A treatise on algebra

GIFT OF
Dr. Horace Ivie

EDUCATION DEPT

TREATISE

ON

A L G E B R 1.

BY ELIAS LOOMIS, LLR,

1 1

PUOFES80B OP NATUBAL PHILOSOPHY AND ASTKONOMY IN YALE COLLEGE, AND ATTTHOB
OF A "COUBSK OF MATHEMATICS."

REVISED EDITION.

NEW YORK:

HARPER & BROTHERS, PUBLISHERS,
327 TO 335 PEARL STREET,

FBANKLIN SQUARE.

1886.

}

*

GIFT OF

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EDUCATION

, according to Act of Congress, in the year one thousand eight hundred and
sixty-eight by

HARPER & BROTHERS,
In the; Clerk's Office of the District Court of the Southern Didtrict of New York.

PREFACE.

THE stereotype plates of my Treatise on Algebra having be
come so much worn in the printing of more than 60,000 copies
that it had become necessary to cast them aside, I decided to
improve the opportunity to make a thorough revision of the
work. I therefore solicited criticisms from several college pro
fessors who had had much experience in the use of this book,
and in reply have received numerous suggestions. The book
has been almost entirely rewritten, nearly every page of it
Laving been given to the printer in manuscript. The general
plan of the original work has not been materially altered, but
the changes of arrangement and of execution are numerous. In
the former editions, in place of abstruse demonstrations, I some
times employed numerical illustrations, or deductions from par
ticular examples. In the present edition such methods have
been discarded, 'and I have aimed to demonstrate with concise
ness and elegance every principle which is propounded.

This book therefore aims to exhibit in logical order all those
principles of Algebra which are most important as a prepara
tion for the subsequent branches of a college course of mathe
matics. I have retained, with but slight alteration, a feature
which was made prominent in the former editions, that of stating
each problem twice: first as a restricted numerical problem,
and then in a more general form, aiming thereby to lead the
student to cultivate the faculty of generalization. At the same
time I have very much increased the number of examples in
corporated with each chapter of the book, and at the close have
given a large collection of examples, to which the teacher may
resort whenever occasion may require.

The proofs of the work have all been examined by Prof.
H. A. Newton, to whom I am indebted for numerous and im
portant suggestions.

CONTENTS,

CHAPTER I.

DEFINITIONS AND NOTATION. f^m

General Definitions 9

Symbols which denote Quantities 10

Symbols which indicate Operations 11

Symbols which indicate Relation 14

Combination of Algebraic Quantities 16

CHAPTER II.

ADDITION 21

CHAPTER III.

SUBTRACTION 25

CHAPTER IV.

MULTIPLICATION.

Case of Monomials — Case of Polynomials 32

General Theorems proved 38

CHAPTER V.

DIVISION.

Case of Monomials — Case of Polynomials 41

When an±6n is exactly divisible 47

To resolve a Polynomial into Factors 50

CHAPTER VI.

GREATEST COMMON DIVISOR. — LEAST COMMON MULTIPLE.

How the Greatest Common Divisor is found 52

Rule applied to Polynomials 55

How the Least Common Multiple is found 58

CHAPTER VII.

FRACTIONS.

Definitions and General Principles 62

Reduction of Fractions 64

Addition and Subtraction of Fractions...., 69

Multiplication and Division of Fractions 72

VI CONTENTS.

CHAPTER VIII.

EQUATIONS OF THE FIRST DEGREE.

Definitions — Axioms employed 79

Transformation of Equations 81

Solution of Equations 84

Solution of Problems 87

CHAPTER IX.

EQUATIONS WITH MORE THAN ONE UNKNOWN QUANTITY.

Different Methods of Elimination 99

Equations containing three or more Unknown Quantities 105

Problems involving several Unknown Quantities 109

CHAPTER X.

DISCUSSION OF PROBLEMS.

Positive Values of x — Negative Values 114

Zero and Infinity 118

Indeterminate Solutions 121

Inequalities 122

CHAPTER XL

INVOLUTION.

Powers of Monomials 127

Powers of Polynomials 130

CHAPTER XII.

EVOLUTION.

Roots of Monomials 133

Square Root of Polynomials 136

Square Root of Numbers 139

Cube Root of Polynomials 143

Cube Root of Numbers 146

CHAPTER XIIL

RADICAL QUANTITIES.

Transformation of Radical Quantities , 150

Addition and Subtraction of Radical Quantities 157

Multiplication and Division of Radical Quantities 159

Powers and Roots of Radical Quantities 164

Operations on Imaginary Quantities 166

To find Multipliers which shall render Surds rational 168

Square Root of a Binomial Surd 171

Equations containing Radical Quantities 174

PAGB

CONTENTS. Vii

CHAPTER XIV.

EQUATIONS OF THE SECOND DEGREE.

Incomplete Equations of the Second Degree 177

Complete Equations of the Second Degree 182

Equations which may be solved like Quadratics 187

Problems producing Equations of the Second Degree 192

Equations containing two unknown Quantities 196

General Properties of Equations of the Second Degree 203

Discussion of the general Equation of the Second Degree 206

Discussion of particular Problems 209

Geometrical Construction of Equations 211

CHAPTER XV.

RATIO AND PROPORTION.

Definitions ,.± 214

Properties of Proportions 216

Variation 222

Problems in Proportion 224

CHAPTER XVI.

PROGRESSIONS.

Arithmetical Progression — General Principles 227

Examples 230

Geometrical Progression — General Principles 233

Examples 237

CHAPTER XVII.

CONTINUED FRACTIONS. — PERMUTATIONS AND COMBINATIONS.

Properties of Continued Fractions 240

Permutations and Combinations , 244

CHAPTER XVIII.

BINOMIAL THEOREM.

Powers of a Binomial 248

Sir Isaac Newton's Binomial Theorem 251

Powers and Roots of Polynomials 255

To extract any Root of a Number 257

CHAPTER XIX.

SERIES.

Definitions — Method of Differences 259

Interpolation 263

Development of Algebraic Expressions into Series 265

Vlll CONTENTS.

Method of Undetermined Coefficients 267

To resolve a Fraction into Simpler Fractions 272

Reversion of Series 273

General Demonstration of the Binomial Theorem 275

Applications of the Binomial Theorem 278

CHAPTER XX.

LOGARITHMS.

Properties of Logarithms 284

The common System of Logarithms 287

Multiplication and Division by Logarithms 293

Involution and Evolution by Logarithms 294

Exponential Equations 296

Compound Interest and Annuities 297

Computation of Logarithms 301

CHAPTER XXI.

GENERAL THEORY OF EQUATIONS.

Definitions and general Properties 306

An Equation of the nth Degree has n Roots 308

Law of the Coefficients of every Equation 310

Number of imaginary Roots 315

Transformation of Equations 316

Descartes's Rule of Signs 320

Composition of derived Polynomials 321

Equal Roots , 322

Sturm's Theorem 323

Solution of simultaneous Equations of any Degree 331

CHAPTER XXH.

NUMERICAL EQUATIONS OP HIGHER DEGREES.

Commensurable Roots 334

Incommensurable Roots — Horner's Method 338

Equations of the fourth and higher Degrees 347

Newton's Method of Approximation 350

Method of double Position 351

MISCELLANEOUS EXAMPLES 355

A L G E B

CHAPTEE I.

DEFINITIONS AND NOTATION.

1. Quantity is any thing that can be increased or diminished,
and that can be measured.

A line, a surface, a solid, a weight, etc., are quantities; but
the operations of the mind, such as memory, imagination, judg
ment, etc. r *are not quantities.

A quantity is measured by finding how many times it con
tains some other quantity of the same kind taken as a stand
ard. The assumed standard is called the unit of measure.

2. Mathematics is the science of quantity, or the science which
treats of the properties and relations of quantities. It employs
a variety of symbols to express the values and relations of
quantities, and the operations to be performed upon these quan
tities, or upon the numbers which represent these quantities.

3. Mathematics is divided into pure and mixed. Pure math
ematics comprehends all inquiries into the relations of quan
tity in the abstract, and without reference to material bodies.
It embraces numerous subdivisions, such as Arithmetic, Al
gebra, Geometry, etc.

In the mixed mathematics, these abstract principles are ap
plied to various questions which occur in nature. Thus, in Sur
veying, the abstract principles of Geometry are applied to the
measurement of land ; in Navigation, the same principles are
applied to the determination of a ship's place at sea ; in Optics,
they are employed to investigate the properties of light; and
in Astronomy, to determine the distances of the heavenly

bodies.

A 2

10 ALGEBKA.

4. Algebra is that branch of mathematics in which quantities
are represented by letters, and their relations to each other, as
well as the operations to be performed upon them, are indi-
catfjc}, by sigtoS *cfr»:'&y rebels. The object of algebraic notation
js.tp abridge ^nd generalize the reasoning employed in the so-
Tu!t.feri.rf^rqi(estiortS''.Tel.ating to numbers. Algebra may there
fore be called a species of Universal Arithmetic.

5. The symbols employed in Algebra may be divided into
three classes :

1st. Symbols which denote quantities.

2d. Symbols which indicate operations to be performed upon
quantities.

3d. Symbols which indicate the relations subsisting between
different quantities, with respect to their magnitudes, etc.

Symbols which denote Quantities.

6. In order to generalize our reasoning respecting numbers,
we represent them by letters, as a, &, c, or cc, y, z, etc., and these
may represent any numbers whatever. The quantities thus
represented may be either known quantities — that is, quantities
whose values are given ; or unknown quantities — that is, quan
tities whose values are to be determined.

Known quantities are generally represented by the first let
ters of the alphabet, as a, 6, c, d, etc., and unknown quantities by
the last letters of the alphabet, as jc, */, z, u, etc. This, how
ever, is not a necessary rule, and is not always observed.

7o Sometimes several quantities are represented by a single
letter, repeated with different accents, as a', a", a!" , a"" , etc.,
which are re£d a prime, a second, a third, etc. ; or by a letter
repeated with different subscript figures, as av a2, «3, «4, etc.,
which may be read a one sub, a two sub, a three sub, etc.
All these symbols represent different quantities, but the ac
cents or numerals are employed to indicate some important re
lation between the quantities represented.

DEFINITIONS AND NOTATION. 11

8. Sometimes quantities are represented by the initial letters
of their names. Thus s may represent sum; d, difference or di
ameter ; r, radius or ratio ; c, circumference ; h, height, etc. All
these letters may be used with accents. Thus, in a problem,
relating to two circles, d may represent the diameter of one cir
cle, and df the diameter of the other; c the circumference of
one, and c' the circumference of the other, etc.

Symbols which indicate Operations.

9. The sign of addition is an erect cross, + , called plus, and
when placed between two quantities it indicates that the second
is to be added to the first. Thus, 5 + 3 indicates that we must
add 3 to the number 5, in which case the result is 8. We also
make use of the same sign to connect several numbers togeth
er. Thus, 7 + 5 + 9 indicates that to the number 7 we must
add 5 and also 9, which make 21. So, also, 8 + 5 + 13 + 11 + 1
+ 3 + 10 is equal to 51.

The expression a-\-b indicates the sum of two numbers, which
we represent by a and b. In the same manner, m-\-n-\-x-\-y
indicates the sum of the numbers represented \)j these four let
ters. If we knew, therefore, the numbers represented by the
letters, we could easily find by arithmetic the value of such ex
pressions.

10. The sign of subtraction is a short horizontal line, — , called
minus. When placed between two quantities, it indicates that
the second is to be subtracted from the first. Thus, 8 — 5 indi
cates that the number 5 is to be taken from the number 8,
which leaves a remainder of 3. In like manner, 12 — 7 is equal
to 5, etc.

Sometimes we may have several numbers Jto subtract from a
single one. Thus, 16 — 5 — 4 indicates that 5 is to be subtracted
from 16, and this remainder is to be further diminished by 4,
leaving 7 for the result. In the same manner, 50 — 1 — 5 — 3 —
9 — 7 is equal to 25.

The expression a — b indicates that the number designated by
c is to be diminished by the number designated by b.

12 ALGEBRA.

11 „ The double sign ± is sometimes written before a quan
tity to indicate that in certain cases it is to be added, and in
others it is to be subtracted. Thus, b ± c is read b plus or
minus c, and denotes either the sum or the difference of these
two quantities.

12. The sign of multiplication is an inclined cross, x. When
placed between two quantities, it indicates that the first is to
be multiplied by the second. Thus, 3x5 indicates that 3 is
to be multiplied by 5, making 15. In like manner, axb indi
cates that a is to be multiplied by b; and axbxc indicates
the continued product of the numbers designated by a, 5, and
c, and so on for any number of quantities.

Multiplication is also frequently indicated by placing a point
between the successive letters. Thus, a.b.c.d signifies the
same thing as axbxcxd.

Generally, however, when numbers are represented by let
ters, their multiplication is indicated by writing them in suc
cession without any intervening sign. Thus, abc signifies
the same as axbxc, or a.b.c.

The notation a.b or ab is seldom employed except when
the numbers are designated by letters. If, for example, we
attempt to represent in this manner the product of the num
bers 5 and 6, 5.6 might be confounded with 5^; and 56
would be read fifty-six, instead of five times six.

The multiplication of numbers may, however, be denoted
by placing a point between them in cases where no ambiguity
can arise from the use of this symbol. Thus, 1.2.3.4.5 is
sometimes used to represent the continued product of the
numbers 1, 2, 3, 4, 5.

13. When two or more quantities are multiplied together,
each of them is called & factor. Thus, in the expression 7x5,
7 is a factor, and so is 5. In the product abc there are three
factors, a, &, c.

When a quantity is represented by a letter, it is called a
literal factor. When it is represented by a figure or figures, it

DEFINITIONS AND NOTATION. 13

is called a numerical factor. Thus, in the expression 5ab, 5
is a numerical factor, while a and b are literal factors.

14. The sign of division is a short horizontal line with a
point above and one below, -=-. When placed between two
quantities, it indicates that the first is to be divided by the
second.

Thus, 24-^-6 indicates that 24 is to be divided by 6, making
4. So, also, a-^b indicates that a is to be divided by b.

Generally, however, the division of two numbers is indi
cated by writing the divisor under the dividend, and drawing
a line between them. Thus, 24-^-6 and a H-6 are usually writ-

ten and .

15. The products formed by the successive multiplication
of the same number by itself, are called the powers of that
number.

Thus, 2 x2=4, the second power or square of 2.
2x2x2 = 8, the third power or cube of 2.

2 x 2 x 2 x 2 = 16, the fourth power of 2, etc.
So, also, 3x3 = 9, the second power of 3.

3 x 3 X 3 = 27, the third power of 3, etc.
Also, a x a = aa, the second power of a.

ax ax a = aaa, the third power of a, etc.
In general, any power of a quantity is designated by the
number of equal factors which form the product.

16. The sign of involution is a number written above a quan
tity, at the right hand, to indicate how many times the quan
tity is to be taken as a factor.

A root of a quantity is a factor which, multiplied by itself
a certain number of times, will produce the given quantity.

The figure which indicates how many times the root or fac
tor is taken, is called the exponent of the power.

Thus, instead of aa, we write a2, where 2 is the exponent of
the power; instead of aaa, we write a3, where 3 is the expo-

14 ALGEBRA.

nent of the power ; instead of aaaaa, we write a5, where 5 is
the exponent of the power, etc.

When no exponent is written over a quantity, the exponent

1 is always understood. Thus, a1 and a signify the same thing.

Exponents may be attached to figures as well as letters.

Thus the product of 3 by 3 may be written 32, which equals 9.

" 3x3x3 " 33, " 27,

" 3x3x3x3x3" 35, " 243.

17. The sign of evolution, or the radical sign, is the charac
ter <|/~. When placed over a quantity, it indicates that a root
of that quantity is to be extracted. The name or index of the
required root is the number written above the radical sign.
Thus,

V/9, or simply V9, denotes the square root of 9, which is 3.

V/64 denotes the cube root of 64, which is 4.

Vl6 denotes the fourth root of 16, which is 2.
So, also,

I/a, or simply -y/a, denotes the square root of a.

I/a denotes the fourth root of a.

I/a denotes the nth. root of a, where n may represent any
number whatever.

When no index is written over the sign, the index 2 is un
derstood. Thus, instead of Vab, we usually write Vab.

Symbols ivliich indicate Relation.

18. The sign of equality consists of two short horizontal
lines, =. When written between two quantities, it indicates
that they are equal to each other.

Thus, 7 + 6 = 13 denotes that the sum of 7 and 6 is equal to
13.

In like manner, a = b + c denotes that a is equal to the sum
of b and c ; and a+b=c— d denotes that the sum of the num
bers designated by a and 5, is equal to the difference of the
numbers designated by c and d.

DEFINITIONS AND NOTATION. 15

19. The sign of inequality is the angle > or <. When
placed between two quantities, it indicates that they are un
equal, the opening of the angle being turned toward the greater
number. When the opening is toward the left, it is read great
er than ; when the opening is toward the right, it is read less
than. Thus, 5>3 denotes that 5 is greater than 3 ; and 6<11
denotes that 6 is less than 11. So, also, a>6 denotes that a is
greater than b; and x<y+z denotes that x is less than the
sum of y and z.

20. A parenthesis, ( ), or a vinculum, - , is employed to
connect several quantities, all of which are to be subjected to
the same operation. _

_

Thus the expression (a-f-5-fc)xa?, or « + /> + cxx, indicates
that the sum of a, 6, and c is to be multiplied by x. But
a + b+cxoc denotes that c only is to be multiplied by x.

When the parenthesis is used, the sign of multiplication
is generally omitted. Thus, (a + Z>-fc)xx' is the same as

21. The sign of ratio consists of two points like the colon :
placed between the quantities compared. Thus the ratio of a
to b is written a : b.

22. The sign of proportion consists of a combination of the
sign of ratio and the sign of equality, thus, : = : ; or a com
bination of eight points, thus, : :: :.

Thus, if a, &, c, d, are four quantities which are proportional
to each other, we say a is to b as c is to d; and this is express
ed by writing them thus:

a: b = c : c?,
or a : b : : c : d.

23. The sign of variation is the character <x>. When written
between two quantities, it denotes that both increase or diminish
together, and in the same ratio. Thus the expression sootv de
notes that s varies in the same ratio as the product of t and v.

16 ALGEBRA.

24. Three dots /. are sometimes employed to denote there'
fore, or consequently.

A few other symbols are employed in Algebra, in addition to
those here enumerated, which will be explained as they occur.

Combination of Algebraic Quantities.

25, Every number written in algebraic language — that is,
by aid of algebraic symbols — is called an algebraic quantity, or
an algebraic expression.

Thus, 3a2 is the algebraic expression for three times the
square of the number a.

7a364 is the algebraic expression for seven times the third
power of a, multiplied by the fourth power of b.

28. An algebraic quantity, not composed of parts which are
separated from each other by the sign of addition or subtrac
tion, is called a monomial, or a quantity of one term, or simply
a term.

Thus, 3a, 55c, and 7a??/2, are monomials.

Positive terms are those which are preceded by the sign plus,
and negative terms are those which are preceded by the sign
minus. When the first term of an algebraic quantity is posi
tive, the sign is generally omitted. Thus a + b — c is the same
as -\-a + b — c. The sign of a negative term should never be
omitted.

27. The coefficient of a quantity is the number or letter pre
fixed to it, showing how often the quantity is to be taken.

Thus, instead of writing a + a + a + a+a, which represents
5 a's added together, we write 5a, where 5 is the coefficient of
a. In §(x+y\ 6 is the coefficient of x-\-y. When no coeffi
cient is expressed, 1 is always to be understood. Thus, la and
a denote the same thing.

The coefficient may be a letter as well as a figure. In the ex
pression nx, n may be considered as the coefficient of x, be
cause x is to be taken as many times as there are units in n.
If n stands for 5, then nx is 5 times x. When the coefficient

DEFINITIONS AND NOTATION. 17

is a number, it may be called a numerical coefficient ; and when
it is a letter, a literal coefficient.

