Vi)Cl/9 + V54).
30. ('l/25+V9XVi35-V375)(7Vl6-3'i/49X3V4+4V7).
31. (5 V500+ V24-6 V256) (V54+5 V243-4 V576).
32. (•1/128 — 3*1/49 + V2000)(V500+V448 -'1/32).
33. (a + 1/6) (a— l/fc). 34. (l/3^+ l/26) (i/3^— l/2fc).
35. (al/6 + »l/y) (a>/6 — xl/y) .
36. (1/7 +1/3) (1/7 -1/3).
37. (3i/5 + 21/11) (3l/5 - 21/11).
38. (l/x + y +l/y)(l^x + y-i/y).
39. (l/x+1 + i/x - 1) (i/x + 1 _ i/x - 1).
40. (l/9x + 5 + 3l/x) (l/9x + 5 — 3i/x).
41. (i/a + fc+x + i/a + 6— x) (l/a + fc + x — i/a + 6 — x).
42. (l/3a - 6 + l/36 — a) (l/3a — 6 — 1/36 — a).
44. [l/(x + 1) (y + 1) + i/(x- 1) (y _ 1)].
[t/(x+1) (y+1) - i/(x-l)(y-l)].
45. l/a + i/fc-l/a-i/6 46. V^e + 2l/5 • 1^6 - 2i/5.
47. V8+ 31/ 7 -"1/8-31/7 48. V21/13 + 5 • V2i/i3 - 5.
49. (-1 + 1/3)". 50. (1/6-1/2)'.
51. (31/2 — 21/3)'. 52. (i/^TT^— i/JITfc)".
53. (a+l/l— a«)'. 54. (l/l + ax— l/l - ax)".
55. (1/7— 5x + l/4x-5)". 56. (al/1 — 6«+ fcl/l + a«)"
362 COLLEGE ALGEBRA [8369
61. (l/a + fc-x+l/a — 6+a;)*.
62. (l/4a + 6 — 4x— 21/36 — a+x)'.
63. (l/3a — 26 — 5x— 1/36 — 2a + 5x)*.
64. []/(a+x) (x + 6) - >/(a — x) (x — 6)] *.
65. (1/2+1/3+1/5) (1/2 + 1/3 -1/5).
66. (1/7 + 1/3 + i/To) (1/7 + 1/3 - 1/10).
67. l^a+l/a*— x« • V^a— l/a« — x«.
68. Vx+i/x'-l • Vx - i/x« - 1.
69. 'l^al/ a + l/a» — x' • ^V aV a — \/a^ — x\
70. (V a + i/a« — b*+Va — \/a* — 6«)'.
71. Vax + a • i/ax« + ax and i/6a — 66 • l/2a« — 26«
72. i/ax — a • l/ax' — a and v x* — x • l/x* — x.
73.
Find the value of each of the following without performing the
actual multiplication:
75. (v^+i/6)' 76. {Va+Viy 77. (l + l/2)'.
78. (2-V3)'. 79. (V3-V2/. 80. ('1/5 -Vi)'.
81. (l/x + l/y + 1/2) (l/x - l/y - l/a) (l/i - l/y + I/2 ).
(y'x+ l/y— 1/2).
82. (a+ l/x). (6 + 1/^).
«370,371] SURDS 363
85. {\/Z+eYb)-{\/a — c*l/b).
• 86. (2v^+3c*i/6)-(i/^+4'i/6).
87. (c *l/^ + d Vft) -(/'v^ + g *l/6).
88. (*i/^+*|/6 + Vc)».
89. l/(a + v^i) • V (c + l/d).
90. V{a + \/b) ■ 'Via - 1/6).
91. (a + 6 + 'l/^ + *l/^) (Va — Vt).
92. "l/(a + V6) • "1^(0 + 'l/rf).
93. V(5 + 2l/6) • l/(3 + v'6).
94. 3 "|/(2 + 4 V3) • 4 V(6 + 2 V9).
95. 5l/2 • 3l/(4 + 6l/2).
96. (a + l/x)* — (a — l/«)* .
97. (a + l/i)' + (a — v/x)' .
Division of Scrds
370. Division of Monomial Surds. The quotient which results
from dividing one monomial surd by another is obtedned by applying
the principle (J866, II).
Va_ la
'Vb " xft'
Example 1. i^= . /l2 _ ^ _ 2
l/3 \3 '^
When the surds are of different orders they should first be
reduced to equivalent surds of the same order.
EXAHPLB 2.
VTS^ "t/l6«aV \3*-b*-aY 1^7" _ P«^
V25^ ~ •'1/25W ~ > 5' • 5»aV " Ns'" ^ " N 26 "
871. The Division of Polynomial Surds, (a.) In case the
divisor is a monomial, divide each term of the dividend (as in the
364 COLLEGE ALGEBRA [8372
division of polynomials) by the divisor and simplify the quotient by
applying the principle of 2370.
Example.
151/105 — 361/10 + 30 V9 -^ 3\/lb =
3 1/15 3 vi5 3 V225 ^ ^ "
(6.) If the divisor is a polynomial it is better to write the quotient
in the form of a fraction and then simplify its terms according to
2364.
372. Type Forms. Many quotients can be obtained easily by
applying the principles of 2356, (iii), and 297 .
Example.
(xl/a— yl/y) -J- (i/x— l/y) =
(l/^_ l/P) ^ (v/x ~l/y) = [(l/x)'- {Vy)"] ^ (l/x- l/^_
EZBBOISB TiXTTT
g=V5 ;n.|l = -l=, 2356,11.
1. How is the root of a quotient found?
2. How is the quotient of two roots with the same exponents
found?
3. How large is the root of the reciprocal of a quantity and
how large is the reciprocal of the root of a quantity?
4. l/Jl; v/jf; |/'5^; l/2ij; VT^.
5. If l/l3 = 3.6055512, how large is |/13 -i- 9?
6- *v1l + Vwi-4 V3f-2 V2H + 3 VlH-
8. . 1?!^+ J^_» /-^ + JJ_.
«72] SURDS 365
"• >|^'-^
6«)» ,(a' - 6')'
12 M I ^ I ^ i ^ f ^
13 V I 1 my-^nt^-'-pV
14. l/a» -^ i/a; 'v/^^tV ^ V^«; V^^^ -^ V^^^^.
16. V(a»6«c'^)*(a*6'c)'^ -^ V(a*6'c*)Ha*6«c)2.
17. V3a«6«c* — 4a*6«c« + ba^b^c^ -^ s^fe-l^ + ^l^.
\a6 6c ca
Calculate the value of:
18 1^12 i/l53 1/304 , 1/105 .
1/3 1/17 1/19 1/2^*
What is the value of:
19. 0. 06 Vl. 7889984 -^ 0. 12 Vo. 0279531.
20. m -4- l/m ; a*h*c* -^ l/a6c ; m 'pV -^ 'l/^iPV-
21. 1*.^.
22. l-*-i/0.04; 1 + 1/0. 01 5625; I+VO.OO8; 1+VO. 001953125.
..^; ,,^|iWM, 1..^
23 ,../,.. , . >v.vv.-„. , . . 10.01357
0.36639
24 1- / "+2^^
•\a'_3ai«+2i''
25. (v/6 + 4l/l8 — 3 — 81/2)^1/3.
26. (31/15 — 1/20+1/10- 7) -s-2>/5.
27. (2|/32 + 3i/2 + 4) -4-41/8. 28. (6+2i/3— *»/l8)^l/6.
'29. (i/8 + *i/]2 + *i/2)h-2i/2.
366 COLLEGE ALGEBRA [M373-376
RaTIONA LIZ ation
373. A surd expression has been rationalized when it has been
freed from irrational numbers.
Example. '^^5 is rationalized by multiplying it by *l/25, since
V5 X V 25 = Vi25 = 5.
If an irrational expression is multiplied by an expression which
gives a rational product, the multiplier is called a rationaliMaig
factor.
Example. 'i/25 is a rationalizing factor for *V\ and vice vena.
374. To Rationalize a Monomial Surd.— A rationalizing factor
of a monomial surd is readily found by inspection.
Examples.
1. A rationalizing factor of *v^x'y* is *l/x*y*; thus
2. A rationalizing factor for "i/x'" is *i/x"~'"; thus
"l/x^ • "i/x*^'- = "i/^ = X.
375. The rationalizing factor for a binomial quadratic sard is its
conjugate (2368).
Examples.
1. (2 + 3i/5) (2 - 3l/5) = 4 — (3l/5)" = —41.
2. (3v^5 - 21/11) (3v/5 + 2]/ TT) = 45 — 44 = 1.
Either of two conjugate surds is a rationalizing factor of the
other.
376. A factor may be found which will rationalize any binomial.
i L 1 1
I. Consider the binomial a^ + 6' . Let x=: aP^ and y = S, and
n be the L. C. M. of p and g; hence x* = a'', and y" = M,tnd
since n is the L. C. M. of p and g, therefore —and— are int^nl.
and therefore a" and y" are rational. Now
(ac+y)(.x«-»— x«-«y+a;«-V— • • • dby""')=x-d=y» [U02, (iii),{ii)]
where the upper or lower sign is to be taken according as « is odd
or even. Thus
x"~* — x"~*y -|- x'^'^y* — ^ y*"*
1 I
is a factor which will rationalize x + y eiz a^ + 6« .
'376] SURDS 367
1 1
II. Consider the binomial a* — 6 «. Make the same supposition
abont Xy y, and n as above.
Since
(a5-y)(a;»-i+x"-V+a;»-V+ +y"-0 = x»-y«, [U02, (i)]
hence x"""* + x»~*y + x"" V + + y""*
is a factor which will rationalize
X — y = a^ — 6« .
Example 1. Rationalize i/3 + V5.
Let X = V 3 = 3l, y = VB = 5l, and n = 4.
Then, by case I, the rationalizing factor is
(3i)*-i - (3l)*-« (5i) + (3*)*-» (5i)« -1 (5i)*-» =
3f~3 -51+ 3i5* — 6*.
The product of this factor by i/3+ V5 is
(y-^y - fi/5)* = 3l - 5t = 3» - 5.
Example 2. Rationalize V2 — 3V5.
Let X = V2 = 2*, y = 3V5 = 3 • 5l, and n = 12.
By case 11, the rationalizing factor,
X— »+x»-«y + x"-y+ + y"-*, becomes forn = 12
(2J)»-i + (2t)"-« (3 . 5i) + (2J)«-» (3 • 5i)« + . . . + (3 • 5i)"-^
= 2V + 3.2¥-5i+3«-2»-5*+ . . . + 3" • 5V.
The product of this factor by V2 — 3 V5 is
(»,/2)«-(3.V5)"=2*-3».5».
Example 3. Rationalize.a* + &t. Here x = a*, y = 6i, n=6.
Thus we have as a rationalizing factor
x' — x*y+ xV — xV+ xy*— y*,
that is ai — a«&l + aM — a5 -{- aibt — ftl.
The product of this factor by ai -f 6^ is
368 COLLEGE ALGEBRA [?J377, 378
377. The rationalizing factor of a trinomial quadratic surd may
be found as follows:
{Vx+Vy+ V'z) {Vx+ Vy-Vz) {Vlc-'Vy+ V'z) (- Vlc-^-Vy^ V~z)
= 2x^4-2x2 + 2^^2— x« — y« — 2«.
Hence the rationalizing factor for any one of the four preceding
factors is the product of the other three.
Example. i/3— i/7 + l/5 has the rationalizing factors
(1/3 + 1/7+1/5) (1/3 +1/7 -1/5) (-1/3 +1/7 +1/5) =
[(v/3+l/7)'-(l/5)T [i/7+i/5-v^3]-
(5 + 2/21) (v/7 + 1/5 -|/3) = 9i/3 + 5i/5-i/ 7+2^/105".
EXEBOISB liXrV
Find the expressions which will rationalize the following:
1. v/5. 2. Vi. 3. 2Vri. 4. V20.
5. V27^ 6. l/^. 7. V^. 8. V9^».
9. "v^^. 10. Viv^. 11. ""V^. 12. 3-V/7.
13. 21/22-1/5. 14. ^i/5+fl/3.
15. 1 + 1/3-21/5. 16. l/2+3l/7-5KlT.
17. i/l_x8— i/l + x«. 18. 1/2^— v«M^*.
19. 1/2 — V 3. 20. %/'3+2V5.
21. al-«^i. 22. I + V2+V4.
23. V2+V6 + VT8. 24. 1/6+1/21-1/10—1/35.
25. 2 V3— V7.
The Reduction op Certain Irrational Expressions
378. The calculations into which irrational expressions enter are
simplified on transforming these expressions by rationalizing their
denominators. • The following examples are some of the transfor-
mations which occur most frequently in Algebra.
I. If both terms of the fraction — - are mutiplied by |/3, the
,^ . 5i/3 ^
result IS
4 4 • l/'3 4i/3
5v 3 5 • 1/ 3 • 1/3 15
J378] SURDS 369
II. Consider the expression,
h+l/c
If both terms of the fraction are multiplied by 5 — i/c, the new
denominator will be the product of the sum and the difference of the
two quantities, which is equal to the difference of their squares,
6' — c, which is rational. Thus
a a{b — Vc) __ ab — a V c .
o-fi'r' (6 + i/c)(6-Vc) 6«-c
similarly
a _. a(h-\-Vc) _. gfe + gV^c
6-i/c (6-i/c)(6+i/c) 6«-c
Example.
3 ^ 3(44-2v^5) ^ 12+61^5 __ 12 + 6 i/5___6 + 3i/5^
4-2 1'5 (4-21/5) (4 + 21/6) 4« — (2 V^S)' 16 — 20 2
III. Consider the more general example, to rationalize the
denominator of
l/6+i/c
Proceeding as above,
g __ a{y/h — ^/ q) __ a^/h — a^/^ _. aVh — gy/c
Vh^^/~c (v^6 + i/c)(i/6--|/c) (Vhf-{y/cf b-c
Similarly, it follows that
a a(l/6+|/?) _ a\^b + aVc
Vb-Vc {Vb-V~c){Vb-\'V'c) b — c
Example.
3 Vl ^ 31^5(21/3 + 41/7) ^ 6 i/l5+12 v/35 _ 6l/l5+12l/36
21/3-41/7 (2v/3-4v/7)(2v/3+4v'7) (2l/3)8-(4l/7)« 12 — 112
_6v^i5+12V35 __ _ 3l/l5 + 6v'35
-100 50 *
The numerator and the denominator were multiplied by the sum
2 1/3 + 4 v/7, which makes the denominator the d&erence of the
squares, 4x3 — 16 X 7 = — 100; and the numerator, 3l/5, when
multiplied by 2i/3 + 4i/7, gives 2l/3 X 3]/ 5 + 4i/7 X 3i/5
= 6 1/ 15 + 12 1/35, since 1/3 x 1^5 = t/3 x 5 = l/l5 ; and simi-
larly, 1/7 X 1/5 = 1/7 X 5 = 1/35. The numerator and denomi-
nator were also divided by 2.
370 COLLEGE ALGEBRA [J378
IV. Consider the expression.
1/6 + i/c + i/d (1/6 + i/c) + \^d
On regarding 1/6+ |/c as a single term, the process is the same
as in 3, and both terms of the fraction are to be multiplied by
(|/6 + i/c) — Vd, Thus,
a _. q(/6+|/c— Vd) __ a i/ft-j-a i/c— q Vd
(i/6+i/c)+i/rf [(i/H-i/c)+i/dJ [(v/fr+i/c)-V d] (i/6+|/c)'-(i/d)'
__ aVb + aVc —a Vd
h + c-d + 2Vhc
The denominator of this last fraction contains but a single irrational
term, 2 \/h c, and it can be rationalized by multiplying both terms
of the fraction by (6 + c — rf) — 2 l/6</, as in II.
Example. —Rationalize -;= 7= ;=•
V2+ 1/3 -21/5
5_ „ 6[(l/2 + y3) + 2|/5] _6l^4-6l/3-flOV^5
V2+Vl-2Vb [.W2+Vl)-2Vb\ [(v'2+l/3)+2l/6] (l/2+l/3)'-(2v^6)*
^ 6 1/2 + 5 1^3 + 10 1/5 ^ 5 1/2 + 5 1/3 + 10 1/6
2 + 3 + 21^6 — 20 —15 + 21/6
_ (6 v/2 + 5 1/3 + 10 Vl){- 15-2 |/6)
(—15 + 2 1/6) (-15-2 Vl)
_ — 105 v/2 — 95 v'3 — 150 1/5 — 20 |/30
(-15)«-(2v/6)'
= - ^ (21 1/2 + 19 1/3 + 30 v/5 + 4 1/30).
BXBB0I8B liXV
Reduce each of the following to equivalent fractions having rational
denominators:
, |/12 1/18 1/54 1/72
2.
1/6'
1/2
1/3
1/6
1/2 X
1/x'
1/7 X
1/6 X
1/2 X
1/48*
1/6^
a-
i/«'
a
a
1
l/«
J378]
SURDS
37
4.
3
v/3'
2
1/2-'
8
1/6'
1
1/5
m
9
10
48
54
5.
2i/3
31/5
61/32'
1/72
6.
a
a
Va''
a
a
Va"-«'
Va
7.
a
a
a
a
Va»'
Va*'
Va»'
Va»
8.
a-i-b
a' — b*
a»-l
x«-l
l/a+b
Va-b
l/a— 1
l/x+1
9.
1
1
3
2
2 + 1/3'
3-1/7"
3+1/6'
2-1/2
10.
1
1
5
1/3
1 + 1/2'
1/2 + 1/3 '
1/2+ V'7'
2-1/3
11.
13
1/2
12
11
7-1/10
1/3-1/2
4-1/7
5+1/3
12.
13
14
12
Bv^i
5 + 2|/3'
8-51/2'
7-31/5'
1/2 +3i/^"
13.
7-1/5
3+1/5'
1/3+1/2
1/3-1/2'
1/5+1/3
1/5 -v/3
9-51/3
7-3l/3
14.
3+1/6
1/3+1/2'
5l/3— 31/5 7l/5 + 5l/7
1/5-1/3 1/7 + 1/5
21/3+1/6
1/3 + 1/6
15.
a
1
1
l/x— l/y
a+ 1/0 '
a-l/fe"
V*- i/y '
l/x+l/»
16.
5 + l/x
5-l/x'
28
3+2l/x
5 + 3l/a;
a+ OVx
c + dl/x '
110
I Q
al/x — i>l/y
c l/x — rfl/j/
17.
3 + l/2+>/7
'" 4+1/5+1/11
21/6
21/15
19.
1/2+1/3 + 1/5
""■ 1/3+1/5 + 21/2
372 COLLEGE ALGEBRA [i378
l + 3l/2->2l/3 6O1/2+ 121/3
21. ""7= 7= 7=' 22.
1/6+1/3 + 1/2 ' 5v 6 + 3l/2 - 2v/3
93 _ 1 _ 24 v'6^v^5-l/3+l2
*" ' 2 + 1/2 + 1/3 + 1/6 • |/6 + 1/5 — 1/3 - l/'2
2 2v
25. -7= 7= • 26. ^
l/a + 1 + l/a — 1 • i/x + y + l/iB - ,
a + X + l/a* + X* 1/0 + X + l/a — x
a + X — l/a',+x* l/a + x — l/a — x
1 /a + i/x
29. — , . 30. V .-•
a>/l + 6« + 6|/l + a« ^a — i/x
31.
\Va±Vh^ 32^ Ja+l/a»-l^
aX/l^h^-hVl-a\ xl/l-x« + j/l/l-y\
' 1/1 - 6« + 1/1 - a« ' xyT^' + yl/l-x*
35 V^q + «) (1 + M — 1/(1 - a)(l - 6)
* i/(l + a)(l + 6) + l/(l-a)(l-fe)'
(g - x)l/?>» + yg - (6 ^ y) |/ gg + x«
36. 7--——-—- ^ 7-^— — ~— '
(6 + y)l/a* + x« + (a + x) l/fe« + y«
|/l + g _ ]/l — g + 1/1 + 6 — i/'l — 6
|/l + g+|/l_-a+v/i+l> + |/l-_6
38.
|/l + g — |/l _ g __ |/1 + b + 1/1 — 6
i/i:fr^+i/i_g+i/r+"6 + |/i — i
1/2-3 V3 5-3 'v 3
39. — = — -=. 40. -7-7=
1/2 + 3 V3 3 V9 + 3
41. 3 - ; ^-' 42.. rT-^ — TT =•
*i/x — *l/v c
43. /.- /,-• 44.
l/jc + Vy ' a+Vfc
CHAPTER VII
IMAGniARY AND COMPLSX NUMBERS
Pure Imaginary Numbers
879. Since no number when raised to an even power can produce
a negative result, it is not possible to express even roots of negative
numbers (rational or irrational) in terms of numbers of our number
system which is still limited to rational and irrational numbers.
E. g., since 5' = 25, then i/-— 25 can not be expressed as a positive
or as a negative number.
A pure imaginary number is an indicated even root of a negative
number; as ^-g_ ,^-r^^ orW~b when r = 1, 2, 3, . . .
On the contrary, all other numbers, rational or irrational, are called
real numbers.
It is necessary either to exclude imaginary numbers from our
consideration or to enlarge our ideas of numbers.
The latter alternative is chosen, because practical use can be
made of imaginary numbers in mathematical investigation, in con-
sequence of a few conventions which will now be explained^
It is assumed therefore that i/—l and in general *'*i/— a, are
numbers, and may be included in the number system of Algebra.
380. The Symbolic Definition of Pure Imaginary Numbers.—
The pure imaginary numbers are defined by the relation
(|/^)' = — 1, and in general ('V^)*'" = —a.
Properties of Pure Imaginary Numbers
881. The Imaginary Unit.^— The study of imaginary numbers
is simplified by considering the properties of l/— 1, which is
called the imaginary unit, t It follows by definition that
(±i)«=-l. _
•QaoM introduced the use of i to repreeent v~ i.
fTbe designation imaginary is not a happy one. because the numbers called Imagi-
nary are no more imaginary in the usual meaning of the word than the rational frac-
tion or negative numbers. The name Neomon for the imaginary tmit^ and Neomonic for
imaginary have been suggested by G. B. Ualstead of Austin, Texas.
S73
374 COLLEGE ALGEBRA t??382-384
382. Multiple and Fractional Parts of Imaginary Units. — It
was found that multiples and fractional parts of the real units + 1
and — 1 obeyed certain laws (??6, 41, 141). Then comes the ques-
tion, what will the multiples and fractional parts of the new unit + t
and — I be if they are made to obey the same laws that the units
of real numbers do.
Just as 3 = 1 + 1 + 1,
so 3i/^= ^^^+ V -T+ v^^,
or 3i*r= t + /+ i;
similarly, just as — 3 = — 1— 1— 1,
80 _3i/irT = -l -Ti- I 31 _i/Zl,
or — 3 / = — i — i — i\
^ . . 4 13
and lust as ^ = . + ^,
5 D 5
4 / — 7 1 / — 7 3 — -
5^-1 = 5^ -^ + 5^ -^»
4 . t . 3i
5^-5 + T
383. Addition and Subtraction of Multiple and Fractional Parts
of Imaginary Units. — The sum and difference of multiples and
fractional parts of the imaginary unit may be combined into a smgle
multiple or fractional part of the imaginary unit. Thus, .
1. 31/ ^ + 41/ -1 = 7v -H [J16]
or 3<' + 4/ = 7i*.
2. 5v III- Di -^= -41/1^1 [J34]
or 5/ — 9/ = ~ 4/.
3. f i/ITT + ||/Z1 = IV^jji/ 3i [il6]
or !'*+|'-1tV*
4. al/irT + Z,v'^ = (« + />)l^^ [{7, LawV]
or ai -\- bi = (a + b)i.
5. aV^-hv'^^{a-h)i':^ [838,5]
or ai — hi z:^ {a — h) t.
384. Multiplication by/.— In order to extend the meaning of
multiplication of real numbers to the product of a real number by
the imaginary unit, t, or by two or more imaginary units, it is
necessary to assume that:
? 2385-387] IMACUNARY AND COMPLEX NUMBERS 375
I. The commutative law (§7, Law III) holds, i. e.,
(V^ — \)n z=zn Y — 1, OT ia = ai.
E. g., /3 = 3/= i+ f+ A
That is, t plays the role of a real factor.
Cor. 1. V ^^ 1 = 11—1, or I • 1 = 1 • t = t.
Cob. 2. t • 0 = 0 • t = 0. [{75]
II. The associative law, (17, Law IV) holds, that is
ai • hi = ah II = aht^.
Thus ai can be given the designation of product in the complete
sense of the term.
(7)x' =
385. Division by /. — According to the definition of a quotient
:«/. [2126]
But a X i = (if\ [Def. product]
hence ( ^ j x i = fi X h [881, Ax. 7]
^ = a. [881, Ax. 4]
Observe that i plays the role of factor in division.
386. In addition to the double series of positive and negative
integers, positive and negative fractions, positive and negative irra-
tional numbers, we have the double series of imaginary numbers;
— 4 1*, — 3/, — 2iy — I*, 0, I*, 2/, 3*', 4/,
Between any two consecutive numbers of this series there are as
many as we please of fractional and irrational numbers of i. For
example, between 2 t and 3 / lie |/, i i/5, etc.
387. Powers of A— The positive integral powers of i are derived
by means of the definition of i, §381 and ^384:
{v^y= v^ITi or t» = i.
(v-T)'=-Jl_ _ _ i«=-l.
(i/-lf={y-l)' • V^ = - V-1 i3 = - i,
iV^*=W-^)^ ' (i/~T)2=(_i) (-i)=i i*=i« • t«=(-i) (-1)=+!.
(V^)^= (i/-T)* -1-1 = i/^ i^ = i^'i = + u
W^y={l ^y -d '-1)2=1 .(_1)=_1 i«=i« -iarrr-l.
(V/^)'=(V/^)*- V/-l=-l/^ ,-7 = i«- i=-i.
(\/^y=(i/~i)' ' (i/zT)«=(_i)(_i)=i if^=if^ . {«=+!.
etc. otc.
From these results is derived the followinjj: rule:
376 COLLEGE ALGEBRA [JJ388, 389
/. Even powers of i are real,
1. If the exponent of the power of i is divisible by 2 and not by 4,
the power is equal to — 1,
2. If the exponent of the power of i is divisible by 4f the power of i
is equal to + 1.
IL Odd powers of i are imaginary.
1. Xf the exponent of the power of i is S less or 1 greater tlian a
multiple of ^, the power of i is equal to -\- i,
2. If the power of i is 1 less or 3 greater than a multiple of ^, the
power of i is equal to — t'.
In general^ if r is any positive integer^ therefore
1. ,.r-.^^i, ,.r^+i L-12 3 er.
IL »*'•-» = i or i*'-+»= f ; t*'^"^ = — i or !*'•+» _, _ j- J'^- ^» ^> ^ • • • 00
388. The Pure Imaginary V—a = l/a • /• — Since by definition
(v/=^)*=-a
and {va • l/^)' = {Vaf {V^Y = -a, [«866, I]
then {V ^y = {va • V -^)'.
\/—a = Va ' y —1 = Va • i.
E. g., 1/-.I6 = 1/I6 • V -1 = 4i*.
K — IT = V TT • 1/ — i = V IT • 1.
889. Addition and Subtraction of Pure Imaginaries. — The
same rules are used in combining imaginary numbers by addition
and subtraction that are used in combining real numbers.
Example 1.
|/ — 16 + v/— 25 = 1/ 16 1 + v/25 1 = 4t + bi = 9i.
Example 2.
7 V — 81 — 5i/— 144 = 7l/81 i — 5i/l44 1 = 63i — 60t = 3t.
Example 3.
V — 1— 2x— x« — l/— 4a5»=l/l + 2x+x«t— v/4xS* = (l+a;)t— 2x1
Example 4.
t«* - 1« = i»(i«- 1^ = (+!)»[- 1 -(-03 = — 1 + 1.
8{390, 391] IMAGINARY AND CX)MPLEX NUMBERS 377
390. Multiplication of Pure Imaginaries.— The products of
pure imaginaries are simplified by the following rules:
L l/a X l/^ = v/a • i/6 . 1 = \/ab t. [?388, J366, I]
11. \/^ X V^ -Va'ixVh 'i [{388]
= l/a-l/6(0«
=— l/^. [{366, I; {387,1,1]
EXAMPLB 1.
— 1/27 X l/— 12 = — 3l/3 X l/l2 i = — 3l/3 X 2^/3 • t
= ~18t.
Example 2.
v/_72 X (-1^ -50) = 1/72 t X (— 1/50 t) = 6 v/2 X (— 5>/2)i«
= 60.
Example 3.
|/-2x3v/-6x(-2l/-24) = l/2tx3v6tx(-2l/24t)
= l/2x3l/6x(-4l/6)i»=-12 • i/2 i/sexC-t*)
= 72 1/2/.
391. Division of Pure Imaginaries.— The quotients of pure
imaginaries are simplified by the following rules:
1. 1^ = ^^ = J? I [{388; {138, 1]
l/6 i/6 A^
n. J^=-lS- = l5.i=J?.-^ = -^|^i. t«388,{131]
III. \^-^^Vj^^J^, [{388, {131]
|/_6 1/6 i Al^
i/_-io i/Tot lio. /=- .
Example 1. ^—--— = — z^ = -% — * = 1/ 5 • t.
l/2 V 2 > 2
„ 2l/-6 2l/6. ^,o ^y-
Example 2. ^- = ^ = -x\- = ^ i/2.
3v^~3 3i/3t 3\3 -^^^•
t' 2 le 2
Examples. -T^ -,•,(! + ,•.)- (_i) (i ^. j) - -|'
378 COLLEGE ALGEBRA [2391
EXBBOISE LXVI
Simplify each of the following expressions:
1.
V- 3«,
V/-4,
l/- «',
5 1/ _ 40,
1/-49+1/-
3
2|.^— 12 + 31/
V —X- V—'X,
I/- 49,
V - h\
I/-I2,
V- 16,
V — 100.
2.
V- 125.
3.
1/— X*",
V-x»".
4.
1/-48,
3 1/- 72,
l/— 90.
5.
2 V- 40,
4 V- 72.
6.
l/-3«x«,
V'-^shj,
2 V -SxV.
7.
04-1^-100
+ 3|/-25-
_. It
-1/-21-
1 -1J-5V
8.
9.
-27, 1/ -
l/_.c- r —y
„%•'' + l/-,r
, V —ny-
— !,■''- 2ab.
-1, v'-""!'-!.
10. 1/3-1/— />, i/_8v^ — 12, Vl5l/— 5, i/_5v— 20.
11. 4]/- 2 • V -3 - 3l/-5 • I -li + i/_2(v -2 + V 3)
-V^^(V -24+ V6 - l/ITj). ^
Ans. _v^6+ 9 + V^^ — 6]/^.
12. aV —aV^^ • K -;f76"^ a^b^V -a-'-^b-' • ]/ ^^^\.
13. (1 - 2V ^) (4 - 5i/Il6) - (7 - 81/ 14)) (10 + 11 ]/^^T2).
14. i", i', iS i^.
15. t»», i", ^l^ i»«.
16. ai ' 6/, j'v a • iV 7>, 2t • 5/, 7 1* • tl/7.
17. iV— «, eV— .r:8, t'|/_.c3^ t l/a • 1/— a.
18. ti/.c • 1/— .r/, 3/v — /^ •l/4/«, 5u'— yw, tV— 5i)«.
19. I**", 1*"+', 1*"-', t*»-«.
20. v/-7«, l/-^^ v^-'S l'^^^
21. (v^Zn + V — 19) • (1 —119- V ^^133). Ans. 21^7.
22. (a + V IT2) (r/ - 1 -^«).
23. (i/II^5+ v'-^?^') • {\/ —a-V —\ —n%''),
24. V^^V^^^- V II5— 1 -ll 1:3 l^^ V'^.
25. I -a-6 -y —ab^ - y -.ab\
{§392,393] IMAGINARY AND COMPLEX NUMBERS 379
V 6 1/—^
27.
28.
29.
30.
31.
V-
6
l/-6
V-
3
1/3
V-
- a
V-
- a
V-
1
V
h
a
V— 3 ]/— 6
\/a V — ax
V —h V — X
a^
111 1
ai h — c dP
32. i^a — b • y h r- a. 33. l/3.c _ 5y • V 5^ — 3x.
34. l/— 176 ~ 1/ 11 — l/— 325 -r- l/— 13 + V 540 -r- 1/-^15.
35. (2i/8 — V^^) -^ (-1/32).
36. (3l/^~2i/iri2+ v/6 — 9) -^ (— 3l/^).
37. (181/ 1^ + 36i/50 - 54v^70) -^ (9i/ -40).
Calculate :
38. {yZ:iy'+{V—lf _ (1/1:1)''+ (!/=!)**+ (i/=lf
_(l/^)^«_(v31)^-
39. (1/Z5)*, (v-3)^ (l/^7)^ (1/Z2)-
Complex Numbers
392. Complex Numbers. — The sum a + 16 is a complex num-
ber; e. g. , 2+ 1/ 5 lis a complex mimber. The terms of the complex
number a-\-tby a and <7>, belong to two distinct systems of num-
bers of which the fundamental units are 1 and t. Since a = a + 1 0
(2384, I; Cor.), all real numbers are included in a system of complex
numbers a-{-ib; similarly, since ib = 0 + ?7>, all pure imaginary
numbers are included also in the system of complex numbers a + ib,
(1) Hence, a+ib is 0 only when a = 0 and b = Q. [§384,Cor.2J
393. Two complex numbers which differ only in the sign of the
imaginary part are called conjugate comj)lex numbers.
380 COLLEGE ALGEBRA [8394
394. The addition^ subtraction, multiplication, and division of
complex numbers are immediately defined by the assumption of the
permanence of the fundamental laws of real numbers.
I. Addition
In particular In general
(2+i3H-(5+t6)=2+6+i(3+6) (a+ib)+((i'+ib')=a+a'+i{b+l/),
= 7 + i9.
For For
(2+i3)+(5+t6)=2+i3+5+i6 (a+ib)+(a'+ib')=a+ib+a<+ii/
[Law II]
=2+5+t3+i6 =a+a'+ib+iy
[Law I]
=7+t9. =a+a'+i(b+bn.
[Laws
n,v]
II. Subtraction
(5+t7)-(3+i4)=6-3+i(7-4)=2+t3. (a+t6)-(a'+iy)=CH-a'+i(6-6^).
[Def. of Subtraction, VI; J88, (&)]
Cor. — The necessary and sufficient condition for the equality of
two complex number a + i'5, a' + ih' is that a = a' and h = h'.
For if (a + ih) — (a' + iV) - a^a' J^ i(h — i') = 0
.-. a— a'=0 and 6— 6'=0, or a=a\ h=h'. Q. E. D. [J892,(l)]
III. Multiplication
(2+i3)(4 + i5) (a+t6)(a' + i60
= (2-4-3-5) + i(2-5 + 3-4) =:{aa' -bb^) + x(a¥+ba^.
= -7+i22.
For For
(2 + t3)(4 + i5) (a+i6)(a' + i6)
= (2+i3)4+(2 + i3)t5 = (a +i6) a' + (a + i6) i!/ [Law 5, {7]
= 8+il2 + ilO+t3-t5 =aa' + i6a' + ai6'+i6i6'[Law6. S7)
= 8 - 15 + i 10 + % 12 ^aa' -bV + tab' + iba' [Laws I-I V,
?«6,7,881]
= - 7 + 1 22. = aa'-bb'+\(ab' + 6aO. [Law V]
Cor. — If either factor of a product vanishes the product vanishes.
For 't X 0 = t(h — h) [Def. of zero]
= ih - ih [J38, 5]
= 0. [Def. of zero]
Hence (a + i6)0 = a x 0 + i6 x 0 [Law V, J7]
= 0+iX6 -0) [J76 and Law IV, 87]
= I 0 = 0.
«395, 396J IMAGINARY AND COMPLEX NUMBERS
381
In particular
2-fi3_ 8 + 21 .12-14
4 + t7~ 16 + 49"^* 16 + 49
For
2 + i3_2+t3 4-i7
4 + i7"~4 + i7^ 4~i7
8 + tl2-il4— i«21
"" 16-(i7)«
(8+21) + t(12-14)
16 + 49
IV. Division
29
65
66'
In general
a + ib _oa^+6y .6a^j;^ay
a'+iy "■ a'« + y« ■*"* o^' + d' *
For
a+ t'6 __ a + t6 a^ - t'!/
a' + ty — a' + t6'^ a' - il/
_ (gg^ + 6y ) + t(a^6 - a!/)
a^-^6'8
+ y«
[HI]
Therefore division is a determinate operation as in the case of real
numbers, except when the divisor is 0 or when a' + tfc' = 0, that
is, when a' = 6' = 0 and therefore a' * + 6" = 0 and the quotient
takes the form
aa' + hh' M'b — ah'
o'*+6'« + V« + 6'«^
0 . .0
0 + V
[?78, 1]
V.
ff the product of two complex numbers u 0, one of the numbers is 0,
The proof is similar to that given in {76.
395. It follows from {375 that a fraction whose denominator is a
complex number can be reduced to a complex number by multiply-
ing both terms of the fraction by the conjugate of the denominator.
Example 1.
2— 3l/-5_(2-3v -5)(-4i/— 10)^-8|/l0t+12l/5 - i/lO t«
4i/irTo "~ (4i/^^IlO)(— 4v/;iT0) "
160
-60v^2— 8v/l0i
=-i^i-.-'-'»
160 » 20
Since the denominator of the given fraction is a pure imaginary it is
necessary to multiply the terms of the fraction by the conjugate.
l/a + ib = l/x + iy/y
\^a — lb =V x-^ iVy*
a + t6 = X — y + 2i l/xy.
a=zx — y
ib = 2i\/xy.
a — lb = X — y — 2 1 \/xy
\/a — ib = l/x — i\^y.
396. If
(1)
then
(2)
From (1)
By {394, II, Cor.
(3)
and
(4)
Subtracting (4) from (3) (5)
• •
(6)
382 CX)LLEGE ALGEBRA [«397, 398
397. The square root of a complex number can be expressed as a
complex number.
Example. Let (1)
1^5 — >/— 11 = V^x — iVy,
By 1896 (2)
Vb+\ —11 = v.r + iv>
multiplying(l) by (2) (3)
V/25-(-ll)=.r + y=C
squanng (1) (4)
5 — V'llt = x — y— 2l/jry t.
By J394, II, Cor. (5)
cr — y = 5.
Adding (3) and (5)
2.r = ll andx = V-
subtracting (5) from (3)
2y = 1 and y = {,
fiT 1
,*.
|/5-V-ll = V2-V2-
In general, let (1)
y'a + ib z=z\ x-\- i\/y
then (2)
V a — ib = \ X — %\^y.
multiplying (1) by (2) (3)
ya}+b^ = x + y
squaring (1) (4)
a+ ?*6 = a — y + 2i |/a?y
by J394, 11, Cor. (5)
a = x — y.
yar -\-b^Jf.a
X=: ! 5
Solving (3) and (5) (6)
«
2
From («), (1), and (2)
898. Let n = 0 in (i), 1307, then
Let now b = 1, then v'{ = *v'^ = JV 2(1 + i),
EXEBCISB IiXVH
Simplify each of the following expressions:
1. 5v -l(i — 2l/-0. 2. 2i/-20-f 3r -45-r— 8<>-
3. 21 _r>.c«+7v -4<r.r«+12v^~36aV.
4. v-^~r-:^+r-u+i/-^
5. t/— (i* + 5i — 9 a* — 2l/— 4 a*.
2398] IMAGINARY AND COMPLEX NUMBERS 383
6. (3 + 50(7 + 40. 7. (7-80(5 + 6/).
8. (11 — 12/)(ll — 10O. 9. (5-2tl/7)(6 — 2iv7).
10. (v'3 — »V'6)(v2 — iVt)).
11. (2i/7 + 3i'v8)(3v 7-10iV2).
12. (v 3 + i V2) (v 2 + I vl),
13. {aVh + ciVd) {av'b — ci\/d),
14. (3+20(3-2i).
15. (3i 3 + 2iv/2) (3v 3 — 2iV2).
16. (2i/5 + 5/ 1/2) {2Vb - 5i> 2).
17. VT+1 vT~i. 18. V 33+56i • i/33-56i.
19. {y^^\/~i)\ 20. (3 + 2iV2)*-
21. (5-2ii/6)'. 22. (va"+~i') + (V i^^')'-
23. (v 4 + 3/+ vT~'^if. 24, (1 + i)'.
0*. 27. (a +60*.
29. (m + H/)'— (m — m)'.
25.
(3
-2/)'.
26. (]
28.
(«
+ 6i)'+(«
-bif.
30.
G
+ .-l/3\'
2 / ■
n. (=L±iU)-.
32. (p+qi)'+(p-qi)\ 33. {p + qi)*-{2n + q)\
34. (^-ii)*- 35. (^:^Y.
36. (l/5 + iViy + (v/5 -* i/'7)V
37. (l+tv5)*+(l-iV5)*.
38. (5+2/v/g)* + (5 — 2iV6)*.
39. (3 + 2/ V 2)* — (3 — 2 1 1/'2)\
40. (0 + 6/)'+ (a — 6i)^ 41. (a + 6/)*-(a — 6/)'.
42. (1 + /)'+(! -/A 43. (1+/V 2)'+ (1-/1/2/.
«. (L-h^L<')'+(L-_.;.-^')".
47. If __ 1^ _ .||. _3 = J> and — .j^ + ^v — 3 = J", prove that
J'» = 1, J*'' = 1, J'2 = ./",
^^ 3n+l __ y /' Sn+8 __ ^ JffSn-i-l __ Jr3H^2 __ ^/
384 COLLEGE ALGEBRA [8398
Reduce each of the followmg to a complex number:
64 29 5
48. :; T— ^- 49. , . „ y—z' 50,
72.
73.
1 + 3V/-7 4 + 7l/— 5 i/2-ti/3
^ _ 5-29tV5 __ l + 33tl/3
|/2 + v/-l 7_3{|/5 4 + 3iV3
1V^ — W^ 83 — 2l/I^ 23~3V^
9- 2v/^ ' 4+51/36* 7 — 61/II2
^^ m+l/— n . m — 1/ — n . 2(m' — n)
65. — ' :^^^+ ;==. Ans. — p-
m — 1/ — » m+l/— n m' + n
56. l±i. B7. l^l+i^^
1 — » 1/3 — 1|/2
69 + 1/^ — 61/^ - 71/15 ^ „ . /-— , /-—
68. — -^-^^ =^-^^ ;^--3-: Ans. 2 +|/— 3—41/— 5.
3__V/__3+3l/-5
59. IZZ^. 60. V^. 61. 1^. 62. =i+^:.
1 — » 1 — 1 (1 + t)' ni — nt
63. }^B±i^. 64. ^■. 65. T^-.+ r^-
l/-_a— l/— 6 at +6 1 + t^l— »
68 ?: - ?^. fio v^^ - y + v'y - ^
X — i 1/ 1 — a:* c + c£i c — di
Vx + tVy _ l/y -f t l/g
V X — * Vy V y — * Vx
|/1 _|> a — tV 1 — a i/l — o — il/1 + a
Evaluate the following square roots:
74. i/3 + 4tzbK3— 4t, |/4+3td=l/4 — 3i.
75. 1^5+211/ 6 d=^^5-2t 1/6, V^ll + 4i v 3d= ^ll-4tV3.
76. 1/5 + 12/, 1/9 - 40t, V2 + 21V3, 1^1-6*1/10.
77. 1/^7+3011/2, 1^3 + 2i>10, 1^1-411/14, Vl-ti/3
78. V'a + I l/ic* _ a« zb l^a — t l/x« — a«.
BOOK IV
CHAPTER I
IHTSODUCTION TO THE STUDY OF THE EQUATION OF THE SECOND
DEGREE
Some principles will be considered here which have already been
learned concerning the formation of squares and the extraction of
roots.
399. Theorem I. — To find the square of the product of several
/actorSy form the product of the squares of the factors.
If the product abed is multiplied by itself, it follows from the
associative and distributive laws of multiplication, that
{abed)* = (abed) {abed) = abcdabcd = aabbccdd = a^b^c^d*.
On regarding a as the coefficient of bed, it follows that the square
of an integral monomial is found by squaring the coefficient and
doubling the exponents of the factors. Consider, for example, the
monomial 11 a^'c*. If it is multiplied by itself the result will be
(lla6V)« = 11 V6V = 121a«6V.
400. Theorem II. — Conversely, the square root of the product of
several factors is found by forming the product of the square root of
each of the factors separately.
Let abc be the product. The square root of this product is
\/abc = }/aVb\/c,
For, on squaring the second member of this equation by Theorem
I, the result is obtained
{l/a • l/6 • l/cj = (l/al/6r/7)(l/al/5l/c) =
= V a\/b\/ cV a\/b\/ c = \/ a\/ a\/b\/b\/ c\/ c = ahc.
which is the square of the first member i/o5c.
38(5 COLLEGE ALGEBRA [2400
It follows that the square root of an integral monomial is foond
by extracting the square root of the coefl5cient and dividing the
exponents of the factors by two. For example,
V 36 a*6V = 6 a«6c»,
because, by Theorem I, the square of the second member is
36 a*b^c\
Therefore, in order that a given integral monomial may be a perfect
square, there must be a second integral monomial which, squared,
will reproduce it; its coefficient must therefore be a perfect square,
and the exponents of its factors must be even.
It may further be obser^^ed that a perfect square may l>e remoTed
from under the radical sign by taking the square root of it before
the radical sign.
For example, by Theorem I, on taking the square root of each
factor separately,
1/(1*6 = 1/ a* • V 6 = aVb,
This result makes it possible to simplify irrational expressions.
Consider, for example, the irrational expression,
1 32a36'c*.
The quantity under the radical can be separated into two factors,
thus,
32 a56V= 16 a«6«c*x 2 «6.
One of these factors is a perfect square. On extracting the square
root of each of the factors as above.
1/32 aH^'c*^ =V 16 a'b^c* • V 2nb = 4 ab^c^ 1/2 ab. [1366, 1]
Converse!}', a factor can be introduced under the radical by
squaring the factor and multiplying the quantity under the radical
by it. _
Thus, ayb z^ y'a*b,
for {v a^bf = a^b,
and (a\ bf = a^ (v^ 6)" = a*b. [Th. I, «S99]
JJ401, 402] THE EQUATION OF THE SECOND DEGREE 387
401. Theorem III. — To square a fraction, take the quotient of the
square of the numerator by tJie square of the denominator.
For, on multiplying ^ by itself,
Theorem IV. — Conversely, to extract the square root of a fraction,
take the quotient of the square root of tlie numerator by the squ>ire root
of the denominator. Thus,
fg _ Va
\^ "^ Vb
because the square of the second member, by Theorem III, is
m=tM-v
the fraction itself.
402. Theorem V. — The square of the sum of two quantities is the
square of the first plus twice, tlie product of the first by the second plus
the square of the second.
By multiplying a + 6 by itself it is found that
(a+ by = (a + 5) (a + 6) = a«+ 2ab+ b\
Thus the squares of the binomials 2 x + 4 and 3 x — 5 are
(2x+4)2=4x2+2 • 2x • 4 + 16=:4x«+16x+16,
(3x— 5)2=[3x+(-5)]*=9x2+2 • 3x(~5) + (-5)«=9x«~30x+25.
Similarly,
(2 ax + 6)' = 4 a«x« + 4 abx + b\
(d.+|y=x« + px+f.
CHAPTER II
SOLUTION OP EQUATIONS OF THE SECOND DEGREE
40S. Solution of the equation ax=b. — Every equation in
which, besides given constants, the first power of the unknown
quantity, x, alone occurs, may by multiplication, addition, and
subtraction, be reduced to the form
ax = h, whence a? = - . [J189]
a
Division is the last step involved in finding the value of x. In
solving an equation of the first degree in x, there are involved only
the four fundamental operations of common Algebra.
404. The Solution of the Pure Quadratic Equation 01* = A.
An equation which involves x" only, in addition to given constants,
may be reduced by the four fundamental operations of oommoa
Algebra, as has been explained above, to the forms,
ax" = 6, and x" = - = -4.
In order to find the values of x, it is necessary to employ a fif^
operation, the extraction of the square root (Theorems 11 and III,
22400, 401), which gives
There will be two values of x, namely,
Xj = + 1/^, and Xg = — l/i;
because (+ l/^)" = il = x"
and (-i/5)* = ^ = x«.
Both values of x will be real when A is positive; butif i ^
negative, the equation x* = — A, can not be satisfied by any real
values of x, since x' must always be positive, and can not be equal
^ 1404] THE EQUATION OF THE SECOND DEGREE 389
tity. Here, however, the <
to a real negative quantity. Here, however, the equation can be
satisfied by the values
with the convention that (v^— ^)* = — -4.
These quantities, x^ and x,, are called imaginary values of x.
The equation, (1) x" = J.
may be solved as follows:
by transposing (2) x* — ji = 0,
whence (3) x«-(v'J)«=(x+v'J)(x-i/l)= 0. [J941
Equation (3) can be satisfied by placing each factor equal to zero.
Thus, x+i/J=0, X — i/J=0
or Xj = —VA, Xj = i/il.
Hence the equation may be written in the form,
X* — J. = (x — Xj) (x — Xj).
Examples.
1. Solve the equation 9x« — 16(1 — x«) = 0.
Multiply out 9x* — 16 + 16x" = 0
transpose and unite 25x' = 16
divide by 25 «' = ^'
25
whence x = it--
5
2. 8olTe^ + (2x«-3) = ^^^2^^-
4 5
Multiply by 20 15x» + 40x« — 60 = 64x« + 36
transpose and unite 9x* = — 96
divide by 9 «. = =9§ = =:32^=4i:J
' 9 3 3
or a5,= +4j:|2_ ^^^_4^.
The values of x are imaginary.
3. Solve i/x« — 5 + Vx^ + 7 = 2x.
Transpose l/x* — 5 = 2x — i/x« + 7
square x» — 5 =4x" — 4x • |/x»+ 7 + x'-J. 7
transpose and unite 4xi/x' + 7 = 4x« + 12
divide by 4 xi/x«+7 = x" + 3
«quare x* + 7x" = x* + 6x« + 9,
x» = 9.
X r= ±3.
2.
X* = 0.074529.
4.
x« = 5.
6.
19x» = 5491.
8.
ya!» = 560.
10.
17*»— 7=418.
12.
«ta!* = o* — nx*.
14.
(u^ — h= ea^ + d.
16.
2x 1050
3 ~ 7» ■
18.
(3«+1.5)(3x-1.5)
= 54.
390 COLLEGE ALGEBRA [«404,
EZBBOISE liXVin
Solve the following:
1. x«=169.
3. «•=«.
5. ax* = h,
ax* c
9, ax* — 6 = c.
11. 9x«+4x«=325.
13. 13x«— 19 = 7x»+5.
,^ 15x 810
17. (x+l)(x-i)=^^.
19. (a + x) (6 — x) + (a — x) (fc + x) = 0.
20. (a + 6x)« + (ax — 6)« = 2 (a«x« + fc«).
21. (7 + x) (9 - x) + (7 - x) (9 + x) = 76.
22. (2x + 7) (5x - 9) + (2x — 7) (5x + 9) = 1874.
23. (1 + x) (2+x>(3 + x) + (l — x)(2 — x) (3 — x) = 120.
24. (2x + 3) (3x + 4) (4x + 5) — (2x - 3) (3x — 4) (4x - 5) = 184.
25. (x + a + 6) (x — a + 6) + (x + a — 6) (x — a — 6) = 0.
26. {a + hx)(h — ax) + {h + cx)(c — 6x) + (c + ax) (a — cat) = 0.
27. (a + x) (6 — x) + (i + ax) (1 - bx) = (a+ fc) (1 +x»).
28. (a + 56 + x) (5a + 6 + x) = 3 (a + 6 + x)«.
29. (9a — 76 + 3x) (9b — 7a + 3x) = (3a + 36 + x)«.
30 ^ + ar _ X + 6 ^^ x + 5a + 6_ x-^a + b
a — X X — 6 ' X — 3a +6 a — x + 36
g — X _ 1 — bx 3a — 26 + 3x _ g — 7g + 86
'1 — ax~ b — X ' a — 26 + x ~~ 3x — 5^ + 46
35 + 3x _ X — 55 7a — 6 + X _ g (g -f 56 + x)
1 + X "" 3x — 53 76 — a + X "" 6 (5a + 6 + x)
ttx + 6 __ ex + f? X + g — 6 __ g (x 4- g + 56)
a4"^^ c+c/x ' X — a+6 6(x + 5a-|-6)
38 25 + X _ 13 + -^ . 3g i7g+6~^x g«(a+176+x)
9 + x""47-x a+176-x 6^(17a + 6+x)
X — 2 3(8-x) (1 + 3x + 5x«) (x« + 3j- + 5) __ 9
3x + 14 ~ 28 - X ' • (1 + 2x + 3x«) (x« + 2^: + 3) 4*
J404] THE EQUATION OF THE SECOND DEGREE 391
1.1 2x
42
l+l/l— rr l_vl— X 9
43. 1/13 + X+1/13 — x=:6.
44. V X + 4 — l/5x — 24 =
45. V' X + a — V" 5x — 3a — 46 =
l/x + 4
26
46. l/x + a — v^x — a =
k'x+ a
X 4" « — ^
47. VZa — 26 + 2x — 2v''3a _ 26 — 2x =
, a + 2 6 + 2x
V 3a — 26 + 2x
48. 2V''5 +2x— v'13 — 6x = v37--6x.
49. v^'Ux — 11+ r3(2x-l) = 2v2x + l.
50. V a + X + Wa — x = V2a.
51. V76 + X + V76 — x = 8.
__ a — Va* — X* a \ a — x \a — x -
52. == = - • 53. — = l/x.
a -f V a' — x« 6 x a
I I4.a:«+|/l_x« a ^^ Vl + x« + Vl - x« a
54. r=zzzr=L = -• DO. I — ;i = = -•
V^+x'— Vl — x« ^ Vl + x« — Vl — x« ^
Vl + x«+*l 1 — x« a
56.
Vi + x^ — V 1 — x«"
6*
57. V0.125x« — 6x = i/0.25x« — 8.
58. (l - i/TZ:^*"' - (l + VI -xf = X-V3.
59. (x + V'2 — X*)"' + (x — V" 2 — x«)"' = x.
m x"
61. V wi • |/x — m = l/7» • }/
X-
62.
65.
X + m — 2n __ n + 2m — 2x
X -f- m + 2fi "" n — 2m + 2x
49/ 7\« 25 ^. 2 , ,^ \
— IX I = — . 64. klO— x = —
64V 9/ 81 X — 10^ 10
a{a — 6) . , , (6 — a) 6
i- ^ a-\-o — X = — — .
X — a — 6 a+6 — x
__ (x+6— c)(x — 6 + c) (g — x) (x — 6) __
^^- "* -(6+c + x)(6+c-x)' • (a-x)-(x-6)""^*
392 COLLEGE ALGEBRA [8406
Thb Solution of thi Equation aa^+bx + c =0
405. An equation which involyes the first and the second powers
only of the unknown quantity x, i. e., x and x*, besides given
constants, may be reduced by the four fundamental operations of
Algebra to the form
(1) ax«+6x4.c = 0,
and finally, by dividing by a, to the form
(2) 2^ + px+q = 0.
In the case of these two equations the problem set is to find what
and how many values of x there are involving a, 6, and c, orp and q
which will satisfy equation (1) or (2). It is to be remarked that
the equality (=) does not exist for all values of x, but for two
only (as will be proved).
It will be found that the final solutions of equations (1) and (2)
are obtained by using only the operations employed in solving the
simple equation of one unknown quantity, which can be reduced to
the form ox = &, and the pure quadratic equation, which can be
reduced to the form x* = ^; i. e., by using the four fundamental
operations and the extraction of the square root.
The method for solving equation (1), which is about to be
explained, will be illustrated by an example. Consider the equation
9 x« + 4 X = 13.
The coefficient of x* is the square of 3 and the equation may be
solved by adding to both members such a quantity as will make the
first member a trinomial square. By Theorem V, J402,
from which it follows that the third term of a trinomial square is
Hence, the third term of a trinomial, which is a perfect square^ i$ the
square of the quotient of the middle term by twice the square root of the
first term. Therefore, in this example,
\2V9W V^-^/ 9'
8406] THE EQUATION OF THE SECOND DEGREE 393
Add - to both members of the equation; then
/'q,_i_2\* 121.
or (3x+-) = — ;
which is of the form a^=:^. Extract the square root and get
^^ + 3 = ^3'
2-1-11 13
Transpose and divide, x = -^ — = 1 or — ^;
i. e. , Xj = 1 and x, = — -^ •
406. Now consider the solution of the general equation of the
second degree, ^^^ „^+6^ + , = o.
The plan of attack, as illustrated by the preceding example, is to
show how the first member of the equation ax'+ 6x = — c (1) may
be made a trinomial square. It follows from the formula for the
square of (y + «) that two things are necessary, namely, to multiply
both members of equation (1) by 4 a, which gives
4 a"x* -}- 4a6x = — 4 acy
and add to both members of this equation,
V2l/4a»W \2 2ax)
which gives finally
4aV-f 4afcx + 6«=6« — 4«c.
After extracting the square root of both members of this equation,
2ax+ 6 = d=l/^'— 4 ac\
and after transposing h and dividing by 2 a, finally
^^^ ^" Ta
*It is possible to solve equation (1) in a direct manner as follows:
Write it in the form fx v'a-f — 7"^ ~ r:; ^-^^ = 0;
V 2 V a/ ^"
the identity of this formula with equation (1) can be easily verified by developing
tbe parenthesis. The following result Is deduced at once:
Ix ya + — —\ = — T- — ;
\ "^ 2Va) *^
hence xy a+ — -=r = -^-^ — -= — »
2 V a 2 V a
or « V a = *^7= •
2 V a
^ „ —ft ±i/ft«— 4ac
or finally « = ^2a
394 COLLEGE ALGEBRA [8407
It will l)e observed that the quantity required to complete the
square of the expression ax^ + J'^r, after multiplying by 4 a, is b\
the square of the coefficient of x in the given equation.
Example 1. Solve the equation 3x«-f lOx = 32.
Here a = 3, i> = 10, c = —32.
Multiply by 4a=4 -3=12 36x«+120x = 384
add fc«=100 36.c«+120x+100=484.
Extract the square root 6x -|- 10 = ±22,
transpose and divide x = — - -^ =z^^-^
*^ 6 3
1 fi
Whence x^ = 2, and x^ = = — 5 J.
Example 2. Solve the equation 2x«— 3x = 14.
Multiply by 4 • 2 = 8
16x« — 24x = 112.
Add 3«, or 9,
16x«-24x + 9 = 112 + 9 = 121.
4x — 3 = dbll.
Whence x, = — i — = — = - , and x, = = — 2.
* 4 4 2 » 4 .
Example 3. Solve the equation 9 x* + 14 x = — 3.
Multiply by 9
81x«+ 12()x = —27.
Add7«
81 x« + 126x + 49 = _ 27 + 49 = + 22.
9x + 7 = dzl 22-
X = ziL^y^
X- ^
407. In formula (3) it is supposed that h^ — 4 oc is positive.
When the equation has been put under the form,
(2ffx + ?;)« = 6« — 4rto,
it is seen that the unknown quantity 2ax -^h ought to have such a
value that its square is equal to the quantity 6*— 4fic;or if i^b^—iac
is taken as the arithmetical square root of the positive number,
6' — 4rtc, i. e., a positive number, commensurable or incommensur-
able, whose square is ^* — 4 ac, then the unknown quantity 2 ox + 6
should be equal to v 6* — 4 ac, affected with the + or — sign.
8408] THE EQUATION OF THE SECOND DEGREE 395
The result of formula (3) is expressed in the following general
rule: the two roots x^ and x^ of an equation of the second degree^ writ-
ten in the form,
aa^ + 6x + c = 0,
are equal to the coefficient of x with the contrary sign, plus or minus
the square root of the result of subtracting four times the coefficient of
x* times the constant term from the sqiuire of the coefficient of a, all
divided hy twice the coefficient of x^; thus
— b + Vh^ — ^ac , ^h — l/b* — 4ac
^3> "' = 2a '-"^d-.^^ 2^
Example. — Solve the equation Sac" — 7x — 6 = 0.
On comparing this equation with aa^ -|- 6x + c = 0, it is seen that
rt = 3, 6 = — 7,c = — 6, and on substituting these values for a, 6,
c in the formulae for x^ and x^, the result is obtained
7^v/49 + 72^7_^n^3^^_^
6 6 ^'
x^ = 3, and ac^ = — |.
408. In case the coefficient of the second term in the equation
ax*-f6a;+ c = 0 is even, say b = 2 6', then the formula (3) be-
comes
— 2 6' dbl/46'« — 4 ac —b'±\/b'* — ac
(^) "^' "« = 2^ = —a
This result may be expressed in the following rule: when the
coefficient of the second term of the quadratic equation is even^ the roots
may be expressed in the form of a fraction whose numerator is minus
one half the coefficient ofx, plus the square root of the difference
between the square of this quantity and the product of the coefficient of
X* by the constant term, and whose denominator is the coefficient of x\
Example. — Solve the equation 3x"— 18x-f 5 = 0.
Here, 6 = 26' = — 18, and 6' = — 9, a = 3, c = 5,
whence x^, x, = = 3 i il/66.
o
In practice formulae (3) and (4) are most used in solving the
equations of the second degree.
NoTV.— This method of completing the square was discovered by Bhaskara (b. 1114)
and Brabamgapta (b. 566), and is known as the Hindoo Method.
396 COLLEGE ALGEBRA [M08
EXBBOisB Tiyrr
Solve the following equations:
1. 25x» + 2 = 30x. 2. 6a;«+a5 = 15.
3. 7x«+25x=12. 4. 6x«+5x = 56.
5. 7aj^ + 9x=100. 6. lix«+10 = 7x.
7. x«+6.51 = 5.2x. 8. x« + 4. 3x = 27.3.
9. 14x«+ 45. 5x+ 36.26 = 0. 10. 7. 82x» — 33. Ix + 35 = 0.
11. 10.85x» + 21. 91x — 10.5 = 0.
12. (x — 7)(x-5) = 0.
13. (x — a+6)(x— 6 + c) = 0.
14. x« — ax = 0. 15. x«+(a — x)« = (a-2a;/.
16. (a — x) (x — 6) + a5 = 0.
17. (a — x) (x — 6) = (a — x) (c — x).
18. a* — x" = (a — x) (6 + c — x).
19. (x — a+ 6)(x — a+ c) = (a— 6)« — x".
20. (x — 6) (x— 5)+(x— 7)(x — 4) = 10.
21. (2x~17)(x — 5) — (3x+l)(x — 7) = 84.
22. (2x-5)»— (x— 6)« = 80.
23. (33 + 10x)« + (56 + 10x)« = (65 + 14x)«.
24. 2x+^ = 3. 25. g^^il4--'^-^=J.fx-l.
ar 9 ^ 5 :f^
26 a;+ll__23;4-l 16-3;_2(x-ll) a:-4
ar + 3 x + b ' 4 ar-6 "" 12 '
Qo a:» — 10Lr« + l o
oQ 3£ _ 3a; — 20 __ o , ac«— 80
• 2 18-2j;~^"r 2(ar-.l)'
34. ?l___10-__t_ = 0 35 5 + a?_8~3a: 2r
a; a: — 2 ar— 3 3— a; a; ~"z— 2
36 g^Jug . ^±i^3a: + ll ^„
^ x-2 ^ x-l- x + 1 '^'•
-i-+-JL_ = _3_ . _2_
x—l ^ x—4 x—2^x—3
2r-l 3j + 1 _&f-14
a; — 2 "*" a?— 3 »— 4
38.
8408] THE EQUATION OF THE SECOND DEGREE 397
7 — a? 6 — 07 5 — a? 4 — a?
40. oo^ — (a« + 1) X + a = 0.
41. a6x« — (a«+ 6«) X + afe = 0.
42. a«(a - x)» = 6*(6 - x)l
43. (a — x)«+(x-6)«=(a-W
44. (a-x)(6-x) = 2(a-6)«.
45. (a— x)«— (a-a:)(x — 6)+ (x--6)« = (a — W
46. (n — |>)x* + (p — »i)x + (m — n) = 0.
47. (a-f-6+c)x«— (2a+6+ c)x+a = 0.
48. (ox — 6)(c — rf) = (a — 6)(cx — rf).c
49. x«—(a+6)x+ (a + c)(6 — c) = 0.
50. x^ — (a — m)x = (a — 1) (m — 1).
51. x» — 2 (rt — 6) x= (a+ c — 6) (6+ c — a).
52. a6x*— (a + 6)x + l = 0. 53. 4x« — 4ax + a« — &• = 0.
54. m"x* — 1» (a — h) x — afe = 0.
55. x» + 2afc (a« + 6») = (a + 6)« x.
56. (a«-6«)(x»+l)=2(a«+6«)x.
57. x + i=a+i. 58. x-i = 5-^.
' X ' a a? 6 a
59. (3x-5)"-8(3x-5)+ 7 = 0.
60. (2x — a)* = 6 (2x — a) + 26».
61. (3x — 2a+6)«+26(3x — 2a + fc)=a» — 6«.
«2- G-5i;=«C-5-1)-i^:
go (g — a:)« 4- (j: - &)'^ q' + y g4 gj^_::^_6^jt_c^c
(a — a:)«-(a: — 6)« a«-&«* da^-ex+f f
65 (a-~ar)« + (a;-6)»^a»-&» gg (,, _ ^)3 + (^ , 6)» ^
(a-ar)-(a:-6) a + 6' (a - a:)« + (x - 6)«
67. 2cg I <^ I CM^+g'^c* I ^(^+2cg).
a: — a a; + a a^ — cfi x" — a*
x-5^^ + 5 ^ a;«-25
69. ai(a*-l)-2(ar+l)«=^,(l — 4a«ar-ar«).
70. afr=4 + l 71. a: = a+l
3 + 1 6 + 1
2+1 c + 1
2 + 1 6+1
3+1 a+x-
4+x.
CHAPTER III
EQUAL ROOTS AITD IMAGINARY ROOTS
409. In the preceding chapter after having put the equation of
the second degree in the form
{2ax+hy=i 6« — 4ac,
it was supposed that the second member was a positive quantity and
then the formula was deduced
Za
which gave the two roots, or the two solutions of the equation.
When the quantity 6" — 4ac =0, the equation becomes
(2 ax + by = 0.
The first member is a perfect square and the unknown quantity
2 X — 6,' whose square is 0, is itself 0; hence
2ax + h = 0;
whence a; = — — •
Here the equation is said to have two equal roots, each equal to
— — • The reason for this is that, as the radical i/6' — - 4ac in (3)
approaches zero, the quantity which is added to — b for one root
and subtracted for the other becomes as small as is desired, the two
roots differ by as small a quantity as we please, and are said to be
equal in the limit.
Imaginary Roots
410. Since the square of positive or negative quantities is always
positive, it follows that it is impossible to take the square root of
negative quantities (1379).
896
8410] EQUAL ROOTS AND IMAGINARY ROOTS 399
Hence, ia the case of the equation of the second degree, if
6* — 4 ac is negative, it will ])e impossible to satisfy the equation
of the second degree. For then
(2ax+h)*= -(6«-4ac),
a positive quantity equal to a negative quantity, which is impossible.
In this case the formulae give fictitious values which have been
introduced into mathematical analysis under the name of imaginary
quantities.
Consider the equation, aja = + 4.
Here the square root of + 4 can be found and the equation will
have the solutions,
Xj = + 2, and x^ = — 2.
If, however, the second member is negative, and
a;» = — 1,
then there is no number, positive or negative, whose square is equal
to — 1, audit is impossible to satisfy the equation. If, however,
1/ — 1 is represented by i and the letter i is introduced into alge-
braic calculations as though it represented a real quantity, with the
convention that its square (t ') is equal to — 1, the equation a* = —1
can be satisfied by the imaginary values + i and — i; for
W=(l/=l)'=-l = -l
(-i)«^(t)«E=-l = -1.
All imaginary values can be expressed in terms of the symbol i
(22381, 882). For example, the equation,
x«=-16
has the two imaginary solutions,
x^ = -f ]/— 16, and x^ = _]/_16.
If it is noticed that — 16 = 16 (—1), and the theorem concerning
the square root of a product is applied (2366), it is possible to write
l/IIIe = V/16(-1) = l/l6 'V^ = ±4i ;
whence sc^ = + 4 1, and x^ = — 4 1 .
Now apply the principle to the more general equation,
x« — 4x+ 13 = 0.
Here aj« _4x + 4 = 4 — 13,
(x-2)«= -9,
whence a — 2 = d= V —9 = ^3i
or Xj = 2+3i, and Xjj = 2 — 3i.
400 COLLEGE ALGEBRA [«411, 412
411. It is necessary to proceed in a similar manner in case of
the general equation,
ax* -(- 6x + c = 0,
if the quantity 6' ~ 4 oc is negative. Because, when this equation
is put under the form
(2ax+ t«) = t« — 4ac
it is impossible, as has been seen, to satisfy it with real values, but
it can be satisfied by the imaginary values.
2ax + b = -ti/— (6«— 4ac),
with the convention, as has already been seen in the preceding
examples, that the square of the symbol i/— (6* — - 4 ac) is always
equal to —(5*— 4ac) 2380. Hence, it will follow that the imaginary
values of X are
_& + |/_(6«— 4ac) _— 6 + 1/6^—4 ac- t
^^~ 2 a " 2 a
^h—V^—ih*—4ac) — 6 — l/6« — 4(
• 2 a 2a
The roots x^ and x, may be written in the form A + A*, where
2 a **""* '" 2 a
The ordinary rule of calculation in common Algebra has been
extended to imaginary quantities as though the symbol t were a
real number with the convention that i*=--l (Chap. VII, Book III).
In view of the preceding consideration, the solution of the equa-
tion ax* -f 6x + c = 0 presents itself under three aspects:
1. If 6» — 4 ac> 0, both roots are real and different.
2. If 6*— 4ac = 0, both roots are real and equal,
3. If 6* — 4 ac < 0, both roots are imaginary and different
Solution op the Equation x* + 2>x + j = 0
412. This equation can be solved by comparing it with equation
(1) 2406, thus:
(1) rtx?+[»x+c = 0,
(2) x«+i)x + gr = 0;
whence a=l, h =py c = q.
On substituting these values in the formulae
- ^b±Vb^-4ac
M12] EQUAL ROOTS AND IMAGINAY ROOTS 401
it is found that the roots of equation (2) are
(5) x„x,=-p±y.^i-g.
The solution of the equation ac* + px -f ^ = 0 will present the
same general cases found in equation (1) {411:
1. If /)* — 4 y > 0, both roots are real and different.
2. If J?* — 4 ^ = 0, both roots are real and equal,
3. If p* — 4 J < 0, both roots are imaginary and different
Examples.
1. Solve the equation x" — 7 x + 10 = 0.
Here p= — 7, ^ = 10, and on substituting in equation (5),
X. , a:. ^7±v^49-40^7±3^ 5 ^r 2.
2. Solve ahix*+l)=z(a*+b*)x.
Removing parenthesis, abx^ -f- a6 = (a" -}- 6*)a
after combining aba^ — (a" + fc')x -f- a6 = 0.
After substituting in formula (3)
_ q»-fy -1-1/0*4. 64 _ 2 a«6«
2ab
"■ 2a6
a«-f y' + q'~y'^2a« __a.
^1"" 2a6 2a6 5'
^8 ~ 2a6 2a6 a
3. Find the value of k in order that the equation,
h* («• — x«) = a« (mx + k)\
may have equal roots.
Develop and arrange with respect to x* and x; then
{a^m^ +b^)a^ + 2a« A;mx + a« A^ — a" 6« = 0.
By the condition for equal roots,
{2a^kmy — 4 (a« m« + 6«) (a* ifc' — a* 6*) = 0. [J411, 2]
Removing the parentheses and dividing by 4,
a*Aj«m« — a* A;«m«+ a*6«m2 — a«6«A^ + a«6* = 0,
a*Z>«m« — a«6«A^+a«6* = 0.
Therefore, after dividing by a't*, A:* = 6« + a" m";
whence A; =db V^^ + «* wi*.
402 CX)LLEGE ALGEBRA [M12
EZSBOISE LXX
1. ^-7S = l. 2. f-| = 9.
3. (a.+ 2)' = 4(x+5). 4. ^=?^±^-^.
X — 1 r-|-5 x-f- 1
F> ^— ? = § a J? — 8 , 2(3: -f 8) _.3r4-10
• 2 3 8* • x-^ "^ ar+4 :r + l *
7. 3(x - D* - 2(a; - 2)* = 0.
8. a:«-2(l + V2)x+ 2l/2 = 0.
9. l/3x + 10+ VX+ 2 = V lOx+16.
10. l/x — 4 + 1 X + 4 = V 2x + 6.
11. (7 _ 4v 3) a^ + (2 - 1/ 3)x = 2.
12. i/82 + x — K82 — x = 2. 13. 4x« + xi/2 = l.
14. v2x+4--J| + 6=l. 15. x«-2ax+t« = 0.
16. x«_.2ax+a«-6« = 0. 17. ? + 5 = f + ^.
a X 6 X
18. ^ + -l_+-i_ = 0. 19. -^-4--5?i|i^-
X— a X — 6 X — c X — 2 x+2 x"— 4
20. 5x_^_^ar + 44 g 21. x-5 + i = 0.
x + 3^4x-8 2^x
oq x + wi I X — m x*4"^^ I ^ — ^*
X — m x + m a:* — m* a;*+ m*'
24. 1 + 1 + 1= 1 .
a 6 X a-|-6 + x
25. {ax — i>) (6x — a) = c*.
26. -«-^ + _A_=-2<L..
X —a X — 6 X — c
2412] EQUAL ROOTS AND IMAGINARY ROOTS 403
27. ^+« t ?±J . ^ii£-3.
X — a "^ X — 6 a; — c
28. -1-+ -1— +- 1— +-^l-; = 0.
a — x b — x c—a c — b
29. If { 2a + (» — 1)^}| is 48 when a = ^ and & = J, find the
values of n.
30. Determine the values of k in order that the equation
x« + 4 (1 + A;) X - 3A; = 0 *
may have equal roots.
31. Determine the value of m in order that the equation
{mx + fc)* r= 4ax
may have equal roots.
32. Find the value of m in order that the equation
mx 4- 6 = \/x* — r*
may have equal roots.
33. Eliminate y between the equations
— — — = 1 and y z=mx-\- b
and determine the value of m in order that the resulting quadratic
equation in x may have equal roots.
34. Solve the equation x« + 1 = ^(J- + J^)-
35. Evaluate |/(5 + |/(5 + |/(5 + . . . oo ))).
36. Separate
x« _ 8x + 15
into a product of two factors of the first degree.
37. Between what limits must x lie, if the expression
x« _ 12x + 27
is to be negative?
38. Between what limits must the values of the fraction
5x— 21
x«+ 16
lie, if X is restricted to taking real values?
CHAPTER IV
THB RESOLUTION OF A TRINOMIAL OF THE SECOND DEGREE INTO
FACTORS OF THE FIRST DEGREE.
418. Consider the trinomial
in which p and q are given constants and x an arbitrary quantity of
any magnitude. The value of this trinomial expression will not be
changed if the same quantity, ^, is added and subtracted; the fol-
lowing identities result:
x* + px+q = a^ + pz + ^+9-^=(^x + £)*-(^^)
-(' + l)"-(=^)'
Since the difference of two squares is equal to the product of the
sum of the quantities by their difference (294), then
Put in this identity
X =-P + ^P'-^7, and a, -p-i^p«-47^
I 2 • 2
Hence, the trinomial x* + P^ + ? ™^y ^ written
(6) a^ + px + q — {x — x^) (x — x,).
This product can be made zero by putting x = x^, when the first
factor is zero, or by putting x = x^, when the second factor is zero,
i. e., x/ + pxj + g ^ (x^ — x^) (x^ — x,) = 0,
^8* + P^t + Q^ (^8 — «i) (^t — ac^ = 0.
Thus by a second method the two roots of the equation
X* 4" p^ + ? = ^)
have been found, as well as the factors of its first member.
4M
JJ414, 415] TRINOMIALS OF THE SECOND DEGREE 405
This decomposition of a trinomial of the second degree into
factors holds in all cases, whether the roots are real or imaginary
and whatever is the value of x.
414. The most general trinomial of the second degree is
ox' -\- hx-\- c^
which can be written
ax* '\-hx-\- c^a(^-\ — x -\- -\ ^^ a{^ '\- px -\- j),
where p = - and g = -• By HIS, (6), this polynomial in paren-
thesis can be decomposed into the factors x — x^ and x — x^ , and,
therefore,
(7) ax* + 6x + r ziE a(x — x^) (x — x^)
whatever is the value of x. Here Xj and x^ are the two roots of the
equation x* + px + j = 0, or, what is equivalent to the same thing,
the roots of
ax* -{'bx-{- c = 0.
Examples.
1. Factor the trinomial x* — x — 6 = 0.
^ 2 2 '
* 2 2 '
Hence, x* — x — 6 ^ (x — x^) (x — x^) = (x — 3) (x + 2),
2. Factor the trinomial 4x* — 4x — 15.
Here « = 4, 6 = — 4, c = — 15.
^ ^ — - ft ±_/6*-4ac __ 4 J: l/l6-h240 _ 4±16.
* * 2a 8 8
whence x^ = - > and x, = — - •
Therefore, according to formula (7),
4x* - 4x - 15 = 4 (x - I) (x + I) = (2x — 5) (2x + 3).
416. Factor ax* + 2/>xy + cy* + 2dx + 2ey + /.
Arranging the terms with respect to x* and x, and equating to zero,
(1) ax* + 2{hy + rf)x + (cy* + 2fy + /) = 0.
Solving by formula (4), ?408, then
406 COLLEGE ALGEBRA [H15
a
Hence, if 7? is put equal to the radical,
(3) a(x-x,)(x-x,) = a(x--(^+^+^)(x^:^^=^)
= i (ax + hy+d+ R){ax + hy + d —Ri
where R = y'{b*—a c)y* + 2{hd — ae)y + (i» — a/.
NoTK.— It has been assumed that a is not zero. In case a Is zero and b not zero, mIt*
for y as we have above for x and proceed In a similar manner.
Example. —Factor 2 x* — xy — y* + 3 ;c + 3y — 2.
On comparing this equation with tlje general equation, it is seen
that:
a=2, 6 = ~|. c= -i; J = |, f = |, /= -2, fc«-ac = |,
hd—ae=i —\^, d^ — a/ =z 5j4, and
\ 4 4^4 2 *
Henue the factora will be
i(aar+ 6y + d+ if) (ar+ 6y + d- iJ)-^| (2x - i j,+ I + i (3y-5))X
(2x-iy+|-J(3.v-5))
= i(2x + y-l){2jr-2y + i)
= {2x + y-l)(x-y + 2).
In case the quantity under the radical in ^ is a perfect squire
(Hll, 2),
{hd — ae)* — (6« — ac) (^ — a/) - 0.
In this case the factors in (3) above are rational. This is the
case in the preceding example ; for
(hd - aer ^(d^- ac) {cP - a/) -r. (- ^)" - (|) (|) =0.
EXEBCISE TiXYT
Investigate whether the following expressions can be separated
into factors or not, and if this is possible, whether the factors are
rational or irrational. If the factors are rational, they can be foond
directly by J418, or indirectly by solving the given equation.
1. x« — 7x + 12. 2. x5+13x + 30.
3. x«-9x+15. 4. x»+12x + 27.
5416] TRINOMIALS OF THE SECOND DEGREE 407
5. x« — 3x — 20. 6. x«+2aj — 35.
7. X* + 4 ax + 3 a«. 8. x* — 6 ax — 30 a\
9. a«_7a6 + 6fc«. 10. a« + 3a6 + 66«.
11. a«— a6— 26«. 12. a«+a6— 26«.
13. 3x« + 4x + 5. 14. 2x>— 7x + 3.
15. 3x*— 17ax +10a«. 16. 4x« — 3ax— 2a*.
17. 6a« — 5a6 — 6 6«. 18. 2a«— 5afc — 3 6*.
Factor the following:
19. 2x»+ xi^ — 3y« — 8x+ 3y + 6.
20. x« + 3xj^+ 2y«+ 3x + 4y +2.
21. 2x«— xi^ — 3y« + 3x— 7y — 2.
22. 6*x« + aV — 2 6«ax + a«6« + 9.
23. 2y* + axy — a^j^ — hx + 2ah^x — h\
416. Quadratic expressions and certain other expressions may
be factored by the artifice of completing the square (2406) and in
connection with i94.
Examples.
1. Factor 9 x« — 24 x — 9.
By {406, the expression 9 x' — 24x will become a perfect square on
adding (4)**, thus,
9j:«— 24x-9^9j«-24a:+16-9-16 ~ (3j:-4)«-25
^ (3j:-4 -5)(ar-4+5) [J94]
= (3a; + l)(3a; — 9)
= 3(3ar + l)(ar-3).
2. Factor5 + 4x — 12x«.
6f 4x-12:t« = -^(36:c«-12x-15) =- i(36a;«- 12ar+l -15-1)
3 o
= - |[(6a: - 1)1 _ 16] = - 1 (6a:-l+4) (ar-1-4) = - 1 (dr+3) (6z-5)
= C2x+l)(5-6r).
3. Factor 6 x« — 11 ax — 35 a\
408 COLLEGE ALGEBRA [W16
Now since it is possible to factor a quadratic equation, the solu-
tion of the equation can at once be found by placing each factor
equal to zero and solving the resulting equations ; thus, in example 2,
5 + 4 a: — 12 x« = (2 X + 1) (5 - 6 x) = 0
2 X + 1 = 0 and 5 — 6 X = 0
X = — ^, and X = |.
4. Factor a* + a»6« + 6*.
a* + a«6« + 6* = rt* + 2 a«6« + 6* — a^b^ = [(a« + fc«)« — a*6«]
= (a« + 6« + ab) (a« + ^>« — a6) [J94]
— (a« + a6 + 6«) (a« — a6 + 6«).
5. Factor 4 x* — 9 x« + 1.
4 X* — 9x«+ 1 =4 x*-4 x«+l— 5 x« £= (2x«— l)'— (xi/5)*
Eiz (2 x'— 1 + xv/5)(2x« — 1 — XV 5).
6. Factor x*+a".
X* + a« — X* + 2 ax« + a« — 2 rtx« = (x^ -h a)« — (x V 2a)"
= (x«+ a — j/2a . x) (x« + a + l/2a • x).
BZEBOISE IXXU
Factor the following:
1. x2 + 3x+l. 2. 4x«+13x + 3.
3. 3x« + 7x — 6. 4. 6 + 5x — 6x«.
5. 6x«— 19X+15. 6. x*+ 1.
7. x* + x«+l. 8. X* — 5x« + l.
9. x*+a*. 10. 4x>+7x«6*+6*.
11. • Solve the equation x* + 16 = 0.
x^ + 16 — x^ + Sj^+\6-Sj^_:{ji^+4f — i2xv^2)*
::::(j4j_^4_|_2jri/2) (x« + 4-2j- v^2)
^rAx^ + 2V2x+4)i3^-2V2x + 4),
Either factor, .r* + 2 V 2 jr + 4 or :c* — 2 1/2 a: + 4, equated to zero will reduce
ar* + 16 to zero. Therefore the roots o( x^-\-16 = 0 will be found by solving
each of the equations
j:« + 2l/2ar + 4 = 0 and a^-2l/2x + 4 = 0
-21^2^: i/8-l() ^. ^ 2l/2j:l/8~16
a-i,a-8= ^ ^1,^8 =
^ ^ — 2l^2±2l^2i ^ 2l/2± 21^21/^
^1, aTi = =^-^^ Ai ^s = - ~^
xi, X2 = V2 (—1 ± i) Xi, x^ = 1^2 (l ± i).
12. Having factored examples 6, 7, 8, 9, 10, solve the equations
found by placing each factor equal to zero
CHAPTER V
PROBLEMS INVOLVING EQUATIONS OF THE SECOND DEGREE IN
ONE UNKNOWN QUANTITY
417. Problem I. — Divide the number 31 into two parts such
that their product shall be 228.
Let X and y be the parts of 31 ; then
(1) x + y = 31,
(2) xy = 228.
From (1) y = 31 — »,
(2) gives x(31 — x) = 228,
x*— 31 X + 228 = 0.
Hence, x = 19 or 12,
and y = 12 or 19.
Here, although two sets of values for x and y are obtained, yet
there is only one way of dividing 31, so that the product of the two
parts shall be 228.
418. Problem II. — Given that the perimeter of a rectangle is
2 p, and its area equal to that of a square of which the side is a.
Calculate the sides of the rectangle.
Let X and y be the sides of the rectangle ; then
(1) 2x+2y = 2p
(2) xy = a\
From (1) yz=zp — x^
and (2) x{p — x) = a*,
or (3) x«— ;>x + a«=0.
Solve (3) X, = P + v-^», ^^ ^p-V^.
^«— 2 '^«"" 2
Discussion. — The solution x^ and y^ is equivalent to the solution
X, and y,; it would be necessary only to interchange the terms
length and breadth of the rectangle.
410
COLLEGE ALGEBRA
[1418
In order that the problem may be possible it is necessary and
sufficient that the sides found are real and positive. In order that
the roots of equation (3) may be real, it is necessary that
(4) p2_4a«>0.
Il this condition is fulfilled, the roots x^ and y^ will both be positive,
since x^ is the sum of two positive quantities, p and v/p* — 4 a*,
and y^ is the difference between p and i/p* — 4 a*, which is + and
less than p. Therefore, the inequality (4) is necessary and suf-
ficient to make the problem possible.
It follows from (4) that the problem is always possible, if
a«<
[M12, 1]
and impossible when a' is greater than ^ ( J412, 3) ; when a* = ^ the
radical yp* — 4 a* is equal to zero, and the two values x^ and y^ are
equal, and the rectangle becomes a square.
From this discussion the two following theorems may be deduced:
1. Of all rectangles which have the same perimeter that which has
the greatest area is a square.
2. Of all rectangles which have a given area that which has the
least perimeter is a square.
Geometric Construction. — Since the two sides of the rectangle
are given by the formulae
it is easy to construct geometrically these sides x^ and y^ by
means of the given dimensions, a and p. In any straight line lay
off a length AB^ equal to
^ , and on AB as a diame-
ter construct a semicircle ;
then about A as a center,
with a as a radius, describe
an arc intersecting the
^ , ^ circumference on AB in C.
Figure 1
From a known theorem of Geometry, it follows that
P/^
BC=^&-^a,
8418] PROBLEMS INVOLVING QUADRATIC EQUATIONS 411
About jS as a center construct a circumference with BC as s, radius,
and let 0 and 0' be the points in which the line AB is intersected
by this circumference. Hence,
AO'=AB -BC=^ — J^ - a».
Therefore, AO and AO' are the sides required. This construction
is impossible if « > f . For in this case the circumference with the
center A will not intersect the circumference on AB, This result
corresponds to the algebraic fact that if a > ^, the radical ^^ — a*
is imaginary, If a =f , BC=Oy then AO=z AO' =^ and the
rectangle will be a square. Thus we are led to the same results as
those which were deduced from the algebraic discussion.
FBOBLEMS
3. Find two numbers such that their sum is 39 and the sum of
their cubes 17199.
4. Find the number such that the sum of the number and its
reciprocal is m.
5. The product of two numbers is 750, and the quotient when
one is divided by the other is 3J; find the numbers.
6. Find the number such that when it is (1) divided by n, (2)
subtracted from n, the result is the same?
7. A number which consists of two digits has this property:
when it is divided by the product of its digits the quotient is 3,
and when it is increased by 10, its digits appear in the reverse order.
Find the number.
8. Divide the number 53 into two parts whose product is 612.
9. Divide the quantity a* + 6* into two parts whose product is
10. Find two factors of 2268 whose sum is 99.
11. Separate ? into two factors whose difference is ? + -.
6 a 6 a
12. Separate the fraction ^ into two factors whose sum is ^^"^ .
13. The sum of the squares of two numbers, one of which is 12
greater than the other, is 1130. What are the numbers?
412 CJOLLEGE ALGEBRA [H19
14. The members of a society each contribute the same amount
to a fund of $336. If there were three members less, each member
would have to contnbute $2 more. Find the number of members.
15. A person purchased a certain number of sheep for $175;
after losing two of them he sold the rest at $2^ a head more than
he gave for them, and by so doing gained $5 by the transaction.
Find the number of sheep purchased.
16. A cask contains 360 gallons of wine; a certain quantity is
drawn and an equal quantity of water is put in ; from this mixture
the same quantity as before is drawn, and 84 gallons in addition; on
replacing the drawn liquid with water it is found that the barrel con-
tains equal quantities of wine and water. How many gallons were
drawn the first time?
17. A number of men pass a certain time in a hotel and on
leaving they have a bill of $12 to pay. Had there been 4 more in
the party and had each spent 25 cents less, their bill would have
been $15. What was the number of men?