In 7ax, 7 may be regarded as the coefficient of ax, or 7a may
be regarded as the coefficient of x.

28 The coefficient of a positive term shows how many times
the quantity is taken positively, and the coefficient of a nega
tive term shows how many times the quantity is taken nega
tively. Thus, -f 4x = + x + x + x + x ;

29. Similar terms are terms composed of the same letters,
affected with the same exponents. The signs and coefficients
may differ, and the terms still be similar.

Thus, Sab and lab are similar terms.
Also, 5a2c and — 3a2c are similar terms.

30. Dissimilar terms are those which have different letters or
exponents.

Thus, axy and axz are dissimilar terms.
Also, Bab2 and 4a26 are dissimilar terms.

31. A polynomial is an algebraic expression consisting of
more than one term; as, a + b ; or a + 2b — 5c + x.

A polynomial consisting of two terms only is usually called
a binomial; and one consisting of three terms only is called a
trinomial. Thus, 3a + 5& is a binomial; and 5a—3bc-\-xy is a
trinomial.

32. The degree of a term is the number of its literal factors.
Thus, 3a is a term of the first degree.

Sab " second "

6a25c3 " sixth "

In general, the degree of a term is found by taking the sum
of the exponents of all the letters contained in the term.

Thus the degree of the term 5ab2cd3 is 1 + 2 + 1 -f 3, or 7;
that is, this term is of the seventh degree.

18 ALGEBRA.

33. A polynomial is said to be homogeneous when all its
terms are of the same degree.
Thus, 3a2— 4ab-}-b2 is of the second degree, and homogeneous.

2a3-f3a2c-4c2d " third " "

But 5a3 — 2ab + c is not homoeneous.

34. The reciprocal of a quantity is the quotient arising from
dividing a unit by that quantity.

Thus the reciprocal of 2 is J; the reciprocal of a is £

35. A function of a quantity is any expression containing
that quantity. Thus,

is a function of a?.

is a function of y.
ax2— by2 is a function of x and y.

Exercises in Algebraic Notation.

36. In the following examples the pupil is simply required
to express given relations in algebraic language.

Ex. 1. Give the algebraic expression for the following state
ment: The second power of a, increased by twice the product
of a and 6, diminished by c, and increased by c?, is equal to
fifteen times x. Ans. «2 + 2a& — c -\-d-l5x.

Ex. 2. The quotient of three divided by the sum of x and
four, is equal to twice b diminished by eight.

Ex. 3. One third of the difference between six times x and
four, is equal to the quotient of five divided by the sum of a
and b.

Ex. 4. Three quarters of x increased by five, is equal to
three sevenths of b diminished by seventeen.

Ex. 5. One ninth of the sum of six times x and five, added
to one third of the sum of twice x and four, is equal to the
product of a, 5, and c.

Ex. 6. The quotient arising from dividing the sum of a and
I by the product of c and c?, is greater than four times the sum
of m: n, x, and y.

DEFINITIONS AND NOTATION. 19

37. In the following examples the pupil is required to trans
late the algebraic symbols into common language.

x m
1 — =•

b c a -f b'

Ans. The quotient arising from dividing the sum of a and
x by &, increased by the quotient of x divided by c, is equal to
the quotient of m divided by the sum of a and b.

Ex. 2. 7a2-\-(b — c) X (d-\-e)=x-\-y.

How should the preceding example be read when the first
parenthesis is omitted?

Ex.4,.

8« — d

Ex. 5. 2a V^? — ac = 6(a + m+x).
v

Computation of Numerical Values.

38. The numerical value of an algebraic expression is the re
sult obtained when we assign particular values to all the let
ters, and perform the operations indicated.

Suppose the expression is 2azb.

If we make a = 2 and 5 = 3, the value of this expression will
be 2x2x2x3 = 24.

If we make a=4 and 5 = 3, the value of the same expression
will be 2x4x4x3 = 96.

The numerical value of a polynomial is not affected by
changing the order of the terms, provided we preserve their re
spective signs.

The expressions a2 + 2a/; + 62, az + l2 + 2ab, and 62-f 2«& + a2,
have all the same numerical value.

Find the numerical values of the following expressions, in
which a = 6, 6 = 5, c = 4, ra = 8, and n-2.

Ex.l. a2+3«6-c2. Ans. 36 + 90-16 = 110.

Ex. 2. a2x(a-h&)-2a£c. Ans. 156.

20

ALGEBRA,

c. 5. -v/62 —

. 6. 3v'c+2aV2a-f-£+2c.
. 7. (3 A/C+ 2a) V2a

— n ra + n

JLx. 9. -

Find the numerical values of the following expressions, in
which a=3, 6=5, c=2, m=4, n=6, and x=:9.

. 10.

x — or— c

Ex. 11. 5,r— 7 Vos.
JB5p. 12. 2x2
JEfc.13. VlO+w-VlO+n.

14. 5(m2 + ?i2)+4ax.
. 15. a4-4a3+7a2-6a.

. 16. V5 VTW + S V^H- Vm+

Ans. 8.

. 175.

..

T/i — n-\-o

n

Ex. 19. — o

^

ADDITION. 2J

CHAPTEK II.

ADDITION.

39. Addition, in Algebra, is the connecting of quantities to
gether by means of their proper signs, and incorporating such
as can be united into one sum.

When the Quantities are similar and have the same Signs.

40. The sum of 3a, 4a, and 5a, is obviously 12a. That is,

So, also,'— 3a, — 4a, and — 5a, make — 12a; for the minus
sign before each of the terms shows that they are to be sub
tracted, not from each other, but from some quantity which is
not here expressed ; and if 3a, 4a, and 5a are to be subtracted
successively from the same quantity, it is the same as subtract
ing at once 12a. Hence we deduce the following

RULE.

Add the coefficients of the several quantities together, and to their
sum annex the common letter or letters, prefixing the common sign.

EXAMPLES.

3a — 3a& 2£ + 3# a— 2x2 2a+ y2

5a — 6a& 5&-f7ce 4a— Bx2

7a — ab b + Zx Sa — 5x2

a —7ab 46+ 3# 7a— x2

The pupil must continually bear in mind the remark of Art.
/26, that, when no sign is prefixed to a quantity, plus is always
to be understood.

When the Quantities are similar, but have different Signs.

41. The expression 7a — 4a denotes that 4a is to be sub*
tracted from 7a, and the result is obviously 3a. That is,

7a— 4a = 3«.

22 ALGEBRA.

The expression 5« — 2a-f 3a — a denotes that we are to sub
tract 2tt from 5a, add 3a to the remainder, and then subtract a
from the last sum, the result of which operation is 5a. That is,
5a— 2a+3a— a — oa.

It is generally most convenient to take the sum of the posi
tive quantities, which in the preceding case is 8a; then take
the sum of the negative quantities, which in this case is Sa ;|
and we have 8a— 3a, or 5#, the same result as before. Hence
we deduce the following

RULE.

Add all the positive coefficients together, and also all those that
are negative; subtract the least of these results from the greater;
to the difference annex the common letter or letters, and prefix the
sign of the greater sum.

EXAMPLES.

-3a 6x+5ay 2ay- 7 -2a2x _6a2 + 2&

4.7^ -Sx + 2ay - ay+ 8 . a2x 2a?-Sb

_j_8a x—6ay 2«y— 9 — 3a*x . — 5a2 — Sb

-_a 2:/:+ ay Say -II 7a2x 4az-2b

-4-1 la (ix+Zay

When some of the Quantities are dissimilar.

42. Dissimilar terms can not be united into one term by ad
dition.

Thus 2a and 3& neither make 5a nor 51). Their sum can
therefore only be indicated by connecting them by their proper
signs, thus, 2a + 3/>.

In adding together polynomials which contain several groups
of similar quantities, it is most convenient to write them in
such a manner that each group of similar quantities may occiv
py a column by itself. Hence we deduce the following

RULE.

Write the quantities to be added so that the similar terms may
be arranged in the same column.

Add up each column separately, and connect the several results
ly their proper signs.

ADDITION. 23

EXAMPLES.

1. Add 2a+3Z>+4c, ft+26 + 5c, 3a-6+2c, an

Ans.

2. Add 2xy-2x2 + ?/2, 3x2-f X7/+4?/2, x2— x?/4-3z/2, and 4x2
-2y2— Sxy. Ans. 6x2— x?/ + 6?/2.

3. Add 5a2x2 — 2xy, 3ax — 4x?/, 7x?/ — 4«x, a2x2 + 5x?/, and
2ax— 3xy. Aws. 6a2x2 4- 3x?/ -f ##•

4. Add 2a2 — 3ac + 36 — ccZ, 4a2 — ac + 2cd—b, 3a2 + 2«c— 4i
+ 3c^, and a2-2«c + 5c^-26. ^4?is. 10a2-4ac-4& + 9«Z.

5. Add 7w-f3rc — 14x, 3aH-9w — llm, 5:r — 4m + 8w, and

6. Add 2a

and 3a2— 2^2x-f 2ax2. Ans. 3a2x— 5x2+2rfx.

7. Add 2a262 — 3ax + 5m2?/, 2w2?/ + 3a2&2-2ax, 4ax-3m2y
— 4a2i2, and «x + 3a262 — tetfy. Ans. ±a2l2.

8. Add 7«£3-12ax2, I3ab3 + ax, Sax? -Sab3, and ~12«&3
4-9ax2-5.

9. Add 4x2+2ax+l, 3ax-2x2 + 5? 3x2-6ax+4, and 5x2

10. Add 2a3x2 — 3a2x3 — a

-|- 2ax2 + 3 ax, a n d 3 a?x2 — ax2 — a2x3 + 4a2x + ax.

11. Add 14a3x - 7a2i2 + 3a2, 5aW + 3a2^2 + 2a2
-5«3x-a2, and 4a2//-9«3x-4a2.

12. Add ax4-

, and 2

43. It must be observed that the term addition is used in a
more general sense in Algebra than in Arithmetic. In Arith
metic, where all quantities are regarded as positive, addition
implies augmentation. The sum of two quantities will there
fore be numerically greater than either quantity. Thus the
sum of 7 and 6 is 12, which is numerically greater than either
5 or 7.

But in Algebra, the quantities to be added may be either
positive or negative; and by the sum of two quantities we un
derstand their aggregate, taken with reference to their signs.

24 ALGEBRA.

Thus the sum of +7 and —5 is + 2, which is numerically less
than either 7 or 5. So, also, the sum of +a and — b is a —b.
In this case the algebraic sum "is numerically the difference of
the two quantities.

This is one instance among many in which the same terms
are used in a much more general sense in the higher mathe
matics than they are in Arithmetic.

44. When dissimilar terms have a common literal part, we
may regard the other factors as the coefficient of the common
letter or letters. The sum of the terms will then be expressed
by inclosing the sum of the coefficients in a parenthesis, and
prefixing it to the common letter or letters.

Thus the sum of ax2, bx2, and ex2 may be written

EXAMPLES.

1. Add ax, 2bx, and 3rax. Ans. (a+2b+&m)x.

2. Add Sax?/2, 2bxy2, and — 5axy2. Ans. (2b— 2a)xy2.

3. Add 2ax+3?/, 5ax— ?/, and x—4y. Ans. (7a+l)x—2y.

4. Add 2x-f 3xy, ax + te/, and bx+3mxy.

5. Add mx + ny, 3ax—%y, and 4bx+ay.

6. Add 4raV^ + 3, 2aV^— 1, and bi/x + y.

7. Add 3ax2 + 2&x— 1, 45x2 — ax + 3, and rax2— nx + 5.

8. Add 2ax4-f 36x3 — 7, 3mx4-nx3 + 2, and 4x4— ax3 + l.

9. Add awx3-fZwx2-hcx, bmx3— a??x2 + ax, and cmx3—nxr

10. Add (a-b)Vx and

SUBTiiACTiON. 25

CHAPTER III.

SUBTRACTION.

45. Subtraction is the operation of finding the difference be
tween two quantities or sets of quantities. The quantity to be
subtracted is called the subtrahend ; the quantity from which
it is to be subtracted is called the minuend; the quantity
which is left after the subtraction is called the remainder.

Let it be required to subtract 8 — 3 from 15.

Now 8 — S is equal to 5; and 5 subtracted from 15 leaves
10. The result, then, must be 10. But, to perform the opera
tion on the numbers as they were given, we first subtract 8
from 15, and obtain 7. This result is too small by 3, because
the number 8 is larger by 3 than the number which was re
quired to be subtracted, Therefore, in order to correct this re
sult, the 3 must be added, and the operation may be expressed

thua' 15-8 + 3 = 10.

Again, let it be required to subtract c-^-d from a— b. It is
plain that, if the part c were alone to be subtracted, the re
mainder would be 7

Ct "~~ 0 — — C.

But, since the quantity actually proposed to be subtracted is
less than c by d, too much has been taken away by c?, and there
fore the true remainder will be greater than a—b—c by c?, and
may hence be expressed thus,

a — b^~

where the signs of the last two terms are both contrary to what
they were given in the subtrahend. Hence we perceive that
a quantity is subtracted by simply changing its sign. In prac
tice it is most convenient to write the quantities so that simi
lar terms may be found in the same column.

B

26 ALGEBRA.

46. Hence we deduce the following

RULE.

Write the subtrahend under the minuend, arranging similar
terms in the same column.

Conceive the signs of all the terms of the subtrahend to be
changed from -f to — , or from — to -f , and then proceed as in
addition.

EXAMPLES.

1. 2. 3. 4. 5.

From I5ax2 5abx2 4abx3 — Babx2 — I2a2bc

Subtract Tax2 llabx2 — Sabx3 — Sabx2 8a2bc

Eemainder Sax2 —ftabx2 7abx3 5abx2 —

6. 7. 8.

From 3a2 -\-4b-2x 5x2-3ax2+ll

Subtract cP + tb-Sx 7x2-Sax2- 2 a?-2ab+b

Remainder 2a2-36 + 6x -2x2+5ax2 + 13 4ab

9. From 3a+5Z>— 2c subtract 2a— b. Ans. a + 66— 2c.

10. From 5aic— 2&— 6 subtract 3aic— 2&+1.

Ans.2abc—7.

11. From 4«2— 7a-h3x subtract a2+3a— 2x.

Ans. 3«2— 10a + 5x.

12. From a — 6-f2ra— x subtract 3x+m— 4b + a.

13. From 2x3— x2y+5xy2 subtract x3—2xy2+y3.

14. From m + n subtract m — n.

15. From m -\-n-\-x subtract —m—n—x.

16. From 5a2 — 3a— 7 subtract — 2a2— 4a+10.

17. From ??i44-37?^3 — 4m2— 2m+l subtract m4— 2m3 + m2
— 3m + 5.

18. From x5-5x4+10x3-3 subtract x5 + 5x4-10x3+3.

19. From 3a2+ax + 2x2— 14a2x subtract x2 — 15a2x -f 2a2

20. From 6a5x— 4m?i+5acc subtract 3mn-}-6ax— 3a&x.

Subtraction may be proved, as in Arithmetic, by adding the
remainder to the subtrahend. The sum should be equal to
the minuend.

SUBTRACTION. 27

47. It will be perceived that the term subtraction is used in
a more general sense in Algebra than in Arithmetic. In Arith
metic, where all quantities are regarded as positive, a number
is always diminished by subtraction. But in Algebra the dif
ference between two quantities may be numerically greater
than either. Thus the difference between -f a and — b is
a-\-b.

The distinction between positive and negative quantities
may be illustrated by the scale of a thermometer. The de
grees above zero are considered positive, and those below zero
negative. From five degrees above zero to five degrees below
zero, the numbers stand thus,

+ 5, d-4, +3, + 2, +1, 0, -1, -2, -3, -4, -5.

The difference between a temperature five degrees above
zero and one which is five degrees Mow zero, is ten degrees,
which is numerically the sum of the two quantities. Ten is
said to be the algebraic difference between -}-5 and —5.

48. When dissimilar terms have a common literal part, the
difference of the terms may be expressed, as in Art. 44, by in
closing the difference of the coefficients in a parenthesis, and
prefixing it to the common letter or letters.

Thus the difference between ax2 and bx2 may be written
(a-b)x2.

EXAMPLES.

1. From ax2?/2 subtract — 3x2if. Ans. (a -\- 3)x2?/2.

2. From 2aa; + 3?/ subtract 5bx—y. Ans. (2a—

3. From mx-\-ny subtract Sax—2y.

Ans. (m — 3a)x

4. From 4mVx + 3 subtract 2aVx— 1.

5. From 2ax4 + 3fcc3 — 7 subtract 3mxA — nx3 + 2.

6. From amx3 + bnx2 -f- ex subtract bmx3 — anx2 + ax.

7. From m + am -\-brn subtract am-i-bm-{-cm.

8. From ! + 3acc2+5a2z3+7a3x* subtract x2— 3az3— 5a2x4.

28 ALGEBRA.

49. Use of the Parenthesis. — If we wish to indicate that one
polynomial is to be subtracted from another, we may inclose it
in a parenthesis, and prefix the sign minus. Thus the expres
sion 7 , .

a — o — (m — n-\-x)

indicates that the polynomial m — n-\-x is to be subtracted from
the polynomial a— b. Performing the operation indicated, we

have a-b-m + n-x.

The expression a— b+(m— n+x)

indicates that the polynomial m—n-\-x is to be added to the
polynomial a— 5, and the result is

a— b+m— n-\-x.

Hence we see that a parenthesis preceded by the plus sign
may be removed without changing the signs of the inclosed
terms ; and, conversely, any number of terms, with their prop
er signs, may be inclosed in a parenthesis, and the plus sign
written before the whole.

But if the parenthesis is preceded by the minus sign, the
signs of all the inclosed terms must be changed when the pa
renthesis is removed ; and, conversely, any number of terms
may be inclosed in a parenthesis, and preceded by the minus
sign, provided the signs of all the inclosed terms are changed.

50. According to the preceding principle, polynomials may
be written in a variety of forms.

Thus, a— b— c + d

is equivalent to a— (b-\-c— d},

or to a—b—(c—d),

or to a + d— (b + c).

These expressions are all equivalent, the first form being the

simplest.

EXAMPLES.

Eeduce the following expressions to their simplest forms.

a&2).

Ans. a3-5

SUBTRACTION. 29

2. a+b+c — (a— b)— (b— c). Ans. b+2c.

3. 4a2— b— (2a—

4. a + 2&-3m-

5. 3a3-
6.

7.

51. Hence we see that when an expression is inclosed in a
parenthesis, the essential sign of a term depends not merely
upon the sign which immediately precedes it, but also upon
the sign preceding the parenthesis.

Thus m+(+n) is equivalent to m + n,
and ra~+(— n) m—n.

But m—('\-n) " m— n,

and m—(—n) " m-f-rz.

The sign immediately preceding n is called the sign of the
quantity ; the sign preceding the parenthesis may be called the
sign of the operation ; while the sign resulting from the opera
tion is called the essential sign of the term. We perceive that
when the sign of the quantity is the same as the sign of opera
tion, the essential sign of the term is positive ; but when the
sign of the quantity is different from the sign of operation, the
essential sign of the term is negative.

52. Use of Negative Quantities. — The introduction of nega
tive quantities into Algebra enables us not only to compare
the magnitude, but also to indicate the relation or quality of the
objects about which we are reasoning. This peculiarity will
"be understood from a few examples :

1st. Gain and Loss in Trade. — Suppose a merchant to gain
in one year a certain sum, and in the following year to lose a
certain sum ; we are required to determine what change has
taken place in his capital. This may be indicated algebraical
ly by regarding the gains as positive quantities, and the losses
as negative quantities. Thus, suppose a merchant, with a cap
ital of 10,000 dollars, loses 3000 dollars, afterward gains 1000

30 ALGEBRA.

dollars, and then loses again 4000 dollars, the whole may be
expressed algebraically thus,

10,000-3000 + 1000-4000,

which reduces to +4000. The + sign of the result indicates
that he has now 4000 dollars remaining in his possession.
Suppose he further gains 500 dollars, and then loses 7000 dol
lars. The whole may now be expressed thus,

10,000-3000 + 1000-4000 + 500-7000,

which reduces to —2500. The — sign of the result indicates
that his losses exceed the sum of all his gains and the property
originally in his possession ; that is, he owes 2500 dollars more
than he can pay.