18. A merchant paid a certain sum for a horse, later he sold the
horse for $144 and thereby gained as much per cent as the horse
cost him dollars originally. What did he pay for the horse?
19. What is the quotient, whose dividend is n times smaller
than its divisor, and the sum of the quotient and its reciprocal is n ?
20. A manufacturer had agreed to pay a capitalist $8,800 after
7 months and $5,940 at the end of 1 year. After how many
months can the manufacturer pay back the capitalist the total
amount, $14,740, if interest at 5% per annum is charged for the
money which he paid later than it was due and if a rebate of 5 %
per annum is allowed for the money paid before it was due?
21. A capitalist lent k dollars at a certain rate per cent and
withdrew each year h dollars; at the end of 2 years there re-
mained k* dollars invested. At what per cent was the money lent?
Problems connected with the Theorem op Pythagoras
419. In this section only those problems will be discussed
which are connected with the Theorem of Pythagoras for right-
angled triangles ; first because it is not desired to make the discus-
sion too extended, and secondly because the Theorem of Pythagoras
for acute- and obtuse-angled triangles has more of a trigonometric
interest.
8419] PROBLEMS INVOLVING QUADRATIC EQUATIONS 413
The first of the topics just mentioned is the most important,
because all problems connected with the second topic depend for
their solutions upon the first and because the larger part of physical
and technical problems are connected with the relation discussed in
the first topic.
We recall that in any right-angled triangle ABC^ Fig. 2, by
Geometry:
B
(1)
a*=e*—h*
b* = c* — a*
or
c = i/o«+6'
^
c/^*^
(2)
a=v/c'-6'
(3)
6 = l/c«-o«
A
b C
FlOUKK 2
Problem I. — In a right-angled triangle the difference between
the longer and shorter legs is equal to the difiTerence between th€
hypotenuse and the longer leg. How long are the sides of the
triangle, if the given difference is 2 inches?
Solution. — Let x = the shorter leg;
then a -|- 2 = the longer leg,
and X -j- 4 = the hypotenuse.
Then it follows from H19, formula (1), that one has the equation
(x+4)« = (x + 2)« + x«
or x«-f 8x-f 16 = x'-f 4x + 4-f x«;
hence x* — 4x — 12 = 0
Xj = 6 and x^ = — 2.
Since the quantities introduced into the equation have the desig-
nation * ^inches", we obtain as the length of the sides of the triangle:
for the shorter leg, 6 in.
for the longer leg, 8 in.
for the hj'potenuse, 10 in.
The solution x, = — 2, has no interpretation, inasmuch as it is
assumed in the hypothesis of the problem that any one side is
measured in but one direction, namely positive direction.
Problem II. — A chord is drawn through a point P, which is 13
inches from the center of a circle of radius 15 inches. The chord
is divided by Pinto two segments, one of which is 10 inches longer
than the other. How long is the chord?
V
414
COLLEGE ALGEBRA
[M19
Solution. — Let x = the shorter seg-
ment FB of the chord AB; hence
the longer segment of AB will be
05+10. Therefore, the entire chord will
be represented by 2x + 10 and the half
of it by x+ 5. Draw OCl AB, then
by Geometry
CB = x+ 5.
Hence, from the triangle OCB,
(1) 0C= V 15«-(x+5)«.
CP=: CB — FB = x+b—x=:b,
Figure 3
Since
then it follows from the A OOP that
(2) 0C= V 169 — 25 = 12.
Accordingly the equation of condition will be,
from (1) and (2), (3) v 15« — (.r + 5)« = 12.
Hence, . (x + 5)* = 81,
a; = -- 5 d= 9,
X, = 4, and
Xj, = — 14.
Therefore, the chord ^IjB = 2 x + 10= 18 or —18 inches. Had the
chord AB been introduced as the unknown quantity, the determining
equation would have been a pure quadratic.
Here, as in Problem I, the negative solution — 14 has no mean-
ing excepting that it is a solution of the same equation of which
Xj = 4 is a root.
Problem IIL — An isosceles triangle whose sides are in the ratio
a : a: h is so inscribed in a square whose side is p inches long that
the vertex of the triangle coincides vnih one comer of the squait"
and the vertices at the base of the triangle lie on the sides of the
square opposite to the common vertex. How long are the sides of
the triangle?
Solution, — Since the sides of the triangle
sought, AEF, Fig. 4, are in the ratio
a : a: b, AE can be represented by ox, AF
by «x, and EF by hx. Since AB=p^
for BE substitute the expression
Figure 4
BE= VaW—p\
2419] PROBLEMS INVOLVING QUADRATIC EQUATIONS
Therefore, EC=^p— Va}x^ — p\
415
In like maimer, CF z=ip—y a*x* — p*
Hence, it follows from the A EOF that
6«x« = (i> — I a'x* — i?«)' + (p — l/a«x«— p«)",
6«x« = 2/>« + 2 a«x« — 2p« — 4pl/aV— p*,
(6« — 2 a«)x« = — 4i>vVx« — p»,
(6« -> 2 a«)V= 16|>«(a«x« — p%
(6« — 2 a«) V— 16 ayx« + 16 p* = 0.
I 64 riV — 16 p*(// — 2 a»)»
(6» _ 2 a«)«
= d=^^^.V2a«d=6l 4a«->.
Considering the value of x with the positive* sign, the side of the
isosceles triangle will be
ax = AEz:^AF= J"^ J2 a« + 6 y 4a«-^«,
6* — 2 a' \ '
and the base
bx = EF= ^i§^,^2«'+tv4«'"^^'.
Problem IV. — If the comers of a square are cut oflf so that a
regular octagon remains, how long is a side of the latter?
Let X = the portion cut off from
one end of a side of the square, Fig. 5 ;
then a — 2 X is equal to a side AB
of the octagon, and
BC'' = BP* + To* = 2x\
But BC=AB = a — 2x,
(a— 2x)«=2x«
or a — 2x = ±x v/2,
2±V2 2
A B X
P
a
X
C
X
\
/
Figure 5
Since the octagon falls within the square, the lower sign must be
taken, for if the upper signs were taken the side of the octagon
would be longer than the side of the square, which is not possible.
* Since any side of a triangle is less than the sum of the other two sides,
2 a > 6, and 4 a* > 6*; hence v^4 «* — 6* is always real.
416 COLLEGE ALGEBRA [8419
Problem V. — The median lines drawn from the vertices at the
acute angles of a right-angled triangle to the perpendiculars are a
and 6. How long are the sides of the triangle?
A Solution, — Let x = the perpendicu-
^^s;;^^^.^^^^ lar A C and a and b respectively the
medians DB and AE. The angle
E
Figure 6 From ^ CDB it follows that
CB = ^a«
4'
and Cir=ija«— ^'.
Hence, it follows from the triangle ACE (Fig. 6) that
AE* = h* = AO* + W
fc« = x» + J(a«-f),
and 15ic« = 16 6« — 4a«:
2
X = ± ,, l/15(4 b* - a«) = AC.
15
Similarly, C5 = -t ^ V^ 15 (4 a« - 6«).
Hence, the h3rpotenuse AB will be
|l/5(a* + 6«).
FBOBIiEMS
6. If a perpendicular of a right-angled triangle is 11 inches
in length and is prolonged beyond the hypotenuse 1^ times the
length of the other perpendicular, and the point thus determined
joined to the other extremity of the hypotenuse, a second right-
angled triangle is thus constructed, whose hypotenuse is equal to the
hypotenuse of the original triangle plus ^ of its unknown perpen-
dicular. How long is the perpendicular in question?
7. How high is an isosceles triangle whose base is a and side
is^^ Ans.: x = d= .^ V^4 i^* — a«.
8. Two parallel chords are drawn in a circle of radius 25 cm.
One of the chords is 1^ times as far from the center as the other.
If the shorter is 16 mm. shorter than the other, how long is each?
iJ
H19] PROBLEMS INVOLVING QUADRATIC EQUATIONS 417
9. A secant is drawn from the vertex of a square ABCD, out-
ing the side BC in E and the side DC produced in F&o that the
segment EF is h inches long. How long is the segment ^^ if the
side of the sq[uare is a inches?
10. If one leg of a right-angled triangle is 5 inches longer than
the other and both legs are produced 3 inches, a new triangle is
formed whose hypotenuse is 4 inches longer than the hypotenuse of
the original triangle. How long are the legs of the original triangle?
11. How long is a side of an equilateral triangle whose altitude
is n inches in length?
12. How long is the longest diagonal of a rhombus whose base
is 100 inches and whose shortest diagonal is 4 inches?
13. The diagonal of a square is a inches longer than a side.
What is the length of the side of the square?
14. The altitude of an isosceles triangle is 3 inches longer than
the base. How long are both if the equal sides are 19 inches long?
15. If an equilateral triangle is inscribed in a square of which a
side is a, how long is a side of the triangle?
16. An isosceles triangle is inscribed in a square of which a side
is a; one of the equal sides of the triangle is h inches longer than
its base. How long are the equal segments cut off from the comer
of the square by the base of the triangle?
17. A circle is inscribed in a circular quadrant, tangent to the
arc and the two perpendicular radii. How long is the radius of the
original circle, if the radius of a new circle is a inches shorter than
that of the given circle?
18. A circular quadrant is circumscribed about a circle of radius
a so that the radii and the bounding arc of the quadrant are all tangent
to the given circle. How long is the radius of the quadrant?
19. In a right-angled triangle, the median drawn to one of the
legs is a inches longer than this leg. How long is this leg if the
other is h inches?
20. One of the legs of a right-angled triangle is a inches and
the median drawn to the leg is 6 inches shorter than the hypotenuse.
How long is the hypotenuse?
418
COLLEGE ALGEBRA
[«420
21. If the altitude of an isosceles triangle is prolonged through
the vertex a distance equal to its own length and the extremity
of the extension joined with an extremity of the base, the joinii^
line will be a inches longer than the side of the isosceles triangle.
How long is the altitude, if the base is ft?
420. Problems Concerning the Area of Plane Figures.
Problem I. — Within a given rectangle whose sides are a and h,
a second rectangle is constructed so that its sides are everywhere
equally distant from the sides of the given rectangle, and that it has
one-half the area of the given rectangle. How long is the pemneter
of the second rectangle?
Solution, — Let x be the distance of
the sides of the second rectangle from
those of the first. It follows from the
annexed figure that the sides of the
new figure are (a — 2 x) and h — 2 x.
^ ^ Therefore its area will be
Figubb7
(6 — 2x) (a— 2x)
and the area of the first rectangle is ah.
Hence, from the conditions of the problem,
{h — 2x) (a^2x)=:^.
X
X
.1
1
•^ a-23>
X
X
or
^_2(a + 6)x + 4x«=0.
x = i(a+6zbl/a'+ h").
Since only the lower sign of the radical can be used, one obtains for
the perimeter 2 (« + 6 - 4 x) = 2 v oM^Ti.
Problem II. — A right-angled triangle whose legs are aandft
is divided into three equal parts by lines perpendicular to the
hypotenuse. Into what parts is the hypotenuse divided?
Solutimt.—Lei ABC be n triangle
right-angled at C (Fig. 8), whose al-
titude is OIL Then
^ ia6=^CJ7va*+6«=area of triangle.
Figure 8
«420i
PROBLEMS INVOLVING QUADRATIC EQUATIONS
ab
419
CH=:
Va'+b^
BH =
Affz=
6«
Therefore the area of the triangle ACH is
an- AH =
ab
6«
a&8
If the perpendiculars ED and GF divide the triangle into three
equal parts, then the area of the triangle ADE is - ah. Since this
triangle is similar to the triangle A CH, one has, if he puts x = AD^
the equation
x»
6*
,=z -ah
1 ab^
a«+6«~6 •2a«+6«
Similarly it follows that •
i>6? = l/^^T^-|(a+ 6) i/3.
Problem III. — A line is drawn from the comer ^ of a square
ABCD to a point E on the non-adjacent side CD so that its length
is a inches longer than a side of the square. The area of the triangle
cut off has to the area of the trapezoidal figure remaining the ratio
h : c. How long is the side of the square?
Solution, — Let x be the length of .
the side, BO, of the square ABCD;
then, according to the condition of
the problem,
BE=z(a + x),
EC=V{a + x)* — x'
r= i/a" + 2 ax.
Since the area of the triangle is to
the area of the trapezoid ABED as j^
h : c, then the ratio of its area to the
square is as
h:{h+ c).
Figure 9
420 COLLEGE ALGEBRA [M20
Therefore the area of the triangle is
b + c'
However, the area of the triangle is also
IbC' CE= %l/2 ax +a\
4 6«x» — 2 a(6 + c)«s6 — «•(&+ c)« = 0,
* = "-^(* + '^±V46'+(6+c)').
4. It is desired to frame a rectangular mirror, whose sides are
a and 6, so that the area of the frame shall equal that of the mirror.
What must the width of the frame be if the mirror framed is similar
in shape to the unframed mirror?
5. Of two sides of a triangle, whose area is 468 square inches,
one is 1 inch longer than the other. If the third side is 25 inches
long, how long are the first two sides?
Explanation, — Let the three sides of the triangle be a, 6, c, and
put
a±h±^
*- 2
Then the area of the triangle will be represented by
]/« (» — a) (» — h) {s — c).
Let X and x + 1 be two of the sides of the triangle ; then
g+fc + c 2xj-Jj^25^^ ^3
2 2 -/-r *«^.
Therefore, « — a = j? — x = 13, etc.
6. Two sides of a triangle are in the ratio 3:5; the third side
is 2 em. long; the area of the triangle is 150 sq. mm. How long
are the other two sides?
Explanation. — Let the two sides a and h be in the ratio 3 : 5,
and put
a = 3 X, 6 = 5 X, c = 20 mm. , etc.
7. If the area of a triangle is d^ and two of its sides are a and 6,
find the value of the third side.
Ans. l/a« + 6« + 1/4 a«6" + 16 (^.
1420] PROBLEMS INVOLVING QUADRATIC EQUATIONS 421
8. A triangle, two of whose sides are a and h and whose altitude
on the third side is h^ is divided by lines parallel to the altitude into
three equal parts. How long are the three segments of the third
side? (Compare solution of Problem II, page 418.)
Ans. AD^^ysr (y a«_A« + y b^ - h^) i/6« - h\
^^ =|l/3 \/{ya^ _ A« + v 6^ — A«) l/a« — A«.
9. From a point without a circle, two tangents are drawn to the
circle ; the two tangents and the radii drawn to the points of contact
form a quadrilateral whose area is a* square inches. How long is the
radius of the circle if the chord which connects the points of contact
is 2 inches long?
10. A frame is made for a mirror. The area of the mirror is a"
square inches, the frame on all sides is h inches wide, and the perim-
eter of the frame is four times as long as the perimeter of the
mirror. How long are the sides of the unframed mirror?
11. Two tangents to a circle intersect in a point which is a inches
from the center, and the radii drawn to the points of contact form a
quadrilateral, whose area is h square inches. How long are the
tangents?
12. How long is the side of a regular decagon whose area is a'
square inches?
13. Within a square whose side is a, a second square is so con-
structed that its sides are at a given distance from those of the first.
Within the second square a third is constructed so that its sides are
at the same distance from the sides of the second square, and a
fourth is similarly constructed within the third. How great is this
distance if the sum of the areas of the second, third, and fourth
squares is equal to that of the first?
14. A circle is inscribed in, and another is circumscribed about, a
square. The ring bounded by the two circles contains a" square
inches. How long is the side of the square?
15. A square is inscribed in, and another is circumscribed about,
a circle. The area of the figure between the two squares is a" square
inches. How long is the radius of the circle?
CHAPTER VI
RELATIONS BBTWEEN THE COEFFICIENTS AND THE SOOTS OF AN
EQUATION OF THE SECOND DEGREE
421. It has already been learned that the equation of the second
degree has two roots, i. e. , that there are two expressions involving
the coefficients, a, b, c, of the equation which will satisfy the equa-
tion. It is next to be proved that the equation ax" + 6x + c = 0
has two roots only.
Let, if possible, the equation
(1) aa^ + bx+c = 0
have three roots r^, r^, r^, all different. Since r^, r^, r^, are roots of
equation (1), they will be connected by the three relations,
ar* -\-br^-\- c = 0,
ar/ + 7>r, + c = 0.
Subtract the second and third equations from the first; then, on
dividing the first equation by (r^ — r^ and the second by (r^ — r^),
which is admissible since r^ — r^ and r^ — r^ are both different from
zero,
«(V- r/) + b{r^ -r,) = 0 or a{r^+ r,) + 6 = 0,
a(r^^ - r,*) + Hr, — r,) = 0 or a(r^ + r,) + 6 = 0.
On taking the difference between the last pair of equations, it is
found that
«(^, — ''3) = ^•
But by hypothesis a is not zero ; hence r^ — r^=z 0, or r^ = r,.
Therefore, an equation of the second degree can have but two roots.
422. By calling x^ and x^ the roots of the equation
x^+px + q = 0,
it was shown that the trinomial,
oc^ + px + q,
can be decomposed into two factors of the first degree,
{x^x;)(x-x;). [{413, (6)]
422
W23] ROOTS OF THE QUADRATIC EQUATION 423
K the multiplication of x — x^ by x — x^Sa performed, the identity
will result.
Since the members of this identity are the same, the coefficients of
X in the two trinomials must be equal and, likewise, the constant
terms; hence,
i> = — (Xj + Xj) and ^ = aci • oc^.
This result can be established also directly from the formulae for
the values of x^ and x^\ thus
12 * 2
By adding »i + ^J. = ^ = — P,
and multiplying the equations member by member,
Such are the fundamental relations which connect the roots of an
equation of the first degree with its coefficients. They may be
stated as follows: If the equation of the second degree is reduced to
the form
x«+i?x+5 = 0,
(1) the sum of the roots is equal to the coefficient of x with the sign
changed*, and (2) their product is equal to the constant term.
The equation ax* -|- 6x + c = 0, on dividing by a, becomes
x« + ^ X + ^ = 0.
'a a
If Xj and Xj are the roots of this equation, then according to the pre-
ceding rule
X, + x_ = — - and x,x_ = -.
423. The practical results of the preceding article are shown as
follows :
Suppose that the sum of two quantities, x^ and x,, is a, and their
product is b ; then x^ and x^ are the roots of the equation
a^ — ax -\- b = 0.
For example, if the sum of x^ and x^ is 7 and their product is 12,
the two quantities sought are roots of the equation
then will x^ = 3, x^ = 4.
424 COLLEGE ALGEBRA [W24
Similarly, two quantities can be found if it is known that their
difference is a and their product b. For, let x^ and — x^ be the two
quantities; then
and on applying the preceding rule, x^ and x, will be roots of the
equation,
x^ — ax — b = 0.
For example, if the difference is 3 and the product 28, it is neces-
sary to solve the equation,
x8_3x— 28 = 0,
whose roots are Xj=7, x^= — 4; the two numbers sought are 7 and 4.
424. Observations on the Properties of the Roots of the Quad-
ratic Equation.
1. If the third term of the equation of the second degree
x* + px -f- g = 0
is negative, the roots are always real, unequal, and opposite in tign:
because in this case the quantity i/p^ — 4q is real and, therefore,
both roots are real and unequal ({412,1). Since the product of the
two roots is equal to the third term — q, they must be opposite in
sign. Thus, for example, it follows that the equation
a^+5x— 14 = 0
has two real and unequal roots with opposite signs; for here
l/p* — 4q = i/25 + 56, and the product of the roots is — 14. Since
the sum of the roots is — 5, the greatest root in absolute value is
negative. These two roots must, therefore, be — 7 and + 2.
2. In case the quantity q is positive, the sign of j)*— 4 g must be
determined before it can be decided whether the roots are real «
not. If this quantity is positive, then the roots are real and haw
the same sign, since j?* — 4 j is less than ji^. Since the sum of the
roots is — p, then the sign of each root is the opposite of that of p-
(a) Consider, for example, the equation
x«->7x+ 12 = 0.
Since jp* — 4 g- = 49 — 48 = + 1, the roots are both real. Since the
product of the roots is + 12, they have the same sign. Since their
sum is +7, the roots are both positive. These roots are + 3 and +4.
«424] ROOTS OF THE QUADRATIC EQUATION 425
(b) Consider an example in which the second tenn also is plus :
ai« + 9 x + 20 = 0.
The quantity // — 4 5= 81 — 80 = + 1 is positive, and the roots
are real. Since the product of the roots is +20, the roots have the
same sign. Their sum being — 9, they are both negative. The
roots are — 4 and — 5.
3. In case the constant term q is equal to zero, the quadratic
equation has the form
x,+ px = 0, or x(x + jp)=:0,
and Xj = 0 and x + ^ = 0, or x^ = — p,
since either factor, placed equal to zero, annuls the product.
Find by inspection the sum and the product of the roots of the
following equations.
1. 4x« + 6x+34=0.
Divide by 4; then ^+i^+^=0.
3 17
Hence the sum of the roots is — x , and their product — -.
2. x»+6x — 11=0. 3. 2x« — 3x+2 =0.
4. 12x— 7 = — 14x«. 5. 8x« + 4=x.
6. 5x« — 8x — 20 = 0. 7. 2 (x— 1) = 3 (x + 2) (x— 3).
8. 9x« — 143 = 6 X. 9. 15 x«— 2 ax — a" = 0.
If Xj and Xj are the roots of the equation x* -[- px + 3' = 0,
10. Prove that H. Find the values of
(*) l + l + ^g=^^ (^) -.-.' + ^^'^r
(b) x«-x/=-i>l/?^. ^^^ ^1*+^,*-
(c) x/-x,x.+ x,«=i^-3(?. ^^> X, +x/
Form the equations whose roots shall be:
12. +.3, -l-
Here put xi = 3 and art = — -.
5
Then, according to the rule, (x — 3 War + ^ =0.
Multiplying by 5, (z — S)(bx+7) = 0.
That is, 5:i:« — 8j:— 21 = 0.
^r
^6 CX)LLEGE ALGEBRA [J424
13. — 5, 3. 14. - 7, — 9.
15. |, |. 16. 0, f.
17. a+h, a^h. 18. ^. A.
19. l + 3l/5, 1— 3i/5. 20. 2 — 3 1/7 • I, 2 + 3l/7-i.
21 ^+ y^ g — 1/^ 22 ^+''* l — m
3 ' 3 ' * / — m' Z + wi'
23. Without solving the equation, find the sum of the squares
and the difference of the squares of the roots of the equation
3x* — 6a;— 1 = 0.
24. Show that the roots of the equation ax*— (a— c)x— (a+!»)=0
are always real if a and h are negative.
25. Find the condition under which one of the roots of
«• -|- px + g = 0 is double the other.
26. Show that (a + 6 + c) x« — 2 (a + 6) x + (a + 6 — c) =0
has rational roots.
27. Find the rational relation which must connect a, 6, c,
a', 6', c', in order that the two equations
ax' -|- 6x + c = 0
a'x« + h'x + c' = 0,
have a common root.
CHAPTER VII
EQUATIONS WHICH ARE REDUCIBLE TO THE SECOND DEGREE
Biquadratic Equations, ox* -f 6x" -f- c = 0
426. Special Case of the Biquadratic Equation
ax* + 4 6x5 + 6 ex* + 4<fx + c = 0.
The solutions of many equations, not really quadratics, may be
reduced to the solutions of a quadratic equation. For example,
suppose that ^4 _25 x« + 144 = 0.
Transpose x* — 25 x' = — 144 ;
u AA'^ 4 oK^ . /25\« /25\« ,,. 625-576 49.
by addition, x* — 25x« + T— j =r— j — 144 = =— ;
after extracting the square root,
^ 25 ^ 7 .
2 2
hence x^ = 16 or 9.
x=:±4 or db3.
Examples of this kind are equations of the general biquadratic
equation type, ax* + 6x« + c = 0.
To solve this equation put
x* = 2 and
hence az* + 62 +
X* = z«;
c = 0;
solve: x* =
— & ± i/6« — 4ac
^ "" 2a '
or
There will be four values for x.
. |-5±i/6»-4ac
„ _ J-6 + v/6«-4ac
*»- \ 2a
'• - - \ 2a
427
428 CX)LLEGE ALGEBRA [M26
If both values of z are reeA and positive, then the values of a;, x^ ,
Xj, a;,, x^, are all real; if both values of z are real, one positive
and one negative, then two values of x are real and the other two
imaginary; in case both values of z are negative or imaginary, then
all the values of x, x, , x^, x^, x^, are imaginary.
Irrational Equations
426. An equation is said to be irrational when the unknown
quantity appears under the radical.
Example. — Solve the equation,
(1) P+VQ = 0,
where i^ is a poljmomial involving x to the first power and Q a
polynomial of the second degree.
If equation (1) is written
and both members are squared, an equation free from radicals will
result
(2) P'= Q or P« - Q = 0,
which in general will be an equation of the second degree in x,
because Q is of the second degree, P of the first, and P* of the
second. Every root of equation (1) is a root of equation (2), but
the converse is not true. For, if the second term of equation (1)
had the — sign, on transposing the result would be
and on squaring both members
(2) P*=Q or P« — Q = 0.
Hence the roots of equation (2) may satisfy either equation
P-j- 1/(> = 0, or i^— i/$ = 0, or both equations. This point will
be illustrated immediately.
As examples illustrating a doubtful solution, consider the fol-
lowing equations:
J426] THE BIQUADRATIC EQUATION 429
1. l/5x+10 = 8 — X.
By squaring 5x+10 = 64 — 16x+ x^,
x«-.21x+54 = 0.
Solve ^^21j;i/441-216^21±i/225^X8or3. '
2 2
On substituting these values in the equation, it is seen that x = 3
satisfies the equation, but 18 does not. However, 18 satisfies the
«<i"««o° _ VbT+To = 8 - ^.
This equation is included in the discussion above, for, on squaring,
the same result is obtained as on squaring |/5x + 10 = 8 — x.
Hence it is not possible to be sure that the values of x which are
finally found will satisfy the given equation; they may satisfy the
equation formed by changing the sign of one or more radicals.
2. Solve the equation.
l/x+ 2+ i/2x+2 = x.
Transpose, then |/2 x + 2 = x — y^x + 2,
square, 2x+ 2 = x«— 2xi/x+2+x+2;
transposing and uniting, 2 x i/x + 2 = x* — x;
on dividing by x, 2 i/x + 2 = x — 1 ;
on squaring, 4x+8 = x* — 2x + l,
whence x' — 6 x — 7 = 0.
x = 3d=l^9+7=3d=4 = 7or— 1.
The value 7 satisfies the equation, but —1 will not; for
v/-l + 2 + l/-^2 + 2=-l;
I.e., l + 0=:-l;
which is impossible. However, x= — 1 will satisfy the equation.
— v/x + 2 + l/2x + 2 = X.
It follows from these two examples that, in case an equation has
been reduced to a rational form by squaring it, it will be necessary
to observe whether the values found for x will satisfy the given
equation in its original form.
This caution applies, for example, to equations like (9), (10),
(12), 8412.
430 COLLEGE ALGEBRA [«427,428
Solution of Equations which can he Reduced to the Form
aa^+2hx+2l\/aa? + 2bx+ c=^p
427. Solve 2a?— 3x+ 8l/2a:« — 3« — 4 = 13.
Put the equation in the quadratic form:
(2x* — 3x-4) + 8(2x»-3a; — 4)1=9,
and, solving, get
(2a:» — 3x-4)l= -4d=l/l6+9 = — 4zfc5 = lor-9.
Hence (2x« — 3x — 4)* = 1 or (2a:« — 3.t- 4)*= -9
2x«-3x-4 = l 2x»-3x-4 = 81
^ __ 3±i/9-h40 ^ __ 3±l^94-680
4 4
x = 5or-l *x = 3±i^.
2 4
Here it should be observed that x = - or — 1 will satisfy the original
equation, but that x =3±X689 ^.jj ^^^ satisfy it but will Bttiafy
the equation 2 x« — 3 x — 8 v/2x« — 3x — 4 = 13.
In general, on adding c to both members of the general equttron
above, it follows that
ax« + 26x+ c + 2Zv/ax» + 26x+ c =j>+ c,
and on solving Vaa^ + 2 &x + c = — ? ± V^ P + (P + ^J-
Hence there are two equations to solve :
aa^+2hx+c = {-^l+V P+(p+ c)y=z2P+p+ c—2lV^+P+^^
and ax«+26x+c=(— Z— T/>+(p+c))*=2P+p+c+2Zi/P+l)+c
One must be careful to select the roots which satisfy the giwn
equation (H26).
SoltUion of the Equation as^^ + 6x* -|- c = 0
428. Here n may be an integer, or a fraction which is positiTeor
negative. Put x" = «, x*" = s*; then the equation ox^'^- 6x"+ csO
^comes az« + i»2 + c = 0.
Solve and get z = x"
_ -6± v^6>~4ac.
2a
whence x = n^ I— 6± i/6>-4flc.
\ 2a
J429] THE BIQUADRATIC EQUATION 431
If n is an even integer, every positive value of «= ^-r
fuTnishes two pairs of real and equal values of x with contrary signs ;
in case one value of the z is positive and the other negative, two
of the values of x are real, equal, and of contrary signs, and the
other two are imaginary; if both values of z are negative or imag-
inary, all the values of x are imaginary. In case n is odd, every real
value of z gives a real value of x of the same sign, and only one.
If n is fractional, say n =^7 [{128]
2a
=->!(-' nr-"')'
1. Suppose, for example, that x + 4 i/x — 21 = 0.
Put |/a = z ; whence x = 2* and the equation becomes
z* +4« — 21 =0;
hence \/x = z = — 2zbl/4 + 21 = — 2zfc5=3 or —7,
x = 9 or 49.
The value x = 49 satisfies the equation x — 4 |/x — 21 = 0.
2. x-«+ 20x-«— 69 = 0.
Put z = x"*, z* = X"*; then the equation becomes
z* + 20 z — 69 = 0.
x-»= 2 = — 10d=l/100+ 69 = — lOzblB =3 or —23;
therefore, ^' = o ^^ ~" o5 '
and x='^| or -'^I-
Reciprocal Equations
429. The solution of a very important class of equations of the
fourth degree (biquadratic equations) can be reduced to the solution
of a quadratic equation. Consider the equations,
(I) ax*+6x«+ cx* + hx-\- a = 0,
(II) ox* 4- 6x» + cx» — 6x + a = 0,
in which the coefficients equally distant from the ends are numer-
432 COLLEGE ALGEBRA [?429
ically equal. Equations such as (I) and (II) are called reciprocal
equcUions, Divide each equation by x* and collect the terms ; then
(ID a(x'+l)+6(x-i)+c = 0.
They may be written in the forms,
(I") a(x + i)V6(a= + ^) + c-2a = 0,
(II") a(ai-l)*+2'(*-;) + c+2a = 0.
Solve and get a; + i = -h :i,VlF^Aac±^j^ = A or B, Bay;
Dividing by x*
and X - 1 = zil±2^^J'<-3^ = A' or B', eay.
Hence there are now the two pairs of quadratic equations to solve,
a^ — Ax +1=0)^ Cx«-^'x — 1=0
a:" — ^x + 1 = 0 j 1 x« — ^'x — 1 = 0.
Example.— Solve x*+2x» — 3x* + 2x+l = 0.
The equation can be written,
(x*+ 1) + 2 (x» + x) — 3 x« = 0.
(x« + I) + 2(x + l)-3 = 0;
(x+iy+2(x+l)-5 = 0.
Solving » + - = — 1 db V^l + 5 = — 1 + |/6
or — 1 — i/6.
Hence,
x + i = — 1 — |/6 andx+i = — l + l/6
x2 + (l+ i/6)^ + l =0 x«+ (l — i/6)x+l =0
_ -(l+i/6) ± V l+6+2v^6 - 4 _ v'O-1 rb V^ 7-21/ 6-4
^ — 2 •'^ " 2
_ -(l + i/6)i: 1^3+21-^6 _ l/6 - 1 zb V^ 3 ~ 2 l/6
The values of x^ and x^ are real, and those of x, and x^ are imaginary.
M29J THE BIQUADRATIC EQUATION 433
BXBBOISB liXXrV
Solve the following examples by the methods used in solving
ax* + 6x* 4- c = 0, and ax*" + 6x" + c = 0.
1. X*— 13x^ + 36 = 0. 2. X*— 21x«=100.
3. (a:« _ 10) (x« - 3) = 78. 4. (x« - 5)« + (x« — 1)« = 40.
5. lOx* — 21=x«. 6. 6x*— 35=llx«.
7. a*+ «>* + X* = 2 a«6« + 2 a"x« + 2 6«x«.
8. 8 x-« + 999 x-» = 125. 9. 2 (i/x _ 3)" — 3 = \/x.
10.
{Vx-iy+v^=Vx.
11. (Vi - 3) (Vx - 4) = 12.
12.
V«' — 2l/x + a; = 0.
13. 2x~»— 3x»+x'= 0.
14.
t 1
a?-\- 8 a^"= 9 x.
l/x — a X
15. .x" + X* = 20 x».
16.
j,j l/x + 2n-5 3a-6
6 ~(a+fe)»"
18.
X + 5 |/37 - a; = 43.
19. 1215 + X = 49 v/615 + X.
20.
X + 2 a v'2 («• -I- 6«) _ X
= 3 a» + y.
21. x+(a + 6)l/a« — a6+6« — x = a«+5».
22. X* — ax« + 6« = 0.
23. x*-4(a+6)x«+16 (a — 6)« = 0.
24. X* — 4 (a»+ «>«) x«+ 4 a«6« = 0.
25. X* — 2 (a«+4a6 — 6«)x«+ (a — 6)*=0.
ofi « I 7. I ^ o7 x* 4- 10 3^ 4- 1 a
26. a = x'+6 + ^. 27. -^^^^-j^ = -
28. ^^'xife^-T^^=I- 29. (x-a)»+-^. = i
30. 4(x — a)*— 4 6(x — a)«+ c« = 0.
31. (x — a)*— (6+ c) (x — a)«+(6 — c)»=0.
no ax — h , cx — d g — 6jr , c — dx
cx — dax — b^c — dx a — bx
qo qx -\-b , cx + d __ ax— b • cr — d
a-\'bxc-\-dx~'a — bx c — dx
34. (a;+l) (x+3) (x-4) (x-7) +
(x - 1) (x - 3) (x + 4) (x + 7) = 96.
35. (1 + x) (2— x) (3+x) (4-x) (5 + x) +
(l— x) (2 + x) (3 — x) (4 + x) (5-x) = 144.
434 CX)LLEGE ALGEBRA [^39
U-\-z . 5 + j: ■ 2 + a?__ll— ar . 5 — a; . 2-
36. ^^^^ + FF+FP = ^^^+^-^ +
7+x ' 3 + x ' 1+a: 7 — a: ' 3 — a; ' 1— a?
ar+2 xfS"^ ar + 5 ar-2 x~3"^ a:-5 '
Solve the following equations, which can be reduced to quadratic form:
38. {j:^ + axy+m{x*+ax)=p. ^
39. X* — 6x»+7x« + 6x— 8 = 0.
40. X*— 10x»+35x*— 50x+ 24 = 0.
41. x^— 2.c»— 7x« + 8x+ 12 = 0.
42. 32x* — 48x« — 10x«+21x+5 = 0.
43. x» — 6 x« + 5 X + 12 = 0.
44. (2x«— 3x+ l)«=22x« — 33x + l.
45. 16 x2 (x - 4)«+ 121 (x - 2)« = 265.
46. (x«— 5x + 7)«— (x — 2) (x - 3) = 1.
47. x« + 5 = 8 X + 2 i/x«— 8x + 40.
48. 2 x« + 3 i/x« — X + 1 = 2 X + 3.
Solve the irrational equations:
I
49. 3x — 7vx+2 = 0. 50. \/x+b = x — l.
51. x+i/x + 3 = 4x — 1. 52. 1— 6x+i/5(x + 4) = 0.
53. 2x — v2x — l = x+2. 54. 3x — 4v'x"^= 2(jf + 2).
55. X — 10 = |(x — 1) — 1/ 2 X — 1.
56. a + V a^ — x^ = x. 57. v a*— ^+V'fe' + JC=o+^•
58. V a — ,r -\- y x — b = y a — h.
59. y a — x+ V^b —X = "|/a + 6 — 2x.
60. V a — bx+ y c — dxz= ya-\- c — (6 + rf)x.
Gl. 1 .r+3+V^2x — 3 = 6. 62. V4x — 3 — l/x-4 = 4.
63. I 5X-1-V 8 — 2.r=i/x — 1.
64. |/4x-3 + | 5.f+l = i 15x + 4.
65. l/x + 7 — V 5(x — 2) = 3.
66. 1/2X+I— 2i/2.r + 3 = l.
67. xi/x— a+ av x+a= v^x^ + a\
2429] THE BIQUADRATIC EQUATION 435
68. 2x" + al/6«+ 4 bx = a(6 + 2 x).
69. |/a(x - 6) + l/6(x — a) = x.
70. l/l + ax — i/l — ax = X.
_- \/a-i-a:+ Va — J a —
I X, — 1=^ — - 33 — •
Va+z—Va — x *
liS. . . =a — 0,
va — x+vx—b
. V g — j: + ^J^ ~ ^ __ la — x
76. l/x H 7^— = -^ H 7= — •
Vb vx Va
77 V^+V^;^ 2Vx __ (:»?4-a)V
l/a— V^a: Va + Vx a(a?— a)
78.
l^g— fej: + Vc — mx |/g —hx — V c — mx
Va^bx+Vnx — d Va — bx — Vnx
Solve the reciprocal equations:
79. x* + x»+x«+ x+ 1 = 0.
80. 6x*— 31a:»+ 51x«— 31x+6=0.
81. x* + l^x»— 8x«+ljx+l = 0.
82. X* — 3jx» + 2x» — 3^x + 1 = 0.
83. X* — 4Jx«+5Jx«-4Jx + l=0.
84. x*+(n-.i)x»— 2n«x«+(n — i)x+l = 0.
85. (x — l)»(x»+l) = a«x«.
86. x*+ax»+6x«+cx + (c«:a«) = 0. Put x = y i/c : a.
87. x*+5x«+10x«+15x + 9 = 0.
88. x*+3x» — 41^»Y**+^^ + ^ = ^•
89. x* + 2x» — 2Hx»+10x + 25 = 0.
90. x»±ax*±ax+ 1 = 0.
Suggestion. a^±a3iP±ax+l=3!^ + l±ax{x-{'l)sss
{x+ 1) (-c»- ar+ l±ax), etc.
436 COLLEGE ALGEBRA H^
91. cc'4-3jx«+3jx+l =0.
92. x» — ljx« — lja; + l =0.
93. x* + ax* + ix' + hx^+ ax + 1 = 0.
94. x« + 3x*+2ix»+2jx»+3x+l = 0.
95. X*— 4jf x* + 4x»+4x« — 4ffx+l=0.
96. x»+ax«+6x + (6»:a») = 0. Put x = -y.
97. x» + 3x« — 6x-8 = 0.
98. x* + 2x«+x+ J = 0.
99. x» + ax* + 6x' + cx« + (c«a : 6») x + (c» : 6») = 0.
100. x*^ — 2lx* + x»+2x« — 20x + 32 =0.
101. x»+2:^*+3x»+6x«+16x+32 = 0.
102. x« + ax*^ + 6x* — 6x« — ax — 1 = 0.
103. x«-5f.x'+9Jx* — 9Jx» + 5tx-l =0.
104. x^+ 4 x«+ 2 x«+ 5 x* + 5 x»+2 x« + 4 X + 1 = 0.
105. x^ + ax»+ ix*^ + (a + 6 — 1) X* + (a + fc — l)x» +
&x"+ax-4-l =0.
i Li
106. Solve the equation (a + x)* + (a — x)* = (/)*.
Here (« + a:)* +(«-«)*= ^ [(a+ x) + (a-x)] *
and, therefore. ("-i-^V 1 = (fj^-if + l/.
Put C'+-^)'=*and.-. ^^=z*;
1
hence, {z + 1) = [^)\z* + 1)*
or (2*+ 4 «» + 6 2» + 4 2 + 1) 2 a = ?(2*+ 1),
i. e., (2a-?) (^+ ^)'+8«(« + ^) + 12a-2(2a-0 = ^'
which is a reciprocal equation and can be solved by quadratics.
Having determined 2, we have the equation ^' ^ = 2* to determiner
Many similar equations can be solved by this device.
107. (5 + x)*^+(5 — x)* = 8*".
108. (x+l)*+ (x + 2)*=17.
CHAPTER VIII
FACTORIZATION
430. If, when all the terms of an equation are brought to one
side, the expression can be factored, then the equation will be satis-
fied by placing each factor equal to zero. Therefore the roots of the
factors will be the roots of the equations. For example, suppose
that
(x — a){7? — (6 — c) X — 6c) = 0;
the left member will be zero when x — a = 0, or when
sc* — (6 — c)x — 6c = 0
and only in these cases.
Solve the equation; then
X — a=0 and «• — (6 — c)x — 6c = 0
x^ = a, (a; — 6) (x + c) = 0 ; [{98, Ex. 4]
whence x— 6 = 0, or x + c = 0;
hence x, = 6, and x^ = — c.
Therefore the roots of the given equation are Xj=a, Xj=6, Xj=— c.
Skill in separating expressions into factors is acquired by ex-
perience,— see Chapter VIII, Book I. Suppose that
X (x — a)" — m{m — of = 0.
It i8 seen that x = m will satisfy the equation. Hence, by 2100,
the first member of this equation is divisible by x — m; therefore
(x — m) [x* -|- xm -f m* — 2 a (x + m) + a*] =0,
(x — «i)[x«+ (m— 2a)x+a«— 2am + m*] =0.
Here, x — m = 0 gives one root and x* -f (m— 2 a)x -(- a* — 2 am
+ m* = 0 furnishes the other roots.
487
438 COLLEGE ALGEBRA [M31
431. The Decomposition of the Trinomial .c^-f ;>x'-|- q into a
product of two real factors. — This trinomial can always be decom-
posed into the product of two trinomials of the second degree,
(1) X* + px« + (/ ^ (ax* + h.t + c) (aV + h'x + c') ;
if the first factor is divided by a and the second multiplied by a the
value of the product is not altered; hence the expression (1) becomes
(2) x* + 2)x« + g — (x« + 7x + m) {aa'j^ + ah'x -|- ac*)
where ? = -> m = - •
a a
The problem is to determine the value of 7, m, a, 6, c, a' , h^ ^
c', in order that the product of the quadratic factors will be
x^ -\- px^ -\- q , If ?, m, a, ft, c, a', 6', c', are properly chosen, then,
on expanding the parenthesis, the coefficient of x* must be equal to
1, that of x* equal to p, the constant term equal to q, and the
coefficients of x' and x each equal to zero, since they do not occur in
the original expression.
Hence, it is necessary to have an' = 1, and (2) may be written
(3) X* +i>x« + 5 = (x' + Zx + m) (x« + I'x + m'),
where V = ah' and m' = ac\
On developing the product of the second member of (3),
(x^ + lx + m) (jc^+l'x +mO = ^i'*+(l+l')^+(tt'+rn+m^)jc'+{lrn'+ym)x+mm^.
Hence, Z, ?' ^ m^ m\ must satisfy the relations,
(4) ^+Z' = 0;
(5) IV + wi + m' = p,
(6) Im' + /'m = 0,
(7) mm' = g.
Hence there results a system of four equations to be solved for four
unknown quantities.
From equation (4), /' = —?, and on substituting this value for
/' in equations (5), (6), (7), the following system is obtained,
(4) Z' = - Z,
(8) m + m' =p+J^,
(9) Z(m'-.m) = 0,
(7) mm' = q,
which is equivalent to the system formed by (4), (5), (6), (7).
W31] FACTORIZATION 439
Equation (9), I (rn' — m) = 0, is satisfied if
(I) / = 0, or (II) (m' — m) = 0.
Case I. HI — 0, therefore, by equation (4), I' = 0, and con-
sequentiy, from equations (8) and (7), m -f- m' =p and mm' = g;
hence m and m' are the roots of the equation,
(10) z^-pz + q = 0. [J422]
Hence, it is possible to write :
and accordingly formula (3) may be written
(11) x«+^x'+gr^(x«+P+^^-'>'?)(a^ + P-^P'-^'?)-
Case II. m' = m. Substituting m' = m in equations (8) and
(7), the result is obtained,
?*-|-^> = 2m, m* =z q.
If q is positive, we have from the second equation, m = ±l/g;
then the first equation gives P = 2vq — p\ and, consequently, if
2]/g — p is positive, Z=± -J'^V q — i?. The plus sign can be
taken before this last radical, because, since m and m' are equal,
equation (3) remains unchanged if I and V are permuted. The
system of equations (4), (8), (9), (7), has therefore the solutions.
I =z y^2l/q — jt), r = — y'2l^q — jt), m = m' = Vq;
and the trinomial x*-^px^-\-q will have, on substituting these values
for J, / ', m = m', in equation (3), the new decomposition.
(12) x*+px»+g= {x*+xV 2V q—p+V'q){x^ -xV 2V q-p+Vq),
where q Q.nd2yq —p are both positive.
When m = — Vq, we find I = ±^ — 2^'^^— j>.
Therefore, when q and — 2i/g— i? are positive, the trinomial is
decomposed in another manner.
(13) 2^+p3*-\-q={a^ + zV-p-^2\^q-\/q){a*'-xV-p-2\/q-'l/q).
440 COLLEGE ALGEBRA [H32
The following table gives a resume of this discussion, where in
the left of each formula is written the condition for which the
decomposition is possible.
It is a very simple matter now to separate x* + px* + q into
linear factors and, therefore, to solve the equation x* + pa^ + ^ = 0
for all cases in which the conditions of formulae I, II, III hold, and
in which the equations formed from the resulting quadratic factors
of the biquadratic, namely,
(14) a^+lx + m=zO and x« + /'x + m' = 0,
have real roots. The roots of these equations will be real when
/ « _ 4 wi > 0, Z' * — 4 m' > 0. Therefore, if x^ , x^ , and Xj , x^ are
respectively the real roots of the equations in (14), it follows that
j^+pji^+q^ix^' + lx + m) (2^+yx + m')~{x-Xi) (x-x^) {x-^x^) (x-x^.
The quantities x^ , x^ , x, , x^ , will be the real roots of the equation
X* + px^ -j- gr = 0 ; for if we put x = x^ , x = x^ , x = x^ , x= x^
the equation will be satisfied.
432. On applying these formulae to a particular case when p
and q are given numbers, it is observed that: —
(1) If g is negative, then formulae II and III do not apply,
for then i/ —qis imaginary; formula I, however, is always appli-
cable, because p' — 4 g is then positive.
(2) In case both p and q are positive, the last formula is not
applicable, because —^x/q — p is negative, and its square root is
imaginary. Smce 4 g — p' ~ (2 x/q — p) (2 \/q + p), the condition
2]/jr_^^0 can be replaced by the condition 4 gr —^> 0. If
4 J — j^is positive, formula II is applicable, but formula I is not.
8432] FACTORIZATION 441
The converse is true when 4g — p* is negative. It follows, there-
fore, that formulae I and II can not hold at the same time in this
case, except in the limit when 2>* = 4 g, when the two formulae
become identical,
x'+px^+q={^ + l)\ l>* = 4g.
(3) Suppose, finally, that j > 0, p < 0. Since 2 \/q —pm this
case is necessarily positive, formula II is always applicable. The
condition —p — 2 Vq > 0, which is the same as — p > 2 Vq, can
be replaced by ( — i))*> 4 q, because both members of the inequality
are positive, since q is positive. Therefore, formulae I and III are
applicable or not according as p* — 4 ^ is positive or negative.
The conclusion is, therefore, that the decomposition of the tri-
nomial X* -f- P^ + q <^(^^ ^^ factored in one way only^ or in three
distinct ways. In order that the trinomial be decomposable in three
distinct ways it is necessary and sufficient that
i><0, g>0, p«~45>0.
These are precisely the conditions which are necessary and
sufficient in order that the quadratic equation,
«* + !>« + ? = 0)
may have real and positive roots ; i. e. , that the biquadratic equation,
may have four real roots.
BXBBCISB LZXV
Factor and then solve the following equations :
1. X* — 9x«+20 = 0. 2. x*-(-9x«+ 81 =0.
3. X* + x« — 4 = 0. 4. X* + x« — 1 = 0.
5. x*~x«+l = 0. 6. x*+l=0.
7. a4_^aJ_2 = 0. 8. x* + 16x2+l=0.
9. 3x* — 2x«— 16 = 0. 10. x*+(2m — /«)x«+m« = 0.
CHAPTER rX
ROOTS OF S0RD £XPRSSSIONS
The Transformation of the Expression
y^a ± Vh
433. Consider the biquadratic equation,
(1) x*+px^+q'^ = 0,
where g'^ = ^, and p and q are both commensurable. In view of
what has been learned, the first member of this equation is easily
factored. For
-[x2+ j'~ V 2q' - i> • a-] [.x« + q' + \/2q' —p • x]
when 2 ?'— i>> 0.
The roots of the biquadratic equation will be found by solving
the equation formed by placing these factors equal to zero.
Thus, solving the equations:
V 2^— P •x+2r' = 0;
x2+ V 2j' — p • x+^' = 0.
Hence their roots are
(2) r2 7^-;)dzl^-i>-2f/ - V'Zq' — /> db t/— P —2q'
- 2 ' 2
Therefore the expressions in (2) are the roots of the biquadratic.
But if the biquadratic equation (1) is solved directly, then its roots
are
(3) ,,^-P±V>'-V
= *Nl-fWf-'^-
44a
JJ434, 435] ROOTS OF SURD EXPRESSIONS 443
Hence, on comparing the expression for the roots of the same equa-
tion, two of four possibilities result:
4^
j— j> ^ Yj)* — 4^1 _ V 2q' —p±: t/— p — 2q'
2 2
It is seen, therefore, that in some cases an expression of the form
y A ± y B may l>e expressed in the form of the sum or the differ-
ence of two simple radicals. This result involves a principle of
practical value. The cases in which this transformation is possible
are determined directly. To investigate the question proposed, the
following lemma is established.
434. L^MMA. — Suppose that a, 6, a', 6' are real and rational and
that neither h nor h' is a perfect square. Thus, if
a+Vb=:a'+ VV,
then will
a =. a' and h =i b\
Suppose that . _ _
a+\/h = a'+ l/6'
where a, 6, a', 6' are real and rational quantities, and h and 6' are
not perfect squares. After transposing,
|/6 z= a' —'a+ Vh\
Square and obtain
i = (a' — a)« + 2 (a' — a) l/F' + 6';
transposing _
2 (a' - a) i/6' = (6 — 6') — (a' — a)*;
i. e. , an irrational expression, 2 (a' — a) i/6, equal to a rational
expression which can be satisfied only when a = a', h=zb'y i. e.,
when both members are reduced to zero.
435. It is proposed now to transform the expression V a-\- 1/6,
in which a and 6 are both rational and h is not a perfect square,
into the sum of two radicals. Put
\ a+\ b = yx+v'y
where it is desired to express the positive quantities x and y
rationally in terms of a and 6. Square both members of this
equation, _
(1) a+ \ h =^ + y+i^4.ry.
444 COLLEGE ALGEBRA [J436
By the lemma (2434),
a = X -|- y» and 6 = 4 xy or xy = -;
i. e. , two equations which have to be solved for x and y. Hence,
X and y must be the roots of the equation ({422)
On solving,
«8_a«+| = 0.
a ± i/a* — 6
y
1-
a
—
l/a» —
6
-4 1
2
^
a -
-iv:
-6
1. e., a? = 2 »
Hence
I. v/^TT7I = ^2+j|i3^^-j^
Similarly, the transformation of V a — Vh may be accomplished.
Thus, put
(2) Va^vl=V~x-\/y,
Square and obtain _
a — 1/6 = x + y — V/4xy.
and 4xy = h or xy = -'
4
The solution of these equations for x and y will be the same as
above and on substituting in (2)
IL V a-vb = J^±^^'J^Z^ _ J« -_v'Z^ .
Formulae I and II will be of practical value when a* — 6 is a perfect
square.
436. Examples.
1. Transform the expression 1^13 + 4 v^3.
Put 4 under the radical sign, 1^13 + 4V3 = V^13 + l/48 ;
here a = 13 and 6=48; [?436,I]
a«— 6 = 169 - 48 = 121 = 11».
8436] ROOTS OF SURD EXPRESSIONS 445
Substitute in formula I, then
i/I^Ti71=^^+^^.
=1/12 + 1/1.
=21/3+ 1.
2. Transform V^&9 — 3i/384.
Here a = 59, 6 = 3456, a« = 3481, a« — 6 = 25.
Therefore,
1/59-31/384 = ^^^-^^^=41/2-31/3.
3. Transform V2m + x — 2l/w« + mac.
Examples 1, 2, 3, etc., may be solved directly.
Put V2m + X — 2l/m« + mx = 1/2 — l/J^.
Square and obtain
2m + X — 2l/m' + mx = 2 + 10 — 2 1/2117. [Lemma, J484]
2 -f- 10 = 2m + X, 2i!7 = m* -f ^^•
Hence,
2* — 2210 -|- m;* = (2 + to)* — 42ti? = 4m* + 4mx + x* — 4m' — 4mx,
and (2— W7)* = x';
therefore, 2 — w? = x;
but 2+to = 2m + x;
after adding and subtracting, 2 = m + x, w = m.
y 2m -\- X — 2l/m* + mx = l/m -f- X — l/m.
4. Fmd the square root of l/27 + 1/15.
Here, l/27+ i/T5 = 3l/3+ l/l5 = l/3(3+ l/5);
thus, 1/(1/27 + 1/T5) = V3 1/(3 + 1/5).
It may be shown as above that
Hence,
v/(l/27 + 1/15) = V3 m + i.) =*^ (1 + V^.
446 COLLEGE ALGEBRA [?J437, 438
437. Sometimes it is possible to extract the square root of quan-
tities of tlie form
a+ Vb + y c + yd,
by assuming that V{a + v/6 + i/c + v^7l) = \/x + y y + \/z.
Then,
«+ V'^ + V^+ V (^ = ^ + y + 2 + 2l/.ry+ 2\ xz + 2Vyz,
Thus it is possible to put
2]/.7^=l/6, 2V^=i/7i, 2>^z^Kc,
and in case the values of x, y, z, which satisfy these three equa-
tions, satisfy also x + y + 2 = «, the required square root will be
obtained.
Example. — Extract the square root of
6 + 2 I 2 + 2i 3+ 2 1/6.
Let 1/(6 + 2v2+2|.3+2 K 6) = \^x + Vy + kI
Then
6 + 2v2 + 2v3 + 2l/6 = .t + y+5; + 2 \'7^+2 Vyi+2 V^,
Put
2v'7cy - 2 1/2, 2Vjz = 2 r 3, 2 l^r^ = 2 |/6, and x + y +2 = 6.
By multiplication V xy '\' yz — \^\
but \/xz = V 6.
By division y = 1 ;
x — 2 and z = 3.
These values satisfy the equation x + y -{- z = 6, Hence, the
required square root is V2 + | 1 -f- | 3 ; that is 1 + v 2 -f- K 3.
438. If 'I'^a + 1 b = x+ Vy, then 'v^« — yb=zx — \/y.
Let 1 a-^}b=x-\'yy'^
then, by cubing,
a+yb = x^+ 3.r2 y y + 3xy + y Vy;
a = x^ + 3xy and 1 b = S.r^ 1 ^y + y y'y. [J434, Lemma]
Hence, a -\ b = x^ — 3./ \y + 3xy c- yv .2/,
and V(a — 1 ^) = X —V ^. Q. E. D.
M38] ROOTS OF SURD EXPRESSIONS 447
Application. — The cube root of a binomial may sometimes be
found.
Let V(a + l/^) = » + 1/3/.
Then V(a - Vh) = x — j/'y. [8438]
By multiplication '|/(a* — h) = x^ — y.
Suppose that a^—h is a perfect cube, and call it c'; then
c = x^ — y or y zzz x^ — c ;
but a = 'j^-\- 3xy. [S488J
On substituting the value of y •= x^ — c,
a = rr' + 3a? (x* — c)
4x' — Scu* = a.
The value of x must be found from this equation by trial; then
the value of y can be found from the equation,
y = x^ — c.
In the first place this method involves the assumption that a* — b
is a perfect cube ; the method is then incomplete, because it is not
possible always to solve the cubic by trial. This proposition is,
accordingly, of little practical value.
Example 1. Extract the cube root of 16 + Si^5.
Let
V(l6+8v'5)=:x+l/Vj
then V(16 _ 8 1/ 5) = .T - Vy. [}488]
By multiplication V(256 — 320) = x^ — y,
that is, — 4 = X* — y.
Also 16 = x» + Sxy = x3+ 3x (x^ + 4);
x'+3x = 4. [{438]
X = 1 is a root of this equation; hence y = 5, and the required cube
root is 1 + 1/5.
Example 2. Find the cube root of 21|/6 — 23 l/5.
2I1/6 — 23^/5 =31/6^7 — f J|)
V(21 1/6 - 23 1/5) = V(3 1/6) V(7 -
The value of '^{'^-J^E)
may be found by the method just illustrated.
23 |5>
448 CJOLLEGE ALGEBRA tW39
The Variation in Sign of the Trinomial of the Second Deoeei
439. It is proposed to study the variation in the sign of the
trinomial of the second degree, ax^ -{-hx -\- c, where x may take nnj
real value. On dividing and multiplying this expression by tiie
same quantity, a, it may be written a{x^+px-\-q), where p = -, y =-•
If a complete square of x^ -f px is formed by adding and snbtracting
^» this trinomial may be written
«x'+ 6a5+c=a p + pa, +^+ J— f] = a [(a: + f)'+ "^^f^
There will be three principal cases to consider:
1. 4q — ^* > 0 or j>* — 4 g' < 0. In this case the roots of the
equation a;*+ px+2'=0 are imaginary (2411, 3). The fraction =^^
is positive by hypothesis; therefore, the sign of the tri-
nomial ax* + ix + c will depend upon the sign of a, since (jf +« I
is necessarily positive, whatever real values x may have.
2. 4q—p^ = 0. This is the condition that the eqaatioo
ax* + 2/x + c = 0 has equal roots, and that ax" ^ hx-{- c becomes a
perfect square (Hll, 2) ; and we may accordingly write
ax^-{-hx-\- c=ia(x+^\
The quantity (^5+^) will be positive for every real value of ac, and
therefore the sign of ax* -[- ftx + c in this case will be the same aa
that of a J and the trinomial will be zero for x = ^.
3. 45'— jp*<0 orp* — 45> 0. The roots of the equatioQ
ax* + 5x + c = 0 are real and different (Hll, 1). Let x^ and x, be
the roots of this equation; then the trinomial ax* -{- hx -{- c max be
written in the form,
ax* + tx + c = a(x — x^) (x — x^),
in which it is supposed that a^j > a^^ • Now, in case x is greater than
Xj and less than x^ , then x — x^ is negative and x — x^ is positive.
And, therefore, the sign of the trinomial will be the opposite of
that of a. For values of x > x^ , the sign of x -— x^ and of x — x^
are both positive and therefore the sign of the trinomial will be the
same as that of a. Finally, if x takes values such that x < x, , the
bigns of X — Xj and x — x^ are both negative, but their product irill
be positive, and the sign of the trinomial will be the same as that of a.
8440J ROOTS OF SURD EXPRESSIONS 449
Hence, (1) in case the roots of the equation ax* + bx -{- c = 0
are equal or imaginary, the sign of the trinomial ax* + 6x + c will
be the same as that of the coeflScient a, for every real value of x ;
but) (2) if the roots of this equation are real and unequal, the tri-
nomial will have a sign contrary to that of a for values of x which
lie between the values of these roots, and (3) the same sign as that
of a for all values of x less than the least of the roots and greater
than the greater root.
The following observation in the preceding discussion is of great
practical value.
440. If, on assigning two different values to x in a trinomial of
the second degree, the results of substitution have opposite signs, it
follows that the roots of the trinomial are real and different, and
that one of them lies between these values. For example, if the
values x = 3 and x = 0 are put in the trinomial 2 x' — 11 x -|- 14, it
is found that the trinomial takes respectively the values of — 1 and
-|- 14; the inference is that the roots of the trinomial are real and
distinct, and that one of them lies between 0 and 3. Because, if
the roots were imaginary, the trinomial would not change signs ; and
further, if x is put equal to 3 the resulting value of the trinomial is
negative, which is contrary in sign to that of the coefficient of x*,
which is here equal to + 2 ; and therefore 3 lies between the two
roots Xj and x^, J439, (2). The value 0, assigned to x, gives a positive
value with the same sign as that of the coefficient of x*, i. e. , as + 2,
and is therefore less than the smallest root x^; hence it follows that
the smallest root x^ lies between 0 and 3, and the largest root x^ is
greater than 3. In fact, the roots of the equation are 3^ and 2.
EXEBOISE LXXVI
Find the value of the expressions in the following examples
from 1 to 26.
1. V^4+2i/3. 2. 1/7 — 21/10.
3. V^18+8l/5. 4. V^75-12i/21.
5. 1/3I + 1/6OO. 6. V|->/|.
Ans. ±(5 + 1/6). Ans. dz (i l/3 - J >/6).
7. 1^11—31/8. 8. V^lOO — 21/2499.
Ans. ±(3-1/2). Ans. d=(l/5T — 7).
450 COLLEGE ALGEBRA [*440
9. l/f+1 2. 10. VV— 2vn— 1.
11. l/2w — 2l m« — n*. 12. Vx + y + 2vl^,
13. V^9w + 25h — 30v^w». 14. Vx+xy — 2^x\ y.
15. V 2i> d= 2 V i>* — ^*. 16. V 2j>« + 3« + 2/> I i^ -f yl
17. l/l32+v24. 18. V I 63— I 35.
19. Vi 27 — 2v(). 20. V^V 1573 + 4V78.
21. i[l 5+1 5 + 1 3— V 5]. 22. i[v3+i 5 + V'5-l/5j.
23. i [1/5+ 1/5+ V 9 — 3 V' 5] +i[v^'l5 + 3|'5-l''3-l/5j.
24. I u: + y + 2 + 2v^x2;+yz. 25. V a«+ 2x V «* — a;".
26. 1 ««+ 5ax — 2al/aar + 4x«.
Find the value of:
27. _JL±i_+^±=LL=wheii. = il
1 + l/l + z l + v'1-2 2
28. ; " - + ,— — when z = -7^.
1+i/l + a i^Vl-z 1^6
29. Vv32 + V30. 30. ^6 + v^S— V 12- 1 24.
Ans. 1 + V 2-1 3.
31. Extract the square root of 8 + 2 v 2 + 2 i '5 + 2 1 10.
32. Extract the square root of 5 + V 10 — V 6 — V 15.
33. Extract the square root of
15 — 2 V 3 — 2 ]/ 15 + 6 v'2 — 2 1 '6 + 2 V 5 — 2 I 30.
34. V *i 4000 + % 221184 + *! 1024000 + \ 3456000=?
Suggestion. V \ 4 (l0 + 2 v'G+ 2 V\0 + 2V\h)=
V2 • I 10+21^6 + 21/ 10+21/ 15 = V2(v^2+v^3 + »'5).
35. Extract the cube root of 10 + v^'108.
36. Extract the cube root of 18 1^3 + 14 1/5.
37. Prove that V ( V '5 + 2) — V (v 5 — 2) = 1.
CHAPTER X
SYSTEMS OF SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS
441. Type I | i^^ ax + hy=c
I (2) Aa^+2Bxi/+ Cy+2 Dx+2Ey-\-F=0,
An equation involving x and y is said to be of the first degree in
these quantities if they occur singly with the exponent unity. If
the exponents of x and y or the sum of the exponents of x and
y in one or more terms of an equation is 2, the equation is of
the second degree.
In the system of equations to be discussed, equation (1) is of the
first degree and equation (2) of the second degree in x and y. The
solution of the most important forms of systems of equations belong-
ing to this type will be illustrated by examples. Every system of
equations of Type I can be solved.
Example 1. Solve the equations
jd) x-2y=z7,
1(2) xy = Sb,
The value of x from the first equation is
(3) x = 2y+7
and if this is substituted in equation (2), then
(4) y(2y+7) = 85
or 2y«+7j^ -85 = 0;
solve and obtain y,, ^^ ^ -7± v/^49 + 680^:^7p^^ ^^ ^.
The given system of equations (1) and (2) is equivalent to the system
of equations (3) and (4). Hence, if the values of y. = 5 and
17
M = — are successively substituted in equation (3), the corre-
sponding values of x will be found as follows:
X, = 2y,+ 7 = 10 + 7 = 17, a-, = 2y, +7 = - 17 + 7 = -10.
451
452 COLLEGE ALGEBRA [{441
Therefore the two proposed equations have the two solutions:
Ut solution: x^ = 17, y^ = 5;
2d solution: x, = — 10, y^ = — ^ *
Example 2. Solve the equations
r(l) y_2x = -4,
I (2) 3x» - 7y«— 6x - 28y + 46 = 0.
Solve (1) for y:
(3) y = 2x-4;
and substituting in (2) obtain
(4) 3x« - 7 (2x - 4)» - 6x - 28(2x - 4) + 46 = 0,
or a:«— 2x=^.
Solving, Xj, x,= Idz^l^Tl.
The corresponding values of y will be found by substituting
successively these values in equation (3) ; thus
+ 21/71-10 +21/71 + 10
Hence, the solutions of the system of equations (1) and (2) are
6+Vn 21/71-10
^1= 6~' ^^ = 6 '
5-1/71 2v^7r+10
x, = — ^— , y,= - ^n^.
Any system of equations of Type I can be solved by the method
illustrated in the preceding example and described in the following
rule:
In general, the solution of two equations in two unknown quantities^
one of the first degree and the other of the second degree^ is reduced to
the solution of a quadratic equation by solving the equation of the first
degree for either of the unknown quantities and substituting this value
in the equation of the second degree.
In all the examples which have just been solved, two pairs of
values of x and y which satisfy the given equations have been found.
This result is characteristic of the following general principle:
If^ of two equations which involve two unknown quantities^ one is
of the first and the other of the second degree^ in x and y, it will
always he possible to find two pairs of values of x and y, real or m-
aginary^ which will satisfy the given equations.
5442] SIMULTANEOUS QUADRATIC EQUATIONS 453
This may be shown as follows: If the equation ax + by = c of
the given type is solved for .v, the value of y will involve only the
first power of x, and since the degree of equation (2) in x and y is
2, the result of substituting this value of y in equation (2) will be
a quadratic equation in x, which will give two values of x. To each
of these values there will correspond but one value of y, determined
by equation (1).
C (1) ax* + bxy + cy* = cf , ") When first members oj
442 Type II k r ^^ equations are homo-
1(2) A^+ Bxy+ c-y'zrA j^e"*"^"-^"^*^'*
The equations in Type II are called homogeneous equations of the
second degree, because the degree of every term involving x or y is
the same and is equal to 2. Such a system of equations can always
be solved by a definite plan explained in the following examples.
Example 1. Solve the equations
1(2) xy=12.
Let y = vx, and substitute in both equations; thus,
(3) xMl— t;») = 7, x«i; = 12;
therefore, by division, ~ =t^;
hence, 12 1;« + 7t; — 12 = 0,
andonsolving .„ .. =-7 ± ^49 + 576^ ^.^^g^^ _4
Substitute these values in either of equations (3) ; then
The values of y which correspond to x = i 4 are real and are
found by substituting x = ±4 in equation (1); thus
16— y« = 7; i. e., y« = 9.
y = ±3.
Hence, the real system of values of x and y which satisfy equations
(1) and (2), are ^.^ = + 4, y,= + 3;
x, = — 4, y, = -3. _
The values of y which correspond to the values x = ± 1/— 3 are
imaginary.
454 COLLEGE ALGEBRA [J442
Example 2. Solve the equations
r(l) x« + i/2 = 58,
1 (2) xif = 21.
Equations (1) and (2) are homogeneous and may be solved by the
method indicated in Example 1 ; but they may be solved also as
follows: multiply equation (2) by 2, and add and subtract the resulting
equations member by member, to and from equation (1); thus
(3) x'+2xy+y^ = 10i),
(4) x«-2.r^ + / = 16.
After extracting the square root of both members of equations (3)
and (4),
(5) . X + y = ± 10
(6) x-t/=±4.
To the four possible combinations of signs there correspond the
four following systems of equations
(7)] (8)] (9)] (10)]
(x— y=4 (x— i/ = — 4 (a-— y = + 4 (x— y=: —
lution of which gives the four systems of values, resp
Cx = 7 Cx = 3 Cx = —3 rx=: —
(y = 3 |y = 7 U=-7 (y = -
x+y=— 10
4
the solution of which gives the four systems of values, respectively,
'x = 7 i X = 3 C X =: — 3 fx = — 7
3
which satisfy the given equations (1) and (2).
Example 3. Solve the equations
((1) 3x«+8y«=14,
1(2) x^ + xi/+4y^= 6.
Let y = IX, and substitute in equations (1) and (2); thus
(3) x2(3+8t;«) = 14 and x«(l + «+ 4t;») = 6;
then, by division, ^±-^^ = 11 = 1.
hence, 4v* + 7v— 2 = 0.
Therefore, (4) r^, r, = zz1±}l31 = | and - 2.
The system of equations y = vx and equations (3) constitute a
system of equations equivalent to the given system. On subatitut-
9442]
SIMULTANEOUS QUADRATIC EQUATIONS
455
ing successively the values of v in the equations in (3), the result is
5'
(5) x« =
Hence,
*-3+-8"V=^ ""'^ ^' = 3 + 8 .,«
(6)
(7)
1 .
'■i= + 2, y,= Vt = |-
The artifice here employed may be conveniently used when both
equations are homogeneous and of the second degree.
In solving examples of this type four pairs of values of x andy,
real or imaginary, have been found, which satisfy the given equa-
tions.
Example 4. Solve the equations
( (1) 2x^ + 3 ory + y^=70,
\ (2) 6 x« + xy — / = 50.
This system of equations has the general form of Type IL
Let y=vx, and substitute in equations (1) and (2); then
(3) x«(2 + 3 t; + v^) = 70 and x«(6 + v -v^) = 50.
Divide the first equation of number (3) by the second, then
Hence
(4)
hence.
2 -1-3. + ,^^; ^^ 12r«+8t._32 = 0.
6+ v— tr 5
3t;« + 2i;— 8--0,
v,,v,= ^^^t^-^. = tor-2;
x* =
3
70
2 + ^v^ + Vi
whence,
3
{^ whence, J ^
Here, t;^ = — 2 gives an indeterminate solution
smce
r« - 70 _ 70 __ _
Nora.— It will always be possible to find four systems of values of x and y, real
or imaginary, wblch will satisfy two equations of the second degree in x and y.
[878, 2]
456
COLLEGE ALGEBRA
C4443
443.
Type HE
etc.
When the members invoiv"
ing the nnknoum quan^
titles are symmeiruxMl tnth
respect to them.
An expression is symmetrical with respect to two letters, x and ^,
when they are involved in the same way, so that the expression is
unaltered in form when x and y are interchanged.
Thus, Ajc^+ 2 Bxy + Ay* is symmetrical with respect to x and y,
because if x and y are interchanged the result is -4y' + 2 Bxy -|- ulic*,
which is identical with the former expression. Similarly,
X* + 4 xV + 5 oi^y* + 4 xy* + y*
is symmetrical with respect to x and y.
Many examples involving symmetrical expressions may be solved
by substituting for the unknown quantities the sum and the differ-
ence of two new variables.
Example 1.
Solve the equations
jd) x* + y* = 82,
1 (2) X + y = 4.
Put
x = tt + t?, y = u — V,
From (2)
{u+v) + (u —v) = 4, therefore, u =
From (1)
(2 + t;)*+(2-t;)* = 82,
therefore,
2(t?*+24t;«) = 50,
or
t;*+24t;«-25 = 0;
t;«=:l or -25,
solve and obtain
t;=±l or ±5l/-l.
Hence, x
= 3, 1 ; 2 ± 5 l/— 1, Imaginary,
y
= 1, 3; 2=f5>/— 1, imaginary.
Example 2.
Solve
|(1) x«+xy + y« = 84,
\ (2) x~v'xy + y = 6.
Put
x = M+v, y=zu—V.
(1) becomes (3) {
[w + t;)» + (ii« - 1;«) + (« - t;)« = 84.
(2) becomes (4) (u + v) — l/tt« — 1;« -f- (w — t?) = 6.
After simplifying (3)
(5) 3u» + t;« = 84,
and (4) (6) 3m«- 24u+ r«+ 36 =0.
8444] SIMULTANEOUS QUADRATIC EQUATIONS 467
After subtracting (5) from (6) 24 m = 120,
u = 5.
Hence from (5) i;« = 84 — 75 = 9,
Therefore, x = tt+t7=:5d=3 = 8or2,
y = t* — v = 5=F3 = 2or8.
Example 3. Solve the equations
1(1) 3(x«+y»)-2xy=27,
1(2) 4(x« + y»)-.6xy = 16.
These sjrmmetrical equations can be solved readily by the following
substitution. Let t; = as* + y' and z = xy.
Hence, (3) 3 r — 2 z = 27
(4) 4 1? — 6 2 = 16.
On solving (3) and (4) it is found that
V = 13 and « = 6,
i. e., (5) x'+y*=13 and xy = 6.
The solutions of these equations are
rx,= 3, f^, = -3, fx3 = +2, U, = -2,
Ui=2, U, = -2, (3/3 = +3, (y, = -3.
444. Irrational Simultaneous Quadratic Equations.— Solve for
X and y the system of irrational equations
; (1) ^ X + y = 58,
(2) i/x + l/y = 10.
Squaring (2) (3) x + y + 2 1/^ = 100,
subtractmg (1) (4) 2 V^ = 42,
(5) xy = 441.
Solving (1) and (5) according to Type I, {441, we get
\ *' = ^' and f *«=*^'
BZBBCISB lOCX-Vn
Solve the following equations:
(sey — x = 0. '(x = 3y.
i:
458
5.
7.
9.
11.
13.
15.
|2x« — 3y« =
{
-3y« = 6
. 19.
0-^ = 12
2x + 3y = 18.
\2x-y = 0.
3x» — 4y = 5x — 2y«
COLLEGE ALGEBRA
5x«+2y = 22
[3444
8
10
, (5x«+2y = 2i
• (3x2—51/2 = 7.
, |x«+y« = 50
• t9x + 7y = 70.
^ I x* — xy + y* = 7
• I2x — 3y = 0.
(x 4- y) (x .
, X ■
2y) = 7
(3x
(3x4-4.v = 10. ""' (x — y = 3.
f (3x-2y)(2x-3y) = 26 ( x« + 2xy -y« = 7(x - y)
(x — 2y + l-0. I2x — y = 5.
I 2 x« — 5 xy + y« + 10 X + 12 ?/ = 100
I2x-3y = l. ^^^ |7(x + 5)«-9(y+4)» = llg
' (x — y = 1.
xy = (3-x)« = (2-y)«.
21.
23.
25.
27.
29.
31.
33.
35.
2x — y+l^8
x — 2y+l 3
X* — 3 xy + y" = 5.
l_±^lb?f=Q
l + 2/ + 2/»
x+y=p.
2x + t/ x + 3
x + y = 10.
r X -|- xy = 35
( y + xy = 32.
C x« + y« + X + y = 18
( X* — y" + X — y = 6.
|5x+y+3=2xy
Ixy = 2x— y + 9.
f(x-l)(y+5) = 100
l(x-2)(y + 6) = 99.
(x - 4) (y + 5) = 0
;x + 2) (y - 3) = 0.
16.
18.
20.
22.
24.
xy = .X* — y« = 2(x + y),
'2a: — 5, 2i/-3_o
^3x-4i/ = l.
|x+l+y+l ^
x2+y« = 2x+y.
;^ + ?/+1^3
I 3/^ + ^ + 1 2
X — y = 1.
1 O.I . 1 r» ^ -l
C4x-\-y—l _4
;j2.r + y-l i
(3x + y = 13.
^ rx(y-l) = 10
'•ly(x — 1)=12.
2 x' — 5 xy + 3 X
2x + y-\2'
( oa-u —
a-y — 2 a;' -|- 7 « •
-2y = 10
8y = 10.
f(x + y)(8-a;) = 10
l(j: + y)(5-y) = 20.
f (x - 2)« + (y + 2)» = 17
• l(.x-l)«+(y + l)« = 13.
ofi ((x + 4)(y_3) = 0
1 (a: + 7) (y - 7) = 0.
32
34
H44]
3^ j3x«-2y«=6(a;-y)
■ I acy = 0.
SIMULTANEOUS QUADRATIC EQUATIONS 459
(4x'-9y« = 0
a;»— 53/*— 3a; — y+22=0
5 = 0
a;-3)(y-2)=y«-3y+2.
■1'
(2x»-3acy+5y-i
• l(x-2)(y-l) = 0.
41. ] 5^2
(.y 3
jx:.y =9 :4
I X : 12 = 12 : y.
fx« + y» = a«
•I X m
rj^+y«=130
• j ^-+-^=: 8.
40 f3(x+y)«-2(x — y)« = 73(x-y)
• l(2x-y) : (4x-3y) = 2 : 3.
50. I
^{x — y) — 5y = 6
3f
rx«+y=y«+x— 18
• U : y = 2 : 3.
45
47
•1^ = 6.
((3x-y)(3y-x)=36
48. j x±ji ^ 5
ax' + (^ — ^) a'y — ^y* = c*
(a; + y) : (x — y) = a : h.
51. i 7-^^='"^
6*y* = acxy + m*.
52.
( X* : y* = a« : 6«
(a — X =z b — y.
53.
3^ y*
1+1 = 1;
X y c
{Vx+Vy_ ___a
Vx—Vy ^
xy = {a^--b^y.
(x^ + xyz=ia
I y" + ^y = ^«
I xy« = ft.
55
57.
59.
54.
56.
hx-\-ay __ m
bx — ay n
^Vx — yV'y ft
{
58 I ^ + ^y* = *
' I y' + x*y = ft.
60.
62,
I x*y + xy« = a
1 x»y — xy* = ft.
( X (x^ + y') = a
• (y(x»+y») = ft.
63.
fxl/x + y = a g^ fxVir2 + y' = a
1 y 1/^ + y = ft. I .V 'v •'•* + y' = ft.
460 COLLEGE ALGEBKA [8444
I (x-y) {x*+ »') = b. **"• Ix-y = 6(«»+ y«).
67. a (x — y) = 6 (x + y) = xy.
68. o(x — y) = 6 (x + y) = x» + y».
gg (x»+y» = (a+6)(x-y)'
' I X* — xy + y* = (o — 6) (x — y).
-n ((a!-y)(x + y)» = a ( (x+ y) (x«+ 3y') = a
U»+ »)(»-»)• = 6. '^•l(x-y)(x«+3y«) = 6.
■(x»_x»y + xy«-y» = ^. ' I ^d -» + y'- »*) = &•
^^ I x* + x«y + xy* + y* = a' (x — y)
■ ( x* — x*y + xy' — y* = 6' (a; + y)-
fx^ + y» = a ^g (x» + y«=130
' ( xy = 6. ' ( xy = 63.
75
77. {^-C=*' 78.j=^" + f
I XV = 21. tccv=6.
79.
81.
, xy = 151. ( xy = 6,
lx*+y*=o. Ix — y=:4.
fx+y = « 82. 1^-^ = '
( xy = 6. I a:y = 36.
(. x* -|- y* = »nxy. Ix^ — xy-|-y' = Zo.
(a?-xy+y» = 39 f x« + y« - 5(x+ y) = 8
Ux*— 3xy+2y« = 43. ( x» + y« — 3(x + y) = 28.
f3xy-2(x+y) = 28 „f,(x+xy + y = 5
""• l2xy-3(x + y) = 2. °°- 1 a? + xy + y» = 7.
so (x+xy + y=ll p„ f(x+y)»-4(x + y) = 45
°''- lx' + x'y«+y' = 49. **"• ((x-y)*- 2(x — y) = 3.
g. ( (2x - y)' - 12(2 x-y)= 189
■ jx^— 4xy + 4y' — 3x+6y = 54.
jx'-.ry + y»=13(x-y) f x» + y»- 2(* - y) = 38
^''•txy=12. '''*|xy+3(x-y) = 25.
p. fx» + y«+x-y = 12 ( x« + y«- 12 = x + y
''*-l2xy = 3(x-y). ' I xy + 8 = 2(x + y).
2444]
96. )
1 + 1 = 5
X y
SIMULTANEOUS QUADRATIC EQUATIONS
97.
461
98.
100.^
X — y = 0.3.
■ 1 , 1_3
a: y 2
1 , 1 _5
.^ »* 4"
f.(l + p=«
'^y(l+9=6.
104
ax + 6y = c
106. {-+.• = -
(ax« + y« = (a«-
l)y.
110.
112.
114.
X — \ __ q — 1
y-1 ""6-1
j»— 1 _ q»-l
(1 -f J.) g -f y) ^
(l-x)(l-y)
(l-h^)(l-V) r
(1+1=1
(a:« + y«=160.
99.
101.
^+-»*=1
m«^ n«
+ ^=1.
5±1 = 2
y + 1
^l±i=5.
l»»+l
l/» + '|/y = «
105.
107.
(Vx + Vi
\x + y = b.
x* + 3 xy* = a
+ 3x«y = 6.
f ax* + by* = ex*
( ex* — <fy* =
109. < ^ — y a —
Ix + y = a + i
111.
113.
115.
6
6.
fx=10.J'^
y+1
— 9 ,x—l
^""2 'x + l"
xy + ^ = a(x« + y«)
2/
xy-^=6(x« + y^.
*116 1^' — ^y + y'= Va« — a6 + fe«
1 y + Vxy* = t. • 1 X + y = 16.
117
^^^ fl/5-3x+x«+>/5-3y + y« = 6
2 = 5
462 COLLEGE ALGEBRA [2*^
{:
120 J^'3-x+Jx»+v3-y+iy' = 3
121 J »" -^^ ~ y^ + ^^^^ — x) = o.
; 6.
_^16
122. ^ y "^ X ~ 12 123. ^ y a; 16
'+5y' = 120
i. Jy^x~12 123. .{y
(x«-y« = 28. (3a;«.
124 f 3x«-8ary + 4y» = 0 ( 2x' - 3xy + y' = 3
U« + y»+13(x — y) = 0. 1 x« + 2a;y — 3^ = 5.
12fi f(2«+3y)(x-y) = 58 ((5a;+ 3y)(3x-5y) = 72
^^'*- l(3x - 2y) (x + y) = 132. ' ( (4x - y) (x + 4y) = 77.
i9« (*»-a!y + y' = 37 | (x + y)« = 3x'-2
(x»-2x^ + 3y« = 3(x-y)
^'*"-(2x»+xy-y« = 9(x-y).
f2x»-3.xy = 9(x-2y) ( (5x - 7y)« = 49(x -y)
^''^- I x« - 3y» = 6(x - 2y). "* * ( (3x- 5y)» = 9 (x-y).
133 i (* + 2y) (^ + 3y) - 3(x + y)
f(2x-
( (3x + 2y) (4x - 3y) = 99(x - 2y).
x + y)(3x + y) = 28(x + y).
-3y)(3x + 4y) = 39(x-2y)
135.j4±f; = ^^ = ^
134
• + y
+ ^'
(14-1 = 14-1 Cx+y z=za+h
I36.ix^y a^b 137. ]x^ + y^^x-y
«=«.-a6+6. f?^ = 4=|
138. ^^ + ,^ ^+,^ 139. I gi;^^!^
f a^ — xy+y*=o' — 06
140. 1 (-+y)' = -(=^'+y') 141. |^^i"+^^ = "
I xy = 6 (x + y). t x» 4- y* = 6xy.
142 I ** + ^^ + ^'^' + xy« + y* = a
• ( X* — x'y + ic^y* — ay* + y* = ^.
f x* + x«y + xV + a^y' + y* = «
^*'*- I x* 4- xV + ^Y + ^y' + y' = h.
{444] SIMULTANEOUS QUADRATIC EQUATIONS •ieS
144 f (» + y)* == « (a=* + y*) ,.. (x*+y* = axy
• I X* + »« = 6(a:'+ »»). *"• I x»+ j,» = 5xy. •
146.
148.
a
5i±Ji? = 6 l(x+3/)(x* + y«) = i.
( (» - y) (x» _ y») (x» — y«) (x« — y') = a
I (* + y) (x» + y») (x' + xy + y«) = 6.
(x_y)«(x'-y»)(x'-y*) = a
(x+y)»(x»+y«)(x' + y<
(« + y) (x» + y*) = axy
149 I ^* - y^* ^*' - y'^ ^** - y*) =
• l(x+y)Mx»+y«)(x' + y«)=i.
150.1;='+^;
(. (« — y) (.ar — y"; = oxy.
151 I (a: + y)Ma:» + a:y + y») = 3a(x'+y»)
^"^- 1 (x - y)» (x« - xy + y») = 3 5 (x' + y«).
-£.„ f (a^+a:y + y')(x+y)« = a(5x»+7xy + 5y*)
'lx« + y* = fc(5x« + 7xy + 5y«).
163. axy (x* + y') = a6 (x» + y*) = 6 (x« + y').
154, a (x' + y») = a6 (x« + y') = 6xy (x» + y»).
155. a (x» + y») =ab(x+y) = bxy (x' + y»).
IRB f(a: + y)»(x»+xy+y«) = 12a
'"'*• U*-y)Mx»-xy+y«)=126.
167. j-; = «- + ^y 158. j^'I^f""-?^
I y* = 6x + ay. C y** = J ay — 6x.
IRQ |(^ + y)(^* + y*) = « ifift ((^ + y)(^-/) = a
((x-y)(x*-y*) = 6. '''"• 1 (x -y) (x» + y») = 6.
— rry« = i(x»+y»)
4- ary* = 6 (x — y).
(x«y — xy« = i(x* + y*) ^^^^ |x«y
( x*y + xy* = 6 (x« — y«). ( x*y
j X* + x«y* + y* = a (x — y)«
' I X* + y* = 6 (x — y)*.
rx*+ y« = a(x4-y) Ta^ + y' = a(x4-y)
-yi~b^"~^^' ljc*-y*
166. ^ ^ + ^
a^ + ^ + y/* = (^ — y)* .
x+y b
CHAPTER XI
SYSTEMS OF SIlfULTAlfSOUS QUADRATIC EQUATIONS OF MORS
THAN TWO UNKNOWN QUANTITIES
Spioial Methods for Solving Sv'stems of Equations of Diobbss
Higher than the Second Involving Two or More
Unknown Quantities
446. The general plan of solving systems of equations of this kind
is to eliminate successively the same unknown quantity from every
different pair of the n equations in the system, and thereby to
obtain a system of n — 1 equations, which involve one less unknown
quantity than the given system. On repeating this process, the
problem can be reduced to solving a system of two equations involv-
ing two unknown quantities, the solutions of which may be found by
the previous sections. But very often special methods furnish the
the most simple solutions. It is only by practice that great skill in
solving such equations can be gained. The student will be best
instructed by the solution of a variety of examples.
Example 1. Solve the equations,
(1) yz + 2(y+z) = ll
zx-\-2(x+ z) = S
xy + 2{x + y) = 16.
From equations (1) and (2),
(4) iy + 2)z = ll-2y
(5) {x+2)z = 8 — 2a;.
By division,
(Q) y±l = n-2y.
^^ x + 2 8 — 2:r
464
8445] HIGHER SYSTEMS OF QUADRATIC EQUATIONS 465
After clearing fractions and transpoeing,
(7) 4y— 5a: = 2.
Multiply equation (3) by 4, and substitute 4y = 2 -f 6 «; thus
(8) (2+5x)x + 8x+2(2 + 5x) = 64,
or (9) 5 x« + 20x — 60 = 0.
Solve (9) and obtain x = 2 or — 6.
On substituting in (7), y = 3 or — 7.
On substituting the corresponding values of x and y in the equationfl
(4) and (5)
,= lk:i22/ = 5^1 or 25^__5
y + 2 6 -5
Hence two systems of values of x, y, z are found as solutions of the
given systems of equations,
x = 2, y = 3, 2 = 1;
and x=— 6, y = — 7, « = — 5.
Example 2. Solve the system of equations,
r (1) x« + xy + X2 = 18
](2) y* + yx+y2 = -12
((3) 2«+2x + 2y = 30.
Add equations (1), (2), and (3), member by member: thus
x« + y«+ ««+ 2xy + 2x2 4-2y« =36
or (x + y+2)»=36;
extract the square root and get
(4) x + y + 2 = ±6.
On factoring the first members of equations (1), (2), (3), and divid-
ing by x + y + z,
(5) x= J« =Jg^=^3,
(7) 2= — ^ — = -?5. = ±5.
The solutions of the system of equations (1), (2), (3), are therefore,
x = 3, y = ~2, 2 = 5;
x = — 3, y = + 2, 2 = — 5.
466 COLLEGE ALGEBRA LJ446
BxAMPLS 3. Solve the equations,
'(1) x* — yz = a*
ii
(2) y« _ xz = 6»
^(3) ^*-xy = c«.