53. 2c?. Motion in Contrary Directions. — Suppose a ship to
sail alternately northward and southward, and we are required
to determine the last position of the ship. This may be indi
cated algebraically, if we agree to consider motion in one direc
tion as a positive quantity, and motion in the opposite direction'
as a negative quantity.

Suppose a ship, setting out from the equator, sails north
ward 50 miles, then southward 30 miles, then northward 10
miles, then southward again 20 miles, and we wish to determ
ine the last position of the ship. If we call the northerly mo
tion + , the whole may be expressed algebraically thus,

50-30 + 10-20,

which reduces to + 10. The positive sign of the result indi
cates that the ship was north of the equator by 10 miles.

Suppose the same ship sails again 40 miles north, then 70
miles south, the whole may be expressed thus,

50-30 + 10-20+40-70,

which reduces to —20. The negative sign of the result indi
cates that the ship was now south of the equator by 20 miles.
We have here regarded the northerly motion as + , and the

SUBTRACTION. 31

southerly motion as — ; but we might with equal propriety
have regarded the northerly motion as — , and the southerly
motion as +. It is, however, indispensable that we adhere to
the same system throughout, and retain the proper sign of the
result, since this sign shows whether the ship was at any time
north or south of the equator.

In the same manner, if we regard westerly motion as -f- , we
must regard easterly motion as — , and vice versa; and, gen
erally, when quantities which are estimated in different direc
tions enter into the same algebraic expression, those which are
measured in one direction being treated as -f , those which are
measured in the opposite direction must be regarded as — .

54. The^same principle is applicable to a great variety of
examples in Geography, Astronomy, etc. Thus, north latitude
is generally indicated by the sign +, and south latitude by the
sign — . West longitude is indicated by the sign +, and east
longitude by the sign — .

Degrees of a thermometer above zero are indicated by the
sign -f , while degrees below zero are indicated by the sign — .

A variation of the magnetic needle to the west is indicated
by the sign +, while a variation to the east is indicated by the
sign — .

The date of an event since the birth of Christ is indicated
by the sign -f ; the date of an event before the birth of Christ,
by the sign — ; and the same distinction is observed in a great
variety of cases which occur in the application of the mathe
matics to practical problems. In all such cases the positive
and negative signs enable us not merely to compare the mag
nitude, but also to indicate the relation of the quantities con
sidered.

32 ALGEBRA.

CHAPTER IV.

MULTIPLICATION.

55. Multiplication is the operation of repeating one quantity
as many times as there are units in another.

The quantity to be multiplied is called the multiplicand;
and that by which it is to be multiplied is called the multiplier.

56. When several quantities are to be multiplied together,
the result will be the same in whatever order the multiplica
tion is performed.

In order to demonstrate this principle, let unity be repeated
five times upon a horizontal line, and let there be formed four
such parallel lines, thus,

Then it is plain that the number of units in the table is
equal to the five units of the horizontal line repeated as many
times as there are units in a vertical column ; that is, to the
product of 5 by 4. But this sum is also equal to the four units
of a vertical line repeated as many times as there are units in
a horizontal line ; that is, to the product of 4 by 5. Therefore
the product of 5 by 4 is equal to the product of 4 by 5. For
the same reason, 2x3x4 is equal to 2 x 4 x 3, or 4 x 3 X 2, or
3x4x2, the product in each case being 24. So, also, if o, 6,
and c represent any three numbers, we shall have ale equal to
lea or cab.

CASE I.

When loth the factors are monomials.

57. Suppose it is required to multiply 5a by 46. The prod
uct may be indicated thus, 5a x 4:1.

MULTIPLICATION. 33

But since the order of the factors may be changed without
affecting the value of the product, the factors of the same kind
may be written together thus,

4 x Sab ;
or, simplifying the expression, we have

Hence we see that the coefficient of the product is equal to the
product of the coefficients of the multiplicand and multiplier.

58. The Law of Exponents. — We have seen, in Art. 16, that
when the same letter appears several times as a factor in a
product, this is briefly expressed by means of an exponent.
Thus, aaa is written a3, the number 3 showing that a enters
three times "as a factor. Hence, if the same letters are found in
two monomials which are to be multiplied together, the ex
pression for the product may be abbreviated by adding the
exponents of the same letter. Thus, if we are to multiply a3
by a2, we find a3 equivalent to aaa, and a2 to aa. Therefore
the product will be aaaaa, which may be written a5, a result
which is obtained by adding together 3 and 2, the exponents
of the common letter a. Hence we see that the exponent of
any letter in the product is equal to the sum of the exponents of
tiiis letter in the multiplicand and multiplier.

59. Hence, for the multiplication of monomials, we have the
following

RULE.

Multiply together the coefficients of the two terms for the coeffi
cient of the product.

Write after this all the letters in the two monomials, giving to
lack letter an exponent equal to the sum of its exponents in the twc
factors.

EXAMPLES.

1. 2. 3. 4.

Multiply lobe 5a2b3c $amx*y 6a263c*

by 5mn 8a&2c ^am^xy3 8a3bc2

Product 35abcp°<n 15a365c2 36a2ra3x5?/4 48a5&4c°

B2

34 ALGEBRA.

5. Multiply 9a3x by 7a3y.

6. Multiply 12aW by llaW.

7. Multiply am by an. Ans. am+n.

8. Multiply 8a™ by 9an.

9. Multiply am by am. .Arcs. a2m.

10. Multiply 9am by 12am.

11. Multiply a™& by abm. Ans. am+15m+1.

12. Multiply 6amxn by 5amcc. J.ws. 30a2ma:7l+1.

13. Multiply 3a2mxn by 4amx3n. Ans. 12a3mx4ra.

14. What is the continued product of 5a, 4m2#, and 9a2m3x?

Ans. lSOa3m5x2.

15. What is the continued product of 7a2&, a&5, and 4ac3F

16. What is the continued product of 3amx, 5ab2, and lalxf

17. What is the continued product of a, a£>, a5c, abed, and

18. What is the continued product of a3, a362, a36c, and

CASE II.

o??e or loth of the factors are polynomials.
60. Represent the sum of the positive terms of any polyno
mial by a, and the sum of the negative terms by b. Then
a— b will represent any polynomial whatever. In like man
ner, c—d will represent any other polynomial whatever. It is
required to find the product of a — b by c — d.

In the first place, let us multiply a — b by c. This implies
that the difference of the units in a and b is to be repeated c
times. If + a be repeated as many times as there are units in
c, the result will be -\-ac. Also, if — b be repeated as many
times as there are units in c, the result will be — Jc, for —b
taken twice is — 2&, taken three times is — 36, etc.; and if it be
repeated c times, the result will be — cb or — be. The entire
operation may be exhibited thus :

a-b
c

ac—bc.
Next let us multiply a— b by c—d. When we multiply

MULTIPLICATION. 35

a_5 by Cj we obtain ac—bc. But a — b was only to De taken
c— d times; therefore, in this first operation, we have repeated
it d times more than was required. Hence, to have the true
product, we must subtract d times a— b from ac—bc. But c?
times a— b is equal to ad—bd, which, subtracted from ac—bc,

gives ac-bc-ad+bd.

If the pupil does not perceive the force of this reasoning, it
will be best to repeat the argument with numbers, thus : Let it
be proposed to multiply 8—5 by 6—2; that is, the quantity
8—5 is to be repeated as many times as there are units in
6 — 2. If we multiply 8—5 by 6, we obtain 48 — 30; that is,
we have repeated 8—5 six times. But it was only required to
repeat the multiplicand four times, or 6—2. We must there
fore diminish this product by twice 8 — 5, which is 16—10;
and this subtraction is performed by changing the signs of the
subtrahend. Hence we have

48-30-16 + 10,

which is equal to 12. This result is obviously correct; for
8— 5 is equal to 3, and 6 — 2 is equal to 4; that is, it was re
quired to multiply 3 by 4, the result of which is 12, as found
above.

We have thus obtained the following results :

•fax (+&)=

+ ax(— b)= — ab,
-ax(+b}=-ab,
— ax(— &)=

from which we perceive that when the two factors have like
signs, the product is positive; and when the two factors have un
like signs, the product is negative.

61. Hence, for the multiplication of polynomials, we have
the following general

BULB.

Multiply each term of -the multiplicand by each term of the mul-

36 ALGEBRA.

tiplier, and add together all the partial products, observing that
like signs require + in the product, and unlike signs — .

EXAMPLES.
1. 2.

Multiply 2a + Sb a2-f-2a&-f&2

by 4a— 56 a+b

Partial (8a2+12a& a3+2a2£ + ab2

products (_ -Wab-15b*
Eesult 8a2 + 2a&-15ia

It is immaterial in what order the terms of a polynomial
are arranged, or in what order the letters of a term are ar
ranged. It is, however, generally most convenient to arrange
the letters of a term alphabetically, and to arrange the terms
of a polynomial in the order of the powers of some common
letter.

3. Multiply a2— ab + b2 by a-\-b. Ans. a3-f &3.

4. Multiply a2-2ab+b2 by a -b.

5. Multiply 3a2-2a+5 by a— 4.

6. Multiply cP — db + W by a2 + ab + b2. Ans. a4 + a2b2+b*.

7. Multiply 2a2— 3a& + 4 by a2-f 2a&— 3.

8. Multiply a3 + a26 + a&2 + &3 by a—b.

9. Multiply a + mb by a + ??5.

10. Multiply 3a+2fcc-3x2 by 3a-2bx+3x*.

11. Multiply together £—5, x-f 2, and x+3.

12. Multiply together x — 3, x — 4, x-j-5, and cc — 6.

13. Multiply together a2+a6 + Z>2, a2— ab + b2, and a2-62.

14. Multiply together a + x, b-\-x, and c-\-x.

15. Multiply a4 + a3& + a2£2 + «&3 + 64 by a-b.

16. Multiply a3-3a2 + 7a-12 by a2+3a+2.

17. Multiply x4 + 2x3 + 3x2 + 2x+l by x2— 2x+l.

18. Multiply 14a3x— 6a2bx+x2 by 14a3x+6a2^— a;2.

19. Multiply x3—x2y+xy2 by x2—xif—y*.

20. Multiply 3x2+8x?/-5 by 4x2-7xi/4- 9.

62. Degree of a Product. — Since, in the multiplication of two
monomials, every factor of both quantities appears in the prod

MULTIPLICATION. 37

uct, it is obvious that the degree of the product will be equal
to the sum of the degrees of the multiplier and multiplicand.
Hence, also, if two polynomials are homogeneous, their product
will be homogeneous.

Thus, in the first example of the preceding article, each
term of the multiplicand is of the first degree, and also each
term of the multiplier; hence each term of the product is of
the second degree. For a similar reason, in the second exam
ple, each term of the product is of the third degree ; and in
the sixth example, each term of the product is of the fourth
degree. This principle will assist us in guarding against er
rors in the multiplication of polynomials, so far as concerns
the exponents.

63. Number of Terms in a Product. — When the product aris
ing from the multiplication of two polynomials does not admit
of any reduction of similar terms, the whole number of terms
in the product is equal to the product of the numbers of the
terms in the two polynomials. Thus, if we have five terms in
the multiplicand, and four terms in the multiplier, the whole
number of terms in the product will be 5x4, or 20. In gen
eral, if there be m terms in the multiplicand, and n terms in
the multiplier, the whole number of terms in the product will
be mxn.

64. Least Number of Terms in a Product. — If the product of
two polynomials contains similar terms, the number of terms
in the product, when reduced, may be much less than mn ;
but it is important to observe that among the different terms
of the product there are always two which can not be combined
with any others. These are,

1st. The term arising from the multiplication of the two
terms affected with the highest exponent of the same letter.

2d. The term arising from the multiplication of the two
terms affected with the lowest exponent of the same letter.

For it is evident, from the rule of exponents, that these iwo
partial products must involve the letter in question, the one

38 ALGEBRA.

with a higher, and the other with a lower exponent than any
of the other partial products, and therefore can riot be similar
to any of them. Hence the product of two polynomials can never
contain less than tivo terms.

65. For many purposes it is sufficient merely to indicate the
multiplication of two polynomials, without actually perform
ing the multiplication. This is effected by inclosing the poly
nomials in parentheses, and writing them in succession, with
or without the sign x . When the indicated multiplication
has been actually performed, the expression is said to be ex
panded.

EXAMPLES.

1. Expand (a+b) (c-{-d). Ans. ac-\-bc + a

2. Expand 9a-7(6-c).

3. Expand and reduce

4. Expand and reduce

28(a-£ + c)+

5. Expand and reduce

24a-6&-9(a+&)

6. Expand and reduce

66. The three following theorems have very important ap
plications.

The square of the sum of two numbers is equal to the square of
the first, plus twice the product of the first by the second, plus the
square of the second.

Thus, if we multiply a-\-b
by a + b

a'z -- ab

we obtain the product a2

Hence, if we wish to obtain the square of a binomial, we
can, according to this theorem, write out at once the term,?

MULTIPLICATION. 39

of the result, without the necessity of performing an actual
multiplication.

EXAMPLES.

1. (3a + Z>)2^ 6. (5a24-7aZ>)2=:

3. (5a-J-3&)2- 8. (2a + i)2-

4. (5a2-f2£)2 = 9. (l+^)2 =

5. (5a3+b}2 = 10.

67. The square of the difference of two numbers is equal to ih&
square of the first, minus twice the product of the first by the sec
ond, plus the square of the second.

Thus, i£ we multiply a — b
by a— b

a2 — ab

we obtain the product a2—2ab-\-b2.

EXAMPLES.

1. (2a— 3b)2 = 6. (7a2— 12«6)2 —

2. (5a— ±b)2= 7. (7a262— 12a^)2 =

3. (6«2_x)2= 8. (2a3-5)2=r

4. (6a2— Sx)2= 9. (2—£)2=

68. Meaning of the sign ±.

Since (a + b)2 = a'2-}-2ab-\-b2,

and (a—b)2 = a2—2ab-\-b'2J

we may write both formulas in the following abbreviated form,

which indicates that, if we use the + sign of b in the root, we
must use the + sign of 2ab in the square ; but if we use the
— sign of b in the root, we must use the — sign of 2ab in the
square. By this notation we are enabled to express two dis
tinct theorems by one formula.

40 ALGEBRA.

69. The product of the sum and difference of two numbers is
equal to the difference, of their squares.

Thus, if we multiply a-j-5
by a — b

a?+ab

-ab-b*
tfe obtain the product a2 — b2.

EXAMPLES.

1. (3a+26) (3a-26)=

2. (lab+x) (7ab-x) =

3. (8a+7bc) (8a-7bc) =
4.

5.

6.

7.

8. (4+|) (4-*)=

The pupil should be drilled upon examples like the pre
ceding until he can produce the results mentally with as great
facility as he could read them if exhibited upon paper, and
without committing the common mistake of making the square
of a + 6 equal to a2 + 62, or the square of a — b equal to a?—b2.
The utility of these theorems will be the more apparent
when they are applied to very complicated expressions. Fre
quent examples of their application will be seen hereafter.

DIVISION. 41

CHAPTER Y.

DIVISION.

70. Division is the converse of multiplication. In multipli
cation we determine the product arising from two given fac
tors. In division we have the product and one of the factors
given, and we are required to determine the other factor.

The dividend is the product of the divisor and quotient, the
divisor is the given factor, and the quotient is the factor re
quired to be found.

When the divisor and dividend are loth monomials.

71. Since the product of the numbers denoted by a and b is
denoted by a&, the quotient of ab divided by a is b; that is,
ab-^-a^b. Similarly, we have abc^-a=bc, abc^rc=ab, abc-^ab
=c, etc. The division is more commonly denoted thus:

abc . abc _ abc _ ,

a ~ b c

abc abc 7 abc

So, also, 12ran divided by 3m gives &i; for 3m multiplied
by 4:n makes 12m?2.

72. Rule of Exponents in Division. — Suppose we have a5 to
be divided by a2. We must find a quantity which, multiplied
by a2, will produce a5. We perceive that a3 is such a quanti
ty ; for, according to Art. 58, in order to multiply a3 by a2, we
add the exponents 2 and 3, making 5 ; that is, the exponent 3
of the quotient is found by subtracting 2, the exponent of the
divisor, from 5, the exponent of the dividend.

Hence, in order to divide one power of any quantity by an
other power of the same quantity, subtract the exponent of the
divisor from the exponent of the dividend.

42 ALGEBRA.

73. Proper Sign of the Quotient. — The proper sign to be pre
fixed to a quotient may be deduced from the principles al
ready established for multiplication. The product of the di
visor and quotient must be equal to the dividend. Hence,

because + ax(+&)=+a

— ax( + b)=—ab. , c }

; ,( 7' therefore

+ ax(— b)=— ab,

+ab^—b=—

-ax(-b)=+ab,

Hence, if the dividend and divisor have like signs, the quotient
will be positive ; but if they have unlike signs, the quotient will be
negative.

74. Hence, for dividing one monomial by another, we have
the following

RULE.

1. Divide the coefficient of the dividend by the coefficient of the
divisor, for a new coefficient.

2. To this result annex all the letters of the dividend, giving to
each an exponent equal to the excess of its exponent in the dividend
above that in the divisor.

3. If the dividend and divisor have like signs, prefix the plus
sign to the quotient; but if they have unlike signs, prefix the
minus sign.

EXAMPLES.

1. Divide 20ax3 by 4o?. Arts. Sax"2.

2. Divide 25&3x?/4 by — 6ay2. Ans. — 5a2xy2.

3. Divide — 72ab5x2 by I2b3x. Ans. — 6ab2x.

4. Divide -77a3//c6 by -Uab3c\ Ans.

5. Divide 48aWd by -I2ab2c.

6. Divide -I50a5b9cd3 by 30aW2.

7. Divide -250a7£V by -5abx3.

8. Divide 272aWx6 by -I7a2b3cx*.

9. Divide -42a663c by 21a63a
10. Divide -SOQa3b*x by -50bx.

DIVISION. 43

75. Value of the Symbol a°. — The rule given in Art. 72 con
ducts us in some cases to an expression of the form a°. Let
it be required to divide a2 by a2. According to the rule, the
quotient will be a2"2, or a°. Now every number is contained
in itself once ; hence the value of the quotient must be unity ;
that is, a° = l.

To demonstrate this principle generally, let a represent any
quantity, and m the exponent of any power whatever. Then,
by the rule of division,

a™ •+• a™ =a'm-m=a().

But the quotient obtained by dividing any quantity by it
self is unity ; that is, a° = l,
or any quantity having a cipher for its exponent is equal to unity.

76. Signification of Negative Exponents. — The rule given in
Art. 72 conducts us in some cases to negative exponents.
Thus, let it be required to divide a3 by a5. We are directed
to subtract the exponent of the divisor from the exponent of
the dividend. We thus obtain

a3"5, or a~2.

a3

But a3 divided by a5 may be written -gj and, since the

ct

value of a fraction is not altered by dividing both numerator
and denominator by the same quantity, this expression is

equivalent to — .

Hence a~2 is equivalent to — =.

az

So, also, if a2 is to be divided by a5, this may be written
a2 1

~~A — — 5 — a
5 3

-3

a" aj
In the same manner, we find

— =a-m-

that is, any quantity having a negative exponent is equal to the
reciprocal of that quantity with an equal positive exponent.

44 ALGEBKA.

77. Hence any factor may be transferred from the numera
tor to the denominator of a fraction, or from the denominatoi:
to the numerator, by changing the sign of its exponent.

Thus, ^ may be written ab~\
g " a^»,

that is, the denominator of a fraction may be entirely re
moved, and an integral form be given to any fractional ex
pression.