After squaring equation (1) and subtracting from it, member from
member, the product of equations (2) and (3), member by member,
the result is
(4) {^ _ yz)% - (y« - xz) (2« - a-2/):Br [(.r' + y» + ««) _ 3 xi^i] =a*- 6«c«.
Similarly
(5) (t/« - X2)« - (22 - j-t/) (^ - yrfev [(J^ + y» + 2») - 3 :n/z] =6*-a«c«,
(6) (2« - :ry)» - (j:« - yz) {if - xz)=z [(x» + y» + ^s) _ 3 xyz] =c*- a«6«.
If the quantity in brackets is called p, then the equations (4), (5),
and (6) may be written,
(7) xp = A^ where ^ = a* — 6*c'
(8) yp^B, «» B = h^-.a^c^
(9) i5p=6; '' C=c^--a*b*
where (10) p = x'4-y'+ z' — 3xy«.
On multiplying equation (10) by p^ and substituting the values of
x'p', y^p\ z^p^j from equations (7), (8), and (9), the value of |) is
found to be
p = VA^ + i?s + (73 _ 3 ABCy
which gives, on substituting the values of Ay B, C^
p = ±i l/a« + Z;« + c« — 3 a^b^c\
The system of equations (7), (8), (9), and (10) is equivalent to the
system of equations (1), (2), and (3). By substituting the values for
p in equation (7), the values of x are found to be
^ '~ i/^ + ^c«_3a«6V
Similarly, on substituting the value of p in equations (8) and (9),
the corresponding values of y and z may be found. Only the real
values of x, y, and z have been given.
446. Special Methods. — The discussion of the solution pi ex-
amples belonging to the Types I, II, III, will be sufficient as a
general explanation of the methods to be employed ; but often special
artifices are more simple and, indeed, necessary. A device which is
frequently used to advantage is to regard the sum, the difference, the
product, or the quotient of the two unknown quantities as a new
{446] HIGHER SYSTEMS OF QUADRATIC EQUATIONS 467
unknown quantity, and to find its value first. Besides these, there
are many other artifices which may be employed to advantage, but
facility in using them can be acquired only by experience.
Example 1. Solve the equations,
f (1) x^ + y'= 126
1(2) x^-^xy+ y^=21.
On dividing (1) by 2, x + y = 6,
or (3) y = 6 — X.
On substituting from (3) in (1)
(4) x3 + (6 - x)» = 126.
After simplifjring (4)
(5) x« - 6 X + 5 = 0.
Solve (5) and get x i= b or 1,
and y = 1 ^^ 5.
Example 2. Solve,
((1) x« + 3xy = 54
1(2) xy + 4i/« = 115.
Add(l) and (2); then
(3)
(x+2y)« = 169,
therefore
(4)
x+2y = d=13,
or
1 = ±13 — 2y.
After substituting in (2),
(5) (±13~2y)y + 4^ = 115
or 2i^«±13y — 115 = 0.
(a) On solving 2 i^« + 13 y — 115 = 0,
On substituting in
x = 13-2
gin)
y )
- 23
y = 5 or — y •
X = 3 or 36.
(b) On solving 2^/' - 13y — 115 = 0,
23
y = ^ or -5.
^ 2
X = — 36 or — 3.
On substituting in )
x=— 13-2y j
The equation (4) with either (1) or (2) forms a system of equations
equivalent to the given system of equations (1) and (2). If care is
taken to observe this principle, one will readily know when all the
values of x and y have been found.
468 COLLEGE ALGEBRA [8446
BxAMPLS 3. Solve the equations,
((1) ^--y^=a^
1(2) x^y^h.
It follows from equation (2) that
6* = x* — 5 x*y + 10x»y* — 10 x«y» + 5a;y* — y»,
or, after rearranging the terms,
(3) «,6 = x*-y»-5xy(x»-y«) + 10x«y»(a;-y);
and 6' = a::* — 3 ac'y + 3 xy* — y',
= x» — y» — 3 x^ (x — y),
on substituting from (2), = x* — y* — 36xy,
i. e., (4) x» — i/» = 6»4-3 6xy.
Substitute the values of x — y, x* — y', and x* — y*, from (2), (4),
and (1) in equation (3) ; thus,
h^=:a^^bxy (6» + Zhxy) + 10x«y«6,
or 5 hx^y^ + 5 h^xy + 6* — - a* = 0.
From this quadratic equation in xy, two values for xy can be found,
sav
(5) xy = m, and xy = w,
where m and n are expressions involving a and 6; then it is neces-
sary to solve the two systems of equations,
\^-y = ^^ and j^-y = ^'
t xy = m, I xy = iu
This example may be solved in another manner. Divide equation
(1) by (2); thus
^i^^ = x*+x»y + x«y« + xy» + y*=^,
x — y b
or (6) ^ + y* + xy(x« + y») + xV« = ^.
Now since x — y = h
(7) x«+y« = 6«+2xy.
Therefore, x* + 2 x«y« + y* = 6* + 4 b^xy + 4 x«y«,
or (8) X* + y* = i^* + 4 6«xy + 2 x«y«.
Substituting the values of x* + y* and x* + y* from equations (7) and
(8) in (6); then
6* + 46«xy + 2xy + 6«xy + 2 xV +^y* = j
or 6 xV+ 5 2»*xy + M — ~= 0,
which is the same quadratic equation as that found above for deter-
mining xy.
J446] HIGHER SYSTEMS OF QUADRATIC EQUATIONS 469
£xAMPLB 4. Solve the equations,
I (1) V{x + y) + ]/(x-y) = l/a
1 (2) l/(x' + y«) + i/(x« - y«) = h.
Square both equations, thus,
X + y + X - y + 2 l/(x« - y«) = a,
x^ + y*+^-y'+ 2l/(x* - y*) = 6«;
after simplifying,
(3). 2y'(x»-.V«) = a-2x,
(4) 2l/(x*-y)=6»-2x«.
Square and simplify equations (3) and (4) ; then
(5) 2y« = 2ax-^
(6) 4y* — 46«a:« + 6* = 0.
Substitute the square of 2y* from (5) in (6): thus,
4aV — 2a»x + ^-46«x2+ 6* = 0,
or (7) 4(a« - 6«)x« - 2 a»x + "i-il^ = o.
4
Solve (7) and get x="'±ff_^-°^.
The corresponding values of y follow from equation (5),
y-^Ul^i"'^'"''-^'^)}-
In regard to the number of solutions which a system of rational
integral equations involving a given number of variables may have,
some principles may be stated as a guide which are proved in
advanced works on the theory of equations.
The only perfectly general case in which the solution of a system of
any given number of equations involving the same number of variables
will depend on a quadratic equation j is that in which one is of the
second degree and the remainder are of the first degree (J441).
If there is a system of three equations of the m***, n***, and p^^
degree respectively in three variables x, y, «, the system will in general
have mnp solutions.
This principle may be generalized for r equations involving r
variables to various powers.
470 COLLEGE ALGEBRA [**«
XZKROISB LXZVm
Solve examples 1 - 123 by special methods.
J (a;l/y + y=40 /V^+V?=20
■ 1 x«y + y»= 1312. ^' I V^ + V'y = 6.
1 x»+ y»=206. ■ I ca:»+ rfy»= i.
fx4-y = 6 |a; + y = a
'•W+y«=97. °-(x*+y«=i.
(a;+y = 4 fx+y = a
"• ( a*+ y»= 244. ( a:«4- y»= 6.
^^' i «• — y« = i. ^^' i x» — y» = 6.
13 I _a'-y = « i4l*~* = ^
■ ( V* — Vy = 6. 1 «•— **= 1023.
15. -j y 16. j y
( xy ■=. 45. (^ ary = 6.
19.
18. -j y+^i+y*
I (a!+l/l+x»)(y+»/l+/)=*'-
ra!»-y»=66 . !r» + y«+ xy (x + y) = 68
^*- j 5 a;«_ 7 y«- 4333. -^^^ | x^ y«i + y«r^ 5.
23 |^'-y' = ^
• ((x + y + a)«+(x-y+a)« = ^.
2/ « 25. ] y "■ a?
3a?y + 2x + y = 485. ( cx^^ + cfx + cy = A.
( 3^ + y _ q* + y ( i/jT 4- i/y _
26. J a:-y~a«-6« 27. J i/i-i/y"''
U46] HIGHER SYSTEMS OF QUADRATIC EQUATIONS 471
bx -{- ay = c {ah — xy).
( xy^ = b.
32.
34.
36.
38.
40.
42.
, fx + y + x« + y« = 2a
• (x — y + x« — y« = 2 6.
f(x«+y«)(x»+y')=455
1 a + y = 5.
((a:* + y')(x» + y») = 4374
( a: + y = 9.
fa; + y+v/^ = 19
1 a:«+y» = 97.
{x*+ xy + y* = a
x«+y« = 6.
a^ — yx = a^y
6«x.
+ y" + ^ + y = i8
xy = 6.
39
41
= xy
+ x' + y* = a.
(ix^ — yx = i
1 xy — y* = I
rx«
' (x+y
r X* — xy z=: a
\ xy — y* = b,
^g r2xy=:(a-6)(x + y)
44. X* + y» = xy = X + y.
45. x» + y« = 10 xy — 5 (x + y) = 5 (xy — 1).
46. x» + y» = 7 xy = 28 (x + y).
47. x» + y» = 4 x» — I xy + 4y« = 13 (x + y).
19 x« — 26 xy + 19 y» = 91
47 x« — 26 xy + 47 y« = 91 (x + y).
49. ^a;«-JT/+s/« 6 50. J a,-* - ary + y*
(x»+y»= 26. (4x*+7x«y«+4y*= 16.
48,
51
(2^+y')(x+y) = a
xy (x + y) = 6.
53. x« + xy + y» =
P.9 fx» + y» = a(x« + y«)
Xx^y + xy^ = b(x^+y*).
b_
xy
^^ + S/«
54. (x + y)(x»+y«) = U(^ + g=a.
r X* + x«y« + y* = a
^^•]x« + xy + y«= A.
f X* — x'y' + y* = 4a
• i x« — xy + y« = ±2 .
172
57.
59.
60.
62.
63.
65.
67.
69.
70.
72.
74.
76.
78.
80.
82.
f(x+y)
COLLEGE ALGEBRA
58.
[3446
l(x-y)(x«-y«) =
61.
lia^+xy + y*) i/x« + y' = a
[ (x« — xy + y«) l/x« + y« = 6.
X» — y» =
(a -x)« - 2 (6 - y)« = (a-x) (6 ~ y)
X — y = 3(a — 6).
(a - x)« + (6 - y) = c
ar + y
r»_y3=a^;
x»+y»=6j^±i^.
L \^-y
{
|(a-x)«+(6-y:
I (a — x) (/> — y) =
C g~~3r . 6 — y_34
Jft — y a — X 16
(x-y=3(a-6).
Cx + y = a
(6-y"^ X 2'
64.
66.
68.
(7 + x)«+(5-y)«= 109
y) = 30.
-8 16
f(7 + x)M
l(7 + x)(5
Js-y^u;-
(x+y=13
3 15
— y = m
6
r (q - jr)g4- (g - :r) y 4- ?/* ^ 49
) (a-x)«-(a-j:)y + y» 19
( X — y = 6.
I x«y = (a — x)'
Uy« = (6-y)».
lx* + y* = 6.
{.y X
T» + t/» _121
u^+x' 13
X + y = 2.
fx+y = o
l(m+a!)»+(n + y)»=6.
fx — y = 3
t(x-4)'+(7-y)»=72.
|x+y = 3
73
75
fx»=(a-«)'(ft-y)
'^- (y» = (a-x)(t-y)».
I x+ y = «
I x» + y» = 6.
rx+y = a
/ ■!*+?/»_ 122
77. -ji^+y* 41
(x+y = 4.
79 iVm + x + Vn+y = o
■ (x+y = 6.
|xH-j^444
°-'- lVx+10 + Vy+ 14=12.
{X — y = 1
{446] HIGHER SYSTEMS OF QUADRATIC EQUATIONS ^^^3
lx+i/ = 2
((x+l)^ + (y-2)^ = 211. ''''
1 X + y = 5.
Jx — y = 50
(V143+X — ViPl8 = l.
86,
87.
j(x«-x+l)(y«-y+l) = 3
l(x+l)(3/+l) = 6.
88 l^+y = 5(^ — y)
• lx + y« = 2(x-y).
90. |^+3/ = « + y' = H^y
89. I
^ + 3^ = « (^ — y*)
x«+y = 6(x« — y«).
91.
{
.x«+y = x+y«=|j(x«+y*).
x« + x»y» + y»=17
92.
93.
95.
97,
99.
x+ X1/ + y = b.
(x+ y) (x + 2y) (x + 3y) = 120
x + 3y = 7.
x*+y*=2a(x« + y«)
xy
^ x« + y« = a
(x* — xy + y* = 2a
(x* — x«y« + y* = 26.
fx(a
l2x
' 1 xy =6.
96.
98.
100.
101 i^+y* = "(^+y)'
' 1 xy = 6(x + y).
r X* — y' = a
Ca; — y j* +
102.
103.
105.
107.
|x« + y« = a
( x* + y* = 26xy.
r xy r= 2 a
^ x« + y« rr= m
•j af 6^: — g V
(y» "" ax — by
|x*+y* = 2a(x + y)«
* lx» + y«=26(x + y).
' X* — y* = a
y + y*"^
2^i^-3i
x+l __ 2/x — 1 \
x«-i-3r+l _ 13/jr— 1\«
I04.j^-H.v»^
ri4-3: + ^ _
106.
108.
x+l __ 3/3: — 1\
y+1 Ajz-i;
ir^-hx+l __ Sl/x^ — x-]-l\
.y' + y + i-sov-y + i/
474
COLLEGE ALGEBBA
[m
109.
111.
113.
116.
(l+x)(l + y)_3
a + 3*)(l + ,i*) 65
(1-^) (!-!/»)" 48
V(l-3^
= 6.
Ui-j^
x-y 2*
jy + l ar
Liy-l-2y'
117..
l+.n,
1 —
•^
119.
2a
l+a«
26
l + 6«'
2a
26
110. J ^+^"^
Ll+ai^ + xy-24l
.3:^1 + 3^)" 9
,1 — xy
' y + y q* — '»*
1 + ^ a* + jn*
3: — y 6*— n*
U~:iy""6« + »/
112.
114.
116.
118.
121.
£±JL
1 — xy
^-y
ll + a;y""l-6«
j 1/^ + 1/(1-^) (1
120.
3^ + y __
X — V
= 31
11
Ll+xy 29
\Vx(l^y) + }/y{l-^
-y) = a
^=6.
122 l^y+l/a-^') (l-y') = a
* Ul/ 1 - y « + y l/ 1 - x» = >/ 1 — 6«.
123. I
l/x(l — y) + l/y(l — x) = a
Vx(l--x)+ v/y{l-y)=6.
Solve the following systems of quadratic equations in thne
unknown quantities.
( ^ + y' + «" = 84 r a:(y + «) = a
124. )x+y+z =14 125. K(a; + 2) = ft
( xy = S, {^z(x + y)z=ic.
(•^ — y +«)(•«+ y — «)(a5 + y+ a) = 4a«(y + «— «)
126. i(y — x+z){x + y.
..... «)(a; + y + a) = 46«(x + «-y)
(aJ — y + «) (y — «+«)(» + y +»)= 4 c*(x + y — «).
Hi6]
HIGHER SYSTEMS OF QUADRATIC EQUATIONS 476
127.-
y+2
x+z
(
'x + y + z=a
129..
xy-h
1
xyz = c.
xz —y*
131.
« + y+« = 2i
.a^+«»-y»=117
X y z
133..
^ x^y^z
x^y^z
135. .
r x"y2 = a
137. -
xy»^ = h
, xya* = c.
'^ ac = oyz
139. -
y = hxz
L 2 = cxy.
' x(y + 2) = a
141. -
y(x+z) = fe
. 2(ic + y) := c.
128.] (6-y)«=|>'
' xy 4- xz + yaj = a
130. ^ X — y = 5
y — « = c.
x« + y« + z« = A!
•1
132
r x« + y» + «» = A!
J ax + a'y + a"z = 0
Vhx+h'y +h"z = 0.
134.
yz
136.
y+z 6+c
_jrz aryz ^
x-\- z a-\-c
a?/ xyz
^ ar + y ~a+6
f x»-ft/»+z»^^
y+2-x
x+z — y
x + y—z
z.
143.
145.
x(x+ y +z) =a — yz
y(x + y+z) = b—xz
z{x+ y + z) = c^yx.
140.
142.
144.
X* = ayz
y' = hxz
z' = cxy.
(x+ y) (x+ 2) = a
(^ + y)(y+«) = ft
(x + z) (y + z) = c.
ic* — (y — zY =. a
y^-{x^zy = h
z« — (x — y)« = c.
1
(x +y — z) (x — y + z) = a
(y + z — x) (y — z + x) = i»
(aj + X — y) (z — X + y) = c.
476
COLLEGE ALGEBRA
[{446
146.
147.
149.
151.
153.
(y+z){2x + y + z) = h+c
(x+z){2!/ + x+z) = a+c
(x + y)(2z+x+y)=a+h.
X = a*(x + y + z)yz
y = h\x + y + z)xz
z = c*(x + y + z)xy.
« y a?
X , 2 __ 6
2 ar "~y
y X z
X + y = at*
X — y = itt
x" + y* = ctt.
X + y =2tt
x^ + y' = 5 f*
x»4. y«=7M*.
X 4- y = au
a : y = y : «
x+y + 2 =19
aj» + y«+«« = 133.
155. \ x«+y« = 5
. x' + y*= c.
148.
150. -
152.
154.
156. J 3-11 = !:?
X
+ y
= 5tf
X
— y
= 2«
x»
+ 2^
= 18511.
X
+ y
= au
x»
+ y'
= bt^
x»
+ y*
= C*tt.
1
2*
^^^
2a
1
1
!/»"
26
1
+^
_ 1
157.
158.
160.
161.
163.
X : y = y : z
X + y + a = 21
(x- y)« + {x-z)* + (y - ;5)« = 126.
x4- y = 2az
a?4.y« = 26^» 159.
«" + y" + «* = c*.
x(y-l)(K-l) = 2a
x8(y« _!)(„_ 1)8 = 4 i,„
L XV - 1) (« — 1)' = 6 CM«.
1x« + (y — 2)« = a
y»+(x-;r)«=fe 162.
2;»+(x-y)* = c.
r y' + 2* — ic(y + «) = a
I »'+y' — 2(a;+y) = c.
(1-xy) (;^+l) = 2
(x-y) (2 + l) = 2a
(x«-y«)(« + l)'=^*'-
X* — yz = a
y* — x« = 6
«• — xy = c.
2x« + y«--y2 + «" = »
2y« + x« — xa + *' = *
2 2«4.x* — xy + y' = *'
J446] HIGHER SYSTEMS OF QUADRATIC EQUATIONS 477
165. -
y-1
166. H^
r 2a;« + x(y+z).
2ac^ + x(y+ z) — zy=.a
167. -! 2y* + .v(x4-2) — 252=6
- ay = c.
(3/* + y^j + 2* = a*
a:«+xz + 2« = 6«
x» + xy + y* = c«.
169. ^
a^ + y'_ tt
(x-y)» 6
(ar— y)«_ c
Solve the following systems of equations in three, four, or more
unknown quantities.
170. -
172. ^
174. -
x + y = 7
V+ 17 = 3
a;+t4«=8
y+t;« = 4.
xy = w»
x + y = 16
tt+ r = 14
? + !^=4.
L t^ y
xy = a
tti7 = a
X -|- i« = &
y + V = c.
171. J
a; + y = 12
w + V = 4
x«+it» = 34
y«+t?«=50.
173.
175. J
' xy = wy
x+ y = a
t4+ V = 6
^±^ = c.
xy = 24
W17 = 6
x+ii = 14
y + t? = 4.
176.
178.
xy = uv
^+y* = a
^ tt« + r« = 6
x+y + «+t?= c.
x-fy = tt
x+ tt = V
tt — t;
a;« -^ y« -I- tl^ + t;« = 3 ml
r x« + y« = a
177.^^'+^* = '
ary 4- ttv = c
xyur = d.
179.^
x + tt = y
3.y— tt
x« 4- y«+ M« + »• = 15 m«.
478
COLLEGE ALGEBRA
[i446
180.
182.
184. J
x + y = 16
u+ V = 12
xy -\- nv := 95
XM -{- yv = 100.
x« + y« = a
ux -|- ^y = <j
rx + wy = rf.
(^ + y)" + (u + t;)« = a
(x + nY + {y+vy=zb
(x + v)* + {y+uy=c
X -\- y -\- u -{- t; = m.
181.
183. -
( X u — g y __ !*r"_^
185. ^ y+2~ a ' a: + z 6
((a+ 6+ c)«(x' + y« + ;?«) +
x«+y«= 17
w« -I- t?« = 13
xy -\- vu-^ 10
x?f -f yv = 14.
2x = y(L+x«)
2y = u(l + y«)
2 M = t? (1 + u«)
2 V = a (1 + tJ«).
ar + y c
4 (x + y + 2)«u« = m«.
186. K+2 a — 2m*
/ x« + y« + 2« =
187.
2 a — M y 26 — M
2 c— u
x+z 6 — 2m a; + y c — 2m
XM = yz
X* + M* = y* -f z'
aj» -(- w» + y» + «» = i».
188. ^
X -|- y = M -(- r
xy = My
X I tt
- + - =a
y »
ac« 4- y« + M« + r« = 6.
189.
191.
r xu z=i yz =. a r xu =: yz z= a
) x + y+z+}t=zb 190. j x + y + z+u = b
( x« + y« + z* + m" = c. (x» + y« + 2« + ti» = c.
•J a; + y-
(x» + y» +
XM = 2X = ^
-|- a -|- M = 6
2^ + m'^ = 31^.
192. 4
xu = yz
x + y+« + w=12
x« + y« + «« + M* = 50
x» + y5 + z» + m» = 252.
193. -I
U -\- V -\- w = 1
MX + ry -J- m:s =0
MX* -|- t-y* -{- wz^ = ^
MX^ + vy^ + wz^ = 0
MX* + vy* + wz* = ^
MX* + ry*^ + wz^ = 0.
194. -I
X + y + « r= a,
X -f- yM + zw = a,
X + y W* + 2 1'* = ttj
X + yw' + zv^ = cr
x4- yw* + sr* = a
«446] HIGHER SYSTEMS OF QUADRATIC EQUATIONS 479
195.
196.
197. ^
198. 4
199.
200.
201.
203.
205. H
xu = yz
X + I* + y + ^= 12
a:« 4- u« + 3/« + 25« = 50
aj*+i**+y*+2*= 1394.
XM = a
yz = b
«+y+ z + u=z c
a^ + y^+z^ + u^z=d,
xy -}- zu r=. a
« + y + « + t4 = ft
» — y + z — u=z d
«* + y* + ^* + ^ = c .
x» + w« = ^— c«
!,• 4- V* = ^ — C?
"^ ' z
»+ t* = c
. xy-^-uv = — c<i.
x + y + 2 + 14 = 5
xz + XI* + ya -}" yw = 6
xyz + xyu + X2m + yzu + 5 = 0
xy + zu+l=z Q.
202.
XI* = yz
« + y + ^ + w=12
«« 4. y« + «• + ti« = 50
x5+y»+«*+ u'*=8052.
xu = y«
«4"^ — y — ^ = ^
X* 4- ^' — y' — 2* = 6
x* 4- ^' — y' — «• = c.
tix 4- vy = 6
wx* 4- vy' = ^»
^±^ = 29
1 + z
?^^ = 8
z-1
£±i^ = 56
Z + M
£ziif^= 1.
2 — tt
204. -
x-l+y
X
206.
a^ — yg
l/(l-2/«)(I-2«)
y — zx
i/(l — 2«)(1— ^)
2—31/
= 5
207.
= c.
-1 = 11.4
^y
+ i. =14.85 4-^4-2^.
xy y '
x + y4- a= 0
t* 4- ^ + ^ = ^
x« 4- y« 4- a« = A:«
M« 4. t;» 4- w;« = p
ttx 4- vy 4- *o« = 0.
208.
x4-y4-t*4-t> = a
x^ + y* — 2 mxy = 4 1**
x* 4- y' -|- 2 »ixy = 4 1;*
ti* 4" ^ — 2 niiw = y*.
Put
^ and obtain
X = «u, y = <M, t> = «tt,
r + - = ±2 ^/l-m«n'
« \ l-m«
480 COLLEGE ALGEBRA [{{447, 448
Problems in Simultaneous Quadratic Equations
447. Problem L — Determine the sides of a rectangle, given the
difference, Z, between tiiese sides and the side a of a square equiva-
lent to the rectangle.
Let X be the longer and y the shorter side of the rectangle, then,
according to the first condition of the problem,
(1) x-y = Z;
by the second condition, (2) xy = a*.
Solve (1) for x, (3) x=l + y.
Substitute in (2) (4) (/ + y) y = a*;
or y*+ly — a*=0;
(5) y^zzl±l^l±^
It follows from (1) that (6) x = ^*^^^"H<
Hence the solution is
Since ^f + «•>!'
the solution given in (7) is always possible.
448. Problem II. — Find two numbers such that the sum of
their squares is 170 and their product is 77.
Let X and y be the two numbers; hence, according to the condi-
tions of the problem we have
(1) x« + y« = 170,
(2) xy= 77.
Add and subtract the equation found by multiplying (2) by 2, namely
(3) 2 xy = 154,
to and from equation (1),
then : (4) x« + 2 xy + y« = 324
(5) x« — 2xy + y«= 16.
Extract the square root of (4) and (5),
(6) x + y = ±18,
(7) X — y = ± 4.
System I is equivalent to system II, or to the systems
^x + y = 18 x + y = 18 x-f-y = — 18 x + y=— 18
■■{
"■{
4 X — y= — 4 X — y= 4 x — y= — 4.
S 3 449, 450] HIGHER SYSTEMS OF QUADRATIC EQUATIONS 481
The solution of these sets of equations in III are respectively
Xj=ll »s= 7 X, = — 7 a;^=— 11
yi= 7 y, = ii ys=-ii ^4 = - '^•
If negative numbers are excluded the only solutions are
^i =11, ^1 = 7; X, = 7, y, = 11.
449. Problem III. — Determine the sides of a right-angled
triangle, given the perimeter, 2p, and the side a of a square which
is equivalent to the given triangle.
Let X and y be the legs of the right-angled triangle and z the
hypotenuse ; then the conditions of the problem give at once for the
perimeter of the right-angled triangle,
(1) a+y+« = 2i>,
for the area (2) xy = 2 a\
for the square on the hypotenuse
(3)
a?+y« = «».
From (1) and (2)
(x+y)
*-2xy = (%p — z)*—Acf
or
(4)
x«+y' = (2p — «)» — 4a».
From (3) and (4)
(5)
z* = {2p-zf-4.a*-
solving (5)
(6)
2« = 4p« — 4p« + a« — 4a«
z-r^-'^.
" •
p
Hence from (1)
(7)
«+» =
2p-. = v-'i^=^,
but by (2)
XXjz:z2a\
therefore, by J422, x and y are the roots of
(8) t««-^^^fc-^i* + 2a« = 0.
460. Discussion, — In order that the values found for x, y, z are
compatible, they must be real and positive.
The value of z, ^~"'
P
is always real and positive if
(9) jo^>a\
Since the values of x and y are the roots of equation (8), they must
be real and positive. In order that they are real, it is necessary that
we have
(10) (£!±^y^ 8a«,
(11) a« — 2apl/2-fi?«^0.
Now the roots of a« — 2api/2 + ^* = 0
482 COLLEGE ALGEBRA [88451,452
with respect to a are ^(v 2 — l), and p{V2+ l).
Hence, in order that a* — 2 apy 2 +^ may be positive or zero as (11 )
requires, it is necessary that we have
(12) a <p(v 2-1)
or (13) a^p(v2 + l), [8489, (3).l
Inequality (13) is incompatible with the inequality
(9) p>a
which is the condition that p is positive; therefore, in order that
the problem may be possible, it is necessary that
(12) a ^j9(i/2 — l)anda<i);
since l/2 — 1< 1
(13) p{i/2-l)<p.
Hence the conditions in (12) are reduced to a single condition,
(14) a^p{\/2^iy,
when this condition is fulfilled the problem has one solution. In the
limiting case
^ a^p(V2-l)
the roots of (8) are equal (8412,* 2) and the triangle is isosceles
since then x=:y,
461. We can deduce the two theorems:
1. 0/ all right-angled triangles which have the same perimeter that
triangle is the greatest which is isosceles,
2. 0/ all right-angled triangles which have the same surface, that
which has the smallest perimeter is an isosceles triangle.
462. Problem IV. — Inscribe in a sphere of radius R a cylinder
of which the total surface is equivalent to the double of the surface
of a circle of radius a.
Let X be the radius of the base of the cylinder and 2y the height
^,y^ ^^^"""^ It follows from Fig. 1. that ac^ + y* = i?*,
that the area of the two bases of the
, cylinder is
J the lateral area of the cylinder is
/ 2 ;rx • 2 y = 4 JTxy ;
' and twice the area of the circle whose
radius is a is 2 n-a'.
^Zr^~^Z^^ Hence, we have, to determine x and y,
Figure 1. ^^ ^^ equations
J453J HIGHER SYSTEMS OF QUADRATIC EQUATIONS 483
1(2) x« + y«=/?,
a system which is equivalent to the following,
(3) y =
II-
2ar '
(4) x« + a*^2cy+^ ^ ^
or 5x*— 2 (a« + 2/?)a^ + a*=:0.
System II is easy to solve, and gives four sets of values for x and y,
468. Discussion, — In order that a set of values of x and y which
satisfy system II may be a solution of the given problem, it is
necessary and sufScient that these values are real, positive, and less
than R, When will these conditions be fulfilled?
Let ^' be a positive value which substituted for x" in (4) will
satisfy this equation. We take as the value of x the positive square
root of AT' and determine the corresponding value of y on putting
K for X in the equation
(3) y = ^-
In order that the corresponding value of y may be positive, it is
necessary that
(5) Z«<a«.
When this is true, the values of x and y are less than R^ since they
satisfy the equation
(2) x«+y«=/?«,
which is one of the equations of system I which is equivalent to
system II, from which these values have been deduced. Therefore,
the number of solutions of the problem is the number of real roots,
positive and less than a', which equation (4) can have, considered as
an equation of the second degree in x'.
In order that equation (4) may have real roots, it is necessary
^^*^ (a« + 2 /?«)«> 5 a\ [{411, 1]
or a«-(-2/?«>a«V^5,
or (6) a« < /?2l^^^|±l.
When condition (6) is fulfilled, the roots of equation (4) are real ; and
they are positive, since their sum and product are positive (2422,1,2).
It remains to determine which of these values are comprised
between 0 and a*. To decide this question substitute 0 and a* for
«• in the first member of equation (4); for x*=0, the first member of
484 COLLEGE ALGEBRA CI 454
eqaation (4) reduces to + a*, which is positive, and for x* = a*, it
becomes ^^^ 5 a*- 2 a*-4 a«/?«+ a* = 4 a«(a« - /?),
which has the same sign as a' — /?.
Two cases can arise:
I. a* — /?• < 0. When a* — R* is negative, one of the roots of
the equation in x' lies between 0 and a* (i440), and satisfies the
condition (5), that K* must be less than a'; the other root is
greater than a' (2440), but will not satisfy the condition (5).
II. a* — /?* > 0. If a* — /?* is positive and at the same time
the condition of reality (6),
z
is fulfilled, that is to say if a* lies between 7P and /?* ^ ' t the
roots of the equation in x^ either both lie between 0 and a* or both
are greater than a' (2439). In order that both roots may be less than
o* it is necessary and suflScient that their half -sum ° \ (2422)
is less than a', that is, that a* is greater than — . But this condi-
tion is fulfilled since we suppose that a^^ R^,
454. Resume, A review of the discussion of this problem leads
to the following results.
1. Problem has one solution when 0 < a' <; i?*.
2. Problem has two solutions when 7? < a' < /?• — ^ti.
3. Problem impossible when R* — ^*— -<«'.
2
4. In the limiting case when a* = /? — ^t— » the two values
of x« are equal to ^-i^ (2412) which, since a« = /?^^%ti,
is equal to ^^tJ^^.
Hence it follows that, of all cylinders inscribed in a given
sphere, that cylinder has the greatest surface which has the radius
10 '
the altitude of the cylinder in Fig. 1,
=^4
2y=:2\/R'^a^ = 2jR'^R» • ^±i^
= 2R^:
6-Vb.
10~'
|5 — v^6
10
H54] HIGHER SYSTEMS OF QUADRATIC EQUATIONS 485
and its total surface is
(8) =2itR»^^^^2kRx^^^ R.
Geometrical interpretation of the result in 4: The surface
expressed by (8) is equivalent to the lateral surface of a cylinder.
Problems leading to Equations of the Second Degree in Two
OR More Unknown Quantities
1. Find two numbers whose product is 576 and whose quotient
i9|.
2. The product of two numbers is p and their quotient is q\ what
are the numbers?
3. Two numbers are in the ratio 11 : 13, and the sum of their
squares is 14210. What are the numbers?
4. The product of two numbers multiplied by their sum is 1820,
by their difference is 546. What are the numbers?
5. The sum of two numbers and the sum of their squares added
gives 686. The difference of the' numbers and the difference of
their squares added gives 74. What are the numbers?
6. The product of the sum and difference of two numbers is a ;
the ratio of their sum to their difference is p : q. W^hat are the
numbers?
7. If the figures in a number of two digits are reversed, the new
number is 18 less than the given number. The produc t of the tw
numbers is 1008. What are the numbers?
8. A grocer buys $44 worth each of coffee and sugar and receives
90 lbs. more of the latter than of the former. He sells 57 lbs. of
sugar and 29 lbs. of coffee at a profit of 20 per cent, receiving $31.
How many pounds of sugar and of coffee did he buy?
486 COLLEGE ALGEBRA [M54
9. A and B together invested $8000 in the same business. A
allowed his money to remain ten months and received for his invest-
ment and gain $4125. B allowed his money to remain eight months
and received for his investment and gain $4590. How much money
did each invest?
10. 60 lbs. of Java coffee cost $4 less than 60 lbs. of Mocha.
A man purchases $8^^^ worth of each kind, and receives 8 lbs.
more of Java than of Mocha. Find the cost of each kind of coffee.
11. Find two numbers whose sum is nine times their difference
and whose product diminished by the greater number is equal to
twelve times the greater number divided by the less.
12. Two workmen were employed at different wages and paid at
the end of a certain time ; the first received $24 and the second,
who had worked 6 days less, received $13^. If the second had
worked all the time and the first had lost 6 days, they would have
received the same sum. How many days did each work, and what
were the wages of each?
13. A vessel can be filled with water by two pipes; by one of
them alone the vessel would be filled 2 hours sooner than by the
other; by both pipes together it can be filled in IJ hours. Find
the time which each pipe alone would take to fill the vessel.
14. The floor of a room has. 273 square feet, one of the walls
189 square feet, and an adjacent wall 117 square feet. How long,
broad, and high is the room?
15. The length, breadth, and height of a stone which has rec-
tangular faces have the ratios 5:3:1. The entire surface of the
stone is 15.94 square feet. What are the length, breadth, and thick-
ness of the stone?
16. Determine three numbers such that the product of the first
and the second is /w, the product of the first and third is n, and the
product of the second and third is p.
17. Determine four numbers such that the products of any three
successive numbers are respectively Z, m, n, p.
2454] HIGHER SYSTEMS OF QUADRATIC EQUATIONS 487
18. The diagonals of the three rectangular faces of a rectangular
parallelopiped, which meet in a vertex of the solid, are respectively
a, by c. What is the area of the three faces?
19. If the first digit of a number containing 6 figures is inter-
changed with the fourth digit, the second with the fifth, the third
with the sixth, a number is formed which, multiplied by the given
number, is 122,448,734,694, and which, diminished by the first, gives
a remainder which is five times the first number. What is the
number?
20. Calculate the sides of a right-angled triangle, given the
perimeter 2p of the triangle and that the volumes of the solids gen-
erated by revolving the right-angled triangle about the two legs of
the triangle is one half of the volume of the sphere whose radius is r.
Suggestion.— let r and y be the legs and z the hypotenuse of the
triangle; then the equations will be
x + y+z = 2p
xy{x+y) = 2r^
x'+yi^zz^ etc.
21. Find the four terms of a proportion, given that the sum of
the extremes is 21, that of the means 19, and the sum of the
squares of the four terms is 442.
22. Find the sides of a right-angled triangle, given the altitude
h (hypotenuse being the base), and the difference, c, between the
legs.
23. If the sum of any two of three numbers is multiplied suc-
cessively by the third, the successive products are respectively 810,
680, and 512. What are the numbers?
24. Four quantities are in a proportion. The product of the ex-
tremes is a, the sum of the first two terms is 6, and the sum of the
last two terms is c What are the four terms?
25. Four quantities are in a proportion. The sum of the first
and fourth is a ; the sum of the second and third is 6 ; and the sum
of the squares of the four quantities is c. What are the numbers?
CHAPTER XII
GRAPHICAL REPRESENTATION OF THE SOLUTIONS OF SYSTEMS OF
SIMULTANEOUS QUADRATIC EQUATIONS
466. Graph of the General Quadratic Function ax' + bx+ c.
This problem is illustrated by the following example:
Example. — Plot the equation y = «* — 4 x — 5.
In the table below are arranged the various values of y which
correspond respectively to values x = 0, +1, +2, ,
— 1, — 2, — 3, etc., in the equation y = x* — 4x — 5.
FoRy = x«
-4x-5
X
y
0
- 6
+ 1
- 8
+ 2
- 9
+ 3
- 8
+ 4
- 5
+ 5
0
+ 6
+ 7
etc.
etc.
— 1
0
-2
+ 7
-3
+16
etc.
etc.
: !
i l\i •
+19 ; ,
,/
! !
: !" V I
1 !
J
1 !
i • 1 '
1
1
: !
: I \ ^
I
; : \ ;
1 '
/!
: i ;\ :
f.5; ;
#■
...
...,
! :
i ■ ' V '
: :
i 1 Mi-
1 :
i i ' u
I ;
' !^
-lb j
-5' ; ; \
-^5
;
^
b2
: ; i \
0 ; ;
; !
' 1 ' it
: :
;
.|..L
-[---Ua
-L-i—
—
--I--
...
-
...
-»; !
y ' i
j
;
-
""r""r"'t'"t"
\r-
/
"
...j...
—
...
...
-.
-
...i.-i.-i-.L.
-'T i "
--
-1...
....
....
...
FiGUKE 1
NoTB 1.— It Is clear that in case the values of x increase beyond those gl^en In the
table, each corresponding value of y will be larger than the one preceding; hence the
values of x and y given in the table are sufficient to determine the ultimate directions
of the curve. In general, when this is found to be the case, one need not compute
more values for the table.
NoTB 2. —If the graph of an equation of the second degree in two variables consists
of a single branch which extends to a part of the plane at an infinite distance from
the origin, it is called a parabola.
S456]
GRAPHS OF QUADRATIC EXPRESSIONS
489
466. Type I.
(J441.)
Graphs of curves of the quadratic form
Ax* + 2 Bxy + Cy*+2Dx +2Ey+ F— 0,
of the straight line ax-\- by -\' c = 0 (1246), and
of the location of the points determined by their
solutions.
Example. — Plot the curves represented by the equations
y = 3x — 12 and a^ ^ y* = IQ
and the points (2243) represented by the solutions of these equations.
For y = 3x— 12
Foi
y=
I y = il^ac'— 16
-l_l/x«— 16
X
y
x 1 y
X
y
0
-12
0
±v^— 16 imaginary
0
± 1/-16
+ 1
- 9
+ 1
± V-\b
— 1
± V"— 15
+ 2
- 6
+ 2
± l/-12
- 2
± 1/— 12
+ 3
- 3
+ 3
± V-1
— 3
±l/-7
+ 4
0
+ 4
0
— 4
0
+ 5
+ 3
+ 5
±3
- 5
±3
+ 6
+ 6
+ 6
± 2/5= ±4.47
- 6
±4.47
+ 7
+ 9
+ 7
± 1/33= ±5.74
- 7
±5.74
+ 00
+ 00
+ 00
±00
— 00
±00
1
2
3
Plotting the points in the first table gives the line BAPST, Fig. 2.
Plotting the points in the second table gives the curve R' ^P'APQR.
Plotting the points in the third table gives the curve on the left,
LA* My which is in every respect equal to the curve on the right.
The curves L^' if and R* AR are called the branches of the graph
represented by the equation ^/^ = x* — 1 6. *
• When the graph of a curve of the second degree In two variables connists of two
branches, each of which extends to infinity, the graph is called an hyperbola. For exam-
pie. f • = x« — 16 is the equation of an hyperbola (Fig. 2).
490
(X)LLEGE ALGEBRA
[2456
In reckoning the values of y which correspond to the values of x,
in case of the hyperbola y' = x* — 16, we notice that for one value of
X there correspond two values of y which are equal and opposite in
sign. The same is true for the values of x which correspond to a
value of y. For this reason the hyperbola is said to be iymmetrical
to the X- and y- axis.
X'
...l.l..L.L_
""
+IP
i / i i
V-
[ yT 1 "i
S
Shl-i-i-
i/i YJ^
SJ-
4 \/^\
\
/^/R i •
\
+ 5
l/^\ i '
1 1 ^
V\ \ \ \
-IC
...;..;..;...
-B
...
...,
...
--
...
1 ^'
p ! 1 ; :
: i ; +10
0
; hA
...;.-;...!.-..
-
.-
r-
-4-
■\
p'l 1 i i
\ ''''*' J
>^'| ; ';'
-5
; Sb'I j_
/
t^i i i
....
—
...
--
—
7
...;.-.
...
..|..p^L
: I !
/
I ; : :
: : 1
A
/
i i
i : :
A
B ;
r : : ;
Y'
Figure 2
The solutions of the equations are
(x = 5
which are represented by the points A and P respectively.
NoTK.— The points corresponding to tbe imaginary results a; = 0, y = d: i/— 14, etc^
are not situated on tbe hyperbola.
8457]
GRAPHS OF QUADRATIC EXPRESSIONS
491
467. Type II.
({442.)
Graphs of curves of the quadratic forms
ax* + hxy -j- cy* = d and Ao^ + ^^V + C^ = /?,
and of the location of the points determined by
^ their solutions.
Example. — Plot the curves repre-
sented by the equations
and
12a?+13y« = 248
x^>^ '
_ ' 1 1 1
■-/QNi--f--
:mx
\'\")r[ INrt""]/'
: ! 4 ; ! ; 1 :
and the points represented by their
solutions.
Figure 3
Plotting the points corresponding to the plus values of x and
both plus and minus values of y in the second table, we get the branch
curve BFAQB'] similarly, for the minus values of x and the corre-
F0K,.-«'^-^>
For
,_ 248-12^
^ - 13
X
y
X
y
0
=h -^—:^ imaginary
0
±V^f-±<.ae...
itl
±0
±2
± 4.2G • • •
± 3.92 • • •
±2
±1.48
dz3
± 3.28 • • •
zt3
=t 3.30 • • •
±4
=t 2.075
±4
zh 4.5-J6
db5
0
± V^ — 4 imaginary
it 00
±00
etc.
1 2
sponding plus and minus values of y we get the branch curve
BSA'RB\ These two branches make up the entire curve BAB'A'B
which is the graph of the first equation.
Similarly, by plotting the values of x and y in the first table, we
get the graph of the second equation, the hyperbola whose branches
are PA^Q and SA'^R. The points corresponding to the imaginary
values of x and y are not points on either of these curves. The
solutions of this system of simultaneous quadratic equations are
492
COLLEGE ALGEBRA
[3458
( :ri=2V'3=3.46 . . ( Xi= 3.46 . . ( Xs=-3.46 . . ( X4=-3.46 . .
( yi=2l/2=2.82 . . ( ^,=-2.82 . . \ys= 2.82 . . ( y4=-2.82 . .
The points corresponding to these pairs of values of x and y are
the intersections of the two graphs (Fig. 3), namely, P, §, i?, S,
KoTx 1.— In case the graph of an equation of the second degree In two rariables Is a
closed curve, the graph Is called an ellipte. For example, 12x< + 18y* = 248 is the
equation of an ellipse.
Notb2.— In case the equation has the form x' + y' = 16 = 4«, the curve is a drele
whose radius Is 4.
468. Graphs of miscellaneous quadratic forms.
Example. — Plot the equations,
(1) x« + y« — 2 a:y — 4 X — 8 y — 20 = 0
and (2) xy = -
The first equation may be written
(1) y«-2(x+4)y+(x«-4x-20) =0.
= -2.
Solving Equation (1)
Solving Equation (2) 1
y = x
+4d=2v/3(x + 3)
y =
— Z
X
X
y
X
y
0
+ 1
4± 2i/9 = + 10 or -2
5 ± 21^12 = + 11.93 or —1.93
0
+ 00»
+ 10
+ 8
+ 2
6 ± 2/15 = + 13.75 or - 1.75
-1
+ 2
+ 3
7 ± 61^2 = + 15.48 or —1.48
— 2
-3
+ 1
+ t
+ 4
2^21^6
8 ± 2v/2r= + 17.16 or — 1.16
0
— 4
-5
-6
— 00
+ »
0
+ 7
+ «
11 ±2V30 = + 21M or +0.04
12 ± 2v/33 = + 23.49 or +0.51
0
+ i
+ 1
— 00*
— 4
-2
-1
-2
3 ±21^6 = + 7.9 or -1.9
2 ± 2i/3 = + 5.46 or —1.46
+ 2
+ 3
+ 4
— 1
— 3
— 4
1 ± 0 = + 1. The two values of
y are equal.
0 ± 2i/— 3 imaginary
+ 5
+ 0
+ 7
+ 00
-1
0
1 2
♦ The student should note that, In equation (2), when x is—, y is +; and that, as x
approaches 0 through uegative values, y is -f- and approaches +ao . Similarly, when x
is -f . y Is—; and as x approaches 0 through positive values, y approaches — «. Thus
as X passes through 0 from positive to negative va'ues, y changes sign from — * to +Q0 .
8458]
GRAPHS OF QUADRATIC EXPRESSIONS
493
1 To plot the points corresponding to pairs of values of x and
y (two values of y for each value of x) in the first table, one point is
located for each pair of values, i. e. , two points for every value of x.
The graph corresponding to the first table is the parabola in
Fig. 4.
X'
1 1 1
ALU...
i ! '
I • I I
/ill
: : : ; p
+.o! I 1 1
I I !
: [ j 1/
III*
J-— r — -f---
•-4-T--i/r-
— ♦-—-»— -t----» —
I : !
\ \ f \
'III
--H/t-ij
+5 1 1 1 :
• 1
\ i \ J
.X ^ ..• ^
; i i
[ I t^'
...i...i...|^
1 ~-
Tr^r-j;^ i
JK ; : [+8
: J i^^
TnV; >
0 ; ; ] ' ^
<t^
\ : :
; ! ; ;
c /"i ; :
1 1 1 •
7: : 1 j
' ' ' !
- 5 ; •' I I
: ! !
: ; i i
1 : : :
Figure 4
The points corresponding to the first values of y are found in
that part of the graph represented by BPA, and to the second values
of y are found in that part of the graph represented by ACD.
The points corresponding to x = 0, y = d= 2i/— 3, are not on the
parabola.
2. The graph of a;y = — 2 is shown in Fig. 4; it has two branches
extending to infinity, one lying in the angle YOX* and the other in
the angle XO V. The curve is therefore an hyperbola.
On eliminating y between equations (1) and (2) we get
X* — 4 x» — 16 X* + 16 X + 4 = 0.
494 COLLEGE ALGEBRA [S458
The values of x in this equation are the abscissae of the points of
intersection of the curves (1) and (2), namely $, P, /?, S\ but
the equation can not be factored and we have not yet had a method
for solving an equation of the fourth degree ; however, a careful
plotting of curves (1) and (2) shows approximately what the values
of X and y are which satisfy the given equations (Fig. 4), namely,
Kx-=.ON, (x=OM, (x^OK, ^\^^(
^ \y = AN, ^ Xy^PM, ^\y = KR, U=.
EZEBOISE TiXXTX
Determine the graphs of the following systems of simultaneous
equations in x and y in Exercise LXXYII, and locate the points
represented by their solutions :
Examples 5-12, 27-32, 37-40.
CHAPTER XIII
GRAPHS OF QUADRATIC EXPRESSIONS AND PROBLEMS IN MAXIMA
AND MINIMA WHICH CAN BE SOLVED BY EQUATIONS
OF THE SECOND DEGREE
469. If a variable quantity y, which, having increased contin-
ually for a given time, then decreases continually, passes through
a value greater than its neighboring values, i. e. , those which imme-
diately precede and those which immediately follow, it is said that y
passes through a maximum.
On the contrary, if a variable quantity 2/, which, having decreased
continually for a given time, then increases continually, passes
through a value less than those which immediately precede and
those which immediately follow, it is said that y passes through a
minimum.
Consider, for example, the sections of a series of ridges and
valleys made by a vertical plane. Suppose this section to be repre-
sented by the curve MN, and that the heights of the different points
of this curve above the horizontal plane PQ are measured. The
summit, A, of a ridge will be a maximum and the bottom, B, of a
valley will be a minimum. If one travels throughout the length of
C
A
Figure 1
the curve, he will ascend till he reaches the point A, then he will
descend till he arrives at B] the height A^A of the point A is greater
than that of the neighboring points which immediately precede or
follow; the point A is therefore a maximum. On proceeding from
495
496 COLLEGE ALGEBRA [M60
A he descends to the point B^ then ascends till C is reached. The
height B'B of the point B^ is less than that of its neighboring
points either to the left or to the right; B'B is therefore a minimum.
Similarly (7(7 will be a second maximum and DD a second mini-
mum ; and so on.
Some simple problems will be investigated which can be solved
by the equation of the second degree or by simple polynomials.
460. Problem I. — Consider the variation of the product of two
quantities whose sum is a constant a.
Let X be one of the quantities and a — x the other, of which the
sum a — x-f- 2c = «; and let y be their product whose variation will
be studied. Here, then,
(1) y = ar(a — x).
It is seen that y = 0, for x = 0 or for x = a. Hence, as x increases
continuously from 0 to a, y increases continuously to a certain value
and then decreases to zero. All of these values of y are finite,
since none of the talues of x is greater than a. It is seen, therefore,
that y must, by definition, pass through a maximum value for some
value of X greater than zero and less than a.
To decide what value or values of x < a and >. 0 will make y
a maximum, write (1) in the form,
(2) y = ax-x'=:+J-(J-ax+x«)
On inspecting this formula, it is seen that y will have the greatest
value for 0<aj<;a, when the least quantity is subtracted from
^ ; i. e., (^— x) =0. This will happen when x=: 5; therefore,
y = - • 0= T* Hence, the product of two factors whose sum is a,
will be a maximum when these two factors are equal to each other.
It is easy to follow the variations of the product y as « varies
from 5c = 0 to X = a. When x increases continuously from x = 0 to
x=z% the term to be subtracted becomes smaller and smaller and
finally becomes 0 for x = - ; the product y increases continuously
from zero to the maximum value — . When x becomes greater than
^ and increases until it becomes equal to a, the term to be sub-
tracted, ^^— x^ or /^x —^) i increases more and more, and y
decreases from ^ to zero.
8461]
GRAPHS OF QUADRATIC EXPRESSIONS
497
B
a-x
Figure 2
Suppose that x increases beyond x = a; then the factor a — jc is
negative, and the product cr(a — x) will be negative and will
Increase in absolute value; therefore the relative value of y dimin-
ishes. The process is similar if x takes negative values.
This problem has the following geometric interpretation: a — x
may be considered the base and x the altitude of a rectangle
ABCD) then y will be the area of ABCD. ^ ^
The condition that ac + (a — x) or AB -\- BChe
a constant a will here mean that the perimeter
of the rectangle is always a constant quantity
2 a. Therefore, the problem may be stated
geometrically as follows: Study the variation
qf the area of a rectangle whose perimeter is a constant 2 a.
The variation of the area of A BCD, as x increases from 0 to a
can be followed easily: when x=0, the base is a and the altitude 0;
therefore the area is 0. Suppose that the altitude increases from 0
to ^; then the area increases continually from 0 to the maximum
value ^' As x increases from - to a, then the base decreases from
^ to 0, and, therefore, the area diminishes from the maximum
value ^ to that of a straight line of length a, and is zero. In this
discussion it has been assumed that x can be neither greater than a
nor negative.
461. Problem II. — Find the largest rectangle which can be
inscribed in a given triangle.
Let a be the altitude and b the
base of the given triangle ABC, and
let X and y be the altitude and the base
of the rectangle to be determined.
The area of the rectangle DEFG
will be . .
But
Hence,
and by (1) (2)
Whence,
and
u =zxy
CB _NA
DE "" MA
■^^= ^;rr» i. e., - =
u = * (ax - sc«) .
tx* — abx -|- a» = 0
ab± Vcfi^—iabn
26
498
COLLE(iE ALGEBRA
[{{462, 463
The largest value which u can have, for which the values of x are real,
is that value of u which will make the radical zero, namely,
a«6«— 4a6u = 0.
ab
Hence,
The corresponding values of x and y will be
ah a
"= = 26 = 2-
and
whence,
6/ a\ h
u = .ry = "- = -(Bxea of the triangle Y
Therefore, the maximum rectangle which can be inscribed in a given
triangle has one-half the area of the triangle.
462. Graph of variation of u =z - {ax — x').
Suppose that 6=9 and a = 3, then it is necessary to consider
the equation
M = 3 (3 ^ — x') = 3x (3 — x).
It is seen that m = 0, when x = 0, or x = 3. For
x = l, «=:6; x=-jm=> x = 2, m = 6.
Hence, as x increases from 0 to 3, u increases
27
from 0 to its maximum value — , then decreases
4
to 0. If X = 4,. M = — 12. For values of x
greater than 3, u is negative, and, as x ap-
* proaches + oo , w approaches — oo . If x = — 1,
u = — 12, therefore, as x approaches — oo ,
Figure 4 u also approaches — oo . The graph exhibiting
these results is shown by the accompanying figure, in which BP
shows the maximum value of ?« ; it z=
"']■■
•--•■
»;..
;-t-6
— f-
-i-l
A i + B
i /
■--+ -
* ! \ i
27
463. Problem III. — How does the area of a triangle vary if the
perimeter and the base are constant?
Call 2 8 the perimeter, and ri, ?>, c, the sides of the triangle, then
it is known that its area is
where
(1) y = \ s {s — a) (« — b) (8 — c),
a-\-b-\-c
8464] GRAPHS OF QUADRATIC EXPRESSIONS 499
By hypothesis two factors, « and s — a, may be regarded as constant,
and the other two factors, s — b and « — - c, as variables whose sum
2 s — b — c=a-4-^+ c — b — c = a, a constant.
Put
then
2
2 2 2*
Hence, it would be necessary to consider the variation of
(2) y = ^'^s {s — a)x{a — x)
or y = l/M^ x(a — x) = M X [x (a — x)]^,
where M* = s {s — a) = constant.
Here it will be necessary to consider the variation of the product
X (a — x), as in Problem I, and when the maximum value of this
product has been found, the maximum value of y will be its square
root times M. Since, by Problem I, the product x (a — x) is a
maximum when x = a — x, i. e. , when x = | j y is also a maximum
when X = ^ > and this maximum value will be ^\? ' 2 ~ 2 ^*
464. The graph of the curve y = Ml x (a — x).
Suppose that JIf' = 4 and a = 5; then y = 2l/x (5 — x).
Consider the points,
, fx = 0
■■1:=
5
= 4
4
x-2
2v/6;
, (x=i
, (x=3
•|y = 2l/6;
, fx = 4^
My =-3;
Y
; ' •
I
45! i^s!
!
-P;/^*
1^
.4.-
1
--^ —
J**'
• 0
-- + --4--
Q
^»!
1
;
-;...
Vi"""*'
-
7
/
- -1 ■ -
-5| |P'
• -*■ " T'"
1
Figure 5
500 CX)LLEGE ALGEBRA [5465
On plotting the points P^, P^, • • • P^, and all intermediate points,
the graph of y = 2 V'x (5 — x) iq the curve P^, P^, P,, • • • P^. If
x>5, then x(5 — x) is negative and the corresponding values of y
are imaginary, and no part of the graph will lie to the right of the
vertical line through P^. If, moreover, x<0, then x(5— -x) is also
negative and no part of the graph lies to the left of the vertical line
through P. Had the sign of the radical been — , then the graph
of y = — 2 i/x (5 — x) would be the curve P^PP^, It is clear
from the figure that the maximum value of the area of the triangle
when if = 4 and a = 5 is
465. Problem IV. — Given that the sum of two numbers is
constant; investigate the question, what is the graph of the sum of
their squares.
Let a be the sum of the numbers, of which x is one and a — x
the other, and the sum of whose squares is y; hence,
y = x' + (a — x)«.
This expression may be written
y=2x«-2ox+a« = 2(x«-ax + |')=2(x» — ax + ^'+ j*)»
i.e.. y = ? + 2(--i)'-
This equation shows that y is the sum of two positive quantities, one
of which, ^, is fixed, and the other, 2 f x — ^) is a variable.
The quantity p will take its smallest value, — , whenfx — -j is
zero, i. e. , when x = ^- If it is considered that x < ^, then x — ^
X — - j will be a positive quantity, which must be
added to ^' to find the corresponding y, which will be greater than — »
ft
the part found for y when x = -• Similarly, if it is considered that
x>^, then fx— ^) is positive, and the corresponding value of
y is greater than — • Therefore, by definition, y is a minimum
ft
for X = -• The geometrical interpretation of the preceding problem^
when a=:4j will now be given.
J{466,467] GRAPHS OF QUADRATIC EXPRESSIONS
466. The graph of y == of + a — xy.
Consider the points:
501
(x = 4 (x = +5
']y=16-, ^•]y = 26,etc.;
(x=+oo ^fx=-l
°|y= + »; My =26, etc.;
ny=l6;
°'L = +
= — 00
CX} .
On plotting the points /\ , -P, , . . . Pqo >
P/, P,' . . . Pqo' and all intermediate
points, the corresponding graph (Fig. 6) is x'
obtained.
■^-
•
~1
f^f-
+25
;
1
— 1
ij
...;.-.
f26
■-j- -
.-+...
...
...
■|-
1
p,
;
/p.
1
+15
I
...
—
...
-y 1---
...
—
i
1
+ii
'1 i
:
...
...
._,
...
Pa\
. — +.4-
4; • 1
! 1 :
i
J--1--4--
...
r —
h--
—1-4-
..-4.4.
.-Lj...;..
— +—
— i—
+ 9!
v
Figure 6
As shown in the figure, the values of y decrease continuously
from +00 to 26, from 26 to 8, then increase from 8 to 26, from 26
to + 00 as a; increases respectively from — 00 to — 1, from — 1 to
+ 2, from + 2 to + 5, from + 5 to + 00 .
467. Problem V. — As was stated in the beginning of this
chapter, the intention was to discuss problems of maxima and
minima whose solutions could be reduced to the solution of a quad-
ratic equation. The problems thus far considered are all special
cases of the more general problem : When has the quadratic expres-
sion aa^ + 6x + c a maximum or a minimum value?
502 COLLEGE ALGEBRA [W68
It can easily be shown that the trinomial
a:x? + 6x -[- c = y
can be written
^=4Hf)'+(^-?)}
where p = -» ^ = -•
The form of the equation can be simplified by putting a (?~ 4 )
= P and X =ix +^; which gives
y = aX^+R
The trinomial, written in this form, consists of two parts, a constant
P and a variable term aX^ which has the same sign as a. If a is
positive, then the value of y will always be greater than P, y will
have its smallest value or be a minimum when X\b zero; i. e., when
x+? = Oorx = — ^. li X increases from — oo to 0, then y will
decrease from -{-co \x) P because X^ is positive ; as X increases
from 0 to + 00 , y will increase from P to + oo . If a is negative,
then the value of y must be less than P, and will acquire its greatest
value or maximum, when X= 0, or x = — ^. If X increases from
— 00 to 0, then from 0 to + oo , it will be seen that y will increase
from — 00 to P, then decrease from P to — oo .
468. Problem VI. — Geometrical representation of the variation of
the trinomial ax* -\- hx-\- c in the several positive cases.
If the trinomial is represented by y, it has already been found
(2467) that
(a) ,=4(.+|y+(,-^)]
where p =-y and j =^ ;
hence lb) , L [(.+A)' + i^].
Consider the variations of y according as 6'— 4ac <0, 6*— 4ac> 0,
6* — - 4 ac = 0, when a > 0.
1. 6« — 4ac <0. The illustration will be more clear if a
numerical example is considered:
(1) y = 2x«-12x + 19,
where a = 2, h = —12, c = 19, and 5« — 4ac = — 8;
hence (2) y = 2(x-3)«+l. [(6)]
M68] GRAPHS OF QUADRATIC EXPRESSIONS 503
Consider the various points of the curve corresponding to
x=: — 00, a;=--2, ac=— 1, x=0, x=l, x=2, x=3, x=4, x=z-\-co .
ix= — 00 j'x= — 2 j'x= — 3
pi- = o rx = i rx = 2
^*)y=:19; "^^1^ = 9; •ty = 3;
= +00
00 ,
It is seen that y has its least, or minimum,
value y = 1 at the point x = 3, because for
all values of x greater than 3 or less than 3,
the corresponding values of y are greater than
1. If the curve corresponding to the points
P^y Pj, . . . P^ and all intermediate points
is plotted, the result is the curve shown in
Fig. 7, as the graph of y = 2x« — 12x + 19.
The ordinate AP^ is the minimum value of y.
2. 6« — 4ac>0.
Consider the curve,
y = 2x«— 12x+ 10.
Here a = 2, 6 = — 12, c = 10, and
6« — 4 ac = 64 >0.
The equation written in the form of (b) is
y = 2[(x-3)«-4].
: ; : ;
p.
Si!!
"
..:..!..;..;..
...
+39 J . .
1 • 1 )
1 • . •
1 1 • !
+25 I . ;
-
...
1 1 • 1
"hi*:":""
+20 I : I
p*
! ; : :
m\±
....
--
'^TV'l"
-
--
:m:\
+»; \; 1 ;/
f \ 1 /
J^jp.
0 '. ! i't ;+5
FlQURS 7
504 CX)LLEGE ALGEBRA [«468
If X = — X , — 1, 0, 1, 2, 3, 4, 5, + 00 , then the points are:
...
-
...
+ i» i ; i
--j-t-t-"!-
..i .
• 1 !
*
p.
+»; j ! !
p'"
• i > !
— .
p_^ ♦-._«._.♦._.,
r
T= ' •' i
f
\i : ; t
*
V I I i
' !
q\ \ : :+8
ft.J. hi
p";
...
— *. --
: \ ! /
;
—
...
--j--
-"^ 1 i !
i
.{::
= —00
+ 00 J
U=-6;
'U = 24;
'ly=0;
fx = 5
U = 0;
fx = 6
U = 10;
etc.
FiGUBE 8
As X varies from — oo to 0, y varies from + oo to 10; as a; varies
from 0 to 1, y varies from 10 to 0; as x varies from 1 to 3, y varies
from 0 to — 8 ; as ;r varies from 3 to 5, y varies from — 8 to 0 ; and
as X varies from 5 to + oo . y varies from 0 to + oo (Fig. 8).
Hence, a negative minimum value PA results for y at P and the
points P^ and P" where the curve crosses the axis of X are found
by solving the equations,
{
y = 0
(x -> 3)« —4= 0,
or
y = 0
'.{'.=1. "'ill
M69] GRAPHS OF QUADRATIC EXPRESSIONS 605
3. 6« — 4ac = 0.
Consider the example,
y=:3x« — 12X+12 or y = 3{x-2)\
Here y has its least value for aj = 2, namely y = 0. On putting
« = — 2, — 1, 0, 1, 2, 3, 4, etc., one obtains the following points:
^»U = 48; ^*U = 27; Uy=18|-,
^ U = 3; U = 12ietc.
Here it is seen that, as x increases from — ao
to 0, y decreases from + oo to 12; as x in-
creases from 0 to 2, y decreases from 12 to
0 ; as X increases from 2 to + oo , y increases
from 0 to + Qo . Hence, the minimum value
of y is zero and the corresponding point P is
on the X- axis (Fig. 9).
Figure 9
469. Problem VII. — An exposition of the variation of the frac"
turn y = — ±-r-'
1. In case a = 0, then y = ^x-f- ory=^x-l-J5, where
a h * ^
J. = — and jB = — I a case which has already been considered (J246).
bi- bi
2. If a^ is not zero, then, by division,
hai — abi
(1)
y = T +
2L.
which can be put in the form
(2) ,^A+j^-^
aix+bi
B
where
^=«, ^=^1-/^, C=-^.
fli
«!*
«1
^^^ CX)LLEGE ALGEBRA LH69
Suppose B, ^, and 6?' > 0. If .r= — oo , then y = ^, a positiTe
quantity, which gives a point F, If .r=0, y = A— ^r which may be
positive or negative, but always < A, which gives a point Q, As a?
varies from — oo to 0, y decreases from ^1 to j4 — ^ a positive or
negative quantity less
than J. Ifir=C, then
y is infinite ; and as Jl
increases from 0 to C*
y decreases from A ,
C
to — 00 . For a value
of X greater than C^ y
becomes positive, and
as X increases from
X=Cto-+aD, y de-
creases from y = + 00
to y = j4. Thus it is
Figure 10 seen that y varies in the
same sense cw X increases j so that y does not pass through a maximum
or minimum value. The form of the curve will be that shown in
Figure 10, which corresponds to the special case when ^ = 4, J5 = 3,
(7=1; in which case equation (2) takes the form
6
-rr-
...
...;
US.
.-,...*...
\ i i
...
♦ 10
iti'"
-•[--
; :
; V! !
; :
I V J
; ;
: N^;
♦a;
H-«/^J
T^_Tr:rS>
Q
^A|-
1 1
l\c_
■-•♦--*--
-8; ; !
...
o\
4-4-1—
♦ ftl ; ! ^10
-\-
; ;
I ! i
i I
' ' i
-sj
y = 4 +
2x — 3
•'-; + N
i '1
> ;
^ J
!" ,
' 1 ' '
* 1
.^. -(_.(..}..
f I p /
'/ '
i J"
t , . t
/\
, !
-t-76>;f?;
* *
¥i< •
■ !
^_ i !
■ -♦.- + .-» .^,
A.-i
j/T-
.;..
tr-lif-
9.1. '
- ♦--■♦-
/.♦?
_;_;_] j-jo
1 1 ' .
' f
t "I't't"
r : !" :
•5l !
LLL'-TTn
Y
/
y-y-| If ^, or ^jfe— a6j,<0, then as x
increases from x = — oo to .'c=C, y
will increase from A to -]- cc and
change signs as x passes .through
X = C\ and as x increases from x = C
toa;= + oo ,y increases from y = — oo
toy = + A The form of the curve
in this case is exhibited in Figure 11,
which corresponds to the special case
-4=4, .fi=— 3, C={; then equation
(2) becomes
3
Figure 11
y = 4
x-l
GRAPHS OF QUADRATIC EXPRESSIONS
507
Problem VIII. — Find the minimum value of the sum of
cors whose product is constant and equal to p*.
calling one of the factors x, the other ^, and y their sum,
3ult is , P*
or
x* — yx + /'^ = 0.
= l±^Vy'-4.p\
for X and get
adding for what values of x, y is a maximum or a minimum,
is restricted to considering real values of x. Here, y* may
jase continuously without limit, but can not take smaller values
i y» = 4i>*, and still have the values of x real. Therefore, the
dlest value which y can have for real values of x i8y = 2|?.
'ref ore, y =:2p is a minimum and j^ = — 2 p is a maximum.
e corresponding values of x are x = -y ; i. e., x =p and x =—p.
4
The graph of the curve y = x + - •
Let X'X be the axis of the abscissae and F !F(Fig. 12) the axis of
coordinates, and construct the points whose coordinates are as follows:
[x = 0
00*;
rx = ^ rx = i
x = 2
y = 4;
X== +00
+ 00 ;
rx=:3
fx = 4
fx=0
(x = — 1
(x=—2 (x=—Z (« = — 4 ,j* =
*'ly=-4; ^qy=_4i; ^•'|y=-5; ^-'}y =
* S«e footnote on f 458,
\U.LL
; ; 1 ;
i'<? : : :
1 i
S?
--•♦- —
-.+_.4--4 T--
k.|.4..|..
^/Y
k
1 ; ! !
III:
\; r 1 :
./
[vT I
1 ! ' 1 !
! ! : :
V jv
'/
^': :
i : : :
'*>^jH/
1 ! j/i
T ! : T
: 1 1 :
; p: ]
1 :
...
; 1 1 !
: ! ! !
V; [ :
: : : !
i i ' i®
1 !
-'9 : : i
-6: 1 ; 1/
^ ! ! ! !*•»
! \ H
! 1 : :
..\..X..j^.\.-
TIT:
L.i.i. r 1
! : : :
1 : ! !
X/\ : [
1 •' 1 I
1 1 '
/T^^N
-ai : • I
I [
\-\X/(-
.-i--;—
: : '
[ ! : M
: i t r
! ! •
! M : 1
-'? i ; j
Figurb12
= — 00
00 .
508 COLLEGE ALGEBRA [JMTl, 472
If all the points /\, /\, • • • /\', P/ • • • are connected with
all the points found by assigning all possible values to x between
0 and ^j ^ and 1, and so on, the graph of the curve in question
(Fig. 12) is obtained.
In accordance with the definitions of minimum and maximum,
the quantity y will be a minimum at P^ and a maximum at P^' ;
because for values of x less than and greater than 2, the values of y
are greater than 4, and for values of x less than and greater than
— 2, the values of y are algebraically less than — 4.
471. Problem IX. — ^When will rational integral fractions of the
form, __ flurg-f 537 + c
~" M + Ci
have a maximum or a minimum value?
By division, 6i»c + q<^i* — hbiCj
a bhi — oci , 61'
61 6i« ^ bix + ci
which can be written in the form,
(1) y=Ax+B + ^.
where ^ = f. 5 = - ^^^f^^, Q^b,!c±ac^*^bb,c,^ k = -^.
Further, Ax^Ak-\- A(x — k) ; hence, for (1),
(2) y = B + Ak + A{x^k) + ^^.
y = B + Ak + A(^x^k + ^y
The variation of an expression of the form.
7. , V/A
x — k
C
has already been studied in Problem VIII where p* would equal
and x=x — k.
472. Example. The complete reduction of a problem of this
character will be illustrated by the example.
y -. -T — --x + 3-i -'
x—1 x— 1
The equation may be written in the last form of equation (2) as
follows: . , r ^v , 2
y = 4 + (x— 1) + -^.
a? — 1
The variation of y will depend on that of
2
(x-l) +
z-l
{473]
This can be determined as in Problem VIII,
2
GRAPHS OF QUADRATIC EXPRESSIONS 509
Accordingly put
M = (x-1) +
hence,
or
2
The least positive value which u can have in order that the values
of 05 — 1 may be real is ia* = 8 or u=z2\/2y and the least negative
value isu=— 2v/2;the corresponding values of x are » = 1 +V^2
and 05 = 1 — ^"2. The former corresponds to a minimum and the
latter to a maximum, because, as x increases from x = — oo to
05 = 1 — 1/2, « increases from — oo to — 2i/2; and, as x increases
from x = l— 1/2 to x = 0, w decreases from — 2^/2 to —3;
moreover, as x increases from x = 0 to x = l, _ is negative
and u will decrease from — 3 to — oo . Hence, the ordinate
u = QA = — 2i/2 is a maximum. Therefore the maximum value
of y is y =4 -f w=4 — 2i/2 = 2(2 —1/2). Similarly, as x increases
- 2 . "
from x = l to x = l+ V2,^-—[^^
positive and, as x passes through
X = 1, u changes signs and de- -
creases from + oo to 2i/2; but, as
X increases from x = 1 + i/2 to x-
+ 00 , n increases from 2\/2 to
-|- 00 , and hence the ordinate
u = PB = 2i/2 is the minimum
value of u. The corresponding
value of Figure 13
y = 4+ii = 4+ 2^^2 = 2(2 +1/2)
is the minimum value of y. The results of this discussion are given
in Figure 13.
478. Problem X. — The maximum and the minimum values in
the general case may now be determined when
(1) y = ^+b^ + c ,
The solution of this problem can be reduced to the solution of
i : i ;
r
y
1 ; I I
+9
I
A
.i. .;..;..;..
\
L
^
\ \ \ \
1 1 : j A
-fl ' • . !
0
B
j+ft
' I ' '
##
\
--
—
..+..
..!..
A\\\
510 COLLEGE ALGEBRA [8474
the problem of the preceding section. By division it is possible to
transform (1) into
(2) y = -+-^2-t-^
ai aix^+bix+ci
where A and B involve given coefficients of equation (1), Now the
study of the variation of y will depend upon the variation of
/Q\ Ax-{- B
w ^- ^^^,+^^^.+,^;
the corresponding values of y will be found by adding — to the
values of 2. It is sufficient, therefore, to study the values of «
when X varies from — oo to -f- Qo . When ar is very large, equation
(3) may be written
A + ^
(4) z = -^
X
The limit of the value of the numerator, as x becomes very large, is
A^ while the denominator becomes as large as is desired. Therefore,
the fraction becomes very small and begins with the value 0 when
X = — 00 only to return to 0 when x = + oo . It is not clear how
the value of z varies in the interval between x = — oo and x = oo ;
but this may be discovered by proceeding as follows: put
(5) V = 1 = «i^+V + ri .
z Ax-{-B '
V is the quotient of a trinomial of the second degree by a binomial
of the first degree. It has already been learned in Problem IX
how to determine the variation of such a fraction.
After it has been determined when v is a maximum or a mini-
mum, it is possible to determine readily when 2 is a maximum or a
minimum. Since rz = 1, z and v must have the same signs. When
the absolute value of v increases, that of z decreases, and conversely.
Therefore, a maximum value of z corresponds to a minimum value
of V and a minimum of 2; to a maximum of v,
474. This discussion will now be illustrated by two numerical
examples.
Example 1.
Let (1) r- ^-^""^
It may be observed first that the denominator has no real roots and
therefore is not zero for any real value of x.
2--S
J474] GRAPHS OF QUADRATIC EXPRESSIONS &ll
Write y in the form,
2 —
X or
and let x now vary from — oo to + oo ; then y begins with the
value 2 and returns to the same value and remains finite during this
interval, since no real value of x will make the denominator of y
zero; therefore, the fraction y will pass through a maximum and
a minimum. In order to determine these, it is necessary to apply
the general method. On dividing the numerator by the denominator
the result is
where ,= 8-^^?^^.
Let «=-,
u
201
then (3) u
_\_a^-Ax-irT ^x 13, 64_
z 8a:— 19 8 64 ' 8ar— 19
-i+5i(«— »+i^)
201
Here, u will be a maximum or a minimum according as 8x— 19+t — --
oX — 19
is a maximum or a minimum. As in problem YIII let
Q in I 201
hence, (8a; — 19)* — « (8x — 19) + 201 = 0;
8x-19=^±^^-^X^<^^
2
Therefore u will be a maximum or a minimum according as
»j = — 2 V 201 and «, = 2 v' 201 ;
and the corresponding values of x will be
Sx — 19 = 5=— ^^201;
whence, (4) x, = ^^-=^ and ., = 1^±^
or iCj = .603 — and x^ = 4.147 +.
The fraction u passes through a maximum for x^ = . 603 — , and
through a minimum for x^ = 4. 147. Therefore z^ and consequently
612
COLLEGE ALGEBRA
y, has a minimum for the value x^ and a maximum for x^
corresponding values of z, i. e., z^ and z^, are
[U75
The
Trh-^k=-^^-^ '■'■
,-.from(2), (5) y, = -.86
1. e.,
and
a, = -2.86
«,= 1.86.
y, = 3.86.
; ;
+«»!:! i
ii
y2)i--;--
; T
' i jT
: :
\ '{
1^
*y * .4...
\ I \
; ;
-a!
X ! 0
t^\^S) :
w. \
.>^
M-^a
o)j !
^ 1 ._!
1 1 •
...(..4 —
. . .t>-4«--
-5: : I
i i
It is now necessary to follow the
variations of y as x varies from
— 00 to + 00 (Fig. U). If X
increases from — oo to ar^, y de-
creases from 2 to — .86.
As X increases from x^ to x^
y increases from y^ = — .86 to
y^ = 3.86; and as x increases
from Xjj to + 00 , y decreases
Figure 14 from y, = 3.86 to + 2. The
ordinate AP is the negative minimum value of the fraction y, and
BQ, the maximum value. It follows from (2) that if x = 0,
y = — .71 = 07?; and if y = 0, then x = ± V 4,
OiV^ = — v\ and OM^ = + v%
475. Example 2. Find the maximum and minimum values of
the fraction -r.\ rr and trace the variations of the fraction when
a* + 4 X + 6
X increases from — oo to + oo .
which may be written in the form,
(2) (2 -y)x« ~4yx+ 3~ 5y = 0.
The condition that the roots of the equation are real is
(3) 4y«-(2-y)(3^5y)> 0,
or _yt+i3y_6>0. _
The roots of the trinomial in y are +13 j: vl45 . these may be
written,
(4) i,^ = ll=jim=0.48
y. = 13±Wil = 12.62,
which are correct within 0.01.
M75] GRAPHS OF QUADRATIC EXPRESSIONS 513
Hence the condition becomes
(5) - (y - 0.48) (y - 12.52) > 0.
In order that this expression may be positive, it is necessary to have
0.48 <y< 12.52.
The two values of x which correspond respectively to the values y,
and y, of y are [from (2)] ,
(6) X, =-2m_ = 0.63 +
* 2 — yi
2.38 +
« 2-
■y%
The fraction y may be written [from (1)],
(7)
y =
^+s
Hence, as x increases from — oo to + oo , y begins with the value
2, and returns to this same value, and is never greater than y, nor
less than y^.
If X = — 4, the corresponding value of y is y = 7 ; if x = ar^,
y = y^; if X = 0, y = |; if X = 5, y = 1.06. Hence, the variation
of the value of the fraction y as x increases from — oo to + oo
will be represented in the following table:
X I — 00 . . . Xj . . . Xj . . . + 00 ,
y I is 2 increases to y, decreases to y^ increases to 2,
where it is necessary to put
x^=0.63 X, = — 2.38
yj = 0.48 y,= 12.48
These results are exhibited
in Fig. 15.
In this figure OM = x^ =
0.63, iVP=yj=:0.48; 07= y
=.6 when x=0; OM' =x,=
— 2.38;.V' P'=j/^= 12.52;
0A = 2= lim. (y)x=oo.
The variation of this frac-
tion is analogous to that of ex-
ample 1, with the difference
that this fraction begins to in- *'
crease, while that in example
1 begins by decreasing as x in-
creases from — 00 to -f- 00 .
...L.
+10
--
...
...
...
--
....
A- h
'l\"\'
"i -r--
t*
...
-
...
...
...
±^
..;-.;.
iVi"
A
...
..
...
...
--
...
^
+ 6
■* ; . i.
'•r
, •
^
M
r-
--
...
Figure 15
514 COLLEGE ALGEBRA [W76
476. Example 3.
It may be observed that the denominator has two real rootsi, x^ = 2,
Xj = 3, and therefore y becomes infinite for these roots. The frac-
tion may be written in the form,
-i+i
If X increases from — oo to + oo , the value of y begins with 1
and returns to 1. For values of x very near 2, but less than 2,
X — 2 is negative, x — 3 is negative, and ar' + 7 is positive, and
therefore the fraction
y = _^+l_ = ^"±1
^ a:«-6j- + 6 (x-2){x — S)
is positive ; but for x very near 2, but greater than 2, x — 2 is pos-
itive, x — 3 is negative, and x* + 7 is positive, and therefore the frac-
tion y changes sign from + oo to — oo , as x passes through x = 2.
Similarly, as x, increasing in value, passes through x = 3, the
fraction y changes from — oo to + oo .
After clearing fractions and arranging the equation with respect
to X, the result is
x«(y-l)-5yx + 6y-7=0.
Solve and get
a. --5y± v^(5.v)»-4(.v-l)(6.v-7)^5y± vV -f- 52 1/ ~ 28
2(y-l) 2(i/-l)
In order that the values of x may be real, it is necessary to have
y«+52i/-28>0,
i. e., (y — Pi) (i/— yj)^0,
where y^ = — 26 -t- 8 l/Il and y^, = — 26 — 8 i/lT.
Hence, y can not take a value which lies between y, and y^, for
then (y—y^ (y— .V^ would be negative and therefore x would be
imaginary; but it can take any value less than y^ or greater than y^.
Therefore, y^ is a minimum and y^ a maximum value of y.
The corresponding values of x are
^ =-^.111-= ^m±JOvTl=^ 2.8b + ,
* 2(yi-l) 2 (-27 +8/11)
• 2(y,-l) 2(-27-8Vll)
M76J
GRAPHS OF QUADRATIC EXPRESSIONS
515
to
A resume of the preceding discussion gives us, as the variation
of y, the following results: as x increases from — oo to x =x^, y
decreases from 1 to y^ = 0.533+ ; as x increases from x^ to a; = 2, y
increases from 0.533 to +oo.
When X passes through the
value X = 2, y changes sign
an<) becomes — oo , and as x
increases fromx=2 to x=x,
y increases from — oo
y :=z y^z=. —52.533; as x in-
creases from Xj to X = 3, y
decreases from y =3/2 ^ — • o^> , rr — ^H*^
and changes sign and becomes
-f 00 as X passes through
x=3; as X increases from
x=3 to + 00 , y decreases
from + 00 to + 1. The re-
sults of this discussion are
exhibited in Fig. 16, in which
the ordinate ^P is the min-
imum value of y and the
ordinate BQ \b the negative
maximum value of y. ""'figubis 16
-50
— i —
;
: ! ! 5
J..
tf i^yai
; I 1 I
-4"
■J""
^ 4..
--■t-— ♦--r- -►- -
; I ] '
BZEBOISB TiXyX
Trace the graphs of
1. y = x? — 5x+6. 2.
3. y = — 4x«+20x — 25. 4
y = — 3x^+12 X — 6.
y = 1/1 + x+ 1/1 — X.
Trace the graphs of the following, and mark in particular the points
wl^re the graph cuts the axes, and the maximum and minimum
values of y.
5. 2 y = (7 - 8 x) ^ (1 _ x).
6. y = (x« — 7 X + 12) ^ (x« + 8 X + 16).
7. y = (x« — 7x+6) -i.(x« — 8x+ 15).
8. y = (x« + 8 X + 16) ^ (x«- 7 X + 12).
9. y = (x«— 5x+4) ^ (x« — 8x+15).
10. y = (x« + 8 X + 16) -^ (x« — 4x + 4).
11. y = (x« — 10 X +27) -^ (x« — 9x + 18).
516 COLLEGE ALGEBRA [2476
12. y = (a^ — 9 X + 18) -s- (x«— 10 X + 27).
13. y = (x«— 10 X + 27) ^ (x« — 14 x + 52).
14. y = (x«— llx + 30) ^ (x» + 2x — 15).
15. y = (x« + 7 X + 9) ~ (2 X + 5).
16. y =(2x«+x-6)4-(2x«+7x-15).
17. y=(x« + «-.5) -^(x«-l).
18. y = 1 ^ (x8+4x+7).
19. Show that the algebraically greatest and least values of
(x«+2x — 2) ^ (x«+ 3x+ 5)
Il2 /l2
are -^— and ""-VTr' What are the corresponding values of x?
Find the maximum and minimum values of y in examples 20-22:
20. y = (x— 2) (x— 3) -^ x*.
21. y = (x — 3)^(x«+ 2x — 5).
22. y = (2 X — 1) (3 X - 2) (x — 3).
23. Inscribe in a square the square of least area.
24. Circumscribe about a square the square of greatest area.
25. Inscribe a rectangle in a circle which has a given area
and determine the greatest such rectangle. Ans. Square.
26. Find the sides of a right-angled triangle, given the perimeter
and area.
27. Circumscribe about a circle the isosceles trapezium of mini-
mum area.
28. Find the sides of a right-angled triangle, given the hypote-
nuse and the sum of the legs.
29. Draw a tangent to a given circle which shall form with two
given perpendicular tangents the triangle of minimum area.
30. A box is made from a rectangular piece of cardboard 1 1
inches by 15 inches by cutting out equal squares at the comers of
the sheet, and then turning up the flaps. Show how to construct in
this way the box of greatest capacity.
31. Find the volume of the greatest cylinder inscribed in a
sphere of radius a.
32. Find the cylinder of least surface, the volume being constant
33. Find the cylinder of maximum volume, the surface being
given.
BOOK V
CHAPTER I
RATIO AND PROPORTION
477. The ratio of one number to another is the quotient formed
by dividing the first by the second.
Thus the ratio of a to 6 is ^ ; and is also written a : h.
478. A ratio of equality is one whose terms are equal, as 4 : 4.
A ratio of greater inequality is one whose first term is greater
than the second; as 7 : 5.
A ratio of less inequality is one whose first term is less than the
second; as 4 : 9.
Inverse ratios are two ratios in which the first term of the one is
the second term of the other, and vice versa; as 3 : 5 and 5 : 3.
The duplicate ratio of a given ratio is one whose terms are the
squares of the terms of the given ratio.
Thus a' : 6* is the duplicate ratio of a : h.
The triplicate ratio of a given ratio is one whose terms are the
cubes of the terms of the given ratio.
Thus, a' : ¥ is the triplicate ratio of a : 6.
The suhduplicate ratio of a given ratio is one whose terms are
the square roots of the terms of the given ratio.
Thus, \/a : \/h is the suhduplicate ratio of a : b.
479. Four quantities are said to be proportionals when the first
is the same multiple, part, or parts, of the second, as the third is
of the fourth ; that is, if a = ?n6 and c = mdy or, what amounts to
the same thing,
a c
- = m - = m,
517
518 COLLEGE ALGEBRA [{{480-485
whence the quotient of a by 6 is equal to the quotient of c by d,
i e --^.
This result is usually expressed by saying a ts to b as c is to d,
and is written
a \ h II c \ d or a:6 = c:J.
480. The terms a and d are called the extremes and h and c are
called the means of the proportion.
The first and the third terms are called the antecedents and the
second and the fourth terms are called the consequents. Thus, in the
proportion a : 6 = c : c?, a and c are the antecedents and h and d
the consequents.
481. A continued proportion is a series of equal ratios, in which
each consequent is the same as the following antecedent; thus,
a: 6 = 6: c z=z c : d =z d : e.
Properties of Proportions
482. Theorem L — When four quantities are proportionals the
product of the extremes is equal to the product of the means.
Let the proportion be f ^ S* [8479]
On multiplying both members of the equation by hd^
ad:= be.
488. If the means of a proportion are equal, either mean is
called a mean proportional between the first and the last terms, and
the last term is called a third proportional to the first and the
second terms.
Thus, in the proportion a : 6 = 6 : c, 6 is a mean proportional
between a and c, and c is a third proportional to a and b,
484. Theorem II. — A mean proportional between two quantities
is equal to the square root of their product.
Let the proportion be ^ ~ ~ '
Then 6« = ac, [{482]
b = V ac,
485. If any three terms of a proportion are given, the fourth
can be obtained from the equation ad = bc: thus,
be J ad ad T be
d c b a
H486-490] RATIO AND PROPORTION 519
486. Theorem III.— (The converse of Theorem I.) If the
product of two quantities is equal to the product of two others^ the four
are proportionalsy the terms of either product hein{f taken as the means^
and the terms of the other product for the extremes.
Let xy = ah.
On dividing hyyh 3^ = ^ or f = i
yb by by
Whence, x : b = a : y, [{479]
In like manner, it may be proved that
X : az=b : y
y : h = a : X, eta
487. Theorem IV. — If four quantities are proportionals, they are
proportionals when taken inversely; that w, the second term w to the
first as the fourth term is to the third.
Let the proportion be a : h = c : d.
Then ad = 6c, [S482]
h:a = d:c, [{486]
488. Theorem V. — Tn any proportion, the terms are in proportion
by alternation; that is, the first term is to the third as the second term
is to the fourth.
Let the proportion be a : b = c : d.
Then ad == he, [{482]
a: cz=b : d. [{486]
489. Unless the numbers are of the same kind, the alternation
can not take place ; because this operation supposes the first to be
some multiple, part, or parts of the third. One line may have to
another line the same ratio as one quantity of grain has to another,
but there is no relation, with respect to magnitude, between a line
and a quantity of grain. In case, however, the four quantities are
represented by numbers or by other numbers of the same kind, the
alternation may take place.
490. Theorem VI. — In any proportion, the terms are in propor-
tion by composition; that is, the sum of the first two terms is to the
second term as the sum of the last two terms is to the fourth term.
Let the proportion be ^ ~ d '
By adding unity to each member,
-+1 = -4 1; thati8,-^=-i-,
a -\- b :b =z c -{- d : d.