This use of negative exponents must be understood simply
as a convenient notation, and not as a method of actually de
stroying the denominator of a fraction.

78. To divide a Polynomial by a Monomial. — We have seen,
Art. 60, that when a single term is multiplied into a polyno
mial, the former enters into every term of the latter.

Thus, (a+b)m=am+bm;

therefore (am-\-bm)~m=a-\-b.

Hence, to divide a polynomial by a monomial, we have the
following

RULE.

Divide each term of the dividend by the divisor, ^nd connect the
quotients by their proper signs.

EXAMPLES.

1. Divide 3cc3 + 6x2+3a:e— I5x by 3x. Ans. a;2 4- 2^+^— 5.
Z Divide 3abc+I2abx— 9a2b by Sab. Ans, c-f 4cc — 3o-

3. Divide 40a3&3+60a262-17a& by -ab.

4. Divide 15a2bc— Wacx2 + Sa&d2 by — 5a2c.

5. Divide 20;r5-35x4-15x3+75a;2 by -5x2.

6. Divide 6a2x4z/6-12a3xy + 15a4cc5?/3 by 3a2x2?/2.

7. Divide xn+l — xn+* + xn+3 — xn+4 by xn.

8. Divide 12ay-16a5/ + 20ay-28ay by -4ay.

DIVISION. 45

79. To divide one polynomial ~by another.
Let it be required to divide

2a6-f a2+62 by a+b.

The object of this operation is to find a third polynomial
which, multiplied by the second, will reproduce the first.

It is evident that the dividend is composed of all the partial
products arising from the multiplication of each term of the
divisor by each term of the quotient, these products being
added together and reduced. Hence, if we can discover a
term of the dividend which is derived without reduction from
the multiplication of a term of the divisor by a term of the
quotient, then dividing this term by the corresponding term
of the divisor, we shall be sure to obtain a term of the quo
tient.

But, from Art. 64, it appears that the term a2, which con
tains the highest exponent of the letter a, is derived without re
duction from the multiplication of the two terms of the divisor
and quotient which are affected with the highest exponent of
the same letter. Dividing the term a2 by the term a of the
divisor, we obtain a, which we are sure must be one term of
the quotient sought. Multiplying each term of the divisor by
a, and subtracting this product from the proposed dividend,
the remainder may be regarded as the product of the divisor
by the remaining terms of the quotient. We shall then ob
tain another term of the quotient by dividing that term of the
remainder which is affected with the highest exponent of a by
the term a of the divisor, and so on.

Thus we perceive that at each step we are obliged to search
for that term of the dividend which is affected with the high
est exponent of one of the letters, and divide it by that term of
the divisor which is affected with the highest exponent of tha
same letter. We may avoid the necessity of searching for this
term by arranging the terms of the divisor and dividend in
ike order of the powers of one of the letters.

The operation will then proceed as follows:

4:6 ALGEBRA.

The arranged dividend is gP+Zab+b* I a-\-b, the divisor.

ft2+ ab _ I a+Vthe quotient.
ab + b2, the first remainder.

0, remainder.

For convenience of multiplication, the divisor is written on
the right of the dividend, and the quotient under the divisor,

80. Hence, to divide one polynomial by another, we have
the following

RULE.

1. Arrange loth polynomials in the order of the powers of the
same letter.

2. Divide the first term of the dividend by the first term of the
divisor, for the first term of the quotient.

3. Multiply the whole divisor by this term, and subtract the
product from the dividend.

4. Divide the first term of the remainder by the first term of the
divisor, for the second term of the quotient.

5. Multiply the whole divisor by this term, and subtract the
product from the last remainder.

6. Continue the same operation until a remainder is found equal
to zero, or one whose first term is not divisible by the first term of
the divisor.

When a remainder is found equal to zero, the division is
said to be exact. When a remainder is found whose first
term is not divisible by the first term of the divisor, the exact
division is impossible. In such a case, the last remainder must
be placed over the divisor in the form of a fraction, and an
nexed to the quotient.

EXAMPLES.

1. Divide 2a2& + &3 + 2a&2 + «3 by a2 + b2 + ab. Ans. a + b.

2. Divide x3 — a3 + 3a2x— 3ax2 by x— a. Ans. x2—2ax-{-az.
B. Divide a6 + «6 + 2a3a;3 by a2— ax+x2.

Ans. «4-f

DIVISION. 47

4. Divide a6-16a3x3 + 64x6 by a2-

5. Divide a4-f-6a2x2— 4a3x+x4— 4ax3 by a2—

.Afts. a2 —

6. Divide 32x5 + ?/5 by 2a?+y.

7. Divide x5 + x3?/2-fx?/4— x4?/ — x2y3 — y5 by x^—

8. Divide x4 + x3-f 5x— 4x2— 3 by a2— 2x— d.

9. Divide a6-b6 by a3 + 2a26-f 2a

10. Divide x6 + l-2x3 by cc2 + l-

11. Divide x* + y*+xzy2 by x2 +

12. Divide 12x4-192 by 3x-6.

Ans. 4x3 + 8x2 + 16x+32.

13. Divide 6x6-6/ by 2^c2-2?/2.

14. Divide a6 + 3a264-3a462-^6 by a3-

15. Divide x6-6x4-f 9cc2-4 by #2-l.

16. Divide a4 + a36 — 8a2^2 + 19«63— 15Z>4 by

17. Divide x3 + ?/3 + 3x?/— 1 by x+y—l.

18. Divide a2&2+2a&c2— a2c2 — 62c2 by ab + ac — bc.

19. Divide a3-63 by a -b.

20. Divide a4-64 by a-b.

81. Hitherto we have supposed the terms of the quotient to
be obtained by dividing that term of the dividend which is
affected with the highest exponent of a certain letter. But,
from Art. 64, it appears that the term of the dividend affected
with the lowest exponent of any letter is derived without re
duction from the multiplication of a term of the divisor by a
term of the quotient. Hence we may obtain a term of the
quotient by dividing the term of the dividend affected with
the lowest exponent of any letter by the term of the divisor
containing the lowest exponent of the same letter; and we
may even operate upon the highest and lowest exponents of a
certain letter alternately in the same example.

82. an — bn is always divisible by a—b. From the examples
of Art. 80 we perceive that a3 — b3 is divisible by a — b; and
ft*_/;4 is divisible by a—b. We shall find the same to hold

4:8 ALGEBRA.

true, whatever may be the value of the exponents of the two
letters ; that is, the difference of the same powers of any two quan
tities is always divisible by the difference of the quantities.
Thus, let us divide a5 — b5 by a— b:

a5 — b5 a — 5, divisor.

a4, partial quotient

The first term of the quotient is a4, and the first remainder
is a46— Z>5, which may be written

b(a*-b*).

Now if, after a division has been partially performed, the
remainder is divisible by the divisor, it is obvious that the
dividend is completely divisible by the divisor. But we have
already found that a4 — b* is divisible by a^b ; therefore a5 — b5
is also divisible by a — b ; and, in the same manner, it may be
proved that aG—b6 is divisible by a— 6, and so on.

83. To exhibit this reasoning in a more general form, let n
represent any positive whole number whatever, and let us at
tempt to divide an— bn by a— b. The operation will be as

follows ;

an—bn a — Z>, divisor.

an — ban-1 I a71"1, quotient.
The first remainder is ban~1—bn.

Dividing an by a, we have, by the rule of exponents, an~l
for the quotient. Multiplying a— b by this quantity, and sub
tracting the product from the dividend, we have for the first
remainder ban~l — bn, which may be written

b(an-^-bn~l).

Now, if this remainder is divisible by a — b, it is obvious
chat the dividend is divisible by a — b ; that is, if the difference
of the same powers of two quantities is divisible by the difference of
the quantities, then will the difference of the powers of the next
higher degree be divisible by that difference.

Therefore, since a4 — b4 is divisible by a— &, a5 — bs must be

DIVISION. 49

divisible by a— b; also a6— Z>6, and so on for any positive
value of n.

The quotients obtained by dividing the difference of the
same powers of two quantities by the difference of the quan
tities follow a simple law. Thus,

(a6 - b5) -r- (a - b) = a* + a3b + a2b2 + ab3 + 64,
etc. etc. etc.

The exponents of a decrease by unity, while those of b in
crease by unity.

84. It may also be proved that the difference of like even pow
ers of any two quantities is always divisible by the sum of the
quantities.

Thus, (a2-b2)^(a + b) = a-b.

(a6 _ b6) -r (a + b) = a5 - a*b + a3b2 - a2b3 + ab* - b5,
etc. etc. etc.

Also, the sum of like odd powers of any two quantities is always
divisible by the sum of the quantities.
Thus, (a3

(a7 + 67) H- (a + 6) = a6 - a5b + a4/>2 - a3b3 -f a2Z>4 - a 65 + 56,

etc. etc. etc.

The exponents of a and b follow the same law as in Art. 83,

but the signs of the terms are alternately plus and minus.

*

85. When exact division is impossible. — One polynomial can
not be divided by another polynomial containing a letter which
is not found in the dividend; for it is impossible that one
quantity multiplied by another which contains a certain letter
should give a product not containing that letter.

C

50 ALGEBRA.

A monomial is never divisible by a polynomial, because
every polynomial multiplied by another quantity gives a prod
uct containing at least two terms not susceptible of reduction.

Yet a binomial may be divided by a polynomial containing
kny number of terms.

Thus, a4— &4 is divisible by a? + a2b + ab2+b3, and gives for

quotient a— b.

To resolve a Polynomial into Factors.

86. When a polynomial is capable of being resolved into
factors, the factors can generally be discovered by inspection,
or from the law of formation.

If all the terms of a polynomial have a common factor, that
factor is a factor of the polynomial ; and the other factor may
be found by dividing the polynomial by the common factor.

EXAMPLES.

1. Eesolve 3a262 + 3a63+3a£2c into factors.

Ans. 3ab2(a + b+c).

2. Eesolve 5a4b2-Wa3b3-5a2b*-5a2b2 into factors.

Ans. 5a2b2(a2-2ab-b2-l).

3. Eesolve Qa2b2c3 — I2ab2c3mx— I8ab2c3y into factors.

Ans. 6ab2c3(a-2mx-Sy).

4. Eesolve 7a3b2 — 7a2b3 — 7a2b2c into factors.

5. Eesolve 8a2bc+l2ab2c— I6abc2 into factors.

6. Eesolve 10ab2cmx—5ab2cy-\-5ab2c into factors.

87. When two terms of a trinomial are perfect squares, and
tke third term is twice the product of their square roots, the
trinomial will be the square of the sum or difference of these
roots, Arts. 66 and 67, and may be resolved into factors ac
cordingly.

EXAMPLES.

1. Eesolre ji2— 2ab-{-b2 into factors. Ans. (a — b)(a — b).

2. Eesolve «2 + 4a^4-4^2 into factors.

Ans. (a + 25) (a + 25).

3. Eesolve a2— 6ab + 9b2 into factors.

DIVISION. 61

4. Kesolve 9a2— 24a&4-1662 into factors.

5. Kesolve 25a4— 60a2&3 + 3666 into factors.

6. Kesolve kmW— 4mn-f-l into factors.

7. Kesolve 49a464— 168a363+144a262 into factors.

8. Kesolve ^3+2n2+w into three factors.

9. Resolve 16a462— 24a2&rax+9?ft2x2 into factors.
10. Kesolve m47i2+2m3^3-|-m2ni into three factors.

88. If a binomial consists of two squares connected by the
minus sign, it must be equal to the product of the sum and
difference of the square roots of the two terms, Art. 69, and
may be resolved into factors accordingly.

EXAMPLES.

1. Kesolve 4a2-952 into factors. Ans. (2a+36) (2a— 3&>

2. Kesolve 9a262— 16a2c2 into factors.

3. Kesolve a5x— 9ax3 into three factors.

4. Kesolve a4— b* into three factors.

5. Kesolve a6—b6 into its factors.

6. Kesolve a8— b9 into four factors.

7. Kesolve 1— TV into two factors.

8. Resolve 4 — TV into two factors.

89. If the two terms of a binomial are both powers of the
same degree, it may generally be resolved into factors accord
ing to the principles of Arts. 82-84.

EXAMPLES.

1. Resolve a? — b3 into its factors. Ans. (a2 -\-ab-\-b2) (a^b).

2. Resolve a3 + b3 into its factors.

3. Resolve a6 — b6 into four factors.

4. Resolve a3 — Sb3 into its factors,

5. Resolve 8a3— 1 into its factors.

6. Resolve 8a3— Sb3 into three factors.

7. Resolve 1 + 2763 into its factors.

8. Resolve 8a3 + 27&3 into its factors.

9. Resolve a16 — b16 into five factors.

62 ALGEBRA.

CHAPTEK VL-r

GEEATEST COMMON DIVISOR. — LEAST COMMON MULTIPLE.

90. A common divisor of two quantities is a quantity which
will divide them both without a remainder. Thus 2ab is a
common divisor of 6a?b2x and

91. A prime factor is one that can not be resolved into any
other factors. It is, therefore, divisible only by itself and uni
ty. Thus the quantity 2a2— 2ab is the product of the three
prime factors 2, a, and a— b.

92. The greatest common divisor of two quantities is the
greatest quantity which will divide each of them without a
remainder. It is the continued product of all the prime fac
tors which are common to both. The term greatest here refers
to the degree of a quantity, or of its leading term, and not to its
arithmetical value.

93. When both quantities can be resolved into prime fac
tors by methods already explained, the greatest common di
visor may be found by the following

RULE.

Resolve both quantities into their prime factors. The continued
product of all those factors which are common to bothj will be the
greatest common divisor required.

EXAMPLES.

1. Find the greatest common divisor of 4a2fce and 6ab2x3.
Kesolving into factors, we have

4a2fce = 2a X 2a x b x x.
6ab2x3 = 2a xSbxbxxXxxx.

The common factors are 2a, 6, and x. Hence the greatest
common divisor is 2abx.

GREATEST COMMON DIVISOK. 53

2. Find the greatest common divisor of 4am2 -f 45m2 and
Ban + 3bn.

Kesolving into factors, we have

4am2 + 45m2 = 2m x 2m (a + b).

3an+3bn=3n(a+b).
Hence a +b is the greatest common divisor.

3. Find the greatest common divisor of x3—^3 and y?—y\

x3—y3 = (x—y)(x2-{-xy-{-y'2).

Hence x— y is the greatest common divisor.

4. Find the greatest common divisor of 35a2bmx2 and
42am2cc3.

5. Find the greatest common divisor of 3a2x—6abx-4-3b2x
and 4a2y — 43)2y.

6. Find the greatest common divisor of 9mx2— 6mx + m
and 9?ix2— n.

7. Find the greatest common divisor of 12a2— 36a5-}-2763
and 8a2-1862.

94. When the given quantities can not be resolved into
prime factors by inspection, the greatest common divisor may
be found by applying the following principle :

The greatest common divisor of two quantities is the same with
the greatest common divisor of the least quantity, and their remain
der after division.

To prove this principle, let the greatest of the two quanti
ties be represented by A, and the least by B. Divide A by
B; let the entire part of the quotient be represented by Q,
and the remainder by R. Then, since the dividend must be
equal to the product of the divisor by the quotient, plus the
remainder, we shall have A— QB + R.

Now every number which will divide B will divide QB;
and every number which will divide R and QB will divide
R+QB, or A. That is, every number which is a common
divisor of B and R is a common divisor of A and B.

Again : every number which will divide A and B will di-

54 ALGEBRA.

vide A and QB ; it will also divide J.— QB, or E. That is,
every number which is a common divisor of A and B is also
a common divisor of B and R. Hence the greatest 'common
divisor of A and B must be the same as the greatest common
divisor of B and R.

95. To find, then, the greatest common divisor of two quanti
ties, we divide the greater by the less; and the remainder,
which is necessarily less than either of the given quantities,
is, by the last article, divisible by the greatest common divisor.

Dividing the preceding divisor by the last remainder, a still
smaller remainder will be found, which is divisible by the great
est common divisor; and by continuing this process with each
remainder and the preceding divisor, quantities smaller and
smaller are found, which are all divisible by the greatest com
mon divisor, until at length the greatest common divisor must
be obtained. Hence we have the following

o
RULE.

Divide the greater quantity by the less, and the preceding divisor
by the last remainder, till nothing remains ; the last divisor will be
the greatest common divisor.

When the remainders decrease to unity, the given quanti
ties have no common divisor greater than unity, and are said
to be incommensurable, or prime to each other.

EXAMPLES.

1. What is the greatest common divisor of 372 and 246?

372
246

246

1

-the first remainder.

126,

1

246
126
126 1 120, the second remainder.

120 IT"

120 | 6, the third remainder.
120 I 20
Here we have continued the operation of division until we

GREATEST COMMON DIVISOR. fc>Q

obtain 0 for a remainder; the last divisor (6) is the greatest
common divisor. Thus, 246 and 372, being each divided by 6,
give the quotients 41 and 62, and these numbers are prime
with respect to each other; that is, have no common divisor
greater than unity.

2. What is the greatest common divisor of 336 and 720 ?

Ans. 48.

3. What is the greatest common divisor of 918 and 522 ?

Ans. 18.

96. In applying this rule to polynomials some modification
may become necessary. It may happen that the first term of
the arranged dividend is not divisible by the first term of the
divisor. Tfiis may arise from the presence of a factor in the
divisor which is not found in the dividend, and may therefore
be suppressed. For, since the greatest common divisor of two
quantities is only the product of their common factors, it can
not be affected by a factor of the one quantity which is not
found in the other.

We may therefore suppress in the first polynomial all the
factors common to each of its terms. We do the same with
the second polynomial ; and if any factor suppressed is com
mon to the two polynomials, we reserve it as one factor of the
common divisor sought.

But if, after this reduction, the first term of the dividend,
when arranged according to the powers of some letter, is not
divisible by the first term of the arranged divisor, we may mul
tiply the dividend by any monomial factor which will render its
first term divisible by the first term of the divisor.

This multiplication will not affect the greatest common di
visor, because we introduce into the dividend a factor which
belongs only to a part of the terms of the divisor ; for, by sup
position, every factor common to all the terms has been sup
pressed.

97. The preceding principles are embodied in the following
general

56 ALGEBRA,

RULE.

1. Arrange the two polynomials according to the powers of some
letter ; suppress all the monomial factors of each ; and if any fac
tor suppressed is common to the two polynomials, reserve it as one
factor of the common divisor sought.

2. Multiply the first polynomial by such a monomial factor as
will render its first term divisible by the first term of the second
polynomial ; then divide this result by the second polynomial, and
continue the division till the first term of the remainder is of a
lower degree than the first term of the divisor.

3. Take the second polynomial as a dividend, and the final re
mainder in the first operation as a divisor, and proceed as before,
and so on till a remainder is found that will divide the preceding
divisor. TJiis remainder, multiplied by the common factors, if any,
reserved at the beginning, will give the greatest common divisor.

EXAMPLES.

1. Find the greatest common divisor of x34-4x24-5x-|-2 and
ic24-5x-f4.

x3+4x24-5x-f2
x34-5x2+4x

x-1

— x24- x4-2
_ x2— 5x— 4

6x4-6

Suppressing the factor 6 in this remainder, we have x+I
for the next divisor.

x2 4- 5x4-4 1x4-1
x2-j- x |x4-4
4x4-4
4x4-4

Here the division is exact; hence, by the rule, x4-l is the
greatest common divisor sought

2. Find the greatest common divisor of 6x3— Tax2— 20a2x
and 6x2+2ax-8a2.

Suppressing the factor 2 in the second polynomial, we pro
ceed thus :

GREATEST COMMON DIVISOR.

5?

6x3-7ax2-20a2x
2- Sa2x

-9ax2-12a2x
— 9acc2— 3a2a;+12a3
- 9a2x-12a3

Suppressing the factor — 3a2,

ax— 4a2

— Sax— 4a2

— a

Hence 3#+4a is the greatest common divisor.
3. Find the greatest common divisor of 4a3— 2a2— 3a+l
and 3a2— 2a— 1.