In like manner a '\- b : a = c + d : c,
since ^ = ^. [{487]
a c
520 COLLEGE ALGEBRA [{{491-494
491. Theorem VII. — In a proportion^ the term$ are in proportion
by diviiion; that is^ the excess of the first above the second is to the
second as the excess of the third above the fourth is to the fourth.
Let the proportion be ^ ~ S *
Subtract anity from each side ; then
a — b \ b =. c — d : d.
In like manner, a — 6:a=c — c?:C.
492. Theobem VIII. — In any proportion^ the terms are in propor-
tion by composition and division; that is, the sum of the first and the
second terms is to the excess of the first above the second as the sum of
the third and fourth is to the excess of the third above the fourth.
Let the proportion be ? = v
6 a
(1) ^ = '^4^, [«490]
(2)
By dividing (1) by (2),
a c
(2) " --^ = '^^. [M91]
a c
a + h _ c + d
a — b c — d
498. Theorem IX. — In a series of equal ratios the sum of all the
antecedents is to the sum of all the consequents as any one antecedent is
to its consequent.
Let the ratios be (1) -^ = ^ = ^ = = r.
(2) a = Ar, b = Br, c^Cr,
By adding the equations in (2) member to member:
a^b-^-c^ = Ar+ Br-\- Cr-\'
= {A + B+C )r.
Hence. ^ + ^ + ^+ =r
' A+B+ C+
and from (1)
a-\-h'\' c-\- a_ h^ ^
A+B+ C+ ~ ^ "" B ■" C ""
494. Theorem X. — When four quantities are proportionals, if
the first and second are multiplied or divided by any quantity^ or if
the third and fourth be so multiplied or divided, the resulting quanti-
ties will be proportionals.
Let the proportion be ? = ^* then ^ = ^,
b d bm dn
or am : bm = en : dn.
In like manner it may be proved that -:—=-:-.
*^ '^ m m n n
NoTB.— Either m or n may be unity.
JM95-498] RATIO AND PROPORTION 521
496. Theorem XI. — When four guantities are proportionals^ if
the first and the third are multiplied or divided hy any quantity ^ or if
the second and the third are so multiplied or divided^ the resulting
quantities will he proportionals.
Let the proportion be f ~ S*
By multiplying and dividing both members of the equation by
m and n,
am _ cm
bn dn
am : bn = cm : dn.
Note.— Either m or n may be unity.
496. Theorem XII. — If the corresponding terms in a series of
proportions are multiplied together their products will he proportionals.
Let the proportions be
a : b = c : d, or
a c
b-d'
b' " d''
and a* \b* -=, c\\ d\ or
Multiplying the equations together,
6 ^ 6^ "■ d ^ d'' bb' " dd''
aa' : bb' = cc' : dd\
This is called compounding the proportions.
This theorem is true when applied to any number of proportions.
497. Theorem XIII. — If four quantities are proportionals^ the
same powers and the same roots of these quantities are proportionals.
Let the proportion be & ~ d *
Tbe. f = |, [i»84]
a" : 6* = c» : (?" and "j/a : Wh = "l/c : V^.
498. If a : 6 = 6 : c, then a : c = a' : 6».
For
a_b
b" c*
on multiplying by f, f X f = f X ^.
^ ___ a a o^
a : c z:^ a* : b*,
NoTB.— The three quantities, a, b, c, are said to be in a continued proportion (S48D.
522 COLLEGE ALGEBRA L««4d9-502
499. In like maimer it may be shown that
H a : b =z h : c = c : dy then a : c? = a' : 6*.
XoTB.— Here the four quantities are said to be In coniinued proportion ($481)-
600. It will be evident to the student from the preceding articles
that, if four quantities are proportional, many other theorems tli&ii
those given may be derived. Thus, for example,
if a \ h z=: c : d^
then ma •{• nh \ pa -^ qh z=. mc -j- nd : pc '\- qd.
For - = -, and .-. -=-.
Add n to both members, m±nk = m±Jii .
o a
SimUarly, S«±^^S£jt^.
On dividing the two equations member by member,
ma-\-nb mc + w<i
pa-\-qb pc-\-qd
ma 4- w^ : pa -{- qb = mc -]- nd : pc ^ qd,
501. It has been assumed in our definition of a proportion that
one quantity is a definite multiple of another, or, what is equivalent
to the same thing, that the fraction formed by making one of the
numbers the numerator, and the other the denominator, is a deter-
minate fraction. This will be the case when the numbers have a
common measure.
Let the common measure of a and 6 be x ; then
a = mx, and 6 = nx, and ^ = ^ = ?5
b nx n
where m and n are integers.
502. Incommensurable Numbers. — But it sometimes happens
that the two quantities do not have a common unit of measure;
that is, both can not be expressed as integers in terms of a common
unit. They are then said to be incommengnrable.
For example, the ratio of the diagonal of a_ square to its side is
the irrational number v 2.
Let AB = a
AC b b b'
Now 1/2 = 1.41421356 which is
greater than 1.414213 and less than 1.414214.
If a millionth part of 6 is taken as the unit
of length, then the value of the ratio ^ is
2S503, 504J RATIO AND PROPORTION B23
1414213 ^g ^1414214 .
1000000^6^1000000'
therefore 7 differs from either of these fractions by less than ^,^^,,^ •
0 1000000
Similarly, if the decimal is carried to the n^ place, the corre-
sponding fraction will differ from the true value of the ratio by less
than 1
W-' "341]
and this fraction can be made as small as one chooses, if n is taken
as large as may be desired, i. e. , by carrying the decimal as far as
may be desired.
JHence, in case two quantities are incommensurable,' there ia no
fraction which will exactly express the value of the ratio of the
given quantities; but it is possible, by taking the unit of measure
small enough, to find a fraction that will differ from the true value
of the ratio by as small a quantity as is desired.
603. Theorem XIY. — In case a and b are incommensurable guan-
titvesy a fraction can be found which will differ from the true value of the
ratio -r by as small a quantity as is desired.
Let b = nx, where n is an integer,
and suppose that mx < a < (m + 1) x.
n o n n n
a m ^\ .
0 n n
and since nx = 6,
if X diminishes, then n increases (6 being constant) and, therefore,
- diminishes. Hence if x is made as small as may be desired, n can
be made as large as may be desired, and therefore - can be made less
than any assigned fraction. Therefore the difference
b n
can be made less than any assigned fraction ({341).
604. Theoeem XV. — If c and d, as well as a and 6, are incom-
mensurable; and if, m ^a ^m , 1
* n<b<^^n'
when !?<^<!? + l,
n a n n
no matter how m and n are increased; then
b d'
524 COLLEGE ALGEBRA [88505, 506
If 7 and ^ are not equal, their difference most be some assignable
o a
qoantttj, since each lies between
n n
I
n
and this difference must be less than - • Now, since n may be made
1 ^
as large as is desired, - can be made less than any assigned frac-
tion, however small: therefore the difference between ^ and 3 can
be made as small as is desired, which can only be tme if
h d'
Hence all the propositions respecting proportionals are tme of the
four quantities a, 6, r, d.
508. The property involved in Euclid's definition follows from
the algebraic definition. £uclid*s definition of a proportion is:
<^The first of four magnitudes is said to have the same ratio to
the second that the third has to the fourth, when any equimultiples
whatever of the first and the third being taken, and any equimul-
tiples whatever of the second and the fourth, if the multiple of the
first be less than that of the second, the multiple of the third is also
less than that of the fourth, and if the multiple of the first be equal
to that of the second, the multiple of the third is also equal to that
of the fourth, and if the multiple of the first be greater than that of
the second, the multiple of the third is also greater than that of the
fourth." (Euclid, Book V.)
For, let a : b = c : d; then r = 3 *
0 a
jm pc
qb qd' ^
Hence pc is greater than, equal to, or less than qd, according as pa
is greater than, equal to, or less than qb,
506. Conversely, the property involved in the algebraic definition
follows from Euclid's.
Let a, 6, c, dj be four quantities which are proportional accord-
ing to Euclid's definition; prove that r = ^ * For, if ^ is not equal
to ^» then one of them must be greater than the other.
Suppose that ^ ^
b^d'
then it must be possible to choose some fraction, ?. t which lies
between them. Then 7 is greater than E and ? is greater than ^ •
0 q q o
«507, 508] RATIO AND PROPORTION 525
Then J«>i'^> aiid qc <^pdy
hence, a, 6, c, d are not proportionals according to Euclid's defini-
tion; which is contrary to the supposition. Therefore ^ and ^ can
not be unequal.
507. Euclid's definition of the ratio and proportion is the prefer-
able one. Straight lines can be represented geometrically, but the
abstract number which expresses how often one straight line is con-
tained in another, can not be represented geometrically. Hence the
common algebraic definition of proportion can not be used in Geometry.
The algebraic definition is, strictly speaking, applicable to commen-
surable quantities only; but it should be noticed that Euclid's defini-
tion is applicable to incommensurable quantities as well. This con-
sideration alone is sufficient reason for the definition which is given
in Euclid.
608. Examples. — 1. Solve the equation,
5x — 3a:5x + 3a = 7a — 5:13 a— 5.
lOar _ 20a— 10
— 6a —6a
10 x = 20a — 10,
X = 2 a — 1.
[2492]
2. If X : y = (x — «)': (y — .:;)-; prove that 2; is a mean propor-
tional between x and y,
y{x-zY = x{y — zy [J482]
or yx* — 2xyz + yz' = xy* — 2 xyz -f xz",
or yx' — xy* = xz* — yz^^
and xy (x — y) = (x — y) 2;',
dividing by x — y, xy =z z^,
.\ z is a mean proportional between x and y (J 483).
3. If l=v
0 a
prove that o» + oc + c» : a'— ac -{■ c* = b* -\- bd -\- tP : h* — hd + d*.
By Theorems XIH and IX, ^ = g = «*±g = ^.
by Theorem V, 5LL£^ = ^±£,
•^ ' ac bd
-by Theorem VIII, -^"^i^; = ^±M±^ .
which was to be proved.
bd
626 COLLEGE ALGEBRA [I5W
Find the ratio compounded of
1. The ratio 32 : 27 and the triplicate ratio of 3 : 4. Abb. 1 : 2.
2. The ratio 6 : 25 and the subdaplicate ratio of 25 : 36.
Ans. 1 : 5.
3. The triplicate ratio oi x:y and the ratio 2y*:3x*. Ans. 2x;Zy.
4. Find a fourth proportional to x', xy^ 5 x^y\ Ans. 5 j^,
5. Find a mean proportional between 4 ox* and 16 a?. Ans. 8 o^
6.
Find
a third
proportional to 6 x" and 5 x*.
Ans.
f'
7.
Find
a mean
proportional between
:r«-
-10x_t2l ^^
x + 5
2«+2r
X —
-15
7
If a: 6
= c ; (f ,
prove that
8.
ac :
6ci= c«
:d^, 9.
a6 : c(^
= a":
c>.
10.
c« :
a» = c«
— t£«:a»— 6«. 11.
a» : 6»
= c»,
(P.
12. a + 26:6=c + 2rf:d[.
13. 2a+5/>;4a — 36=2c + 5rf:4c — 3rf.
14. If a, 6, and c are proportional, and a the greatest, show that
a + c > 2 6.
JT — V y — 2 _ z —a?
c
a + 6 + c = 0.
15. If ^ = ^ = 1 and x^ ^, 2 are unequal^show that
Solve the equations:
16. 3x — 2a:3x+2a = 5a — 3:15a + 5.
17. 3x — 1 : 6x— 7 = 7x— 10 : 9x+10.
18. y«— 16 : y* — 25 = y« — 2y — 24:y«— 3y — 10.
19. 2— l/l — X : 3+1/I — x=l/a— l/a — 6 : i/a+ V'a-ft.
20 ^^~^ — ^^^ 2x— 1
jr + 4 "" a;»-fj; + 4*
21. Find x and j/ when
r3x— 5y:5x+3y = — 16:15
(xy = 3» ^^
22. Findxwhenx«— 2x + 3 : x» — 3x + 5 = 2x — 3 : 3x-5.
23. Find x : y, if given x" + 6 y« = 5 xy. Ans. 2 or 3.
te08] RATIO AND PROPORTION 527
24. Find two numbers in the ratio 3 : 4 {suggestion^ 3 x and 4 x)
of which their sum is to the sum of their squares as 7 : 50.
25. Find two numbers in the ratio of 5 : 4, such that their sum
has to the difference of their squares the ratio of 1 : 18.
26. Find two numbers such that if 7 is added to each they will
be in the ratio of 4 : 3; and if 11 is added to the greater and sub-
tracted from the smaller the results will be in the ratio of 5 : 2.
27. If7x — 42:8x — 32; = 4y — 72:3y— 8z, prove that z
is a mean proportional between x and y.
28. 3 l/y + a : 3 l/y — a -=, m : n] find x.
29. If mx -|- ny : px ■\- qy = my -\- nz : py -\- qz, show that
n I q z=z m I p,
30. If2a + 36:3a — 46=2c+3ci:3c — 4(£, prove that
a : b =z c : d.
31. If 2 men working 9 hours a day can do a piece of work in 32
days, in how many days can x men working y hours a day do the
work?
If a : 6 = c : </, prove that:
32. a:a+c = a+b:a+h + c+d.
33. a*c + ac^ : hH + i(£2 = (a + c)» : {h -f d)\
34. a«+6*:--^=c»+cZ«:-^.
' a+6 ' c + d
35. (o+6+c+fi) {a—h^c+d) = (a— 6+c— rf) {a-{-h^c^d).
36. Show that, when four quantities of the same kind are propor-
tional, the sum of the greatest and the least is greater than the
sum of the other two.
OT T* ay — bx cx — az bz — cy 4.l^_ i
37. If -^ = — I — = -1 then x : a = y : b =z z : c.
c b a ^
38. Each of two vessels contains a mixture of wine and water; a
mixture consisting of equal measures from the two vessels contains
as much wine as water, and another mixture consisting of four
measures from the first vessel and one from the second is composed
of wine and water in the ratio of 2 : 3. Find the proportion of wine
and water in each of the two vessels.
Ans. In the first the wine is ^, in th'e second |.
39. If the increase in the number of male and female criminals
is 1.8^, while the decrease in the number of males alone is 4.6^
and the increase in the number of females is 9.8%, compare the
number of male and female criminals respectively.
Ans. Number of female criminals four-fifths the number of male
criminals.
528
COLLEGE ALGEBRA
[2509
Application of Quadratic Equations and Ratio and Pbopos-
TioN TO Geometry
609. Example 1. If the sides of a triangle are divided bj a
line drawn parallel to the base, so that the upper segment on one
side is eqaal to the lower segment on the other side, how large is
this segment, if the other two are respectively 15.125, and 8 feet?
Soluiion, —Let AI>=zEO= x.
By the conditions of the
problem Z)jB= 15.125 feet
and ^ JS; = 8 feet
Then, byGeometiy,
AD: DB=z AE : EC
or X : 15| = 8 : x.
^ By 8482, x«=15jx8=121
Figure 2 x = zfcll.
The negative value has no meaning in this problem.
Example 2. A line DE drawn parallel to the base BC of a
triangle ABC meets the side AB in D and the side AC in E. The
upper segment of one side is 5 feet, and the lower segment of the
same side is as much greater than the upper segment of this side as
the upper segment of the second side is less than the upper segment
of the first side. If the second side is 2. 8 feet, how long is the first
side?
Solution 1. — Let x be the first side ; then the upper segment is
AD = 5, and the lower DB = x — 5.
Hence
and
but
^Z> — Z)-4 = (x — 5) — 5 = X — 10,
XD — ^JS;=x — 10;
Z>^ = 5.
A /. 5 — >lJ^=x — 10,
\E .'. AEz=zlh — X,
and i;C7 = 2. 8 — (15— xl
^ But, by Geometry, it is known
that AD: AE=z DB : EC.
Figure 3
Hence, by substituting the values of AD, AE, DB, EC,
5 : 15 — x = x — 5 : 2.8 — 15 + x,
(15 — x) (x — 5) = 5(x — 12.2).
Xj = 14, and x^ = 1.
Remark.— The second value, o-g = 1, is not a solution, becaase the entlra aide AB
would be less than a part of It, AD = 6.
J509]
RATIO AND PROPORTION
529
Solution 2, — Let x be the amount in feet by which the lower seg-
ment DB is greater than the upper segment AD] then the first lower
segment DB will be 5 + x, and the second upper segment AE will
be 5— X, and the second lower segment will be 2.8— (5— x)=x— 2.2,
and the following proportion is obtained.
5:5 — x = 5 + x:x — 2.2
x« + 5x — 36 = 0.
Hence, x^ = 4, and x^ = — 9.
Therefore the first lower segment is x + 5 = 9 feet, and the side
^J5 = 9 + 5 = 14.
iP, = — 9 is not a solution.
Example 3. From a point
without a circle a tangent is
drawn to it. From the point of
contact a perpendicular is
drawn to the line joining the
center with the external point.
Find the radius of the circle if
the tangent is double the length
of the perpendicular.
Figure 4
Solution,
Let BQ = x (Fig. 4). By hypothesis, AB=\AP=^y
and by Geometry,
BP= ^AP"- AB' = Ja^ _ A' = al^ .
FQ= PB^BQ = a^^x,
2
By Geometry,
PRxPQ= ^31^j or PR =
V3
Now,
RB=PR ^PB=z
By Geometry
'' 2 (a 1/3-2 x)
2a»
' ^Vl =
2a^
fl«-f 20x1/3
aV3-2x 2 ' 2(al/3-2a7)
BR'.PR^BQiPQ)
a»4-2a3:i/3 aVS — 2x^ 2a«
— — i— = r • X -.^ •
av^3 — 2a?
530
COLLEGE ALGEBRA
[1509
By removing fractions,
(a + 2 X 1/3) (a l/3 — 2 x) = 8 ax.
By multiplying the second parenthesis and the term on the right bj
V3 and patting 2x1/ 3 = z, the equation is obtained
(a + 2) (3 a — z) = 4az,
and hence
That is,
«i = — 3 a, 2, = + a.
X, = -|V3, x, = +^v3.
PE =
av'3
• Xg —
?l/3,
2 6 3
.2«! 2^ ^ a|/3.
(il/3-2a;i al/3-^V3
3
Hence the diameter of the circle will be
RQ = RP— QP= a 1/3 — ^ = ^Vl\
o «5
and the radius,
|v/3.
Example 4. Place a rectangle whose sides are in the ratio a : h
in a circular sector whose central angle is 90°, so that two of the
comers of the rectangle lie on the arc and the other two on the radii
of the sector. How long are the sides of the rectangle?
Solution,— Let the L POQ (Fig. 5) be 90° and the radios
of the circle be r; then the length of the chord PQ will be found by
the theorem of Pythagoras to be ry2. Hence,
Since the sides of the required rectangle
are in the ratio a : b,
let
and
BC=ax and AB = hx
FiGUBB 6
. •. 0R=: OT—RT =Ji'— (^V-^*
«509] RATIO AND PROPORTION 531
Since the A ORD is isosceles, it is possible to form the equation,
OR=RD,
>/'^-(f/-'^ =
ax
2"'
x«(2 a« + 4a6 + 4 6«) = 4r«.
Taking the positive sign,
BC= ax = «rl/2a« + 4afe + 46«
<i« + 2a6 + 26«
AB= 6x = ^^2a« + 4a6 + 4y^
a« + 2a6 + 26«
For a = 6, 6 = 1, it follows that
5 6
5. From the right angle of a right triangle a perpendicular is
drawn to the hypotenuse, dividing it into two segments in the ratio
3 : 4. How long is the hypotenuse if the perpendicular is three feet?
6. From the right angle of a right-angled triangle draw a per-
pendicular to the hypotenuse ; the perpendicular divides the hypote-
nuse into two segments, one of which is six inches longer than the
other. How long is the hypotenuse if the perpendicular is four
inches?
7. Two circles are tangent externally, and two tangents, common
to both circles, are drawn. The distance of the intersection of the
tangents from the point of contact of one of the tangents with the
larger circle is 2.4 times as large as the radius of the smaller circle;
the radius of the larger circle is 5 inches longer than the radius of
the smaller circle. What is the distance of the intersection of the
tangents from the center of the larger circle?
8. A line of the length a is to be divided harmonically, so that
one of the external segments is the fourth part of the other external
segment What are the lengths of the segments?
532 COLLEGE ALGEBRA [1509
9. If a line is divided hannonically, so that one external seg-
ment is one foot longer than the middle segment, and the other
external segment is one foot longer than seven times the middle
segment, how long is the entire line?
10. How large is the radius of a circle that is circamscribed
about an equilateral triangle of which a side is a?
11. In a circle of radius r a chord of length a is drawn and &
radius is drawn perpendicular to this chord. How long is the chord
one of whose extremities lies on an extremity of the given chord
and the other extremity lies on an extremity of the perpendicular
radius?
Anfl. -J2r»-2H >/'" " (|)'.
12. How long is the radius of a circle which is drcumscribed
about a regular pentagon with a side a?
2a
Ans.
VIO -2^/5
13. A square is inscribed in a circular sector whose angle is 60^
so that two comers lie on the arc and the other two on the two radii
of the sector. How long is the side of the square if the radios of
the circle is r?
Ans. r l/2 — ^/^
14. From a point P without a circle a tangent is drawn to the
circle, the point of contact being at Jf, and a secant is drawn
meeting the circle at A and J?, A falling between P and B, How long
IS the tangent, if AB\%1 inches and PA is 3 inches longer than PMf
15. Two vertices of a rectangle whose dimensions are in the ratio
a : 6 lie on the arc and the other two on the bounding radii of a cir-
cular sector whose central angle is 60°. How long are the sides of
the rectangle if the radius of the circle is r?
16. How long is the radius of a circle which passes through two
of the vertices at the oblique angles and one of the vertices at an
acute angle of a rhombus whose diagonals are respectively 18 and 80
inches in length?
CHAPTER II
VARIATION
510. The following chapter consists of a series of propositions
related to the definitions of ratio and proportion, stated in new
forms which have some practical applications.
611. One quantity is said to vary directly as another when the
two quantities are so related to each other that, in case one is
changed, the other is changed in the same proportion.
Non.— It is customary to omit the word **<Urecily** and to say that one quantity varUt
as another.
512. For example, suppose that a laborer receives $3 per day for
his work. For rn days' labor he will receive $3 w, and for n days $3 n.
wages for m days __ $3 m ^_ m
wages for n days $3 n n
Hence the ratio of the wages received for any two periods of time is
equal to the ratio of the corresponding number of days worked, that
is, the amount which the laborer receives varies as the number of
days he is at work.
513. I/A varies directly ew B, then A = m^ where m is a constant.
Let a and b be corresponding values of two quantities, and also let
A and ^ be a second pair of corresponding values of these same
quantities; then
- = ^. [Def., 8613]
a o
(i) A = ^B, A = mB.
b
where m is equal to constant ^ •
NoTB.— The symbol a is used by some writers to express variation; then. Act. B
Is translated , A varies as B.
588
534 COLLEGE ALGEBRA [J8514-518
514. One quantity is said to vary inversely as another token the first
varies directly as the reciprocal of the second (2515).
Let A and B be the two given quantities,
then, by (i), (ii) ^ = m (i ) = | , [{618]
where m is constant.
A is said to vary inversely as B,
515. For example, if a railway train doubles its speed it will
travel a given distance in one-half the time ; i. e. , the time required
for a train to travel a given distance varies inversely as its speed.
516. One quantity is said to vary jointly as tvoo others when it
varies directly as their product.
Thus, the area of a rectangle varies jointly as its base and altitude.
Area C2 =z ah. ^
Hence, by (i) {513, if A varies jointly as B and (7,
(iii) A = niBC,
where m is constant. Fiqurb 1
617. A quantity is said to vary directly as the second and inversely
as the third, when it varies jointly as the second and the reciprocal of
the third.
Thus, according to Newton's law, the attraction of a body varies
directly as the quantity of matter in the body and inversely as the
square of the distance.
Thus, if
..V g tnB
Ov) A=--,
where m is constant, A is said to vary directly as B and inversely
as a
Problems in variation are readily solved by converting the varia-
tion into an equation by aid of formulae (i)-(iv).
518. Examples.
1. If 5x -f 7 a 6y — 23, and x = 6 when y = — 3, what is the
value of X when y = 10?
If 5x-f7a6y — 23, then by formula (i),
(1) 5x-f 7 = m(6y-23).
8518] VARIATION B35
And if X = 6, when y = — 3, it follows from (1) that
30 + 7 = m (~ 18 - 23) = — 41 «i
37 = - 41 m,
(2) 5x + 7=-. |?(6y-23).
41
If y = 10, then,
5 .c + 7 = _ ?Z (60 - 23) = - ??,
^ 41 41
5x= -287-1360 __ _ 1656
41 41 '
— 1656 _
X = —
\S'
205
2. The area of a circle varies as the square of its diameter. If
the area of a circle whose diameter is 10 is i^^, what will be the
the diameter of a circle whose area is 200?
If the area of a circle varies as the square of its diameter df, then, by
formula (i),
(1) the area of the 0 = md*.
But the area of the 0 = ^ when <£ = 10;
hence
III = 100 »»,
.-.
"'-200'
Hence (1) becomes,
(2)
0 = ^^d\
200
If 0= 200, it follows from (2) that
200 = ^^iP and d^ = ^^ ,
200 157
l/157
3. If ^ a C; and B ex C, then A±zB <x (7, and VAB a C,
Let A=tmCj and B = nCj where m and n are constant ({618).
Hence A+ B = {m + n) C^ and ^ — jB = (m — n) (7,
A^B <x a
Also AB = nmC* and \/AB = i/mn • C,
VAB oc a
536 COLLEGE ALGEBRA [1518
PBOBI.E1CS
1. If y a 2c and is equal to 50 when x = 5, what is the value
of y when aj = 1 1 ?
2. li y oc X and is equal to 64 when x = 4, what is the value
of y when x = 10?
3. If X varies inversely as y and is equal to | when ^ = |, find
the value of y when x = ^.
4. If z varies jointly as x and y^ and is equal to 15 when x=2
and y = 3, find the value of z when x = J and y = ^.
5. If z varies directly as x and inversely as y, and if x = 3
and y = b when z = 2, find the value of z when x=:15 and y=16.
6. Suppose that y ocx -|- c, where c is a constant; then if y = 4
when X = 2 and y = 10 when x = 4, find y when x = 5.
7. If z OCX and yoc x, prove that 2fy OCx*; also that 2*-|-y* a 3f*.
8. If5x+llya5x + 13y and x = 5 when y =2, find the
ratio of X to y, and then show that x varies as y.
9. If J a ^, and C a A then AC(X BD,
10. li A<xB, then ui" a ^".
11. The distance in feet that a hody falls is proportional to the
square of the number of seconds occupied in falling. If a body
fall 16 feet the first second, how many feet will it fall in 5 seconds-
12. With the same supposition as in the last problem, find the
height of a tower if a stone dropped from its top reaches the ground
in 3 1 seconds.
13. The surface of a sphere whose radius is 5 feet is 314.16
square feet. What is the surface of a sphere whose radius is 9 feet,
if it is known that the surface of a sphere varies as the square of
the radius?
14. The volume of a sphere whose radius is 6 feet is 523.6 c»ibic
feet; what is the volume of a sphere whose radius is 10 feet, if it is
known that the volume of a sphere varies as the cube of the radius?
8518] ' VARIATION 537
15. If A en By then AFqc BP^ where P is any quantity variable
or invariable.
16. H A olB when C is invariable, and A oi C when B is inva-
riable, then will A oc BC when both B and Care variable.
17. If the velocity of a falling body is 160 feet per second after
falling 5 seconds, what will the velocity be after the body has fallen
12 seconds? (Velocity is proportional to the time, v = mt.)
18. How long must a body have been falling to have acquired a
velocity of 400 feet per second? (See 17.)
19. The volume of a cylinder of revolution is represented by the
formula V=mr*h, where m is a given constant, r the radius of the
base, and h the altitude of the cylinder. The volume of a cylinder
of revolution whose altitude is 10 inches and whose radius is 5
inches is 787.9 cubic inches. What is the volume of a cylinder of
revolution the radius of whose base is 8 inches and whose altitude is
15 inches?
20. The total surface of a cylinder of revolution varies as the
square of the radius of the base and jointly as the radius of the
base and altitude. The total surface of a cylinder whose altitude
is 10 and whose base has a radius of 5, is 471.239 square inches;
also the total surface of a cylinder whose altitude is 15 and whose
base has a radius of 10, is 1570.797. What is the total surface of
a cylinder if it has an altitude of 11 inches and base with radius
of 6 inches?
21. If the illumination from a source of light varies inversely as
the square of the distance, how much farther from a lamp must a
book which is now 20 inches off, be removed so as to receive just
one-fifth as much light?
22. At what distance will a light of intensity 10 give the same
illumination as a light of intensity 12 at a distance of 20 feet?
CHAPTER Til
PK06S£8SI0n : ARITHMETIC, GEOMETRIC, AND HARMONICAL
Arithmetic Progression
619. Definition. — An arithmetic progreMton is a series of nambers,
which is so constituted that each number is equal to the preceding
increased, or decreased, by a constant number, which is called the
common difference. The various numbers of the series are the terms
of the progression.
The progression is increasing, when the terms continually in-
crease. In this case the common difference is positive. The increas-
ing progression 3^ 7^ ^^^ ^5^ ^9^ 23, etc.,
has the common difference -f- 4. On the contrary, the progression
is decreasing, when the terms continually diminish. In this case,
the common difference is negative. The decreasing progression
21, 16, 11, 6, 1, —4, —9, etc.,
has the common difference — 5.
Thus the common difference is regarded as positive when the
progression is increasing, and as negative in the contrary case; for
this reason every term of the progression is the algebraic sum of the
term which precedes it and the common difference.
E. g. , in the first example above,
19 = 15 + 4,
and in the second example
6 =11 + (-5), and —9 = -4 + (—5).
620. If a series of numbers a, 6, c, c?, . . . . is in an arith-
metic progression, this fact may be indicated by the abbre^atioa
A. P.
521. PnoBLEBf I. — Given the first term and the common differ-
ence of an arithmetic progression ; calculate the value of the n^ term.
Let a be the first term and d the common difference of the A P.
538
JS522, 523] ARITHMETIC PROGRESSION 539
Bj' definition,
the 2"* term is equal to the 1** term increased hj d, or a -|- (/
the 3'* ** ** ** ** »* 2°'* ** *< **c?, ora+2ef
the4'*' << u u <( it 3rd u u ''dyOTa + 3d
the 5"* ** ** ** ** ** 4*** ** *< *< {;?, or a4-4c?
and so on; so that any term in the progression is equal to the first
lerm plus the product of the common difference hy the number of
the term preceding the required term. Hence if / is the »*** term of an
A. P. , that is the term which has n — I terms preceding it, I will be
determined by the formula
(i) I = a+ {n — l)d.
522. Application. — A body falls at Cincinnati 4. 902 meters in the
first second, and in any succeeding second 9. 804 meters more than in
the preceding; if the body falls for 6 seconds find the distance
through which it falls during the sixth second.
According to formula (i) the distance will be
4.902m + (6 — l)9.804m^or 53.922m.
What is the 12th term of the A. P. i i -i, . . . . ?
2 6 6
Here ^ ~ i' ^ = "" q» »* = 12,
'=l+(>^-0(-i)=i-j=T-
623. In an increasing A. P., the terms increase without limit;
that is, n can be taken so large that the n^^ term is greater than any
given quantity A, For this purpose it is sufficient that
a + (ii — l)rf>il, or (?i — l)(f>^ — a;
and if d is positive n — 1 > ~"
A —a
or finally, (ii) n>H-
In case of an increasing A. P. , if the n^^ term is greater than Ay
then it follows that every term which follows the u'*» term is also
greater than A.
Example. — In the A. P. 5, 15, 25, 35, . . . etc., what values
must n have in order that the n^^ term may be greater than 5000?
According to formula (ii)
«^ 1 I 5000 — 5
i. e. n> 1 + 499.5 or 500.5;
it is suflScient therefore if n is equal to or greater than 501.
540 CX)LLEGE ALGEBRA [J«524, 525
624. Theorem I. — In an A. P. corfipoaed of a limited number of
terms the sum of two terms equally distant from the extreme terms is
constant, i. e., independent of d.
Consider the A. P.
(a) a, 6, c, jt>, . . . . y, A, Ar, Z,
composed of n terms in which the common difference is d. If the
order of the terms of the A. P. (a) is reversed, a new A. P. is
formed, namely
(b) /, kj A, ^, . , . . i>, c, 6, a,
which consists of n terms, in which the common diflference is — d.
Consider, in the A. P. (a), the terms p and g which are equally
distant from the extremes; the first, p, is preceded by three terms
and therefore by formula (i)
p = a + 3 f / ;
the second, ^, has three terms following it, and, consequently, has
three terms preceding it in the A. P. (b), therefore by formula (i)
g = l+3{—d) = l—3d',
adding, the sum of these two terms is
p + gz=a+3d+l — 3d = a + l
In general, consider in the A. P. (a), two terms which are equally
distant from the extremes, such that the first has r terms before it,
and the second r terms following it. By formula (i), the first of
these terms is a -\- rd-
the second can be regarded as a term in the A. P. (b) which is pre-
ceded by r terms; it is therefore equal to
I + r{—d) = l — rd]
therefore the sum of the two terms is likewise equal to
a -{-rd -\- {I — rd) = a -\' I,
626. Problem II. — Given the first term and the last term of an
A. P. ; find the sum of the first n terms.
Let a, t, c, ^, . . . . g. A, k, l,
be the first n terms of an A. P. Call S the sum of these terms;
then it follows, on reversing the order of the terms in the second
member, that
S = a + h+c+p+ ^ g ^ h+ k + I
and S = I + k + h + g + +P + c + h + a.
.-. 2S = {a+l)+{b+k) + ic + h)+{p + g)+ . . .
.. . + ig + P) + (h +c) + (k + b) + H + a).
«526, 627] ARITHMETIC PROGRESSION 541
But by the preceding theorem
a -f Z, ^ + Ar, c 4- ^»
are each equal to a -f- l\ the number of these sums is evidently the
same as the number of terms of the A. P. to be summed, that is n ;
therefore it follows that
(iii) 2S = n[a+l) or S=l{a+l),
Formula (iii) expresses the sum of the first n terms of A. P. in terms
of the number of terms, the first, and the last term. By formula (i)
Z = a + (n— 1) (Z.
Therefore formula (iii) can be written
A^=|(a + a+(n-l)4
or (iv) ^=|(2a + (n~l)(f).
526. Applications. — 1. Find the sum of the first n integers
1, 2, 3, 4, , w —1, w.
These numbers form an A. P. ; the first term is 1 ; the last or n***
term is n; and therefore by formula (iii) the sum of the first n
terms is
2. Find the sum of the first n odd integers
1, 3, 5, 7,
These n numbers form an A. P. ; the first term is 1 ; the common
difference is 2 ; therefore, the sum of the first n odd integers is
^^= 1(2 + (n -1)2)= I -2 n=n«.
527. The formulae
Z = a + (n — 1) d,
A'=|(a+0,
give two relations connecting the five quantities a, /, c?, n, S\ these
two equations enable one to calculate any two of these five quantities
if the other three are given. There will, therefore, be ten different
problems, according as the quantities comprised in one of the fol-
lowing groups are the unknown quantities:
(a, 0; (a, ^\ (a, n); (a, ^); (Z, c£);
(Z, 71); (/, Sy, {d, n); {d, S)] (n, S).
542 COLLEGE ALGEBRA LJ528
In case a and n, or I and n are taken as the unknown quantities, the
problem leads to an equation of the second degree ; in all other cases
the problem is solved by working out equations of the first degree.
Example 1. Find the first term and the number of terms in an
A. P. in which the common difference is 2, the sum of the series is
72, and the last term is 21.
Here d = 2, ^=72, and / = 21 ; hence from formulae (i) and (iii)
(1) 21 =a+(n -1)2
(2) 72= |(a + 21).
From (1) (3) a = 23 — 2n.
From (2) and (3), (4) 72 = ^ (23 — 2 ?i + 21) = 22 n — n\
Solving (4) n = 18 or 4.
From (3) a = — 13 or 15.
Hence there are two solutions.
Example 2. Sum 100 terms of the A. P, whose third term is 5,
and tenth term 75.
According to formula (i)
I z=z a -{- {n — \)d\
if Z = 5 when n = 3,
(1) 5 = a + 2rf;
and if Z = 75 when n = 10,
(2) 75 = a + 9 d.
Solving equations (1) and (2)
f? = 10 and a = — 15.
628. Problem III. —Insert n arithmetic means between two
given numbers a and h ; that is, form an A. P. composed of n -f- 2
terms, whose extreme terms are a and h.
Let the common difference be d^ then by formula(i)
(V) d =
6 = a+ {n+\)d,
b-a
n + 1'
in which 6, n, and a are given numbers.
J529]
GEOMETRIC PROGRESSION 6
Hence
!■* term =
«;
2»«*tenn=a + ^=^ =
b-\-na
n+l'
(1)
n+l n + l
26+(n-l)a
n + l •
4*»» term = 2A±i?LzzlI«.+ ft- « :
n + l n + l
_36 + rn-2)a.
"+1
543
rt- tenn = (r-l)h4-\n -(r-2)^a
n + l
(» + 2)»>tenn=^"-^)^ + ^-<'-^>1". whenr=n + 2;
n+l ' '
_ (n + 2 - 1) ft + [n - (n + 2 - 2)] q .
n+l
= <"+,^>/ = 6, Q. E. D.
n+ 1 '
Thus the series is
(2)n H-nq 26+(n-l)« 36+(yi-2)a (r-l)6+[n— (r-2)]a n6+a ,
^^ 'n+l' n + l ' n+l ' " n+l '"'n+l*
Example. — Insert 11 arithmetical means between 5 and 41.
ByfomuIaW . = ^^J = |1^ = | = 3.
.-. A. P. is
5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41.
Geometric Proqression
629. Definition. — A geometric progression is a series of numbers
so constructed that each is equal to the preceding multiplied by a
fixed number, which is called the common ratio.
The progression is increasing or decreasing^ according as the ratio
is greater or less than unity.
Examples. — The numbers
4, 12, 36, 108, 324, .... etc.,
form an increasing G. P.,* of which the ratio is 3.
The numbers
- o 9 27 81_ 243
^' "^^ 6' 25' 125' 625'
form a decreasing G. P. of which the ratio is |.
• For brerity O. P. is used instead of the phrase, geometric progression.
544 COLLEGE ALGEBRA [Ho30-53!
530. Problem I. — ^Given the first term and the common ratio
of a geometric progression, find the expression for the n^^ term.
Let a be the first term and r the common ratio; by definitioii
the 2~* term is equal to the 1** multiplied by r, or to ar,
the 3"* ** ** ** 2*** ** <« r, ** ai*,
the 4"* <* ** ** S'* ** ** r, ** (If',
the 5*** ** ** ** 4*** *« ** r, ** ar*,
and so on ; therefore, any term of a G, P, is equal to the fint
term multiplied by the common ratio raised to a power of whitk tk
exponent is one less than the number of the term.
Let I be the n^ term, that is, the term which has n — 1 tenns
preceding it; hence it is determined by the formula
(i) I = ar'^-l
Example. — Find the value of the eleventh term of the G. P.
whose first term is 63 and the ratio |.
By formula (1) the eleventh term is equal to
531. In case the ratio in a G. P. is greater than 1, and the fint
term positive, it is clear, in view of the law of the formation of the
terms of the progression, that the successive terms increase con-
tinuously; on the contrary, if the ratio is less than 1, the successive
terms of the progression decrease continuously. In the first case,
as n approaches infinity, the terms become larger than any assigned
quantity, while in the second they become smaller than any assigned
quantity. In order to demonstrate this proposition the two follow-
ing lemmas must be proved.
532. Lemma I. — The successive powers of a number greater than 1
become greater than any given number.
Let ^ be a number greater than 1, and put
(1) q = l+k
where k is any positive number. Compare two consecutive powen,
(1 -J- A;)*, (1 + A;)"+^ of this number; their difference
(1 + A;)»+* — (1 + A;)"
is equal to
{i+kra + k-i) = (i + kyk;
1533] GEOMETRIC PROGRESSION 545
and, since (1 + ky is evidently greater than 1, this difference is
greater than k. Since the difference of two consecutive powers of
1 4- ^ i^ greater than k, the following equation and inequalities can
be formed:
l+k=l+k
(l+ky^(l+k)>k
a + k)' -(i + ky>k
{i + kr-^{i + ky-'>k,
whence, on adding the equation and these inequalities member to
member and simplifying, it follows that
(1 + Aj)» > 1 + nk.
In order that 1 -{- nk may be greater than or equal to any assigned
number A, it is sufficient that
{l + nk)>A,
or (2) ^>^-
Hence (1 + Z:)" is greater than any assigned number A when 1 -\- nk
IB greater than A, that is when inequality (2) holds.
533. Lemma II. — The succeBsive powers of a number less than 1
can he made less thftn any assigned number.
Let r be any number less than 1, then
1 + *
where kis a positive quantity. It follows that
r" = — i — ;
and in order that r** may be less than any assigned small number, e,
however small, it is sufficient that we have
or that (1 + A;)« > 1 ;
and by the preceding lemma, in order that this condition may be
satisfied, it is sufficient to have
1-1
546 COLLEGE ALGEBRA [ i 2534-531
634. Consider now an increasing G. P. : let a be the fiist tern
and r the common ratio. By formula (i) {590, the (n+ 1)^ tenD is
equal to ^^».
in order that this term may be greater than a given namber J, it is
sufficient that j
ar^^ A or r" > - .
a
and, according to Lemma I, n can be taken sufficiently large to satisf j
this condition; i. e., when
„>iL_ or n>^-.
Since the progression is increasing, if the n^ term is greater than J,
then it follows that every succeeding term is greater than A. There-
fore, in an increasmg G. P, the successive terms increase^ andJUail^
become larger than any assigned quantity.
535. Similarly, consider a decreasing G. P. ; let a be the fiist
term and r the common ratio. The {n -f- 1)^^ term is
ar^.
In order that this term may be less than any assigned quantitT, c
however small, it is sufficient that
ar^ <re or r" < - J
a
and, according to Lemma II, n can be so chosen that this ooaditaoi
is fulfilled. Therefore, in a decreasing G. P. the suceessiue term
decrease, and finally become less than any assigned number.
536. Theorem II. — In a G. P. composed of a limited niiiR6^ *f
terms, the product of two terms which are equally distant from tit etJ
terms is constant.
Let the G. P. be
(a) a, b, c, d, g, h, k, 7,
composed of n terms, of which the ratio is r. If the order of tfc^
terms is reversed, a new G. P. is formed,
(b) I, k, hy g, (/, c, b, a,
composed of n terms of which the ratio is - •
In the progression (a), take the terms d and g which are eqvoBf
distant from the extreme terms ; the term d, which has three tei^
preceding it, is equal to
ar^',
U537, 538] GEOMETRIC PROGRESSION 547
the term g, which has three terms after it in the G. P. (a) and three
preceding it in the G. P. (b), is accordingly equal to
The prodnct of these two terms is equal to
(«^) ?•-=«? = constant
In general, consider, in the progression (a), two terms equally distant
from the extreme terms, such that the first has m terms before it,
and the second m terms after it. The first of these terms is equal to
ar"^', [2530]
the second, which can be considered as a term in the G. P. (b) pre-
ceded by m terms, is equal to
« (r)«'
and the product of the two terms is equal to
537. Problem II. —Find the product of the first n terms of a
6. P. , knowing the first and the last term.
Let a, h, Cy dy ffy h, h ^
be the first n terms of a G. P. and let Q be their product. Then
it follows that
(1) Q = abed ghkly
or, on reversing the order of the factors of the second member,
(2) Q = lkhg dcba.
On multiplying equations (1) and (2) together, member by member,
(3) Q* = al'bk'ch hc'kb'la.
Since the products bk, cA, . . . are all equal to al, by 2636, and
there are n of these products which equal in number the number of
terms in the G. P., therefore, it follows that
nvhence
(4) Q = v^(^.
638. Problem III. — Find the sum of the first 7i terms of a G. P.,
^Ten the first term and the ratio.
I^et (1) a, b, Cy dy ^, A, k, ly
l>e the first n terms of a G. P. in which the ratio is r. Let ^be the
0imi of the first n terms ; then it follows that
548 COLLEGE ALGEBRA [11539, 540
(2) S=a+ b+ c + d+ ^ g^ h^k+l.
Multiply both members of (2) by r, and it then follows that
(3) Sr = ar + br + cr + dr+ +gr + hr+ kr -\- Ir.
Subtract equation (2) from (3), hence
Sr-^ S = lr — a.
Since h =. ar^ c = 6r, etc., I = kr^ [1530]
(ii) s = ^i^.
r— 1
Substitute in formula (ii) ar""'^ for I [2580, (i)], and obtain the
formula sought
(iii) ^=«(r»_=il).
r — 1
This formula enables one to calculate S when the first term a, tlie
ratio r, and the number of terms «, of a G. P. are given.
539. One can easily verify formula (iii) in 2538 on observing tlist
the quotient of r" — 1 by r — 1, or, what amounts to the same
thing, of 1 — r" by 1 — r, is
1 + '-+^+'^+ • . . . +»•"-';
hence it follows that ^^^ "^, ' can be written
r— 1
a -|- ar + ar^ + ^^ +••••+ ^'^"S
that is, as the sum of the first n terms of the G. P. where the first
term is a and the ratio r.
540. Discussion of the Preceding Formula* — Consider the
formula / ^^x ^ _ a{r^ — \)
If the ratio r is greater than 1, r" increases while n increases, and
consequently S increases, which is evident a priori. But, moreover.
n can always be taken so large that r* will be larger than any
assigned quantity, and consequently S also becomes greater than
any assigned quantity (Lemma I).
If r is less than 1, the formula may be written as follows,
1 — r
which is equivalent to the difference between two fractions with
positive numerators, namely,
S^^ :^.
1 — r 1 — r
In this form it is seen that the sum of the first n terms of a Q. P. is
composed of a constant part ^
1— r
S541] GEOMETRIC PROGRESSION 549
minus the part _2^r__ ^
1 — r
which varies as n varies. As n increases, the fraction , ^^ dimin-
1 — r
ishes, and consequently S increases; since as n increases, less and
less is subtracted from the fixed quantity —^ — But, when n
increases without limit, the variable fraction - — can be made
1 — T
less than any assigned quanitity, siuce the denominator of the frac-
tion 1 — r is fixed and the numerator ar^ can be made less than any
assigned quantity (Lemma II).
Since
1 — r 1 — r
and since J''^ can be made as small as is desired, it follows,
1 — r ' '
therefore,- that S becomes as near equal to , as is desired.
1 — r
This result is expressed by saying that, as n increases without limit,
the sum S approaches the limit r-ff- Thus,
(iv) ^^ = lim(-^ ^\ =-^, r<l.
541. Applications. — 1. Find the sum of the first fifteen terms of
the G. P.,
1, 2, 4, 8, ... .
By formula (iii) 2538,
o_a(r"-l)
»'-l
where a = 1, r = 2, and n = 15.
>Sr = ^^1^^^ = 32767.
2. Calculate the limit of the sum of the decreasing G. P.,
.111
^' ?. 9' 27' • • • '
supposed to be continued indefinitely.
By formula (iv) J 540,
lim ;S' =
1— r
where a = 1, r = J.
3. Calculate the limit of the sum of the decreasing G. P. ,
1 1 1 1 1
^' ? ? 8 16' • • • •
550 COLLEGE ALGEBRA [SI542-544
Consider the formula
S =
l-r 1-r'
here a=l, r = J, ?i=as large an integer as is desired = oo .
Hence ^ = _X_ _ .(ll!L = 2 - J^.
Since the progression is composed of as large a number of terms as
is desired, 2""^ can be made larger than any assigned quantity and
^-^ therefore smaller than any assigned quantity. Hence it may be
said that 6y taking n large enough^ the sum of n terms of the series
can he made to differ from 2 hy as small a quantity as is desired.
This is abbreviated into the following: The sum of an infinite num-
her of terms of this series is 2,
542. Theorem III. — In a decreasing G. P. continued to infinity ,
each term hears a constant ratio to the sum of all which follow it^ the
common ratio heing supposed less than unity.
Let the series be a + ar -f- «r* + ar* + . . . . ; then the n*"»
term is ar^'^] the sum of all the terms which follow this is
ar"(l + r+r«+ .... ) = f}-^^ r<l.
Therefore, the ratio of the n*** term to the sum of all which follow it is*
„_i . ar^ 1 — r
1 — r r
This is constant, whatever n may be. If it is desired to determine r
so that this ratio may have a given value A;, we put = k\
from which r = — ,— -.
543. Recurring decimals are cases of what are called infinite
geometrical progressions. Thus, for example, .5343434 ....
denotes A+R+M + ^^+ Here the terms after ^
constitute a G. P. of which the first term is — ^» and the common
ratio is — • Hence it follows that the sum of an infinite number
S4 / 1 \ 34
of terms of this series ^^ ^ -r- M — — j» that is, --. Therefore
the value of the decimal is — + <^^- The general rule for such ex-
amples will be discussed iu the next section.
644. To Find the Value of a Recurring Decimal. — Let i?^ denote
the figures which do not recur, and suppose that they are n in number;
let Q denote the figures which do recur, and suppose that they are
81545,546] GEOMETRIC PROGRESSION 551
m in number. Let *S^ denote the value of the recurring decimal ; then
(1) ^=-I^QQQ . . . . ,
(2) lO-S=r.QQQ . . . . ,
(3) lO^^^S =FQ,QQ . . . . ,
. •. from (3) and (2), (4) (10«+»»— 10")^ =FQ^F,
But lO"*"' — 10" = 10"(10'" — 1); and 10*" — 1 expressed by figures
in the usual way is composed of m nines. Hence follows the rule
for finding the value of a recurring decimal: Subtract the integral
number consisting of the non-recurring figures from the integral number
consisting of non-recurring and recurring figures, and divide by a
number consisting of as many nines as there are recurring figures
followed by as many ciphers a^ there are non-recurring figures,
645. The formulae
(i) I = ar^-^
r — 1
express two relations between the five quantities, a, l, r, n, and S; the
two relations enable one to calculate any two of these five quantities,
when the other three are given. One is therefore led to ten
different problems, according as the quantities comprised in one of
the following ten groups are taken as unknowns:
(a, D] {ay r); (a, n); {a,S): (I, r)]
(/, n); (;, ^); {r, n); (r,^^); (n,^).
646. Problem IV. — Insert n geometric means* between two
given numbers, a and b ; that is, form a G. P. composed of n 4- 2
terms, such that a and b are the extreme terms of this progression.
Let r be the ratio of the G. P. Since the term b is preceded by
n + 1 terms, it follows that
b=ar-^\ [«680,(i)J
whence (1) r = "^*^l-
Therefore the required G. P. will be
<2) «. <!)"• K-^r- «(!r- • • • • . ^-
r
For, the (r + 1)"» term is a (-Y and hence the (n + 1)"» term is
a(^y^\ and finaUy the (n+ 2)»>» term is a(^)"^' ="*(«)= ^•
" " Q. E. D.
«The abbroTlatlon 0. M. Is used for the phrase, geometrical mean.
552 COLLEGE ALGEBRA [11547-550
ExAMPLjB. — Insert 5 G. M. between 2 and 1458.
By formula (i) r = •%/'^ = VV729 = 3.
The progression is therefore
2, 6, 18, 54, 162, 486, 1458.
Harmonical Progression
547. Three quantities are said to be in harmonic&l progression
when a: c = a — b : h — c. Any number of quantities are said to
be in H. P. * when any three consecutive quantities are in H. P.
548. The reciprocals of quantities in harmonical progression are
in arithmetical progression.
Let a, 6, c, be in H. P. ; then
a : c =za — b :6 — c [1547]
a{h — c) ^ c{a — b).
Divide by a6c, thus
^'^ c 6-6 a
which was to be proved.
549. The property demonstrated in 2548 is sometimes taken as
the definition of an H. P. and the definition stated in 2547 proved as
property of an H. P. The second definition of harmonical progres-
sion may be stated thus: quantities are said to be in harmonical pro-
gression when their reciprocals are in A. P.
The term harmonical is derived from a physical property of
musical sounds. Suppose there is a set of strings of the same
substance, and whose lengths are proportional to 1, J, ^, J, J, and J;
and let them be stretched tight with equal forces. If now any two
strings are sounded together, the effect will be harmonious to the ear.
The property of an H. P. derived in 2648 enables us to solve some
problems related to harmonical progression, although there is no
formula for the sum of any given number of quantities which are
in H. P. , a being the first term and c the last
550. Insert n harmonical means between two given terms a and h.
The meaning of this problem is that one is to find n -j- 2 terms in
an H. P. of which a is the first term and b the last term. Hence the
problem can be reduced to the following:
Insert n arithmetical means between - and 7 .
a 6
* The abbreyiation H. P. U used for harmonical progression and H. M. for bannoo-
Ical mean.
8551] HARMONICAL PROGRESSION 553
Let d be the common difTerence; then
l = l+{n+l)d, [1521, (i)]
therefore d = (\--) H- U+ l) = T ^ r
\h a) ' ^ ^ ^ {n + l)ah
Accordingly, the A. P. is
H + '^' ^ + 2'^' \+-^^ I'
that is
(IX 1 6(?i+l)+(«~6) 6(n+l)+2(a-6) 6(n+l)4-n(a~6) 1
^ ^ a ah(n+l) ' a6(n+l) ' • • • • a6(n+l) fc'
Hence the H. P. is
/9x ^ g6(n-|-l) Q6(n+1) a6(n+1) r
^"^ » 6(n+l)+(a-6)' 6(n+l)+2(a-t)' ' ' ' ' fe(n+l)+n(a-6)'
551. Let a and h be any two given quantities and A their A. M.,
G their G. M., and J7 their H. M. Then by definition
^-a=6-^; ... ^=«±i. [{628,(2), n = l]
^^) \ a: G=G : h', .\ G = Vol. [1646,(2), n=l]
a : h=a-H: H-b- .', ff =~^' [1550,(2), 7i = l]
It follows from these equations that:
A ' li — — T- . — —- = ah
2 a+ h
and G^ = ab,
(iii) 6!^« = ^ • //and ^ : G= G.H.
Hence, 6^ lies in magnitude between A and //; and A is greater
than /T, for
J _ ^ _ a -f6 __ _2«6_ _ (g 4- fe)^ - Aah _ (a-6)V
2 a+6 2(a+6) ""2(0+ 6)'
since r^ "^ ^ is a positive quantity, therefore
A:>ir.
Hence from (ii) and (iii) it follows that
(iv) A>G>rf.
Examples.
I. Continue the series 7 + ^^ + } two terms if the terms of the
series are in H. P.
According to {548 7 h' 7
is an A. P. Therefore the common difference is
^ - A _ I - 3
^ " 14 7 ~ 14
554 COLLEGE ALGEBRA [5551
Hence the A. P. eontiiiued two terms is
1 5 4 11 1
and therefore the corresponding H. P. is 7, "T' 7» jp 1 ; ^^'^ ^
seriesisT + ^ + ^ + J^+l.
5 4 11
2. The sum of two contiguous terms in H. P. is ^^^, and tiieir
product is -^j. Find the series.
Let a and h be two terms in H. P., then according to the conditions
of the problem
ah = -V
52
The solutions of these equations arc the roots of the equation,
^«_ 29^^+_L— 0, [1422]
2 1
which are a=z— and h =--
\o o
Since a and h are two contiguous terms in H. P. then - and - an
a 6
two contiguous terms in A. P., i. e.,
13 8
T 1
in A. P. Therefore the common difference is
2 ' 2
and the A. P. is y, 8, ^, 11, J. 14 ... .
Hence the corresponding H. P. is
A 1 2^ JL A Id
13' 8' 19' ll' 25 • • • •
EXEBCISE XXXXn
Solve the following examples and problems in A. P., G. P., and H.P.
1. Find the sum of 23 terms of the A. P. 4 + 7 + 10 + etc.
2. Find tlie sum of 1 + 3 + 5 + . . . + (2» + 1).
3. What is the fourteenth term of the A. P. whose fiftii tenn
is 11 and whose ninth term is 7?
4. Sum the following arithmetical progressions:
(a) 5+9+13+ ... to 15 terms; (b) 'l+i+ ... to 16 terms;
(c) i + ^-^-^ + . .. ton terms; (d) I±j+-1L + . .. to Z terms.
8551] PROGRESSIONS 555
5. How many of the natural numbers, beginning with unity,
amount to 500500?
6. Insert
(a) 15 A. M. between 3 and 30; (b) 10 A. M. between —3 and+3;
(c) 30 A. M. between 5 and 90; (d) 36 A. M. between 8 J and 2 J.
7. Find the sum of n terms of the series obtained by beginning
with the !•*, r*^ 2 r^, 3 r'^ etc., terms of the A. P. whose first
term is a and whose common difference is d.
8. Sum the following geometrical progressions:
(a) 6 + 18 + 54+ .... to 12 terms;
(b) .333 .... to 71 terms;
(c)
... _|_ r u + ... to 20 terma:
»/3 + l 1/3 + 2
(d)
1 — ^H-^— to 17 terms;
(e)
6 — 4+ .... to 10 terms;
(f)
6 — 18 + 54 .... to 12 terms.
9. The third term of a G. P. is 2 and the sixth term is — ^;
what is the tenth term?
10. The sum of the first 10 terms of a certain G. P. is equal to
244 times the sum of the first 5 terms; what is the common ratio?
11. Insert
(a) 2 G. M. between 8 and —1; (b) 3 G. M. between 2 and 18;
(c) 4 G. M. between | and 128; (d) 5 G. M. between*||. and 4f
12. What common quantity must be added to a, h, c, to bring
them into G. P. ?
13. Show that the product of any odd number of terms in a
G. P. will be equal to the n"* power of the middle term, n being the
number of the terms.
14. Show that 4, 6, 12 are in H. P., and continue the progression
both ways.
15. Find the H. P.
(a) whose third term is 5 and whose fifth term is 9 ;
(b) whose p^^ term is P and whose q^^ term is Q.
16. Find the H. M. between the A. M. and G. M. of a and 6.
17. Insert
(a) -2 H. M. between 1 and 3; (b) 5 H.M. between 6 and 8;
(c) n H. M. between a and h.
556 COLLEGE ALGEBRA [J551
18. A servant agrees to serve his master for twelve months, his
wages to be one dollar for the first month, two dollars for the second
month, four dollars for the third, and so on; what did he receive for
the year's service?
19. The population of a certain town is P at a certain epoch;
annually it loses ^ % by deaths, and gains ft % by births, and
annually a fixed number E emigrate; find the population after a
lapse of n years.
20. Show that if a, 6, c are in H. P.,
(a) ? = ^ ■ '
6 6~a ' 6-c'
(b) 1^ + 1^ = 2;
(c) a, a — c, a—h are in H. P. ;
(d) «-|. |. c-|areinG.P.;
(e) c^ c — Gy c — b are in H. P. ;
(f) rrr^ , -r^' -r! are in H. P.
6+c— o c+a—b a+b—c
21. Show that, if a, 6, c, d are in A. P., a, c, /, d in Q-. P.,
and a, y, ^, d in H. P.,
22. The series of natural numbers is divided into the following
groups: 1;»2, 3; 4, 5, 6; 7, 8, 9, 10; and so on; find the sum of the
numbers in the A;"* group.
23. If a, 6, c are in A. P., and a', 6», c* in H. P., show that
— ^ » 6, c are in G. P. , or else a=ib :^c.
24. Show that any even square, (2 n)', is equal to the sum of n
terms of one series of integers in A. P. , and that any odd square,
(2 y^+ 1)*, is equal to the sum of n terms of another H. P. increased
by unity.
25. If the same number of geometric means are inserted in each
interval between any two consecutive terms of a G. P. , and if the
progressions then found are arranged in a series one after the other,
so that the last term of one progression is the first term of the one
following, a single G. P. is formed.
26. What is the sum of the decreasing geometric progression
1 1 1
n+1 (n+l)« (n+lf
(n>0)?
CHAPTER IV
LOGARITHMS
662. Definition. — 1. If y = a^, the mode of dependence of y
upon X is as the square of x, and the values of y are readily caleh-
lated if we know those of x\ thus, when
a; = l, y = l;
x = 2, y = 4;
X = 3, y = 9, etc.
553. 2. The mode of dependence of x upon y can also be
expressed by extracting the square root, thus
x = Vy\
and the values of x are readily found when those of y are known;
thus when y^^^ ^^^/l^l.
y = 2, x = v'2 = 1.414 . . . ;
y = 3, x= 1/3 = 1.732 . . . ;
y = 4, X = i/4 = 2, etc.
554. 3. In the equation N= 10* the values of N which cor-
respond to the values of x can also be found by involution or
evolution, thus, when
x = l, iNr=10;
x = 2, ^^=100;
X = ^, N= i/lO = 3.162 ... etc.
However, the inverse problem of finding x when N is given can
not be solved by algebraic operations as was the case in Examples
1, 2, and the first part of 3.* The mode of dependence of x upon
iV, by definition is called a logarithm, which is defined by the follow-
ing equations:
If N = a* then by definition x = log ^ AT.
*In a subsequent chapter tt will be shown that equation lO' = 1, lO' = 8 . . . can be
solved approximately^ that Is, for example, although a value of x that will make lO' = 2
exactly can not be found, yet a value of a; which will make lO' differ from 2 by as smaU a
quantity as is desired, can be found.
567
558 COLLEGE ALGEBRA [H555^7
555. Definition of a Logarithm. — The logarithm of a gitm ■■«•
her IS the exponent denoting the power to which a fixed number cnlki
the base must be raised in order to produce the given number.
Example 1. Since 4' = 64, 3 is the logarithm of 64 to tk
base 4.
Example 2. Since 10* = 10, by definition 1 = log,, 10
10* = 100, ** ** 2=log,olOO
10» = 1000, ** " ' 3 = logi9lO«X»
etc. etc.
556. A System of Logarithms. — ^The logarithms of all positive
numbers to a given base constitute a system of logarithms. Now, if it
is supposed that a remains fixed while iV takes in succession all Tallies,
from 0 to -f- oD , the corresponding values of x will constitute a «y«fm
of logarithms to the base a, and a is called the ba^e of the systm.
Any number might be taken for the base a of a system, and cone-
spending to any such base a system of logarithms of all numbeis
could be found.
Thus, suppose a = 7.
Since 7» = 7, 1= logy 7.
7« = 49, 2= log 7 49.
7' = 343, 3 = log 7 343.
Thus, the integers 1, 2, 3, . . . are respectively the logarithms of
7, 49, 343, ... in the system whose base is 7.
Properties of Logarithms
557. 1, The logarithm of 1 is 0 whatever the base may be.
For, a* = 1, for all values of a\
logo 1=0, whatever a may be. [iSMj
2, The logarithm of any number to that number as bcue is 1,
For a^ = a.
Logo a = 1. [I5»:
3. The logarithm of 0 in any system whose base is greater tka*
1 is minus infinity.
For a-"=4s = - = 0
LogaO = — 00, a> 1. [1658]
8557] LOGARITHMS 559
4. The logarithm 0/ + 00 in any system whose base is greater than
I IS plus infinity.
For a* = ao
loga(+ao)=+oo, a>l. L?855J
5. The logarithm of a product of several factors is equal to the sum
of the logarithms of the factors. Thus
log a (wi • n) = loga m + loga w.
Proof — Suppose that mn is the product and a the base of the
system.
Let (1) m = a%
and (2) n = a^ ,
(3) a; = logom,
and (4) y = log a «• [ J555]
Multiply equations (1) and (2) together
mn = a' - a^ = «'+».
(5) Loga (ww) z=x + y. [{555
Substitute for x and y in (5), the value of x and y in (3) and (4).
(6) Log a (wiw) = loga^+ log a n,
CoROLLABT. — ^Put itp for w in (6), then;
Loga (fnnp) = loga m + log a ^/^
Loga i'^»P) = loga Wi + loga « + logai>. [(6)]
B. g. Loga 42 = loga (2 X 3 X 7)
= loga2+loga3 + loga7.
6. The logarithm of a fraction is equal to the logarithm^ of the
numerator minus the logarithm of the denominator. Thus
loga^ = loga »» — loga «.
Proof — Let - be the fraction, and suppose
n
(1)
m = a*,
and (2)
n = ay.
By {666 (3)
x = log„w,
and (4)
y =log„n.
Divide (1) by (2), (5)
- = — = a*-».
n ay
.-. by «666, (6)
log.
.(")= '-»■
Sabstitate in (6) the value
X and y from (3) and (4),
(7) loga- = log„ m — loga ^-
560 COLLEGE ALGEBRA [{{558-560
COBOLLAEY. — Log« ^ = log„ m - log„ {np) [(7)]
= log„ m — log„ n — log« p. [{557, 5]
= log^ 2 + log« 3 + log„ 5 - log, 7 —log, 11.
558. The logarithm of any power , integral or fractional, of a
number is equal to the product of the logarithm of the number and the
index of the power. Thus
log a (m») = nlog,m.
Proof
Let (1) m = a*
(2) cc = log,tn. [{555]
Hence, from(l), (3) m» = {a'Y = a'«, [{313]
(4) log,(m«)=nx. [{555]
Substitute in (4) the value of x in (2),
(5) log„(m'') = n log,m.
E. g., log, [(21)^ = 5 log, 21 = 5 (log, 3 + log, 7).
Corollary. —Put n = - forn in (5), then
(6) loga (m'-) = ilog,
That is, the logarithm of a root of a number is equal to the logarithm
of the number divided by the index of the root.
E. g., Log/VIT9 = -ilogall9.
559. It follows from (5), (6), and (7) that by the use of loga-
rithms the operations of multiplication and division may be re-
placed by those of addition and subtraction respectively, and the
operations of involution and evolution by those of multiplication and
division.
560. Comparison of Two Systems of Logarithms. — Given the
logarithm of a number to the base a, find the logarithm of the same
number to the base b.
Thus Logftm =l5Sj!J?L.
logab
For, let m be any number whose logarithm to the base b is required.
Let (1) m = a'' also . m=i*',
then (2) x = \ogam and y = log i,m. [{555]
From (1), (3) a' = 6" ;
2561] LOGARITHMS 561
hence (4) a^ = b or a = 6*
(5) ^=logaft or 2^=:loge,o.
y X
Hence (6) y = x log ^ a, and y = — -^ — ^
log a b
i. e., logftwi = J2£«J^ = logaWiX log^a.
logo 6
Therefore the logarithm of a number to the base b may be found by
multiplying the logarithm of the number to the base a by
log ft a, or by —J—-,
log a 6
Finally, multiplying together the equation in (5)
(8) logfta X loga& = 1.
561. Applications.
1. Express the logarithm of ^-^ in terms of log a, log b, log c.
Log.^; = log , A = log . a* -log.(c»6')
= |log«-(31ogc + 21og6)
2.
= - log a — 3 log c — 2 log 6.
Log (^+y)"^ = log (x + y)«2« - log (x + y)xo^
= n log (aj + y) + m log 2 — log (x + 2/) — | log w?.
EXEBOISE LXXXm
Apply the principles of logarithms to the following examples.
1. log a6c. 2. log 3 rtx(x + y).
3. log _4_. 4. log ^^^.
5. log ab'^. 6. log (a6)«.
7. log(a« + 6«), 8. log(a»-6«).
9. log a Wb. 10. log 1/^.
*NoTi.— When the same base Is used throughout a set of problems, the Indication of
the base is omitted. For example, the base here might be anything.
562 CX)LLEGE ALGEBRA [8561
11. log5a«6Vc. 12. log7xV^3.
13. log31x(7u:-8)'. 14. log 8 a«6(6 c — c/)>.
15. log5xVa(8i/ — «). 16. log 9 xt/' i V + fc')c.
17. log ^. 18. log ^.
19. log ^^"^;'-?^^^^> 20. log ^;^^t^-^^ .
21.
23.
324V((u--t/)« (^+2) Vex - rf
log J^^. 22. log (^^^)^
logj^^l- 24. .og«5^,^^
6*
1
2^- ^«S (^+6V- 26. log ^,^^^=-
27. log -"^"^-^^i 28. log i^A'*-
29. log V^3 + l/2. 30. logV^ + Vc.
31. log«4±|. 32. log ^^1
Find the expressions whose logarithms are
33. log a + log h — log c.
34. log a — log b + log c — log d,
35. log a — (log b + log c) + log (Z.
3G. 31oga+ 21og 6 — 41og c.
37. ^logx — ^logy+ Jlog^;.
38. 21oga-ilog6 + ilogx-31ogy.
39. 7 log (a + 6) - f log (a— 6) + ^log X — 4 log y,
40. J log (ax — 6) — . I log (rx — (/) + ^ log (wx — n).
41. J log (a« + 6«) - J [log (a + 6) + log (a - 6)] .
42. 2 log (x- 1/)- Jlog (x2-xi/ + /) - ilog (x + y).
43. logf+log^+log^-log^J.
44. log - + log (xy) - 3 log (x-ij) — log J^.
5562] LOGARITHMS 563
Common System op Loqarithmb
662, It is possible to have any number of systems of logarithms
but common usage has adopted only two, viz. , the Napierian sys-
tem and the Common system.
The Napierian system, named after its inventor, John Napier, is
used in theoretical investigations; its base is
' = ^ + ^ + 21 + 3! + ^+ .... =2.7182818....
The Common system of logarithms is used in numerical calcu-
lations; the base in this system is 10.
Now.— The odTantage due to the use of the base 10 will be seen in the rules for the
Characteristics which immediately follow and the Characteristics given in the next
paragraph.
If the base is 10, then
10^ = 1 .-. 0=logl
101 = 10 .-.1 = log 10
10» = 100 .-. 2 = logl00
10^=1000 .-.3= log 1000
log of numbers between 1 and 10,
such as log 7, = 0 + decimal.
log of numbers between 10 and 100,
such as log 89, = 1 + decimal.
log of numbers between 100 and 1000,
such as log 749, = 2 + decimal.
log of numbers between 1000 and 10000,
such as log 6979, = 3 + decimal.
10* = 10000 .-. 4 = log 10000
etc. etc.
It follows from this table that the logarithms of numbers greater
than 1 consist of two parts, an integral part and a decimal part.
The integral part is called the Characteristic and the decimal part
the Mantissa.
It follows also from the preceding table that if a number is ex-
pressed by one digit the characteristic of its logarithm (e. g. , log 7)
is 0 ; if it is expressed by two digits the characteristic of its loga-
rithm (e. g. , log 89) is 1 ; if it is expressed by three digits the char-
acteristic of its logarithm (e. g. , log 749) is 2 ; and so on, the char-
acteristic being one less than the number of digits in the number.
Thus, the characteristic of log 11749 is 4, of log 6748.63 is 3.
Therefore the characteristic of the common logarithm of any
number greater than 1 can be written down by the following rule.
564 COLLEGE ALGEBRA [J {563, 564
563. Rule I. — 7%e characteristic of the logarithm of a number
greater than unity is positive and is equal to the number of digits \%
its integral part less one.
564. Similarly, if the number is less than 1 and greater than 0,
10* =1, hence O=logl
10-1
_ 1
10
io-«
_ 1
100
io-»
1
1000
= .1, " — l = log.l
For example
log .6 = — 1. + decimal
= 9. decimal — la
For example
log .07 = — 2. -U decimal
= .01. " -2= log .01 ^ =8. decimal -la
For example
log .008 = — 3. + decimal
= .001, " -3= log .001 =7. decimal -la
For example
log .0006 = — 4. + decimal
10-* = -J— = .0001, " -4=log.0001 ^ =6.d©cimal-ia
10000 ' ^
etc. etc.
From an inspection of this table it is clear that, in the common
system, the logarithm of any number between
1 and .1 is some number between 0 and — 1,
e. g., log .6 = — 1 + decimal = 9. decimal — 10;
.1 and .01 is some number between — 1 and — 2,
e. g., log .07 = — 2 -|- decimal = 8. decimal — 10;
.01 and .001 is some number between — 2 and — 3,
e, g., log .008 = — 3 + decimal = 7. decimal — 10; and bo on.
In other words, the logarithm of any decimal with no zero between
its point and first figure, is equal to 9 plus some decimal, minus 10 ;
the logarithm of any decimal with one zero between its point and
first figure, is equal to 8 plus some decimal, minus 10; the loga-
rithm of any decimal with (wo zeros between its point and first
figure, is equal to 7 plus some decimal, minus 10; and so forth.
In General. — Let 2> be a decimal with n zeros immediately slier
the decimal point; then
7) L_ 1 0+d«<^™»l 10-(ii+l)+ decimal
— 10"-»-i ~"
log i> = — (« + 1) + decimal. [IMS]
Therefore the characteristic of any number less than 1 and greater
than zero is determined by the following rule.
«565,566] LOGARITHMS 565
666. BuLE II. — If the number is less than 1 the characteristic is
found by subtracting the number of zeros between the decimal jwint and
the first significant figure from 9; writing — 10 after the mantissa,
E. g. , the characteristic of logarithm . 00679 = 7. decimal with
— 10 written after the mantissa; and the characteristic of logarithm
.3796 = 9. decimal with — 10 written after the mantissa.
In practice it is customary to omit —10 after the mantissa; it is
however a part of the logarithm, and should be allowed for and
subjected to exactly the same operations as the rest of the logarithm.
Beginners will find it useful to write — 10 in all cases, and in many
problems it can not well be omitted.
NoTX.— Many writers. In using logarithms less than 1, combine the two parts of the
ehaiacterlstic and write the result as a negative characteristic before a positive mantissa.
Thus, instead of the logarithm 7.578803 —10, the student will frequently find 3.573968, a
minus sign being written over the cbaraoteristic to show that it alone is negative,
the mantissa being always positive. A well-founded objection to this notation la that it
18 inconvenient to use numbers partly, positive and partly negative.
Use of the Table
566. Calculations have been made for the common logarithms of
all integers from 1 to 200,000, and the results tabulated. For
general use, tables of logarithms give six decimal places, but those
used especially for astronomical and mathematical calculations give
seven or more decimal places. In the examples we use the common
table, which gives the mantissas of the logarithms of all integers
from 1 to 10,000 calculated to six decimal places. For convenience,
the logarithms of integers from 1 to 100 are given on the first page,
but the same mantissas are to be found in the rest of the table.
1. Tlie characteristic of a logarithm, can be written according to
the rule^ of 22563, 565, and the mantissa looked up in the tables.
2. Hie mantissas of the logarithms of all numbers which have the
same sequence of figures is the same; so only the mantissas of integers
are given in the table.
For, let iV be a number of any sequence of figures, then
log,o(iV'xlO») = logioAr+logiolO»=logioiV+n,
log io(^-4-10») =log ,0^-log iolO"=log ,0^-n.
Thus only the characteristic of the logarithm is affected.
Examples.
1. Given log^o 3296.78 = 3.518090,
then log ,0 329678 = log ,o (3296. 78 X 10«) =log io3296.78+2
= 3.518090+2=5.518090.
566 COLLEGE ALGEBRA [H567,568
2. Logio3.29678=logio(3296.78^10»)
=log,o 3296. 78-3=3.518090— 3=0.518090.
3. Log 10.00329678=3.518090—6=7.518090-10.
That is, in the common system of logarithms, if the logariikm. of any
number is known, we can immediately determine the logarithm of tkf
product or the quotient of that number by any power of 10, hy adding n
to or subtracting n from the chara^eteristic, axxording as the number is
multiplied by 10^ or divided by 10'*.
567. To find the Logarithm of any Number Consisting of Four
Figures* — Find, in the colamn headed N, the first three figures of
the given number. The mantissa required will be found at the
intersection of the horizontal line through these three figures and
the vertical column headed by the fourth figure. If only the last
four figures are found, the first two figures may be obtained in the
same vertical column from the firat mantissa above consisting of
six figures. Prefix the proper characteristic (2563 or 1565).
For example, 1<^ 579.8=2.763278,
log .006847=7.835500—10,
log 9899=3.995591.
To find the mantissa of a logarithm of a number consisting of
two figures, use the first page of the table ; for a number consisting
of three figures, look in the column headed N and take the maotisn
from the column headed 0.
For example, log 8.95=0.951823.
568. To find the Logarithm of a Number of More Than Four
Figures*— For example, find the logarithm of 356.478.
From the table, the mantissa of 3564 = .551938
mantissa of 3565 = .552060
Hence, the change in the mantissa correspond- ) __ 000199.
ing to a change of one unit in the number )
therefoi*e the change in the mantissa due to a change of .78 units
in the number is .78 x .000122 = .000095
correct to the sixth decimal place.
Hence, mantissa of 3564 = .551938
Correction for ,78 is .78 X .000122 = .000095
.-. . log 356.478 =2.552033
Note.— In making tho correction in the mantissa corresponding to .78 IneiMM is
the number, It was assumed that the differences in logarithms are proportional to i^
differences of the corresponding numbers, which is not exactly true but is auifieicntll
accurate for practical purposes.
«569]
LOGARITHMS
567
569. The following rale is derived from the preceding discussion:
Take from the table the mantissa of the first four figures without
regard to the position of the decimal point.
Subtract this mantissa from the mantissa of the next higher number
of four figures; (this difference is the tabular difference^ and is found
in the column headed D on each page).
Multiply the tabular difference by the remainder of the figures of the
given number^ with a decimal point before them.
Add the result to the mantissa of the first four figures.
Prefix the chara4:teristic (2663 or 2566).
Example.— Find the logarithm of .003426098.
Tabular difference = 127 mantissa of 3426 = .534787
.098 12
1016
1143
mantissa of 3426098 = .534799
12.446 = 12 nearly. Ans. 7.534799—10.
EXEBCISE liXXXIV
Find the logarithms of the following numbers:
1.
Log 369,
log 58. 6,
log 2.390.
2.
1x^476.9,
log. 03788,
log. 8673.
3.
Log 77860,
log 54327,
log 13084.
4.
Log 99286,
log 90801,
log 55080.
5.
Log 10010,
log 99991,
log 10001.
6.
Log 1851273,
log 14459809,
log 10134761.
7.
Log 7095137,
log 506860900,
log 3. 614699.
8.
Log 84.827567,
log 211447.39.
9.
Log 703.84,
log 73. 084,
log 0.0073084
10.
Log 0.008765,
1(^0.00987,
log 0.00003.
11.
Log 0.87701,
log 368. 13,
log 5. 0009.
12.
L<% 0.000875,
log 1.0001,
log 0.00173.
13.
Log 8. 0808,
log 0.3769,
log 0.070707.
14.
L<^ 0.0003599547,
log 75907 J,
log32116f
15.
Log 2528811^,
log 522076, ij,
log 80325 1-f
16.
Log 0.0013514133.
568 COLLEGE ALGEBRA 181570-573
570. To Find a Number corresponding to a Given Logaritimi.
Example 1. Find the number whose l<^arithm is 2.713S47.
Since the characteristic of a number is determined entirely by tix
position of the decimal point in the number, the mantissa alone
detejmines the nature of the sequence of the digits in the number
without regard to the position of the decimal point in the number.
We find in the table the next less and the next greater mantissa:
Mantissa of 5171 = .713575
Mantissa of 5172 = .713659
Correction for change of unity = . 000084, tabular diflTereooe.
Thus a change of 84 in the mantissa produced a change of 1 in the
corresponding number. Now find the difference between the given
mantissa and the next less:
.713647
.713575
.000072.
Hence an increase of 72 in the mantisa will produce an increase of
^of 1 in the number, or - • Therefore, since the characteristic is 2,
the number corresponding = 517 (1 +7)= 517.1857.
571. The following rule is inferred from the preceding opera-
tions:
Find in the table the next less mantissa and tabular differenkot.
Subtract the next less mantissa from the given mantissa and divide l&e
remainder by the tabular difference (the quotient is usually only
correct to two decimal places, note 2568).
Annex to the first four figures all of the quotient except the deci-
mal point and point off according to the rules which are the reverse
of the rules in IS563, 565.
Rules foe Pointing off
572. I. If — 10 is not written after the mantissct^ the number of
the digits to be pointed off will be one greater titan the chanzcteristic
II. If — 10 stands after th^ mantissa^ the characteristic subtracted
from 9 gives the number of zeros to be placed between the decimal poini
and the first significant figure.
8573] LOGARITHMS 569
Example 2. Find the number whose logarithm is 7.115658—10.
.115658
Next less mantissa = .115611, four figures corresponding are 1305.
Tabular difference 332 ) 47.00 ( .14 nearly,
332
1380
1328
~^
therefore the number corresponding = .00130514.
EXEBOISE IiXXXV
Find the numbers corresponding to the following logarithms:
1. 0.903090. 8. 0.00008376.
2. 2.397940. 9. 7.069907.
3. 0,724030. 10. 0.6260096 — 1.
4. 3.908190. 11. 8.234560 — 10.
5. 8.389910 — 10. 12. 0.02020 — 2.
6. 0.003176849. 13. 6.321434-10.
7. 0.0387695. 14. —5.8794362.
15. Given log 2 = .301030 and log 3 = ,477121, find the log-
arithm of .05, 5.4, .006, 36, 27, and 16.
16. Given log 648 = 2.81157501, log 864 = 2.93651374, find
Ic^ 3 and log 5.
17. Given log 2, find log l/l. 25.
18. The logarithm of 7623 is 3.8821259: write the numbers
whose logarithms are .88211259, 4.8821259.
678. Hints. — The following hints will be useful in performing
the fundamental operations on logarithms whose characteristics are
negative or expressed by writing — 10 after the logarithm.
1. Addition. — If, in finding the sum of several logarithms, it
is found that — 10, — 20, — 30, etc., are written after the mantissa,
and the characteristic written before the mantissa is greater than 9,
subtract from both parts of the characteristic such a multiple of 10
that the part of the characteristic before the mantissa is less than 10.
For example, 17.466478—20 should be changed to 7.466478—10;
35.604762—40 should be changed to 5.604762—10.
570 COLLEGE ALGEBUA [8573
2. tSabtractton. — In subtracting a larger logarithm from a smaller
or in subtracting a negative logarithm from a positive, add each •
multiple of 10 to the characteristic of the minuend, writing the same
multiple of — 10 after the minuend, that the characteristic of the
minuend shall be greater than the characteristic of the subtrahend
Examples. Subtract 4. 617325 from 2. 145631.
Arrange the work thus : 12. 145631 — 10
4,617325
7.528306-10
Subtract 9.946534—10 from 9.352062—10.
Arrange the work thus: 19.352062—20
9.946534—10
9.405528-10
3. Multiplication. — To multiply a logarithm by an integer, simplify
the positive and negative parts of the characteristic as in addition.
To multiply a logarithm by a fraction, first multiply by the
numerator and divide the product by the denominator.
Kg., 3x5.216347—10=15.649041-30-^5=35.649041—50-^5
=7.129808-10
4. Division. — On dividing a negative logarithm, add such a mul-
tiple of 10 to the characteristic before the mantissa and the same
multiple of —10 to the part after the mantissa, that the quotient of
the l&tter by the divisor is —10.
Example.— Divide 6.123456—10 by 7.
Arrange the work thus : 7 ) 66. 123456—70
9.446208—10
BZEBOISE LXZXVI
1. Add 5.013789 — 10, 9.114679 — 10, 7.556688-10.
2. Add 4.673749—10, 3.245789 — 10, 9.567482-10.
3. Subtract 0.794684 from 0.469108.
4. Subtract 8.702143—10 from 2.009901.
5. Subtract 9.864732—10 from 9.795544—10.
■ 6. Multiply 9.112345—10 by 4.
7. Divide 8.608047—10 by 7.
8. Divide 9.879647-10 by 12.
9. Multiply 9.778837—10 by f
«574] LOGARITHMS 571
TuE Solution of Numerical Problems by Logarithms
674. To find the value of any complex numerical quantity by
logarithms, find first the logarithm of the quantity, as in i561, by
means of the table ; then find the number corresponding to the result.
Examples. — 1. Calculate the product
X = 87.56348x0.00628240.
The sum of the loga- f log 87.56348 =1.942323
rithms of the factors ^ log 0.0062824=7.798126-10
gives log X (J667, 5). |^ ., log x =9.740449-10
The number corresponding to 9.740449—10 is x = 0.550109.
2. Calculate the product
X = 0.08756348 x 0.00628241.
log 0. 08756348 = 8. 942323 — 1 0
' log 0.00628241 = 7.798126-10
log X = 6.740449—10.
Hence x = 0.000550109.
The addition of the logarithms gives
16.740449-20 = 6.740449-10;
find, then, by means of the table, the number corresponding to the
result.
3. Calculate by logarithms
__ 42.567X521.62
^~ 9.6843X0.005(37'
Applying the principles of the logarithm of a quotient and product
(S557, 6, 5), log X = log 42.567 + log 521.62
— log 9.6843 — log 0.00567
Arrange the calculation as follows:
log 42.567 =1.629073 log 9.6843 =0.986068
log 521.62 =2.717354 log 0.00567 = 7.753583—10
—log 9.6843 =9.013932—10
—log 0.00567=2.246417 107)72.0(673
642
log x =5.606776 jy ^ 107
780
mantissa 4043 =0.606704
749
72 310
X = 404367.3 321
572
COLLEGE ALGEBRA
[U575, 576
676. Powers. — We know that the logarithm of a number raised
to a given power is the logarithm of the number multiplied by the
exponent of the power.
Examples.
1. Calculate x = 5*.
Here log x = 8 log 5
log 5 =0.698970
8
Hence,
log X =5.591760
, *,
X =390625.2
2. Calculate
X = 0.4326'.
Here
log X = 3 log (0.4326)
log 0.4326 = 9.636087-10
3
Hence,
log
3. Calculate x
X
X
2
10
Hence,
= 8.908261-
= 0.080958.
=©'
log X = 5 (log 2 -
log 2 = 0.301030
— log 37 = 8.431798—10
^og~ = 8.732828—10
'' 5
log
log 37)
= 3.664140-10
X = .000000461466.
676. Roots. — The logarithm of a root of a number is the loga-
rithm of the number divided by the index of the root
Examples.
1. Calculate x = V7239.812.
log^ = jj; log 7239.812
log 7239.812 = 3.859728.
Hence log V7239.812 = .3508844.
X = 2.24328.
2. Calculate x = V-0. 00230508.
The sign of x will be minus.
log (— x) = I log 0.00230508.
log 0.00230508 = 7.362686-10
21577, 578] LOGARITHMS 573
Hence log (—x) = 9.472537— 10.
— x=. 296850.
3. Calculate x = s / ^-^^ .
\tn-ar 8.427
log X = J(log 4.528 — log fTT — log 8.427)
log 4. 528 = 0. 655906 log | ;r = 0. 622089
-logjT =9.377911 — 10 log 8.427 =0.925673
— log 8.427 = 9.074327 — 10
3 log x = 9.108144— 10
log x = 9.702715— 10
.-. x = . 50433.
677. The Arithmetical Complement of a logarithm, or, briefly, the
Cologariihm of the number, is the logarithm of the reciprocal of
that number.
Thus the colog 225 = log ^\^ = log 1 — log 225. Since log 1 = 0,
it may be written in the form 10 — 10 and then subtract log 225,
which gives
colog 225 = (10 — 2.352183) - 10 = 7.647817 — 10
Hence
Rule. — To find the cologarithm of a number^ subtract the loga-
rithm of the number from 10 and write — 10 after the result.
578. The advantage gained by the use of cologarithms is the
substitution of addition for subtraction.
Example. — Find by the use of logarithms the value of ^'" ^^ »
^ ^ 6 87 X. 079
= log 5.37 + colog 6.87 + colog .079
= .729974 + (9.163043 - 10) + 1.102373
= .995390.
The number corresponding to this logarithm is 9. 8944.
To find the colog .079 = log ^ = 10 — log .079 — 10
= 10 - (8.897627 - 10) — 10 [2577]
= 10 - 8.897627 + 10 — 10 = 1.102373.
574
COLLEGE ALGEBRA
[»78
BXEBOISE LXXXVH
Galcalate the value of the following expressions bj aid of Ic^aiithmB:
1.
319 • 765
138
2.
213 • 7.655
3145 ■ 718
3.
3.5347 • 2.685
137.65 • 5944
4.
47 0.663 -121
3576 • 1620
5.
0.765-0.0018
31457 • 567,*,
6.
0.0ia594 • 763f|
7t>->4.3 • 79i
7.
3».
8,
(1.1768)'.
9.
(0. 69038)*.
10.
a^i)'^.
11,
14.
(3H)*-".
12.
15.
V783.
13.
V 906. 80.
VO. 01764.
\97
16.
, |71
■^ 93406
17,
20.
tt 1 1»
\ 18706
18.
21.
V9fi.
50864 (O.OOOSTSDf*
98017(0.0019843/*
19.
V98765VV.
V3iV.Vl'V-
22.
2019 • (0.008715)'
3051 • (0.000631)*
23,
26.
29.
32.
Sl^lO
17
24.
27.
30.
33.
109 l76
716\U3*
25.
0.0875 1 78
5076 Vo.tX)? 109
9384Vd,00(»3i8"
809(H-V0.031
9.83M\0.007616'
64081 -VO.©!?
28.
.^0.09.. ^1
, |76 l75l
"\93 ■ -^olS
/318 • Va045y
V 43,0798 / '
31.
J 87 • 1^7194
\ 9807654
V27>/ A43/
34.
m"-
35.
(71) •
36.
(0.0009)*^««.
37.
(0.0378)'"".
38.
(0.3768)^'.
39.
(0.00893)-^.
40.
(-8.5768)-«-^.
41.
(-7.05873)-»^».
42.
(0.637803)**
43.
(0.0237998 A
44.
/3806\'^
47. Vsi
45.
-7
/1201\"S-
\2940/
46.
V78 + V31.
'K 0.947.
48.
V5-4.38v8r
76. 49. V2.
51. 7'--'l
1.24202.
). 53. 'Vr
0.961863.
3 — ;
19 VO. 07031.
50.
V4V5432I.
Ans.
/7'l
K'T.
Ans. 599392.
52.
'V2V2-1-I/IC
Ans.
Ans. 1.2268751
54.
(V3)""~.
Ans. 2.478061
«579] LOGARITHMS 575
Find the value of the following logarithms:
55. log {ah + ac + be), if log a = 0.75643, log h = 0.87254,
log c = 0.49832. Ans. 1.92440.
56. log l/a« + b^, if log a = 0.78241, log b = 0.63575.
Ans. 0.87174.
57. log v^a* — b\ if log a = 2.87655, log b = 2.79287.
Ans. 2.62898.
58. log (a*-6i), if loga = 1.28643, \ogb = 0.85794.
Ans. 1.81746.
59. log J A (a + 6 + Vab), if log h = 0.87432, log a = 0.47655,
logh :=z 0.36954. Ans. 1.29956.
CO. log i A IT (r« + p« + rp), if log h — 0.87456, log ir = 0.49715,
1<^ r = 1.75846, \ogp = 1.48763. Ans. 4.67237.
Exponential EquItions
579. An Exponential Equation is one in which the unknown quan-^
titj appears as an exponent.
Certain equations of this character can be solved by taking the
logarithms of both members, which gives an equation of the first
degree that can be solved by the usual methods. Thus:
1. Solve for x, 13' = 7.
X log 13 = log 7. [J658]
Hence x = >g^ = ^-^"^^ = .75865.
log 13 1.113943
The value of the fraction ,' , 'o can be found by logarithms or by
division.
2. Solve for x, 5*-» = 8«'+».
(x-3)log5 = (2x + l)log8,
^^31og64-31og2
log5 — 61og2
_ 3 (0.698970) -f 3 (0.301030) __ 3.000000
0.698970 — 6 (0.301030) —1.107210
= - .27095.
576 COLLEGE ALGEBRA [1579
*B3QSHCISB T.'nr-irviii
Solve the following equations for x\
1. a*+' = a". 2. fc'-' = h\
3. y*»+« = y«-to 4. ni»<'-5) = m« <'-*'.
5. a« • a'o^') = a • a«*-» 6. m -m'^*-^ = m*"^-* • m*-».
7. (a»-5)*-« = (a»-«)*-i.
8. a' • (a*-»)*^* = «'"• • (a'-«)*'-^
11. -V^ = "Vi?. 12. '-"i/^ = •'"V^^.
13. i/^^=:'"V^. 14. V^*^» = *^VV.
15. Va'-"* = '" Va«. 16. "l/o^^ = "Va-.
17. Va^^ • Va**"^*' = a" . l/o**.
18. i/i^^ • V^^ • V^^^ • W^F^ = 1.
19. 5* = 25, 3' = 27, 2* = 1024.
20. 2' = 16, 2* = — 16, 2-' = 16.
21. (- 2)* = 16, (— 2)* = — 16, (— 2)-' = — 16.
22. (— 2)* =32, (— 2)' = — 32, (— 2)"' = — 32.
23. 27* = 81, 27*=— 81, 27"* = 81.
24. (-27)' = 81, (—27)* = — 81, (-27)-' = 81.
25. 16' = 8, 16'=— 8, (-16)' = -8.
26. 32' = 8, 32' = - 8, (- 32)' = - 8.
27. 10' = 1, 100' = 1000, 1000' = 100000.
28. 10' = 0.01, 100' = 0.001, 1000' =0.01.
'•■ (ir=ar- '^- (iir=(fD"-
33. (0.25)' = 2*^ 34. 4' = 0.125.
35. (0.05)«'-»=20>'-«. 36. 8«'+* = (0.125)*-^
8579J LOGARITHMS 577
37. (^y^' = (OJb)"^-''. 38. (?)"'"' = (0.765625)«*-«.
39. 4* — 3'~* = 3"'^* — 2«*-».
40. 32'-7 = 0.25 -128'^.
41. 10* = 3, 100* = 0.005736, 1000* = 0.093768.
42. 2* = 10, 7* = 100, 0.025229* = 1000.
43. 3.111*= 1.7497. 44. 10* = 1.3713*>.
45. (l,04952*)i»=:(100»»)*««». 46. 10** =5.7544.
47. 7.8886* = 9.92126. 48. V9977 = 2.511308.
49. V7692.3 = 0.00013. 50. (0.088308)»*+» = (88.308^
51. 3.9345«*-*=5(1.2708)**-». 52. 25"* = 11.
53. (1)"^' =(!)''-'. 54. 179 (11^*-"= 356 (I)
ifac-S
55. 21* = 1.78. 56. ^'1/7*^+^ = '*l/5^*+^
57 /725\«'-» /351 \*-» _. / 87_\*-« /675\«^-«
V936/ ' \575/ U84/ ' V351/
\\&J7J \V532/ V1547/
59. 3* _ 5*+t = 3^;+* _ 5X+S gQ 5te+i _ yx+i = 5to _^ 7*
61. 7**"i 3*^"* = 7*«+i 3'*+*.
62. 5* + 5*+» + 5*+* = 3* + 3*+* + 3*+".
63. 2* + 2*+^ + 2*+» + 2*+* + 2*+* = 3* + 3*+* + 3*+» + 3*+' + 3*+*.
64. ap**+" + a^p*^»»i + a^p**+% = 6grte+« ^ h^i^+\ + 6j^**+«8.
65. 5(^) = l^'^l 66. o^'') = 3. 694575 ('••^^'^'l
g^ I 5* • 8i' = 512000 gg ( "y 777 • V
t x+y = 7. ■ (7x+5y =
g^ j 3* • 4* = 15552 ^^ ( 5* = 18. 0
1 4* • 5" = 128000. ' 1 55* = 18.
1/555 =9.33525
2xy,
5*= 18.0690 -3*
2347 • 6*
71 i'^l"^ ' ''^^^ = 7.429765 ( Vll = 1.825209 • VO
( '|/7 • V9 = 5. 1 05798. ^' \ Vll = 0.
0.907936 • V6.
578 COLLEGE ALGEBRA [J}580,581
Compound Interest
580. Interest is money paid for the use of money. The sum lent
is called the principal. The amount is the sum of the principal and
interest at any time. Interest is of two kinds :
Simple Interest, which is interest of the principal alone, and
Compound Interest. We say that capital is placed at compound
interest when at the end of each period of time, of suitable duration,
the interest is added to the capital and both bear interest during the
following period.
The rate of interest is the money paid for the use- of a certain
sum for a certain time. In practice the sum is usually $100, and the
time one year; and when we say that the rate of interest is 4, 5, or
6%, we mean that $4, $5, or $6 is to be paid for the use of $100
for one year. In theory it is more common to use a symbol to
denote the interest of $1 for one year.
681, To find the amount of a given sum at any time aJt compound
interest.
Let P be the principal in dollars, n the number of years for which
interest is charged, r the interest of $1 for one year, A the amount
Since r is the interest of $1 for one year, a capital of Pdollare
earns 7^ dollars in one year and acquires, therefore, at the end of
one year the value P+ Pr, or
dollars. Therefore, in general, knowing the capital at the beginning
of a year, we obtain the value which this capital has acquired at the end
of the year, by multiplying this value by 1+r.
Hence the value which P dollars will acquire at the end of 1, 2,
3, .... n years will be respectively:
at the end of the
1'* year P{ 1 + r),
2-'* year P{l + r) x d + O = PH + rf,
3'* year /'(l+r)«x d + r) = P{l+r)^,
n^ year P(] + r)»-»X (1 + r) = P(l + r)\
We have, therefore,
(i) A= P{l+r)\
8 J682, 583] LOGARITHMS 579
582. Solution of (i). — Formula (i) involves four quantities, -4,
Pj r, n. Any one of these four quantities can be found if the other
three are given. These four problems can be solved by applying
logarithms to both members of formula (i), thus
r log ^ = log P + n log ( 1 + r);
(ii) I transposing
( \ogF = log A- n log (1 + r);
by transposition and division
log(i + .) = '-^2^^=^^=^
log(l+r)
Application, — Problem I. Calculate the value acquired in 19
years by capital of $5689 at 4^ ^ per annum, compound interest
According to the first formula in (ii),
log ^ = log P + n log (1 + r ),
where P = $5689, r = $0,045 and n = 19. Thus we have
log 5689 = 3.755036,
1910^1.045 = 0.363204 w 1 045 - 0 019116-
log ^ = 4.118240 ^"^ ^'^ - "•"^^"^'
A = $13129.24.
Capital is sometimes loaned at a certain rate per annum, compound
interest after a given number of years and a fraction of a year; or
the period of time for compound interest may be less than a year.
These facts lead to the following considerations.
583. First Convention. — Suppose that the principal is placed at
compound interest at the rate r per $1 for one year and for n years
and - of the year following.
The value of P at the end of n years will be
(1) P(l+r)«. [2681, (i)]
The amount of $1 at the rate of r for -- of a year is
P
and therefore the amount of P ( 1 + r )» at the rate of r for - of a
year is
(iii) ^=P(l + r)-(n-|r).
580 COLLEGE ALGEBRA [1^
ApplicatioH, — Problem IL Find the valae acquired at the end
of 12 years and 4 months by the principal $7654 drawing 4^^ interest
per annam, the interest being compounded at the end of each year.
The capital at the end of n years draws simple interest during the
interval ? years.
According to formula (iii), we have
^ = />(l + r)-(l+|r),
whence log ^ = log P + n log ( 1 + r) + log (l + |rj,
p 1
where we take P= $7654, r = 0.04, » = 12, and ^ = 3*
log 7654 = 3.883888
12 log (1.04) = 0.204400 log 1.04 = .01703334
log (1.0133) = 0.005738
log ^ = 4.094026
A = $12417.26.
584. Second Convention. — Since the value of $1 at the end of 1
year is 1 + r, we may suppose that after each period of a fraction -
of a year, the value of $1 is increased by a quantity r'jSuch that, it
the end of a year, the value acquired by $1, after one, two, three,
.... of these periods, will be, 1 + r', (1 + r')', (1 + r^', ....
etc. ; and, consequently, after q of these periods, that is to say after
a year, the value acquired by $1 will be (1 -|- r')'. But this value
is also 1 + r, therefore we have
(1 + r')' = 1 + r;
whence (iv) 1 + r' = (1 + r)«.
After the time ?i the value acquired by $1 will be
(1 + r')P = (1 + r)«.
During the same time the value acquired by a capital Q will be
0(1 + r)\
Since the value acquired by a principal P during n years is PiX -f rf,
the value acquired by the same principal during n years plus tbe
fraction ? of a year, will be
P{1 + r)- X (1 + r)« = I\\ + r)""'
5585] LOGARITHMS 581
Let n' represent the whole time, n-{- E years, and A the amount at
the end of the time n\ then vre have
A=P{l + rr,
and, consequently, under the second convention we apply the formula
(v) A = P{1 + r)^
in all cases whether n is integral or fractional.
Bankers as a rule adopt the * ^second convention, " which involves
more simple calculations than the first.
Application, — Solve the problem in the preceding paragraph by
this method. We have, according to (v),
A = I\1 + r)»;
whence log A = \ogP + n log (1 + r),
where P = $7654, r = 0.04 and n = 12 + 1 = y-
Thus we have log 7654 = 3.883888
^log 1.04 = 0.210078 l^g 1-^4 = -01703334
log^ = 4.093966
therefore A = $12415.55. The difference in the results of the pre-
ceding paragraph and this is $1.71.
685. Problem III. — Calculate the principal P which it is neces-
sary to put at interest for n-(- ? years at r% per annum that it may
amount to A dollars.
If we adopt the first convennon we have from formula (iii)
P= d ,
{i + rr{l+^r)
and log P = log j4 — n log (1 + r) — log (l + - rV
P can now be calculated as in the preceding problems.
The Time Unknown
Applying logarithms to the relation
A = P(l + rr (l+-^),
it follows that log ^ = log P+ w log (1 + r) + log (l +EA ;
and, on solving this equation for n,
m ^-log^-logi^ log(l+fO
^^^ **-- \og(l+r) log(l + r) •
Since the time is to be determined, n and E are to be determined.
This can be' done as follows: — Since E is less than 1, Er is less
9 9
than r; and, consequently, log (l+EA is less than log (1 + r);
582 COLLEGE ALGEBRA [8586
therefore,. -.jll-TJ^-^ is less than 1. Since the integer n is the
excess of l2£AzL^g_f over the fraction ^^f^~- 'which is less
log(l + r) log (I -fr)
than 1, the integer n is the integral part of the quotient of the divi-
sion of log ^ — log P by log (1 + r). Let k be the integral part
of this quotient and R the remainder of the division, then we have
n =z k
and (vi) n = * + ^^f:p5-i^il+l;j>;
whence log(l +^r)= ^.
Application. — Problem IV. Calculate the time during which the
principal $5435 will acquire the value $12840, interest compounded
at 5 ^ per annum.
Adopting the first convention and using formula (vi), 2585,
__ log yl -logP logU-K|rJ
"* - rog(l + r) Tog(l + r)'
we take n for the integral part k of the quotient ^? — ,"~.^ — , and
log(l + r)
determine log^l +^r\ by writing log ^1 +^r\ equal to the re-
mainder of this division, 2585, (vi).
[n this case A = $12840, P= $5435, and r = 0.05;
whence log^ = 4.108565 log(l + r)= 0.021189
logP = 3.735200
log ^ — log P= 0.373365
log A - log P __ 0. 373805 _ 1 7 i 0.013162
log (1 + r) 0.021 189 "*" 0.021189
and A: = 17, /? = 0.013152;
therefore n = 17
and log (1 +|r) = 0.013152.
Hence l+^r= 1.03075
and ?r = .03075; but r = 0.05;
% = :^^.6l5.
q 005
Whence the time is 17. 615 years or 17 years, 7 months, and 11 days.
586. The case vrhen k is taken as the unknown is less simple ; r is
calculated by the method of successive approximations.
It follows from
log^ = logP+ nlogd + r) + log(l + ?r) [J688, (iii)]
8586] LOGARITHMS 583
that (1) log(l + r) = '2E.i=i^!iZ_Ml±il).
Since r usually lies between .03 and ,06 and ?r is less than 1, the
number 1 +?r differs from 1 by a small quantity; consequently,
logH +?rjf and hence ^ ^ K are very small numbers. For
example, in case r = .045 and ? = -, ^1'+-^^ = 1.01125 and
log 1.01125 = .004858; and if n = 5 years, then
log(l+*fr) _ .004858
.0009716, a very small number.
Neglecting this very small number, a first approximation is
(2) log(l + r.) = H^-loKf.
the number r^ is a little larger than r, since the negative term in
equation (1), ^— ^ » has been omitted.
Put (3) log (1 + r.) = IggjL^.''^- '"g(^+f '•0 ■
Since r^ is a little greater than r the fraction subtracted in (3) is
greater than the fraction subtracted in (1), and therefore r^ is a little
less than r. The two numbers r^ and r^ are the approximate values,
the one greater and the other less than r. On taking the one or the
other of these values for r, the error made is less than the difference
Tj — Tj. On taking the arithmetic mean ^^ "^ ^^» the error made is
less than the difference ^^ "" ^*- For,
let rj = r + Aj,*
and r^= r — h^\ h^ and h^ are positive,
it follows that ^-^4^ =r+^'-^'
2 ' 2
and ^^i-=^ = ^i-=t^ .
2 2
On taking ^' ^ ^^ for r, we commit an error equal to ' "~ ^
whose absolute value is less than ^ ^ ^, and therefore less than
2
Application. Problem V. — At what rate is it necessary to place
a principal of $2543 in order that it may acquire after 24 years, 5
months, and 10 days, the value of $7460; the interest being com-
pounded annually?
584 COLLEGE ALGEBRA [1586
Here, according to equation (2),
log (1 + rj - ^og^-^Qg^- 3-872739 - 3.i05346
= 0.019475.
Hence, 1 + r, = 1.04587,
r^ = 0.04587.
But we know that r^ y> r.
According to equation (3),
log(l + r.) = loK^-»oRJ'-''^(^+f'-').
= 0.019475-'°'^^,?^'"
24
S-- f = ^ + aofi2 = l ^^ f^ =.02039.
hence log (1 + rj = 0.019475 - .000365 = 0.019110,
l + r,= 1.04498,
r,= .04498,
and r^ < r.
If the arithmetic mean between r^ and r^, or .04542, is taken, the
error is less than ^^ "T ^^> and therefore less than .0005. The rate is
therefore 4.545, with a possible error of .05 of 1 %.
A nearer approximation for the rate can be found by carrying the
process a step further; put , .
' n n
= 1.019475 - log(lOl^) = .019117;
whence 1 + r, = 1.045002.
r,= .045002.
If ^' + ^» , or 0.04499, is taken for r, the error is less than^^^=^' that
is leas than .000011. The rate is therefore 4.499% with a possible
error of .0011 of 1%.
If the second convention is followed, the formula
A = P{1 + /•)»
is taken, and if 24 + 1 or ^^ is substituted for », it follows thit
log(l+.) = «I!o£A^JojLPi
= 0.0191206;
whence, 1 + r = 1.045011,
and r= .045011.
Hence the rate is 4.5011 %. The difference of the two results is
4.5011% — 4.499% = .0021 of 1%.
8J587-589] LOGARITHMS 585
Annuities
687. When a person receives every year a certain sum of money,
say $Ny he is said to possess an annuity of $N. The right to receive
this annuity may continue a certain number of years and then lapse,
or it may be invested in him and his heirs forever; in the first case
the annuity is terminable^ in the second, perpetual. An example of
a terminable annuity is a common arrangement in lending money
where A lends B a certain sum, and B repays by a certain number
of equal annual installments which are so adjusted as to cover both
principal and interest. An example of a perpetual annuity is the
case of a freehold estate which yields its owner a fixed income of $N
per annum.
In valuing annuities it is customary to speak of the whole sum
which is paid annually, yet, in practice, the payment may be in
semi-annual, quarterly, etc. installments ; and this must be taken
into account in calculating annuities.
588. Contingent Annuity. — In some cases the annuity lasts only
during the life of a certain named individual, called the Jiominee, who
may or may not be the annuitant. In this and similar cases an esti-
mate of the probable duration of human life enters into the calcula-
tion, and the annuity is said to be contingent,
589. Value of a Forborne Annuity. — An annuitant B^ who had
the right to receive n successive pa^^m'ents in ji annual installments,
has for some reason or other not received these payments. The
question is, what sum should he receive in compensation?
Let P be the value of each payment and r the rate of interest.
It is clear that the whole accumulated value of the annuity is the
sum of the accumulated values of the ?( installments and that com-
pound interest must in equity be allowed on each installment The
first installment will draw interest compounded annually f or n — 1
years and will acquire the value
P(l + r)«-^ [J681, (i)3
the second installment in n — 2 years acquires the value
i"(l + r)»-*;
the third, in n — 3 years acquires the value
/>(1 + r)n-»;
586 CX)LLEGE ALGEBRA [18590-593
and so on until the n^ installment, which will not draw interest, since
the whole annuity is due at the time the n^^ installment is due.
Hence the whole accumulated value of the annuity is
^ = P(l + r)»-» + P(l + r)-«+i^(l + r)-»+ . . .+ P(l + r) + J^;
or, better,
A = Fll+{l + r)+{l + ry+ +(l + r)-»].
The quantity within the brackets is a G. P. in which ti=l, r=l-}-r,
n z= n\ hence, according to formula (iii), S640,
(1) A = pS^ + '-)"-K
r
690. Since the four quantities Aj P, r, and n are connected by
equation (1), any one of these four quantities can be calculated if
the other three are given.
Calculation of A, P, n, and r
691. Formulae for calculating the values of A, P, n, and r by
means of logarithms readily follow from equation (1), S889, thus:
692. I. Given P, r, n ; calculate the value of A, Applying loga-
rithms to equation (1), 2689, it follows that
(i) log ^ = log P + log [(1 + r)« - 1] - log r.
NoTB.— T&e value of (1+r)" can be calculated by logarithms, and then subBtituted
in the bracket.
Application. — A father wishing to provide for a dowry for one
of his children, invests $1250 each year for 20 years; what will the
amount of the dowry be, including the last payment, if the interest
is compounded annually at 4%?
Use formula (i^ and put P = 1250, r = .04, and n = 20. Thus,
log^ = log 1250 + log [(1.04)» — 1] — log .04
= 3.096910 + log (1.19109) — (8.602060 — 10)
= 3.096910 + .075945 — 8.602060 + 10
= 4.570795.
^ = $37221.58.
693. II. Given A, r, ?i; calculate P. Solve formula (i) for
log P^thus log P = log ^ + log r — log [(1 + r)" — 1]
(ii) = log Ur) - log [1 + r)" - 1] .
Application, — How much money must a person invest annually
for 15 years at 4^% compound interest in order that a capital of
$24000 may be accumulated with the fifteenth installment?
\
15594, 595] LOGARITHMS 587
Put in formula (ii) A = 24000, r = .045 and n = 15, then
log F = log (24000 X .045) — log [(1.045)" — 1]
= 3.033424 — (9.970934 — 10)
= 3.062490.
P= $1154.75.
694. III. Given A, P, r; calculate the value of n.
Solving equation (1), ?589, for n,
^r = P(l+r)"-P,
(l + r)»=l + ^,
nlog(l+r)=log(l+^);
whence (iii) n = ML±Jp) .
log (1 + r)
Application, —YoT how many years must 11400 be invested each
year at 3^% compound interest in order that a capital of $45260
including the last investment, may be accumulated?
Substitute in formula (iii), A = $45260, P = $1400, and r= .035;
then , /, , 45260 X .085\
^ _^"g (^-1 1400 ) __ log 2.1315
log (1.035) 0.014940
^0328685^ 22 years.
0.014940 ^
696. IV. Given J, P, n ; calculate the value of r. Equation (1),
2589, can be written
(iv) 4^(L±^.
When (iv) is expanded, it involves r to the power n, and lower
powers; the value of r can not be found immediately except
when n is 2, in which case the values of r are the roots of a quad-
ratic equation; but the values of r can be derived by a method of
successive tests to any desired degree of approximation. This
method is illustrated by the following problem:
On investing $1150 edch year for 25 years at compound interest,
the accumulated value of the total investment, including the last
installment, is $50000 ; what was the rate of interest?
In this case
(1) 4 = ^^ = 43.4782 and w = 25.
Jr 1150
Suppose for the first trial that r = .04, and calculate the value of
(l+r)"-l .
r
588 C50LLEGE ALGEBRA [8596
when n = 25, this fraction is found to be 41.63 . . . <-; therefore
r = .04 is too small.
Assume, similarly, that r = .05; for r = .05, and n = 25 it
follows that
(2) (1±^ ^ 47.7 . _ ^A,
Since the result is greater than — , the number r = . 05 is too larg^.
Suppose, moreover, that r =.045; then for n = 25, it follows that
(3) 'lif^ = 44.5 ...>j,.
A
Since the result is greater than - (2), the number r =.045 is too large.
Suppose, further, that r = .044; then since n = 25, it follows that
(4) (kiirl!=l^ 43.96 . . . > 4;
the number r = .044 is still too large. Suppose r = .043. Then it
follows that ^^^ (l + r)»-l ^ 43.37 ... < 4.
r P
therefore r r= .043 is too small. It follows from (4) and (5) that
.043 < r < .044
and • "^ — r= .0435 is the value of r with an error less than
. 0005 ; the rate is therefore 4. 35 % , with a possible error of . 05 of 1 %.
On continuing this process, as close an approximation as desired
may be found.
Repundino of a Debt by Annuities
596. Sinking Fund. — To make the calculations for a sinking
fund is to calculate the purchase price of a given annuity.
Suppose that B desires, b}" paying down at once a sum of tE^
to secure for himself and his heirs the right of receiving n annual
payments of $ Peach, the first payment to be made m years hence.
E is the sum of the present values of the n payments. The first
payment is due in m years hence ; its present value is, therefore,
-— - — ;^ — The second is due in m+ 1 years hence: therefore the
(I + rr P
present value of $Pis-— — . and so on. Hence
*- (I + r)'"-^^
/1\ EJ P I P. I I £_ .
^^ (1 + r)»» ^ (1 + rr+i "^ * • * "^ (1 + r)"»+"-i'
Hence ^ = ^-^. ( 1_ ^^) ^ (l _ ^-J-)
(2) = ^ • il±l)l-J.
(l + r)" r(l + r)»-»
5 5597-601] LOGARITHMS 589
Corollary. — It the annuity is not **def erred/' but begins at
once, i. e., the first payment is due in one year, then m = 1, and
697. The Calculation of E or of P, — These two problems do
not cause any difficulty. If the unknown is E^ then from (3), 2596,
(4) log i;= log P + log [(1 + r)»- 1] - n log (1 + r) - log r.
Applieation, — What is the value of a loan E which can be refunded
by 34 annuities of $1500 each, the rate being 4^% and the first
annuity being paid at the end of one year?
Put P= 1500, r = .045, and n = 34; using logarithms one finds
(1 + r)" = 4.46626;
\ogE= log 1500 + log (3.46626) — n log (1.045) — log .045
= 3.176091 + 0.539862 — .649944-8.653213 X 10
= 4.412796;
^ = $25870.
698. If P is unknown it can be found by solving equation (4),
5697, for log P, etc.
699. To find w, solve equation (3) J696, for (1 + r)", giving
and ^^1okP-]ok(P-^^),
log (1 + r)
600. To calculate r, use the formula
P _ Til + rV'
^""(l + r)"-l*
and determine r by successive approximations, employing the
method of 2595.
601. The Period of Time a Fraction of a Year. — Suppose that
the interest is compounded every six months instead of every year
(as was the case in 2597), and in this case suppose that P* is the
sum paid each period of six months, r' the interest on tl for six
months, and n' the number of half-yearly payments ; then it follows
that (\ A.iJ\nf 1
A^F IL+Jll 1 . [J689,(i)]
But here P'is-r» / is-, and n' is 2»; hence the compounding
formula is /- , r^n__ ^
Az=P)lZM 1.
r
A similar formula holds for payments made at the end of other
fractional parts of a year.
590 COLLEGE ALGEBRA [WOl
FBOBLEXS
1. Find the amoant of $100 in 50 years at 5% interest^ com-
pounded annually.
2. In how many years will a sum of money double itself at 4|^
interest, compounded annually?
3. If in the year 1776 $1000 had been left to aocnmulate for
124 years, find the amount in the year 1900, reckoning compound
interest at the rate of 5% per annum.
4. If a sum of money doubles itself in 40 years at simple interest
find the rate of interest.
5. Find the present value of $10000 due in 10 years hence at 4^
interest, compounded annually.
6. Find the amount of an annuity of $100 in 15 years, allowing
compound interest at 4% per annum.
7. What is the present value of an annuity of $1000 due in 30
years, allowing compound interest at 5% per annum?
8. What sum of money at 5% interest, compounded annuallv,
will amount to $1000 in 16 years?
9. In how many years will a sum of money treble itself at 3}^
interest, compounded annually?
10. A person borrows $1225 to be repaid in 5 years by annual
installments of $220 ; find the rate of interest if simple interest is
allowed on the payments.
11. A person lx>rrows $60025; find how much he must pay annu-
ally that the whole debt may be discharged in 35 years, allowing
simple interest at 4%.
12. A merchant marks his goods with two prices, one for ready
money and the other for a credit of 6 months; find the ratio tlA
two prices ought to bear to one another, allowing 5% simple interest
13. Find the amount of an annuity of $200 in 25 years at 4J<^.
compound interest.
14. A county treasurer borrows $50000, and wishes to repay it
in 25 annual payments, the first of which should be paid one year
after the loan was made: rate 4% compound interest: what ought
the amount of each annuity to be? (Compare 2598.) Ans. $3200.59.
15. How often should one pay $8869.90 in order to refund t
debt of $100000, the first annuity being paid one year after the debt
was contracted and the rate being 5% interest, compounded annually?
(Compare §599.) Ans. 17,
BOOK VI
CHAPTER I
MATHEMATICAL INDUCTION
802. In simple cases already considered, but more particularly in
subsequent parts of this work, it is convenient to use a method of
proof which is called Mathematical Induction. This method will
now be illustrated.
603. Sum the First n Integers.
Consider the following equations:
_ 2(2+1)
"■ 2
(2) 1 + 2 + 3 = ?lMli,
(1) 1 + 2 - ^
2
(3) 1+2 + 3 + 4 = liiJiJJ,
(4) 1 + 2 + 3 + 4+5 = 51A±JL).
Fint Step, — The following theorem holds in cose of equations
(1), (2), (3), (4):
Theorem. — Tf a number of the first consecutive intei/ers are added^
their sum I'j one-half of tlie number of integers added times the number
of integers plus 1.
Second Step. — Assume that this law holds for the sum of the
first n integers, then
(5) 1 + 2 + 3+ . . . . +n :^ »(» + !).
Add n + 1 to both members of this equation,
(6) 1 + 2 + 3+ ... . +n+(n + l)=^-^".tL) + n + l
_ (n + l)r(n + l)+l]
2
The same law is expressed in (6) that is expressed in (5) ; i. e. , if
the theorem holds for n integers, it holds for n + 1 integers.
fiOl
592 C50LLEGE ALGEBRA [«604, 605
Third Step, — But in (4) it is noted that the theorem is tme when
n = 5, therefore it follows from the Second Step that the theoTem
holds for n = 6, then for n = 7, and so on. The sum of the first
n integers has been found in 2526.
604. Sum the First n Odd Integers.
Suppose it is observed that
(1) 1 + 3 = 4 = 2«,
(2) 1 + 3+5 = 9 = 3»,
(3) 1 + 3+5 + 7 = 16 = 4»,
(4) l_|.3+5+7 + 9 = 25 = 5«.
JfHrst Step, — ^For these four cases, one sees that the following
theorem holds:
Theorem. — 7%c ium of a number of the fint odd integers it the
square of the number of integers added.
Second Step, — Assun^e that this theorem holds for it odd integers,
then
(5) 1 + 3 + 5+ ... . +2n—l = n\
Add the ( w + 1 )"» odd integer, 2;i + 1, to each member of (5),
(6) 1+3+5+ .... +2/1—1 + 2/1 + 1 = »« + 2»+l
= ( n + 1 )\
It follows from equations (5) and (6), that if the theorem holds for
the sum of n odd integers, it holds for the sum of » + 1 odd integers.
Third Step. — But the theorem holds when n = 4, therefore it
holds when n = 5, hence when n = 6, and so on.
605. Suppose that one desires to prove the formula,
(i) l« + 2«+3«+4«+ . . . . + n« = riin + l) {2r.-^l\
6
This formula is true in the case n = 2, 3, 4; thus,
l« + 2« = 2(2+1) (4 + 1) ^ ^^
l«+2* + 3« = 3(3 + 1) (6 + 1) ^ j^^
l« + 2«+3« + 4« = 4(4+1) (8 + 1) __ 3(j.
but it is desired to show that the formula holds universally. Sup-
pose that formula (i) is true for any number of terms, say r; then
(1) l + 2«+3»+ . . . . +r« ^r(r + l)(2r + l)^
6
?8606, 607] MATHEMATICAL IXDUCTION 593
Add (r + 1 )» to both members of this equation, theu
l + 2» + 3'+ .... +^ + {r + l)*== '•('•+1M2H-1) + (r + 1 )«
= (r + l)rr(2r + l)+6(r+l)1 _ (r + 1) (2r« + 7r+6)
(2) - (r + l)(r4-2)(2r+3) _ (r+ 1 ) [ (r + l) + 1 ] [2(r+ 1) + 1]
Thus we see that {r+1) is involved in the same manner in the
second member of equation (2) as r is in the second member of
equation (1). That is, if formula (i) holds for any number of terms,
whatever that number may be, it holds when the number is increased
by one. But the formula does hold by actual calculation when 2, 3,
or 4 terms are taken, therefore it holds when 5 terms are taken, and
so on. Hence the formula must hold universally.
606. The three theorems which have been proved by the method
of induction may be established otherwise. The first and second
theorems are examples in A. P., and have been proved in 2526.
There are many other theorems which can l^ proved readily by induc-
tion. The theorem proved in H02, respecting the divisibility of
X* db «" by X it «, may be proved by induction. For example,
x^ — a^ _ ii-i _L_ g ( x""^ — a"~^) .
— X -j- 1
x — a a: — a
hence x" — a" is divisible by x — a, when x**"^ — a""* is. But
X — a is divisible by x — a, therefore x* — a' is divisible by x — a,
and so on; hence x" — a" is always divisible by x — a when n is a
positive integer.
Similarly, other cases may be proved. As another exercise the
student may consider the theorem in {268.
607. Proof of the Binomial Theorem for Positive Exponents.—
Asa last example illustrating the method of mathematical induction,
a proof of the Binomial Theorem, for a positive integral exponent,
stated in 2265, is here given.
I^irst Step, — It was proved in §265 that the Binomial Theorem
holds in the particular cases :
((x + a)« = x« + 2xa + a«,
(x + a)» = x» + 3x«a + '^xa» + a«,
I (x + a)* = x^ + 4x»a + H^xV -^. ll?-!^ ara»+ a*.
The symbol 3 ! or [3 = 1 -2 • 3; and the symbol n ! or[n = 1 • 2 -3
....(» — 1) • n, and is called **n factorial".
594 CX)LLEGE ALGEBRA [1607
Second Step. — It is assumed that the theorem holds for <njr
positive integral exponent it, then
(2) (x+ a)" = x» + tui-'ki+ " <"^7 ^)jc-'*a* + n(n-lKn~2) ^.y ^ . . .
_!_ n(n — 1). . . (n — r + 1) -p*-r^r
, n (n ~ 1) . . . (n — r) M,r-iq..4.i
^ (r+1)!
2!
Multiply both members of (2) hy x-\-a, then,
(3) (x+a)«+i=x»+i + rix-a + ?i^£^^x*-^^
2 ! o I
I n(n — 1). . . (n — r+1) r-T+i^r
^ r!
, n(n-l). . . (n — r)^,^ -4.1 ,
^ (r+l)! "^ ^•"
_L n (n - 1 ) ^^^_, _^ nA»-i + xa«
2 !
+ i*a + i2"-ia« + ^2^^|=^x^^^
. n(n — l)...(n— r + 1) ju-r^r-n |
r !
+ "^^""^^ j^a«-i + lura- + a»+i;
combining similar terms in (3),
(4) {X + a)"^>=x"-^H-(n+l)jr»a + ^^±^x-^ ^in-\'hn(n-l)^,^^ ^
. (n4-l)n(n-l)...(n~r + l) ^^-r^r+i ♦
^ (r + D!
+ ^" "^ j^^ '^ P*""-* + (n + 1) j«» + a-+i.
Equation (4) shows that if the binomial formula in (2) hokls for the
exponent n, the binomial formula holds for the exponent /i + 1.
Third Step, — Therefore, on combining the results of the first and
second steps, the binomial formula in (1) is true for any positiw
integral exponent, — because if it holds for any exponent, it holds
for an exponent one greater; but it holds for the exponent 4, hence
for the exponent 5, then for 6, and so on for any exponent
• The (r + 2)th term of the expansion (4) Is fonnd as follows:
(r + l): X a -+ ^, X" a
— n(n-n ■ . . .(n-r-l-1) ^n-r ^^^ ^.^ ^^^
= (nf l)nfn-l). . . . (»*-r^l> „,^ ^^^
(r-fl)! ^ « •
18608,609] MATHEMATICAL INDUCTION 595
608. The method of mathematical induction may be divided into
three parts: First, it is ascertained by observation or trial that the
theorem under consideration is true in some particular cases ; in the
second step, it is proved that if the theorem is true in any case.it is
true in the neaU case] in the third step, the conclusion is deduced
that the proposition holds in any case by combining the results of the
first and second steps, i. e. , if the proposition is true in any arbitrary
case it is true in the next case (second step) but the proposition is
known to be true in some cases, therefore it is true in the next case,
and so on for any case.
609. There is, in general, a marked distinction between the
method of mathematical induction and the inductive method of
reasoning used in the natural sciences, for example, in Chemistry
and Physics. In these sciences a law or theorem is formulated
which is observed to be true in a number of particular cases,
verified by experiment or investigation. The investigator however
guards his conclusions by verifying as many test cases as possible.
Such a law thus established is accepted as true until the discovery
of a fact (hitherto not taken into account) compels a modification.
The second step in mathematical induction is entirely omitted in the
inductive method of reasoning as used in the natural sciences. The
method of mathematical induction is just as rigorous as any of the
direct methods of mathematical demonstration.
BXBBOISE liXXXIX
Prove by mathematical induction:
1. 2 + 4 + 6 + 8+ . . . . +2n = nin+l),
2. 2« + 4« + 6»+ .... +(2>»)«^^^^^» + ^)<^^ + ^)
3- A + 2-3 + 3^+ • • • • to n terms = ^-^.
4. 2+2« + 2»+ .... +2" = 2(2» — 1).
5. 1»+ 2» + 3» + 4» . . . . n» = ^'^'^^ ^ ^'.
6. Prove the general formula in {268 for expanding
(a, + a, + a3+ .... + a J».
CHAPTER II
ARRANGEMENTS AND COMBINATIONS
610. One supposes that there are n distinct objects at hand. The
arrangements of these n objects taken r at a time are the different
dispositions which can be made with these n objects, by taking them
r at a time in all possible ways, and placing them side by side in a
straight line. Two arrangements differ either by the nature of the
objects which compose them, or only by the order in which they are
placed.
For example, in the case of three letters, a, 6, c, taken two at a
time in all possible ways, we can form the six following arrange-
ments:
a6, aCy ba^ 6c, ca, ch.
The first and the third, the second and the fifth, the fourth and the
sixth, differ respectively only in the order in which the letters are
placed.
In general, let the n different objects be represented by the fol-
lowing symbols,
«1» «2» «3» • • • • ^n-U ^ny
and let ^A* designate the number of arrangements which can be
formed with these n objects taken r at a time and in all possible ways.
611. The number of arrangements of n objects taken one at a
time is found evidently by taking each of them separately; which
gives n arrangements,
«i> «8» «s» • • • • «!••
Here one has (1) ^A^ = n.
The arrangements of n objects taken two at a time are found by
placing after the first symbol, a^ each of the other symbols suc-
* n is not a factor but a part of the symbol.
596
«611] ARRANGEMENTS AND COMBINATIONS 597
cessively; after the second symbol, a^, each of the others succes-
sively, and so OD ; which gives us the following table of arrangements:
«.«„
°lS>
«,«!, . .
• • a,««-i.
«,««
«.«!'
°.«J.
«,«., . •
• • «,««-!.
«,«.
«s«i.
"a"*.
«s«.. • •
• • «j«»_i.
Oj*-
«»«!> «««j, «n«8, .... a„a„_2J «».«*-i-
The first horizontal line contains all the arrangements which begin
with the symbol a^, the second all of those which begin with the
symbol a,, etc. ; thus are formed all the arrangements of n symbols
taken two at a time. Since each horizontal row contains n — 1
arrangements, and there are n rows, the table contains n{n — 1)
arrangements; therefore,
(2) ,^, = n(n-l).
Similarly, if after each of the arrangements taken two at a time,
each of the other n — 2 symbols is placed successively, the following
table of arrangements taken three at a time is formed:
a^a^Qj^ «I«8«4' .... «l«,«n)
a^a^a^, «i«s«4» .... Oi«3««,
• ••• ..^. . .... ....
«««i«8» S«i«4- .... «,Oia«,
^9^3 ^V S^8^4» .... «8«3«n.
Each of the other symbols a^y a^, «» ^^ ^®®^ placed after
the first arrangements of two symbols, a^ a^; similarly, after the
second, a^a^, each of the other symbols, a,, a^ . . . a^, etc., has been
placed. Therefore, all the arrangements of n things taken three at
a time have been tabulated ; because an arrangement of three letters
is successively composed of an arrangement of two letters followed
by another letter. The same arrangepaent is not repeated, because the
arrangements of the same horizontal row differ by the third symbol,
and two arrangements of two rows differ by the arrangement of the
first two letters. Each horizontal row contains n — 2 arrangements ;
and there are n(n — 1) horizontal rows, just as many as there are
arrangements of n things taken two at a time ; therefore, the num-
ber of arrangements of n symbols three at a time is
(3) .^, = n(H-l)(/i-2).
698 COLLEGE ALGEBRA . [«612
By contlDuing the same reasoning the general formula is obtained,
(i) ,^ = n(n-l)(n-2) .... (n-r + 1).
I%e number of arrangements of n objects taken r at a time is equal to
the product of r consecutive decreasing integral numbers beginning
with n,
612. It remains to prove that the formula for ^A^ is general.
Suppose that the arrangements of n symbols taken r — 1 at a time,
have been formed and that it is desired to form the arrangements of
n symbols taken r at a time. One places after each of the arrange-
ments takenr— lata time, each of the remaining n— (r— 1)ih» — r-fl
symbols successively. Thus are formed all the arrangements taken
r at a time; for an arrangement of r symbols is composed of r — I
Sjrmbols followed by another symbol. The same arrangements will
not be repeated, because any two arrangements thus formed differ
either by the last symbol, or by the arrangement of the first r — 1
symbols. Each of the preceding arrangements will furnish n — r + 1
new arrangements. Hence, it will follow in general that
n^r-n^T-\ ( W — r + 1 ).
If the values 2, 3, 4, .... n are given successively to r, the
result will be
,^, = „^,(n-2 + l) = ,^,(n-l)=n(n-l),
,^3=,^,(n-3+l) = „^,(n-2),
^^^=„^,(H~4 + l)=„^3(n-3),
«^ = «^r-i (n — r+ 1.)
From multiplying these equations together member by member the
following result is obtained:
. . .^^_^n(n-l)0i-2)(ii-3) . . . (n-r+1);
or, after dividing out the equal factors,
„^,= n(«-l)(n-2) . . . (n-r + 1).
Applications. — 1. What is the number of the arrangements of
nine symbols taken three at a time? It is the product of three con-
secutive decreasing integers beginning with 9,
,^, = 9- 8-7 =504.
88613,614] ARRANGEMENTS AND COMBINATIONS 599
2. How many different words of five letters can be formed from
nine letters? The number of words will be the same as the nomber
of arrangements of nine symbols taken five at a time,
,^5 = 9-8-7-6-5 = 15l20.
3. How many numbers of four digits each can be formed from
the first eight digits? There will be as many as there are arrange-
ments of eight symbols taken 4 at a time,
^^=:8-7-6-6 = 1680.
Permutations
618. By the number of permutations of n objects is meant the
number of different dispositions which may be made of n objects by
placing them side by side in a straight line. Each permutation con-
tains all the objects, and two permutations differ only in the order
of the objects.
For example, two permutations can be formed from two objects,
a, b,
ah J ha.
The permutation of n objects is, in general, represented by /\.
It follows from the definition that the permutation of n objects is
simply the arrangement of n objects taken n at a time.
Therefore,
(ii) Pn=n^ =n(n-l)(n-2). . . . 3-2-1,
or, if the order of the factors is changed,
P„=l-2-3 . . . . n = nl
The number of permutations of n symbols is equal to the product of the
first n positive integers.
614. Applications. — 1. How many different words of four letters
each can be formed from four given letters? There are as many as
there are permutations of four letters:
P; = 1 • 2 • 3 • 4 = 24.
2. In how many ways can a troop of eleven soldiers be disposed
of in a line? In as many ways as there are permutations of eleven
objects:
Pj, = 1 • 2 • 3 11 = 39,916,800.
600
COLLEGE ALEGBRA
[{615
3. In how many ways can eight guests
be seated at a circular table? The num-
ber will be the permutations of seven
Sjrmbols:
^,= 7-6-5-4-3-21 = 5040.
NoTB— By some authors no distinction 1b made between permutations and arramge-
mentt: by them the term permutations is given the definition which l8 given to
arrangements.
615. The formula for permutations has been derived, from the
formula for arrangements as a particular case. It can, however, be
derived independently.
The number of dispositions which maybe made of a single letter,
a, is clearly one ; thus,
/>,= !.
Evidently two permutations can be formed from two letters, a
and />,
and it follows that
P, = 1 • 2.
If a third letter is placed in every possible position in each of
the preceding permutations, at the beginning, in the middle, and at
the end, one obtains for the permutations of three letters a, 6, c,
cahy achy ahcy
chdj hca, hac.
Hence, all the permutations of three letters have been formed,
because a permutation of three letters is composed of the permuta-
tion of the first two letters, a and 6, to which is added the third
letter c in all possible ways. The same permutation is not repeated,
because every two permutations differ either by the position of the
letter c or by the disposition of the two letters a and h . Hence,
each of the preceding permutations furnishes three new ones, and
one gets
^3=7^3 = 1
2 -3.
Similarly, if a fourth letter, d, is placed in each permutation of
three letters, a, 6, c, in all possible ways, it will have four positions,
two mean and two extreme positions ; this gives the permutations of
8616] ARRANGEMENTS AND COMBINATIONS 601
four letters, a, ?>, c, rf, and since each of the permutations of three
letters gives four new permutations, we have
P,=P^ -4=1 -2 -3 -4.
Assume now that the law holds for n objects, then
(1) Pn = n\
Now with one of these n ! permutations of n things, place an object
in all possible positions, first before the first object, then between
the first and second objects, and so on until finally it is placed after
the last or n*^ object: then w+1 acts have been performed, and each
act results in a permutation of n+1 objects. Hence, for any
permutation of n things there have been formed n-|-l permutations
of « + 1 things. Hence, from n! permutations of n things are
formed
(2) P„^^=n\x{n+l) = {n+l)\
of n -f 1 things at a time.
Therefore, if the law expressed by (1) holds for n things at a
time, it holds for n -|- 1 things at a time ; but it holds for 4, there-
fore it holds for 5, etc.
Combinations
616. By the number of combinations of n objects taken r at a
time is meant the number of different groups which can be selected
from these objects, taken r at a time, in all possible ways, with the
convention that two groups differ at least by the character of a single
object. In the formation of combinations, no attention is paid to
the order of the objects.
For example, but three combinations of two letters can be made
from three letters; namely:
ab, ac, be,
while six arrangements are possible.
Let, in general, „(7rbe the number of combinations of n things
taken r at a time. The formula for combinations may be deduced
from that for arrangements and permutations. Suppose that the
combinations of n symbols, taken r at a time, have been formed.
If one gives to r symbols which compose each of these combinations
all possible dispositions, that is, if one forms the permutations of
these r symbols, he will have the arrangements of n sjrmbols taken r
at a time. He will thus have formed all the arrangements; because
every arrangement is a combination in which the n symbols which
602 CX)LLEGE ALGEBRA tWH
compose this combination have a given order; and no two arrangie-
ments will be alike, because the arrangements furnished by the
same combination differ in the order of the symbols, and those which
are famished by different combinations differ at least in one symbol
The number of arrangements furnished by each combination is
represented by P^; hence
whence,
and on substituting the known values for the numerator and denom-
inator the result is
(iii) ,fi="<»7/j:3.-/»-;+^>. [«i6ii,(i)]
Applications, — To apply the formula first write in the denomi-
nator the first r integral numbers and then write in the numerator
as many decreasing integral numbers beginning with n.
1. Number of combinations of 7 objects, 2 at a time:
2. Number of combinations of 11 objects, 4 at a time:
fy __ 11 • 10 • 9 • 8 _ OOA
3. The number of combinations of n symbols, one at a lime, b
n, as is evident a priori,
4. The number of combinations of n things, n at a time, is
p _ n(n-l)(n~2) .... 3'2'1 _ ,
nW - 1.2.3 _ _ (n-l)n ""
Moreover, it is evident if all the symbols are taken at one time, bnt
a single combination will result.
Two theorems in combinations which are useful in the following
discussions will now be proved.
617. Theorem I. — The number of combinattons of n tyxM*
taken r at a time is the same as the number of combinations of *
symbols taken n — r at a time.
Suppose that one has n objects in an urn; if he takes r of these
from the urn, there will remain n — r in the urn; that is, for eveir
combination of r things taken from the urn, there remains a combi-
nation of n — r things in the urn, and conversely. Therefore,
«618, 619] ARRANGEMENTS AND COMBINATIONS 603
The equality of these two symbols can, however, be formally
verified,
n(n-l)...(n-r + l)
• W- ^,
C _n(n-l)...(r + l)
- "-''^ (n-r)!
Multiply both numerator and denominator of the first and of the
second fractions by (>i — r) I and r I respectively, and obtain
Cf n (n — 1 ) . . . (n — r + 1) ' {n — r)\ y? !
(n — r)! -r! t\' {yi — T)\
^ _n(n-l). ..(r4-l)-r!_ n!
n ^ n—r '
(n — r) ! • r! \n'-r)\ ' r\
and the two expressions are equal.
For example, the number of combinations of 7 objects taken 4
at a time is equal to the number of combinations of 7 objects taken
3 at a time ; i. e. ,
X\ = -J^ = ^ and ^C -.^l_=t:-5_l«^ = 5ji.
'*4!-3l 1 ^»3!-4! 24 1
618. Theorem II. — The number of comhinationt of n objects taken
r at a time is equal to the number of combinations of n — 1 objects
taken r at a time^ plus the number of combinations of n — 1 objects
taken r — 1 at a time.
The following is a formal proof of this theorem. By the preced-
ing discussion,
(2) , C = — ("-1)' = ("r-Jili"^ ri
^ ' ""' ' r!(n-r-l)! r!(n-r)!
(V\ C = (n-D! ^ (»-!)! r
^' ""' '"' (r-1)! -(n-l — r+l)! r!(n-r)!
fA\ . r A. r — (n— l)Hn— r+r) _ n!
This last fraction is equal to the value given by the first equation
for ,C,
619. Thus far, in the formulae which have been proved, the
things have been regarded as unlike. When things are spoken of
of as dissimilar, different, unlike, it is assumed that the things are
visibly unlike, so that they are easily distinguishable from another.
On the contrary, things are said to be alike when they can not be
distinguished from one another by the eye.
604 COLLEGE ALGEBRA [8620
620. Pkoblem I. — ^Find the number of ways in which n thin^
can be arranged among themselves, when all are taken at a time, if
p of these things of one kind are alike, q of them of another kind
are alike, r of them of a third kind are alike, and the remainder
all are different from one another and the other kinds.
Let n = the number of letters;
and p of them be a
q of them be 6
r of them be c,
and the remainder all unlike.
Let X be the number of permutations required. Suppose now that in
any one of x permutations the^ letters a were all to become unlike
letters and different from any of the remaining letters. Then from
this single permutation could be found, without changing the posi-
tion of any of the remaining letters, p I new permutations. Conse-
quently, if this change were made in each of the x permutations, it
would produce x *p\ new permutations.
In like manner, if the q letters 6 should all become unlike, then
from one of the x -p ! permutations could be formed q ! new permu-
tations; and, if this change were made in each of the xp\ permuta-
tions, it would produce xplql new permutations.
Similarly, should the r letters c all become unlike letters, the
number of permutations would be
X'jfl'ql'rl,
But now the n things are all different, and may, therefore, have
n ! permutations among themselves. Hence,
X'pl-q \ 'r\ = n !;
that is, X = — — '- — :»
' pi q I ri
which was the original number of permutations sought.
I^OTs.— An analogous formula holds for the case In which the number of groups of
letters which are alike is greater than 8.
Example 1. How many different permutations can be made out
of the letters, taken all together, of the word Mississippi?
Of the 11 letters in this word, 1 is m, 4 are i, 4 are «, and 2 are^.
Therefore the number of permutations
llj
"" II 41 4! 2!'
= lMO-9-7-5,
= 4950-7 = 34650.
8621] ARRANGEMENTS AND COMBINATIONS 605
Example 2. How many numbers can be formed with the digits
3, 2, 5, 6, 5, 2, 3, 4, so that the odd digits always occupy the odd
places?
The odd digits 3, 5, 5, 3, can be arranged in their four places in
(1) 2-|il ways.
The even digits 2, 6, 2, 4, can be arranged in their four places in
(2) 2-7771] -'^y^- .
Each of the ways in (1) can be associated with each of the ways in
(2). Hence, the required number of ways is
4> x^_^41^ =3^x^=72.
2! 21 ^ 2! 1! 1! 4^2
621. The number of arrangements of n things r at a time, if each
thing may be repeated once^ tvoice, . , , , up to r times in any ar-
rangement, IS n**.
Here the number of ways in which r places can be filled up,
when there are n different things at one's disposal, and when each of
the n things is used as often as one pleases in any arrangement, is
to be considered.
The first place may be filled in n ways, and when it has been filled
in any one way, the second place may also be filled in n ways, since
one is not barred from using the same thing again. Hence, the first
two places may be filled n x fh ^^ ^' different ways. The third
place can also be filled in n ways, and hence, the first three places
can be filled in n' x ^, or n' different ways.
Since at any stage of this process the exponent of n is the same
as the number of the last place filled, the number of different ways
in which the r places can be filled will be
Example. — In how many ways can the following prizes be given
away to a class of 21 pupils: the first and the second Mathematical,
the first and the second Classical, the first Science, and the first
Spanish, if no pupil may receive a first and a second prize in the
same subject?
The first Mathematical prize may be given in 21 ways, and for
each way the first Mathematical prize can be given, the second
Mathematical prize can be given in 20 ways; hence, the first and the
second Mathematical can be given in 420 ways. Similarly, the first
606 COLLEGE ALGEBRA [2622
and the second Classical can be given in 420 ways, since they may
be obtained by a boy who has already received a prize. Thus the
Mathematical and the Classical prizes can be given in
420 X 420 = 176400
ways; but the first Science may be given in 21 ways, and the first
Spanish may be given in 21 ways. Hence, all the prizes may be
given in
420 X 420 X 21 X 21 = 77792400
ways.
622. How many selections can he made of n things hy taking some
or all of them 9
Each thing may be taken or left; i. e., it can be dealt with in
two ways. But either way one thing is dealt with may be associated
with either way each of the other things is dealt with. Therefore,
the number of selections is
2-2-2-2 to w factors = 2".
After rejecting the case in which all the things are left and none
taken, the total number of ways is
2"— 1.
Example. — There are 10 books on a table. In how many ways
can 1 or more of them be taken from the table?
One must take some or all of the books; and, therefore, the
number of ways is
2>«— 1 = 1023.
This result can be verified as follows : the books may be taken
singly, in twos, in threes, etc. ; therefore, the number of possible
selections
= 10 + 45 + 120 + 210+252 + 210 + 120 + 45 + 10+1,
= 1023.
PBOBLBMS
1. Find the number of permutations that can be made out of the
letters of the words, (1) rector^ (2) oculist ^ (3) algorithm,
2. How many arrangements can be made out of the letters of each
of the words in problem 1, taking (1) two letters at a time, (2) three
letters at a time, (3) six letters at a time?
8622] ARRANGEMENTS- AND CX)MBINATIONS 607
3. How many combinations can be made out of the letters of the
word diplomat^ taking (1) five letters at a time, (2) seven letters at
a time, (3) eight letters at a time?
4. Find the number of permutations that can be made out of the
letters of the words (1) phenomenon, (2) Oskaloosa, (3) concatenation.
5. Of the permutations that can be made out of the letters of
the word quadrilateral, in how many will the r immediately follow
thed?
6. In the following equations find n :
(1) ,,6;:„6;=44:3. (3) „6; = ,C..
(2) „A,^, : „4,^, = 30800 : 1. (4) 3,0, = 5-,_,C..
7. Find the number of combinations three at a time of the letters
a, b, c, d, when the letters may be repeated three times.
8. Find the number of ways n books can be arranged on a shelf
BO that two particular books shall not be together.
9. Prove that
(1) .^. = „-^^ + i>r,.^^,_, + ^^^Kr-l)„.^., + . .
. . +r(r-l). . . . {r-p+l),^^A,^.
(2) nCr^^^Cr-Pn.,C^l + ^^^n.,C^,+ • •
10. In how many ways may a product of m factors be formed
out of the product a^ a, . . . . Omn ^
11. A man has four diflferent coats, seven diflferent vests, and
five different pairs of trousers. In how many different suits may he
appear?
12. How many different arrangements can be made out of the
letters in the product a^6* c*?
13. From three cocoanuts, four apples, and two oranges, how
many selections of fruit can be made, if at least one of each kind
is taken?
14. If a guard of r men is formed out of a company of m men,
and guard duty is equally distributed, show that no two particu-
lar men will be together on guard r (r — 1) times out of m(m — 1).
608 COLLEGE ALGEBRA L?622
15. A man puts his hand in a bag containing 71 different things.
If he may draw 0, 1, 2, or any number up to n, how many d^awulg!^
can he make?
16. Find the sum of all numbers greater than 10,000 formed by
using the digits 1, 3, 5, 7, 9.
17. In how many ways can 7 persons form a ring? In bow
many wa^'s can 7 Englishmen and 7 Americans sit down at a round
table so that no two Americans shall be t<^ether?
18. How many different sums of monej^ can be made with the
following coins: a cent, a dime, a quarter, a half-dollar, and a
dollar?
19. In how many ways can five things be distributed among two
persons?
20. In a lottery 5,000 tickets are issued. In how man}* ways
may 450 tickets each win a prize?
21. How many numbers of six digits may be formed out of the
numbers 0, 1, 2, 3, 4, 5, 6?
22. In how many ways may twelve balls be distributed among
three boxes, so that three balls are in the first box, four balls in the
second, and five balls are in the third?
23. A polygon is formed by joining n points in a plane. Find
the number of straight lines, not sides of the polygon, which can
be drawn joining any two angular points.
Solution. -C. = number of lines which can be drawn between n
points. Of these n are sides of the polygon; hence the number of
diagonals is „C\ - «, or '±-^1^=^ .
24. How many lines of limited length may be formed by the
intersection of n lines?
25. In how man}' points can n lines intersect if /? of them are
parallel?
26. If / straight lines pass through a point Ay m through J^. and
n through (7, and no one of the straight lines contains more than oae
of the points, A, i?, C, and no three meet in any point except A, B,
or C, find how many triangles are formed by the lines.
4622] ARRANGEMENTS AND COMBINATIONS 609
27. Of » straight lines, p pass through one point and q through
another; ia how many points may all the lines intersect?
28. There are p points in a plane, no three of which are in the
game straight line, with the exception of q of them, which are
all in the same straight line; find the number (1) of straight lines,
(2) of triangles which result from joining them (q <Cp).
29. How many diflferent ?i -sided polygons may be formed by n
straight lines in a plane?
30. The streets of a city are arranged like the lines of a chess
board. There are m streets running north and south, and n east and
west. Find the number of ways in which a man can travel from the
northwest corner to the southeast corner, going the shortest possible
distance.
31. In how many ways may 2 7i persons be seated at two round
tables, n persons being seated at each?
32. Show that n planes through the center of a sphere, no three
of which pi|ss through the same diameter, will divide the surface of
a sphere into ?i' — « + 2 parts.
33. Find the number of parts in a sphere when it is divided by
a -^b -\- c , . . planes through the center, a of the planes passing
through one given diameter, h through a second, c through a third,
and so on; and no plane passing through more than one of these
given diameters.
34. Show that n straight lines, no two of which are parallel and
no three of which meet in a point, divide a plane into - n (n + 1) -^ 1
parts.
35. Show that n planes, no four of which meet in a point, divide
space into - (»' + 5 » + 6) different regions.
36. Find the number of combinations of 3 » things, n at a time,
when n of the things and no more, are alike.
37. Find the number of ways mn things can be distributed
among m persons so that each person shall have n of them.
38. There are 2? suits of cards, each suit consisting of q cards
nombered from 1 to j ; find the number of sets of q cards numbered
from 1 to ^ which can be made from all the suits.
CHAPTER III
FORMULA FOR THE EXPANSION OF A BINOMIAL
623. The product of two polynomials is eqnal to the snm of the
products which are obtained by multiplying each term of the molti-
pllcand by each term of the multiplier. In general, the product of
several poljmomials is the sum of the products which are obtained
by taking in all possible wa^^s a term in each of the given poly-
nomials.
Suppose that it is desired to find the product of n binomial
factors,
(x + ttj) (x -f a,) (Jr + a J,
arranged with respect to the decreasing powers of or. According to
the law which has been stated, the product of these binomials is the
sum of the products which one obtains by taking in all possible wtjs
a term from each of them. The first term of the product will be
found by taking the product of the n first terms, that is, x". If one
takes the second term a^, of the first binomial, and the first term x,
of all the other binomials, he obtains the product a^x""' ; similarly,
on taking the second term a^ of the second binomial with the first
term x of all the other binomials one has a^x^"^ ; a second term of
any of the binomials combined with the first term x of all the other
binomials furnishes a term involving x"~^; if one adds together
all the terms of the degree n — 1, he sees that the coefiScient of
x"~^ is
which, for brevity, is called S^, Hence, the second term of the
product is /SY'*""^
Form now the products of the two second terms of any two bino-
mials with the first term x of all the remaining binomials; then the
terms of the degree 71 — 2 of the product will be obtained. Bach as
J623] THE EXPANSION OF A BINOMIAL 611
a^a^^"*^ ajAjX""*, a^a^"'*, etc. ; on adding these terms together, it is
Been that the coefficient of x"~^ is the sum of the products of the
quantities Oj, a^, . . . a^ taken two at a time, which is represented
by jS'j. Hence the third tenn is JS^x""'^,
On forming all the products of the second terms of any three
binomials and the first term x of all the other binomials, the terms
of the degree w — 3 of the product are obtained, such as, a^a^a^x^'^,
aja^a^x*"^, etc. On adding these terms together, and calling S^ the
sum of the products of a^, a^, a^, ... a„_j, «„ taken three at a time,
the fourth term ^S^^x""' of the product is obtained.
In general, on taking the second term of any r of the binomial
factors and the first term x of the other remaining n — r terms,
the term of the degree n — r is formed; on adding these terms
together and calling S^ the sum of the product of the n quantities
Cj, a^, . . . a^ taken r at a time, the general term JS^x"""^ of the
product is obtained.
The term of the first degree will be found by forming the product
of the second term in all the binomial factors, excepting one, with
the first term x of this remaining factor, these terms added
together will give the last term but one of the product, S^.^x. The
last term of the product required will be the product of the second
terms of the binomial factors, namely a^, a^, a,, . . . a„, which we
call -ST..
Hence the product of the n binomial factors is expressed as
follows:
x-+6>«-* + A>"-«+ . . . +A>«-'-+ . . . +^„-jX + ^„.
Suppose that the quantities a^, a^, a^, . . . a„, are all equal to
a, then the product of the n factors
(x + a^) (x + rtg) . . . (x + a J
takes the form (x-f-a)". Moreover, the sum JS^ of the quantities a^, a^,
Oj, . . . a„ is na, since each of these quantities, n in number, is equal
to a. The symbol S^ represents the sum of the products of these
same quantities taken two at a time; every such product is a*, and
the number of them is the number of the combinations of n things
taken two at a time, or ^'^^~ ^ ; therefore their sum is equal to 77 'a\
Similarly, S^ designates the sum of the products of these same n
quantities taken three at a time ; since each of them is equal to a*
612 COLLEGE ALGEBRA [«34
and their number is ^^^^^f^^^^^ , their sum is »^n-lM»-2)^
In general, S^ represents the sum of the products of n quantities
^v S' ^s> ' - • ^ni taken r at a time ; since each of the quantities is
equal to a, each of the products is equal to a**; since their number is
the number of combinations of n things taken r at a time,
o _n(n-l). . . {n^r+D
Finally, the product of the n quantities «,•«,'. . . . . «» is «*•
Therefore the formula
(i) (x+a)-=^af^ + f-^ ar-i + ^^ a«x"-« + n(n-l)(n-2)^^„^
which is known as the hinomvil formula. It is of very great practical
use, serving to form the development of any positive integral power
of a binomial. It will be shown later that the law of expansion
exhibited in formula (i) holds for any exponent. The general tern
which is called the (r -j- l)*"* step in the development is, as has been
seen,
(1) n(n-l)..^.^(n-r+l)^,^,..
The development for (x — a)" is deduced from formula (i) on
substituting — a for a in (i), thus
(2) (x~a)» = x--^ar»-»+«^'^a«r«-«- ± a»,
in which the signs alternate.
624. Characteristics of Development of the Binomial (x-f «}",
where n is a Positive Integer.
1. The exponent of x in the first term is », and decreases in
each succeeding term by unity.
2. The exponent of a in the second term is one, and increases
uniformly by unity.
3. The sum of the exponents of x and a in every term is the
same and equal to w, the degree of the binomial.
4. The number of terms in the expansion of (x + a)* is n + 1;
because the exponents of x form the series of the first n integral
numbers plus the exponent zero of the last term,
n, 71 — 1, n — 2, . . . 2, 1, 0,
in all n+1 terms (189, VIII, 1, 2, 3, 5).
4625] THE EXPANSION OF A BINOMIAL 613
The coefficients of the tnnns equally distant from the extreme terms
are equal. The coefficient of the second term in the expansion of
{x 4- a) in formula (i), {623, is „6'j, and of the third term ^C^ etc. ;
hence
(3) (x + a)"=x»+^6\x"-ia + „CyK"-*a»+ . . .
Both the first and the last terms have the same coefficient, unity,
the second and the term before the last have the coefficients „C^
and „C._i; but from the theorem proved in {617, it follows that
these two expressions are equal. Similarly, the third terms, count-
ing from the extremities of the expansion, have as coefficients the
equal numbers, „(7, and „6^„_2, etc.
The coefficients of the expansion for (x + a)" are connected by a
very simple law : the coefficient of any term multiplied hy the exponent
of X in tliat term and divided hy the number of the term will give
the coefficient of the next term.
The coefficient of the third term is
n(n-l)
2! '
which, multiplied by the exponent {n — 2) of x in that term and
divided by 3, the number of the term, gives
w (n — 1) (n — 2) _ n (n — l)(n-2)
2! • 3 ~" 3!
the coefficient of the fourth term.
In general, the coefficient of the r*** term is
(A\ n{n — A) . , . (n—r-\-2) .y
^*^ (r-l)! -«W-i,
which, multiplied by the exponent (ii— r+1) of x and divided by the
number of the term r, gives the coefficient of the (r-|-l)*** term,
(5X n(n-l) . . . (n-r4-2)(n~r-f1) n .
Suppose that we put x=l, and a=x in equation (3) {624, then
(6) (l + x)"=l + ,<7,x+,6;x« + „6;x»+ . . .„C„_,x«-^+x».
626. It is important to be able to develop rapidly any power of
a binomial. The following illustrations will much assist the calcu-
lation.
1. (x + «)• = x« +6 x«a + 15 x*a« + 20 xV + 15 xV + 6 xa^ +a\
The coefficient of the third term is found by multiplying 6 by 5
and dividing by 2; the coefficient of the fourth term is 15 times 4
614 CX)LLEGE ALGEBRA t*626
and the product divided by 3. Since the exponent of the binomial
is 6, the number of terms in the expansion will be 6 + 1? and the
coefficients of the remaining terms will be the same as those which
precede the coefficient of the middle or fourth term in reverse order.
2. (x + a)»= x» + 9 x»a + 36 x^ a« -f 84 xV -|- 126 x^a* + 126 x'(t
+ 84 x^ a^ + 36 x« a^ + 9 xa« + a».
The development contains 10 terms; it is only necessary to calcu-
late the first five terms; when the fifth term 126x'^tt* has been calcu-
lated, the coeflScients are reproduced in reverse order.
3. (x — ay^= x'^— 10x»a + 45 xV- 120 xV+ 210 xV— 252 x*a^
+ 210 xV — 120 x'a^ + 45 x«a» — 10 xa» + a^\
Since the number of terms is 10-|-1, which is odd, the last term
will have the sign -|-, and the terms equally distant from the ends
will have the same signs.
4. (x — a)»= x" — 11 x% + 55 x»a«— 165 xV-|-330 xV— 462x*a*
+ 462 xV — 330 xW + 165 x» a»— 55 xV+ 11 xu'^ — a".
Since the number of terms is even, the last term will have the
sign — , and the terms which are equally distant from the ends will
have contrary signs.
626. The coefficients increase from the beginning to the middle of
the development^ and diminish from the middle to the end.
We have already seen that the ratio of the (r + 1)*** term to the
(7) ng:,^n-r+l [1624, (4), (5)]
nOr-l r
The coefficients will continue to increase so long as the multiplier is
greater than unity they begin, on the contrary, to decrease as soon
as this multiplier is less than unity. If one puts
and solves this inequality, he will have
(8) -<'^-
The fraction r <^^— represents one half of the number of terms
of the development; hence the terms increase from the first term
till the middle of the series ; after the middle term, i. e. , after the
inequality is reversed, the coefl3cients decrease.
2627] THE EXPANSION OF A BINOMLVL 615
There are two eases to consider:
1. When n is even, the number of terms in the expansion is odd;
Hnd the middle tenn is greater than any other term. For example,
■vhen n is even, the coefficients of the development of (1 + x)* are
1, 6 15, 20, 15, 6, 1;
hence, the coefficient 20 is the largest.
2. When n is odd, the number of the terms is even, and the two
coefficients equally distant from the extremities of the development
are the greatest For example, the development of (1 + xY has
the coefficients
1, 9, 36, 84, 126, 126, 84, 36, 9, 1,
of which the two coefficients 126, 126, are the largest.
627. The preceding discussion gives a property of combinations
worthy of notice. Suppose that one desires to know, for example,
in what way six objects must be combined in order to obtain the
greatest number of combinations. It is evident that the six objects
should be taken three at a time, because the coefficients of the
development of (1 + x)*, beginning with the second, are the number
of combinations of six objects taken respectively one, two, three,
etc. , at a time ; the largest coefficient being the fourth. Whence it
follows that the number of combinations of six objects taken three
at a time is the greatest of all possible combinations of six objects.
Similarly, in case of nine objects, the greatest number of combina-
tions is obtained on taking four or five at a time.
Suppose that Ji is odd, and equal to 2 p + 1 ; then
n + 1^2p + l + l^^ . 1
2 2 ^ ^
Now for all values of r up to p inclusive, ^'^ is greater than r,
J626, (8), but if r = 7> + 1, the multiplying factor
n-jr±l ^ 2P + 1-P-1-H = 1 , [;j626, (7)]
r p-i-l
and therefore from (6)
Therefore, the number of combinations is the greatest in case
the things are taken ^"1" ■ or ^*~ ■ at a time ; the result, however, is
the same in both cases.
616 CX)LLEGE ALGEBRA L«628-630
628. Suppose x = 1 in equation (6), 2624, for the expansion of
(1 + ac)"; tlien,
(9) (i + i)« = 2" = i + ,(7^ + ,c;+,c;+ . . . +«c;.,+ i
l^n^n(n-l)^ . . . + 1.
Therefore, it follows from (9) that the sum of the coefficients of
the development of (x + a)*» is
(10) 2- — 1.
It follows also that the total number of combinations which can be
made with n objects by taking them in all possible ways, one at a
time, two at a time, etc., is 2" — 1 (J622).
629. If one puts a =1, x=l in the development for (x — a)",
(2) S623, he obtains
(l-l)" = 0 = l-,C,+ ,^,-„6;+ . . . d=»C,;
whence it follows that
(11) nO,+ nO,+ . .. =l+^nC,+ .C^+ ....
Hence, when all possible combinations of n objects are formed,
the number of combinations which can be made in case odd numbers
of them are taken is greater by unity than the number of combina-
tions which can be made when even numbers of them are taken. Let
these two numbers of combinations be v and w] then by (9), {628,
(12) t; + M7 = 2"— 1,
formula (11) v — w=l]
whence, it follows, from (12) and (11), that
(13) V = 2"-», w = 2"-i ~ 1.
For example, with 11 objects one can form in all 2"— 1, that is,
2047, combinations. Of these combinations, there are 1024 which
are composed of an even number and 1023 of an odd number of
objects.
630. Summation of the Same Powers of Numbers which Form
an Arithmetical Progression.
The solution of this problem is attained by another application
of the binomial theorem.
Let (1) aj, a,, . . . . a„_j , a„ ,
be n terms of an arithmetical progression ; call d the common dif-
ference of the successive terms and represent the sum of the r***
powers of all the terms by S^.
8630] THE EXPANSION OF A BINOMIAL 617
Then
(a,+rf)-^'=a5+«+'±laJrf+('+l)r«J-'d»+. . . +!±la,d'+(r+>,
Let the equations be added member to member, and the equal terms
(aj+(f)'-+i and aJ+V(«8+^)''"^^ and a;+^ . . . (a„_i+rf)'*+i and a;+»,
which occur respectively in the two members, be suppressed; on re-
placing a^+d by a^+wf/, it follows that
(2) {a^+ndr^^=a['^^+''-±^dS,+ ^-^
On putting in (2) r = 1, one obtains the value of the known sum tS^
of the n terms of progression (1), since in this case series (1)
consists of three terms; then
(a^ + nd)* = «i + 2 dS^ + wc^;
(3) S^=l{2a^+{n^l)dy
If one puts r = 2 in (2), the value of S^ is found by means of ^S'^, thus
{a^ + ndy=- a? + 3 dS^ +3d*S^+ nd\
Hence, replacing ^S^^ by its value,
af+3aJ/icf+3ayc/«+e^i»==aS+3c7/S,+3e?«r^J2aj+(n— l)rfn + Mrf».
(4) 3S^=3a1n+3n\d+ d'n^—^2a^d— | n(n— 1) d'—nd'
= 3a\n+(3n\—3naj)d+ in^— | n(n — 1)— n id^
= 3a2» + 3na/n — 1M+ -V2n»— 3n + 3 — 2^"
= 3a?n+ 3n(w- Dajci + ^{n-i)(2n-^l)d\
On putting r = 3 the values of S^ are found by means of the
values of jS\ and S^ , and so on.
618 COLLEGE ALGEBRA [iK^l 632
631. In particular, if one desires the sum of the first, second,
etc., powers of the first n integral numl)ers, he has only to puta,=L
and c/ = l in equation (2) and then proceed as has just l)een descrilHfd
L If a, = l, cZ = l, r=l;
(1 + w)» =V+ iii S^ + n • V, whence S^ = ?^^^-
1 2
II. When a = 1, d = 1, r = 2; then,
l+3it+3«« + n'=l + 3<S; + 3 "'"+^> + it,
whence, S, = "'»+M(2»-fl) ." IIW6M
The results of I and II can be deduced by putting 0^ = 1 and rf=l
in equations (3) and (4).
III. When, a = 1, d = 1, r = 3, then
632. The sum of the squares and the sum of the cubes of tlie
first n integral numbers may be obtained by direct methods, without
using the binomial theorem.
1. In order to find the sum of the squares, consider the foUow-
ing table, which contains n — 1 columns and n — 1 rows:
1
1 2
12 3
12 3 4
12 3 4 5
12 3 4 5 (»— 1).
If the sum of the numbers of this table be taken by horizontal rows,
it follows that their sum is
1(1 + 1) 2(24- 1), 3(3 + 1) , , (n-l)(n-l + lj.
2"'"2"^2"*"'*''"^ 2
+ (l+i+2+l+---+^-7-)=
2 4
8632] THE EXPANSION OF A BINOMIAL 619
Reckoning the sum of the nnmbers of this table liy vertical
columns, one obtains the sum
(n — l) + 2(n-2)+3(w-3)+ . . . +(»-!) {n-in-1));
i. e., n[l-|-2+3+ . . . +(»_l)J_Ll + 2'+3«+ . . , (»-!)']
or (2) 5*<!^>_ LI + 4 + 9+ . . . +(»-l)«]. '
On equating expressions (1) and (2), one finds
l_,_4+9+ . . . + («-l)'=^'<» =il) _fel} ^"(»- 1) (2" - 1),
o o o
which, on adding ?i* to each member, gives
g = *L(n + l)(2n+l) , (-jg^g^ ^j)-,
2. The snm of the cubes of the first n numbers may be found
by a method due to M. Barbier. Construct a table by the multiplica-
tion of the first n numbers : —
1 X 1, 1X2, 1X3, . . . 1 X n,
2X1, 2X2, 2X3, . . . 2Xn,
3X1, 3X2, 3X3, . . . 3Xn,
n X 1, n X 2, w X 3, . . . n X n.
The sum of the numbers of this table found by adding the columns
by horizontal rows is the square of the sum of the first n numbers,
for the sum of the first column is the sum of the first n numbers,
which we call JV, multiplied by 1 ; the snm of the numbers of the
second column is iV x 2, and of the third column is ^V x 3, and
so on. The entire sum is therefore
JV (1 + 2 + 3 + . . . +7i)=X\
This sum may be formed in another way. Group the products in
the following manner:
1X1,
1X2,
2X1, 2X2,
3X1,
3X2,
1X3,
2X3,
3X3,
4X1,
4X2,
4X3,
1X4, . . . .
2X4, . . . .
3X4, . . . .
4X4, . . . .
620 COLLEGE ALGEBRA [1633
In general, the p^ group is composed of all the products in the p^
row and the p^^ column, till the column and the row meet in a com-
mon element inclusively. The sum iS' of these groups may be
arranged as follows : —
1^ .1X1 = 1,
2°'>. 2 X 2 (1) + 2* = 2* + 2« = 2>,
3"*. 3x2(l + 2) + 3« =2-3« + 3* =3',
4'^ 4 X 2 (1 + 2 + 3) + 4« = 3 • 4« + 4« =41
p^. 2i>]l+2+3+ . . . +{p-l)\ + p'=^-^f^+p'=p'^
n'\ 2 II-; 1 + 2+3+ . . . +(n-l)}+««= n".
These n groups added together are equivalent to the sum of the
numbers in the table, or
j^ ^ rn(n+l)-[» .
but they are also equal to
l» + 2»+3'+4»+ . . . +p»+ . . . +«';
i. e., the sum of the cubes of the first n numbers is equal to the tquare
of the sum of these numbers^
633. Example 1. What is the tenth term in the expansion of
/m _ 2r\"^
\ n m /
Comparing the expression f- ^V with formula (i), J688, it
follows that
x = rn, a=-2£, „= 15;
n m
substituting their values and r = 9 in (1), 2683, we have
(9+1,. u™ = .o.«™ = ■'-^s^;yr'(-a'(:)'
= — 2562560 -^ •
Example 2. Find the first five terms of the expansion of
(S»<-?.V)"
{633] THE EXPANSION OF A BINOMIAL 621
We have: (3n*-|nV')" = (3n*)"-ll • (3«*)" • (|«»y-«)
+ imi=Ji(3..)'(|„v-.)'
_„,.-.„„-.,(3.,)..(|..,..).
+ "'"-i'.T.I.'i'""'"(^"')'-(l'-''-)'
=3" »V _ 22 . 3» nV y-« + 220 • 3^ • n"y-*
— 1320 • 3*-n'^y-«+ 6280 • 3» nVy-*.-., +. . .
Example 3. Write the term of f^2 a* ^ which contains a*.
\ ara/
On comparing the expression ^2 or \ with formula (i), 2623,
it follows that x =2o* a = f V n =20; substituting their
values in (1), 2623, we find for the (r + !)**» term of the expansion
of the given binomial
(Ij 20-19.. .(20-^ + 1) (_ X.J |-2 ^^j-r
Now, it is required that we find the term in which a' appears ;
100-7r
hence a « z^a*,
therefore 13?-=iZl = 8 and 7r = 84;
.-. (2) r = 12.
Substituting back in (1) the value r = 12, the required term is the
la^'term
f'^f^ ,1 (- D" • 2" • x-« • a" = 125,970 • 2« • a:"" • a».
Example 4. Find the cube root of 121 to five places of decimals.
(121)^ = (5»-4)*=5(1-|)*
=:5/l_li — 11!— 1^14»_ \
\ 3 ' 58 9 ' 5« 27 ' 6 ' 5» /
_ 5 _ 1 i. __ 1 4' _ 1 i' __
3 ' 25 9*5* 3* ' 5^
= 6 —.0533333 — .00056888 — .000010113 - . . .
= 4.94608, correct to five places of decimals.
622 COLLEGE ALGEBRA H^^
BZEBC3ISE XO
Develop the following to five terms:
1. (x - 2 yy, {Sx + y)», (2 a; + 3 y)\ (5-2 1)*.
2. (1 + x»)«, (1 - j^y, (1 + a^y, (x« - syy.
3. (|. + 2/. (lx-3.)', (|-|.y.
5. (2 a;- 1/3 y)S (ay*+ 26)», (v .r- V>)», (,-^^ - ?|^)'.
«.,s.<-,.„-,..(i-i^.)',S+,-^y.
8. What is the fifth term in the development of (^ — ^)\ the
"iT" I ' *°^ what terms have
in every case the same coefficients as those required and what are they?
r 11"
9. Find the coefficient of ;r* in the development of -r* — - '
[a 2 r'l^
- — ^ , and the term involv-
10. What is the term involving a"* in (- — ^S), and the term involv-
mgx-m (•---) ^
Develop the following expressions by means of the binomial theorem:
11. {la+V ly + {ya - Vb)\ (1 + Vxy - (1 - I x)\
12. (1 + iy + (1 - lY, (1 + 0* + (1 - 0*.
13. (1 + lY^ - (1 - 0^ (1 + 0" — (1 - 0".
U. (3 + IV 5)' + (3 - iV'^y, (3 + iv by - (3 - 1*1 5)'.
15. (1 + iv iy + (1 - iv §)•, (1 + hsy - (1 - II 3)'.
16. (2+3iy + {2-3iY, (2i+3)»+(2+30*.
17. (4 + 3iy + (4 — 3i )», (4 + 30* — (4 — 30«.
Find to four places of decimals the values of:
18. h 123. 19. V630. 20. ^ 251.
21. 1.1^ 1.02«^, 1.003« 1.0007''.
22. 0.9», 0.98l^ 0.997«S 0.9995»
-(s)' (i)'. (i)* (i)'
BOOK VII
CHAPTER I
LIMITS
634. Constants and Variables. — A constant number is one that
always remains the same throughout the investigation. A variable
number is one that changes its value, so that at different stages it
requires different numerals to express it. In the following pages,
the word number will usually be omitted, and the words constant
and variable will be used alone.
Constants are represented by the first letters of the alphabet a,
6, c, . . . and by numerals; variables by the last letters of the
alphabet x, y, 2, . . .
635. Limits. — When a variable takes successive values which
approach nearer and nearer to a given constant, so that the differ-
ence between the variable and the constant can be made smaller
than any assigned number, the constant is called the limit of the
variable (J 346). Suppose that a point moves from 0 toward ^Y ac-
cording to the following law ; during the first second the point moves
one-half the distance from 0 \x) X and arrives at^^j during the
second second the point moves one-half the remaining distance ^^A'
and arrives at ^g; during the third second one-half the remaining
distance a^Xand arrives at 8^\ and so on indefinitely.
H'^ \L Li ii — Li — \2
•-'/
Fic.irRE 1
623
624 COLLEGE ALGEBRA [1635
Suppose that OX is two feet. Let «j , «, , «, , etc. , be respec-
tively the distances of the point from 0, and a^', a^', «,', etc. , the
distances of the point from X at the end of the first, second, third
second and so on, then:
after one second »j = 1, a^' = 1
after two seconds a^ = 1 + ^, V "= i
after three seconds «3 = I + ^ + J, V =^ i
after four seconds «4 = 1 + ^ + i + J, V ~ i
after n seconds «„ = 1 + i + i + • • ^i> «n' = ^,-
If the values are represented on a line, it is easy to see the law by
2 Ji ft ?i f# X
Figure 2
which any «„ can be obtained from its predecessor «„_i, namely;
«„ lies half way between a„_j and 2.
If therefore n is increased without limit,
l»°i «„=2 and ^i"^ « /= lim / i.^\ _. q. [8635]
The same result could have been derived arithmetically from the
formula for the sum «„ of the first n terms of the geometric series
a + ar + at-* + . . . . + ar^''\
«» = "-^- [«640,(ui)]
Here a = 1, r = ^;
2
If M increases without limit, --_- approaclies 0 as a limit ({636), and
therefore
lim «„=lim /2_ 1 )=2. [J6S6]
{1636-638] LIMITS 625
636. Test for a Limit.— The definition of a limit illustrated by
the preceding example furnishes a test for a limit; to prove that a
variable approaches a constant as a limit, it is necessary and sufficient
to prove that the difference between the variable and constant can
be made less than any assigned quantity, but can not be made abso-
lutely equal to zero, i. e., their difference approaches the limit 0.
637. Infinitesimals and Infinities. — A variable which approaches
zero as a limit is an infinitesimal. For example, the difference
between a variable and its limit is a variable whose limit is zero.
E. g., *n=^^i (2636) approaches the limit zero as n is indefi-
nitely increased, and is accordingly an infinitesimal.
The reciprocal of an infinitesimal is a variable that can become
larger than any assigned quantity and is called an infinite variable.
E. g., the reciprocal of the infinitesimal — — given above is 2""^
which is an infinite variable, if 7i is allowed to increase indefinitely.
REMABR.~In all cases, whether a variable actually becomes equal to Us limit or not,
the important property is that their difference is an Infinitesimal. An infinitesimal Is
not at all times during its existence a very small number. Its virtue lies In the fact
that it decreased numerically through positive numbers or increases algebraically
through negative numbers, approaching zero as a limit, and not in the smallness of any
constant value through which it may pass.
Fundamental Theorems Concerning Infinitesimals and Limits
IN General
638. Theorem I. — The prodvct of an infinitesimal e hy any
finite constant c is an infinitesimal.
For brevity we shall express symbolically the fact that a variable
X approaches a limit a, thus, x = a.
Since e is an infinitesimal, then by definition ({637).
e=0,
and similarly for any other infinitesimal. The theorem requires us
to prove that if c = 0
then ce = 0.
For, let k be any assigned number; then, by hypothesis, e can be
made less than _ , i. e. , ce can be made less than any assigned
number. A;, and is, therefore, infinitesimal.
626 COLLEGE ALGEBRA L«639-642
639. Theorem II. — The algthraic sum of a finite number^ w, of
infinitesimals is an infinitesimal; i. e., if
c, =0, 6,= 0, ^3 = 0, . . . e„=0,
then e^+e^+e^+ , . . +e„=0.
For, the sum of n variables does not numerically exceed the protluct
of n by the largest of these ; but their product by theorem I is an
infinitesimal ; therefore the sum of the n infinitesimals is an
infinitesimal.
640. Theorem III. — The product of two infinitesimals is an
infinitesimal; i. e., if
Cj = 0 and e^ = 0,
then e^ e^ = 0.
For, let k be any assigned numl)er <! ; then Cj , e^ can each be made
less than k (§637) ; hence e^e^ can he made less than k\ which is
less than k, since /c <; 1 ; that is Tj r^ can l)e made less than any
assigned number, and is, therefore, infinitesimal.
641. Theorem IV. — If two variahles, x and y, are continually
equal and if one of fhetn^ u*, approaches a limit, a, then the other
approaches the same limit; i. e., if
a5 = y and ar = a,
then y = a.
Since the difference between a variable and its limit is an infinitesi-
mal (1637), then
X = a + e where c = 0;
hence y = a -)- c and y — a = e\
y=a since e = 0. [3635]
642. Theorem V. — The limit of the sum of a constant, c, and
a variable, x, equals the sum of the constant and the limit of the
variable; i. e. , ^
lim {c -\- .r) = c + lim x.
For, let X = a,
then X = a + e where r = 0 ;
c-^x=c-\-a-{-e and (c^x) — (c-|-a) = e,
and c + X = c -{- a since f = 0; [{636]
i. e., lim (c+jr) = c -\- a = c -{- lim x.
JJ643-645] LIMITS 627
643. Theorem VI. — The limit of the prodvct of a constant, c,
and a variable, x, is equal to the product of the constant by the limit,
a, of the variable; i. e. ,
lim {ex) = c lim(.T).
As in theorem V, J642,
X = a + e where e r^ 0
and hence ' ex = ca + ce]
but ce = 0, [J638]
ex = ca]
h e., \im{cx) = ca = c lim x,
644. Theorem VII. — If the sum of a finite number of variables
(a^i, Xj, . . . .T„) is variable, and if each variable approaches a limit,
then the limit of their sum is equal to the sum of their limits; i. e.,
lim(urj+ Xg+ . . . a-J = lim x^-{- lim x^-^- . . . limx„.
For, let ^i = ^v ^i = ''v • • • ^« = «";
then Xj =«!+«,, x^=a^-\re^, . . . .r„=a„+e„,
where e^ = 0, e^ = {}, . . , e,,=^Q ; [?637]
hence x^+x^-\- . . . +x^ = {a^+a^+ . . . +aj^{e^+e^+ . . . ej;
but ^1+^2+ . . . +«n=0; [1639]
lim m-x^+ . . . +x„) = ia^+a^+ . . . +aj
= lim Xj + . . . + lim x„.