We first multiply the greater polynomial by 3, to render
its first term divisible by the first term of the other polyno
mial.

12a3— 6a2— 9a-f 3 I 3a2— 2a— 1
12a3— 8a2— 4a | 4a, +2

2a2-5a+3
6a2-15a+ 9

-lla+11

Here we multiply the first remainder by 3, to render the
first term divisible by the first term of the divisor. As the
two partial quotients 4a and 2 have no connection, they are
separated by a comma.

Kejecting the factor —11 from the second remainder, we
proceed as follows :

3a2— 3a

3a+l

a-l
a-l

Hence a— 1 is the greatest common divisor.

4. Find the greatest common divisor of a2— 3a6-f2&2 and

a*_a6— 262. Ans. a— 25.

02

58 ALGEBRA.

5. Find the greatest common divisor of a3 — azb-\-Bab'2—Bb9
and a2— 5a5-f452. Ans. a-b.

6. Find the greatest common divisor of Sx3 — 13cc2-f 23£— 21
and 6z3+#2-44x+21. Ans. 3z-7.

7. Find the greatest common divisor of

x4_7o;3+8z2+28z-48 and a3-8x2+19x-14.

Ans. x—2.

98. To find the greatest common divisor of three quantities. —
Find the greatest common divisor of the first and second, and
then the greatest common divisor of this result and the third
quantity. The last will be the greatest common divisor re
quired.

EXAMPLES.

1. Find the greatest common divisor of 3a2m2, 6&2m2, and
12ra3ce. Ans. 3m2.

2. Find the greatest common divisor of 4x3— 21x2 + 15x+20,
x2— 6^+8, and xz— x— 12. Ans. x—4.

3. Find the greatest common divisor of 6x4 -\- x3 — x^
4z3— 6x2— 4x + 3, and 2x3 + cc2+x— 1. Ans. 2x— 1.

4. Find the greatest common divisor of 4x4-f 9x3+2x2— 2x— 4,

— x+2, and x3 + ^2— ^+2. Ans. a+2.

" LEAST COMMON MULTIPLE.

99. One quantity is a multiple of another when it can be di
vided by it without a remainder. Thus 5ab is a multiple of 5,
also of a and of b. When one quantity is a multiple of an
other, the former must be equal to the product of the latter by
some entire factor. Thus, if a is a multiple of bt then a=mb,
where m is an entire number.

100. A common multiple of two or more quantities is one
which can be divided by each separately without a remainder.
Thus 20a3Z>2 is a common multiple of 4ab and 5«2&2.

101. The least common multiple of two or more quantities is
the least quantity that can be divided by each without a re-

LEAST COMMON MULTIPLE. 59

mainder. Thus 12a2 is the least common multiple of 3a2.and

102. It is obvious that the least common multiple of two or
more quantities must contain all the factors of each of the quan
tities, and no other factors. Hence, when the given quantities
can be resolved into prime factors, the least common multiple
may be found by the following

EULE.

Resolve each of the quantities into its prime factors ; take each
factor the greatest number of times it enters any of the quantities;
multiply together the factors thus obtained, and the product will be
the least common multiple required.

EXAMPLES.

1. Find the least common multiple of 9x2y and ~L2xyz.
Eesolving into factors, we have

9x2?/ = 3 x Sxxy, and I2xy2 — 3 X 2 x 2xyy.
The factor 3 enters twice in the first quantity, also the fac
tor 2 enters twice in the second; x twice in the first, and y
twice in the second. Hence the least common multiple is
2 x 2 x 3 x Sxxyy, or 36x2?/2.

2. Find the least common multiple of 4a262, 6a26, and
We have 4a262 = 2 x 2aa56,

=2 x 5aaaxx.
Hence the least common multiple is

2 X 2 x 3 x 5aaabbxx, or 60a?bW.

3. Find the least common multiple of a2x— 2abx+b2x and
y—tyy.
Here we h ave a?x — 2 abx -f b2x = (a —b)(a— b) x,

Hence the least common multiple is

(a — b) (a — b) (a + b) xy, or a3xy — ab2xy — a2bxy + b3xy.

60 ALGEBRA.

4 Find the least common multiple of 5a262, 10a&8, and 2dbx,

Ans. lOaWx.

5. Find the least common multiple of 3a&2, 4ax2, 552x, and
6a2x2. Ans. 60a2b2x2.

6. Find the least common multiple of x*— 3^+2 and x2— 1.

Ans. (x + I)(x— l)(x— 2), or x3— 2x2— cc + 2.

7. Find the least common multiple of a3x-|-63a? and 5a2— 562.

or 5a4x— 5a3bx-{-5ab3x — 564ic.

103. "When the quantities can not be resolved into factors
by any of the preceding methods, the least common multiple
may be found by applying the following principles :

If two polynomials have no common divisor, their product
must be their least common multiple ; but if they have a com
mon divisor, their product must contain the second power of this
common divisor. Their least common multiple will therefore
be obtained by dividing their product by their greatest com
mon divisor. Hence, to find the least common multiple of two
quantities, we have the following

RULE.

Divide the product of the two polynomials by their greatest com
mon divisor ; or divide one of the polynomials by the greatest com
mon divisor, and multiply the other by the quotient.

EXAMPLES.

1. Find the least common multiple of 6x2 — x — 1 and

The greatest common divisor of the given quantities is 2x—
1. Hence the least common multiple is

2) or

2. Find the least common multiple of x3— 1 and x2-{-x—2.

Ans. (x3-I)(x + 2).

3. Find the least common multiple of x3— 9x2+23x— 15
and x2— Sx+T. Ans. (x3 — 9o;2+23x-15)(^— 7).

LEAST COMMON MULTIPLE. 61

104. When there are more than tivo polynomials, find the
least common multiple of any two of them ; then find the
least common multiple of this result, and a third polynomial ;
and so on to the last.

4. Find the least common multiple of a2+2a — 3, a2—!, and
a-1. Ans. (a2-!) (a +3).

5. Find the least common multiple of 4a2-f-l, 4<x2— 1, and
2a-l. Ans. 16a4-l.

6. Find the least common multiple of a3— a, a3+l, and
a3-!. Ans. a(a6-l).

7. Find the least common multiple of (x-\-2a}3, (x— 2a)3,
and sc2-4a2. Ans. (#2-4a2)3.

62 ALGEBRA.

CHAPTER VII.

FRACTIONS.

105. A. fraction is a quotient expressed as described in Art,
71, by writing the divisor under the dividend with a line be
tween them. Thus r is a fraction, and is read a divided by b.

106. Every fraction is composed of two parts: the divisor,
which is called the denominator, and the dividend, which is
called the numerator.

107. An entire quantity is an algebraic expression which has
no fractional part, as a2 — 2a&.

An entire quantity may be regarded as a fraction whose de
nominator is unity. Thus, a2 = y.

108. A mixed quantity is an expression which has both en
tire and fractional parts. Thus «2 + - is a mixed quantity.

109. General Principles of Fractions. — The following princi
ples form the basis of most of the operations upon fractions :

1st. In order to multiply a fraction by any number, we must
multiply its numerator or divide its denominator by that number.

Thus the value of the fraction —^ is b. If we multiply the
numerator by a, we obtain — , or ab •; and if we divide the de
nominator of the same fraction by a, we obtain also ab; that is,
the original value of the fraction, b, has been multiplied by a.

Zd. In order to divide a fraction by any number, we must divide
its numerator or multiply its denominator by that number.

Thus the value of the fraction ~ is ab. If we divide the

FRACTIONS. 63

numerator by a, we obtain ^, or b ; and if we multiply the de
nominator of the same fraction by a, we obtain ^ , or b ; that
is, the original value of the fraction ab has been divided by a.

3d The value of a fraction is not changed if we multiply or di
vide both numerator and denominator by the same number.

ab abm abmx T

Thus. — = - = -- = b.

a am amx

110. The proper Sign of a Fraction. — Each term in the numer
ator and denominator of a fraction has its own particular sign,
and a sign is also written before the dividing line of a fraction.
The relation of these signs to each other is determined by the
principles already established for division. The sign prefixed
to the numerator of a fraction affects merely the dividend ; the
sign prefixed to the denominator affects merely the divisor;
but the sign prefixed to the dividing line of a fraction affects
the quotient. The latter sign may be called the apparent sign
of the fraction, while the real sign of the fraction is the sign
of its numerical value when reduced.

The real sign of a fraction depends not merely upon its <7/>
parent sign, but also upon the signs of the numerator and de
nominator. From Art. 73, it follows that

ab — ab

-ap - = +&,

a —a

, — ab ab

and • - = — =—b.

a -^-a

Also, since a minus sign before the dividing line of a frac
tion shows that the quotient is to be subtracted, which is done
by changing its sign, it follows that

ab _ —ab _
a~ — a~

and

a — a

Hence we see that of the three signs belonging to the numer-

64 ALGEBRA.

ator, denominator, and dividing line of a fraction, any two may
be changed from -+- to — , or from — to -f-> without affecting the
real sign of the fraction.

111. When the numerator or denominator of a fraction is a
polynomial, it must be observed that by the sign of the numer*
ator is to be understood the sign of the entire numerator, as dis
tinguished from the sign of any one of its terms taken singly.

m, a + b + c . , —a—b—c
Thus, — is equivalent to -\ — .

00 00

When no sign is prefixed either to the terms of a fraction or
to its dividing line, plus is always to be understood.

Reduction of Fractions.

112. To reduce a Fraction to its Lowest Terms. — A fraction Is
in its lowest terms when the numerator and denominator con
tain no common factor ; and since the value of a fraction ia
not changed if we divide both numerator and denominator
by the same number (Art. 109), we have the following

RULE.

Divide both numerator and denominator by their greatest com
mon divisor.

Or, Cancel all those factors which are common to both numerator
and denominator.

EXAMPLES.

azbc
1. Eeduce -=- to its lowest terms.

a2bc c x a2b
We have ~ 070 =

x azb'

Canceling the common factors a25, we have

c

CX \ X2

2. Reduce „ 9 to its lowest terms.

have

x(c-\-x) x

=

FRACTIONS. 65

~7a& 7a

3. Eeduce ? - ^r- to lts lowest terms. Arcs. ^-.

lOac — ooc oc

£.2 _ tt2 i

4. Eeduce —. - , to its lowest terms. Ans. — - «.

a4— a4 85*+ a*

2x2—16x _ 6

5. Kedace .2_94a;— 9 tO itS lowest terms-

6. Eeduce 0 ? ^ -- 7^3 to lts lowest terms.

—

7. Eeduce -5 — ^ , ,2 to its lowest terms.

8. Eeduce -^ — ^ - 5 to its lowest terms.

a2—

x2 — 16
9. Eeduce -5— — ^ to its lowest terms.

x2— cc— 20

3cc3-16x2-f-23x-6 . . ,

10. Eeduce ^-^ — ^ , . 1f7 - 5 to its lowest terms.

— 6

_ __ .

11. Eeduce ~-^ — ^— —5 to its lowest terms.

JjOC — 3C — X-f- 2i

— 3

12. Eeduce 75-= — g . ' c— — r to its lowest terms.
3x3 + 5x2 —

113. To reduce a Fraction to an Entire or Mixed Quantity. — •
When any term of the numerator is divisible by some term in
the denominator, the division indicated by a fraction may be
at least partially performed. Hence we have the following

RULE.

Divide the numerator ly the denominator, continuing the opera
tion as far as possible ; then write the remainder, if any, over ihi
denominator, and annex the fraction thus formed to the entire part*

EXAMPLES.

ax x^

1. Eeduce -- to an entire quantity.

7 O2

2. Eeduce - = - to a mixed quantity.

66 ALGEBRA.

3. Reduce to a mixed quantity. Ans. a-f x-\ .

a — x a—x

4. Reduce - —2. to an entire quantity.

x—y

5. Reduce — — to a mixed quantity.

ox

6. Reduce - — ^r— — to a mixed quantity.

7. Reduce = — ~ o to an entire quantity.

x2 — 3x+2 J

8. Reduce - — 2 9 , ,2 — ' - to an entire quantity.

114. To reduce a Mixed Quantity to the Form of a Fraction. —
This problem is the converse of the last, and we may proceed
by the following

RULE.

Multiply the entire part by the denominator of the fraction ; to
the product add the numerator with its proper sign, and write the
result over the denominator.

EXAMPLES.

y-y2 _ ^ rrQ yy*

1. Reduce x-\ -- to the form of a fraction. Ans. — .

x x

rtnf _L nr*&

2. Reduce x+- — to the form of a fraction.

2a Sax+x*

Ans. — n - .

2x-7 2a

3. Reduce 5H — 5 — to the form of a fraction.

ox

/v» _ /y _. "1

4. Reduce lH — - to the form of a fraction.

x—3
--r
ox

_ gc2
6. Reduce 7-\ — 5 — jj- to the form of a fraction.

x—3

5. Reduce 1-f 2#-f--r — to the form of a fraction
ox

FRACTIONS. 67

7. Reduce 2a— 7 — ~ „ to the form of a fraction.

8 Eeduce (a— I)2— - - '- to tlie form of a fraction.

A (a-
Ans. -

115. To reduce Fractions having Different Denominators fo
Equivalent Fractions having a Common Denominator :

* ct (* rrn

Suppose it is required to reduce the fractions 7, -3, and — to

a common denominator. Since, by Art. 109, both terms of a
fraction may be multiplied by the same quantity without
changing its value, we may multiply both terms of each frac
tion by the product of the denominators of the other fractions,
and we shall have

a_adn c _bcn , m__bdm

b bdn d bdn n bdn'

The resulting fractions have the same value as the proposed
fractions, and they have the common denominator bdn. Hence
we have the following

BULE.

Multiply each numerator into all the denominators, except its
own, for a new numerator, and all the denominators together for
the common denominator.

EXAMPLES.

1. Beduce T and - - to equivalent fractions having a

0 C

, A ac a

Common denominator. Ans. y-,

, .

be be

2. Eeduce ~-, ~-, and - to equivalent fractions having a
common denominator.

o 0,-v, A™.

3. Reduce p -jr-j and a+-^- to equivalent fractions having
a common denominator.

68 ALGEBRA.

4. Keduce -> -y» and - - to equivalent fractions having

a common denominator.

x x-\-~L , 1 — x

5. Keduce -> — ^— , and — — to equivalent fractions hav«

o O 1 -\-X

ing a common denominator.

116. Fractions may always be reduced to a common denom
inator by the preceding rule ; but if the denominators have any
common factors, it will not be the least common denominator.
The least common denominator of two or more fractions must
be the least common multiple of their denominators.

Suppose it is required to reduce the fractions ^ and -:- to

ox *±x

equivalent fractions having the least common denominator.
The least common multiple of the denominators is 12#2. Mul-

tiply both terms of the first fraction by -0-5-, or 4, and both

12x2
terms of the second fraction by -- , or 3x, and we shall have

4:X

Sa , I5bx
{ l

which are equivalent to the given fractions, and have the least
common denominator. Hence we deduce the following

RULE.

Find the least common multiple of all the denominators, and use
this as the common denominator.

Divide this common denominator by each of the given denomina
tors separately, and multiply each numerator by the corresponding
quotient. The products will be the new numerators.

6. Keduce ^- and -^^ to equivalent fractions having the

4ac x

least common denominator. Ans. -- and

_ j

7. Reduce - — 7 and ° 19 to equivalent fractions having

a — b az — b2 , 7N2 ,

A (a + b)2 , c+d

the least common denominator. Ans. ~ — =£• and -5 — ^.

a2 — 62 az—b2

FRACTIONS. 69

8. Eeduce 4» — » and - to equivalent fractions having the

x3 x2 x

least common denominator.

9a 7b lla , 7(a+5) . . , . f

9. Eeduce ?— , ,7^-, OQ— » and -—-1 to equivalent frac-

$7n 86m 28m 4m

tions having the least common denominator.
o g 2x _ 3

10. Reduce -, ^ - 7, and -r-z — T to equivalent fraction^

cc 2cc— 1 4cc2— 1

having the least common denominator.

r^

Addition of Fractions.

117. The denominator of a fraction shows into how many
parts a unit is to be divided, and the numerator shows how
many of those parts are to be taken. Fractions can only be
added when they are like parts of unity ; that is, when they
have a common denominator. In that case, the numerator of
each fraction will indicate how many times the common frac
tional unit is repeated in that fraction, and the sum of the nu
merators will indicate how many times this result is repeated
in the sum of the fractions. Hence we have the following

RULE.

Reduce the fractions to a common denominator / then add the
numerators together, and write their sum over the common denomi
nator.

If there are mixed quantities, we may add the entire and
fractional parts separately.

EXAMPLES.

/JT* IT o

1. What is the sum of o and gf

Reducing to a common denominator, the fractions become

jX

Adding the numerators, we obtain — .

It is plain that three sixths of x and two sixths of x mako
five sixths of x.

70 ALGEBRA.

2. What is the sum of T, -, and — ?

o a n 7 , , 7

. aan + ben 4- ft#m

J.715.

,

3. What is the sum of - 7 and

a-j-ft a—b

t TTTT- A ' ^ £ K %a J a 0

4. What is the sum of 5x. -^—^j and — -. - ?

' Sxz 4x

5. What is the sum of 2a, 3a+-> and &+••

. 6a-|--r=-.

45

6. What is the sum of a+x, - , and — ^- ?

a-x a 2

Ans.

a — b a— ax

7. What is the sum of — and ^~- ? Ans. a.

2i A

o -rrn ^ • xi r a «— 2m , a+2m0

8. What is the sum of ^, — j — , and — I— — ?

n TTTI, 4. • 4.1 r ma—b •> na-\-b~

9. What is the sum of - and -- — ?

m-\-n m-\-n

•in TTTI, 4 - xi. c x—n y—z . 2z+n „

10. What is the sum of — — , — — — , and — — — ?

x+y+z x+y + z x+y+z

11. What is the sum of = and

-- - , -

12. What is the sum of -=-. - ^-, — -=-, - j-, and —-=-7 - -^
b(a—b) 5(a— ft) 5(a— ft)

. 9.

r + aj — a; — c

13. What is the sum of - -- , —— , -- 1 , , -- T—--I
1— a; 1 + x 1+x2 1— a;3

and — 1? -r

Subtraction of Fractions.

118. Fractions can only be subtracted when they are like
parts of unity ; that is, when they have a common denomina
tor. In that case, the difference of the numerators will indi
cate how many times the common fractional unit is repented
in the difference of the fractions. Hence we have the following

FRACTIONS. 71

BULK

Reduce the fractions to a common denominator ; then subtract
Hie numerator of the subtrahend from the numerator of the minu
end, and write the result over the common denominator.

EXAMPLES.

2x . , Sx
1. From -£- subtract — .
o o

Beducing to a common denominator, the fractions become

.
15- and IS"

IQx 9x x
Hence we nave -jg — 15 "15"'

and it is plain that ten fifteenths of cc, diminished by nine fif
teenths of x, equals one fifteenth of x.

Sx

2. From -=- subtract

/ o

9a— 4x , 5a— Bx

3. From — = — subtract -- 5 — .

/ o

It must be remembered that a minus sign before the divid
ing line of a fraction affects the quotient (Art. Ill); and since
a quantity is subtracted by changing its sign, the result of the
subtraction in this case is

9a— 4x 5a — 3x
~T~ ~3~;

which fractions may be reduced to a common denominator,
ai>d the like terms united as in addition.

ax , ax 2acx

4. From 7 — subtract — -. Ans.

b— c

, ,-, 0 , 2 + 7x 5x-6 355.T-6

5. From 2x-\ -- 5 — subtract x -- ^ — . Ans.

21 168

/v» />* ___ , /

6. From 8^+^:7 subtract cc

Zo G

a-\-b a—o

7. From -~- subtract -— .

72 ALGEBKA.

, - . t -

8. From - r — subtract — - — . Ans.

6o-76 , . . 5a 64a*-15a2-63i2

- From 3^=25 SubtraCt 95'

10. From ,+ / subtract unity.

4:dO

11. From —^- subtract =- — -^.

lly Ix—by

12. From —?— subtract -^—n.

v* — * r|

Multiplication of Fractions.
119. Let it be required to multiply ^ by ~.
First let us multiply | by c. According to the first princi
ple of Art. 109, the product must be T\

But the proposed multiplier was ^ ; that is, we have used a

d

multiplier d times too great. We must therefore divide the
result by d; and, according to the second principle of Art.
109, we obtain

CtO ' . i , • Uj . L CiC

r-;," that is, rx~7— r~7«
Jc? & d bd

Hence we have the following

RULE.

Multiply the numerators together for a new numerator, and the
denominators for a new denominator.

Entire and mixed quantities should first be reduced to frac«
tional forms. Also, if there are any factors common to the nu
merator and denominator of the product, they should be can.
celed.

EXAMPLES.

_.. . x , 2cc x2

1. Multiply g by --. Ans. .

x ,

2. Multiply - by

FRACTIONS. 73

bx , a
3. Multiply 6 + — by -.

_

4. Multiply -- by

a a_

5. Multiply ^-^ by — ^.

of 4x

6. Multiply together --, and

7. Multiply together — , — , and

cc-f 1

8. Multiply together x, ~ — , and

ct

v /j/y^Y*

9. Multiply s by

10. Multiply 463c6Ua763cM ^ a2&2c*-

11. Multiply together ^^, ^^, -^^ and — .

Ans. — XT.

a-

- ,

12. Multiply together -- , = — ' and
7(m — ?ij 39 (a— b)

3dn 3bm 5mn , \\abc
13. Multiply --- 1 — =--— TH— by
r J 4ac 7 66c

a2— cc2 bc + bx , c—

14. Multiply together — :—, — - -, — - , ana
r J 22 2

c—x a2^-ax a — x

_^ _y

15. Multiply together - — , - -~-~, and
1-f?/ x + x2

A
Ans. —.

4y

ic nr i.- i
16. Multiply
rJ

17 . -

17. Multiply — — ^-j — r. by -r- — 7. -dn*.

1 J — 2 J a2 + ab a

T)

74 ALGEBRA.

18. Multiply xz-x+ 1 by ^4-jfl Ans.

-in TVT u- i a + ^ a~ b 462 a 4-6

19. Multiply —---- by --. An,. 2.

20. Multiply V--+1 by -9-f-- + l.

a2 a ; a2 a a* ' a* '

120. Multiplication of Quantities affected with' Negative Expo
nents. — Suppose it is required to multiply -^ by — .

According to the preceding article, the result must be — .

1 1 a

But, according to Art. 76, -3 may be written a~3 ; — may be
-» a a

written a~2; and -g may be written a~5.

Hence we see that a~3xa~2=ia-s;

that is, the rule of Art. 58 is general, and applies to negative as
well as positive exponents.

EXAMPLES.

1. Multiply —x~2 by x~\ Ans. — x~5, or — —

2. Multiply a-2 by -a3.

3. Multiply a~3 by a3.

4. Multiply a~m by an.

5. Multiply «x~m by a~n.

6. Multiply (a— 6)5 by (a— b)~3.

Division of Fractions.

121. If the two fractions have the same denominator, then
the quotient of the fractions will be the same as the quotient
of their numerators. Thus it is plain that f is contained in -J
as often as 3 is contained in 9. If the two fractions have not
the same denominator, we may perform the division after hav
ing first reduced them to a common denominator. Let it be

required to divide j by -,

FRACTIONS. 75

Eeducing to a common denominator, we have ^ to be di
vided by j-|. It is now plain that the quotient must be repre-
sented by the division of ad by be, which gives j- ;

a result which might have been obtained by inverting the
terms of the divisor and multiplying by the resulting frac
tion ; that is. , ,
a^€__a_ d_ad

b ' d~ b c~bc
Hence we have the following

RULE.

Invert the terms of the divisor, and multiply the dividend by the
resulting fraction.

Entire and mixed quantities should first be reduced to frac
tional forms.

EXAMPLES.
X Zx

1. Divide ^ by -^-. Ans. 1J.

2. Divide -=- by -7.

b J d

8' Divide 7^3 bJ

, -p.. ., x+1 Zx
L Divide — ^— by -^-.

D O

• TV -j x—b . 3cx

). Divide -^-7- by -rr. Ans.

6. Divide ~T '7 by

C — X

a

7- Divide ^b + ^Tb ^ l^b-aTV Ans' Unit^

8. Divide 7a2-3cc+- by &2--.
n J 3

— an

76 ALGEBRA,

,. T>.. . , 3a6 .. a — b

9. Divide 15x2— -=— by x . Ans. = = =7.

5c J c 5cx— 5a + 56

ce3 a6 ax2 — <

10. Divide x2-{ 7 by 7 — ^. Ans.

a—b J a—b

27 (a- 6)

11. Divide ^ — -/ by OQ . -^r.

32(m+6) J 128w(m4-6)

12. Divide ^T+- by -9 h-.

y* x > y2 y x y

13. Divide -+- by . Ans. Unity.

1 1

14 Divide x2-f-+2 by a-f—

X2 CC

15. Divide

' a+b-

Ans. a2— 62+c2 — 2ac.

Divide a2-62-c2-26c by ^
J a+6— c

122. Division of Quantities affected with Negative Exponents. —

Suppose it is required to divide —s by -3. According to the

preceding article, we have

1 a3 a3 1

But, according to Art. 76, —5 may be written a~5 ; -3 may be

written a~3 : and — may be written a~2. Hence we see that

a2 J

that is, the rule of Art. 72 is general, and applies to negative as
well as positive exponents.

EXAMPLES.

1. Divide or5 by — a~2. Ans. — or3, or — ^

2. Divide —a2 by a~l.

3. Divide 1 by a~4.

4. Divide 6«n by — 2a~3.

FRACTIONS. 77

5. Divide bm~n by bm.

6. Divide 12x-2y~4 by — 4xy2.

7. Divide (x— y)~4 by (x— ?/)-6.

123. The Reciprocal of a Fraction. — According to the defini
tion in Art. 34, the reciprocal of a quantity is the quotient aris
ing from dividing a unit by that quantity. Hence the recipro

cal of \ is . a b_bf

D JL -7- 7 — IX — — — /

b a a'

that is, the reciprocal of a fraction is the fraction inverted.

a r) 1 o^

Thus the reciprocal of ^ - is - ; and the reciprocal of
_! b+x a

is b+°:

It is obvious that to divide by any quantity is the same as to
multiply by its reciprocal, and to multiply by any quantity is the
same as to divide by its reciprocal.

124. How to simplify Fractional Expressions. — The numerator
or denominator of a fraction may be itself a fraction or a mixed

quantity, as -j-. In such cases we may regard the quantity

above the line as a dividend, and the quantity below it as a
divisor, and proceed according to Art. 121.

Thus, 24-r.£=ix£=Y=3£.

The most complex fractions may be simplified by the appli
cation of similar principles.

EXAMPLES.

1+f

1. Simplify the fraction - 2

!+!

b+a a+b
This expression is equivalent to — -, — • -- — »

or to — r-X— — T' which is equal to r, Ans.

b a b o

78 ALGEBRA.

i a—

2. Simplify _£=!2. Am. -a-±^.

^ a_ a—m

a+m

24

3, Find the value of the fraction H. Ans. 2.

4. Find the value of the fraction , f^

lla&

a+b a—b

~ ci- f/. c-4-d c — d ac — bd

5. Simplify — —= r. Ans. j-,,

^-h^ a— 6 ac+6^

c—dc+d

a+x a—x

„ „. ,.« a—x a+x

6. Simplify - — . Ans.

a+x a—x

a—x a+x

— m

n mz — n2

1-

n m

7. Simplify — ^ ^ x~3~i — §• Ans. m.

a

8. Simplify - -7- adn + am

b -i Ans.

bdn+bm+cn

EQUATIONS OF THE FIRST DEGREE. 1\)

CHAPTEK YIII.

EQUATIONS OF THE FIRST DEGREE.

125. An equation is an expression of equality between two
algebraic quantities. Thus Sx=2ab is an equation denoting
that three times the quantity x is equal to twice the product
of the quantities a and b.

126. The first member of the equation is the quantity on the
left side of the sign of equality, and the second member is the
quantity on *the right of the sign of equality. Thus, in the
preceding equation, Sx is the first member, and 2ab the second
member.

127. The two members of an equation are not only equal
numerically, but must have the same essential sign. If, in the
preceding equation, x represents a negative quantity, then the
first member is essentially negative, and the second member
must also be negative ; that is, either a or b must represent a
negative quantity.

128. Equations are usually composed of certain quantities
which are known, and others which are unknown. The known
quantities are represented either by numbers, or by the first
letters of the alphabet; the unknown quantities are usually
represented by the last letters of the alphabet.

129. A root of an equation is the value of the unknown,
quantity in the equation ; or it is any value which, being sub
stituted for the unknown quantity, will satisfy the equation.
For example, in the equation

See— 4 = 24— a,

suppose x — 7. Substituting 7 for cc, the first member becomes
3x7—4; that is, 21—4, or 17; and the second member be-

80 ALGEBRA.

comes 24 — 7; that is, 17. Hence 7 is a root of the equation,
because when substituted for x the two members are found to
be equal.

130. A numerical equation is one in which all the known
Quantities are represented oy figures; as, x3-f 4x2 —3x4-12.

131. A literal equation is one in which the known quantities
are represented by letters, or by letters and figures.

Thus x3 + axz + bx = m. )

and rf-sox'fc^ are llteral e(iuatlons-

132. The degree of an equation is denoted by the greatest
number of unknown factors occurring in any term.

If the equation involves but one unknown quantity, its de
gree is denoted by the exponent of the highest power of this
quantity in any term.

If the equation involves more than one unknown quantity,
its degree is denoted by the greatest sum of the exponents of
the unknown quantities in any term.

Thus ax-\-b = cx+d is an equation of tine first degree, and is
sometimes called a simple equation.

4:X2—2x=5—x2 and 7xy— 4x-f-?/=40 are equations of the
second degree, and are frequently called quadratic equations.

x3 + ax2 = 2b and x'2-{-%xy'*+y = m are equations of the third
degree, and are frequently called cubic equations.

So also we have equations of the fourth degree, sometimes
called bi-quadratic equations; equations of the fifth degree, etc.,
up to the nth degree.

Thus xn + axn~l — b is an equation of the nth degree.

133. To solve an equation is to find the value of the unknown
quantity, or to find a number which, being substituted for the
unknown quantity in the equation, renders the first member
identical with the second.

The difficulty of solving equations depends upon their de
gree, and the number of unknown quantities they contain.

EQUATIONS OF THE FIRST DEGREE. 81

134. Axioms. — The various operations which we perform
upon equations, in order to deduce the value of the unknown
quantities, are founded upon the following principles, which
are regarded as self-evident.

1. If to two equal quantities the same quantity be added, the
sums will be equal.

2. If from two equal quantities the same quantity be sub
tracted, the remainders will be equal.

3. If two equal quantities be multiplied by the same quanti
ty, the products will be equal.

4. If two equal quantities be divided by the same quantity,
the quotients will be equal.

135. Transposition. — Transposition is the process of changing
a term from one member of an equation to the other without
destroying the equality of the members.

Let it be required to solve the equation

x-\-a=b.

If from the two equal quantities x-{-a and b we subtract the
same quantity a, the remainders will be equal, according to the
last article, and we shall have

x+a— a = b — a,
or x=b — a.

Let it be required to solve the equation

x— a — b.

If to the two equal quantities x—a and b the same quantity
a be added, the sums will be equal, according to the last arti
cle, and we have x — a + #— b-\-a,

or x

136. Hence we perceive that we may transpose any term of an
equation from one member of the equation to the other, provided we
change its sign.

It is also evident that we may change the sign of every term of
an equation without destroying the equality ; for this is, in fact, the
same thing as transposing every term in each member of the
equation.

D2

82 ALGEBRA.

EXAMPLES.

In the following examples, transpose the unknown terms to
the first member and the known terms to the second member.

1. 5z+12 = 3x+18. Am. 6x~ Bx=lS -12.

2. 4x— 7 = 21 — Sx. Ans. 4x + 3x=2l+7.

3. 2x— 15 = — 7x+30. Ans. 2z-f 7x=30+15.

4. ax+bc=m— 2x. Ans. ax+2x=m—bc.
6. 4ax— b+2c=3x— 2ab— Bmx.

Ans. 4ax— 3x -\-3mx =b—2c—2ab.

6, 4ab— ax— 2c=bx— 3m. Ans. ax+bx=4:ab—2c+3m.

7. ab—cx — 2mx=3ax — 4b.

Ans.

137. To clear an Equation of Fractions. — Let the equation be
?=& If we multiply each of the equal quantities - and b by

the same quantity a, the products will be equal by Art. 134,
and we shall have x~ab.

, x x
Suppose the equation is — \-j-=m.

If we multiply each of the members of the equation by a, we

shall have ax

x-\ — T-=am.
o

If we multiply each of the members of this equation by 6,
We shall have bx+ax=abm.

Hence, to clear an equation of fractions, we have the follow
ing

RULE.

Multiply each member of the equation by all the denominators.

EXAMPLES.

x x_S
3~5~i

x x 3

1. Clear the equation -— -=- of fractions.

Ans. 20x- 12^=45.

O™, O™ O

2. Clear the equation ————-= of fractions.

Ans. 63*- 70^ =45.

EQUATIONS OF THE FIRST DEGREE. 83

3. Clear the equation -=- — —-\---6 of fractions.

i 4 o

Ans. 40z-105x+28z=840.

4. Clear the equation o+z+g = 10 °f fractions.

138. An equation may always be cleared of fractions by muL
tiplying each member into all the denominators; but some'
times the same result may be attained by a less amount of mul
tiplication. Thus, in the last example, the equation may be
cleared of fractions by multiplying each term by 12 instead
of 6x4x2, and it is important to avoid all useless multiplica
tion. In general, an equation may be cleared of fractions by
multiplying each member by the least common inultiple of all the
denominators.

2x Bx 7

5. Clear the equation — +—= — of fractions.

The least common multiple of all the denominators is 20.
If we multiply each member of the equation by 20, we obtain

The operation is effected by dividing the least common mul
tiple by each of the denominators, and then multiplying the
corresponding numerator, dropping the denominator.

6. Clear the equation y — Tj— oT °^ ^ract^ons-

x— 4 1

7. Clear the equation Bx— — -A — =^ °f fractions.

^ 4 12

It should be remembered that when a fraction has the mi
nus sign before it, this indicates that the fraction is to be sub
tracted, and the signs of the terms derived from its numerator

must be changed, Art. 118. 0 , ^ 0 .,

Ans. 36x— 3^+12 = 1.

a— x 3x—2b x-\-ab

8. Clear the equation — = =- — = — - — .

b ab a2

Ans. a3 — azx—

84 ALGEBRA.

139. Solution of Equations. — An equation of the first degree
containing but one unknown quantity may be solved by trans
forming it in such a manner that the unknown quantity shall
stand alone, constituting one member of an equation ; the other
member will then denote the value of the unknown quantity,
Let it be required to find the value of x in the equation
4x— 2 5x_3x
~~5~ ^""T4
Clearing of fractions, we have
32;c-16 + 25z
By transposition we obtain

Uniting similar terms, 27#=216.

Dividing each member by 27, according to Art 134, we have

x=8.

To verify this value of x, substitute it for x in the original
equation, and we shall have

32-2 40 24

or

that is, 11 = 11,

an identical equation, which proves that we have found the

correct value of x.

140. Hence we deduce the following

EULE.

1. Clear the equation of fractions, and perform all the opera
tions indicated.

2. Transpose all the terms containing the unknown quantity to
one side, and all the remaining terms to the other side of the equa
tion, and reduce each member to its most simple form.

3. Divide each member by the coefficient of the unknown quan
tity.

There are various artifices which may sometimes be em-

EQUATIONS OF THE FIRST DEGREE. 85

ployed, by which the labor of solving an equation may be con
siderably abridged. These artifices can not always be reduced
to general rules. If, however, any reductions can be made be
fore clearing of fractions, it is generally best to make them ; and
if the equation contains several denominators, it is often best to
multiply by the simpler denominators first, and then to effect
any reductions which may be possible before getting rid of the
remaining denominators. Sometimes considerable labor may
be saved by simply indicating a multiplication during the first
steps of the reduction, as we can thus more readily detect the
presence of common factors (if there are any), which may be
canceled. The discovery of these artifices will prove one of
the most useful exercises to the pupil.

EXAMPLES.

1. Solve the equation

_

~1~ ~T~
Clearing of fractions,

Transposing and reducing,

25x=150.

Dividing by 25, x—6.

To verify this result, put 6 in the place of x in the original
equation.

Solve the following equations:

2. Sax— 4:ab—2ax— 6ac. Ans. x= 46— 6c.

. Ans. x=9.

a(dz-\-x*) ax d

4. -A— - - J-—ac-\--j-. Ans. x—-.

dx d c

- x_5 284-x

5. — j — \-6x= — - — . Ans. x=9.

a ab 7 171 ab — 1

6. — = bc+d+-. Ans. x=

. . ,

x x fa +

Ans. x~

ALGEBRA.

0.

9.
10.
11.

13.
14.
15.
16.
17.
is

UU-JL ^U-\-^±l>JL, AA-\-UC.

3x-5 2x-4

T \ I '

£LiUS. u; — -=. —r: ^,

5a+46— 2

t/O ' f — _L 4— o *

^j O

8-r 11 E\r E\ Q7 TV
i/. ~~- J- JL t^tX/ 1~^ c/ t7 i ""™" 4 */>*
•y 1 _l_ 1

^o ^ 4a2-35

1 ^i R 9 *

Q x— 4 5x+U 1

4 3 "12-

a+x 5 — x

o a
3x x 2x

8a4-3ac— 3*
a free/

a b~~ d'

a2c
(a-\-x)(b-\-x) — a(b-{-c)= —--{-x2

~"3&d+ad-4a&d-2a&'
ac

-d.715. iC=T-«

O^, O OQ q> 6^ 8

"~~s~"*

7x+16 x4-8 _x
~~21 4x-ll~~3'
6x4-7 7x-13 2x4-4

19. §

9

(^ , fac— Icx=:fac+2a5— 6cx.
6534

J.ns. 05=

7Qq5— 3ac

320c *

22.

T"+ 3 4

= 7.
+^+1.

6

5 8

11

73:_6 x-5 =x
35 6x— 101 5'

8a
Ans. x=^.

Ans. x=5.

Ans. x=S%.

Ans. x=l±.

Ans. xr^ll.

EQUATIONS OF THE FIEST DEGREE. 87

27-

xx a a2(b—a)

28. __u_ - — - - . Ans. x=-j-7j—t — r-.
a^b-a b+a b(b+a)

29. x+x-4--(x+5)(x-3). Ans.x=l2.

25— ^-x 16x+44 23

-••30.- TT-+ o To =~TT+5- Ans.x=S$.

x+1 3x+2 x+1

V

Solution of Problems.

141. A problem in Algebra is a question proposed requiring
us to determine the value of one or more unknown quantities
from given conditions.

142. The solution of a problem is the process of finding the
value of the unknown quantity or quantities that will satisfy
the given conditions.

143. The solution of a problem consists of two parts :

1st. The statement, which consists in expressing the condi
tions of the problem algebraically ; that is, in translating the
conditions of the problem from common into algebraic lan
guage, or forming the equation.

2d. The solution of the equation.

The second operation has already been explained, but the
first is often more embarrassing to beginners than the second.
Sometimes the conditions of a problem are expressed in a dis
tinct and formal manner, and sometimes they are only implied,
or are left to be inferred from other conditions. The former
are called explicit conditions, and the latter implicit conditions.