Corollary. — If the sum of a finite number of variables
(uj^+Xg + . . . x„) is constant and if each variable approaches a limit,
then this constant, c, is equal to the sum of their limits; i. e., if
then lim x^+ lim iTg + . . . + lim x^=c.
Transposing, x^+ . . . x„ = c — x^;
hence, by theorems IV, VII, and V,
lim Xj+lim.Xj+ . . . + lim x„=lim (c — x^)
= c — lim Xp
lim Xj+lim x^-{- . . . +lim x„ = c.
646. Theorem VIII. — If the product of a finite number of
variables (x^, x^, . . . x„) is variable and if each variable approaches a
limit, then the limit of their product is equal to the product of their
628 COLLEGE ALGEBRA * [8646
limits; i. e. , with the same relations as in 2645 it is to be proved
^^^ lim {x^x^ . . . xj = ttj • a, • a, . . . a„.
For, XjX, = a^a, + a^c, + a,Cj + e^e^
where a^e, = 0, a^e^ = 0, e^e ^ 0 ;
lim {xyX^ = a,aj = lim ucj • lim a?,. [{635]
Since lim {x^x^ = a^a^^ and x,Xj may be considered as a single
variable and a^a^ as its limit, then we have
lim [(aria;^)xj = a,a, -a,;
i. e., lim (x^x^x^) = lim x^ * lim x^ • lim x^.
On continuing this mode of reasoning, it follows that the theorem is
true for any numl)er of variables.
Corollary. — If the product of a finite number of variables
(xjXg . . . x„) is constant, then this constant is equal to the product
of their limits ; i. e. , if
XiXj.Tj=c, then, lim Xjlimx, limXj = c.
The proof is left to the student.
646. Theorem IX. — 7/ the quotient of two variables^ x and y,
which approa<ih limits^ is a varidhle, then the limit of their quotient is
equal to the quotient of tlieir limits, provided their limits are Jinite
and different from zero; i. e.,
lim^ =
y
lim X
limy
For
hence.
by
X r=
theorems IV and VIII,
i-
lim X =
:lim^
y
,
lim 5 =
lim X
lim y;
* * y liiii y
Corollary 1. If the quotient of two variables, x and y, is a con-
stant, c, then c is equal to the quotient of their limits ; i. e. , if
X .1 ^„ lim X
- = c, then r: = c.
y lini y
Here x = cy
and lim x = lim (cy) = c lim y; [2643J
lim X
lim y
8647] LIMITS 629
Corollary 2. The limit of the quotient of a constant, c, and
a variable, x, is equal to the constant, c, divided by the limit of the
variable; i. e.,
lim- = -
X lima:
For, let - = y, then c =zxy
and c = lim X- limy; [J646, Cor.l]
limy = 77^—; that is, lim - = ,t^— .
lima:' x lima:
647. Theorem X. — The limit of a power of a variable equaU
the same power of the limit of the variable.
Let X = a ;
then it is to be proved that the
lim (x") = (lim x)*
where n is positive or negative, integral or fractional.
I. When » is a positive integer.
By 3646, lim (x • x • x . . . to n factors) = lim x- lim x* lim x . . . .
to n factors.
lim (x") = (lim x)".
II. When n = positive fraction, ? .
Let
^'^=y,
then
(1)
« = y';
by case I
(2)
lim X = lim y< = (lim
from (1)
(3)
X^=}f''.
By case I
(4)
lim (x j« = (lim y)»*,
but from (2)
(5)
(lim a;)« = (Um y)P,
.-. from (4), (5),
(6)
lim (^ ac4 j = (lim x)?.
m. When n is a negal
sive number, — «.
a
and lim (x"0 = .-: — -— = jrr-
lim (x* ) (lim x)* '
lim (x~*) = (hm x)"*.
CHAPTER II
C0NVSR6BNCB
648. Definition of an Infinite Series. — Letu^, u^, «,, . . .
be any set of values, positive or negative, or both, and form the
««"«« (1) «,+„,+„,+ ...
Represent the sum of the first n terms of (1) by «„:
«n = *'o + "i + «8 + • . • + «*n-l •
Suppose that n increases without limit. Theiji either (!) s^ will
approach a limit i\ jjj^ __
or (ii) 8^ does not approach a limit. In either case, (1) is called an
infinite series, because n takes values larger than any assignable
number. In case (i) the infinite series is said to be convergent and
to have the value IT, or to converge toward the value If. In case (ii)
the infinite series is said to be divergent.
The geometric series
1+1 + 1+ +2^+... [{635]
is an example of a convergent series.
The sum of the first n terms of the arithmetic progression
1 + 2+3 + 4+ ....
is *» ~ 9 ^^^ + (n—l)d] where fi = 1, </ = 1, w = w ;
hence cr= Jim g (2 + n _ 1)] = Ji-„ (^) - « ^
therefore the series is divergent. Only convergent series can l)e
used in mathematical investigations.
The series i/^+Wj+m^+ . . . ad infinitum
is sometimes used instead of the limit U, or again
U= W0+W1+ Wj+ . . .
Rkmark.— The student must remember that U is not the sum of the series but tha
litnit of the sum of the series. Similarly lUHtead of finding "the sum of an Infinite
number of terms *' one finds the limit of the sum of n of these terms as n increases
without limit.
ciao
JI649, 650] CONVERGENCE 631
649. Series in which all the Terms are Positive.
Examine the convergence of the series
(2) e = H-f + X + _J_+... +1.,+ ...
Omitting the first term of (2), compare the next n terms
1 1-2 • 1-2-3 ' • • • ' 1.2-3. . . . n
with the corresponding sum
*« "^2"^2-2"^ * • • '^2-2-2. . . . (i»-l) factors
= 2-2;^<2. [?640,IV]
Disregarding the first two terms of s^ and «„', each term of «/ is less
than the corresponding term in «„ , and hence
.„'<^,<2;
and 1 + «^' < 3.
If 8„ is the sum of the first 74 terms of series (2), then the sum
of its first (h + 1) terms is
l+'- = '-=l+I + F2+rV3+ • • • +l-2-3.\..n<^
no matter how large n is taken. Thus it is seen that »„ is a variable
which increases continually as n increases, but which never takes as
large a value as 3.
650. Graphical Representation of These Results.
Plot the successive values of «„ as points on a line,
», = 1 " =1
*, = 1 + 1 =2
',= 1 + 1+2! =2-5
»,= l + l+2-! + |, =2.6667
*.= l + l+|^^ + |i + 5! =2.7083
«.= l + l + |^^ + ;, + [, + |! =2.7166
s=l + l + l + L + l + L + l =2.7179
..= l + l + 2-, + 3-, + 4-,+|i + e-! + |! =2-7181
632 COLLEGE ALGEBRA [H65L652
^-^.5 e
The preceding table shows that, as n increases by 1, the point
represented by s„^^ moves continuously to the right but never
moves BO far to the right as the point 3. Therefore, there must be
some point, e, to the left of 3 (i.e.,6<3), which s^ approaches as a
limit, but never reaches (2650). The table shows that the value of e
correct to the third decimal place is 2.718.
651. Fundamental Theorem. — It has been shown in {650 that
the variable «„ approaches a limit as n =£: oo ; and although we do
not as yet know how to calculate the numerical value of the limit e.
the reasoning by which the existence of the limit e is proved is
of great importance. It can be formulated as follows:
J[f a variable «„ (i) always increases when n increases, i e.,
but (ii) is always less than some definite fixed number N, i. e.,
for all values of w, then s^ approaches a limit U; i. e.,
n=Qo *
U N
The limit iZmay be coincident with JVor some value less than N:
Example. — State the principle for a variable which is always
decreasing and always greater than a certain fixed quantity, and draw
the corresponding figure.
652. I. Comparison Test for Convergence. — The following test
for the convergence of an infinite series is based upon the theorem
in the preceding paragraph:
Let it be required to test the convergence of the infinite series
(i) «o + «i + ^+
all of whose terms are positive. Suppose that we can find an infinity
'^** (ii) «. + «.+ «.+ .■'...
whose terms are all positive and which is convergent; then, if the terms
of series (i) are respectively, less than {or at the greatest equal to) tie
corresponding terms of series (ii), tJie series (i) is a convergent serifs^
and its value is not greater than that of series (ii).
W52] CONVERGENCE 633
For let «n = t*o+«, + «, + +««-i,
and J\°^ S^ = K.
n=oo *
Then, since by hypothesis «^ < 5„ and iS;< JV(§686), it follows that
s^ < Wand therefore, by J661, s^ approaches a limit less than or at
most equal to JV.
L— In BtDdying the convergence of a series it is often convenient to omit a
fixed number, say m, of tbe first terms of the series and to consider the new series thus
arising. The convergence of the new series is necessary and suflicient for the con-
vergence of the given series, for,
•n =(«0 4- «! + «,+ 4- «m-l) + («m+«m+l+. • • + «„_i)
= *m + *«_,«.
' By hypothesis t^ is a constant, and therefore <^ will approach a limit if 9^-^ does, and
conversely ({048).
BXBBOISB ZOI
Prove the convergence of the following series:
1 l4-i4-i4-l4-
2 2« 2» -2*
2- ' + i + i> + T^+
3. X + X* + X* + a" + 0 < X < Z < 1.
Write S^ In the form
Urn j3 _ llm /i LA _ i
llm o _ llm /, 1_\ _ .
n-ioo «n - n-ao V n+U "
m X 3tt X^ X^
* 1-2 ^3-4^5-6^7 -8^
8- h+h+h+
634 COLLEGE ALGEBRA 1^653
Standard Series for Comparison Tests op Contsroenci
653. It follows from example 10,' Exercise XCI, that the series
(3) i+ij;+|. + i+
is convergent when /> > 2. It will now be proved that it is oon-
vei^ent when /> > 1. Let /S^ be the sum of the first *i terms of (3),
then 5.= l + ^ + |;+.... + i.
and ^,=1+1; + 1 + . . . .+^1 + ^ + . . . .+-U
, O Cr 1 I 1 I t 1
The second member is the sum of u terms and each of them is less
than - y hence it follows that
^2n— ^n < ^^p '
Put n = 2~ and make successively w = 1, 2, 3, . . . (r— 1); then for
m = 1 we have n = 2 and «""* = 2^"*
m = 2 *< *» n = 2» ** n^-^ = 2**^-^
VI =3 ** '* M = 2' *« «**-* = 2*'"*»
I 2u =2'" )
and therefore A^^ — S^ < — -^
^8 - *^4 < 2i^]FI^
y ^ ^ 1
16 *" 8 "^ 2^ "^1>
Adding these inequalities member by member, we get
'^2'' " ^^^2 \ ^];31 + 2^<l'->> « • • • • 2<r-l) <j>-i) *
The terms in the second member form a G. P. whose ratio is r-.
which is less than 1 so long as p is greater than 1, and therefore
i i 654, 655] CONVERGENCE 635
their sum remains less than a fixed number N no matter how large
r is (2540, IV). Hence, whatever r is, we have,
s^r-s^<:N and .-. ^,r<.s;+iv;
As r increases indefinitely, S^r increases indefinitely and approaches
a limit. The case when r increases indefinitely according to any
law whatever is reducible to the preceding case. It is sufficient to
consider the highest positive integral power of 2 which is contained
in II ; if r is the exponent of this power
S^r <*S'„ < *Sy+i n in general not integral,
when n increases indefinitely, r increases indefinitely and S.f and
S^r+i approach a common limit, and aV„, which is comprised between
them, approaches the same limit. Therefore the given series is
convergent when j> > 1.
654. Series (3) is useful as a test series, for many series which
can not be proved convergent by means of the geometric series can
l)e proved convergent by using (3) . For example,
1 , 1 , 1
3V3
i + ri^ + rr7^ + 7^ +
is convergent since here^^ = | >1. [2653]
655. The Harmonic Series.
(4) i + | + | + 4+ • • • +1+ • ' • is clivergent.
The harmonic series is derived from series (3) by putting p = 1.
If iS^ is the sum of the first n terms, then
V^ o 1 , 1 I 1 ^±_
There are n terms in the second member, the smallest of which is
the last, — • Therefore their sum is greater than ?/ - , hence
*%» — *Sf„ > _ .
2n'
1
Let n = 2**, and if we put successively ?h = 1, 2, . . . r — 1, then
<SV _ ,Sf,r-j > - ;
636 COLLEGE ALGEBRA [1656
henoe, by adding these inequalitiefl, we get
S^-S,>'-^ and S^>S, + ^-^-
Therefore, since n, and consequently m, increases without limits r
increases without limit; hence S^r increases without limit But if we
take n > 2'", we have
Therefore S^ increases without limit when n increases indefinitely,
and the series is divergent.
656. Test for Divergence. — ^The test for the divergence of a
series may be established in the same way as the test for conTer-
gence was derived in 3648-650, thus:
Lit (i) ' u^^u^+u^+ ....
be a series of positive terms which is to he tested /or divergence.
Suppose that a divergent series
(ii) a^ 4- flj + flf^ + . . . .
cmi he found such that every term of (i) is greater than or equal to tkt
corresponding term of (ii), then series (i) is divergent
Examples. —1. The series
is divergent. For the square roots of 2, 3, 4, .... are respo-
tively less than 2, 3, 4, .... , and therefore the terms of (i) tf«
respectively greater than the corresponding terms of the hannonic
series
(ii) 1+1+1+
which is known to be divergent, and therefore, according to the
previous test, series (i) is divergent
2. i+2^ + |; + fp+ -P<1-
3- 1+I + I + I+----
4. With the series
l+i+^+ --h'+l+l+i+ >' •
which is divergent, compare the following: [WMj
M657, 658] CONVERGENCE 637
667. II. Ratio Test For Convergence. — Let it be required to
test the series
for oonvergence. Form the test ratio ^^5^\ This ratio will in
general approach a fixed limit or increase without limit when n is
indefinitely increased. If the ratio has a limit let the limit be r.
658. Theorem. — If r <; 1, the, series is convergent; if r > 1, itu
divergent; (/" r = 1 the series can not he said to he convergent or
divergent without further examination,
lim !^i=r<l, convergent;
fl=O0 Un
(( (( — y^l^ divergent;
(( n =ry--i^ test fails.
Case I. r < 1. Then, as n increases, the points corresponding to
the value of ^^^^^^ will arrange themselves about the point r, and
hence if a fixed point p is chosen at will between r and 1, the point
!^!Lti will, in case n is taken large enough (i. e., for n equal to or
Un
greater than a certain fixed number m), lie to the left of p and we
shall have
— <P^ n:^m
or n^m "^ <P^ ««+i < «ml>
Um
n = w + 1 1^' < p, w„+a < u^^^p < u^p^
n = TO + 2 ^<p, «^+, < M,+,i) < u^p»
um+2
Adding t of these equations, we get
W«+l + ^+2+ Um+t<Um(p+P^+P^+ +P^)
or
The sum of the terms of the series beginning with u^^^ can never
be so great as ^ , however large t is, i. e. , however many terms
are taken. Therefore the u series is convei^ent.
638 COLLEGE ALGEBRA [W58
Case II. r = 1 . The series can not be said to be convergent or
divergent. For consider the series (3). Then
lim / Un+\ \ _ lim / n \P_.lim /n+ 1\"^
«=<»V t4„ / n=xU + l/ «=«\ n )
Then r = 1, no matter what jy is. But when jj > 1, (3) converges
(J668); and when/) < 1, (3) diverges (§656, Ex. 2). That is, r may
equal 1 both for a convergent and for a divergent series.
Note.— The student should note that the theorem requires that the UmU of the ntk)
series.
is alwaj's less than 1, in case the series is convergent. Thas, in caseof the harmonic
Urn ^n±i _ lim 9n-^ _ lim ^ S' = i
«_ao u^ »-« 211+1 '*^^jo_i;
L ^
Here the ratio — — is less than 1 for all values of n, yet the series is divergent ($655; ,
but the limit of the ratio is not less than 1 but equal to it.
EXEBCISB ZOU
Are the following series convergent or divergent?
Here »,..= ^+!y' «» = "-; -".««^d — = (^-)'rXr/
"*■ (n+1)! " n! u. \ n J (n-\-\j'.
. lim «.+j ^ lim /i + ly _i.^ _ 0 if ^ ig finite.
n=oo Wn n=« \ n/ n+1
Therefore the series is convergent for all finite values of x,
2 i -I- — 4- — -I- — 4-
2 2' 2' 2*
^- •^ + 2!^3! + 4!^
± 1,1 2,1 2 3 ,
^' 3"^3*5"^3'5*7'^
^- i'^ + A'^3^"^A"^ for positive value of J.
6. 1 + 3 X + 5 x« + 7 .x^ + 9 x* + for positive value of i
7. ^ + ^ + _2»_+.
2*** 3^** 4^**
a ^j.J! -I-JL4.
100^100«^100»^
8659] CONVERGENCE 639
9. Show that the series
^ + 2! + 3! + 4l+
is convergent for all values of p,
10. 1 + |. + |^+J|x»+ +1^--"-+
12 ^ 1 ^ I ^ I
13. Show that the series
1+ — + - + — +
^ 2" ^ 3" ^ 4» ^
is convergent if n is greater than 2, divergent if n is less than
or equal to 2.
14. Suppose that in the series v^ + ^'i + «8+ ^'s + ®^^^
term is less than the preceding; then show that this series and
the series w^ + 2mj+ 2*1/3 + 2« m^ + 2*u^ + are both
convergent or both divergent.
Series with Positive and Negative Terms
669. Alternating Series. — Tbeovleu.— Suppose that tlie terms of
the given series are alternately positive and negative^
(5) u^ — u, + 1/^ _ Wj + . . . . ;
and that each term is less than or equal to the one which precedes it,
and let ^_^^u^ = 0; then the series is convergent.
Throughout the steps of the proof which is to follow, consider
as an example the series
(6) i-.i4.1__l + l_.lj- ....
^ ^ 2^3 4^5 6^
Outline of the Plan of the Proof of the Theorem
I-et s^ = u^—u^+u^—u^+ .... +(— )»-1m„_,
and plot the points «j, «j, <j, Then we shall show
*• *« *« r* ^^ *« «* »f
that the points <i , *,» *fi> • • • • *2r+i> • • • move to the left, but
1
640 CX)LLEGE ALGEBRA H^^^
Bever advance as far to the left as #,. Then according to the theorem
in 2651 they approach a limit, U^ :
Urn . — 77
r=oo *^+
Likewise the points «,, «4, «,,.... «ft., ... . continually move
to the right but never advance as far to the right as s^ ; hence
according to the same theorem they approach a limit, U^ ;
lim . — IT
T =00
Since, «^^i = «^ + u,,.
we have 1^«* «^^, = ^J™ «"* + l^i" t/^ ;
r=oo ^■*"* r=Qo r=ao
but, according to the third part of the hypothesis of the theorem,
lim „ — 0
hence t/j = JT^,
or let us say V, That is «, approaches a limit [7, continually oscil-
lating from one side of its limit to the other.
\L \1 1^ I I I I I* I^ i£ -
It is now required to establish analytically the facts on which
the plan of the proof rests. It is required to prove:
First, «2r+i^«jr-i and v^'ar-a-
For «2r-l = ^— K— «b) • • .— (M2r-» — «*2r-j)
«2r+l = Wo — (^1 — **«) • • • — (^ar-l — **2r-2) — (<*Jr-l — «ar)
«2r-2 = ^%— W,) + . . . + (lV.« — t<^.,)
«2r = (^0 - t*i) + . . . + (Uj,.,, - W^. J + (iV.j — tl^,)
^^ *2r-2 + \**2r-2 W2r-l) J
where the parentheses according to the second part of the hypothesis
are all positive (or zero). Hence, the values of «, , «^ , ^^ ,
continually increase, and the values of *j, «j, *j, . . . continually
decrease.
Second, ^^r+i > '2 and *jr < '1 •
For, according to the second equation above,
*2r+l ^^ *2r-l W2r-1 "T '^ >
hence, «2r+l = V + "jr ^ *2 + «2r > *a >
and «2^ = «2^^, — W2r^ «i — ^r <*l ,
results which complete the proof.
2659J 1
CONVERGENCE
Example.— Calculate the value of
111,1
2 2'2«"'"3
1 1
■2» 4'
2*^5
i-.+.-
» •
correct to four places of decimals:
^ = .50000
i(i)'= 04167
^(^)» =.00625
|(^)' =.00112
^(^)* =.00022
TV(ir= -00004
i(\y =
.12500
.01563
.00260
.00048
.00010
.00002
Ti,(i)"=. 00000
.14383
.54930
.14383
641
.40547
or to four places, the value of the series is .4055.
The Limit of Error in an Alternating Series
In calculating the value of an infinite series, it is an important
matter to know that the value of the series is correct to a given
number of decimal places, say to four places.
In order to determine the value of an alternating series correct to
the fourth decimal place, it is not suflQcient to know that the series is
convergent, and that therefore enough terms can be taken so that
their sum, #„, will differ from the limit ^of the series by less than
.0001, since the series might converge so slowly that it would be
necessary to take n = 100,000 or greater, so that it would be prac-
tically impossible to compute so many terms.
Rule. — 7%c sum of the first n terms of an alternating series, (5),
2659, «^, differ from Uy the value of the series, by less than <^(n-|-l)***
term. Hence we can stop computing terms as soon as a term is
reached which is numerically less than the proposed limit of error.
The proof of this rule follows from the discussion connected with
the figure on page 639. «„^^ is determined from «„ by adding ± w„,
a quantity which is greater than the distance from «„ to U, But u^
is the (n+l)*** term of the series. This proves the rule.
642 COLLEGE ALGEBKA [*660
BXEBOI8B ZCm
Are the following series convergent or divergent?
X x + o x-\-2a a;+3a
where x and a are positive numbers.
JL , 1 1
3«
3- 1-^ + ^2-^-
. 1 , 1 , 1 1 1 ■
where x and y are positive numbers.
^ 2 3,4 5,6 7 ,
^- i" 2 + 3-4 + 5-6+'-'
6. Compute correct to three decimal places the value of
1-1. i + 1. 1-?:. 1 + Ans. .288.
3 2 3« 3 3» 4 3*^
7. Compute the value of the series in Example 3 correct to foui
decimal places.
680. General Theorem.— L^^
«o + ^1 +
be a convergent series 0/ positive and negative terms.
Then J\"i t*n = 0 ,
or more generally
nTloi^n+Un+l+ +««+r-l) = 0
where r is an integer which is a constant or varies with n.
Proof.— Let «„ = w, + w^ + . . . . +m„-i ,
and plot the points «i , «2 , J»3 , . . • . Now, when we say that
the u series is convergent we mean that s„ approaches a limit U;
that is, that there is a point U about which the values of s^ arrange
themselves as n increases. In all the series thus far discussed s^
always came nearer to U as n increased ; this is not necjjBSsarily
U-e s, f s, ^\e s,
required by the hypothesis of this theorem. Thus ^3 may be farther
away from U than s^. But the hypothesis of this theorem does
require that ultimately «„ may be made to differ from IT by as small
S660] .CONVERGENCE 643
a quantity as one chooses. Thus, let e be taken at pleasure
^~ 10 000 000* ®^y^ ^^^ ^*^' ^^ ^"^ interval 6 extending from CTin ieach
direction, tf—e and U-\-€\ then for all values of n taken sufficiently
large, say n>/), *„ will lie within this interval. This can be formu-
lated algebraically as follows:
^— c<«„ < t/'+e, when n>j>.
Having explained what is meant by "«„ approaches a limit ^," the
proof of the theorem is given. The sum
If n >j?, the points corresponding to s^ and «^+^ will lie in the
interval bounded by the points U —e and t/'+ c, and the distance
between the points «„ and »„^^ is less than 2 e. Hence
— 2€<t«„4.t*„^.,+ .... +w„^.^,<2c
for all values of r. Thus the quantity
which depends upon n, can be made to remain numerically as small
as one pleases by increasing n, hence it approaches 0 as a limit,
when n = 00 , which proves the proposition.
It should be noticed that the condition ^^™ t« = 0, is a neces-
sary condition if the series converges but is not a sufficient condition
for convergence. For example, in the harmonic series,
but the series is divergent. However, the harmonic series does
not satisfy the general condition of the theorem ; for put r = n,
I I I 1.1. _1 \1
and does not therefore converge toward the limit 0. Thus we have
a new proof of the divergence of the harmonic series.
It can be proved, however, that the condition
J^(w„ + w,^,+ . . . + 1*,^,.,) = 0,
where r can be taken larger than any assigned quantity,is a sufficient
condition for the convergence of the series.
644 ' COLLEGE ALGEBBA [*^1
On applying the teste of the theorem proved to the series
the first test of the theorem gives
Um lim /1^\ _ /»
the second test gives for r = n
_ 1 . 1 , , I
(n + l)« • {n + 2f {n + nf
n 1
lim { ^i _, . ^^ ^m / ^ \
That is, the series is convergent as the tests of the theorem require.
661. Conyergence: General Case.
Let (i) «• + «!+
be a series consisting of an infinite number of positive and negative
terms. Let the positive terms in (i) be denoted by the series
t'.+ ^+
and the negative terms by
taken in the order in which they occur in the series.
For example, if the u series is
^^2« 3« 4« 6» 'T^''*'*
the V series is (ii)
1+^+^+
and the to series (iii)
1 1
2« 4« ' ' • '
Let
«»
= «0 +«!+... +««-! ,
d,
= ^ +^+ . . . +«^*-l,
Pj
= «'0 + ^I+ • • • +«!/-! •
Hence whatever value n may have
»n = A— P^;
where t is the number of terms in $^ and J^ their sum. When s
increases without limit both i and y increase without limits aDd
two cases may arise.
1661] CONVERGENCE 645
Case I. Both cf^ and p^ have limits, thus
lim T TT Hrn nr
that is, both the d and p series are convergent and therefore the u
series is convergent; then
and U=V-W.
The series (ii) and (iii) belong to this case. [2653,(3)]
Case II. One at least of the variables d^^ p does not approach a
limit Let the u series be
^ 2«^3 4*^5 6«^7 >"1-> • • • •
Here the d^ series is divergent and the pj series is convergent; there-
fore the u series is divergent. In the series
2^3 4^5 6^7 • • • • '
both the d^ series and the pj series are divergent; but the u series is
convergent, (Exercise XCllI, example 1, x=l). These examples
show that the u series may be divergent or convergent.
Let now a series whose terms consist of the absolute values (224)
of the terms of the u series be formed and represent this series by
«'o+<+ ....
If u^ is positive then u\ will be a certain v and in case u^ is nega-
tive m', will be a certain w.
In case one puts
it follows that
^'n = di + Py
Hence it follows from this relation that the u* series in case I is a
convergent series.
Conversely, if the u' series converges^ then both the series df and pj
converge, which gives I.
Since the df series and the 2)j series are composed of positive terms,
then no matter how many terms are added in either series their sum
can not exceed the limit IT toward which »'„ converges. Therefore,
according to the theorem 2651, both the seriea r/^ and^^ converge.
646 COLLEGE ALGEBRA [{{662-604
Definition. — A series «„ is said to be absolutely convergent
if the series «'„ formed by taking the absolute values of the terms
of the series «„, n taken as large as one chooses, is convergent; all
other series which are convergent are called non-ahsolutely conver-
gent or coiuUtwnaUy convergent.
663. Resume of the Test for Convergence. — The u series is
absolutely convergent if the m' series is convergent; and, since the
m' series consists of positive terms, the convergence of the u' series
can be tested by the rules in ({{648-658) and therefore the con-
vergence of the u series can be inferred ({661). Most of the series
in common use in elementary analysis either belong to this class
and can be tested for convergence in the manner just described, or
they belong to the class of alternating series discussed in {659.
664. Simpler Formulation of the Ratio Test.— If the ratio
!*'^ has a limit, ^, the rule is this: — If t is numerically hss than 1,
Un
the series converges absolutely; when t is numerically greater than 1,
the series diverges; when t is numerically equal to 1 another test must
be employed,
r — 1 < < < 1 convergence;
nSo-^f^= ', j ^ > 1 or f < - 1 divergence;
yt z=\ovt =z —\ test fails.
Proof I. Convergence. — The test ratio '^^^^ is numerically equal
to the test ratio ^^^. Hence ({641)
lim / ^<^n+i \ __. fp
where T is the numerical value of t. If therefore — 1 < f < 1, it
follows that 7* < 1 and the u' series converges. The u series is
absolutely convergent.
Example. — Consider the series
1 2 "^3 4 "^ • • • •
J^«ii - (_ 1)»?!^. ^? (_i)_JL-^
Un "^ ^ n+1 (- l)»-ia:»' "" ^ ^n-fl '
lim «j!!±j. _- lim / L_a.\ _: _ a-^^.
n=^oo «M n=^« \ jiX /
2664] CX)NVERGENCE 647
Hence the series converges when the absolute value of x is less than
1, i. e., when
-1<2;<1.
When X = 1 or — 1, this test furnishes no evidence that the series
is convergent or divergent; but we know that the series is conver-
gent when a: = 1 (§669), and divergent when a: = — 1 (1666).
Proof II. Divergence, — To prove the divergence of a series
^+^ + ^+ • • •
which has positive and negative terms it is not sufficient to prove
that the v! series is divergent. Thus the series
^ 2+3 4+6 '+'•••
is convergent, but the v! series
1 + 1 + 1 + 1 + 1+ . . .
is divergent.
It is sufficient to show that the terms of the t«-series do not
approach 0 as a limit; or at times it will be most convenient to show
that the terms of the u' series do not approach 0 as a limit. That
is, if <> 1 or « > — 1, then T > 1 and ^^"/ ^>1, when n^k, where
k is some given quantity. Hence
<+i > «'fc
<+3 > w'fc+2 > ^\
or O^^'fci «>^;
that is, all the terms u\ from n = A; on are greater than a certain
fixed quantity T = \Ck and therefore u\^ can not approach 0 as a
limit when n =oo .
Example. — In the series in case I, « = — x; hence the series
diverges for all values of x numerically greater than 1. The regions
in which the series is convergent and divergent are represented
graphically as follows:
—10 1
, 1 I •
piver^ence Convergence Dlyergence
648 CX)LLEGE ALGEBRA [W64
Caution. — It is not desirable to extend furtlier the discussion of
the convergence of infinite series, but it is desirable to caution the
student against a common error in discussions connected with
applying algebraic operations in infinite series. For example, the
algebraic sum of a finite number of terms is independent of the
order in which the terms are arranged, but this is not true in case
of an infinite series which is not absolutely convergent This point
is usually illustrated by taking the following series:
Its value is less than 1 — ^- + 1 = ^ . [«659]
2 3 6
Rearranging the terms of (i),
The general relation which these consecutive terms satisfy is
4n — 3^4?i-l 2n
Putting the positive terms in parentheses we have
»"> (' + 5)-|+(5 + ))-J+a+n)-5+---
which is an alternating series which satisfies the relation
«„^,^»„ and Jr^«,^,=0, [JM9]
since we have
4u — 3 ' 4n— 1^ 2n^ 4n+l ' 4n + 3
Therefore series (iii) converges toward a value greater than
(1 -j_ i\_ 1 = 2. The series (ii) and (iii) have the same value.
Therefore the rearrangement of series (i) whose value is less
than - » has led to series (iii) whose value is greater than '- . It
can be proved, then, that it is possible to rearrange the terms of
an infinite series which is conditionally convergent so that two diflfer-
ent arrangements of the series will have two different values.
Theorem. — An infinite series which is absolutely convergent may
have its terms rearranged at will without altering the value of the series,
(Proof left to the instructor. )
8664J CON\"ERGENCE 649
sxEBcisE zorv
Determine the values of x for which the following series are con-
vergent and the values for which they are divergent and give a
graphical representation of the results. Compare table in closing
chapter of this book.
1- i+H^.+4.+ 2. i-i+i;-i;+
3-M-+S^+S^+
^- *-2 + 3-4+'-
5 x+— + — + Ans. — 1 < as < 1, conv. ;
»/2 v^3 a-M.1, x< -1, div.
'■ F2 + 2^ + 3-^ + A+ • • • «• ^^.f + ? + ? + .• • • •
•1-2 3-4^6-6 7-8^
10. x + S'x'+S'xS + T'x' +
11 ij.2'.3',4«, 12 1 ,1',2«,3»,
1^- l+2! + 3i + 4!''" ^^- r« + 2i + 3"» + 4i+ • • •
13. 10x + 10»x'+10'x»+ ... 14. l+^ + | + |!+---.
15. 1— 2x + 3x»-4x» + , — ,
16. i-^+i-i+i.-'^
17. Show that 2H^ + 3(2-J+T? + 5("2:H^i7+----
is convergent for all positive integral values of x,
.1... Ll3^+L^6 ..
18. l + |-'+2^^+§^6-*+
19- l-|^-2-^*'-2-H^^-
20. i+x+g+g+ 21. .-g+|;-|;+,-,
CHAPTER III
UNDETBRMINED COEFFICISNTS
686. If the equation - ^ ' ^ =x — a is cleared of fractions
X — a
we obtain an equation of the second degree that has an important
property which the equations of the second degree previously treated
(23404-454) do not have. Every equation of the second degree we
have thus far discussed has two roots ; but this equation is satisfied
by all possible values of x.
The reason for this is that after the equation is cleared of frac-
tions, both members of the equation are the same expression in ar.
THE0RE3I I. — J/ two polynomials in x of the «"* decree,
A^+ A^x + + A,x- and B^+ £^x+ B,x\ art
equal for every value of x, the coefficients of the same powers of x in
the two polynomials are equal; thus, A^ = B^, A^ = J5j, ^, = B^, etc.
Since the two expressions are equal for all values of x we have
(1) A^+A^x+A^a^+ + A,x-=B^+B^x+ +B^x^
and since equation (1) is true for all values of ar, we may regard x
as a variable, varying in any way we choose.
Let X be a variable which approaches a limit, then each member
of (1) is a variable which approaches a limit. Hence we have in (1)
two variables which are always equal and each of which approaches
a limit, therefore their limits are equal (§641). Let the limit of x
be zero; then (3642)
\im{A^ + A^x + + ^x") =^,,
lim(J?^+^,x + +B,x^) =5,;
hence
(2) A^=B,.
Subtracting equation (2) from equation (1), we obtain
(3) A.x + Ax* + + ^„x»= J5,x + « x» + + B^x\
650
«667,668] UNDETERMINED COEFFICIENTS 651
Since x can not be zero (2636) divide both members of (3) by x,
(4) ^^+ ^^+ + A^x-'^^B^-y^B^Jr • + ^n^""'
Let X approach the limit 0, then
lim(^j+JjjX+ +^„ac»->)=^,
and lim(5j + j5,x+ + J5„x«-^) = J5^,
therefore (5) A^ = B^.
Subtracting equation (5) from (4) and dividing the resulting equation
by x, since x can not be zero, we get
(6) ^,+ V+ + ^X'-»=^,+ ^3^+ + ^n^"-'.
Similarly it follows that A^ = B^, A^ = B^, A^ = B^, ^«=^«,
which was to be proved.
667. Theorem II. — The limit, as x approaches zero, of the sum
of the first n terms of the infinite series l+x+x'+x'+
as n increases without limit, is 1.
The given series is a geometrical progression and the sum of the
first n terms is i — :r'« r •« >./.n
** 1 —a: ^
lim o _ 1 lim :r" __ 1
*^n ^ ; — . ; — ; '
n =00 l—x n =00 1 — a: l—x
lim « f\
smce . x" = 0;
n=ao
hence the sum of the series under the conditions of the hypothesis is
lim/_l_\ = i.
668. Theorem III. — The limit of the infinite series A^+ A^x
+ A^x' + A^x^ + ...., a» fAc number of terms increases without
limit and as x approaches zero, is A^.
Let iV be a positive number and equal to the greatest of the co-
efficients A in the given expression. Then A^X'\-A^a^+A^x^+ ....
<J\r(x+x«+x5+ ). By J642,
^^!^(l+x+x«+x'+ ....)= 1 + lim(x+x«+x'+ ....).
But by §667 the limit of the first member is 1, hence
1 = 1 + J'S(a=+x'+x»+....)
652 COLLEGE ALGEBRA [1669
Hence, x'^^{^+ ^ + ^ + ) = 0, [i64S]
Since the left member of the inequality is always nnmericallj less
than the right member, the left member can be made as near zoo
as we please.
Hence, J^{Ax + Bx* + Cx' + ....) = 0.
x!L°0 (^0+ A^ + ^^+ )= ^0. [»«]
669. Theorem lY.-^If the infinite series A^+A^x-{-A^'\- ....
and ^^-|-^jjr+ J5jX*+ .... are equal for all values of x for tchidi
they are convergent^ then the coefficients of like powers of x in the two
series are separately equal.
For the values of x for which the two series are always con-
vergent they are finite variables both of which approach a limit:
and by hypothesis these variables are always equal for all values
of X for which the series are convergent Hence their limits are
equal (1641) for the same values of x ; but the two series are con-
vergent when X approaches zero as its limit, since the limits of the
series, for jca^O, are respectively ^^^ and B^{i66S)\ for
(1) A^+A^x+A^ot^+ =B^+B^x+B^+
or
A+ xi"S('V+^r-* + • • •) = a+x5K='+^^+- • •)•
But the limit of the second term in the first member and also in the
second member of this equation is zero ({668); hence
(2) ^0 = ^0-
Subtracting equation (2) from (1) we get
(3) A^x+A^x*+ =B^x+B^a^+
Now divide (3) by x, which can not be zero (J686), and we get
(4) A^+A^x+A^j^+ =zB^+B^+B^^+
As before, we have two series which are convergent, equal, and each
approaching a limit when x approaches zero as a limit. But as x
JJ670, 671] UNDETERMINED COEFFICIENTS 663
approaches 0, as a limit the first member of (4) approaches the limit
Ai and the second member the limit B^,
Hence, {641,
A, = A.
Similarly it may be shown that
etc.
Development or Expansion of Functions
670. A function of x is said to be developed or expanded when it
is expressed in the form of a series the sum of whose terms in case
the series has a finite number of terms, and the limit of the sum when
the number of terms is unlimited, is equal to the given function.
671. The Method of Undetermined Coefficients is the name
given to the particular method for the expansion of functions
which makes use of the theorems described in J!2665-669.
This method is illustrated by solving the following example.
Expand ^^"7] in ascending powers of x.
Assume
(1) 1 ^ "^/ ""i ^f^ = A + Bx + Cx^+ Dx^ + Ex^ +
1 — 4a? -|- oar
where A^ B, C, A ^i etc. , are quantities which are not dependent
upon Xy and are to be determined by the theorem of undetermined
coefficients. Clearing fractions and collecting the terms in the
second member involving the powers of x, we have
(2) 2+x—S2^=A+ B
— 4A
x+ C
+ bA
x*+ D ] .r'+ E
+ 5^1 +5(7
x*+
The second member of (1) expresses the value of the fraction for
every value of x which makes the series convergent (J648).
Hence, equation (2) is satisfied when x has any value which
makes both members convergent; and by the theorem of undeter-
• The yertical line, called a bar, is sometimes used instead of a parenthesis.
Thus -\' B\x\a equivalent to {B—\A)x.
654 CX)LLEGE ALGEBRA [«672,673
mined coefficients the coefficients of like powers of x in the series
are equal.
Then A = 2, .\ A= 2
B-4Az=l', .-. ^ = 4^ + 1 =9
C_4^+5J=-3; .-. C7=4^-5^-3 =23
D-4C+bB = 0 .-. D = 4C—bB =47
£-'iI)+bC=Q .-. E = 4D — bC =73;elc.
Substituting these values in (1), we have
2 + .r-3.r* ^^^ 2 + 9 x + 23x« + 47 a^ + 73 x* +
Kora 1.— This resalt may be verified by division.
NoTs 2.— The series is an equivalent expression for the fraction onlj' for those viliies
of X for which the series is convertsent.
672. In case both terms of the fraction involve only even powers
of X, the process can be abridged by placing the fraction equal to
a series which involves only the even powers of x,
1 -8 ^ r*
Thus if the fraction was ~_ ' . we would assume it equal to
A+ Bx*-\- 6V + Dx^+ Ex^+
Similarly, if the numerator involves only odd powers of x and
the denominator only even powers of x, then assume the expression
equal to a series of odd powers of x.
673. If any term of the numerator contains x and the first term
of the denominator begins with a constant, we may assume a series
whose first term begins with the lowest powers of x in the numerator.
If every term in the denominator contains x, the series will begin
with a term in x formed by dividing the lowest power of x in tiie
numerator by the lowest power of x in the denominator, and the
exponent of x in the succeeding terms in x will increase continually
by unity or otherwise according to the powers of the terms of the
fractions.
9 -I- r
Expand "|T in ascending powers of x. Dividing 2 by 3 jt*
2
the quotient is -x~'; assume then
o
(1) ^ + ^ := Ax'*+Bx-^+ C+Dx+£x'+
Clearing fractions
2 + x = 3^ + 3^
A
x+3 6"
B
a^+ SB
C
x' + 3i?
D
1 674] UNDETERMINED COEFFICIENTS
Equating coefficients of like powers of x we get
2
3
656
3^ = 2 .
•. A =
3^+ A = \ .
. B =
3C+ J5 = 0 .
c =
32)+ C=0 ,•
. D =
3i;+ /) = 0 .
. E =
-B
3
3
— D
3
-1
27
±1
81
■ 243'
Substituting these values in (1), we have
3j* + x«""'3 "•" 9 27 "^81 243"^
2-h:r _2j-« , jr:i 1
9
NoTE.~This expansion can be verified by Long Division.
etc.
Ans.
EXEBOISE ZOV
£xpand each of the following to five terms in ascending powers of x.
1. ^ I ^^
10.
1-2j-
2 — 3x + 4x«
8j^-3j»
1-2j«
2.
11.
13.
3 + x
2 — a:-j^'
2 + .r-3j-«
i— 4xH-5j:*'
lH-J'--^+J^
^-2j^
1
1 + a.r — ajc^ — j^
a-{-bx-\- cj^
{1-xf
3.
6.
9.
12.
1-8t
1 H-5a; — 2j«*
3-.r
3j:»-6-c'*
(l-x)«'
Development of Irrational Expressions
674. Not only fractions but surds and some irrational expressions
can be developed by the method of undetermined coefficients.
Example. — Expand i/l — x in ascending powers of x.
Let
(1) V n^ = ^0+ ^^1-^ + V + ^a^'+ V* + • • •
Squaring each side we get
l-z=^o'+
k+ A^^
+2AoAi\ +2^<k1«
•^+
^+ A,*
x*+
+2A,A,
+2AtAt
+2AtAi
+2AtA,
+2AtA, +2.M.
+2AtA,
^+
656
COLLEGE ALGEBRA
U675
Equating coefficients of like powers of x, we get
A,* =1
2A^A, =-1
2V. + A' =0
2A^A^+2A,A^ =0
2A^^ + 2A,A^ + A^^ =0
A ~=^
A =
-2 .4iA
JL£!«
A =
2il^,+ 2^4^^+ 2^,^, = 0
2.4o
2A^
A _-'{2ArA,^2AM
. __--(2Ji^5+2.M4-is')
etc.
2A^^+2A,A^+2A^A^+A^^ = 0
etc.
The law of the series can be seen from these expressions. Solving
the last set of equations in order, we find the values of the unde-
termined coefficients to be the following:
Substituting these values in equation (1)
2 8 16 128 256 1024
675. The expansion of an expression irrational in form msy
result in a series of a fixed number of terms.
Example. — Expand v^l + 2 x -|- x^ into a series in ascending
powers of x.
(1) Vl + 2x + a:» = ^o + V + ^^ + -*s^+ • • • '
Put
Squaring both members of (1) we have
l + 2x+x^i4o*+ \x+ Ai*
+ 2A^Ai\ +2A^t
+ 2^0^,1 +2A^,
+ 2 ^1^1 +2^1^,
x* + .
Equating coefficients of like powers of x,
A* =1
2A^ =2
2^,^,+ 2V. =0
^' + 2J..4. + 2V,= 0
we have
A^ =
.4^ = 0; etc
•The minus sign (vlo=— 1) will give the same series on taking the radical —
plus sign (ilo = 1) does, taking the radical -}-.
,astte
25676, 677] UNDETERMINED COEFFICIENTS 657
All the coefficients succeeding A^ are zero. Hence we have
676. The development of some irrational expressions begins with
negative powers of x.
Example. — Expand ^P- — i in ascending powers of x. The ex-
pansion should begin with x'^ for when x"^ is squared we have
.«"* = -^» the lowest power of the quantity whose square root is
desired; then put
4-
L-l=A_,x-' + A^ + A,x+ ^,x» +
Note.— If one desires the expansion of the algebraic sum of several ezpretstons, it
may be found by finding the expansion of each expression separately and taking thetr
sum.
EZEB0I8E JLQVT
Expand each of the following to five terms in ascending powers of x:
1. Vl-3x. 2. Jl-2x+^
2. Jl-2x-
j^
3. l/l — X + x«. 4. Vx + X*.
5. (1-x)*. 6. Vl-Sx.
7.
^yjf^^ 8. x(l+x-.rM
9. ^1-^ 10. ^..-1+i/l-a..
11. .|H-'^+7^*H-f^+f6^*.
12. ^.^-2.r + 5 + ?-l. 13. (l-x)«-^.
14. (5 + 8x-llx»+14x'^-21x^*.
Rbversion of Series
677. To revert a convergent series y=a^-{-a^x-\-a^x^+a^^+ .
is to express a: in a convergent series of ascending powers of y.
Example 1. Revert the geometrical series
(1) y = x — jt^-\-x^ — x*+x*— ,+, ....
which is convergent when — 1 < x < 1.
658
COLLEGE ALGEBRA
im
Here ^e constant term in the series is zero; and y is zero when
X is zero; further, when y=0, one value of :e is 0, since from (1)
0 = x(l — x + jr«— , +,
.).
Hence, the series for x in the ascending powers of y begins with y.
Therefore assume
• (2) x = Ay+Bt/'+Cy^+lY+
and determine the values of A, B, C, A . . . b}' means of the theo-
rem of undetermined coefficients. Substituting in (2) the valoes
of y given in (1)
.)
X = A(x — x*+2^-^x*+
+ C(x— x«+x»— x*+
_|.2>(a._x«+x»— x*+ . . .)♦ J [ +D{x*+
Equating coefficients of like powers of x.
A =z 1
— A + B = 0
A - 25 + 67 = 0
— ^ + 35 — 3C + D = 0
.4(x — x«+a:^— x*+,-J
+5(x«-2x»+2x*+x*+,-,)
+ ax»-3x*+ )
.)
B = 1
C = 1
D = 1
Substituting the values of A, 5, (7, 2>, .... in (1),
(3) a^=y+y«+y»+y*+
which is convergent when — 1 < y < + 1.
Example 2. — Revert the series
(1) y = l-2x+3x«-4x»+,-,
The reversion of this series which has a constant term can be ma<te
to depend upon the principle used in example 1. Transposing, (1)
may be written (2) 1— y = 2x— 3x»+4x'— , + , ....
Puta = 1 — y;
then (3) z=2x— 3x2+4x»— , + , ....
Assume (4) x=^Az + Bz*+ Cz^-\-Dz*+ ....
«677] UNDETERMINED COEFFICIENTS
Substituting in (4) from (3)
659
x=^(2a:— 3x«+4a;»— , + , . . .)
-f^(2x— 3.r«+4x»— ,+, . . .)'
+ 6X2x-3x«+4a:»-, + , . . .)»
+i>(2x-3x«+4x»-, + , . . .)*
^(2x—3x«+4x»— ,+,...)
+5(4x«— 12x»+ 9x*+ 1 6x*)
+C(8x'-36x*+ . .
[ +i>(16x*+ )
.)
^ = -2-
„ 3.4 3
p . -^+3B
i.etc.
£k;[uating the coefficients of like powers of x,
2^ = 1 .'
— 3^+45 = 0
4A — 12B + SC=0
Substituting these values of ^, JB, C, . . . . in (4)
but z = 1 — y;
(5) x = |(l-y)+|(l-y)«+i(l_y)»+....
Rkmabil— It Is assamed in the discussion of example 2, that the series (3) and (4) are
convergent.
EXEBOISE XOVn
Revert each of the following series to four terms:
1. y=x+x"+x8+x*+x*+
2.
3.
4.
5.
6.
7.
y = X — 2.c« + 3x»— 4x*+ 5x«— . . .
y = X + 3x« + 5x«+ 7x* + 9x«+ . . .
V =X4-- + - + - + -+
^ ^2^3^4^5^
^ ^ II ^ 2! ^ 3! ^ 4! ^ 5! ^
y — -^ — 9! "T" 41 6! "^ 8! » "^» • ■
2!
8. y = x-^+-
^ 3! ^ 5!
- + - - +
7! ^ 9! ' ^'
9. y = a + fcx + ex* + cZx* + ex* +
CHAPTER IV
THB BinOMIAL THEOREM FOR ANY EXPONSKT
678. It has already been proved (2607, 2 ; x = 1, a = x) that the
expansion of (l-|-x)", when n is a positive integer, is
(l+x)-=l+««+ 2i^^ + n(n-lM>.-2)^^ . . .+^,
which may be written in the form
(1) (l+x)»=l+«x +Ax?+Bix^+
If n is a positive rational fraction ^ ^ we have
(1 + x)^ =V(i + x)p= V1+PX+
which on extracting the q^ root to two terms (8802) may be written
(2) (l + x)«= l4.?x+ Ac« + iBx»+
Again, suppose that n is a negative number — m, then
(1 + x)-« = (f+^ = i + „^+...'
which becomes, on applying the rule for division,
X3) (1 + x)-'"= 1 — wix + ^«+ Bq^+
We notice in the expansion (2) and (3) that the coefficient of ar in
the second term is the exponent of the binomial ; hence, whether s is
a positive or negative rational fraction or integer, we may write
(4) (l + x)"=l + «x+^x»+5x»+ . . .+ C^+ C^iX^'+ . . •
where A, B, 6% C^+i are to be determined and r 13
a positive integer.
660
8678] BINOMUL THEOREM FOR ANY EXPONENT 661
Substitutiiig - for x in (4), we get
Multiply both members of this equation by a" and we obtain
(5) (a + xY = a- + «a"-^x + -la"~«x« + ^a»-»a^ +
Let C> be the coefficient of x^ and 6^^+^ the coefficient of x'*+*, in the
second member of (4).
Then (4) and (5) may be written
(6) (l+x)»= l + nx+ + 6>'-+ 6;^ix'-+i+
(7) (a+x)"=a"+na"-^x+ . . . + 6;a»-'-x'*+ 6V+ia"-'-^x'-+*+ . . .
Now change x into z + y in (6) and obtain
(8) (l + z+y)"=l+n(z+y)+ . . .+ 6;(2 + y)'-+ 6;^jU+y)'-+i+. . .
Again, change a into l-f^^ and x into 2; in equation (7), and obtain
(9) (l+y+«)»=(l+y)«+n(l + y)»-i;j;+ + 6;(l+y)"-'-«''
Expanding the powers of 2+y in (8), we have
(10) (l+y+z)-=l + w(z+y)+....C;U'-+r2'-V+ )
+ CUiU'*^^+(''+l);5'-y+. ...)+....
Since the first members of (9) and (10) are identical, their second
members are equal for all values of z for which these series are
convergent, hence the coefficients of z"" in the two series are equal
(8671), thus
(11) 6;(l+y)"-'-=C,+ C;+,(r+l)y+ terms in y\ y\ etc.
According to (6), the expansion of (1+y)""'" for any value of n for
two terms is
(l+y)"-''=l+(«-r)y+
and 6;(l+y)»-'-=6V+6;(/i— r)i/ + terms in y\ y»,
Hence from (11)
(12) CV+6>-r)y+ = 6;+6;^i(r+l)y+
662 COLLEGE ALGEBRA [«679-681
This equation holds for all values of y for which the two series ire
convergent, and hence the coefficients of like powers of y are eqiui
That is 6;(u— r) = 6Vn(r+l)
and (13) 6Vn = C;
n — r
r + 1"
679. Rule. — If the coefidetU of any term tn the expantiom of
the binomial {a -|-- sc)** ti multiplied by the exponent of a in that
term and divided by the number of the temi^ the result is the
coefficient of the next succeeding term.
We know that the coefficient of the second term is n and the
exponent of a is tt — 1, hence according to the rule the coefficient of
^^ 3«>tennisn(n-l)-.2 = ^^^^^
4*htermis?^i^(u-2)-3 ^ n(n-l)(n-2)^
o! 4!
Therefore we have the following formula for any rational exponent h:
(14) (a+x)-=a«+fia-»x+ ^^^^j=^-V+ "^'^r-.yin^ZJ ^,.^^
The expansion extends to x"* when n is a positive integer; but in all
other cases the expansion continues indefinitely, since the factors io
the numerators of the coefficients are found by subtracting integers
from the exponent n, and hence none of them can ever be 0.
Bbmark.— It will be obeerved that all the laws of the blnomtal theorem lUtcd tn
189, VIII, excepting the first law (which refers to the numberof terms In the eximnsionl,
hold as well when the exponent to a negative number or a fraction as when It isa posi-
tUe Integer.
CONVEROENCE OF THE BiNOMIAL EXPANSION
680. It has just been pointed out that in case the exp(ment of
the binomial is fractional or negative the expansion may be extended
indefinitely, thus expressing the expansion in the form of an infinite
series. It has been remarked that an infinite series can not be nsed
in numerical calculation unless it is convergent.
681. When is the expansion
(o + x)» = o" + na»-»x + ?^i^=^a«-«x" + . . .
22682, 683]BINOMIAL THEOREM FOR ANY EXPONENT 663
convergent? Let r be a positive integer, then the (r 4- 1)*** term is
and «, = "<''~^\^L-/)"~'' + '^> a""*' * '"'
«r-n _ n — r 4- 1 ^^"^ _, fn + l
■a)-(^-');
lim/lr+i\=^ limr/n+1 i\^l= ^.
r=co\ Ur / r=x L\ r / a J a
Hence the expansion will be convergent when
-l<x<l. [J658]
The regions of convergence and divergence are shown in the fol-
lowing diagram:
—a 0 a
1 1
Divergence Convergence Divergence
688. The laws relating to the exponents and coefficients of the
terms in the expansion of the binomial (a -f x)" (2678) have been
proved to hold for any rational exponent w, under the conditions for
which the expansion is convergent. Therefore, we are now able to
apply the binomial theorem to examples in the expansion of bino-
mials which have positive or negative rational exponents.
It will often simplify the calculation to put the terms of the
binomial in parentheses, especially terms which are negative.
683. Example 1. Expand (l — |) to four terms.
On comparing ^1 — ^\ with the binomial (a + x)*,
we obtain a = 1, x = — f and n = — 3 ;
substituting in (14) 2678, we get,
(l-|)-'^X.K-3) (-!) + <- ^)^-^-^)(-|)'
^ 31 \ 3)^ ' ' '
= l + x + 6f+10g +
= l+»+|x« + g^+. . . .
664 COLLEGE ALGEBRA [J684
1
Example 2. Expand s ,: — i _o^\» ^ ^^^ terms.
,-_^ =(x-t-3yl)->
=Mr'+(-i)(--»)"*-'(-3.v»)
+ ^ 3n^ 3 \,-i)-t-'(-3^l)'
^ H)Hz^XiL)(,-,)-M (_3,i)V . . .
o
EzsBcisB xcvm
Expand each of the following to five terms:
1. (1-x)-". 2. (1 + x)-^ 3. (A-^x)".
4. (l-2x)*. 5. (3a — 2,r)*. 6. (x'-ay)"*.
7. (a«-2ax)*. 8. (5-7x«)"^ 9. (l-x)-^.
1 1 1
10. - — -' 11. r~7 7VT' 12,
Vl— 2ir ' >(l+ar)« ' "i x»-na
684. The formula for the (r+ 1)"» term of (a + a-)" [J623(l)] holds
for all values of w, since it was deduced from an expansion which
has been proved ({678) to hold in all eases.
Example. — Find the (r + 1)*** term in the expansion (1 — 2 jf'
and the 11^*" term. Comparing (1 — 2.r)~* with (a + x)", we have
tt = l, x = (— 2 .r), n = — J; substituting these values in the formuU.
(r + ly^ term = n{n-l) i^ - r + 1) ^n-rj.r [§628.(1)]
r!
r!
= ^ V2A2>> V 2 ; (_l)r2v
r!
^ (-!)'-• 7-9 (7+2r-2) gr^-r
2T!
= 70 (2»- + 5)^r
rl
«685,686J BINOMIAL THEOREM FOR ANY EXPONENT 665
Let r = 10, then r+ 1=11 and
ll"»term =^'^ ^^x'<>
10!
_ 11 -ISMS- 17 '19- 21 •23-25 ^w
"" 1 • 2 • 3 • 4 • 6 • 6 • 8 • 10
_ 5 ' 7 ' 11 ' 13 • 17 • 19 • 23 ^,o
2« * •
685. The formulae derived in J 627 for the greatest coefficient in
the expansion of the binomial {a -\- x)" hold when n is any rational
number, since the law of the formation of the expansion has been
proved to hold universally.
686. Extraction of Roots by the Binomial Theorem-
Find the value of ^V 128 approximately to five decimal places.
Vl28 = (5» + 3)* = 5(1 + |)^
= 5/lJ_l !L_1 ij__10 27 80_ 81, \
\ "^3*5' 9*5«"''6 -27' 5» 24 -81* 6""*" )
= 5 + .04 - .00032 + .000004267 — .0000000683
= 5.039G8.
BXSBOISB XCIX
Find the following:
1. 7'** term in the expansion of (a — x)'\
2. 9'** term in the expansion of (1 — 5 x)~^
3. 12'*^ term in the expansion of (1 — 2.r)"'.
4. 8'^ term in the expansion of (4 — 7 jry .
Find the (r + 1)'*^ term in the expansion of:
5. {l-x)-\
6. (l-rr)'.
7. (2a -3x)"*.
8. ' ._:.
Vl-X
9. (.t« - 2 ^<i)'.
10. {l—px)P.
666 COLLEGE ALGEBRA iieS&
1
11. (4a — 8 x)"'. 12. /x"— -5L\"» 13. ^
14. Show that only two terms in the expansion of (1 — x)^ hare
positive coefficients.
15. Find the first negative t^rrn in the expansion of ^1 + -x\
16. Prove that the coefficient of x*" in the expansion of (1 — 4 x)~" is
(2r)!
17. Find the greatest term in the expansion of:
(i) (1 — x)~" when x = ^;
(ii) (1—7 x)"'' when x = ^;
(iii) (1 — |x)~« when x = |;
(iv) C2x — 5y)~" when x = 8, y = 5;
(v) (1 — x)"" when x = ^Ty and » = f ;
(vi) In J when n is a positive integer.
18. Find the first term with a negative coefficient in the expansion
of (l + ix)V.
Find the approximate value of each of the following to five places
of decimals:
19.
l/l.OOOl,
1/0.9,
1 1.01,
20.
Vl.0004,
21.
VO.9996,
VO.98,
22.
l/905,
1/8,
23.
V65,
V7ai,
24.
1/3,
1/7.
25.
1/19,
V2,
26.
•l/5.
V244,
27. V2190.
1/0.9999.
vro2.
l/26.
j/253.
V1OO3.
l/l3.
V4.
V727.
CHAPTER V
PARTIAL FRACTIOUS
687. An algebraic fraction may sometimes be decomposed into
the sum of two or more simpler fractions; for example
3-2r _ 1 . 1
2 — Sx + j^ l—x 2 — x
It is useful to be able to separate systematically certain classes of
fractions into more simple partial fractions.
Case I. When the factors of the denominator are of the first
degree and all different ^ and the degree of the numerator is less than
the degree of the denominator.
Expand into partial fractions -, z—^ — "i^, -i where a, 6, c,
'^ '^ (x — /) (x — m) (jF — n)
/) m, n are numerical numbers and the degree of the numerator is
one less than the degree of the denominator.
Let (1) f^'r^'t; =^+-^ + _g-
{x—l){x—m){x — n) X — I x —m x — n
The problem is to determine Ay B, C, in terms of the given con-
stants a, b, Cy I, m, n so that (1) is an identity which is true what-
ever values X may have. Multiply equation (1) by (x — I) {x — m)
(x — n) ; then all we require is that the following is an identity,
a3^-\-bx-{- c = A{x — m){x — n) + B{x — r){x — n)-\- C(x — 0 (x — m)
(2) = (^ + J?+ C) aJ» — [(m + m) ^+ (w + 0 ^+(/ + wi) C'] x
— (^ mn + Bin + Clm).
Equating ooefficients of like powers of x (2666)
t A+ B-^ C= a
(^)){m+n)A + {n+T)B + {l+m)C=--h
( Amn + Bin + Clm = — c.
These three equations determine one set of values for A^ B, C,
W7
668 COLLEGE ALGEBRA li6S9
We next illustrate this metb
Separate into partial fractions :
\
We next illustrate this method by a numerical example.
ar— 2
(^-l)(j;-2)(x-3)
T J. / 1 \ gj* — 2 A I B I C
^ ^ (jr_l)(x — 2)(x-3)'":r — I'^ar — 2"''ar-3*
Clearing of fractions
(2) 3x-2= ^(x-2)(x-3)+JB(x— l)(a;-3)+ C(x— l)(x— 2)
= J(x«-5.r+6)+^(x«— 4x+3)+6'(x«— 3^+2)
hence, equating coefficients of like powers of x:
/ 6-4+ 35+ 26^= -2 (^ = J
(3)] 5^ + 45+ 36'= -3 .-. ] 5=-4
(. ^+5+c = o Ic^i.
Therefore
3x - 2 1
+
(^ — l)(x-2)U-3) 2(x-l) x — 2 • 2(x-3)
The values of A^ B, and C may be found in another manner,
which is very simple, by putting successively in equation (2)
x = l, « = 2, x = 3,
then 3 - 2 = ^ (1 — 2) (1 — 3), .-. -A = J,
6-2 = 5(2 — 1) (2 — 3), .-. 5 = — 4,
9-2 = 6^3-1) (3 — 2), .-. 0=1
NoTB l— The method Illustrated above can be applied to any fraction, the denomi-
nator of which Is the product of distinct factors of the first degree, and the numerator
of a lower degree than the denominator.
Nora 2.— The method however Is not complete because we have not shown that
equations (3) are independent and consistent ($288)' However In a treatise of this kind
simple examples only are solved and the results can be easily verified. The place
for more extended discussion of the subject will be found in integral calculus and
theory of equations.
689. Case II. When one or more of the factors of the denomi-
nator are of the second degree and the other /actors are of the first
degree.
In case the denominator of a fraction can be resolved into fac-
tors partly of the first and partly of the second degree, or all of
the second degree, in u^, and the degree of the numerator is less
than the degree of the denominator, we can separate the given frac-
tion by means of the theorem of undetermined coefficients into
partial fractions whose denominators are respectively the factors of
the denominator of the given fraction and whose numerators are
J690]
PARTIAL FRACTIONS
669
independent of x in ease of fractions corresponding to factors of
the first degree, and of the form Ax -^^ B in case of fractions cor-
responding to factors of the second degree.
Example. — Separate
Assume (1)
x^-%
(ar+2)(:i^+l)
a«-3 _ A , Bx-S-C
into partial fractions.
(x + 2)(^^ + l) ^ + 2 a^ + 1
Clearing fractions,
— 3 + x« = ^ (x« + 1) -f (5x + (7) (x + 2)
= U + 2C) + (2^ + C)x + U + B)x\
Equating coeflBcients of like powers of x^
^ + 2(7= -3,
2i? + C = 0,
^ + ^ = 1
Solving these equations we have
Substituting in (1)
A = \B=%C^-\.
3
(x+2)(^^ + 2)
5\x + 2^a^+l )
690. due III. In case the factors of the denominators of the
fractions in Cases I and II are repeated.
Separate
Assume (1)
,, , ' ^"'". — — into partial fractions.
■ B , Oc+D
1—x "^ (l-x)«"*" l-a;+a;«'
— 2 — 3j:-|-j« _ ^
Ex + F
(l-X+3^f
(i—xf (l-X + J^j*
Clearing fractions,
(2) — 2— 3x+x«=^(l— x) {l—x+x^)^+B{l-x+x?)*+
+ iCx+D) (l-x)« (l^x+a^) + {Ex-\-F) (l-x)«.
Arranging the second member of this equation according to the
ascending powers of x,
-2— 3x+x*=^— 3^
B-2B
C
D-SD
F+(E+2F)
X +bA
x> —5.1
x" +3 A
X* —A
+3B
-2B
+ B
-3C
+4C
-367
-\-c
+4/>
-3D
+ D
-2E+F
+ E
X*
670 COLLEGE ALGEBRA [«W91, 692
A+B + D + F= —2 A=-l
-^3A—2B+ C^ZD+E— 2F=— 3 ^=-4
5^+3J?-3<7+4/>-2J5;+F =1 C = - 7
— 5^— 25+4C— 3Z>+i; =0 2) = 4
3^+5— 3(7+ /> = 0 f;=— 3
—^+6^=0 F = b,
Hence, substituting the values of A^ B, C, />, E, F^ in (1), we get
-2 — 3j: + :g« ^ -7 4 7j~4 8j-5
691. The following artifice will sometimes be found useful.
Example.— Separate ^/^+,ff' + f^f into partial fractions.
Assume (1) 54:»+6x«-f 5x^_^ . J^xL
Assume u; ^^ _ ^^ ^^. ^ ^^^ ^_^ -h ^^_l_^^^
where ^ is a constant and / {x) a function (1. e. , some expresuon
involving x) to be determined.
Clearing fractions
(2) 5x»+ 6.x«+5a:= ^(x+l)*+/(a:) (j-1).
Since the identity is to hold for all values of x, we may put x=l
in (2), .-. 16 = ^(2)* = 16^ or ^ = 1.
Substituting .^ = 1 in (2), and transposing
(a;— l)/(j;) =5ar»+6x«+5x— (x+l)*=— .i;*-f x»+x— 1;
hence /(x) = — x'+ 1.
To determine the partial fractions corresponding to— rj— putx+l=x;
then I^^2-3z+3z«-,«^^ g g^, " ^Case IIi:
2 2* 2' 2*
__:^ , __3 3 L 2
•' (.,_l)(:r+l)* x-1 l + x"^(l+x)» (l + x)*"^(l+x}*'
692. In all the examples thus far discussed it has been assumed
that the degree of the numerator is less than that of the denominator.
In case the degree of the numerator is greater than the d^ree of the
denominator, divide the numerator by the denominator until the de-
gree of the numerator of the remainder is less than the d^ree of
the denominator.
J 693] PARTIAL FRACTIONS 671
Example. — Separate 7 . tT/ — -fe~ ^^^ partial fractions.
By division — ;^ — — rr- ^^ — = X — 2 + . -— — , -— ,
4-7 T — 17 , 17 11
and — '
(x+l)»(x~3) 16(x-3) • 16(J + 1) 4(j:+1)«
a4_3j3_33<4.io _i7 17 11
{x+lf(x-S) "" "^16(2-3)"''l6(x+l) 4(jr + l)«'
693. The expansion of a rational fraction in ascending powers of x
may be expedited by resolving the fraction first into partial fractions.
Example 1. Find the general term of ; — ^^ . —when expanded
in the ascending powers of x.
By {688 we have — - rtl_ = ^-^ + ^-^^ .
1 1 4
5 2 + x 6(3-a)
=fo(i-i+r-r+'---(-^)t+----)
Hence the coefficient of x*" in the expansion of the given fraction is
10^ M*- 15 "S*- 5V 2'-+* 3'^+V'
Example 2. Expand in ascending powers of x and
find the general term.
By 2689
x* — H ^ _1 , _4^-8
(x+2)(>+l) b{2+x)^b(i + x^] '
=^o-i+2:-S+'----(-^)"£+ 0
+^(x-2)(l-x«+-x*-x*+. . . +(-iyx«'-+. . .).
To find the coefficient of x**:
I. If r is even, the coefficient of x** in the second series is
— - (— l)* ; therefore in the expansion the coefficient of x*" is
II. If r is odd, the coefficient of x** in the second series is
- (— l)^~ and the required coefficient is - (— l)^~ — r-. ij;r^i '
672 COLLEGE ALGEBRA LW93
Resolve into partial fractions:
1. «^-«7 2.
(ar-9) (x-ll) • (a; ~1)« (-«:*+ 1)
3 1 + J^ . 3j« — 3jr-l-2
(l-jr)(2--rM3-.r)* ' (x - 1)« (-r« - x + 1) '
5. _ ^. ^. 6. ^
(jr-l)(x-2) • a:3-l
1+j-f .T* 9 2^-24x«4-48jr
• aJ_4a4+i- + (j' • (j:_2}*(jtr+ir
9. e£±JZ io_ ^^.-t^f
(j: -a) {X- h) {X -c) {x- l)*(a-« + 4)
11 ^ + 2j-1 -2 l-:r+j-«
• (x-3)(j:-2)«* • (x-l)«(x-2)(l+Jt«)'
Find the coefficients of ac" in the expansions of:
• (2 + z) (l--:r) • (j + 2)(x-l)»
iR 1 1/5 54-ftr
17. ^ + 3^ + 2^ ■ 18. i
(l-ar)(l+x-2j») (1 — or)(l — JatMI-")
19. Show that the coefficient of x^"'' in the expansion of
'-"> isl^
{j^-'\){x-\-2) 2«"
Decompose into partial fractions:
99 ^ 93 3:^4- 73^+ 13 J-f 3
24. Decompose ^ ^ i i\t ^^^ proper fractions whose denomi-
nators are (x* -|- x + 1)* and (ar* — x + 1)*.
25. Find the coefficient of x" in the expansion of ^ ' -3-
26. Show that the coefficient of x''+'' in the expansion of
;-[±|f. i8(-2)-(r_2;,+ l.)
27. Prove that the coefficient of x*""^ in the expansion of
J(l_.) (1_„,) (1_„.,) (l_„3,)r' i, a-ar)a-ar^^)n-ar\
i ) (1— a) (1— a*) (1— a*)
CHAPTER VI
EXPONENTIAL SERIES, LOGARITHMIC SERIES, AND
DE MOIVRE'S THEOREM
694, Expand a' in a series of ascending powers of x.
Write a'=[l-|- (a — 1)]'; expanding by the binomial theorem,
we have
[l+(a-l)]'=l + x(a-l)+ £(^(a _ 1)«
+ y_(r-l)(x-2)(.r-3)(^_l). ^
4 !
(1) .-. a'=l + x[(«-l)-l(a-l)«+|(a-l)»-l(a-l)*+ . .]
+ terms involving a:*, x*, and higher powers of a*.
Series (1) is convergent provided
— l<a-l<l. [1681]
Kquation (1) shows that «' can be expanded in a series beginning
with 1 and proceeding in ascending powers of x; assume, therefore,
that
(2) rt' = 1 + A^x + .lj.r« + .4,x» + A^jc* +
where ^1,, vl^, A^ are quantities not involving x and there-
fore do not change however x may be changed; it follows from (1)
and (2) that
(3) A, = {a^ 1) - \{a - 1)« + \{a - 1)» _ J(a-l)* + . . .
where ^4^, A^ are as yet unknown.
The condition that (1) is convergent is also a sufficient condition
for the convergence of (3), ?664. The coefficients .1^, A^^ etc., are
easily determined by the property
073
674 COLLEGE ALGEBRA [2694
where y is assumed to be independent of x and finite. Changing x
into X + y in (2), we have
a'+» = a»+' = 1 + A,{y+x) + .1, {y+xY + A^iy+x^ +
r\+A,y + Ay+Aj,^+
(4) = j + ('^ + 2.1^y + 3 jy + )x
y. + terms in liigher powers of x ;
but a''ay=ia^{l + A^x+A^x^+A^€'+ . . . .)
(5) =««'+^jrt»x + ^ja^^x'^- . . . .)
The series (5) is convei^ent in case series (2) is convergent since the
multiplier a^ is finite ; and series (4) is convergent under the same
conditions since it is equal to series (5).
Since the two expansions of a'+"'.are identically equal, then the
coefficients of like powers of x in the two series are equal, thus
(6) J, + 2.4^y + 3^y + 4.4y+ .... [ie69j
= A^ay = A^{\ + A^y +Ajf^+ A^ + ....).
These series are convergent since they are respectiveh' the coeffi-
cients of convergent series (4) and (5) ; hence the coefficients of like
powers of x are equal (3669) ; thus
2^, = A^ ; therefore A^ = '^,
3.4, = .4,^, ; therefore ^3 = "^ = %
4.4^ = A^A^ ; therefore ^1^ ='^^^^? = -^^*.
Substituting the values of ^j, .4^, . . . in (2), we have
(7) a'=\+A^. + ^ + ^+-^+ . . .
Series (7) is convergent for all finite values of x\ since we have
lira ««Jii - lira /^i^ + ii^licllI^W lira ("idlf \=0.[|«4]
n=« «« 7i-'oo\ n\ (w — 1)! / t.'=oo\ n /
Since expansion (7) is true for all values of a:, take x so thtt
AyT =1 ; then ^ = y and
\_
this series is usually denoted by c; hence a 1 =e, therefore a = <
{695] EXPONENTIAL AND LOGARITHMIC SERIES 675
and xlj =: log^a, hence from (7)
(9) a- = 1 + (log.a)a, + <i2^f)^ + Uo^f>^+
This result is called the exponential series. Put « = c in (9), and
we get an Important particular case of the exponential series, thus
(10) e'=l + l+f,+g + g+
696. Expand log ^ (1 + cc) in a series of ascending powers of x.
In {694, (8) we had A^ = log « a which, substituted in equation (3), gives
log.a = (a-l)-|(a-l)« + l(«-l)»-l(a-l)*+
For a put \-tx\ hence
(11) log^(l+x) = x-| + ^''-^+
which is convergent when — 1 <a; ^ 1. Hence, if x is unity or a
proper fraction, (11) may be used to calculate the value of log^ (1-p x) ;
but in case x is not very small the terms in the series diminish
so slowly that it would be necessary to retain such a large number of
them in order to secure perfect accuracy in any given decimal digit
as to make the formula an impractical one for calculation. If x is
greater than 1 the formula can not be used because then the series
would be divergent. Hence an expansion which converges more
rapidly is derived.
Change x into — a; in expansion (11) ; thus
(12) log.(l-a!) = -a:-|-J-^-
which is convergent for — 1 ^ a; < 1.
Subtract (12) from (11) and oijtain log^(l + x) — log ^ (1 — x) =
log. l^-^ 2 (x + f+f +....)
ICoTE.— It has been assumed, in finding the difference of the second members,
that we were dealing with sums. This is not true; we have simply the limit of bums.
This step requires that the following theorem Xhq proved: If a aeries is absoUitely
convergent (p. 648, caution and theorem) the terms of the series may be arranged as one
chooses uithout altering the value of the series.
Substitute in the preceding equation
, ^^ = ^^t_- ; i.e., x = - — — - , and we obtain
1— j: n' 2?i + r
log,5±i = log>+l)-log,„=2[-l^^-+3-^,+g^+ . . .]
or (13) log>+l) = log.n+2[^^l^- + 3-^, + g^.+ ....]
which is convergent for all values of n greater than 0 (Ex.XCIV,17).
678
COLLEGE ALGEBRA
£2696
The Table op Napierian Logarithms
696. The log^ is found by putting n=l in expansion (13); thus
^' V3^3 3»^r> 3»^7 37^" 'V
(J) = .33333333 ... .-. (^) = .33333333 .
(1)' = .03703704 ...... J (J)' = .01234568 .
(])' = .00411523 ... .-. i (J)' = .00082305 .
(lY = .00045725 ... .-. \ (jy = .00006532 .
(!)• = .00005081 ... .-. i (0* = .00000565 .
(1)"= .00000565 ... .-. TJr(ir= 00000051 .
(;)"= ,00000063 ... .-. -^jQ)"= .00000005 .
(^)"= .00000007 ... .-. ^ty(iy*= .00000000 .
.34657359
Though the term- - (-\ has no effect on the eighth decimal place,
this does not justify the stopping of the calculation at this point
It must be shown that the remainder of the series from this point
on can not influence this place either. The remainder of the series is
1 1,1 1,1 1
15
JL+1
JL+1.
3" ^19
+
lo 3»^^ ^17 3«^1
319
15 I
19 * 3*
+ .
.)
The value of the series in the parenthesis can not be easily calcu-
lated ; but this is not necessary, because this series is evidently less
than the G. P.
1 ^9
8
^3»^3*
1-i
and therefore the remainder of the series is less than
which does not affect the seventh place. Hence we obtain for log, 2 the
value
2 X .346 573 5 (9?) = .693 147 1 (8?)
or to seven places,
log, 2= .693 147 2.
2697] EXPONENTIAL AND LOGARITHMIC SERIES 677
Logg 3 is found by putting n = 2 in expansion (13); then
log.3=log.2+2(i + ^ + ^, + ^+...)
= .693 147 2 + .405 465 2 = 1.098 612 4 . . .
Proceeding in this way the logarithm of any number to the base e
can be found to any desired degree of approximation.
Similarly we obtain
log,li= .223 143(4?)
log.,5=1.609 437(8?);
and
log. 10 = log, (2 • 5) = log,2 + log, 5 = 2.302585092
log,« m = i^£i^ = ^ log.m
^** logao 2.302585092 ^*
= (0.434294481 )log,m;
thus a table of common logarithms can be made from the table of
Napierian logarithms by multiplying each logarithm by the multi-
plier, M = .434294481 ; this multiplier is called the nwduius
of the common system.
NoTB.~0n account of tbe great importance of the numbers^, JT, and tbe logarithms
to the base 0 of 2. 3, and 5, in numerical calculation, they have been calculated to more
than 280 places of decimals. (Proceedings of the Royal Society of London, Vol. XXVII,
pagesa)
697. The Number Called e in Mathematics is Incommen-
surable.— Let us suppose, for a moment, that e is equal to the com-
mensurable number — , where m and n are integers. Hence we will
have
^n!U + i^(/i+l)(n + 2)^(n + l)(n + 2)(u+3)^ J
Multiply both members of this assumed equality by n!. The first
member will become the integer m-l-2*3 n — 1. In the
second member beginning with 1 and ending with "|, the fractions
are transformed into integers whose sum we shall represent by S^^ ,
and the multiplier of the bracket becomes 1. Therefore we have
(15)
m-(n-l)I = .Y+ r ^ + 1 + 1 +. . .1
678
COLLEGE ALGEBRA
[12698, 699
Excepting the first, the terms of the bracket in (15), are respecdvelj
less than the terms of the progression
(16)
n
Hence the bracket in (15) lies between ,-_ and - and is there-
n-|- 1 n
fore not an integer. It follows, therefore, from equation (15) that
the parenthesis, which is not an integer, is equal to the first member,
an integer, minus N an integer, i. e. , a number not an integer equal
to an integer, which is impossible. But the series is convergent
that is, the sum has a limit, and since this limit can not be a com-
mensurable number it is necessarily an incommensurable number,
which mathematicians have denoted by r.
698. The Remainder of the e Series.
Let R^^y represent the remainder of the e series, 1694, (8), after
the first n terms, then we have
1 . 1 I 1
(17) R^
nlln+
+
+
»+^ nlbi+l ' (n+l)(n+2) ' {n+l) {n+2) (n+3)
It has just been shown, 2697, (16), that the parenthesis is less
than - , therefore
R.
<-i L.
699. The Calculation of a. — It follows from the law of the
formation of the expression of e that
the sum of the first three terms is
4**» term z= J of ^ . . . . . .
b^^ term = J of the 4"* term
6'^ term = ^ of the 5*** term
7^ term = J of the 6*** term
8'** term — \ of the 7*** term
9*»» term = ^ of the 8"* term
10'^ term = | of the 9"^ term
ir»» term = j\ of the lO*** term
12'*» term = ^\ of the 11**^ term
13'** term = ^^ of the 12*^ term
14*^ term = j\ of the 13'»* term
2.500 000 000 0
0.166 666 666 7
0.041 666 666 7
0.008 333 333 3
0.001 388 888 9
0.000 198 412 7
0.000 024 801 6
0.000 002 755 7
0.000 000 275 6
0.000 000 025 1
0.000 000 002 i
0.000 000 000 2
2.718 281 828 6
«700, 701] EXPONENTIAL AND LOGARITHMIC SERIES 679
The calculations of this table are subject to two errors.
The first error is due to the omission of all the terms of the
expansion of e after the 14th term; hence the error due to this fact
is (2698)
1. e.,
1* "^ la isf I
■ 13 13! 80,951,270,400
^u<rTJro = 5 of .000 000 000 1.
'8 10" 8
The second error is due to adding in the nine cases of the final
dotted figures less than half a unij;, or in all less than 4^. The error
made, therefore, in the last digit 6 due to both errors can not sur-
pass 4 units of the same order. Hence the value of e correct to the
ninth decimal place is
e= 2.718 281 828 ... .
The value of c to twenty decimal places is
e = 2.718 281 828 459 045 235 36.
700. The Error in the e Series. ~The error made in taking the
sum of the first (u -f 1) terms of the series
"^ l'^2!'^3!"^4!'^ • • ■ ' '^n!'^(n+l)l'^ ....
as the sum of this series is
"-^»^ n! n + l — x
The proof of this result is similar to that given in (§698).
Special Exponential Forms
701. It is desirable in this connection to derive the following
exponential forms:
(i) lim(l + 0' =lim(l-0"^=c
1^0 f =0
(ii) lim (1 + I)' = lim (1 — /)~^ = e'.
1^0 IdsO
680 COLLEGE ALGEBRA [2701
It follows from the binomial formula, J 679, that
(1) (l + 0"'=l + 7>i/+^^^"^-^^^' + ^<^-^H'"^^>/»+ . .,,
(2) (l-Z)— = 1 + ^aj^m{m-^l)p^m{m+l){m-j-2)p ^ .
or
(3) (l + O-l + nW+f (!-!) + f(l-i)(l- 1)+....
(4) (l_0-".l + n.Z+^(l + i)+f (l + i)(l + |) +.
If now m is made to increase without limit and at the same time /
is made to diminish indefinitely, but in such a manner that the product
ml approaches a fixed limit x, the various terms of the second
members of (3) and (4) will approach respectively the terms of
the series ^ , .-2 r»
1 + X + ^ + 1?- +
2! 3!
which is convergent for all values of x (J664). It follows, there-
fore, from (3) and (4), remembering that
(5) lim (ml) = .r,
that we have
(6) /im (l+0"'=/i^„(l-/)-" = l+f, + g + g+. . . .;
and further, in order that (6) might be deduced by means of (5),
it is not necessary' that m, as it is made to increase indefinitely, be
restricted to taking integral values. For if we take a positive varia-
ble, w, which increases without limit while I diminishes indefinitely,
but always so that we have
.(7) „"?;("0=-r,
and m is the number next less than m; then, since u is inclosed
between m and m-\-l, the ratio ^ » which is comprised between 1 and
1_|_ will have the limit 1 as m t^oo . Since
m
(1 + 0"= [(1 + i)-f an<l [(1-/)-"] = [(1-0-f
we therefore have
llm (ii/)u^llm (! + /)'» and »»" (i_Z)-«=iim (l-_?)-«
iiO /=iO /-O f-0
Therefore we still have
(8) uT. ^^ + ^r = .T^ (1 -^""=i + «' + i+i+
2702] EXPONENTIAL AND LOGARITHMIC SERIES 681
Equation (7) will be satisfied, if one supposes
since we will always have ul = x. Hence equation (8) gives
(«) /ro (1 + '>' = iTo (1 - ^>"' = 1 + ^ + 2! +3i +
Putting a; = 1 in (9) we get
(10) ^lim (1 + /)! ^ nja^ (1 _ jyl ^ e.
Equation (9) gives
(11) /Jf^ (1 + 0^ = ,>l?o (^ - ^"' = ^»
the two results which we desired to establish.
• De Moivre's Theorem and Certain Series which may be
deduced from it
702. DeMoivre's Theorem, (cos x + i sin jt)" =cos nx + i sin nx,
is proved by the inductive method. First is proved the following
Lemma:
(cos x^ -\- I sin x^ (cos x^ + i sin x^ (cos x„ + i sin x„)
=cos (^i+Xj+ .... +a;J+i sin {x^+x^+ .... +x„)
where n is a positive integer and i = V —I.
Proof, — It follows from the product of two imaginary factors
and the addition theorems in trigonometry that
(cos x^-\-i sin x^) (cos x^+i sin x^
= cos Xj COS x^ — sin x^ sin x^-^i (sin x^ cos uc^+cos x^ sin x^)
= cos (Xj+iCjj)-f I'sin (iCj+Xjj).
Similarly, [cos {j'^+x^)']-i sin (fj+J^g)] (cos ^3+/ sin x^)
= cos {x^+x^+x^)+i' sin {x^+x^+x^),
Carrj'ing out the second and third steps in the inductive method
of proof we have
(a) (cos Xj+i sin x^) (cos x^-\-i sin x^) . . , (cos x„+i sin xj
= cos {x^+x^+ . . . +x„)+t sin {x^+x^+ .... +x„).
Put in (a) Xj=arj= . . . =x„=:x, then we have for positive integral
values of w,
(b) (cos x+i sin x)"=cos ux+i sin //x.
682 COLLEGE ALGEBRA C«7^
It follows from (b) that
\
cos x-^t sin X ="j/co8 nx-\-i sin nx ;
put nx = y or X =1^. and we have
cos ? -Lt sin ? = "v^cos v+ i sin y .
Substitute back x for ^ and we have the equation
J.
(c) "l cos x4-i sin .r — (cos x+i sin x)"= cos - + i* sin - •
' n n
Hence one of the n^^ roots of cos x+i sin x has been found. It will
l>e shown that there are n roots of this binomial. We have from
(c) and (b)
(d) (cos x+ i sin x)"= (cos - +i'sin - ) =cos — + t sin "^'
\ n n/ n n
Hence De Moivre's theorem holds also for fractional exponents.
Suppose that n is a negative integer or fraction, then by i309 and (d)
(cosx+isin j')""= —
(cos x-\-i sin or)" cos nx -\- % sin nx
C08 nx— t sin nx __ coe (—nx) -j- i Bin (— n.r)
(cu8 nx-\- i sin »u) (cos nx—i sin r?j*) cos* ru-j-sin* nx
= cos ( — nx) +1* sin ( — nx),
which was to be proved.
703. The binomial theorem, proved for any rational exponent,
gives us
(cosx + isinap)''=co8"jc+ '?cos''"*x(isinx) + -^"^^cos''""x(i8inx)'
+ ni^zilL(n-2) co3"-»ir (/sinx)' +
or
= cos" x+ i- cos""* X sin x — ^^""~^ ^ cos""* x sin* x
1 2!
_^.n(/i-l)(7j-:2)^^„-3^gin,^ .
i. e.,
(cos X + /sin x)"=: /"cos" x — ^*^"~^^ cos""* x sin* x
+ ^* ^"-^) ^"--^ ^"-''^^cos^-^xsin^x -- )
+ i7^ cos""»x sin X - !L(!i=ilH^rr2) cos"-»x8in« x
\1 3!
+ n(»-l) (»-2U»-.S) (»-'^)cos"-^xsin'^x —...).
8704] EXPONENTIAL AND LOGARITHMIC SERIES 683
But we also have
(d) (coax + I sin a;)* = cos nx + i sin nx, [ J702]
Hence the second members of the two preceding equations are
identical. Therefore the real parts, so also the imaginary parts, are
respectively equal (2394, II Cor.) Therefore, for rational values of w,
(e) cos nx = cos" x — tii^zd} cos""* x sin* x
21
+ ^ (^~^) (^-2) (^-'^) cos"-* X sin* X ~
4!
(f) sin nx = - cos""* x sin x — n(yi— 1) (n— 2) ^.^gn-s 3. gj^s ^
^ n(n-l)(»-2)(«-3)fn-4) ^n-, ^ gi^. ^ _
5!
704. In the figure
AB = sin X,
and
^'6'=tanx.
From Geometry
AR<AC<A'C,
^ or
B C
AB ^AB
0^ ^ sin jr
v^^^^ j\n ^ yix> ^^1 _„ ^yo ^. Bill -K ^ -x
hence _<_<l or --^<_-<l,
where x is measured in radians. Let x = 0, i. e., the point A
approach C^ then the lim OB^OC=\, Therefore
x^O ("i^) = ^ ^^ ®^° x=x+ex where e =0.
When X is infinitesimal, J 637, then the arc can be substituted for the
sine of the arc.
cos x=v/l— sin* x=:(l— sin* x)*,
therefore, cos" x=(l — sin* x)*.
Hence by the binomial formula for any rational value of n , since
sin* X is always less than 1, we have
n(n i\ nin ,\ (n n\
COS" x= l-^ sin* x+lAjLziJ/sin* x^^?"^^ ^y-^W« x+. . .
2 2! 3!
and consequently
cos"
n 2 n 2! n 3! n
684 COLLEGE ALGEBRA [1705
If n = 00 , then we can substitute - for sin - , etc. , and obtain
cos- ^-1 n.^ ■ ;(g-l).:t^ ;(n^l)(g^2).^ ■
^ n"^ 2 n«^ 2! n* 3! n*^* ' • *
a n "^ 2! n« 3! n* "^ " " * '
This series converges very rapidly for large values of n and small
values of ^•
n
If n = 00 , then all the terms to the right of unity disappear and we
have,
cos"- = 1.
n
705. Ih view of the preceding discussion, put in (e) nx = y, or
X = y., and it is transformed into
n
cos y = cos" ? — " ^^—y cos"-« l sin« H
n 2! n n
+ n (n-1) (n-2) (n-S) ^gH-4 y gin* ? __
4 ! n n
This formula is simplified very much in case n-^ co ] then we have
cos" Y = 1 ;
n
similarly
cos"-«L^= ?^=1=1;
^* cos' I 1
also
cos"-'?' =1, etc.:
n
and further the arc may be substituted for the sine. Then it follows
from the preceding equation that
cos V = 1 — ^ ^""^) 2^ -4- n (n-l) (n-2) (n-S) y*
. 2! n« ^ 4! n*
_ n (n-l) (n-2) (n-S) (n-4) (n-b) ?/• ,
l.(l_l) (1^1) (i_.2) (i__3)
2\ ^ ^ 4! ^
(l_t)(l_2)(l_3)(l_i)(l_5)
6!
8706] EXPONENTIAL AND LOGARITHMIC SERIES 685
12 3
Since - » - » - » etc. , for n= co , have the limit zero and one can
n n n
Tvrite X instead of y, we have
(g) cosx=i-g+|;-g+....
Similarly we can derive from (f)
(h) sin aj = X — ^ + ^* — , + . . . .
The series (g) and (h) are convergent for all values of x, but the
smaller x is the more rapidly they converge (Ex. XCIV, 22, 21).
We notice that the series for the cos x contains only the even
powers of x and the series for the sin x/jontains only the odd
powers of x. These two facts are compatible with the properties of
the sine and cosine, namely, that cos ( — x) = cos x, and the
sin (— j-) = — sin x. These series may be used to calculate the values
of the sines and cosines of angles.
706. It follows from (g) and (h) that
cos x-l-t sin x= 14- — — •?? — H?4-Ef 4. 1^ _ _ . ,
=1-1-^ + ^^-"^ + ii-^)' + ^^+ [1387]
^11^ 2! ^ 3! ~ 4! ^ L»wj
Moreover, according to the formula for (^1 + -\ for positive integral
values of n which are infinitely great,
(1 +»;)"= l + |2+lii*+f-' + . . . . [2701, (6)]
Since the second members of these equations are the same we have
cos x+t sin X = (l-\- — \ =e*^-
Since t-^-. — = cos x — i sin x, it follows also that
C08 jc-\-i 8in X
cos X — t sin X = 7 = e"**
cos x + i* sin x=e*^
COS X = — ^
2
Sin X =
2%
686 CX)LLEGE ALGEBRA [«707, 708
707. The Value of ^ Calculated by means of Gregory's Series*.
We have from J 706 cos x + i sin x = e**.
log (cos x-fi sin x) = tx — ^<^x < sr
hence log [cos x(l+e tan x)] = tx
or log cos x-|-log (l + i tan x) = ix
i.e., logco8x+(^tan*x-— Jtan*x+, — . . .)
+ /(tan x-i tan»x + ^ tan*x — , + . . . ) = i x [J695, (1 1 ) J
when —1 =tan ar ^ 1,
or _^=^=^.
Since cos x is +, log cos x is real, and we have [2394, II Cor. J
log cos X = — ^ tan' x — | tan* x -f . . . •
(1) X =K tan X— J tan'x +itan* X — , + . . .
Series (1) is called Gregory's series, and holds if
— 1 ^ tanx = 1.
The value of ^ can now be found to any desired degree of ap-
proximation by means of series (1). Then put x = -^ in (1) and we
l^ave ;r_ 11 1
^"^ 4-^"3 + 5~7+
which, however, converges much too slowly to be of any practical
use in the calculation of rr,
708. Euler's Series.— Use the identity j = tan-» I + tan-» \ and
1 1
substitute the values of tan"* - and tan"* - found by means of series
^ o
(1); we then get
<')i-Ma)'+Kr--+i-i(i)'+sa)'---
which converges quite rapidly. Euler gave the series
«) J-,«{'+i(r^)+l-5Q"+ }
A. _75S4_ 1^,2 / 144 \ . 2jJ / 144 y , 1
^loooool "^auooooo/^a •sViooooo/ ^* * */'
which he deduced from the identity
n = 20 tan-*^ + 8tan-*|-.
Series (4) converges much more rapidly than series (3).
NoTB 1.— The value of rr has been calculated to many decimal places by various
mathematicians; but W. Shanks has carried the calculation to the most decimal places,
707. (See Proc. Royal Soc. ; Vols. XXI. XXII.)
NoTB 2.— It can ^ proved as in the case of e, that n is also incommensurable.
* This problem is also known as the (quadrature of the circle.
2708] EXPONENTIAL AND LOGARITHMIC SERIES 687
SXZBCISE CI
Verify the relations:
1. ^^^z£l=x+^+^+
2 ' 2! 4! 61
3. l_l = a;-^+^-
e' 21 31
V ^11^21^3!^ A 1!^2! 31^ /
5 g^ I I 1 + 2 I 1+2 + 3 .14-2 + 3 + 4 .
6. ^"' = r^ + . o^o .+ • • • + ^
l-3'l-2-3-5' ' l-2-3...(2n-l)(2?i+l)
7. ^=1 + 2_^J_^4^
2 ^3!^ 5! ^7!^
■ *-l .1 + A+1+....
2! 41 8!
10. 5e=i+|;+|;+i;+ —
1+1 + 1+1+ ....
• e«+l 1+L + 1 + 1+
12.
1 i: , 1 a:« 1 x* , 1 r«
■ 137^=^+2 11 + 6 2!~36 41+45 6!- • • ' f^""""!
13. Log [(1 + x)'« (1 - x)'-] = 2 (-^- + g2l^ +-^ + . . . ).
14. The limit of ^1 +-\i for »i =00 is f i.
15. The coefficient of x" in tlie expansion
^ II ^ 21 ^ 3! ^ • • • nl
16. Log.(l + 3x+2x«) = 3x-5^ + |?-l^+.. .
+ (-l)"-«?^x"+ ....
688 COLLEGE ALGEBRA [1708
17. 21og.x-log.(x+l)-log.U-l)=l^ + A.4.A.+ ...,
a;>l.
1 Q e^ + e-' , , 1 2«ji» 1 2*r« . 1 2«^
19. FincUUevalueofi-2-l- + -L,_-J^ + ^-
20. Prove that the coefficient x" in the expansion jlog^ (1 + -^ !'-
21. Draw the curve y = log^ x, discussing the grapli for the inUr-
vals — cx) <J^ <0, 0<ic^l, l<.T^oo.
22. Draw the curve y = «'.
23. Derive from formulae (e) and (f), J703, the values of cos 3 w,
cos 4 X, cos 5 X, . . . also the values of sin 3 x, sin4x, sin5 x, . .
24. Calculate the sine and cosine of an angle whose arc is ^i^ of a
circumference of radius 1, using five terms of the expansions
(g) and (h), §705.
25. Prove that
g«fx _ COS x + i sin x , e^ca? _ 14-»tanr
cos X — i sin X 1 — i sin x
26. Given tan ^ = tan 45° = 1, show that
4
1-*
27. Show also that
2*f-log,|±i and ;;=.J-.log,L±i.
4 1 — t 42i l — %
28. By putting x = sin"^ i/ in Gregory's series prove that
X 1 .r3 , 1 .?*
sin"*x :
(l-x2)i -^(l-a^jg 5(i_j4J^J
jJ}r+l
2r+ 1 (l^^^l(2r+l)
^2*3 2-46^"^ 2-4-6...2r 2r+l"^ '
I 2 V 2
CHAPTER VII
THE SUMMATION OF SERISS
Recurring Series
709. Definition.— The series n^-\- y^-]-n^-]- .... is called a
recurring series, if from and after some fixed term each term is
equal to the sum of a fixed number of the preceding terms multi-
plied respectively by certain constants. These constants are the
same, whatever terms are considered.
Examples. — The G. P. ^r -|- ar + nr* -|- . . . is a simple example
of recurring series; for each term after the first is r times the term
which precedes it. If ?<„_, and u^ are respectively the {n — 1)**» and
the {iiY^ term, then ?/„ = ru^^^ or ?/„ — ru„_^ = 0; the algebraic sum
of the coeflScients of w„ and m„_j, i. e., 1 — r, is called the scah of
relation.
710. Scale of Relation. — In the series
l + 3x+ llx« + 43x» + 171x*+ . . .
each teon after the second is equal to the sum of the two preceding
terms multiplied respectively by the constants 5x and — 4x'.
Thus llx« = (5x) 3x — (4x2) j
that is, (1) «j = 5x^2 — 4x*//j | ( u^ — bxit^ + 4x'Mj = 0.
(2) n^ = SxMg — 4x*/fj [ \u^ — 5xWj + 4x^u^ = 0.
In general, when 7i is greater than 2, each term is connected with
the two terms which immediately precede it by the equation
"n = 5x//„_i — 4xhi„_^
or (3) u„ — 5x?f„_, + 4x2m„_2 = 0.
In equation (3), the coefficients of w„, w„_„ u„_^y taken with their
proper signs, form the scale of relation.
690 COLLEGE ALGEBRA LSJ711-713
Thus the series
l+3x+llx« + 43x»+171x«+ . . .
is a recurring series in which the scale of relation is
1 _ 5x + 4x*.
711. To Find any Term when the Scale of Relation is Given.
If the scale of relation of a recurring series is given, then any tenn
of the series can be found provided a sufficient number of terms is
given. Suppose that
1 — px — qx^ — rx?
is the scale of relation of the series
then, a^x^ = px -a^^j.c""* + 5x**a„_2x"~* + rx'-rt„_3ic"~',
that is, any coefficient can be found when the coefficients of the
three preceding terms are known.
712. Order of a Recurring Series. — A recurring series is said
to be of the Jirst^ second^ or r^^ order ^ according as each term depends
respectively upon one^ two, or r preceding terms.
713. To Find the Scale of Relation.
Example. —Required the scale of relation of the series
(1) 4x + 14x» + 40x» + llOx* + 304x« + 854x« + 2440x^ + . . .
This series is evidently not of the first order since the series is not a
G. P.
Suppose that series (1) is of the second order and let the scale be
1 — px — q:x^ ;
we should then have the following equations:
40 = 14p -f 4^ ^ These relations give
110 = 40i>+14(^j p = ^. ?=-|
But the fifth coefficient 304 is not equal to
^^.110-;^. 40 = ^ = 300,
6 o o
as it would be if the series was of the second order; hence our
assumption is false.
S714J g^HE SUMMATION OF SERIES 691
Suppose next that the series (1) is of the third order and let the
scale be
1 — px — jx* — rx^\
then the equations for determining p^ q, and r are :
110 = 40/> + 14^ + 4r
304= 110/>+405+ 14r
854 = 304/) + llOg + 40r
These relations give
jp=6, 5= — 11, r=6.
According to these values for p^ q, r, the seventh term should be
(6-854 — 11 -304 + 6-110)x^ = 2440x^
which is correct Hence the scale of relation is
l_6a; + llx«— 6x«.
Thus if the series is of the third order, six terms at least must be
given in order to determine the scale of relation.
In general, to find a scale of relation of a series of the m*** order,
we must know at least 2m consecutive terms.
714. As in the preceding example the order of the recurring
series may be determined by trial by assuming the series to be first
of the second order, then of the third order, and so on, until the
true order is found. In the footnote is given LaGrange*s general
method for finding the order of a recurring series. *
Example. — To test whether the series 1, 3, 6, 10, 15, 21, 28,
36, 45, . . . is recurring or not Introduce x, and write:
5^=l+3x+6x«+10aj»+15x*+21a:*+28x«+36x^+45x«+ . . .
•Let the series be
flTsr.l + Jte-HCte'-h/^aj'-f, etc
Divide unity hj 8m far as two terms of the qnotlent. which will be of the form a-\-bx,
and write the remainder in the form 8's^, S' being another infinite series of the same
form as /&
Next divide i9lqr fi^ as far as two terms of the quotient and write the remainder in
the form of 5"a**
Again, divide 8' by 6^', and proceed as before, and repeat the process until there
is no remainder after one of the divisions.^ The series will then be proved to be a
recurring series, and the order of the series wiU be the same as the number of divisions
which have been effected in the process.
692 COLLEGE Ai^GEBRA [5715
Then we have v ~ ^ — 3 x + . . . .
with a remainder
3x« + 8a^»+ 15x*+ 24x»+ 35u-« + 48x' + 63 x« + . . .;
hence ^' = 3 + 8 x + 15 a:« + 24 x» + 35 u-* + . . .
^ = - + - with a remainder
^(x»+3x»+6x*4-10a^+. . . .).
Therefore we may take
^'" = 1 + 3 X + 6 x« + 10 x' + . . .
Finally 4^7 = ^ — x, without any remainder.
Thus it follows that the series is a recurring series of the third
order. It is the expansion of
l—Sx + Sx'^-j^
715. To Find the Sum of n Terms of a Recurring Series.
Let the series be u^^n^x--\-u^j^'{- .... and the scale of relation be
1— ^x— jx*, so that for every value of n greater than 1
Represent the sum of the first n terms of the series by S^, then
S„ = u^ + u^x+n^a^+ . . . +w„_,x"-*
qx^S^= %qx^+n^qx^+ . . . + w„_3^^""*+"«-2?^'*+«*m-i?a:"-^^
hence, on subtracting the second and third equations from the first,
we have
aS'„— J>X^'„— yx2iS'^=M^+WjX— M^i>x— i/„_i;>x"— ?/„^2yx"— M^_i^x-+»,
since all the other terms of the right members disappear by virtue
of the relation (1) which holds for any three consecutive terms of
the given series; therefore
" 1 — j)X — qx*
If the given series is an infinite convergent series, the expression
88716,717] THE SUMMATION OF SERIES 693
approaches the limit zero as n increases indefinitely (3660), and
Iience for an infinite converging series we have
716. If the recurring series is w^-f w^-f-iij^-f . . . . , and the
scale of relation 1 — p — q, it is only necessary to make ic=l in the
results in the preceding article, in order to find the sum of n terms,
or of an infinite number of terms.
The expression
^o + a;(t/^ — j>u^
1 — px — jx*
is the sum of the series
«o+^i^+^«^+ • • • •
only when the series is convergent
If we develop the fraction
1 — px — qx^
according to ascending powers of x (2670) we shall find that the
expansion will be the series
whether this series is convergent or divergent. Accordingly, the
fraction
1 — px — qa^
is called the generating function of the series
^0 + "i^ +u^x^+ , . , ,
But the generating function of a recurring series is not the sum of
the series unless the series is convergent.
717. General Term of the Recurring Series.— If 1 — px — gx«
can be resolved into two real factors of the first degree in x, the
expression -^ ' f- can be decomposed into partial fractions,
1 — px — qx^
each having for its denominator an expression containing only the
first power of x (?688). In this case each partial fraction can be
developed into a geometrical progression whose general term is
readily found, and the algebraic sum of these general terms will be
the general terra of the recurring series.
694 COLLEGE ALGEBRA [WIS
Example. — Find the generating function, the general term, and
the sam of the first n terms of the recurring series
(1) l_7x — aj«-43j:*— . . .
Let the scale of relation be 1 — px — qjc^\ then
7p — y — 1 = 0, i> + 7<? — 43 = 0;
whence p = 1, j = 6 ; and the scale of relation is 1 — x — 6x?.
Hence the generating function is
1 — X — 6x*
Separating the generating function into partial fractions,
l__.r — 6x« 1 — 3x"'"l + 2x'
Expanding these fractions we haye respectively
(3) j-5ij=-l[l + (3x)+(3x)« + (3x)»+ . .. +(3x)'+...];
(4) rhr-= 2 [l + (-2x)+(-2x)«+(-2x)»+. . . +(-2xr+ . . .]•
Hence the (r+l)**» terms of the expansions (3) and (4) arerespectiyely
(— l)3'^x' 'and 2 ( — l)'2'"x'';
therefore the (r+l)**» term of the expansion of the given series (1) is
[(-l)'•2'•+'-3'•]a•^
If we add the sums of the first n terms of the series (3) and (4) we
shall obtain the sum of the first u terms of the given series, namely,
1 _ 3"x" ( — l)*2"+^x» — 2
3x— 1 "^ — 2x — 1
718. The student can readily show that if the series
Wa+Wi^+ w^+ • ■ • •
is a recurring series of the third order, whose scale of relation ia
1 — px — 5^ — rx*,
and is convei^ent, the sum of the first n terms is
^_ a^ + (n^ — pajx -\- (a, — pa^ — qa^a^
1 — px — qj^ — rx*
5719] THE SUMMATION OF SERIES 695
EXBBOISE on
Note.— In tho following exercises such values only of x are considered for which
the series are convergent.
Find the generating function of the following recurring functions :
1. 2 - 5 + 29 - 89 + ... .
2. 2 + 5 .r + 10 x« + 17 x» + 26 .c* + 37 u-5 + . . . .
3. 1 + X+ 2.1:2+ 7.r»+14u-* + 35.r»+ ... .
4. l+x + 2.r«+2.r'+3x* + 3a:5+4.r«+4.r7+ ....
5. 4.r+ 14.r« + 40x» + 110.c*+ 304x»— 854x«+ ....
6. 1*+2*j: + 3*x«+4*x»+5*jj*+ . . . .
7. Find the sum of the infinite series
l-|-4.r+ 11 x«+26jr»+57.r*+120u:*+ . . . when ;r < J.
8. Find the n^^ term and the sum of the first n terms of the
recurring series:
(i). 3 + 11 + 32 + 84 + . . .
• (ii). 1 — 2.C + Ox^ — 4x' — 4.r* — 12x* — 20x« — . . .
(iii). 1 + 5 + 17 + 53 + 161 + 485 + . . .
(iv). 1 +2x + 3x«+4x' + 5x*+ . . .
(v). l + 3x + 6x^+ 10.c» + 15x* + . . .
(\i). 4j: + 14.T:* + 40x5 + llOx* + 304x* — 854x« + . . .
(\ii). 1 + 3x + 6.r2 + 10x«+15x* + 21x*+28x«+ . . .
(viii). 2 — X + 2.r« — 5x3 ^ jqx* — 17x'^ + . . .
9. Show that the series P + 2''x + 3''x« + 4'-.c' + .... is the
expansion of an expression of tlie form -J*-* — ^-r-* — , and
(1 — x)'"'"*
also that «p = 0 and n^^ = «_ j.
Summation by Undetermined Coefficients
719. Thus far it has been shown how to sum three particular
kinds of series, namely, arithmetical progression, geometrical
progression, and recurring series. The method of undetermined
coeflQcients can be employed to sum a fourth kind of series which
usually has the character that the sum of the first n terms of the
series is an expression containing n, which expression has the same
form no matter what value n has.
696 COLLEGE ALGEBRA [1130
In order to illustrate this method, find the sum of the series
1« + 2« + 3« + + n«.
Assume
(1) l«+2«+3« + +n* = A^+A^n+A^n^+A^ +
where the ^'s are constants to be determined. Since this eqnatioo
is to remain true for all values of n, change n into n-|- 1 ; thus
(2) l«+2»+3^+ . . . +n«+(n+l)«=^, + J,(,, + 1) + J,(«+l»«
+ ^,(i. + l)»+
Subtract (1) from (2),
(3) n*+2n+l=A^+ A^{2n+1) + AJiZ ««+ 3h + 1)
+ ^^ (4 »' + 6 n« + 4 n + 1) +
On equating the coefficients of like powers of n, A^ will be 0 since
n' does not occur in the first member of (3) ; similarly all sacceedLog
coefficients A^, A^ etc. , are zero. Hence, equating coefficients of like
powers of n in (3), we have
A^+A^ + A^=l, 2^,+ 3.1, = 2, 3^, = 1,
and ^3 = ^, J, = \, ^1 = i- •
Hence substituting in (1), we get
l« + 2' + 3' +..... +„« = ^.+ l„+l««+^««.
Since this equation is to hold for all positive integral powers of n,
put n = 1 ; then A^ = 0. Hence the required sum is
The same method may be applied to find the sum of the cubes of the
first n positive integers or the sum of their fourth powers, 'and so on.
Series which may be given the Form or the Alternating Series
720. It has been shown, ?669, that a series of the type
(1) Wj— «,+ «j— «5+tt3 — W^+ tt^ — , +,
is convergent in case
lim (w„) = 0 and u^+j ^ m„.
Since the sum of series (1) is i/j, the sum of every series which cm
be reduced to the form of (1) can be found.
8721] THE SUMMATION OF SERIES 697
Example. — Find the sum to infinity of the sertes
. x(x + ay(x + a)(x + 2a)^(x+2a){x' + Say
where x and a are both positive.
Series (2) may be written
\x~ x + a)'^\x + a^ x + 2a)'^\x + 2a'^x+~3a)'^
Hence according to the principle just explained the sum of (2) to
infinity is - •
721. The sum of the first n terms of a series which can be re-
duced to the form of (1) can be readily found.
For example, the sum of the first n terms of (2) is from (3)
equal to
(4) i ?_ = _M_-.
X x+iia x-\-na
BZBBOISE cm
Show by the method of undetermined coeflScients that the sum of:
1. i« + 3'+5«+... +(2»-l)» = ^"-M<'^"+^>.
2. 1-2 + 2-3+3-4+4-5+... + n (>t+ 1) = "^" + ^^ ^" + ^).
3. 2« + 4'+6«+. . ■ 2«=^"<"+^><^" + ^)-
4. 1-3 + 3-5 + 5-7+ . . . + (2«-l) (2/1 + 1) = f(4„t+6«-l).
5. l'+2»+3» + 4'+ . . . +»'=^^^^J^.
6. 1 • 4 • 7 + 2 • 5 • 8 + 3 • 6 -9 + . . . + «(« + 3) (n + 6)
= n(« + l)(»+6)(n + 7).
7. l»+3' + 5»+ 7'+ . . . + (2ft— 1)»= n«(2n«-l).
Using 22720, 721, sam the following series to n terms and to infinity,
ft J_-l-_L_j._l_4._l_4.
9. -^+-^+ '
1 • 4 ' 4 • 7 ' 7 • 10 '
^^- F4"^2^+3MJ"*T^+- • • •
1 1 1 I 2 , 3 , n 1
21 ' 3! ' 41 n + ll • » ~ n! n+1 !
698 CX)LLEGE ALGEBRA H 8722, 723
The Summation op Arithmetical Series op Higher Order
The Method of Differences
722. Order of Differences. — If the first term of a given series
is subtracted from the second, the second from the third, and so on,
these differences, arranged in the order of their formation, constitute
a series called the first order of differences.
If the series of the first order of differences is treated as just
described the resulting series is the second order of differences of the
given series.
The first order of differences of the second order of differences is
the third order of differences of the given series, and so on.
Hence we have, if the given series is
1, 6, 21, 56, 126, 252, 462, 792, .
first order of differences 5, 15, 35, 70, 126, 210, 33U, .
10, 20, 35, 56, 84, 120, .
** 10, 15, 21, 28, 30, .
5, 6, 7, 8, .
1, 1, 1, .
0, 0, .
723. To Find the /ith Term of a Given Series.— Suppose that
the terms of the given scries are
second
third
fourth
fifth
sixth
(1) M,, w,, w,, . , . n„
"-, w„
*j, «j, «,, . . . "„«i, "„, "„+i, . . . ,
then the successive orders of differences will be
Ist, Wj— «i Ws— W2 W4 — Wj ... tl»-f 1 — u,,
2nd. Mj — 2m2+m, , Hi — 2«3 + ««, W5 — 2u|+ Wj . . .
3rd, Mi — 3W3+ 3tt8 — Ml, «5 — 3m4 + 3m8 — t/j, ... ...
4th, M5 — 4«4+Gj/3— 4m2+«i ... ... ...
It is customary to represent the first terras of the various orders
of difference respectively by d^^ d^, d^y </^, etc.; hence we have
d^=u^—2u^+u^=u^^2{u^+d^)+u^, .'. t/3=Mi+2(/,+(?„
d^=u^ — 3m3+3wj— w,,
= u-3(u^+2d^+d^) +3{u^+d;)^u^, .-. n^=y,+3d^+3d^+d^
The coeflScients of the terms in u^ are the same as those of the
expansion of the binomial (l + .r)*and the coefficients of the terms
in H^ are the same as the coefficients of the expansion of (l+x)'.
It will now be proved by induction that this law holds for the
expression for the n^^ term of the series in terms of the first term of
the given series and the first terms of the vanous orders of differences.
8724] THE SUMMATION OF SERIES 699
Hence, assume that the law holds for the n*** term of any series*.
thus we have
(i) .,=«,+(n-iH+i-» ^^!^;"-^ ,1, + (" - ^> ^-- ^' (»-«)^,+ . .
If the law of formation of the expression of u^ holds for any
series it holds for the n^^ term of the first order of differences; here
lij, d^, d^, d^, , . . correspond respectively to (fj , d^, rf^, d^, etc.,
and the ii*** term of the first order of differences in u^^^ — u„\ hence
we have
/o\ ^ If tx^ I (n~l)(n— 2)^ , (n— l)(yi— 2)(n— 3) ^ ,
Adding equations (1) and (2), we get
r(n-l)(n-2)(n^3) , (n -^) (/Lr: 2)1 ^ .
L 31 "^ 21 J s"^ • • •
or
(3) t.„,.=«,+nrf.+ tLilL^ </, + »(» - 1) ^(» - 2) ^^ ^ . . .
One sees that the same law of formation of the expression for w„
holds for the formation of the expression m„^,-, l)ut we know that this
law holds for the formation of u, , therefore it holds for u^ and so on.
724. The Sum of the First n Terms of a Given Series.— Suppose
that the given series is
(1) 1/,, «2, Wg, M^, . . . w„, w„+,, . . .
and that the sum of the first n terms of this series is S.
Consider the series
The fourth term of series (2) is the sum of the first three terms
of series (1), ftnd the (w + l)^^ term of series (2) is the sum of the
first n terms of series (1); and, further, series (1) is the first order
of differences of series (2) ; therefore, the sum of the first n terms
of series (1) may be found by formula (i) §723. To find the (n-|- !)*»>
term of series (2), i. e., the sum of the first n terms of (1), we must
substitute in (i), „ ^ ^ 7 r t
^ for «„ and d^ for d^
0 for tij * < d^ for d^
i«j for c^j << etc.
Hence we have
/::\ o 1 n(n — 1) J , n(n — l)(n—2)j .
700
COLLEGE ALGEBRA
[«I725,726
In case one order of diflferences, say the r**, is found, all of whose
terms are 0, then d^, ^/^^.j, ... are 0 and S will have a finite
number of terms and be finite.
725. An Arithmetical Series of the /ith Order is a seiies in
which the «*** ortler of diflferences are all equal. For example, tiie
arithmetical progression
(1) - a, a+<£, a+2(f, a+^d, . . .
is of the first order; hence, u^=a, d^=d^ d^=d^= . . . 0.
Substituting these values in (i), the n*^ term of (1) is
(i) «+(»— ly. [1S21]
Substituting the same values in (ii), the sum of the first n terms of
(Dis
;S=na+ --^^^= I [2a+(n_lM]. [J625,(iv)]
726. Some of the important classes of series which can be
summed by the method of diflferences are:
1. Fignrate Numbers.
Identical case
1,
1,
1,
1.
1,
ete.
Natural
2,
3,
4,
5,
6,
ti
Trigonal
3,
6,
10,
15,
21,
<f
Pyramidal
4,
10,
20,
35,
56,
II
Trigonal-Pyramidal
5,
15,
35,
70,
126,
II
etc.
. Polygonal Number*.
Triagonal 1,
3,
6,
10,
15,
21,
28,
etc
Quadrantal 1,
4,
9,
16,
25,
36,
49,
ti
Pentagonal* 1,
5,
12,
22,
35,
51,
. 70,
u
Hexagonal* 1,
6,
15,
28,
45,
66,
91,
ct
etc.
• Euler's Algebra, Vol. 1, p. 250; ed. 1TB6.
KlQURB OF PBNTAGOKAL NyMBKBS
Figure of HEXAao^'▲I. ^'rx99>8
«727]
THE SUMMATION OF SERTRS
3. Power* of Numbers.
It is supposed
in each case
that a
= 1,2,
3, 4, . .
. .
a* 1,
1,
1,
1,
1,
1,
etc.
o» 1,
2,
3,
4,
5,
6,
((
a* 1,
4,
9,
16,
25,
36,
it
a* 1,
8,
27,
64,
125,
216,
it
«♦ 1,
16,
81,
256,
625,
1296,
li
etc.
701
727. Miscellaneous Series. — Most series are summed by special
methods. There is do general principle that can be used to sum all
series or any very extensive class of series.
For example, sum the series
a -|- (a + d)r -|- (a + 2 d)?* + . . . to » terms.
Put
(1) S=a+(a+d)r+{a+2d)r^+{a+3d)r'+. . . +[a+(n-l)d]r«-'-
multiply (1) by r,
(2) rS=ar+{a+d)T^+{a+2d)f^+, . . [a+(H-2)rf]r»-H-[a+(«-l)<^]r».
Subtracting (2) from (1),
S—rS=a+dr+dr^+dr^+ . . . rfr""*— [a+(?i— l)ci]r»
=a— [rt+(n— l)(/]r«+rfr(l + r+r"+ . . . + r"-«)
— q — [<i+ (n — ly]?-** drjl — r"-')
1 — r "^ (l-r)«
Example. — In an incomplete rectangular pile of shot the top
layer has 5 shot on one side and 21 on the other, and the bottom
layer has 50 shot on its shortest side. Find the number of shot in
the bottom layer, and the number of shot in the pile.
The number of shot in the several layers is:
5-21, 6-22, 7-23, 8-24, 9 • 25, . . .
or 105, 132, 161, 192, 225, . . .
First order of difference 27, 29, 31, 33, . .
second ** *' ** 2, 2, 2, . . .
third ** ** ** 0, 0, . . .
Here 11^ = 105, d^=27, </,= 2, ^,=0, n = 46;
hence, from (i) u^ = 105 + 45 • 27 + ^'^-^^ 2 =3300,
and, from (ii) S r= 46105 + ^^^ 27 + ^^-^^^-H^ 2 = 63135.
702
COLLEGE ALGEBRA
[«72B
728. To Interpolate a Term between Two Terms of a Series by
tbe Method of Differences.
Example.— Given log 61, log 62, log 63, log 64; find 1<^'62.73
from the various orders of differences from the given logarithms as
in 1722
log 61 log 62 log 63 log 64
Wj 1.785330 1.792392 1.799341 1. 806180
d^ .007062 .006949 .006839
d^ —.000113 -.000110
e/j + • 000003 considered to vanish.
Log 62. 73 must be regarded as an interpolated term, the number of
its place being 2. 73
Hence put n=2.73 in formula (i) {723,
ti^= 1.785330, rfj = . 007062, ^7^= —.000113, cf,= 0,
and find log 62.73 = 1.7974759.
EXEBOISE OrV
1. (i) Sum by means of the method of differences the first 20
terms of the polygonal series of the 3'^, 4***, 5^, and 6^ orders.
Order
PoLYtJONAL Numbers
n** Tkrm
l.t
1111 11...
1
2nd
2 3 4 5 6 7 . . .
n
3rd
3 6 10 15 21 28 .. .
|«(« + 1)
4th
4 9 16 25 36 49 .. .
n«
5th
5 12 22 35 51 70 . . .
|«(3«-1)
6th
6 15 28 45 66 91 . . .
(2» — l)n
n + --^{r 2)
^th
1,
r,3+3(r-2),4 + 6(»-_2),
5+10(r-2), . . .
(ii) Verify the expression for the w*** term in the annexed table, and
show that the series of the r'** order has the form given in th^
table.
S728]
THE SUMMATION OF SERIES
703
(iii) Show that the sum of the first n terms of a polygonal series of
the r'** order is
n(yi.-fl) I n(yi — 1) (n+1) (r — 2)
2. (i) Sum by the method of differences the first 20 terms of the
figurate series of the 3"*, 5***, and 6^** orders.
Order
Figurate Numbers
w**»Term
1st
1 1 1 1 1 . . .
1
2nd
2 3 4 5 6 . . .
71
3rd
3 6 10 15 21 . . .
n(n4-l)
2!
4th
4 10 20 35 56 . . .
n(n + l)(n-f2)
31
5th
5 15 35 70 126 .. .
n(n-hl)(n + 2)(/i + 3)
41
6^
6 21 56 126 252 .. .
»(r^4-l)(n4-2)(«-f3)(n+4)
5!
(ii) Verify the expression for the w*** term given in the annexed table
(iii) Prove that in case of a series of the r*** order the n^ term is
T?(n+1) {n-\-r-2)
r—V.
(iv) Show that the sum of the first n terms of. a figurate series of
numbers of the r*** order is
n{n+]).
. (»4-r-l)
3. Given log 71, log 72, log 73, log 74; show by interpolation,
using the method of differences, that log 72.54 = 1.8605777.
4. Given log 9, log 10, log 11, log 12, log 13; show by inter-
polation that log 11.834 = 1.073132.
5. Given the h 91=4.49794, V92=4.51436, V93 = 4.53066;
find V 91.54.
6. Given ]/ 51 = 7.13979, 1/52 = 7.21110, 1 '53 = 7.88012,
1/54 = 7.34847; find l/53.47.
7. Given tan 54*^= 1.376381, tan 55*^= 1.428147, tan 56*^
= 1.482562, tan 57° = 1.539868, tan 58° =1.600337; show that
the tan 56° 32' 44" = 1.513403.
704 COLLEGE ALGEBRA 11728
8. How many shot are contained in 15 layers of an incomplete
pile of shot whose base is a rectangle with 24 shot in one side and
35 in the other?
9. Show that the number of shot in a complete pile whose base
is a rectangle, which contains m and n shot in its sides, is
J m (m + 1) (3 n — m -|- 1) if n'^m.
10. Prove that the sum of the first n terms of the series
l-f3x+5x*+ . . . .ton terms
is -s = 1 - ^-4^ + ^irfg-- tm?]
11. Prove that the sum of the first n — 1 terms of the series
{n — l)x +(n — 2)x« + (» — 3)x^+ . . . + 2 x»-« + x-»
i3 ^,^(n-l)x-n^+^^i, ^j„7j
12. Show by means of the theorem of undetermined ooeflScients
that the sum of the r*** powers of 1, 2, 3, 4, .... m is
where the -4's are obtained by substituting successively p = 1, 2, 3.
etc. , in the equation
I ^ _|_ ^t 1 £^j I
2{p + l)\ (p + 2)!^r(p)!^r(r-l)-(p-l)!^- * *
^r(r-l)...(r-p+l)
13. Prove that the n*** term of the series
5 -7 -9 + 7 -9 11 + 9 -11 -13 + 11 13- 15+
is (a + nd) [a + {n + l)d] . . . [a + (ti + f»_ ly]
where a = first factor in the first term, n == the number of terms,
and d = the common difference of factors.
14. Prove by induction or by the theorem of undetermined
coefficients that the sum of the first n terms of the series in 13 may
be found by multiplying the last term by the next highest factor
and subtracting from this product the product of the first term by
the next lowest factor and dividing the difference by (m+ l)d. Thus
if S^ is the sum of the first n terms of the given series,
o _ 11 -IS- 15 17-3 '5 7 -9
(i+l)2
CHAPTER VIII
LIMITING VALUES OF INDSTSRMINATE FORM
729. It has been shown, 273, that the numbers which take the
form - are indeterminate, i. e. , there is no particular number which
represents the value of this fraction.
ExAMPLB 1. For example, the fraction — — ^ "^ takes the form
- forx = 2, since
L j:«-4 Jar=2
If the fraction is written in the form
.r« — 5.r-f 6~[ 4— 10-f-<> — 0.
4—4 ~"0'
^r- 53--f-ft — (-g— 2)(.r- 3)
j^~—A' ^{x — 2){x + 2)'
and the numerator and denominator are divided by x — 2 (which
operation is admissible only in case x is not zero), we have
J* — 5j:-|-6_j— 3
^-4 x+2
which is equal to .r-r-^ = ^ 7 when x = 2.
But the division of the terms of the fraction by x — 2 is not admis-
sible when X = 2, i. e. , we are not allowed to divide both terms of
the fraction by 0. Hence it is necessary to define what we mean
by the value of the indeterminate form - •
780. Definition of the Value ^- —Let /(.r) and F (x) be respec-
tively the numerator and the denominator of a fraction, and suppose
that/(f#) = 0 and F(a) — 0, i. e., the numerator and denominator
are respectively zero in case x is replaced by a\ then we have
^ ^ LF(x)Jx=a"~ jF(a)~0
706
706 OOLLEGE ALGEBRA [i731
We define the value of the fraction ;,— for x = a as the
lim fix)
x~aF{jc}'
Thus in the preceding example we have
lim rr» — Sx + ei lim (x — 2) ( z — 3) ^ lim t — 3^ 1
x=2L x* — 4 J x_12(j- — 2)(j; + 2} jrA2x4-2 4*
NoTB.— The division of the terms of the second fraction by x—2 is admissible since
l^ definition of a limit, $686. x can not take the value 2.
731. Further Examples. — All the algebraic and transcendental
operations used in the previous chapters may be employed to give
the expressions which are indeterminate for a particular value of
X, another form which is not indeterminate as x approaches this
value. To illustrate:
Example 2. Find the value of the fraction
^— 15:r» + 24x— 10
£lr=lii_+2i^l_ fo, ^ ^ 1.
This fraction = - for x = 1. Put x = y -\- 1, then for x = 1,
y = 0, and the value of the fraction is
lim ^-2-r'4-2r-l _ii„, (y + 1)* - 2(v4- l)»-h2 (y + D -1
or— la:«-15.i* + 24x-.10 y^O (y+ 1)«— 15(i< + 1)« 4-24(i/ + l) — 10
-lim It + ^ll ^-Cy+2) 3^2^!
y^-0y« + 6y*+15i^+20y8 j^^^i (i^' + 6i/>+15y +20) 20 10
[JW6J
Example 3. The fraction — — y^~ — ^ becomes - ioTx=a.
j;_ 1/2.1^ — «« 0
Multiply both numerator and denominator respectively by their
complementaiy surds, then the fraction becomes
lim [4 x« - (5 j» - (!»)] [x -f v^2.r - a«] ^ ijm (a«- j^) [x-f V'2^-a']
rF=^-a[^_(2a;« — a«)][2x+v'5x«— a«] ^^^{(^ — a^)[2x+Vbj^-tf]
= lim G:-fv^22^-a^)
"x=^(2x+Vbj*--(^)
^ a + Va* ^l
2a+\/4^ 2
«732] LIMITING VALUES OF INDETERMINATE FORM 707
Example 4. — — i takes the form ^ f or x = 0.
X 0
Here
a'=l +xlog,a + ^(log,a)« + g(log,a)»+ . .
lim «^jzl1 =
x^^ X
= log . a + ^ [^=^^f^ + ] = log.a.
Example 5. ««in^=«l>l_^ becomes ^ for x = 0.
To find- the true value of this fraction, substitute for sines and
cosines their expansions, 2705, and the fraction becomes
■■(-,-|?-)-(-r^3+-)-°i:'-PF3-'+r^-]
^- I (— 1 + a*) + terms in x*, x*, etc. J
- I ( _ 1 ^ a*) + terms in x', x*, etc. J
Dividing both terms of the fraction by ^ the value of the fraction is
lim ^ (g' — 1) + terma in a-*, etc. __ a (a' — 1) __ a
JT^O 3 (a« — l)+term8in.««, etc. 3(a« — 1) 3*
732. The Indeterminate Form g^-— Any number which takes the
form g is indeterminate. For, by the definition of a quotient, we
must have (^) oo = cjo ; but x • oo = oo , where x is any number
not zero, i. e., any finite number x multiplied by a number larger
than any assignable number, or cjo . Hence ^ can have any value
that X can, and therefore ^ is indeterminate.
To evaluate an indeterminate form of this kind let the fraction be
--,'--: where /(a) = CO and F{a)—'-si ;
■t (x)
708 COLLEGE ALGEBRA [1732
but ve may write
for 35= a.
Thmt IS, fractMHis which take the indetenninate form ^ may be
reduced to the indeterminate form -. Hence, if a given fraction takes
tlie indeterminate form ^ reduce the fraction to the indetenninste
0 *
form - and proceed as in 2731.
Find the limiting valnes of the following fractions:
^ ^""^ forx=l. Besult li
forx=2. «* 2|.
M.
ji»-l
9
j*-M
jr*-32
3.
jr*-j*
2*-l
4.
<i*-jr*
a*-x»
5,
6.
T» - 4j^ -4- :lr - 6
T.
if — la — Ti*
s.
ut — h,*— xa — Tf*
9,
10.
1 1-x
11.
1 1 ^ r - 1 ii
for a:=l.
for ac=a. <« to*.
forx=:— 1.
forx=2.
for x=0.
52
(D-
» .S -f X — 1 5
for ,=5. .. J^+iJ.
forx=l. «« I
forx=l. ** 0.
forx=2. " \VtL
9
J732] LimTING VALUES OF INDETERMINATE FORM T09
12. y3^^-l forx=2. Kesult 1
13 g-V'a'-^ forx=0. " i
aj 2a
a^ + ai^-7x^^27x-18 f^r^^3, u 10.
16. ^""^ forx=l.
1
n
16. j£z^ forx=a. «* 0.
17. — =7^— == forx=0. ". Va. "
Va+x—va—x
18. t ~ -^ . for x=a. " >- . •
1
19 i^x~-v'a±Vx^^ forx=a. " ,-.
20 — ^— forx=0. ** 1.
«* — 1
21. °^-^ forx=0. '« log^.
22. ^-^-f^ forx=0. " -|-
23. ^^^ forx=0. ** -^•
1 — COS mx "*
24. ^-^"' forx=0. " 2.
ainx
25 cotanx for x=0. «* 1.
cosecx
|+l + lag(l + .)
26. 1 fora==-l. «' 0.
«*+ 1 - 1
710 COLLEGE ALGEBRA [J733
788. Table Indicating the Region of Conrergence of Some Important 8am
The region of oonvergence is Indicated by the heavy line.
y^=l+x+x»+a^+ ....
-1 0-1
1 1
a — bx a a* a* a*
— r 0 r
1 1_
r=j numerically.
6
log(l+:r)=a.-f + f-J+. . .
— 1 0 1
— I 1-
-1 0 1
1 1
(l + x)''=l+nx + fe^^ + n(n-l)^ ^
-10 1
1 1
v/r~?"^''"2^^2-4^ ^2-4-G-^ ^
—1
— H
v/rni = i-ix.--i-x-^-i^..-
-10 1
—I h-
^= 1+3, + ?* + ?! + ^ +
31^5! 7!^
^ ^rA ^
««* = l-2i + 4!-6!+
BOOK VIII
CHAPTER I
INTRODUCTORY CHAPTSR IN THE THEORY OF DETERMINANTS
I. Determinants op Two Rows
Equations of the First Degree in Two Unknown Quantities
734. Determinants of Four Elements. — The expression
is written in various ways, for example
and is read the determinant of the elements (o^ , h^^ (o^ , h^. The
term a^ h^ is called the principal term of r; the horizontal rows
(for example a^ h^ are called simply rows (numbered first, second,
etc., downward), and the vertical columns (for example M are
called simply columns (numbered first, second, etc. , from left to right).
735. The Principle of Development of r. — The determinant r
is equal to the diflference between the products of the elements in
the two diagonals, in which the principal term is + . Accordingly
(i)
(ii)
7 8
-3 2
= 7-2 - (—3-8) = 14 + 24 = 38
\oga\ogh
= 7 log a — 3 log 6 = log a^ — log 6' = log ^.
Rbmab«.— Every term of a detenninant of four elements contains one clement from
Mcb row and jeach column.
711
712 COLLEGE ALGEBRA
BXBBOISB OVI
Calculate the value of the determinants:
1.
5. Find a determinant which is 0 if a : h = c : iL
Develop the determinants:
[81736, 737
20 70
If 41
ma^ mb^
«i
*,
2.
3.
4.
SA
6 5
H H
«. 6.
",
6.
7.
^ —a;, y — y.
8.
736. The Elimination of One Unknown Quantity from Two
Equations of the First Degree.
Let the given equations be
(1) a,x ^ h,
(2)
a^x = 6,
or
a^x — 6j = 0
a^ — feg = 0
in which it is assumed that the coefficients a^ and a, of the unknown
quantity are not 0. It is desired to know what the necessary and
sufficient conditions are that the same value of x satisfies both equa-
tions. Multiply equation (1) by a^and equation (2) by —a^ and add,
(3)
hence (4)
a^{a^x — \) — a^ (a^ — 6^^) = 0 ;
or 1 ^ = 0.
r or aj6g — a^ft^
[I734J
Equations (1) and (2) lead to equations (3) and (4). Equation
(4) is a necessary condition in case equations (1) and (2) are not con-
tradictory. Equation (4) is also a sufficient condition ; since (2) can
be derived from (1) and (4), thus
from (4) «! = ¥*;
substituting in (1), ^ x = 6j, whence a^x=h^y equation (2). Q.E.D.
Similarly equation (1) can be derived from (2) and (4).
737. The Eliminant. r is called the eliminant of the system of
equations (1) and (2), or of the equivalent system
(10
(20
8738]
THEORY OF DETERMINANTS
713
(1') and (20 are found by patting x = — ^in equation (1) and (2) of
y
2736. Hence, according to what precedes, the eliminant of two
equations of the first degree, (1) and (2), is the determinant of the
coefficients of the unknown quantity and the constant terms, or of
the coefficients of the unknown quantities in case the given equations
are written in the form of homogeneous equations of the first degree.
The resultant is found by placing the eliminant equal to 0.
Rbhabk.— The results can be described as foUowi: If the equations (1) and (2) are
compatible, or what amounts to the same thing, in case these equations are eatisfled by
tlae same values of a;, then the Identity (S) exists among the coefficients of these
equations.
738. The Solution of two Equations of tixe First Degree in the
Case when the Determinant of the Unknown Numbers is not 0.
Let the equations be
(1) a^x+h^yz=z c^
(2)
•+6^ =
Multiply equation (1) by h^ and equation (2) by — 6^ and add; then
we get (3) {afi^ — a^^x = cfi^ — c^^.
Multiply equation (1) by — a^ and equation (2) by a^and add, and get
(4) (afi^ — a,6j)y = a^c^ — «,<?!;
whence it follows from (3) and (4) that
(5)
X =
"l
\
«.
K
«1
^
«t
h
y =
«l
«l
«.
«.
«l
h
«t
h
Rule. — 7%e denominator of x and y is the determinant of the co-
efficients of the unknovm quantities of equations (1) and (2); the num-
erator of X is found by substituting in the denominator the terms c^
and Cj in the second members of (1) and (2) for the coefficients a^ and
a^ of x; similarly the numerator of y is found by substituting c^ and
c^for the coefficients b^ and b^ of y.
The correctness of the solutions in (5) can be verified thus:
Substitute in (1) and (2) the values of x and y in (5), and they will
become
ai(ci6a - cM 4- b^iaiCt - a^Ci) _ ^
aib% — a^bi *'
OiiCibt — fgfci) -i- h^laic^ — a^Ci) __^ .
aibi—Oibi «'
714 COLLEGE ALGEBRA
or arranging the numerators with respect to the c%
Qibt — atbi ^ II'
aibt — aj>i ' 9 r
That is, the values in (5) satisfy equations (1) and (2).
Example. — Solve the equations
9x + lly = 5
8x+ lOy = 4.
[{739
Equations (5) give
X =
5 11
4 10
-^
9 11
8 10
y =
9 5
8 4
-
9 11
8 10
= (50 — 44) -5- (90 — 88) =3,
= (36 — 40) -5- (90- 88)=-2.
739. Homogeneous Equations.
1. With two Unknown Quantities. — If c^ = c^ = 0, tlien equatiom
(1) and (2), §738, are homogeneous and give aj = 0, y = 0 as solu-
tion of the equations when afi^ — aj^^ ^ 0.
2, With Three Unknown Quantities. — Substitute in equations (1)
and (2), {738, x Y a ^^ ^
' ' X = — -^ y = — -y and they become
(6) a,X+6,F+ c,Z=0,
(7). , a,X+6,r+c,Z=0.
Then, instead of equations (5), we obtain the equations
X -T Z
(8)
a. 6.
Hence JT, — Y^ Z are proportional to the determinants of the table
\ «=.
a, 5, c,
found by striking out the first, then the second, and finally the
third column.
W40]
THEORY OF DETERMINANTS
715
II. Determinants op Three Rows
740. Determinants of Nine Elements. — The expression
is written as follows
A =
«i \ <^i!
a, 6, c, -SzhaiV8^KVs) = K^2«8]
«3 h ^8
and is called the determinant of the elements («pti,Cj), {cL^h^^c^^
(ajj,6j,Cj). The term o^J^^c^ is called the principal term of A.
The definition of columns and rows given in 2734 is adopted in case
of a determinant of nine elements.
Formation of the Determinant A. — Place the first and second
columns to the right and next the ^ven determinant or the first and
second rows immediately below the third row of the given determi-
nant thus:
Now form the six products of three elements which lie on lines
drawn through the two diagonals and the lines parallel to them,
neglecting lines which pass through but one or two elements. Place
the sign -f before the principal term and the remaining products of
lines running from the left above toward the right below, and before
the other products place the sign — .
Examples. — 1.
2 -1
—3
=2-6-3+(-l)(-3)+l-4'2-l-61
— 2( — 3)2-(— 1)4-3=65.
716
COLLEGE ALGEBRA
[H40
3.
c
h
c
0
a
a h g
h b /
ff f c
=2abc.
= ahc — af* — bg^^ ch* + 2/gK
0 — c b
— c 0 — a
b -•a 0
= 2abe.
a b
— a b
— a — b
c
m
c
= abc + (bm) {^ a) + c (— a) {— b)
— cb{ — a) — am{ — b) — 6( — a)c.
Rbv ARK.— Each of the six terms of a determinant of nine elements contains an
element from each column and each row.
BXEBOISB OVU
Calculate the value of the determinants:
1.
4.
7.
10.
a b
b c
c a
5
7
2
3
4
1
0
a
b
6
1
3
2.
0
2
5
3.
d
0
c
1
5
2
0
1
6
e
/
0
0
a
b
0
a
0
a
0
b
a
0
b
6.
b
0
c
6.
0
c
0
a
b
0
0
d
0
d
0
c
0
a
b
X
a
b
X
0
c
— a
0
c
8.
—a
X
c
9.
-1
X
b
-b -
— c
0
-b -
— c
X
0-
-1
a
2741J
THEORY OF DETERMINANTS
717
Write the following polynomials in the form of determinanto:
11. o^5,_6^a^+a,6,-6,a,+a,6,-aj6,. 12. a» + 6» + c» - 3atc.
13. abc -\- am* + bn* -f- cj^. 14. 2p*q— p*—pq*.
15. Find the value of x in the equations
(1)
X — 2
3 4
^:
(2)
1 1
5 6
(4)
a -4 1
—6 3 —2
X 2 1
h
h
X
= 0
(3)
1 1
a X
b h
= 0
X
a
a
a
X
a
= —
a
a
X
X
b
16. Prove that
0
b.
741. Relations Between Determinants of the Second and Third
Orders. — We have
0 a,
and in particular
X
y
«1
^
o.
b.
mm
—^
1
1
«•
h.
a
b
s
%
^t
8
m 0
z a
u a
I
X
y
1
0
0
0
«i
^
=
z
«i
\
0
«.
^
u
".
h
for arbitrary values of x^ y, z^ u.
One can therefore write a determinant of the second order in the
form of a determinant of the third order and in some cases conversely.
Suppose that one uses stars (*) instead of the arbitrary elements;
then one has
a; * ♦
0 y
0 0
y *
=x
0 z
— xyz.
718
COLLEGE ALGEBRA
[1742
Hence, in ease all the elements on the same side of the diagonal
of a determinant are 0, the determinant is equal to its principal tenn.
Conversely, the product xyz can be written in the form of a
determinant of the third order;
X
*
*
X
*
*
0
y
♦
=
0
y
0
0
0
z
0
*
z
II L Properties op Determinants
742. First Property. — In order to multiply or divide a deter-
minant of the second or the third order by m, one multiplies or
divides all the elements of any row or any column by m,
Verificati(m, —One can test any case by developing the determin-
ants. Proof: Each term in the development of a determinant con-
tains an element from each column or row ; hence each term of the
determinant expanded is multiplied by m if any column or row is
multiplied by w, and therefore the determinant is multipled by m.
Similarly for the case of division.
Corollary 1. — In case all the elements of a column or a row
have a common factor, then the elements of the column or row can
be divided by this factor provided the determinant is multiplied by
the same factor.
Corollary 2. —In order to multiply or divide a determinant by
(__1)^ it is sufficient to change the sign of the elements in any col-
umn or row.
Examples. —
1.
ma^ mb^ mc^
% ^8 S
28 18 24
12 27 12
70 15 40
= {+mafi^c^+mb^c^a^+mc^nJj^^mcfi^a^
— wif/jCjfej — mb^a^c^ = m A.
114 9 12
= 2-3-5! 4
114
= 2-3 5-2 3-4
3
3
1
= 2-35-2-3-4(— 13).
58743, 744]
THEORY OF DETERMINANTS
71d
EXEBCISB CVni
Verify the following identities:
This result can be verified by multiplying the columns and rows by (—1).
«1
\
"l
«.
^
«*
=
«»
h
s
2.
he 1
ca 1
ah 1
1 a a'
1 h h^
1 c c2
3.
«,-l 0
6, «,-l
=
0 K ".
«, -l/6, 0
^. a, -1/6,
0 VK
4.
a
h
c
1 1
1
1
»
y
z
= xhc
aye
ahz
^
B
C
A
B
h
C
c
743. Second Property. — In a' determinant of the third order,
one can interchange columns and rows and rows and columns without
altering the value of the determinant. For
^ ^ h
«l ^l «l
a, 6, c,
«, i, c,
Rexabk.— The following (so-called seml-sjmmetrtcal) determinant
0 c b
-c 0 a Is equal toO:
-b—a C
for the value of this determinant Is not changed if we multiply all columns or rows by
(—1), I.e.. If we multiply the determinant Itself by (— 1)» = — 1, a result which can only
be true in case the determinant is 0.
744. Third Property. — A determinant has its sign changed if two
columns or two rows are interchanged. For
(c, 6j a.
<^» \ ««
S h "j
+ «i''t«, + ^1%"^ + °l«.^!
'^j - '•t««''»
^t«2"»
°i \ «i
a, 6, c.
% h ''si
Similarly for the five other possible interchanges of rows and columns.
720
COLLEGE ALGEBRA
[1745
746. Fourth Property. — A determinant of any order z= id if it
hat two columns or two rows that are exactly alike.
First Proof: One can verify the theorem for any case.
Second Proof: Let the determinant with two columns alike be
(1)
A =
By interchanging the first and second columns we get (§744) :
(2)
A =
ma, ma^
= -A;
hence
2A = 0
and
A = 0.
Similarly it can be proved that a determinant of the fourth or
higher orders is zero if it has two rows that are exactly alike.
This theorem may be stated as follows: If the elements of a
column or a row of a determinant of any order are replaced respec-
tively by the elements of a parallel column or row the determinant =0.
For example, if the first row (1, «, y) in the determinant
1
X y
1
«=i y.
1
*. Vt
is replaced by the elements of the second row (1, Xj, y^ or those of
the third row (1, x^, y,), the determinant = 0 (J746).
Corollary. —A determinant of any order is equal to 0, if the
elements of a column or row have the same common factor whose
removal leaves a column or row the same as one of the remaining
parallel columns or rows. For, on removing the factors from the
elements of the given columns or rows we multiply the resulting
determinant respectively by these factors, but the resulting determi-
nant has two columns or rows alike and is therefore zero (2746).
2746J
For example:
THEORY OF DETERMINANTS
721
6 7 5
6 7 5
18 21 15
= 3
6 7 5
2 3 7
2 3 .7
= 30 = 0.
Without expanding the determinant prove that a and b are roots
of the quadratic equation in x written in determinant form,
1 1 1
a X a
h h X
= 0.
rV. Properties op Minor Determinants
746. Definition.— The coefficient of an element of a determinant,
or more precisely, the entire multiplier of an element in a determi-
nant, is called the minor determinant of this element of the given
determinant. The minor determinants of
(1)
A =
«i ^1 ^1
a, 6, c,
% K ^3
(2) or A = 5, {c^a^ — c^a^) + \ (a^c, — a^c^) + b^ {c^a^ — c^a^),
(3) or A = c^ (0,63 - aj>;) + c^ (6^a, - ft^a^) + c, (a^6, - a,6^),
with respect to the nine elements
are usually represented respectively by the symbols
A^, A^, J,; ^„ ^„ ^,; ^3, ^3, 6%.
Hence it follows from (1), (2), (3) that
(4)
A =
A =
\c,=
a, 6j
«8 ^
A,= -
B,=
^; = -
6j
Cj
63
C3
a^
Tj
e?3
Cj
^1
6^
a^
6,
^4,=
^,=
<'; =
722 COLLEGE ALGEBRA L « J 747, 748
Rule. — To find the minor determinant of any element strike omt
the column and row having the element in common. The minors of
the elements in the two diagonals (.4^, B^, C^\ A^, B^y C\) of the given.
determinants are pluSy the signs of the other minors are minus.
Corollary. —The minor of any element is independent of the
elements of the column and row to which this element belongs, and
therefore remains unchanged if the elements of the column and
row to which the element belongs are replaced by other numbers.
747. Properties of the Minor Determinants of A. — These arc
expressed in the following 18 equations:
A = a^A^ + h^B^ + CjC, A = a^A^ + a^A^ + a,.4,
0 = a^A^ + \B^ + c/; 0 = b^A^ + b,A, + 6,.4,
0 = a^A^ + b^B^ + c^C^ 0 = c^A^ + c, J, + c,^,
0 = a^A^ + b^B^ + c^C, 0 = afi^ + a,B^ + a^,
(5) ^ A = a,A, + b,B, + c,C\ A = b^B^ + b,B^ + b^B^
« = %^\ + \^t + ^,^; ^ = ^ A + ^.^. + ^ A
0 = a^A^ + bfi^ + c^t\ 0 = a^t\ + a^C, + a^C^
0 = a^A^ + b^B^ + c^(\ 0 = b^(\ + 6,6; + 63C;
I A = a^A^ + b^B^ + C36; A = c,t\ + c,6; + c,C;
The truth of these equations follows readily from direct calculation.
They can also be proved without direct calculation.
748. First Property. — The determinant ^ is the sum of the pro-
ducts of each element of any column or row by its corresponding minor
determinant J i. e.,
(6) A = a^A^+ b^B^ + c/.\.
According to the definition of minor determinants, a j-4j is the algebraic
sum of all the terms of the determinant A which contain the factor
fi,; b^B^ is the algebraic sum of all terms which contain the factor 6,
and c, C^ of all terms which contain the factor c^. Hence (i) eveiy
tarm of the determinant is contained in the sum (6), for each term
contains an element of the first row, hence either a, or 6, or Cj ; and
(ii) no term is counted twice in this sum, for no term contains two
elements of the first row, i. e. , does not contain at the same time
a, and b^ , or a^ and Cj , or b^ and c^ . In order to prove that
A = OjJj + a^A^ + a^A^ it is necessary and sufilcient to prove that
every term of the determinant contains but one element of th^ finst
column of A.
88749, 750] THEORY OF DETERMINANTS 723
749. Second Property. — The sum of the products of each element of
any row or column by the minor determinant which belongs to the
corTesponding element of a parallel row or column, is zero, that is
Substitute in the determinant ^ the elements a^ , h^ , Cj of the third
row for the elements CTj , 6, , c^ of the first row, then the minor de-
terminants with respect to the elements of the first column will
remain the same (J746, Cor.) A^, B^, Cy "The new determinant \
is, therefore, as we have proved
This determinant = 0, because the first and third rows are the same
(2746). Therefore
In order to prove that
one may use in a similar manner the theorem that a determinant
with two equal columns = 0.
V. Solution op Equations op the First Deoree in Three
Unknown Quantities
750. The Elimination of Two Unknown Quantities from Three
Equations of the First Degree.
Determine the conditions under which the three equations
(1) a^x + h^y = Cj or a^x -j- h^y — Cj = 0,
(2) a^x + t^y = Cg or a^x -(- Z,^ — c, = 0,
(3) a^x +h^ = c^ or a^x + bj/ - c^=0,
are compatible with each other (J233).
Put A = (abc) and represent the minor determinants of A by
A,, B„ Ci, ^1, etc. Multiply (1) by C^, (2) by C\, (3) by C\ and
add the equations; then we have
(4) ( C,a^+ C,a,+ 6>,)x+( C\b,+ C,b,+ Cjb^)y-{ C,c,+ C\c,+ C,c^ = 0.
But according to the properties of minor determinants the coeffi-
cients of x and y are zero (J749) ; hence from (4)
(5) A=c;c,+ 6;c3+c;c3=o.
724 COLLEGE ALGEBRA [J751
From equations (1), (2), (3) we have deduced equation (4), which is
the necessary condition that (1), (2), (3) are compatible.
Conversely, if this condition exists, i. e. , if equation (4) or (5) is
true, then any equation of (1), (2), (3) can be deduced from the
other two; hence this condition is sufficient.
The determinant A is called the eliminant and A =0 the resultant
of the system of equations (1), (2), (3), or of the equivalent system
of homogeneous equations
(!') a,X+b,r+c,Z=0,
(2') ««^+ ^^+ CjZ=0,
(30 a^X+bJ+c^=(^,
which are found by substituting
-AT -r
""= Z ' ^ = T
successively in equations (1), (2), (3).
Hence, the eliminant o/ three equations of the first degree^ (1), (2), (3),
is the determinant of the coefficients of the unknown and the known
terms, or of the coefficients of tlie equations written in homogeneous
form, (r), (2'), (3'). The resultant is found by placing ^ = 0.
761. In case C^ = 6, = 6, = 0, equation (4) is satisfied, and
hence the necessary condition for the compatibility of the equations
is satisfied; but since equation (4), combined with two of the equa-
tions (1), (2), (3), is not sufficient for the derivation of the third, no
conclusion can be drawn in this case.
Example. —The equation of a straight line has the form
y = ax+6 or y— ax — 6 = 0. [i245]
If the points (x^,y^), {Xi^y^^ lie on this line, the coordinates of these
points must satisfy the equation y = ax + 6, i. e.,
Pi — «^i — 6 = 0
and y^ — ax, — 6 = 0.
The elin^ination of a and 6 from these equations gives
X
y
1
X,
Vi
1
.T,
Vt
1
= 0,
which is the equation of a straight line passing through (.z,, y^ (x,, y^.
2752]
THEORY OF DETERMINANTS
725
752. The Solution of a System of Three Equations of the
First Degree in Three Unknown Quantities when the Determinant
of the Coefficients is not = 0.
Suppose that the three given equations are
(1) -| a^x -{. h^+ c^z = c?,,
I a^x +hj/+ c^z = d^.
Multiply the equations in (1) respectively by A^, A^, ^i, and add
{a^A^+a^A^+a^A^)x+(biAi+b^A^+h^^)y+{ciAi+c^A^+c^^)z
= d^A^+d^A^+d^A^.
The coefficients of y and z are zero (2749); therefore
(2)
d^A^+d^A^+d^Aj^
d,
6,
«i
d.
h
«•
d.
6,
«.
o,
6,
«i
a,
6,
«t
«8
6,
«s
Similarly, multiply the equations in (1) respectively by 5^, B^, B^
and again by 6',, Cg, 6,, and add; whence we get
(3)
y =
a,
'^i
Cj
a.
d.
Cg
«3
d^
Cj
«i
K
Cj
a,
K
Cj
«.
h
c
z =
«l
^
d^
a.
6.
d,
«»
K
d.
«i
\
c,
«t
f>»
c,
«»
h
Cj
The determinant of or, y, z is the determinant whose elements are the
coefficients of the equations; if in this determinant the coejfficienti of the
several equations are replaced respectively by the second members of the
eqitationSj one obtains the numerator o/x, and tV* a similar m/znner the
numerator of y and of z.
726
COLLEGE ALGEBRA
imi
Non 1.— Here it is dear that the denominator, A, of the ralue at or, y. and s ahoQjd
not be lero, for then the values of x, y, and s would be « in case tbo numerators wet
not lero, and Indeterminate In case they were zero.
NotbS.— That the values forx, y, and g are correct can be tested by subsUtDtiac
them in one or more of the equations in (1) and verifying the identlUeo.
EzAMPLB.— Solve the equations
505 — 8y-f2« = 8,
4x-f5y — 8« = 21.
6aj — 2y — 8« = —12.
We have
8-S 2
5 8 2
21 6-3
4 21-3
-12-2-8
5-12-3
- - - 2, v=
=5, z =
5-3 2
5-8 2
1
4 6-3
4 6-8
6—2—3
5 —2—3
6-S 31
A S til
6 —2 —12 1
= 4.
5-8 21
4 6 — sf
5-2-8
BXBBOI8BOIX
Solve the equations:
'i
X + 2y + 3« = 14
3x+ y + 221 = 11
2x+ 3y+ z z= 11.
2 r3a; + 2y = 118
1 x+5i/ = 191.
3jc+12y — 52 + 43 = 0
4x — 17y+22 — 23 = 0
5x — 3y — lOz + 76 = 0.
/• 505 + y — 42 = 0
4. ]3x + 2y+32 = 110
(2x — 3t/+ 2 = 0.
{ (6+c)x— a(y+2) = 6— c
5. } (c+a)y — 6(2-|-x) = c — a
\ (a'\-b)z — c{x-{'y)=za — 6.
rax+ y+ 2 = 1
7. ^a + ay-j-2=m
(. X + y + a2 = n.
Particular cases where (1) a = 0, (2) a = 1.
3. ]3ix +
I2|x +
.2x+ .3y+.42 =29
.3x+ Ay+ .52 = 38
.4x+.5y+.72 = 51.
2 Jx + 3^y + 4\z = 140
4jy + 5i2 = 175
2§x + 3}y + 4|2 = 157.
ax -}- ^i/ + cz = d
a«x + 6«y + c*2 = c?.
r ax + 5y — cz =
10. ^ — ax+ by — cz =^
V ax — by -{- cz z=
— cz = 2ab
2bc
2ae.
ru
11. }u
tt + av + ^''^ = <*'
+ bv + bho = i'
+ Cl? + C*M? =- C^
5 8753,754]
THEORY OF DETERMINANTS
727
753. Homogeneous Equations.
(1) With three unknown quantities.
If c?j = (£jj = {/j = 0 in equations (1), ?762, i. e., if these three equa-
tions are homogeneous, then in case A is not zero the only solution
these equations have isx = y=5j = 0.
(2) With four unknown quantities. But in equation (1), {762,
x=-.^, y=--^, 2=— -|^, we get
a^X+ h^Y+ c^Z+ d^W= 0
a^+ b^Y+ c^Z+ d^W= 0
a^+bJ+c^Z+d^W=0,
Here TTis supposed to be not zero. Hence we will have from (2)
and (3), 1752, after simple transformations,
X —Y Z —W
K
«i
«/»
«i
<^i
'^
«i
6.
rf.
«i
K
«i
\
«.
d^
«•
"*
d.
%
^
d»
««
K
''i
h
*=.
d.
«.
^x
d.
%
h
d»
«.
h
«»
Hence it follows, that X, — Y^ Z, — TT are proportional to the
determinants, which are found by striking out the first, second,
third, and fourth columns of the table,
«»
^
«i
d^
«.
K
«.
<it
«8
h
't
^,
Principle op Addition of Rows oa Columns
764. Fifth Property. — 7%€ elements of a row or column can be
multiplied by any number and added to or subtracted from the
corresponding elements of any other row or column of a determinant
unthout altering the value of the determinant; thus we have
a^ + m6, — nc,
K
«i
«.
^
''i
a, + mi, — nc,
6.
c.
= A =
o.
i.
c.
«s + ^K - "«,
h
''i
«.
^
«8
728 COLLEGE ALGEBRA [W54
Proof.— The first determinant is, according to 2747, equal to
But a^A^ + a^ + a^^ is, according to the first property of determi-
nants, J747, equal to A. The parentheses h^A^ + b^A^ + b^A^ and
c^A^ + c^A^ + Cj^j are, according to the second property of determi-
nants, 2748, equal to zero. Since rows and columns can be inter-
changed without altering the value of the determinant, the theorem
applies to rows as well as to columns. This is a very important
principle in the reduction of determinants. For example:
5
-3
2
4
5
5
-2
-3
1
1
1-1 2! I 1 0 0
I
10 2 —3 = 10 12—23
11 __5 —3
111 6-25
12 -23
6 -25
= - 162.
The second determinant is derived from the first by multiplying the
elements of the third column by 2 and subtracting them from the
first column, and adding the elements of the third column to the
elements of the second column; the third determinant is derived
from the second determinant by adding the first column of the
second determinant to the second column, and subtracting twice the
elements of the first column from the third column. The next step
in the reduction follows from {747.
Example.
1 o a'
16 6'
=
1 c c«
a
a*
b-a
6«— a*
=
c—a
c'-a*
6— a 6«-
(6 — a) (c — a)
= (a-b) {a-c)(c+a-a-b)={a-'b) (]b-c){e-a).
1 6+a
1 c+a
First step, subtract the first row from the second and third rows;
second step, 1747; third step, ?741; fourth step, {784.
Remark.— The principle of addition can be used to give a determinant
of the second order the form of a determinant of the third order.
Thus we have
I
0
0
1
X y
Ti — X
yi-y
==
1
Xi-X
yi-y
^
1
xi y,
x^ — x
y% — y
I
x^ — x
y«-y
1
xt yt
«754]
THEORY OF DETERMINANTS
729
BXBBOISB OZ
a^^ m -\- p b^+m + IT 1
«J + W + i> ^, + » + TT 1
1 10
a, 6, 1
a, 6, 1
1 1 0
a — h in — n x — y
b — c n — 2^ y — ^
c — a p — m z — X
= 0
3.
X a 6-(-c
X b c+o
X c a+6
= 0
C a(b — c)»+ 6(c -a)»+ c(a- 6)» |
I _(a+6+c)(a— 6)(6— c)(c— a)| ~
a(6 — c) 6(c — a) c(a — b)
a — b b — c c — a
c—a a—b 6— c
5.
««+«s-«i ^+^~^ ^«-S-^i
"8+«l^»« ^3+^1-^2 ^3+^1-S
«l+«2-«S ^'l+^-^ ^1+^2-^3
= 4
a^ 6, c,
«2 ^ S
^3 ^ ^3
I •'■3+ '3/3+^*^3 a^j+Zyj— wi«3 x^—ly^—mz^
z^^hn
x^ yi 2,
^i Pi «3
Calculate the values of p^ 5, r, 8 and prove that q* = p (r+*), if
II a
p = 1 b b' , q =
Ice
1 a «»
1 6 i»
1 c c»
1
a
a*
1
b
¥
,« =
I
c
c«
1 a« a^
1 b* 6»
730
COLLEGE ALGEBRA
[J755
The Product of Two Deteeminants op the Third Order
755. The rule for forming the product of two determinants of any
order will be found by verifying the following identity whicb
expresses the product of two determinants of the third order:
(a)
«1
«.
«3
''.
K
h
«l
«.
«.
(0
2 I
This identity may be verified by means of the addition theorem,
2753. If the last determinant is developed by the addition theorem,
we obtain 27 determinants of the following types:
(1)
a^wij 6jWj CjWij
«i^i ^^ ^i^'i
(2)
a^Wj tjWj ^jWijI ,
a^tij ^^ S^
(3)
a^Wj ^j^m^ CjWij
^«1 ^"2 ^s^
(4)
(5)
a^m^ h^m^ c^m^
«3"3 ^^'2 ^3^3
^2 h ^1 h ^3 '3
^2^2 ^1»^1 ^3^^3
S"2 ^I^'l S«3
etc.
Of these determinants, (l)is zero, since on removing the factors a^,
i, , Cj the columns are identical ; similarly, on removing the factors
Oj, 6g, Cj two columns of the determinant (2) are the same and
therefore the determinant (2) is 0 ; determinants (3) and (5) may be
written
(3) «.6.c,
h
h
h
h
h
h
m^
m,
"»s
, (5) a^\c^
m,
m,
^»
"i
«,
"t
".
"1
">
= — ^1«2^3 ^h^n^^*
8756]
THEORY OF DETERMINANTS
731
i?v'liile determinant (4) is zero for the same reasoD that (3) is zero.
If, now, determinant (a) was developed by the usual method,
determinant (3) would be the product of the determinant (/) times
tbe element of the leading diagonal a^ b^ c,, and (5) would be the
product of (0 by the second negative term, b^ c^ a^, of the
development of (a), etc.
Rule. — To form the product T^of two determinants T^ and T^,
first connect with plus signs the constituents in the rows of both the
determinants T^ and T^ . Then place the first row of T^ upon each
row of T^ in order and let each pair of constituents as they touch
become products. This is the first column of T^,
Perform the same operation upon T^ with the second row of T^ to
obtain the second column of T^ and so on.
Determinants of the Fourth Order
756. Consider the system of four linear homogeneous equations
(1)
f a^x + bjf + Cj2 + c^jt? = 0
%x + ^^y + c^z + dfjV = 0
L a^x + 6^ + c^2 + d^v = 0.
Solving the last three equations for x, y, and z by J762 we have
(2) X -y
^.
««
rf.
"%
Cf
d.
^
'»
d>
«j
''z
ds
K
"t
d.
«4
«4
d.
Z
«»
6.
d.
«»
h
^
«4
K
d.
— V
«t
K
Ct
«»
\
<^>
«4
h
''4
The eliminant of the given system is found by substituting these
values of x, y, and z in the first equation; it is
(3
6, c, d^
6j c, d^
K ^ ^*
—b.
Cg c£,
«8 c, e/j
% ^4 ^i
+ ^1
a, 6, rfj
«3 ^8 ^8
% ^4 ^4
-d.
a, 6, c.
a, 6j c,
a b c
4 4 4
= 0.
732
COLLEGE ALGEBRA
[2J757, 758
The first member of this equation is called the determinant of the
fourth order and is written
(4)
6,
c^
d.
= 0.
The coefficients of a^ , 6j , Cj , d^ in equation (3) taken with their
proper signs are respectively the witnor« of o^, 5^, c^, d^ f ound bj
striking out the row and the column which respectively have their
constituents common, and are represented by A^y B^y C,, i>j.
767. Since it is not the purpose of this text to study determi-
nants of an order higher than the fourth we call attention to the fact
that the properties of determinants which have been established for
those of the second and the third orders are general and may be
extended to determinants of any order.
768. The determinant of the fourth order in (4) can be developed
by means of equation (3) ; the development will be found to be
«i ^ S ^i- «i ^
+ «i ^s ^4 ^t— «i ^
-\-a^ hi Cj d^ — a^ h
+02 ^s Ci d^-~ a^ b
-|-aj b^ Cj d^ — a^ h
-j-tt, ^2 c^ d^ — a^ fcj
^4 ^8— «i ^8 ^« 'h+ «i ^i ^% ^s
^8 ^^— «» ^1 ^8 ^4+ S ^1 ^4 ^8
^2 '^3 — O3 by c^ c?2+ a^ 6, Cj d^
c, c^3— a, 62 Ci rf^+ a^ 62 Cj </,
Cj (/j — ^2 ^8 ^4 '^l4~ ^2 b^ C, f?,
^^-
^ ^2 dy+'a^ b^ C2 rf,.
We illustrate the usual method for reducing a determinant of the
fourth order by an example.
Find the value of
3
2
1
4
Subtract the first col-
3
1
1
1
15
29
2
14
umn from the fourth,
15
27
2
-1
16
19
3
17
and the third from the
second, and obtain
16
16
3
1
33
39
8
38
.
33
31
8
5
8«759, 760J THEORY OF DETERMINANTS 733
Subtract the fourth column from the second and the third and obtain
[J766, (3)]
3 0 0
1
0 0 0
1
15 28 3
— 1
18 28 3
— 1
18 28 3
16 15 2
1
=
13 15 4
1
= -1
13 15 4
33 26 3
5
18 26 3
5
18 26 3
Subtract the third row from the first and find
0 2 0,
'I
13 15 4
18 26 3
= 2
13 4|
18 3|
= -66.
Elimination
769. We have already learned how to find the eliminant of a
system of linear equations of two, three, and four unknown quanti-
ties connected respectively by two, three, and four equations.
760. Another important problem in elimination can be solved by
the use of determinants, namely:
By means of ISylvesters Dialytic Method we can find the rational
eliminant for the case when two equations are of any degree.
Example 1. Find the eliminant of the two equations of the
second degree,
(1) rtx« + 6x + c = 0 and (2) a'x« + h'x + c' = 0.
Multiply both (1) and (2) by x', x\ then we have the following system
of equations:
ax* -j- 1)3^ -j- ^^' = ^)
ax? -\- bx^ -j- ex = 0,
a'x* + 6'x' + c'x« = 0,
«'.T»+ h'x^+ c'x=0.
We may regard these equations as linear and homogeneous in the
four variables x*, x^, x*, x considered separately. Hence their
eliminant is
a h c 0
0 a b c
a' b' c' 0
0 a' b' c'
= 0.
734
Example 2.
(1)
(2)
CX)LLEGE ALGEBRA
Given
ax' + ^^ + cx-^d = 0
px* -|- yx -f- r = 0.
[«760
Multiply (1) successively by x*, x, and (2) by x*, x", x. Then we
have the following systems of equations and their eliminant:
ax* + 6x* + cx» + (fx« =0,
ax* + 6x* + c^' + '^*^ = ^>
j>x* + jx* + rx' = 0,
px^ + S'^' -^ ra? =0,
/>x' + yx* -f- rx = 0.
a 6 c ef 0
0 a 6 c <;
;> y r 0 0
0 i> ?
0 0 p ^
r 0
r
= 0.
1. Show that
BXBK0I8B OXI
2. Show that
9
13
17
4
18
28
33
8
30
40
54
13
24 37 46 11
3. Show that
0 7 14
2 3 12
5 0 6 5
8 5 3 10
5. Verify the relations
X X X X
^ y y y
X y c c
X y c d
= -15.
= - 384.
6
15
21
9
8
9
1
4
2
5
7
3
= 0.
Expand
20 2 7 8
7 4 13
0 11 0 5
6 0 18
= x(y— x) {c^y)(d^c)]
110 1
10 11
0 111
1111
=1.
8760]
a — a — a
h h -h
C X c
d X X
THEORY OF DETERMINANTS
1
1
= 2\ahcd\
6. Verify the relations
«0^+«,^+«2^ + «3=
X
0
0
a
-1
X
0
«i
0
— 1
X
«
0
0
-1
a
X X
I X
I m
I m
1
0
1
1
X
X
X
n
735
= -1.
where /, w, n are roots of the equation
a^jt^ + a^x' + a^x + a^ = 0,
Prove that
1111
abed
a« h^ c« d"
a'
* b*
1, a + d, a^-\.a*d+ad^+d^
1, b + d, b^+V^d+bd* + d^
1, c + d, c»+ c«J+ crf»+ J»
where m = (a — d) (b — <£) (c — rf); finally
( (a-ci) (6~J) {c^d)]
=z(a + b+c + d)\ \
8. Eliminate x from Zx* -|- wix + wi = 0 and 7)ix*-|-«x -f- w=0.
9. Eliminate x from ox' -^ 6x -f c =0 and px" -|- yx -f- ^ = 0.
10.
0
1
2
1
0
1
1
3
2
1
1
0
0
2
1
1
= 4
11.
X
x«
X
1
y
1
=
2
1
X* — xy+y', x* — xz-\'Z*
= x«— xy+y«, y',
CHAPTER II
THE CUBIC EQUATION
761. Definition of tiie General Equation of tiie /ith Degree. —
We represent the rational integral expression
X" + p^x-' + p,x"-« + . . . + p^,,x + p,, [1100]
of the n}^ degree in x by the symbol f{x) ; then a rational integral
equation of the «"* degree in x is defined by the equation
(a) f{x) - X" + PjX"-> +i>,x"-« +....+ p^^^x + i>„ = 0.
In the following discussion the coefl3cients Pi, p^- - * are supposed
rational unless otherwise stated.
NoTB 1.— If X** in the equation fix)=0 is multiplied by a constant, po, the equation
can be reduced to the form in (a) by dividing both members of the given equation
by po.
Note 2.— If any of the terms of equation (a) are wanting, the equation is called
incomplete, but if the terms are all present it is called complete.
762. A Root.— Any number, r, which substituted for x makes
f{x) identically 0, i. e, /(r) = 0, is called a root of the equation
/(x) = 0.
763. The Existence of a Root.— It is assumed in this treatise
that every rational integral equation of the n^^ degree, f{x) = 0, has
one root, real or imaginary. The proof of this theorem is too diffi-
cult for a text of the scope of this work, and is given in treatises on
the theory of equations.
764. If r is a root of J\x) = 0, then f{x) is exactly divisible
by X — r. For, if r is a root of the equation f\x) = 0, then, by
{762, if r is substituted for x, /{r) = 0 and the theorem follows
at once from the factor theorem, UOl.
73e
23765-768] THE CUBIC EQUATION 737
766. Conversely, if the first member of the equation /(x) = 0
is divisible by x — r, r is a root of f{x) = 0.
For, if $ is the quotient of f{x) by x -— r, we have
/(x) = (x-r) (?.
Substitute in this equation x = r and we have
/(r) = (r-r)(2^, = 0.
Therefore r is a root of the equation /(x) = 0 since Q ^Ir =7- oo .
766. Definition. — The cubic equation is a rational integral equa-
tion of the third degree and is written
(b) x» + p^x^ + p^x + i>, = 0.
767. The Simplest CasCi Cube Roots of Unity.
Solve the equation
(1) x»— 1 =0 orx = Vl.
The equation x* — 1=0, may be written
(x~l)(x«+x+l) = 0;
. '. either x — 1 = 0, or x'-(-x-(-l=0;
whence Xj = 1, or x^, x^ = — — ^ — ^^—
It is easy to show that
Xj» = x,» = x,» = l;
then unity has three cube roots
^' 2 ' 2 '
the last two of which are imaginary numbers.
The last two of these, x^ and x^ are the roots of the equation
x« + X + 1,
therefore, their product x^ x^ is equal to unity ; then we have
^2 ^8 = 1»
x/x3 = x/.
Since x,' = 1, then x^ = x,'.
Similarly, x^ = x^*.
768. Since each of the imaginary cube roots of unity is the
square of the other root, it is customary to represent the three cube
roots of unity by 1, w, w*.
738 COLLEGE ALGEBRA [H769-7T1
769. The Equation jr^ = a.— Here we have
(2) x'~-a = (x — V^ (x« + Vax+ Va«) = 0.
X — '|/a = 0, and x*+ Va «+ V^" = 0;
hence, x= ya and x= =*=-r = =^^ . v
2 2 ^"-
Therefore the cube roots of a are
]/a, o> y a, «' 'l/a.
770. The Symmetrical Cubic Equation,
(3) ox* + 6x«+ 6x + a = 0,
has already been solved, in 3429.
771. The Cubic Equation with One Rational Root.
Definition, — A real root which is either an integer or a fractioa
is called a rational root or a commensurable root.
It was shown in J764 that if r is a root of /(x)=Q^ then /(jt) is
divisible by x— r. Hence r must be a factor of the constant term
in/(x), (3101, note). Therefore to solve a cubic with one rational
root, find the factors of the constant term and determine which of
these substituted for x will make the first member of the equation 0;
then divide the first member of the given cubic by x minus this
factor (3764). The roots of the two factors equated to zero will be
the roots of the equation.
Example. —Solve the cubic x'+5x«+7x + 2 = 0, which has
one rational root
The factors of 2 are 1, 2, — 1, — 2; of these four numbers, —2
only will make x' + 5 x' + 7 x + 2 = 0. Therefore the first member
of the given equation is divisible by x— (— 2) or x-f-2 (3101, Ex. 1),
Hence we have
x»+ 5 x« + 7 x+ 2 = (x + 2) (x« + 3 x+ 1) = 0;
therefore the roots of the given cubic are those of the factors
equated to zero, or of
x + 2=:0 and x«+3x + l=:0-,
X- 2 ^^-^±^^
• • X — — 4, x= •
i'^72] THE CUBIC EQUATION 739
BXBB0I8B OZn
Find all the roots of the following cubic equations, which have at
least one rational root:
1. x8__4x«_^ x+ 6=0. 2. «» — 6a;^+llaj— 6=0.
3. x» + 8a?+ 5 X — 50=0. 4. x* + 2x«— 23 »+ 6 = 0.
5. x« — 4x«— 15x — 42 = 0. 6. x»— 4x«+ x— 4 = 0.
7. x«— 5x«+ 8x— 6=0. 8. 6x» - 29 x« = 45 — 53x.
9. x»-2|x«+ 2|x-l = 0. 10. x»- |^'-J« + | = 0.
11, 70 + 71x = 47x« — 6x».
In case the cubic equation does not belong to any of the classes
previously discussed and does not have a rational root, then other
considerations are necessary for its solution. In this case we begin
with what is called the reduced form of the cubic equation
(c) a^ = px + q,
where p and q are arbitrary real numbers which may be integral or
rational, positive or negatiire.
A. Cardan's Solution
772. Reduction of the Cubic to the Reduced Form.— The cubic
written in the general form is
(b) , ^+Pi^^ + P2^ + Pi=0. [2766]
The cubic (b) can be reduced to the form (c) by the transformation
X = y — f* . Hence we have
Pi^= Pi}/'-lp*y + ^
p,=«= p,y—^
■ Pz= Pr
Adding together the several terms on each side of the sign of
eqnalitj, we find
«'+l'.x'+;,.x+p,=y- (f -i).) y - (- ^+^'^ - P») = 0;
740 COLLEGE ALGEBRA [2772
or, if we replace ^ — p^ by p and — ''i +^^ — P, by q
we get
y' = /^y + ?>
which is the form required in equation (c) where x takes the place
of Jf.
For example, the equation
is reduced, by replacing y by x— 2, to
a^— 3x + 2 = 0.
The equation
a:» + 21x*+ 146 x+ 335 = 0
is reduced, by replacing x by y — 7, to
y»-y — 1 = 0,
The equation
^ ^ 4 ^ 16 ^ 192
is reduced, by replacing y by Jc— J, to
Hence we may assume that the given cubic has the reduced form
(c) x'=^x + 5.
Put (4) x = V« + V'«
where x equals the sum of the positive real cube roots of s and t
From (4) we have
x* = 3 'r 7t(h/s + 'v 7) + s + t
(5) = 3%^<f •x+«+ f,
replacing 'i « + 'v / by x; but if the values of x and of x* are the
same in equations (4) and (5), then
3 ^v7t ' x + s + t
and P^ + 9
are identical expressions, and we have
3V«^=J>, and 8 + t = q', [2086]
or *' ~ ^ *°^ « + ' = ^-
ill 2] THE CUBIC EQUATION 741
Hence s and t are the roots of the quadratic equation (2422)
(6) ^•-2«+g = 0.
The solution of (6) gives
Hence from (4) we have
(7) X = *^^ + '^^ where r = yjf-^p^
In order to obtain in this way the values of the three roots of the
given equation (c), we put for the real values of the cube roots,
and obtain
(8) Xj = tt + r, X2 = — -^~ + —^- 1/— 3,
If r*, i. e. q* — ^ 2^\ is positive, which is always the case when p
in equation (c) is negative, then the equation a:^ -{- px^q = 0 has
one real and two imaginary roots. *
If r*=0, then we have three real roots, two of which are equal;
one of the equal roots must have one-half the value of the third root,
with the opposite sign.
If r* is negative, r is imaginary, and all of the roots according to
Cardan's formula appear to be imaginary, but in fact in this case all
of the roots are real. This is called the irreducible case, because
we do not yet have the means to reduce the imaginary form to a
real form.
• Cardan's formula does not often give the rational root of a cubic, If it has a mtlonal
root, In a rational form. In order that this may happen. It Is necessary, according to a
discussion due to K. Liebknecht, that the cubic equation shall have the form
a;' = 3 mnx -|- m' -|- n>,
wnere m and n are arbitrary rational numbers. Cardan's formula then gives a; = m + a*
742 COLLEGE ALGEBRA [W73
EXEBOI8S OXm
The following equations have at least one rational root and are
of such a kind that Cardan's formula gives this root in rational form.
Find all the roots of the equations by means of Cardan's aolntion:
1. a:» = 3x + 2. 2. x* = 36x + 91.
3. x^ = dx — 28. 4. sc» + 9x + 26 = 0.
5. a^ - 18x = 35. 6. ar» - 72x - 280 = 0.
The following equations have one real rational or irrational root
Cardan's solution gives also the rational root in an irrational fonn.
In this case the square and (ful)e roots must be found, since fre-
quently the irrational form can only be brought to a rational form
by complicated calculations.