144. It is impossible to give a general rule which will enable

88 ALGEBRA.

us to translate every problem into algebraic language, since the
conditions of a problem may be varied indefinitely. The fol
lowing directions may be found of some service :

Represent one of the unknown quantities l)y some letter or sym
bol, and then from the given conditions find an expression for each
of the other unknown quantities, if any, involved in the problem.

Express in algebraic language the relations which subsist between
the unknown quantities and the given quantities ; or, by means of
the algebraic signs, indicate the operations necessary to verify the
value of the unknown quantity, if it was already known.

PROBLEMS.

Prob. 1. What number is that, to the double of which if 16
be added, the sum is equal to four times the required number?

Let x represent the number required.

The double of this will be 2x.

This increased by 16 should equal 4x.

Hence, by the conditions, 2x+W=4:X.

The problem is now translated into algebraic language, and
it only remains to solve the equation in the usual way.

Transposing, we obtain

I6=4:x-2x=2x,
and 8=x,

or x—S.

To verify this number, we have but to double 8, and add
16 to the result ; the sum is 32, which is equal to four times 8,
according to the conditions of the problem.

Prob. 2. What number is that, the double of which exceeds
its half by 6?

Let cc = the number required.

Then, by the conditions,

Clearing of fractions, 4x—x=~L2,
or 3* = 12.

Hence cc=4.

To verify this result, double 4, which makes 8, and diminish

EQUATIONS OF THE FIRST DEGREE. 89

it by the half of 4, or 2 ; the result is 6, according to the con
ditions of the problem.

Prob. 3. The sum of two numbers is 8, and their difference
2. What are those numbers?

Let #=:the least number.

Then x+2 will be the greater number.

The sum of these is 2os+2, which is required to equal 8.
Hence we have 2x4-2 = 8.

By transposition, 2x=8 — 2 — 6,

and x~B, the least number.

Also, #+2=5, the greater number.

Verification. 5 + 3 = 8)

K Q 0:f according to the conditions.

O — O = Z ;

The following is a generalization of the preceding Problem.

Prob. 4. The sum of two numbers is a, and their difference
I). What are those numbers?

Let x represent the least number.

Then x+b will represent the greater number.

The sum of these is 2x-\-b, which is required to equal a.

Hence we have 2x+b=a.

By transposition, ^x—a—b,

a— b a b , . ,

or x=— — = - — ft the less number.

lj a &

Hence x+b=^— -+&=- + -, the greater number.

£t A A A

As these results are independent of any particular value at
tributed to the letters a and b, it follows that

Half the difference of two quantities, added to half their sum,
is equal to the greater ; and

Half the difference subtracted from half the sum is equal to the
less.

The expressions |+| and §— ~ are culled formulas, because

LA A A

they may be regarded as comprehending the solution of all
questions of the same kind; that is, of all problems in which
we have given the sum and difference of two quantities.

90

ALGEBRA.

Thus, let a=8 \ .

1—2 \ as m tne Precedmg problem.

Then o^~2~~~2~~^' ^e Sreater number.

a & 8—2 0 ,, , ,

And -—-=—^— = 3, the less number.

10;

12;

23;

100 ; their difference -
100;
5;

10;

6;

2;
ii;

50 ; required the numbers.

1;
*;
ii

Prob. 5. From two towns which are 54 miles distant, two
travelers set out at the same time with an intention of meet
ing. One of them goes 4 miles and the other 5 miles per hour.
In how many hours will they meet?

Let x represent the required number of hours.

Then 4# will represent the number of miles one traveled,
and 5x the number the other traveled ; and since they meet,
they must together have traveled the whole distance.

Consequently, 4x+ 5#— 54.

Hence 9x=54,

or a;=6.

. Proof. In 6 hours, at 4 miles an hour, one would travel 24
miles; the other, at 5 miles an hour, would travel 30 miles.
The sum of 24 and 30 is 54 miles, which is the whole distance.

This Problem may be generalized as follows :

Prob. 6. From two points which are a miles apart, two bod
ies move toward each other, the one at the rate of m miles per
hour, the other at the rate of n miles per hour. In how many
hours will they meet?

Let x represent the required number of hours.

Then mx will represent the number of miles one body moves,
and nx the miles the other body moves, and we shall obvious
ly have mx-\-nx—a.

EQUATIONS OF THE FIRST DEGREE. 9}

Hence x—

This is a general formula, comprehending the solution of all
problems of this kind. Thus,

(150;
Let the J 90; onebody

15;
210; 20;

Eequired the time of meeting.

We see that an infinite number of problems may be pro
posed, all similar to Prob. 5 ; but they are all solved by the
formula of Prob. 6. We also see what is necessary in order
that the answers may be obtained in whole numbers. The given
distance (a) must be exactly divisible by m-{-n.

Prob. 7. A gentleman, meeting three poor persons, divided
60 cents among them ; to the second he gave twice, and to the
third three times as much as to the first. What did he give
to each?

Let £e=the sum given to the first; then 2x=ihe sum given
to the second, and 3x— the sum given to the third.

Then, by the conditions,

x-f 2cc+3x=60.

That is, 6x=60,

or 05=10.

Therefore he gave 10, 20, and 30 cents to them respectively.
The learner should verify this, and all the subsequent results.

The same problem generalized :

Prob. 8. Divide the number a into three such parts that the
second may be m times, and the third n times as great as the

first.

A a ma na

Ans. -r ; : ; .

L -\-7n-\-n L-\-m-\-n \-\-m-\-n

What is necessary in order that the preceding values may
be expressed in whole numbers?

Prob. 9. A bookseller sold 10 books at a certain price, and
afterward 15 more at the same rate. Now at the last sale he

92 ALGEBRA.

received 25 dollars more than at the first. What did he re
ceive for each book? •* Ans. Five dollars.

The same Problem generalized:

Prob. 10. Find a number such that when multiplied success
ively by m and by n, the difference of the products shall be a.

A a

Ans. .

m — n

Prob. 11. A gentleman, dying, bequeathed 1000 dollars to
three servants. A was to have twice as much as B, and B
three times as much as C. What were their respective shares?

Ans. A received $600, B $300, and C $100.
Prob. 12. Divide the number a into three such parts that the
second may be m times as great as the first, and the third n
times as great as the second.

. a ^ ma mna

l + m+mn' \-\-m-\-mn'1 \-\-m + mri

Prob. 13. A hogshead which held 120 gallons was filled with
a mixture of brandy, wine, and water. There were 10 gallons
of wine more than there were of brandy, and as much water as
both wine and brandy. What quantity was there of each?
Ans. Brandy 25 gallons, wine 35, and water 60 gallons.
Prob. 14. Divide the number a into three such parts that the
second shall exceed the first by m, and the third shall be equal
to the sum of the first and second.

a — 2m a-f-2m a
ns. —J—\ — |— ; 2'

Prob. 15. A person employed four workmen, to the first of
whom he gave 2 shillings more than to the second ; to the sec
ond 3 shillings more than to the third ; and to the third 4 shil
lings more than to the fourth. Their wages amount to 32 shil
lings. What did each receive ?

Ans. They received 12, 10, 7, and 3 shillings respectively.

Prob. 16. Divide the number a into four such parts that the
second shall exceed the first by m, the third shall exceed the
second by n, and the fourth shall exceed the third by p.

a — 3m — 2n—p . ,

Ans. The first, z L ; the second,

EQUATIONS OF THE FIRST DEGREE. 93

^ Ai_- j — , f ,

the third, — ' - — - ; the fourth,

Problems which involve several unknown quantities may oft
en be solved by the use of a single unknown letter. Most of
the preceding examples are of this kind. In general, when we
have given the sum or difference of two quantities, both of them
may be expressed by means of the same letter. For the differ*
ence of two quantities added to the less must be equal vo the
greater ; and if one of two quantities be subtracted from their
sum, the remainder will be equal to the other.

Prob. 17. At a certain election 36,000 votes were polled,
and the candidate chosen wanted but 3000 of having twice as
many votes as his opponent. How many voted for each ?

Let x=thevnumber of votes for the unsuccessful candidate;
then 36,000 — x = the number the successful one had, and
36,000— x+ 3000 = 2x. Ans. 13,000 and 23,000.

Prob. 18. Divide the number a into two such parts that one
part increased by b shall be equal to m times the other part.

. ma — b a + b
Ans. - — r ; - -.

772 + 1 ' m + 1

Prob. 19. A train of cars, moving at the rate of 20 miles per
hour, had been gone 3 hours, when a second train followed at
the rate of 25 miles per hour. In what time will the second
train overtake the first?

Let x=the number of hours the second train is in motion,
and 05 -f 3 = the time of the first train.

Then 25x=:the number of miles traveled by the second train,
and 20(cc + 3) = the miles traveled by the first train.

But at the time of meeting they must both have traveled the
same distance.

Therefore 25x = 20x -f 60.

By transposition, 5x=60,

and 05=12.

Proof. In 12 hours, at 25 miles per hour, the second train
goes 300 miles ; and in 15 hours, at 20 miles per hour, the first
train also goes 300 miles; that is, it is overtaken by the sec
ond train.

94 ALGEBRA.

Prob. 20. Two bodies move in the same direction from two
places at a distance of a miles apart ; the one at the rate of n
miles per hour, the other pursuing at the rate of ra miles per
hour. When will they meet ?

Ans. In — J— hours.
m — n

This Problem, it will be seen, is essentially the same as
Prob. 10.

Prob. 21. Divide the number 197 into two such parts that
four times the greater may exceed five times the less by 50.

Ans. 82 and 115.

Prob. 22. Divide the number a into two such parts that m
times the greater may exceed n times the less by b.

ma — b na+b

Ans. - ; - .

m+n m-\-n

"When w = l, this Problem reduces to Problem 18.
When 5 — 0, this Problem reduces to Problem 24.

Prob. 23. A prize of 2329 dollars was divided between two
persons, A and B, whose shares were in the ratio of 5 to 12.
What was the share of each ?

Beginners almost invariably put x to represent one of the
quantities sought in a problem; but a solution may often be
very much simplified by pursuing a different method. Thus,
in the preceding problem, we may put x to represent one fifth
of A's share. Then 5x will be A's share, and I2x will be B's,
and we shall have the equation

and hence x =137;

consequently their shares were 685 and 1644 dollars.

Prob. 24. Divide the number a into two such parts that the
first part may be to the second as m to n.

ma na
Ans.

m-{-n
Prob. 25. What number is that whose third part exceeds its

fourth part by 16 ?
Let 12x=the number.

Then 4x-3x=l6.

EQUATIONS OF THE FIKST DEGREE. 95

or x=I6.

Therefore the number - 12 x 16-192. ^

. Prob. 26. Find a number such that when it is divided suc

cessively by m and by n, the difference of the quotients shall

be a.

amn
Ans.

n —

Prob. 27. A gentleman has just 8 hours at his disposal ; how
far may he ride in a coach which travels 9 miles an hour, so as
to return home in time, walking back at the rate of 3 miles an
hour? Ans. 18 miles.

w Prob. 28. A gentleman has just a hours at his disposal ; how
far may he ride in a coach which travels m miles an hour, so
as to return home in time, walking back at the rate of n miles

an hour?

amn .,
Ans. — - miles.
m+n

- Prob. 29. A gentleman divides a dollar among 12 children,
giving to some 9 cents each, and to the rest 7 cents. How
many were there of each class?

-Prob. 30. Divide the number a into two such parts that if
the first is multiplied by m and the second by n, the sum of

the products shall be b.

o — na ma — o
Ans. - : - .

m — n m — n

-Prob. 31. If the sun moves every day 1 degree, and the
moon 13, and the sun is now 60 degrees in advance of the
moon, when will they be in conjunction for the first time, sec
ond time, and so on ?

- Prob. 32. If two bodies move in the same direction upon the
circumference of a circle which measures a miles, the one at
the rate of n miles per day, the other pursuing at the rate of m
miles per day, when will they be together for the first time, sec
ond time, etc., supposing them to be b miles apart at starting?

AM. In ~> £±*, 2*t*, etc, days.

m — Ji in — n m—n

It will be seen that this Problem includes Prob. 20.

96 ALGEBRA.

~ Prob. 33. Divide the number 12 into two such parts that the
difference of their squares may be 48.

* Prob. 34. Divide the number a into two such parts that the

difference of their squares rnay.be b. a2— b a2 + b

J i, Ans. — — ; -- — .

2a 2a

Prob. 35. The estate of a bankrupt, valued at 21,000 dollars,
is to be divided among three creditors according to their re
spective claims. The debts due to A and B are as 2 to 3,
while B's claims and C's are in the ratio of 4 to 5. What sum
must each receive?

- Prob. 36. Divide the number a into three parts, which shall
be to each other as m : n: p.

ma na pa

Ans.

When _p = l, Prob. 36 reduces to the same form as Prob. 8.
^ Prob. 37. A grocer has two kinds of tea, one worth 72 cents
per pound, the other 40 cents. How many pounds of each
must be taken to form a chest of 80 pounds, which shall be
worth 60 cents?

Ans. 50 pounds at 72 cents, and 30 pounds at 40 cents.

Prob. 38. A grocer has two kinds of tea, one worth a cents
per pound, the other b cents. How many pounds of each must
be taken to form a mixture of n pounds, which shall be worth

c cents? n(c—b)

Ans. — - — j-Z pounds at a cents,

n(a — c) ,

and — - — — ^ pounds at b cents.
a — b

-Prob. 39. A can perform a piece of work in 6 daj^s; B can
perform the same work in 8 days; and C can perform the
same work in 24 days. In what time will they finish it if all
work together?

~Prob. 40. A can perform a piece of work in a days, B in b
days, and C in c days. In what time will they perform it if all
work together? ^ «fe _ d

Prob. 41. There are three workmen, A, B, and C. A and
B together can perform a piece of work in 27 days; A and C

EQUATIONS OF THE FIRST DEGREE. 97

together in 86 days; and B and C together in 54 days. In
what time could they finish it if all worked together ?

A and B together can perform -5-7- of the work in one day.

AandC " A

B and C * ^

Therefore, adding these three results,

2A + 2B + 2C can perform -^V + ^V S-^V iri one day,
=1-^ in one day.

Therefore, A, B, and C together can perform -^ of the work
in one day ; that is, they can finish it in 24 days. If we put
x to represent the time in which they would all finish it, then
they would together perform | part of the work in one day,
And we should have

Prob. 42. A and B can perform a piece of labor in a days ;
A and C together in b days; and B and C together in c days.
In what time could they finish it if all work together ?

A -,

Am. -T— — y- days.

This result, it will be seen, is of the same form as that of
Problem 40. ^

Prob. 43. A broker has two kinds of change. It takes 20
pieces of the first to make a dollar, and 4 pieces of the second
to make the same. Now a person wishes to have 8 pieces for
a dollar. How many of. each kind must the broker give him?

Prob. 44. A has two kinds of change ; there must be a pieces
of the first to make a dollar, and b pieces of the second to make
the same. Now B wishes to have c pieces for a dollar. How
many pieces of each kind must A give him ?

^.725. — — 7— of the first kind ; ~ — ^ of the second.
a — b a — b

Prob. 45. Divide the number 45 into four such p^rts that
the first increased by 2, the second diminished by 2, the third
multiplied by 2, and the fourth divided by 2, shall all be equal.

In solving examples of this kind, several unknown quantities
are usually introduced, but this practice is worse than super-

E

98 ALGEBRA.

fluous. The four parts into which 45 is to be divided may be
represented thus:

The first =a-2,

the second =cc-f 2,

the third =f,

the fourth = 2x;

for if the first expression be increased by 2, the second dimin
ished by 2, the third multiplied by 2, and the fourth divided by
2, the result in each case will be x. The sum of the four parts
is 4^x, which must equal 45.

Hence x=10.

Therefore the parts are 8, 12, 5, and 20.

Prob. 46. Divide the number a into four such parts that
the first increased by m, the second diminished by m, the third
multiplied by m, and the fourth divided by m, shall all be

equal.

ma ma a m2a

Ans'

(m + 1)2 ; (ra + 1)2^ ; (m + 1)2' (m+lf

Prob. 47. A merchant maintained himself for three years at
an expense of $500 a year, and each year augmented that
part of his stock which was not thus expended by one third
thereof. At the end of the third year his original stock was
doubled. What was that stock?

Prob. 48. A merchant supported himself for three years at
an expense of a dollars per year, and each year augmented
that part of his stock which was not thus expended by one
third thereof. At the end of the third year his original stock

was doubled. What was that stock ?

USa
Ans. — .

Prob. 49. A father, aged 54 years, has a son aged 9 years.
In how many years will the age of the father be four times
that of the son ?

Prob. 50. The age of a father is represented by a, the age of
his son by b. In how many years will the age of the father be

n times that of the son ? a— nb

Ans. =-.

n — 1

EQUATIONS WITH MOKE THAN ONE UNKNOWN QUANTITY. 99

CHAPTER IX.

AQUATIONS OF THE FIRST DEGREE CONTAINING MORE THAN
ONE UNKNOWN QUANTITY.

145. If we have a single equation containing two unknown
quantities, then for every value which we please to ascribe to
one of the unknown quantities, we can determine the corre
sponding value of the other, and thus find as many pairs of
values as wevplease which will satisfy the equation. Thus, let

2a+4y=16. (1.)

If y—\, we find cr=6; if y=2, we find x=4, and so on;
and each of these pairs of values, 1 and 6, 2 and 4, etc., sub
stituted in equation (1), will satisfy it.

Suppose that we have another equation of the same kind,
as, for example, 5x+3y=19. (2.)

We can also find as many pairs of values as we please which
will satisfy this equation.

But suppose we are required to satisfy both equations with
the same set of values for .a; and y ; we shall find that there is
only one value of x and one value of y. For, multiply equa
tion (1) by 3, and equation (2) by 4, Axiom 3, and we have

6x+12?/=48, (3.)

20x+12y=76. (4.)

Subtracting equation (3) from equation (4), Axiom 2, we have

14^=28; (5.)

whence x=2. (6.)

Substituting this value of x in equation (1), we have

4+4y=16; (7.)

Whence y=3. (8.)

Thus we see that if both equations are to be satisfied, x must
equal 2, and y must equal 3< Equations thus related are called
simultaneous equations

100 ALGEBRA.

146. Simultaneous equations are those which must be satisfied
by the same values of the unknown quantities.

When two or more simultaneous equations are given for so
lution, we must endeavor to deduce from them a single equation
containing only one unknown quantity. We must therefore
make one of the unknown quantities disappear, or, as it is
termed, we must eliminate it.

147. Elimination is the operation of combining two or more
equations in such a manner as to cause one of the unknown
quantities contained in them to disappear.

There are three principal methods of elimination : 1st, by ad
dition or subtraction ; 2d, by substitution ; 3d, by comparison.

148. Elimination ly Addition or Subtraction. — Let it be pro
posed to solve the system of equations

5x+%=35, (1.)

7x-3y=6. (2.)

Multiplying equation (1) by 3, and equation (2) by 4, we have

(3.)
(4.)
Adding (3) and (4), member to member (Axiom 1), we have

43^=129; (5.)

whence <r — 3. (6.)

We may now deduce the value of y by substituting the value
of x in one of the original equations. Taking the first for ex
ample, we have 15+4^=35;
whence 4y=20,
and y=§>

149. In the same wa}^, an unknown quantity may be elimi
nated from any two simultaneous equations. This method is
expressed in the following

RULE.

Multiply or divide the equations, if necessary, in such a manner
iliat one of the unknoivn quantities sliatt have the same coefficient in

EQUATIONS WITH MORE THAN ONE UNKNOWN QUANTITY. 101

both. Then subtract one equation from the other if the signs of
these coefficients are alike, or add them together if tJie signs art
unlike.

In solving the preceding equations, we multiplied both mem
bers of each by the coefficient of the quantity to be eliminated
in the other equation ; but if the coefficients of the letter to be
eliminated have any common factor, we may accomplish the
same object by the use of smaller multipliers. In such cases,
find the least common multiple of the coefficients of the letter
to be eliminated, and divide this multiple by each coefficient;
the quotients will be the least multipliers which we can employ.