7. ap» = 2jc + 3. 8. x^ = x — 7.
9. x»+ 5^—4 = 0. 10. x'=4x+15.
11. x'+7x — 8 = 0. 12. x»=26x + 60.
The following equations have likewise one real rational or irra-
tional root, but must first be more or less transformed in order that
the Cardan solution can be applied to them :
13. 4x5 — 5x — 6 = 0. 14. 7x» + 3x — 100 = 0.
15. 15x»+ 13.r* — 2= 0. 16. lllx» = 5x*+4.
17. x' — 3x* + 4x-4 = 0. 18. 5x» + 10x« + 7x — 2 = 0.
19. 3x5 ^ i3^« ^ iij. _ 14 = 0. 20. 28x5— 126x*+195x-139=0.
B, Thb Trigonometric Solution
773. The equation
(c) x^=px+q
may be solved in the following manner, by Trigonometry, in 'the
4
irreducible case, when q^ — —p^ is negative.
Assume x = n sin $. Divide the given cubic by n^\ then
Substitute x = n sin ^ ; hence
sin'd — 2. sinfl— ^=0.
But, by Trigonometry, sin' ^-|8in 0+^^—^= 0.
17 74 J THE CUBIC EQUATION 743
Equate the coefficients of the two equations ; the result is
£=f, and a=-?i^;
hence, for o^
IT— -H=.
(1) n = — ^ \Ipj we have sin 3^ = U
where ^ must be found by aid of the trigonometric tables. Having
found $j the three roots of the equation are given by the equations:
(2) x^ = - J|^sind, a:^=- J^p 8in(60-d),
x,= ||p8in(60°+d).
The roots of the equation x^ znpx — q are the same as those in
(2) with opposite signs. It follows from the solution of the equation
a:^=px— q that its roots are all real. The case 7? z=i^px-^q does
not belong to the class of cubics discussed here, since if p is nega-
tive then ^ — —p^ is positive, it belongs to the Cardan solution;
4
but the class of equations discussed requires that j'— 07^^' "< ^*
The following cubic equations have three real roots, some of
them rational and some irrational. It is required to solve the
equations by the trigonometric formulae in (2). If the equation is
not in the given reduced form, then it must first be reduced to
this form.
. Example. — Solve the equation x' = 7x — 5.
Here i> = 7, g = — 5, and sin 3d =-^l=,
7v/|7
.-. — 3 d = 44° 32' 21", and x^= — V^|7 sin d = + 0.7828;
similarly .t^ = + 2.1660; and J", = — 2.9488.
1. ^_7a;__6 = 0. 2. x'=12x+14.
3. x» — 19x+30 = 0. 4. 30x»— 61x«+36 = 0.
5. 4.T'— 13x+6 = 0. 6. 2x» — 5x«— 13x + 30 = 0.
7. 8x»+12x« — 4x- 1 = 0. 8. 27x'- 54 x«+ 25 x + 1 = 0.
774. In solving a cubic equation, one must determine first whether
the cubic has one rational root or not. If the equation has a rational
root it can be readily solved by J771; if it does not have a rational
root, it must be reduced to the form x' = j^x + 3 by J 772, in case it
does not already have this form.
744 COLLEGE ALGEBRA [?775
Then the sign of n^ — ^ «»
must be determined. If this sign is -f-, then Cardan's solution is to
be applied; if this sign is — , the trigonometric solution must be
applied.
C. Trigonometric Solution of Cubic Equations with Two
Imaginary Roots
776, The case in which one only of the roots of the cubic is real
can be handled trigonometrically. This will be the case when the
cubic is written in the form
(1) x» + 3 7x + 2 y = 0,
and I is positive under all conditions ; or in case the cubic is written
in the form
(2) «»— 3?x+2(7 = 0,
where ^ < J* ; hence VP-^q < 1. [?772, (8) ff.]
In the first case put cos 2P = ^ and tan P = '|/tan tf, then
after trigonometric transformations of the values found by Cardan's
formula, 8772, we obtain the following solutions of equation (1):
a;^= — 2Vlcot2Py
X, = Vl cot 2P+ { -^y 0-.. = vl cot 2P- * 4%.
" ' 8m2P ' 8in2P
v> ^^J 4.-^ D_ s.
In case of equation (2) put sin 2 ^ = — — and tan P= ytand,
and find 2 r
EXEBOISE OZrV
Test whether the following equations have one or three real roots
and solve the equations by the trigonometric method which is
applicable.
1. x» + 2x = 357. 2. 4ar»— .7x=87.
3. x3-5x = 12. 4. ar»— 9a; + 10 = 0.
5. x»+ 3 x2__ 7 a. _ g4 g 2 x»+ 5 X* = 36.
7. 3x5 — 2x«=i^. g g3j»__8x« = 9.
9. 20x»+llx = 8. 10. 7x»+9x«— 2 = 0.
CHAPTER III
THE EQUATION OF THE FOURTH DEGREE
776. Equations of the fourth degree which can be solved by
means of the quadratic equation have for the most part been solved
in Chapter VII, Book IV. Their solution offered no difficulty.
If an equation of the fourth degree has one rational root, it may
be found by the same method used in connection with the cubic
equation, J771. If this root is r, then the equation of the fourth
degree can be divided by a; — r and depressed to an equation of the
third degree, which we have already learned to solve in its most
general form. If the equation of the fourth degree has two rational
roots, then it can be depressed to an equation of the second degree,
which can be readily solved.
Example. — Solve the equation ar* + 4x^ — 6a^ + 24x — 72 = 0.
The factors of 72 are 2,-2, 3,-3 and combinations of these
by twos, threes, and fours. On trial it is found that 2 is a root of
the given equation, and on dividing it by x — 2 the depressed equa-
tion is the cubic
ar^ + 6x« + 6x + 36 = 0.
The factors of 36 are 2, — 2, 3, — 3 and combinations of these by
twos and threes. On trial it is found that — 6 is a root of the
cubic. Dividing the cubic by x +. 6, the depressed equation is
.T« + 6 = 0 or a; = ^i/ITe.
Hence the roots of the given equation are 2, — 6, i/-— 6.
745
746 COLLEGE ALGEBRA [1777
BXSBOISB OZV
The following equations are of the character just described; they
have at least one or two rational roots. These may be found
according to the rule illustrated in J771 ; then the remaining roots
are to be found by depressing the equation as illustrated in J776.
1. X* — 3x» — 34x« + 18a; + 168 = 0.
2. x*+4x»— 6x« + 24x — 72 = 0.
3. X* + 4x» — ll.x« — 12x + 90 = 0.
4. ar* — 13x« + 48.r - 60 = 0.
5. X*— 6x»+ 24^ — 16 = 0.
6. X* + 3x5 _ 52^« _ 60x + 288 = 0.
7. X* + 7x' — 13x« ~ 175x — 300 = 0.
8. X* — 1 7x3 + 95^2 __ 199j; _^ 120 = 0.
9. X* + 19x» + 123x« + 305x + 200 = 0.
10. X*— 2x«— 7x«+ 19x — 10 = 0.
11. x* + 5x'»+ 12x»+52x-40 = 0.
12. X* — 9x'+20x«— 13x+6 = 0.
13. X*— 17x« + x + 20 = 0.
14. 2x* — 13x'+ 16x«-9x+20 = 0.
15. 3x* — 8x' — 36x« + 25 = 0.
16. 6x* — x3— 8x* — 14x + 12 = 0.
17. 6x*— 13x3+ 20.c»—37x+ 24 = 0.
18. 6x* — x3 - 49x« + 55x — 50 = 0.
19. 10x*+ 17x»— lGx* + 2x — 20 = 0.
20. 12x* + 5x» - 23x2 _ 5x + 6 = 0.
21. 26x* — 108x' + 323x« — 241x + 60 = 0.
22. 36.r* — 72x' — 31x« + 67x + 30 = 0.
777. Resolvent Cubic. — In order to solve the equation of the
fourth degree in general, one must first reduce it to the solution of
a cubic equation. Then, by means of the roots of this equation,
which is called the resolvent of the given equation, we have to deter-
mine the roots of the given equation.
«778] THE EQUATION OF THE FOURTH DEGREE 747
778. In order to solve the equation of the fourth degree,
(1) ax* + 4 6x» + 6cx« + 4 cfx + c = 0
in general, multiply it by a ; thus
(2) a«x* + 4 a6x» + 6acx« + 4 adx + ea = 0.
This equation can be written in the form
(3) (ax« + 2 6x + c + 2 0* — (2 Px + 0« = 0,
where f, /*, Q are to be determined. Equation (3) developed and
arranged with respect to x is
(4) a«x*+4a6x'+(4«^+4a^+2ac-4/^)x»+(46c+86«— 4P§)x
-|-c«+4<«+4c^— ^=0.
On comparing like powers of x in equations (2) and (4) we have
(5) P=Vb^-ac + at, (6) Q :=zV{c + 2t)* - ae,
(7) />$ = 6c + 2 6« — ad.
Eliminating Pand Q from (5), (6), and (7) we have the resolvent cubic
(8) 4 ^~ (a€ — 46J + 3€«) t+ace + 2hcd^ ad^— h*e — c^z=i 0.
The biquadratic equation (3) is equivalent to the two equations:
(9) ax* + 2(6 — P)x + c + 2<— Q = 0,
(10) ax« + 2(6+P)x+c + 2f+ Q = 0,
If Xj and Xj are the roots of (9) and x, and x^ are the roots of (10),
then we have
(11) u-.+ x,= -?(6-i'), x, + x. = -?(6 + P).
Since there are three values of P corresponding to the three values
of /, equation (11) includes the following systems:
^. + ^, = - ' (^- p.) «=,+ ^. = -! («'+ ^.)
*»+ *3 = - ! (^-■p.) ^.+ ^. = -!(6+i',)
748 COLLEGE ALGEBRA [I77S
Whence follow the values of the root x^, ar^, x^^ x^.
^. = ; (- ^ + A + ^. + ^J
^. = 5 (- «» + ^x - A - A)
•^3 = i (- ^ - A + ^. - n)
(12)
Siace the P's in (12) are square roots, they may have positive as
well as negative signs; the choice of signs which must be made
is determined by the circumstance that the relations
(13) x^x^^x^= ^, {x^ + x^) x^x^ + {x^ + x^) x^x^ = — ^,
between the roots and the coefficients of equation (1) are satisfied
under all conditions.
The relations in (13) are due to the following theorem, which is
true for any rational integral equation in x (2796):
Theorem. — In any rational integral equation in x, the coefficient
of whose highest term is unity, the coefficient p^ of the second term
with its sign cJianged is equal to the sum of the roots.
The coefficient p^ of the third term is equal to the sum of the
products of the roots taken two by two.
The coefficient p^ of the third term with its sign changed i$ the
sum of the products of the roots taken three at a time in all possi-
ble icays; and so on, the signs of the coefficients taken alternately
negative and positive, and the number of the roots multiplied together
in each term of the corresponding expression of the roots increasing by
unify, till Ji nail y that expression is reached which consists of the pro-
duct of the n roofs.
If all of the roots x^, x<^, Xg, x^ are real, which can happen only
in case all the F's are real, and consequently if the three values of
t are real, then the given equation can be separated in three
ways into two quadratic factors with real coefficients. The imagi-
nary roots must of necessity be conjugates and enter in pairs, because
an imaginary root arises from the extraction of a square root, and
the root may be positive or negative. If Xj and x.^ are a pair of con-
2778] THE EQUATION OF THE FOURTH DEGREE 749
jugate roots, then Xj -[- x.^ and also x, x.^ are real, and equation (7)
has also in this case real coefficients. Hence one decomposition
(and one only) in two quadratic equations with real coefficients is
possible where the roots of the given equation are all or in part
imaginary.
If, therefore, the resolvent cubic has three real roots, there is
one of them which gives a real P and, on account of (7), a real Q,
If the resolvent cubic has but one real root, then this real root must
also furnish a real P\ for only in such a case can there be a decom-
position in two real quadratic factors which have real coefficients.
In case the biquadratic equation has i = 0, then the resolvent
cubic is
(c — t) {ae — (c + 20*) = <«/* and P = Va(t — c).
EZSBCISE OXVI
The following equations of the fourth degree (1 — 34) may in at
least one way be separated rationally into quadratic factors. T*)e
resolvent cubic has therefore at least one rational root, which can be
determined simply in the usual way; the expression for Pcan be at
once calculated, and the calculation of the roots of the given
equation follows without any great difficulty.
1. X* — 4x3+20x — 25 = 0.
2. X* — 2x» — X + 2 = 0.
' X* — 4x» + 6,r« + 12x —15 = 0.
a* — 6x» + 12x« — 14x + 3 = 0.
X* — 4x» + 3x« + 4x — 4 = 0.
^. X* — 4x'+9x«-8x + 14 = 0.
7. X* — 12x«+51x« — 88x + 48 = 0.
8. X* — 8x»+4x2+ 24x — 21 = 0.
9. X* — 8x» + 8.c« + 40x — 32 = 0.
.0. X* - 12x3 _(. 47 ,.« _ 72.r + 36 = 0.
11. X*— 16.^3+ 70x« — 60x — 88 = 0.
12. X* — 37x* - 24x + 180 = 0.
13. X* — 6x»+ 17x«-24x+ 18 = 0.
14. X* - 8x8+ 21x«—34x+ 20 = 0.
15. X*— 6x'+ 16x2 — 20x + 12 = 0.
16. x^ + Sx' — 7x'^— 50x + 48=0.
17- a*- 12.r3+ 43.r2-42x+10 = 0.
9
10
760 COLLEGE ALGEBRA 1*778
18. ie* — 6a:»+13x« — 30.T + 40 = 0.
19. a^+x» — 14x» — 2a; + 24 = 0.
20. aj* — 3x8 — 5x« + 29x — 30=0.
21. a*— 6a;» + 3x«+26x— 24 = 0.
22. X* — 2x»— 10x«+6x+ 45 = 0.
aj* + llx» + 35x«+ 13x— 60 = 0.
a^ _ 3.r» — 5x« + 29x — 30 = 0.
x4_llx»+ 47x« — 97x+84 = 0.
X* — 9x» - 5x« + 153x — 140 = 0.
X* — 13x»+ llx« + 337x - 840 = 0,
23.
24.
25.
26.
27. ^ , _^ , ^^_ ^..
28. 4x* — 8x»+ 5x« — 8x + 1 = 0.
29. 4x* — 12x« + 31x« — 60x + 55 = 0.
30. 4.C* — 16x» + 15x« + 5x — 7 = 0.
31. 9x* + 15x« — 143x« + 41x + 30 = 0.
32. 16x* — 48x« + 80x» — 60x + 27 = 0.
33. 16x* — 80x« + 136x« — 108x + 45 = 0.
34. 16x* — 32x» — 32x« + 40x + 15 = 0.
The following equations can not be decomposed rationally into
quadratic factors, hence the resolvent cubic does not have anj
rational root The cubic equation must be solved by the method of
Cardan or by the trigonometric method.
35.
36.
37.
38.
39.
40. x'
41. x'
42
43
a;*_4.T»+ 12 = 0.
X* — 24x + 37 = 0.
3x*— 5x» = 31.
4x* + 7x+ 100 = 0.
o;4_4aj8_|.6x«— 8=0.
;* _ 8x» + 4x — 7 = 0.
^4 _ 12x» — 16x + 41 = 0.
3x* — 2x8 — 7x -(- 20 = 0.
2x* = 7x8 + 5x« + 30.
44. 5x*=3x» — 14x — 100.
45.
46
47.
OX' = iJX" — l-lx — lUU.
7x*— 8x8 _^ 24x8— 4x+ 37 _ q
3x* — 2.c8 — 21x8 L. 4x+ 11 = 0.
4Y. 2.C* — 3x8 — 7x8 + 8x+ 21 = 0.
48. 5x* — 7x8 - 30x« + 8x + 28 = 0.
-" 7x* + 8x8 _ 12x8 _ 5x — 16 = 0.
llx* + 7x8 _ 44a:« — 8x + 23 = 0.
8x* — 62x8 -)- 162x8 — 172x + 63 = 0.
49.
50.
51.
CHAPTER IV
THE n ROOTS OF UNITY
779. Given the equation
cos n A + I sin « ^ = (cos .1 -f i sin AY^ i = l/— 1,
it is proposed to determine the values of w^ from the sine and cosine
of w^l. It will be found that there are n different values of A which
ffive different values of ....
^ cos A -{- I sin A,
and which equally satisfy the equation.
For, by Trigonometry, we know that the terms of the two series
nA, 2 7r-)- « j4, 4t +?^ A, . . . 2(u — 1) «•+ nA, 2« tt -\- nA,
V — w -1, Stt — nA, bv—n J, . . . 2(h — 1) v—nA, (2 n -\-l) ir—nA,
and no others have the same sines, and those of the two series
n A, 2ir+ nA, 4 tr + n A, . . , . 2 {n — 1) v + n A, 2 /i tt -(- n j4,
— n A, 2ir— n A, 4 ir — n ^1, . . . . 2 (/i — 1) ir — » vl, 2 « ir — w ^,
and no others have the same cosines; hence it follows that the terms
of the series
n A, 2'!r+nA, 4w + n A, , . . . 2 {n — l)+ n A, 2 w ir + ?i A,
and no others have simultaneously the same sines and cosines, and
hence each gives the same value for the expression
cos n A -)- I sin n A
when both cosine nA and sine nA are given. Therefore the n terms
of the series . , o / n ^
^» n +^*' n ^^^' n ^ ^'
and no others, give different values for cosine A and sine A (one or
both), and hence different values for
cos A -j- i sin .1 ;
but thoy give the same values to both terms of the expression
cos n A -\- i sin n A.
751
752 COLLEGE ALGEBRA [W80
Therefore n values of this series are the only values of A which
satisfy the equation
(cos A -{- i sin A)*=cos tiA + i sin n A.
Hence it follows that any one of the n expressions
(1) cos ^ + 1* sin J,
(2) cos (^ + J. ) + I sin(^ + .i),
(3) cos(i^ + ^) + I- sin i^-^ + ^y
(«^1) cos{2(-^- +Al+ ,-sm|2i^^ +^j.
{ n S I n i
raised to the n^ power will give the same value for the expression
cos n A + 1* sin n A,
provided n A is to be determined from the value of its sine and cosine.
Note.— Since sin (2t— y)= — sin x and cos(2x — x) =cos x, it follows ihit
cos [^A'^Pl- +^] + i sin p<^-P)^ + a]
= cos (^^JL-a) - i sin (^P^^a), p=U2 i»,
and therefore that the n different values In the preceding table can be reduced to
equivalent expressions, which involve no angle greater than 180».
780. Solution of the Equation jr''= 1.— The roots of the equation
X"— 1=0 may readily be found from the equation
(1) [cos(^ + A'j+ i sin (2^ + Ayj"= cos n .4 + i* sin n A.
where, according to the preceding table, ^=0, 1, 2, . . . . n— 1.
Put yl = 0, and hence cos n A = l] sin n A=:0, and we have
Ccos^^ + t8in2Jl£y = i,
\ n n /
and therefore
(2) ' cos^i^ + *8in2i^=(l)i,
n n
where 7> = 0, 1, 2, ....« — 1, giving n values which are different,
and no more.
«781-784] THE 71 ROOTS OF UNITY 753
781. The Cube Roots of Unity.— Let n = 3 and .1 = 0, then it
follows from (2) and table in 2779 that the cube roots of unity are
expressed by
(1) cos 0 + t sin 0 ~ 1,
(2) cos ^+ iQin^^ cos 120° + % sin 120° :=z ~ ^ "t^**
(3) cos ^ + I sin ^ = cos 120° — i* sin 120° — ~- -7-^^-
These results agree with those already given in {767.
782. The Biquadratic Roots of Unity.— Let n = 4 in (2), {780,
and table in {779, and the four roots are expressed by
(1) cos 0 + I* sin 0 = 1,
(2) cos^ + isin ^=0+1 = 1,
(3) cos -j^ + t sin -j^ = — 1,
(4) cos ^ + / sin ^ = cos 270° + i sin 270° = — t.
4 4
783. The Quinary Roots of 1.— Let «=5 in (2), {780, and table
in {779, and the required roots are
(1) cos 0 + I sin 0=1,
(2) cos^+tsin^^- = cos 72° + / sin 72°= .309 + VX.951,
5 5
(3) cos ^ + I sin ^ = cos 1 44° + / sin 1 44°,
o 5
= _cos 36° + /sin 36° = — .809— tX.588,
(4) cos^' — isin^ = cos 72° - / sin 72° = — .309 — ix.951,
o o
(5) cos"^,' — i sin^ = — cos 36° — i sin 36° = —.809 — eX.587.
5 5
Geometrical Representation op Complex Numbers by Points
784. The Correspondence Between the Complex Number Sys-
tem and the Points of a Plane. — All the numbers included in the
system of complex numbers a+i'6 can be represented by the
points in a plane.
754
COLLEGE ALGEBRA
LH785-787
Let XOX' and TOY' be two
perpendicular lines lying in the same
plane and intersecting in O,
Suppose that XOX' is the axis of
real numbers and let the real numbeis
be represented by the points of XOX*
as described in 331, and suppose that
YOY' is the axis of pure imaginary
numbers, representing ih by the point
of 0 y whose distance from 0 is 6
in case h is positive, and at the same
distance from 0 on OY' when h is
negative.
To construct the point corresponding to the complex number
a + tft, where a and h are positive, lay off on 0 JT, OM = a, and on
the perpendicular to OX at M, MP = h.
There is a one-to-one correspondence between the numbers of the
complex system (a + ih) and the points of the plane. To every
complex number there corresponds one and but one point in the
plane, and to each point of the plane there corresponds one and bat
one complex number.
If the point P is made to move about in the plane OM^ and PM
varies, then a-\- ih varies and is called a complex variable^ usually
written in this case x -|- ly.
786. Modulus,— The length of the line OP, which is l/a* + 6", is
called the modulus of a + ih. Represent it by r, i. e.,
(1) r = V a'+h\
786. Argument. — The argument of the number a + t6 is the
positive angle XOP. Represent its numerical value by A.
The angle A is always measured in the positive sense from XO
toward the modular line OP.
787. Sine. — The ratio of PA^ the perpendicular from any point
P in the modular line to the axis of real numbers, to the distance
of P from 0 is called the sine of A and is written sin -1, L e.,
(2) ^ = sin .1 .
r
Sin A is by definition positive when P lies above the axis of real
numbers, and negative when P lies below.
81788-790] THE U ROOTS OF UNITY 755
788. Cosine. — The ratio of the distance of M from 0 (or of
PN ± OF) to OP is the cosine of A, written cos.i, i. e.,
(3) ^ = cosA.
r
The cos^ is positive by definition when M lies to the right, and
negative when M lies to the left, of the axis of pure imaginary', YOY*,
789. The expression of a -j- ib in terms of the modulus r and
argument A \% r (cos^ + i ^XuA).
For, by J788, - = cos^ .-. az=:r cosA ;
and, by J 787, = sin.l .-. hz=z r sin J.
Hence, a + t6 = r cos A + ir sin^ = r (cosX + i sini4.) If r
is fixed and A varies from 0 to 2 tt, the point P will move about the
circumference of the circle PBAP counter clock-wise.
Whenyl = ^,
(4) cos| + i8in^ = ?;
and when ^ = tt ,
(5) cos ^ + 1 sin ?r = _- 1.
In the first case, a = 0, 6 = 1, and P is at ^; in the second case,
a = — 1, 6 = 0, and P is at .4 '.
The sign of r in (1) is always taken positive in the expression
a + t6 = r (cos.i + i sin yl), and is called the absolute value of a -\-ih
and represents the distance of the corresponding point P from O.
Of two complex numbers, that is the greater whose corresponding
point is at the greater distance from 0.
The expression cos A + / sin A has the same kind of geometrical
meaning as + and — , which are simply particular cases of it, namely,
in the first case ^ = 0, cos 0 + t sin 0 = + 1> *^^ ^^^ second case
^ = », cos » + I sin IT = — 1.
790. Geometrical Representation of the n Roots of 1 (Fig. 2).
1. Two roots of 1, X = V 1.
Here cos -4-f i sin -1~ zfc 1, hence, ^4 = 0 or ir and r = 1.
The roots + 1 and — 1 are represented by the points f\ and P^,
which bisect the circumference.
756 COLLEGE ALGEBRA [1790
2. Cube roots of 1, a = Vl. Here the roots are ({781)
cos^ + i smA
r= 1
^ = 0,
cos J!! 4- V sin y
r = 1
A = 120°
r = 1
^ = 240^
which correspond respectively to the points P^, Q^y Q^,
The points /\ , Q^, Q^ divide the circumference into three equal parts.
3. The biquadratic roots of unity are (2782)
r +1
r -1
{ + *■
— »■
^ = 0 -
A = -r ■
A=l ■
i _3jr
^-2
[ r=l,
[ r = l,
[ r=l,
[ r=l,
which correspond respectively to the points P^, By P^y B\ which
divide the circumference into four equal parts.
4. The n roots of 1 are given in the table in J779. It will be
noticed on comparing the ai^uments for the various roots that the
difference between any two consecutive arguments is \ . Therefore
the n points which correspond to the n roots will divide the circoxn-
ference into n equal parts.
EZEBCISE OZVn
1. Show that the septenary roots of 1 are
1, . 623 + i . 782, - . 222 + t . 975,
—.893 — % .450, —.222 - * .975,
2. Find the six roots of 1.
3. Show that n roots of — 1 are found by giving the following
values tor, ^^q, 1, 2, 3, .. .r.-l,
in the expression cos ^ ^ ' ^^ + % sin ^ ^\ '^.
4. Give a geometrical representation of the quinary, sextenaiy,
and septenary roots of unity.
—.893 + 1.450,
.623 — t. 782.
CHAPTER V
THEORY OF EQUATIONS
Pbopertieb of Equations
791. First Property.— 7/* r is a root of f(jr) = 0, then /{x) is
exactly divmble by x — r (J764).
792. Second Property. — If the first memher of the equation
f{x) = 0 M divisible by x — r, r is a root of f\x) = 0 ({766).
798. Third Property. — Every equation of the n^ degree lias n
roots and no more. Consider the equation
Ax)^x-+p^x-'+p^x-*+ . . . +p„-ix+p„ = 0.
If a^, real or imaginary, is a root of f{x) = 0, then f{x) is exactly
divisible by x — a^. Let the quotient be Q^ {x), then one may write
f{x) = {x-a;) Q^{x),
If a, is another root, real or imaginary, of /(.r) = 0, it must lye a
root of Q^{x) = 0, hence Qj{x) is exactly divisible by x — a, and
therefore we may write
fix) = {x — a^) {x — a^) Q^ (x)
where the second quotient Q^ (x) is of the degree n — 2.
Continuing this process we obtain
fix) = (x — a^) ix — ttj) . . . . (x — a„).
Since fix) vanishes if x is made successively equal to the n values
a^y Oj^y a„, fix) has n roots ; and the equation has no more
than n roots, for, if any value of x, which is not one of the n
values a^, a^, a^ . . . a„, say x=:a, is substituted for x, we have
fia) = (rt — a^) (a — rig) .... (a — rrj,
which can not vanish since none of the factors of the product
vanishes.
757
758 COLLEGE ALGEBRA [«794-796
794. Fourth Property. — To depress an equation. If one of
the roots of the equation f{x) = 0 is known, the equation may by
division be depressed to an equation of the next lower degree, which
contains the remaining roots. Thus, ^^(x), in ?793, contains all
of the n roots of /(x) = 0, except a^. Thus, if r rootd are known,
the degree of the depressed equation is («— r). If all the roots but
two are known the depressed equation is a quadratic, which can be
readily solved.
796. Fifth Property. — To form an equation whose root* are
known. Since we have
/(x )= (x -^ a^ (x - a,) .... (x - a„), [J798]
it is evident that an equation can be formed by subtracting each
root from x and placing the continued product of the binomial
factors equal to 0.
Example. —Form the equation whose roots are 1, 2, — J, —J.
(x-l)(x-2)(x + p(x + J) = 0,
• 8x*— 14x'— llx« + llx+6 = 0.
796. Sixth Property. — The relations between tJie coefficients oj
/(x) and the roots o//(x) = 0. Let the equation be
Ax) = x-+p^x-'-'+p^x-'*+ +p, = 0,
whose roots are «i ,«,,.... a„ ; then we have
/{x) = (x — a,) (x - a J. ... (x - a J [?793]
^x"-.S>"-^+.S;x-2_,+, . . . +(_l)-i ^;.,x+(-l)-^,.
[«666]
Equating the coefficients of like powers of x (3666),
where
'S^i = «! + «,+ • • • +«n; [5623]
and S^ is the sum of the products of the // roots taken two at a time
and so on till aV„ = a^a^ . . . a„.
82797-799] THEORY OF EQUATIONS 759
Thus. The sum of the roots equals the coefficient of the second
temi with the sign changed.
The sum of the products of the roots taken two at a time is the
coefficient of the third term.
The sum of the products of the roots taken three at a time is the
coefficient of the fourth term tcith its sign changed, ami so on.
The product of the roots equals the last term, taken with the positive
or negative sign according as the number of roots is even or odd,
797. Although the relations just derived will not in general
determine the root^ of any proposed equation, yet by means of them
we can derive many relations which are of value in solving various
problems.
Let a^ , GTj , a^ , be the roots of the equation
x^ + p^a^+p^x+2>^ = 0;
hence, we have — p^z= a^-\- a^-^- a^^
Thus i>,« - 2i>, = a,« + a/ + a^* + 2o,a, + 2a^a^ + 2a^a^
^2{a^a^ + a^a^+a^a^),
= «,*+«,' + <.
Hence, if in a cubic p^ — 2/?^ is negative, the roots of the equation
can not all be real.
798. The relations connecting the roots and coefficients of an
equation sometimes enable one to find all the solutions of the
equations in case the roots are required to satisfy assigned relations.
799. The Sixth Property. — In an equation with real coefficients^
imaginary roots enter in pairs.
Let the coefficients in the equation
f{x)^x^ + p^x+p^x^+ .... +i>n=0
be real; then, if a + 6i is a root of /(x) = 0, we are to prove that
a — hi is also a root.
Since a + hi is by hypothesis a root of f{x) = 0, then on
replacing x by a -j- 6i, this equation will take the form P-j- ^ 6t = 0,
where P and Q involve even powers of h. Because, for example,
if x^ is expanded, where x = a + 6i, the even powers of hi give rise
760 COLLEGE ALGEBRA [«800, 801
to real terms, so that i can occur only in connection with odd
powers of h ; thus
x8 = (a + 60» = a'+ 3a« • hi + Za{hi)^+{hif
= (a»_ 3a6«) +(3 6 a« — fe») t".
By hypothesis the coeflScients of /(x) are real and hence i can not
occur except with some odd power of h. The result of substituting
a — hi for X in f{x) is therefore obtained by changing the sign of
h in the expression found on substituting a + hi for x in f{x)\
the result is therefore P— Qhi, Since we had
P+Qhi = {),
and, since P and Q are real, therefore, P=0, ^ = 0. [i394,CoE.j
Hence, P — Q ht^Q-,
therefore, a — hi is also a root of /(x)=0.
800. According to the preceding article, if a + ih is a root of
/(.r) = 0, a — hi is also a root and f{x) is divisible by x — (a + i6),
also by x — {a — hi)^ that is by
[x_(a+6i)] [x-^{a-hi)] = [{x-a)+hi'] [(x-a)-ti]
a quadratic factor which is always positive for real values of x.
Let a it ^h ^ ± ^'i ^ ±/'» ^® ^® imaginary roots of /{x) — 0,
and /(x) be the product of the quadratic factors which correspond to
the pairs of imaginary roots; then
/(x) = [(x-«)» + i«] [(x-c)«+cf«] [(x - e)«+/«].
Since each of these factors is positive for all real values of ar,
therefore /(x) is always positive for real values of x.
801. A proof exactly similar to that given in 2799 shows that
some roots of the form a ± i/ 6 enter in pairs into equations whose
coefficients are rational.
Example. — Solve the equation x*+ 2x'— 5x'4-6x+2=0, which
has the root — 2-|-i^3.
Since — 2 + v 3 is a root, then — 2 — i/3 is also a root; and
therefore the equation is divisible by
[x-(-2+v 3)] [.r-(-2-V3)] = [(x+2)-l/3] [(x+2)+V3]
= (x+2)« — 3 = x« + 4x+l.
JJ802-804] THEORY OF EQUATIONS 761
Hence the remaining roots of the equation are obtained by solving
the equation
thus ac, = 1 + V — 1 and x^ = 1 — j/— 1.
I, Fractional Roots. — A rational fraction can not he a root of
an equation whose coefficients are integers and the coefficient of the first
term is unity.
For, let the rational fraction ^ i which is in its lowest term, be a
root of the equation
which multiplied by i""* becomes
This equation requires that a fraction, — ^ , which is in its lowest
term, is equal to an integer, which is impossible. Therefore a rational
fraction can not be a root of the ^ven equation.
Transformation op Equations
803. A given equation may be transformed into another whose
roots are related in some definite manner to those of the given equa-
tion, and, as will be seen as our discussion is developed, we need
not know the roots of the given equation. Later illustrations will be
given, showing that such transformation may be used in the solutions
of equations.
804. First Transformation. — To transform a given equation into
another whose roots are equal to those of the given equation tcith opposite
signs. JjetfXx) = 0, ?761, be the given equation and put x = — y,
so that for every value of r, y has an equal value with contrary
signs ; hence, substituting x =— y in f{x) = 0, we have
f{x)-A^tj)^(-yr+p^{-yy+p^{-yr-'+ . . . .-p^_^y+p^ = 0,
that is y«_p^y«-i-f p^7/»-«__ . . . ±i'„-iy=Fi>„ = 0.
Thus the transformed equation is obtained from the given equation
by changing the sign of ever}' other term, beginning with the second.
762 COLLEGE ALGEBRA [«805-807
806. The preceding rule is applicable to a complete eqnation.
If the equation is not complete, for example suppose that some of
the coefficients are 0, as in the equation
a;«_|- 4x« — 5x» — 3x + 5 = 0,
then to transform this equation into another whose roots have signs
contrary to those of the given equation, we may write it thus
aj« + 4x* ± Oa^ — 5a:»± 0.r« _ 3a: + 5 = 0.
Therefore, according to the rule, 3804, we have
x« _ 4x^ -t Ox* + 5ar» ± Ox« -f 3x + 5 = 0,
or «• — 4x* + 52:* + 3aj + 5 = 0.
806. Second Transformation. — To tranaform an equation into
aiiother whose roots are equal to those of the given equation mulfipiicd
hy a given number.
Let the given equation be /{x) = 0 and the given multiplier be k\
hence, if we put y = A-.r, then for any value of x the corresponding
value of y is A; times as large.
Solving y ^= kx for x, we get » = ? and substituting this value
for X infix) = 0, the transformed equation is
% + P^^.+P»'[^^+ • ■ • • +Pn-rl + P. = 0,
or on multiplying by k* we have
y-+p,ky-'+pJ.^y-'*+ +;>»-iA:"-^y+i>n^"=0.
Hence, the transformed equation is derived from the given equation
by multiplying the coefficient of the second term by k, that of the
third term by A;*, and so on.
807. The Transformation of an Equation with Fractional
Coefficients into another whose Coefficients are Integers is one of
the most valuable applications of the preceding principle.
Example. —Transform the equation
(1) ar'— — + — — - = 0
into another whose coefficients are integers.
Put X = ? in (1), t]
k
formed equation is
Put X = I in (1), then, according to the rule in {806, the trans-
(2) i/'-Y/+fA:V-|A:' = 0.
{J808,809] THEORY OF EQUATIONS 763
Put in (2) A; = 2 • 3 = 6, . •. Aj» = 8 • 27 and equation (2) becomes
y'-|(2-3)y'+|(2«-3«)y-?(2.3)»=0,
or (3) y»-9y»+45y — 48 = 0.
The roots of equation (3) are thus six times as large as those of
equation (1).
808. Third Transformation. — To diminish or increase the roots
of an equation hy a given number.
Let the given equation be y(j;)=0; then, to transform this equation
into another whose roots are diminished 1)y a given number, k^ put
;y=x — k. Therefore, for any value of a*., y is less than x by k.
From y = X — /j we have x=y+/c, and the transformed equation is
(1) yT[y + A-) = 0.
Similarly, in. case the roots are to be increased by A;, put y = x + /c,
whence x=y— A; and the transformed equation is
(2) /(y-A:) = 0.
809. When the degree of the equation f(x) = 0 is greater than
3, the calculation indicated in the preceding article becomes laborious
and a more simple mode of effecting the transformation is desirable.
(1) Letyi:x) = x»+p^x»-»+;>,x"-«+ . . . + Pn-i ^ +i>n = 0,
and suppose that the transformed polynomial in y is, by 8808, (1),
(2) Ay-\-k)-y- + q,y''-'+q^y''''-\- • • • +qn-iy + yn\
since y = x — A-, expression (2) is equivalent to
(3) (x-^-)"+gl(x-A:)»-*+(7,(.r-A:)"-«+ . . . +^„_, {x^k) + q,,
which must l^e identical with the given polynomial in (1). Hence if
the given polynomial is divided by x — Aj the remainder is j„ , and
the quotient
{x-kY''J^q^{x-kY-*+ . . . +g„_^;
and if this polynomial is divided by x — A:, the remainder is g„_,
and the quotient
(x^kY-^ + q^{x^ky+ . . . +^„_,.
Thus, it follows that by the repetition of these arithmetical opera-
tions, the successive remainders are
?n» S'n-U qn-2 - - - -
Hence, the transformed equation may be determined by the rule;
764
COLLEGE ALGEBRA
[«810
Divide /{x) by x^k according as the roots of the given equation
are to he diminished or increased by k, and the remainder is the last
term of the transformed equation. Divide the quotient just found by
X — kf and the remainder is the coejicient of the term next the last nf
the required equation; and so on.
The numerical calculation involved in the application of this role
is much abbreviated by synthetic division.
810. Synthetic Division Divide x'— 5x* + 6x» + 3x« — 8x — 3
by X* — Sx" -f- 4x -}- 5, using detached coefficients.
l + O — 3 + 4+ 5) 1+0 — 5 + 0 + 6+3-8- 3( 1 + 0 — 2— 4
—1+0+3-4—5
0—2-4+1+3 —8
2+0-6 + 8 + 10
— 4-5+11 + 2 -3
4 + 0 — 12 + 16 +20
— 5-1 +18 +17.
Quotient x* — 2x — 4. Remainder — 5x' — x* + 18x + 17.
NoTB.— Tbo signs of the partial dividends are changed and then added.
Syntubtic Division
Using the last example, the work is arranged as follows :
Divisor
1
— 0
+ 3
— 4
— 5
Quotient
1 +
0 —
5 +
0
0 +
3-
4
+
0 +
0
+
0
1+0-2—4
+6+3-8-3
- 5
+ 0 + 0
_ 6 + 8 + 10
+ 0-12+16 + 20
5— 1+18 + 17 Remainder.
NoTB.— The coefficients of all the powers of the unknown number are written down
and the signs of the terms of the divisor excepting the first are changed.
Example.— Find the quotient of «* -H 4a:' — a« -f 11 by « — 3.
Il-f-O-h 4- 1-h 0-h 11
-f3[ -+3 4 94-39+114 + 342
1 -H 3 -\- 13 +38 f fl4 \~m
Hence the quotient is aH -f- 8x' + ISz^ + sar +114 and the remainder 863.
Explanation.— The coefficients of the dividend are written' in the first row. To
the left is written the second term of the divisor with its sign changed. The first term
of the divisor is not written. The first term, 1, of the third horizontal row Is the quotient
of 1, the first term in the dividend, by 1, the first term of the divisor. The first term,
3, of the second row is the product of the divisor 3 by 1, the first term of the third row;
it Is added to the term above It for the second term in the third row and so on.
?811] THEORY OF EQUATIONS
To illustrate 2809, transform the equation
2x* — 13x»+16a^ — 9x+20 = 0,
into another whose roots are diminished by 3.
765
2
-13
6
+ 16
-21
— 9
—15
-24
+ 20
-72
+3
2
— 7
- 5
—52
6
- 1
— 3
— 8
-24
2
-48
2
6
+ 5
+ 6
+ 15 .
+ 7
2
+ 11
Therefore the transformed equation is
2y+ llj/» + 7 3/« — 48y — 52 = 0.
811. Fourth Transformation. — To transform an equation into
another which lacks any assigned term.
Let the equation be
(1) i>o^"+i>iX"-i+^^x'»-«+ .... +i>n-,a^+l>n=0.
Put x = y-\-k in (1) and it becomes
The development of the binomial in the several terms of (2) gives
' pJ<y+kY = p^{y-+ny-''kJr-''-^y'''''1^+ )
i>i(2^+^)""'=i>i(y"-l+(«-l)3/"-'A;+ )
Adding the second members of these equations and arranging the
terms with respect to the descending powers of y, we have for the
transformed equation,
(3) ;>oy"+(^^i^o^^+i>>"-^+[^^^^^
766 COLLEGE ALGEBRA L?812
The second term of equation (3) will be wanting if we put
npjc+p. = 0 or k= — J^y
^^ ' ^^ npo
and the corresponding substitution is x = y — ^•
Example. — Thus the solution of the equation
192ic»+144x« + 132«+91 =0
can be made to depend upon the solution of the simpler cubic,
6y'+3x+2 = 0,
by substituting x = y — \ in the first equation.
If one desires to remove the third term in equation (3) put
''-^'^y.^-* + ('^ - Di*.* + A = 0,
from which k can be found by solving a quadratic equation.
Similarly other terms may be removed.
812. Fifth Transformation. —/?cc>)>roc(i^ roots and reciprocal
equations. Let the equation be
(1) x- + p^x-'^ + p^x-'^+ .... +i>,_,x + p,= 0;
then, to transform equation (1) into another equation whose roots
are the reciprocal of those of equation (1), put x = - and obtain
(2) ln+l\^r+P.^t+ ■ ■ ■ +^"f +i'. = 0,
or, cleared of fractions,
(3) p„5^" + P„-,y"-^+ .... +/>,y' + i>iy+i = o.
Definition, — If equation (3) takes the same form as equation (1).
then equation (1) is called a reciprocal equation. The necessary
and sufficient condition that this may be so is that
^^ Pn ^^ Pn ^^ Pn ^'^^ Pn ^"^ Pn ^ " fn
The last of these equations gives p^ = 1, or 7>„ = di 1, hence
there are two classes of reciprocal equations according as />„ is equal
to + 1 or — 1 :
I. If ^„ = -|- 1, then we have the relations
Thus, in the first class of reciprocal equations the coefficients reckoned
from the beginning and the end are equal in magnitude and have the
same signs.
22813,814] THEORY OF EQUATIONS 767
II. J£ p^z= — 1, then we have the relations
Pn-l = — Fv Pn'2 = — Pv • • • • y Pi= — Pn-V
Tlius, in the second kind of reciprocal equations the coefficients reckoned
Xrom the beginning and the eml are equal in magnitude and have con-
trary »igns.
In case n = 2m ; then one of the conditional equations becomes
_p^ = — ■ i>m > ^' ®- ) i'm = 0 } thus, the middle term is absent in a recip-
rocal equation of the second class and of an even degree.
If r is a root of a reciprocal equation, then - is also a root, for -
is a root of the transformed equation and by definition it is identical
^ith the given equation ; hence, the roots of a reciprocal equation
enter in pairs r, - ; ', - ; etc.
818. The Standard Form of Reciprocal Equations.
Lety^x) = 0 be a reciprocal equation.
If /(x) = 0 is of the first kind and of an odd degree, then it is
evident from the form of the equation that — 1 is a root. Hence,
the equation is divisible by x -|- 1 and the depressed equation is of
• the first kind and of an even degree.
If y^x) = 0 is of the second kind and of an odd degree then it is
evident from the form of the equation that + 1 is a root. Hence
the equation is divisible by x — 1 and the depressed equation is of
the first kind and of an even degree.
If f(x) = 0 is of the second kind and of an even degree it is
divisible by x* — 1 since the equation can be written in the form
x« — 1 +2>i^(^""'— 1)+ .... =0.
By dividing by x' — 1 the depressed equation is of the first kind
and of an even degree. Therefore all reciprocal equations may be
reduced to those of the first kind whose degree is even, and hence it
is regarded as the standard form of reciprocal equations,
814. The Reciprocal Equation of the Standard Form can have
its Degree Diminished One-Half. — Let the equation be
(l)p^x*"'+pjX"»-i+^,x«'"-«+ . . . i>^x'«+ . . . +i>,x'+i?iX+ J)^=0.
Divide this equation by x*", and regrouping the terms we have
(2) p, (x" + i;) +^, (a:-- + ^.) + . . . + i>„ = 0.
768 CX)LLEGE ALGEBRA [«815, 816
For we have the identity
(3) --• + -J^. = (x^ + h)i^ + l)- (--• + i^.>
Put y = X -f. - ; hence for r = 1, 2, 3, . . . . »i we have
^+?=('+l) i'^+D- (1+1) =^-2.
and so on. Thus we see that x" + -- is of the m dimensions in >
and hence (2) is of the dimensions m.
Descahtes's Rule of Signs
815. Definition. — If bcth the signs of two succeeding terms are
4- or — a permanence is said to occur. If the signs of two sue-
ceeding terms are respectively + and — , or — and + , a vartatiun
is said to occur.
816. Descartes's Rule of Signs for Positive Roots This rale
enables one, by merely inspecting the signs of the terms of an equa-
tion, to assign a superior limit to the number of positive roots of
the given equation. It may be stated as follows:
The number of positive roots of a given equation can not he greater
than the number of variations of the signs of its terms^
This rule is but a particular case of the more general theorem hy
Budan and Fourier. The usual proof of this celebrated theorem of
Descartes is given, though it amounts to but little more than a veri-
fication instead of a full demonstration.
Suppose that the given equation is f{x) = 0, and that the sigia
of the polynomial /{x) succeed each other in the following order:
+ + + +-+.
If a positive root a was introduced into the equation f(x)=0, we
would multiply the equation hj x—a. There are six variations in
the signs of the given sequence, and it is proposed to show that if
the positive root a is introduced into the equation f(x) = 0,
there will be at least seven variations in the signs of the resulting
equation, (.r — a) fix) = 0.
«817] THEORY OF EQUATIONS 769
Writing down only the signs which occur in the operation, we have
+ + --- + -- + - +
+ -
+ + --- + "- + " +
-" + -!- + - + + - + -
The double sign dc occurs whenever there are two terms with dif-
ferent signs to be added. On examining the product it follows that:
I. The ambiguous sign occurs whenever -f- follows + and — fol-
lows — in the original sequence of signs.
IT. The signs before and after an ambiguity, or set of ambiguities,
are unlike.
III. A change of sign is introduced at the end.
Take the most unfavorable case, that in which all the ambiguities
in signs are taken as continuations; then it follows from II that the
number of changes in signs is the same whether the upper or lower
sign is taken; e. g., take the upper sign, then the number of changes
of sign can not be less than the number in
+ + --- + -- + - + -
which has the same arrangement of signs as the original poly-
nomial excepting a change of signs at the end.
Suppose now that a polynomial is formed of the factors cor-
responding to the negative and imaginary roots of an equation;
the result of multiplying this product by each of the factors
X — a, x—hj x—c, etc., corresponding to the positive roots a, i, c,
etc., is that at least one variation in sign for each root is intro-
duced; therefore an equation can not have more positive roots than
variations in sign.
817. Descartes's Rule of Signs for Negative Roots. — If — x is
substituted for x in the equation /(x) = 0, the resulting equation
/{—x)=0 has the same roots as the equation /(.c) = 0, excepting that
their signs are changed. This follows from the identical equation
J{x) = {x — Oj) (.c — Oj) . . . . (x — a„)
from which we deduce
A- a-) = (- ir {x + a^) {x + «,).... (x + aj.
770 COLLEGE ALGEBRA [J §818, 319
Hence the roots of /(— ar) = 0 are
— «i, — «8» • • • • — «»•
Therefore the negative roots of f{x) = 0 are the positive roots of
/(— x) = 0, and hence follows Descartes' s rule for negative roots:
The number of negative roots of the equation f{x) = 0 can not he
greater than the number of variations of signs mf{ — x).
Example. — The equation x* + 15 x* + 7x — 11 = 0 has one
variation of sign and can not have more than one positive root.
Again /( — - x) = x* + 15 x" — 7x — 11 = 0 has one variation of
sign and therefore /(x) can not have more than one negative root.
818. Determination of the Existence of Imaginary Roots by
means of Descartes's Rule.
In case the sum of the maximum number of positive and negative
roots is less than the degree of the equation we are sure that the
equation has imaginary roots.
Example. — The equation x* + 15x*+ 7x — 11 = 0 has, J817, at
most one positive root and, 8817, at most one negative root. Hence
the given equation can not have more than two real roots, and
therefore must have at least two imaginary roots.
819. Derived Functions.
Let /[x)=p,x"+i>^x"-» + ;?,x"-2+ . . . +i>,_iX+/>„;
then
f(x+h)=p^{x+hr+p^(x+h)-'+pJix+hr'*+ .... +Pn.i(x+h)+p^.
Expanding (x -|- A)", (x -\- A)""*, .... by the binomial theorem
and arranging the whole result according to the ascending powers
of /«, we have
/(x+A)=p^X"+J>^X«-»+7?^X~-«+ . . .+Pn-iX+Pn]
+ ^ [^^i>o^"-'+(n-l)i)jX»-«+(/i-2)i>,x-»+. . . +!>„.,]
+ |j ['i(n-l)i>o^"-«+(n-l)(n-2)p,x"-»+ . . . +2i>,.J
+
+ ^^,[.(«-l)("-2)...;2.1;>J.
The first line of the expression is evidently /(x). The coefficient of
h is represented by /'(x), of ^ by /"(x), and in general the coefficient
of — by /^{x), and so on. Hence
S820] THEORY OF EQUATIONS 771
/;x+A)=/(x)+V(x)+|j/" (x) + |j/"'(x)+ . . . ^^r{x)
y'(^), /"(«), /'"(aj), are called respectively the first, second,
and third derivatives of f{x) with respect to x. It follows that
f(ac) is obtained from f(x) by multiplying each term in f{x) by the
exponent of x in that term and diminishing the exponent of x by 1,
Example. —Find the derivatives of 3 a;* — 2 x' — 5 a; + 7.
/(rc)= 3x* — 2ic»— 5x+7
/(x) = 12x» — 6x«-5
/'(.t) = 36x«— 12x
f'\x) = 72 X — 12
/'"(x) = 72 x»
/""(x) = 0.
820. Equal Roots. — If the equation /(x) = 0 has p roots, equal
to a, then f(x) = 0 has p — 1 roots equal to a.
If the equation /(x)=0 has p roots equal to a it is divisible p
times by x — a or by (x— a)^; let the quotient of f{x)=Q by (x— a)^
be jP(x) ; then we have
(1) /{x) = (±-^a)PF(x).
Put in (1) xi^x-|-A; hence we have
(2) y\x + A) = [(x - a) + h]pFix + h).
Using the expansion forytac+ A) and /'(x-|-A),?819,and the expansion
of [{x — a)-[-A]p by the binomial theorem, we have
Ax)+ fj/'(x) +|-j/'(x) + ^ [{x-a)P+p{x-a)P-'h+ . . .]
x[F(x)+^^F'(x)+f^F-(x)]
=(jr— a)''J^(x) + [(x— a)P/"(x)+j>(x— a)P-»/'(x)]A+terms in A«,A', . . .
Thus we have two integral polynomials which are identically equal
for all finite values of A ; hence the coefl3cients of like powers of A
are equal, thus
f'{x) = {x—ayP(x) +p{x—a)P'^F(x),
Therefore, /{x) contains the factors x — a repeated p — 2 times;
that is, f{x) has p—l roots equal to a. Similarly, if 6 is a root
of f(x) = 0 repeated r — 1 times, then 6 is a root of /'(x) = 0
•The notation /'(Of), f"(x), f"(x)..,^ becomes Inconvenient when the number
Of accents Is large, and hence /^(x) Is used for the coefficient of —
772 COLLEGE ALGEBRA [JI821-323
repeated r — 2 times, and so on. Therefore, /(x) = 0 has or has
not equal roots according as J\x) and f{.t) have or have not thr
common faxitor^ x — a, or some power of x — a.
821. Hence it follows that the equal roots of /(x)=:0 are given
by finding the greatest common divisor of /(x) = 0 and /^(jr) = 0,
and placing it equal to zero, then solving the resulting equation.
Example. — Find the equal and non-equal roots of the equation
yi^a-)— x«— 5x» + 5x*+9a^— 14x«— 4x + 8= 0.
Hence /'(a-) : : 6 a:^ — 25 x* + 20 x» + 27 x« — 28 a; — 4.
The G. C. D. oijlx) and/(x) is x+1 and (x— 2)«; hence (x+1)-
and (x— 2)' are factors of /(x). The remaining factor of /(x) is x— 1.
Therefore — 1 is a double root, 2 is a triple root, and 1 a single
root of ytx) = 0.
822. Continuity of a Rational Integral Function of x If /[x)
is a rational integral polynomial in x, and x is made to vary by in-
finitesimal increments (1637) from a to a larger numl)er 6, we shall
prove that/(x) at the same time varies by infinitesimal increments;
in this case it is said that/(x) varies continuousily with x.
Let c and c + A be any two values of x lying l}etweeA a and h.
Hence we have, ?819,
(1) Ac+h) - Ac)=hf'{c)+ IJ /-(c) + |j/'"(c) + . . . + JJ|/« (r).
Since y^x), /'(x), /"(x), . . . are all finite for x = c, the limit of
the second member of equation (1) is 0 for h ^(S (?668). Therefore,
the j^^Q \f{c + h) — /(c)l = 0. Hence, to infinitesimal changes in
X there correspond infinitesimal changes in /(x), and as x changes
continuously from a to ^, the function f{x) changes continuously
from/(fi) to/(^).
823. When /'(c) is positive, f{,r) increases with x; and when f'{c)
is negative, /(x) diminishes as x increases.
For,
^/'U)+|/"(c)+...+g/«(c)
-A [/'(o) + |/"(c) + ....+ (;i^/-(c)].
The limit of the bracket for /* ^ 0 is /'(c) (?688).
2824]
THEORY OF EQUATIONS
773
This theorem may be made more perspicuous by aid of the
graphical representation of the following example.
f(x) = 10x» — \l7? + a; + 6 = 0.
z
A^)
— 4
-910
— 3
— 2
-420
— 144
— 1
- 22
0
0
6
1
1.1
0
— .16
1.2
2
3
0
20
126
etc.
In the annexed figure we notice
that to x= — 1= OQ corre-
sponds/(— 1)= 0^= — 22, and
to X = 0 corresponds /(O) = 6.
One root, 06' = — |, of A^) = 0
lies between a;=0 and a;=— 1,
for which the corresponding val-
ues of /(x), namely —22 and -f 6, ^,
have opposite signs, and the
hfo\f^^+ A)-/(c)]= (lim^)/'(c);
hence the sign of the difference
[/(c + A) -/'(c)] for any h de-
pends upon the sign of /'(c).
Q. E. D.
824. Theorem. — Iffip) and f{Jb) have con-
trary signs then the equation f{x) = 0 must have
at least one real root situated between a and b.
ir
■ [ ; I
;tt4
. .. .4-' --
....I...]..
.^-.^^-J-.
-J, ..4-H-.
-i"'4-
.,. ..;...
■:--r-hr-
.1. .i..^.
i : [ ^
-I ..i... .
-t-l-
'-H
"■;::::.::r
■ ■
■l-i-j-t
J.....
., '.t\"-i'^
-^ |.4.
,..;,.
■; f- ■;■■
p
U,-,
r^r
-4-,
-J r" '
"p^".
' r :■■ i-
^iS^^"
■■:-\ --
'/~:"r7 "
. .. ,.
■if''"
-^.
-h-'i-
^i--
"^ -
^- '■^■1
l-:f -
—44-
— h-f-
1- 1 .
1
tt-ll-
;.•' ■■:■■!■-
Figure 1
For, since /(ac) changes continuously from /(a) to /(6), and there-
fore passes through all intermediary values as x changes continuously
from a to 5, and since by hypothesis one of the signs of f{a) and
J\b) is positive and the other negative, it follows that/lfa;), for some
value of X between a and ^, must take the value zero situated between
/(a)and/(6).
Caution, — It does not follow that f{x) = 0 has but one root
between a and />; nor does it also follow, in case /(a) and/(6) have
the same sign, that /(x) = 0 has no root between a and 6.
774 COLLEGE ALGEBRA [?i825-827
825. Theorem. — Every equation of an odd degree has at least one
real rooty xchose sign is the opposite of that of the last term.
Let the equation be f\x) ez x" + p^ x"~* -|. . . . -|. p^_^ x-\- p^ = 0.
Then we have
for a; = — 00 , /(x) is negative, n being odd,[RKiprocal«fJ668J
f or X = 0 , /(x) has the same sign as p^,
f or X = + 00 , JXx) is positive. [Reciprocal of 1668]
If p^ is positive then /(x)=0 must have at least one real negative
root between — oo and 0 (1824); if p^ is negative then/[jc)=0 must
have a real positive root between 0 and oo (S824). Q. E. D.
Example.— Show that /(x) = x^ + 5 x* — 20 x* — 19 x — 2 = 0
has a root between 2 and 3 and a root l>etween —4 and -—5. Here
we have /(2)= _8 yi;- 4) =^ + 10
and
/(3) = + 399 /(- 5) = — 407. Q.E,D.
Thus/(x) = 0 has a positive root between +2 and +3, whose sign is
opposite to that of the last term, —2.
826. Theorem. — If the last term of an equation of an even degree
is negative the equation has two real roots j one positive and one negative.
Consider the equation in §825, and for n even let
X = — 00 , then f(x) is + ,
X = 0 , then /(x) is — ,
X = + 00 , then /(x) is +.
ffence tliere is at least one real root between — oo and 0, and one
between 0 and +qo (1824). Q. E. D.
827. Theorem. — If two numhers, a and 6, are substituted for x
in /(x), and f{a) and f(fi) have contrary signs, then an odd number
of roots of /(x) = 0 lies between a and b; if f{a) and f(b) have the
same signs then either no root or an even number of roots of f{x)z=.Q
lies between a and b.
The theorem in 2824 is a particular case of this theorem. We
give a proof for the first part of the theorem and the second part
may be established in a similar manner.
Let a be less than 6, and suppose that of the roots of y^x)=0,
a^, ^j, . . . a,., and no others, lie between a and 6.
Let the quotient of /(x) divided by (x— a^) (x — a,) . . . (x— a,),
be -F(x), then we have
/(x) ~ (x — a^ (x — a,) . . . (x - a^) F\x).
«828, 829] THEORY OF EQUATIONS 775
Put .15 = a, X = h and obtain
/(a) = (a — ttj) (a — a,) . . . (a — a^) /1(a),
/(6) = (&-a^) (6-a,) . . . (6 -a,) /'(fc).
/X«) ancl F(h) have the same signs; for if they had opposite signs
then there would be one root at least of F{x) = 0, between a and />,
which is excluded by hypothesis. By hypothesis /(a) and f{b)
have opposite signs. Therefore the two products
(a — a,) (a — a,) ... (a — a^)
(/,-a^)(6-«a,).. . (Z»-a,)
have opposite signs; but h is greater than any of the a*s; hence the
second product is positive and consequently the first product is
negative. Since a is less than a^, n^, a^, the first product
can be negative only when r is odd. Q. E. D.
We have a geometrical illustration of this theorem in the graph
in 3828. From the table we have
for X =-1, f{x) =/(— 1)=— 22 )/(-!) and f(2) have opposite
and (Aree roots, — i, 1,
}?T!ie
for x=2, f(x) =/( 2 ) = 20) i.l, lie between -1 and 2;
for a: = 0, /(x) = /(O) = 6 ) /(O) and f{2) are both positive,
., V ^/«x «/v c^^^ ^wo roots, 1, 1.2, lie be-
f or X = 2, Ax) = /(2) = 20 ) tween 0 and 2.
Sturm's Functions and Theorem
828. Algebraic solutions of the equations of the third and fourth
degrees have been found. A method will now be given which
enables one to obtain approximately the value of the real roots of
an equation of any degree. This part of the subject is begun by
proving Sturm's theorem, which has for its object, the determination
of the situation and number of the real roots of an equation.
Sturm's Functions. — Let /{x) = 0 be an equation whose
equal roots have been removed, and let /^(x) be the first derived
function of Jlx), Apply the process of finding the G. C. D. of
/(x) and/^Cx), with the exception that signs of the remainders are
changed, until the last remainder does not contain x.
Let fj{x)y /jCx), . . . /^(x) be the series of modified remainders
thus derived. They are called JStnryn's Functions,
Let ^1, J21 Js, . . . S'n-i ^® ^^^ successive quotients which arise
in performing the indicated operations; hence, we have the following
relations:
776 COLLEGE ALGEBRA [1830
(1)
Three inferences may be drawn from these relations:
I. The function /„(x) is not zero. For, if /„(x) is zero ihenfix)
and /j(jr) have a common divisor, Sill, and therefore the eqaation
/(jr) =0 has a pair of equal roots, which is contrary to the hypothesis.
II. Two consecutive functions can not vanish for the same value
of X, For, if /j(x) =fj,x) = 0, thenfjix) = 0, and so on until /.fj-i
is zero, which is impossible by I.
III. If any function excepting /(.r) vanishes for any value of or,
the two adjacent functions have opposite signs. Thus if /,(x) = 0.
then we have r/^\ rf^\
830. Sturm's Theorem. — I. I/y as x increases, J\x) passe*
through the value zero, JStu mis functions lose one change tn sign.
For let X, as it increases, pass through c, a root of /(x) = 0, then
we have for x = c + A,
y(<: + 70=/(c)+/(c)'ij + /'{c)|^+ . . . [«819]
Suppose that h is infinitesimal (?637); then, since /(c) = 0, the
sign of /(c -|- h) depends upon the sign of f(c) • h (J668) ; thus we
^^'^^ (1) Ac+h)=f{c)h',
hence /{c -\- h) and/'Cc) have the same sign where A is positive.
Therefore the function /[.r), just after x passes through a root, c, has
the same sign as /^(x) at a root.
Changing h into — h we have from (1)
(2) /(c^h) = ^hf{c);
hence the function /(x), just before x passes through a root, c, has
the opposite sign of /^(x) at a root. Q. E. D.
IL Sturnis functions neither lose nor gain a change of sign when
X passes through a value which makes one of them except f\x) vanish.
Let/,.(6) = 0, then/_j(c) and/,.^.j(c) have contrary signs, and hence
just before x = c and also immediately after x = c, the three func-
tions /r-i(a:), frM, fr+iM havc one permanence of sign and one
variation of sign; for, when f^iix) and /^(x) have contrary signs,
/^(x) and/^^j(x) have the same signs, and reciprocally. Q.E.D.
JJ831,832]
THEORY OF EQUATIONS
777
Theorem. — The number of roots of f{x) between a and b is equal to
the difference between the number of variations of signs in Stumi^s
functions when x ■=. a and x = b.
Corollary. The total number of roots of /(.r) = 0 is found by
taking a = + x.and b = ~ ao , since the sign of each function is the
same as that of its first term (^825).
831. When the number of functions is greater by unity than the
degree of the equation, the following theorem can be proved:
I. All the roots of /(.r) = 0 are real, if the first terms of all the
functions are positive,
II. There will be a pair of imaginary roots off{x) = 0 for every vari-
ation of sign in the first term of the function, if they are not all positive.
Proof. — Use Descartes's rule (2816) after putting ac = + oo and
— 00 , examining" the number of changes of signs in each case.
Example. — Locate the roots of the equation
x» — 10 x« + 35x + 50 = 0.
Sturm's functions, calculated according to the preceding rule, are
here tabulated.
Since there is one variation
in the signs of the first term
of the function there is one pair
of imaginary roots.
The changes in sign of the
functions as x passes through
integral values are exhibited
in the adjoining table. There
is one variation of sign lost
while x passes from —2 to
— 1 ; and no other.
/(.t) = x» — 10 x« + 35 X + 50.
//.x) = 3a^-20x + 35.
/^(.r) = -x-80.
20835,
/s(-^) =
X =
-2
-1
0
1
2
3
4
/(.r) =
—
+ +
+
+
+
+
/,(x) =
+
+
+
+
+
+
+
A(^) =
■—
—
—
—
—
—
—
/,(') =
—
—
—
—
—
—
—
No. of Clunges
of Sign .
2
1 1
1
1
1
1
Calculation of Incommensurable Roots by Horner's Method
Solve the equation (1) a:*— 7 x-|- 7 = 0. This example was
selected by Lagrange on account of its difficulty.
According to Sturm's theorem this equation has three real roots ;
pne lies between — 4 and — 3, and two between 1 and 2, that is,
between 1.3 and 1.4, and 1.6 and 1.7; of these we shall calculate the
second.
778
COLLEGE ALGEBRA
[1832
DiminishiDg the roots of the equation (1) x* — 7x+7 = 0 b j 1.3,
we put x=« + 1.3, and have (S810, Ex.),
1-t 0 - 7 +7
+^1.3
Hence, the transformed equation is
(2). z»+ 3.9 2« — 1.932+097=0.
1.3
1.69— 6.903
1.3
1.3
-5.31
3.38
2.6
1.3
/1,.097
.4,-1.93
Since a root of this equation is so many
hundredths we may neglect the powers of z
A^ 3. 9 above the first in determining the next figure in the
root; thus — 1.932 + .097 = 0 or s = .05. Hence 5 is the
next figure of the root. Diminishing the root of (2) by .05 we have
3.9
.05
—1.93
.1975
.097
—.086625
3.95
.05
-1.7325
.2000
.4. .010375
4.00
.05
CI 4.05
-1.5325
Hence the transformed equation is
(3) y»-f 4.05 y^ - 1.5325 y + .010375 =0.
Hence the next figure of the root is y =t^^=-006.
1.5«)25
Continuing this process in a more compact form we have
4.05
-1.5325
.024336
.010375
-.009048984
.05689
.006
4.056
.006
-1.508164
.024372
^1, .001326016
.0011844291
>68
4.062
.006
^3-1.483792
.00325504
A^ .000141586432
6;4.0680
.0008
4.0688
.0008
4.0696
.0008
6^4.0704
-1,48053696
.00325568
^^_1. 47728128
The fifth decimal 9 is found from the equation
.000141586432
1.47728128
■=.00009.
The root correct to nine decimals is 1.356895867.
{833]
THEORY OF EQUATIONS
779
833. Second Illustration of Horner's Method. — Find the numer-
ical values of the real roots of the equation
This equation has two roots between 2 and 3 (J762). The method
of calculating the least of these roots is given below :
1
—4
2
—2
2
0
2
2
A
40
4
44
4
+ 1
--4
-3
0
-3
4
(\ +100
176
276
192
+ 6
—6
0
-6
B^ -6000
1104
—4896
1872
B^ —3024000
68161
48
4
468
20^
C^ 67600
-2955839
68723
52
4
2>, 560
1
561
1^.
562
1.
563
1^
Z>, 5640
4^
5644
4^
5648
4^
5652
4^
A 5656
561
68161
562
68723
563
C\ 6928600
^22576
6951176
22592
6973768
22608
C; 6996376
11
J?, -2887116000
27804704
700
-28285470
700
B^ —28284770
21
69974
11_
69985
n
(\ 69996"
-2828456
21
B^ —2828435
+2 I _2. 41 42 13
0
^.
20000
■19584
A
4160000
-2955839
A
12041610000
11437245184
A
604364816
-666003348
A
38361468
-28285470
A
10075998
-8485368
-2859311296
27895072
B^ —2831416224
139948
-283001674
139970
B^ -282861704
A.
1590630
Root=2. 414213562372.
282843)1590630(562372
1414215
28284) 176415
169706
2828) 6709
5657
282) 1052
848
28) 204
197
2) 7
5
2
780 COLLEGE ALGEBRA [«834
Rule. — Diminish the roots by 2 according to the method of {810.
The coefficients of the transformed equation are A^, B^^ C^, D^.
— ^ is an approximation to the remainder of the root (2832}.
This gives .3 for the next figure of the root; but the highest figurt
must be taken which will not change the sign of A ; this will be
found to be . 4.
Diminish the roots by .4. To do this, annex zeros to A^,B^, ^p^,»
as shown above, and use 4 instead of .4. Havinor found A,, and
noting that its sign is -f , retrace the steps and try 5 instead of 4.
This gives -4g with a minus sign, hence the root lies between 2. 4 and
2.5. The new coefficients are Jjj, -Bg, C^^ D^. — ^ gives 7 for the
next figure of the root.
Annex zero as before and diminish the roots by 1, representing
the new coefficients by A^^ B^^ C\, D^,
The signs of A and B must remain unchanged. If a change of
sign takes place it shows that too large a figure has been used.
834. The Abridgement of the Calctilation. — After a certain
number of figures of the root have been found (sa}' four), instead of
annexing zeros, cut off one digit from B^^ two from (7^, and three
from D^. This is equivalent to annexing tiie zeros and then dividing
by 10000.
Continue this work with the numbers so reduced, and cut off
digits in the same manner at each stage until the D and C columns
have disappeared.
Then J^ and B^ alone are left, and six more figures of the root
are correctly determined by the division of A^ by B^
The second root, which lies between 2.7 and 2.8, may be found in
a similar manner.
EXEBCISE CXVin
1. Show that 2 is a root of x» — 7x + 6 = 0.
2. Show that — 3 is a root of 2x' + 5a;« + 9 = 0.
3. 1 is a root of x* — 3x* -f- 4x — 2 = 0 ; find the others.
4. — 1 is a root of 3.r* — x' = 5x* — x — 2 ; find the others.
5. Solve the equation x* + 2x' — 5x' + 6x + 2 = 0, which has a
root — 2 + 13.
6. One root of y^ + 2a/ + ha^y + 4a' = 0 is — a; what are the
others?
2834] THEORY OF EQUATIONS 781
7. Form the equations whose roots are:
(a) 1,2,3,4.
(b) -1, +2, +3, -4.
(c) l/±^,zbU/"7.
(d) 2, 3,^,i.
(e) —0.2, -0.5, 2.2, 3.3.
(f) _ 4, - 3, 3 zb 1 5.
(g) ^(l=b/l 3),zbV'5.
8. Transform the equation 12x* — 34x« + 33.7; — 1 = 0 into
another which shall have the same roots with opposite signs.
9. Transform the equation x' — 15x* -j- 7x + 125 = 0 into
another whose roots are less by 5.
10. Transform the equation x^ — 3.5x* + 7.5x — 1.25 = 0 into
another whose roots are double those of the given equation.
11. Transform the equation x' — 12x* — 18x -(- 135 = 0 into
another whose roots are J of the roots of the given equation.
12. Solve the equation x* — 9x* -(- 14x + 24 = 0, two of whose
roots are in the ratio of 3 to 2.
13. Solve the equation x' — 9x' + 23x — 15 = 0, whose roots are
in A. P.
14. Solve the equation 27.r'+ 42x* — 28x — 8 = 0, whose roots
are in G. P.
15. Transform the equation x' — «.r* — bx + c = 0 into another
whose ropts are the square of the roots of the given equation.
16. The equation 3.r« — 2dx^ + 50x2 _ 50x + 12 = 0 has two
roots whose product is 2 ; find all the roots.
17. Show that the equation x' — x* — 1 = 0 has one real root
only.
18. Show that .r' — 2x + 2 = 0 has a real negative root.
19. Show that x* + 2x' ^x' + x — 1 = 0 has two real roots.
20. Discuss the roots of x* + 2j^ _ x« — 1 = 0.
21. Find the inferior limit to the number of imaginary roots of
the equation x* — 3x* — x + 1 = 0.
22. Find the nature of the roots of the equation x* + 15x*+ 7x
— 11 = 0.
782 COLLEGE ALGEBRA [«834
23. Find the multiple roots of the equations:
(a) jr» + y*- 16^^+20 = 0.
(b) y*^6i^»+12y«-10y+3 = 0.
(c) y*-10y*+15y~6 = 0.
(d) y*-2i^»-lly«+12y + 36 = 0.
24. Determine the number and situation of the real nxits of the
equations:
(a) x»-2x«-10 = 0.
(b) aj»— 9x + 5 = 0.
(c) 5x» — 7x« + 3x + 9 = 0.
(d) X* — 19x + ll = 0.
(e) X* - 2x» + 3x* — 20x _ 47 = 0.
(f) X* + 8x» — 30x« — 210x + 241 = 0.
(g) X* — 7x« + 33x« — 55x + 80 = 0.
Determine the real roots of the following equations by Homer s
method:
25. x'+ X = 20. Ans. 2.5917.
26. x» — 8x — 24 = 0. Ans. 3.7866.
27. x» — 10x« + 35x + 50 = 0. Ans. — L067S.
28. 2x« — 12x«+ 9x + 24 = 0. Ans. 4.3098; 2.7155; -1.0253.
29. The equation 2x» — 650. 8x« + 5x— 1627 = 0 has a root
between 300 and 400; find it. Ans. Commensurable root 325.4.
30. Find the root between 20 and 30 of the equation
4x» — 180x« + 1896x — 457 = 0. Ans. 28.52127738.
31. X* — X = 60. Ans. 2.8809; — 2.8193
32. X* — 10.i;« + X = 61. Ans. 3.7509; — 3.8048.
33. 2x* — 4x» + 3x« — 1 = 0. Ans. +1 ; — 0.4406.
34. 3x* — 2x» — 21x« - 4x + 11 = 0.
Ans. 4.0071; 0.6339; — 0.9503; — 2.0241.
35. lOx* — 7x» — 15x« + 2x + 3^ = 0.
Ans. 1.5055; 0.5367; - 0.5397; — 0.8025.
36. The equation x« + 2x* + 3x» + 4x« + 5x = 321 has one real
root; find it Ans. 2.638605803327.
INDEX
PAGE
Addition, sum of two integers,
Bum-groiip 17
operation of 17
I^wsof, I, II 18
of similar monomials . . 24, 82
of pol^'nomials of positive terms 24
ofpositive and negative numbers 27
of algebraic numters .... 31
rule for addition of algebraic
numbers 31
Argument 754
Arithmetic, fundamental postulate
of 11
Arrangement ...... 696
Axioms, 1—8 72
Binomial, square of ... . 80
Theorem 82
Binomial theorem, proof for posi-
tive integral exponent . 593
proof also 610-612
form of development of . . 613
greatest coefficient of ... 615
Barbier's theorem .... 619
for any exp<jnent 660
convergence of binomial expan-
sion 662
extraction of roots by. . . . 6()5
Braces 50
Brackets 50
Combinations ()01-6O7
Continuity of integral functions. 772
Coefficient, numerical, literal . 23
Convergence of series, all of whose
term:? are positive . . . 6;^1
comparison test for .... 632
standard series for comparison
test of 634
of the Harmonic series . . . 635
the ratio test of .... 637, (U6
of alternating series .... <J39
limit of error in alternating series ()41
general theorem of ... . 644
Counting 15
DeMoivre's Theorem .... 681
Descartes' 8 Rule of Signs . . 768
Determinants :
principle of development . .711
the eliminant 71*2
solution of two cKjuations by .713
of homogeneous equations . 714
PAGE
Determinants, (continued):
of three rows 715
properties of .... 718-721
minors 721
properties of 722
elimination by 723
solution of three equations by 725
of homogeneous equation . . 727
prod\ict of two 730
of fourth order 731
Divei]gence of Series .... 636
Division, numerical .... 57
definition of 67, 58
formal rule of 59
the index law of ...>.. 59
of monomials 60
of polynomials by monomials . 61
rules 1 and 2 61
by polynomials 62
dividend and divisor ... 63
when inexact remainder . . 64
third and fourth rules of . . 67
by zero, indeterminate ... 69
determinateness of symbolic 69
Divisor, definition of .... 58
greatest common of two expres-
sions 101.102
rule for G. C. D. of two or more
expressions 108
Elimination, see elimination under
"Equation*'
Sylvester's dialytic method . 733
Equation, definition of . . . .14
symboHc 45
Equation of the first degree . .71
simple 71
of condition 71
identical equation .... 71
unknown quantity in and root of 71
rules for simplification of . . 73
degree of 146
equivalent equation, theorem for
transforming an eq\iation
into an equivalent equation 147
application 148-158
formulae for the solution of an
eqnatirm of the first degree 158, 159
indeterminate solution . . . 160
problems which lead to a simple
equation 164
as
784
CX)LLEGE ALGEBRA
Equations, numerical and literal 173
pntblems leading to . . . .180
infinite, solution of . ... . 193
of the first degree 198
simultaneous system of linear 198
indeterminate 198
independent 199
incompatible 200
equivalent system of . . . 201
solution of two equations of the
first degree 204
elimination:
by addition and subtraction . 204
by substitution .... 207
by comparison 212
by indeterminate multipliers 218
literal si multaneous equations 2 1 5
fractional equations . . . 215
exponential 575
general solution of system of two 2:i0
the composition of the formula 231
symmetry 232
discussion 233
case a//— a'6^^ .... 233
case a¥—a^b=0 234
case ab*—a'h-ch'—hc*=ac'—a'c=0 235
r^sum^ of discussion .... 236
homogeneous equations . . 236
two equations which have com-
mon root 237
general solution of a system of
three 238
rule for solution of three equa-
tions 241
Equations, n linear equations of the
first degree, solution of . . 242
problems involving three or more
linear equations .... 251
graphs of solution, see "Graph** 265
Equations, diaphantian equations
and problems .... 268
indeterminate equations of the
first degree 268
indeterminate equation a.r-f-/iv=r,268
indeterminate equation rur— 6.i/=c,272
general solution of two indeter-
minate equations . . . 274
Equations of the Second Degree . 385
introduction, theorems I -X, 3a>387
solution of 388
pure Quadratic aj^=h . . 388
complete quadratic,
aj^+px-\-y=() , 392
factors of 405
roots of— equal roots . . 398, 771
imaginary roots .... 398
real and different . . . 400
real and equal .... 40t)
imaginary and um^u il . 40)
solution of 3^+px-\-ij . . 400
Equations, (continued):
factors of ..... 404
problems in ...... 4Ul»
concerning the theorem of
Pythagoras 412
concerning the area of plane
figures 418
has two roots only .... 422
relation between the roots and
coefficients 42:?
properties of roots of . . . . 424
Equations which are biquadratic 427
roots of biquadratic ar*4-/>J:*-Hr=0,4i7
solution of <u:*'*"|-/>r»4c=0 . . 430
Equations which are irrational . 428
solution of
ar«+6j*4-2 iVa^^x^^p . 430
Equations of the form
ac*"4-6j:''-}-r=0 .... 430
Equations which are called recip-
rocal 431, 761
solution of 432
Equations, simultaneous quadratic
in two unknown quantities 451
type I ........ 451
type II 45,3
type III 45i
irrational quadratics . . . .457
in three unknown quantities . 4<>4
special methods of solution 464-47i»
problems in . . . . . . 480
graph of y=ifu^-\-hx-{-c . . . 488
type I 489
type II 491
Equation, the cubic .... 736
a root of 736
cube roots of unity .... 737
the symmetrical cubic . . . 738
Cardan's solution 73? J
the irreducible case .... 741
trigonometric solution . . . 742
the biquadratic 7 '5
resolvent cubic ..... 746
Equations, theory of ... . 757
properties, 1 — 6 . . . 757-761
transformations of . . . 761-767
Evolution, definition of a root . 283
the radical sign, radicand . 2S3
the index of a root .... 283
like and unlike roots . . . 284
law of signs of roots .... 284
odd and even roots . . . 285
principal root 286
theorem in 287
square root of compound quanti-
ties 290
square root of arithmetical num-
bers 295
cube root of a polynomial . . 299
INDEX
785
PAGE
Evolution, (continued):
cube roots of arithmetical num-
bers 303
n*** root of a polynomial . . 307
Exx>onent, law of formation , . 21
Exjjonenta, integral, see table . 309
rules, distributive formulae
rt'«-t.a" = a'"+» .... 22
a"»-f-a" = a'»-» .... 60
associative formula
(a"')» = a»«-" 79
distributive formula
(a6)» = a''6" 79
Exponent, (f)'' = ^-" .... 131
Exponents, fractional, definition of 309
principles 1, 2, 3, 4 . . . 312-314
theorems I — V . . . 313-317
Factor, definition of .... 19
theorem 94
Factors of ji^-\-px-\-q .... 404
ax^-{-bx+q 406
ai^-^2bjnf+cf+2dx+2ey+f . 405
Factoring:
case I, to factor a polynomial 86
case II, trinomial .... 87
case III, the difference of two
squares 88
case IV, the sum and difference
of two cubes .... 89
case V, j^-\-px-^-q .... 90
case VI, polynomials of more
than three terms ... 92
case VII, compound expressions 96
case VIII, polynomials of four
terms 98
Factoring, solution of equations by 98
Factorization, example of ... 437
factors of :r*+pjc«+g . . .438
different cases 440
Fractions, definition of . . . 113
numerator, denominator, terms 1 13
Fractions, rational .... 113
rules of signs for terms of 114,115
reduction of fractions to lowest
terms 115, 121
to reduce a fraction to an integral
or mixed quantity . . .118
reduction of mixed quantity to 119
reduction of fractions to lowest
terms 121
reduction of fractions to com-
mon denominator . . . 122
addition and subtraction of . .124
multiplication 129
powers 131
division of 133
reciprocal of 133
complex 134
contmued 136
PAGE
Fractions, (continued):
special theorem, application 139,141
partial 667
Functions, expansion of . . . 653
derived 770
continuity of 772
Graph, graphical representation of
points and lines . . . 256
grapnical representation of a
point 256
of solution of a conditional equa-
tion 259
of the straight line y=mx-\-b . 262
of the solution of y=mx'i-b and
its intercepts 264
intersection of pairs of lines . 265
resum^ of discussion .... 266
Groups, equality of, one-to-one
conrespondence of ... 12
Horner's Method 777
Indeterminate forms .... 705
Induction, mathematical . . 591
as used in the sciences . . . 595
Inequalities, greater and less . . 15
convention concerning . , 323
general definition 324
between two algebraic expres-
sions 324
equivalent 324
theorems I, II . . . . 325,326
solution of an inequality . . 330
important theorem .... 331
Intercept, see intercept under
*'Graph of Lines"
Interest, compound .... 578
annuities 585
refunding a debt by . . . 588
Interpolation 702
Involution, index law of . . . 277
the law of signs 277
powers of a fraction .... 278
powers of binomials . . . 278
powers of polynomials . . . 280
Limits, definition of . . 343,623
four rules of 348
test for 62.>
theorems concerning . . . 625
Logarithms, definition of . 557,558
systems of, and base of systems 558
properties of ... . 558-560
comparison of two systems . 560
the characteristic and mantissa
of 563,564
use of tables of . . . 565-5(59
addition, subtraction, etc., of . 569
arithmetical complement of . 573
cologarithms 573
Napierian 676
value of e 678
786
COLLEGE ALGEBRA
PAOK
Maxima and minima values of a
quadratic expression
(u*+bx+e 495
definition of 495
problems for illustration . 496-505
^P^^^y=:^M • • • -^
gniphof,= ^;;-^f-^ . . 508
gn.phof,= --f-^ . .509
Modulus 754
Monomial 23
rational and integral ... 79
positive integral jwwer of, rule 79
Multinle, least common . . . 109
L. C. M. of several expressions 110
ruleforL. C. M Ill
Multiplication, operation of, product,
multiplicand, multiplier . 19
law of. III, IV, V . . . . 20
index law of 21
of monomials 24
of polynomials of positive terms
by a monomial .... 25
of positive and negative num-
bers ...... 53, 54
Negative, the, rules of calculation
for the symbol —d and 0 43 - 45
Number, notion of 11
representation by symbols . . 13
positive integer 14
natural 28
positive 29
negative 30
absolute value of .... 30
imaginary 99
conjugate 359
Numbers, irrational and limits . 337
rational numbers insufficient 337
complex numbers .... 337
geometrical representation . 753
introduction ...... 338
the irrational '^l^ . ... 339
definition of special case of irra-
tional, theorem II . 340,341
geometrical illustration . . 342
properties of the series which
define '*! ^ 343
the irrational a limit . . . . 343
SBneral definition of irrationals 344
efinition of zero, positive and
negative 345
the fo\ir fundamental operations
with 345-347
resume 347
equality of two 483
Numbers, imaginary .... 373
Numbers, (continued,:
pure, imaginary, and symbolic,
definition of 373
the imaginary unit .... 37:5
multiple and fractional part of 374
addition and subtraction of
multiple and fractional parts .
of imaginary units . . . 374
division by i .' STo
powers of t . . . _:_ • • 375
the pure imaginarv \/—a . . 370
addition and subtraction of 376
division of 377
complex numbers and
conjugate 379
addition, subtraction, multi-
plication, and division of . 380
square root of .... . 382
incommensurable 522
Parentheses, use of 50
Partial fractions, cases I, II, III 667
Permutation 599
Polynomial, definition of . . 24
arranged in ascending and de-
scending powers ... 65
Principle of permanence, perma-
nence of form .... 46
Problems, positive and n^ative,
solution of 187
zero solution of ... . IIM)
infinite solution of . . . .193
Product, definition of ... 19
vanishing of 70
Progression, arithmetic . . . 538
increasing and decreasing . . b3H
last term / = a-f (rt—l)d . . 5;^>
theorems I and II 640
sum of 6'= I (a + f), etc., . . 541
insertion of arithmetic means . 542
geometric, definition of . . 543
common ratio of .... 543
increasing and decn asing . 54 J
Lemmas 1, II 544
(«4-l)"' term &4fi
problems I, II, III . . . 547
sumof ^=^^,*i;''^ . . .548
infinite decreasing G. P.
S=Y^ .... 549-550
application 550
value of recurring decimal . 550
problems of G. P. ... 547
insertion of geometric mean . 551
harmonica!, definition of . . 552
insertion of harmonical means 552
Radical, definition of .... 349
Ratio and proportion .... 617
INDEX
787
PAOS
Hatio, (continued):
riitio of equality, of greater ine-
quality, inverse ratio, dupli-
cate, triplicate, and subdu-
plicate ratios .... 517
extremes, means, antecedents,
con8i»quent«. continued pro-
portion 518
properties of proportion, mean
proportional, third propor-
tional, theorems I-X.I1I 518-521
special theorem 522
incommensurable numbers 522
theorems XIII-XV . . 522,523
Euclid's definition of . . 524-525
application of quadratic equa-
tions and ratio and propor-
tion to geometry . . . 528
Rationalization, see **Surd8."
Itenminder, definition of , . .64
theorem 93
Root8,extraction of ,666 * 'Evolution
of Ec|uations," see "Equations."
imagmary roots 398
of surd expressions . . 442-447
Root of an equation .... 736
equal roots ...... 771
Roots of unity, cube roots of 737-753
n roots of 751
solution of a:"=l .... 752
geometrical representation of n
roots of 1 755
Scale of relation 690
Series, infinite geometric . 549,(iv)
reversion of 657
exponential 673
logarithmic 675
exponential forms .... 679
Gregory's 686
Euler's . 686
summation of 689
recurring series 690
genera! terms of . . . . 693
Sine 754
Signs, unlike 30
plus 17, 28
minus ........ 28
of aggregation, rule for their
removal 50, 51
the double sign, ± . . . .80
Sturm's Functions 775
Theorem .777
Subtraction, definition of . . 35
of algebraic numbers, nile . . 36
of similar monomials . . . 3()
• generalized discussion ... 39
equation of, VI, minuend and
subtrahend 39
determinatenesa of numerical
subtraction 40
PAOK
Subtraction, (continued):
formal rules of and proof . 40-44
numerical subtraction, limita-
tion of 45
of polynomials 47
Summation of series, definition of
sum 689
of recurring 689
scale of relation .... 690
the sum of n terms . . . 692
general terms of .... 693
by undetermined coefficient . 695
the alternating series . . . 696
by method of differences . . 698
of arithmetical series of n'*» order 700
miscellaneous series ... 701
Surds, definition of 349
orders of 350
reduction to simplest form . . 350
distributive formulae,
Vo • "i/6 = V«*» etc., . 351
associative formulae,
^l/a'' = (Va)«, 6tc. . . 351
rules 351,352
addition and subtraction of . 355
similar surds 355
reduction of surds of different
orders 368
multiplication of .... 359
monomial surds .... 359
polynomial surds , , . 359
conjugate surds 359
typ^ forms 360
division of 363
monomial 363
polynomial 363
type forms 364
rationalization of 366
monomial 366
binomial and polynomial 366-368
reduction of certain irrational
expressions 368
roots of surd expressions, see
"Roots" 442-447
Synthetic division .... 764
Terms, positive 23
similar or like and unlike . . 23
dimension and degree of . . 23
homogeneous 23, 84
Transformations, see "Transform-
ations" under "Equations."
Trinomial, variation in sign of . 448
Variation, definition .... 53.3
A varies directly as ^ ... 533
A varies inversely as B . . 534
A varies jointly as B and C . . 534
A varies both directly as B and
inversely as C .... 534
Vinculum 50
Zero, definition of and operation on 345
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