150. Elimination by Substitution. — Take the same equations
as before : 6x+4y= 35, (1.)

7a-3y=6. (2.)

Finding from (1) the value of y in terms of x, we have

y=^. (3.)

Substituting this value of y in (2), we have
105-15$

7x — rTrtv>> •

Clearing of fractions,

whence x=3.

Substituting this value of x in (3), we have

y=6.

The method thus exemplified is expressed in the following

RULE.

find an expression for the value of one of the unknown quan
tities in one of the equations; then substitute this value for that
quantity in the other equation.

151. Elimination by Comparison. — Take the same equations
as before: 5z+4?/=35, (1.)

7x-3y=6. (2.)

102 ALGEBRA.

Derive from each equation an expression for y in terms of a\
and we have 35— 5x

y= -— ,

-
and y=— g— . (4.)

Placing these two values equal to each other, we have

3 4

Clearing of fractions,

28x-24=:105-15z;
whence 43x-129,

and x=S.

Substituting this value of x in (3),

y=5.

The method thus exemplified is expressed in the following
4-

KULE.

Find an expression for the value of the same unknown quantity
in each of the equations, and form a new equation by placing these,
values egwal to edch ether.

In the solution of simultaneous equations, either of the pre
ceding methods can be u&eri, as may be most convenient, and
each method has its advantages in particular cases. General
ly, however, the last two methods give rise to fractional expres
sions, which occasion inconvenience in practice, while the first
method is not liable to this objection. When the. coefficient of
one of the unknown quantities in one of the equations is equal
to unity, this inconvenience does not occur, and the method
'Jby substitution may be preferable ; the first will, however, com-
irnonly be found most convenient.

EXAMPLES.

f llsr+3?/=100) x
1. Given j 4 _</ _4 f to ^d the values of x and y.

Ans. x=8: ?/=4.

EQUATIONS WITH MORE THAN ONE UNKNOWN QUANTITY. 103

2. Given

3. Given

4. Given

1+!-'
H=8

x + 2

y+5

4

to find x and y.

Am. x — 6; y=12.
to find x and y.

to find x and y.

a o

5. Given < * ^
c d

--] — — n
x y

3 2

i

to find x and y.

be— ad be— ad

Am. x=—, ,; y=

nb — mo?' me — na

j 5x-7y=20 ) 4 G ,
6. Given -j n - r . . [• to find x and £/.
( 9x— ll?/=44 )

7. Given

-=m

/y> rt y

8. Given <J V to find x and ?/.

-=i-» 2 2

y x J Am.x=- -; y=

m — n

T^^o. n

9. Given <J ^~ \ to find x and y.

\LX-\-v( .

f~ i~i-r ' i >4 775 x=:21 * v — 3^

10. Given

to find x and

Ans. vc=na—a; y =

104

11. Given

12. Given

ALGEBRA.

to find x and -

11— x 4x+8y—2
~~2~ ~~9~

13. Given

6 3

to find x and

™ — 1 • ?/ — 9

<L — i, y—&.
to find x and y.

14. Given

13

— 5?/+6
19

6x— oy+4 3x+2?/+l

to find cc and T.

15. Given \ ? [ to find x and y.

( x—y—l \

Ans. x=— ; y=

a-b*

16. Given

to find x and .

17. Given

5 ~10156
a;_8y_1_y-a? a; 1
2 20 15 ^6^1

to find
and .

EQUATIONS WITH MORE THAN ONE UNKNOWN QUANTITY. 105

Equations of the First Degree containing more than Two
Unknown Quantities.

152. If we have three simultaneous equations containing
three unknown quantities, we may, by the preceding methods,
reduce two of the equations to one containing only two of the
jnknown quantities; then reduce the third equation and either
of the former two to one containing the same two unknown
quantities ; and from the two equations thus obtained, the un
known quantities which they involve may be found. The
third quantity may then be found by substituting these values
in either of the proposed equations.

Take the system of equations

3x+2?/-5z= 8, (2.)

§x-§y 4- 3z rr 6. (3.)

Multiplying (1) by 3, and (2) by 2, we have

6x+9z/-fl2z=48, (4.)

6x4-4^-102=16. (5.)

Subtracting (5) from (4), 5z/+222 = 32. (6.)

Multiplying (1) by 5, and (3) by 2, we have

"lO;c+15y 4-202=80, (7.)

10:c-12y4-6* = 12. (8.)

Subtracting (8) from (7), 27y 4-142 = 68. (9.)

Multiplying (6) by 27, and (9) by 5, we have

135 ?/+ 594s = 864. (10.)

135?/+702 = 340. (11.)

Subtracting (11) from (10), 5242=524 ;
whence z = l.

Substituting this value of z in (6),
5y4-22=32;
whence 2/=2.

Substituting the values of y and z in (1),

2^4-64-4=16;

whence <c=3.

E 2

106 ALGEBRA.

153. Hence, to solve three equations containing three un
known quantities, we have the following

RULE.

From the three equations deduce two containing only two un*
known quantities ; then from these two deduce one containing only
one unknown quantity.

154. If we had/owr simultaneous equations containing four
unknown quantities, we might, by the methods already ex
plained, eliminate one of the unknown quantities. We should
thus obtain three equations between three unknown quanti
ties, which might be solved according to Art. 152. So, also,
if we \\2idifive equations containing five unknown quantities,
we might, by the same process, reduce them to four equations
containing four unknown quantities, then to three, and so on.
By following the same method, we might resolve a system of
any number of equations of the first degree. Hence, if we have
m equations containing m unknown quantities, we proceed by
the following

RULE.

1st. Combine successively any one of the equations with each of
Hie others, so as to eliminate the same unknown quantity; there
will result m — \ new equations, containing m — 1 unknown quan
tities.

2d. Combine any one of these new equations with the others, so as
to eliminate a second unknown quantity ; there will result m — 2
equations, containing m — 2 unknown quantities.

3d. Continue this series of operations until there results a single
equation containing but one unknown quantity, from which the
valne of this unknown quantity is easily deduced.

4th. /Substitute this value for its equal in one of the equations con •
taining two unknown quantities, and thus find the value of a second
unknown quantity] substitute these values in an equation contain
ing three unknown quantities, and find the value of a third' and
so on, till the values of all are determined.

Either of the unknown quantities may be selected as the one

EQUATIONS WITH MORE THAN ONE UNKNOWN QUANTITY. 107

to be first eliminated. It is, however, generally best to begin
with that which has the smallest coefficients ; and if each of
the unknown quantities is not contained in all the proposed
equations, it is generally best to begin with that which is found
in the least number of equations. Sometimes a solution may
be very much abridged by the use of peculiar artifices, for
which no general rules can be given.

EXAMPLES.

Solve the following groups of simultaneous equations :

Ans. =

2. x+z=b

Note. Take the sum of the three preceding equations.

[ x+y+z=29\ (x= 8.

3. < x+2y+3z = 62V Ans. |yi 9-

+z=10) (*=12.

5. 1 x-?j+z= 654V

( -X + 7/ + 2=-12)

Ans. =

7. J

0 7

108

ALGEBRA.

8. -

9. -

M=«

x y

-4-- = b

x z~

l-+l-=c

1 l_l=a
a y z~

i_i+u

x y z

12

= 1

30 t 37 =3

222

10. 4

p2cc+5y-7z=-288
11. i 5x-y+Zz = 227

:s= 297

12. J

x y Iz
3 + 5+Y

2,^8^5'

13.

4y—3u+2v= 9

Ans. -

Ans. -<

x=

y—

a+ti

2
a-f-c*

2

r*=l.

J.n.'j. -! y = 2.

U=3.

x-13.
y=24.

x= 12.
y= 30.

= 50.

z = S.
M=3.

EQUATIONS WITH MORE THAN ONE UNKNOWN QUANTITY. 10§

f x = 3.

1 x+y+t+u+v =

14.

Note. Take the sum of these six equations.

Ans.

v=8.

Problems involving Equations of the First Degree with several
Unknown Quantities.

Prob. 1. Find two numbers such that if the first be added
to four times the second, the sum is 29 ; and if the second be
added to six times the first, the sum is 36.

Prob. 2. If A's money were increased by 36 shillings, he
would have three times as much as B ; but if B's money were
diminished by 5 shillings, he would have half as much as A.
Find the sum possessed by each.

Prob. 3. A pound of tea and three pounds of sugar cost six
shillings ; but if sugar were to rise 50 per cent, and tea 10 per
cent., they would cost seven shillings. Find the price of tea
and sugar. Ans. Tea, 5s. per pound; Sugar, 4 pence.

Prob. 4. What fraction is that to the numerator of which if
4 be added the value is one half; but if 7 be added to the de
nominator, its value is one fifth ? Ans. -^.

Prob. 5. A certain sum of money, put out at simple interest,
amounts in 8 months to $1488, and in 15 months it amounts
to $1530. What is the sum and rate per cent. ?

Prob. 6. A sum of money put out at simple interest amounts
in m months to a dollars, and in n months to b dollars. Ke-
quired the sum and rate per cent.

na—mb ., b — a

Ans. The sum is

n — m

the rate is 1200 x

Prob. 7. There is a number consisting of two digits, the sec
ond of which is greater than the first ; and if the number be
divided by the sum of its digits, the quotient is 4; but if the
digits be inverted, and that number be divided by a number

110 ALGEBRA.

greater by two than the difference of the digits, the quotient is
14. Kequired the number.

Let x represent the left-hand digit, and y the right-hand
digit.

Then, since x stands in the place of tens, the number will
be represented by IQx+y.

Hence, by the first condition,

=4;

x+y
by the second condition,

Whence x— 4, y=8, and the required number is 48.

Prob. 8. A boy expends thirty pence in apples and pears,
buying his apples at 4 and his pears at 5 for a penny, and aft
erward accommodates his friend with half his apples and one
third of his pears for 13 pence. How many did he buy of
each?

Prob. 9. A father leaves a sum of money to be divided
among his children as follows: the first is to receive $300 and
the sixth part of the remainder ; the second, $600 and the
sixth part of the remainder; and, generally, each succeeding
one receives $300 more than the one immediately preceding,
together with the sixth part of what remains. At last it is
found that all the children receive the same sum. What was
the fortune left, and the number of children ?

Ans. The fortune was $7500, and the number of children 5.

Prob. 10. A sum of money is to be divided among several
persons as follows: the first receives a dollars, together with
the nth part of the remainder; the second, 2a, together with the
nth part of the remainder; and each succeeding one a dollars
more than the preceding, together with the nth part of the re
mainder ; and it is found at last that all have received the same
Bum. What was the amount divided, and the number of per
sons? Ans. The amount was a(n — I)3;

the number of persons =r? — 1.

Prob. 11. A wine-dealer has two kinds of wine. If he mixes

EQUATIONS WITH MORE THAN ONE UNKNOWN QUANTITY. Ill

9 quarts of the poorer with 7 quarts of the better, he can sell
the mixture at 55 cents per quart; but if he mixes 3 quarts of
the poorer with 5 quarts of the better, he can sell the mixture
at 58 cents per quart. What was the cost of a quart of each
kind of wine?

Ans. 48 cents for the poorer, and 64 for the better.

Prob. 12. A person owes a certain sum to two creditors. At
one time he pays them $530, giving to one four elevenths of
the sum which is due, and to the other $30 more than one
sixth of his debt to him. At a second time he pays them $420,
giving to the first three sevenths of what remains due to him,
and to the other one third of what remains due to him. What
were the debts ?^_

Prob. 13. If A and B together can perform a piece of work
in 12 days, A and C together in 15 days, and B and C in 20
days, how many days will it take each person to perform the
same work alone?

This problem is readily solved by first finding in what time
they could finish it if all worked together.

Prob. 14. If A and B together can perform a piece of work
in a days, A and C together in b days, and B and C in c days,
how many days will it take each person to perform the same

work alone? rt , 0 7

. . 2abc -p, . 2abc ,

Ans. A in — —, -- 7 days-; B in - days;

ac + bc—ab ab + bc—ac

n • -,

C m -j— — j- days.

ab -\-ac- be

Prob. 15. A merchant has two casks, each containing a cer
tain quantity of wine. In order to have an equal quantity in
each, he pours out of the first cask into the second as much as
the second contained at first; then he pours from the second
into the first as much as was left in the first ; and then again
from the first into the second as much as was left in the second,
when there are found to be a gallons in each cask. How man;y
gallons did each cask contain at first ?

lla , Sa
Ans. -rr- and -—.
'4l &

112 ALGEBRA.

Prob. 16. A laborer is engaged for n days on condition that
he receives p pence for every day he works, and pays q pence
for every day he is idle. At the end of the time he receives a
pence. How many days did he work, and how many was he
idle?

Ans. He worked — days, and was idle -£• days.

p + q p + q

Prob. 17. A certain number consisting of two digits contains
the sum of its digits four times, and their product three times.
What is the number?

Prob. 18. A father says to his two sons, of whom one was
four years older than the other, In two years my age will be
double the sum of your ages ; but 6 years ago my age was 6
times the sum of your ages. How old was the father and each
of the sons?
Ans. The father was 42, one son 11, and the other 7 years old.

Prob. 19. It is required to divide the number 96 into three
parts such that if we divide the first by the second the quo
tient shall be 2, with 3 for a remainder ; but if we divide the
second by the third, the quotient shall be 4, with 5 for a re
mainder. What are the three parts? Ans. 61, 29, and 6.

Prob. 20. Each of seven baskets contains a certain number
of apples. I transfer from the first basket to each of the other
six as many apples as it previously contained ; I next trans
fer from the second basket to each of the other six as many
apples as it previously contained, and so on to the last basket,
when it appeared that each basket contained the same number
of apples, viz., 128. How many apples did each basket contain
before the distribution?

Ans. The first 449, the second 225, the third 113, the fourth
57 the fifth 29, the sixth 15, and the seventh 8 apples.

155. When we have only one equation containing more than
one unknown quantity, we can generally solve the equation in
an infinite number of ways. For example, if a problem involv
ing two unknown quantities (x and y) leads to the singh equa
tion

-EQUATIONS WITH MORE THAN ONE UNKNOWN QUANTITY.

we may ascribe any value we please to x, and then determine
the corresponding value of y. Such a problem is called inde
terminate. An indeterminate problem is one which admits of an
indefinite number of solutions.

156. If we had two equations containing three unknown quan
tities, we could, in the first place, eliminate one of the unknown
quantities by means of the proposed equations, and thus obtain
one equation containing two unknown quantities, which would
be satisfied by an infinite number of systems of values. There
fore, in order that a problem may be determinate, its enuncia
tion must contain as many different conditions as there are un
known quantities, and each of these conditions must be express
ed by an independent equation.

157. Equations are said to be independent when they express
conditions essentially different, and dependent when they express
the same conditions under different for 7ns.

Thus \A ~*y~ \ are independent equations.
I 2x+y=W )

But •! ' 9 _ 14.C are n°t independent, because the one
may be deduced from the other.

158. If, on the contrary, the number of independent equa
tions exceeds the number of unknown quantities, these equa
tions will be contradictory.

For example, let it be required to find two numbers such that
their sum shall be 8, their difference 2, and their product 20.
From these conditions we derive the following equations :

ay =20.

From the first two equations we find
tc=5 and ?/=3.

Hence the third condition, which requires that their product
ehall be equal to 20, can riot be fulfilled.

114 ALGEBBA.

CHAPTEK X.

DISCUSSION OF PROBLEMS INVOLVING SIMPLE EQUATIONS. — •
INEQUALITIES.

159. To discuss a problem or an equation is to determine the
values which the unknown quantities assume for particular hy
potheses made upon the values of the given quantities, and to
interpret the peculiar results obtained. We have seen that if
the sum of two numbers is represented by a, and their differ
ence by 5, the greater number will be expressed by — — , and
the less by — — . Here a and b may have any values whatever,

JU

and still these formulas will always hold true. It frequently
happens that, by attributing different values to the letters which
represent known quantities, the values of the unknown quanti
ties assume peculiar forms, which deserve consideration.

160. We may obtain jive species of values for the unknown
quantity in a problem of the first degree:

1st. Positive values.

2d. Negative values. ~

3d. Values of the form of zero, or -r.

A

4th. Values of the form of — .

5th. Values of the form of -z.

We will consider these five cases in succession.

161. 1st. Positive values are generally answers to problems in
the sense in which they are proposed. Nevertheless, all posi
tive values will not always satisfy the enunciation of a prob
lem. For example, a problem may require an answer in whole

DISCUSSION OF PROBLEMS. 115

numbers, in which case a fractional value of the unknown quan
tity is inadmissible. Thus, in Prob. 17, page 93, it is implied
that the value of x must be a whole number, although this con
dition is not expressed in the equations. We might change
the data of the problem, so as to obtain a fractional value of
x, which would indicate an impossibility in the problem pro
posed. Problem 43, page 97, is of the same kind ; also Prob,
7, page 109.

If the value obtained for the unknown quantity, even when
positive, does not satisfy all the conditions of the problem, the
problem is impossible in the form proposed.

162. 2d. Negative values.

Let it be proposed to find a number which, added to the
number 6, gives for a sum the number a. Let x denote the re
quired number; then, by the conditions of the problem,

whence cc = a — b.

This formula will give the value of x corresponding to any
assigned values of a and b.

For example, if a = 7 and b = 4:,
then x = 7— 4 = 3,

a result which satisfies the conditions.

But suppose that a —5 and b = 8,
then x — 5— 8=— 3.

We thus obtain for x a negative value. How is it to be in
terpreted ?

By referring to the problem, we see that it now reads thus:
What number must be added to 8 in order that the sum rn;iy
be 5? It is obvious that if the word added and the word sum
are to retain their arithmetical meanings, the proposed problem
is impossible. Nevertheless, if in the equation 8 + x = 5 we
substitute for +x its value —3, it becomes

8-3 = 5,

an identical equation; that is, 8 diminished by 3 is equal to 5,
or 5 may be regarded as the algebraic sum of 8 and —3.

The negative result, x=— 3, indicates that the problem, in a

116 ALGEBRA.

strictly arithmetical sense, is impossible ; but, taking this value
of x with a contrary sign, we see that it satisfies the enunciation
when modified as follows : What number must be subtracted
from 8 in order that the difference may be 5 ? The second
enunciation differs from the first only in this, that we put sub
tract for add, and difference for sum.
If we wish to solve this new equation directly, we shall have

8-z=5;
whence x=S— 5, or 3.

163. For another example, take Problem 50, page 98. The
age of the father being represented by a, and that of the son

by Z>, then - — - will represent the number of years before the

age of the father will be n times that of the son.
Thus, suppose a =54, 5—9, and ?z=4;

54-36 18
then x = —^- = -g = 6.

This value of x satisfies the conditions understood arithmet
ically ; for if the father was 54 years old, and the son 9 years,
then in 6 years more the age of the father will be 60 and the
son 15 ; and we see that 60 is 4 times 15.

But suppose a =45, £=15, and n=4;

45-60 -15
then x— — - — = — —- = —5.

O O

Here again we obtain a negative result. How are we to in
terpret it?

By referring to the problem, we see that the age of the son
is already more than one fourth that of the father, so that the
time required is already past by five years. The problem, if
taken in a strictly arithmetical meaning, is impossible. But
let us modify the enunciation as follows :

The age of the father is 45 years; the son's age is 15 years;
how many years since the age of the father was four times that
of his son ?

DISCUSSION OF PROBLEMS. 117

The equation corresponding to this new enunciation is

4

whence 60— 4x=45— x; and x—5,

a result which satisfies the modified problem taken in its arith
metical sense.

From this discussion we derive the following general prin
ciples :

1st. A negative result found for the unknown quantity in a prob
km of the first degree indicates that the probkm is impossible, if
understood in its strict arithmetical sense.

2d. This negative value, taken with a contrary sign, may be re
garded as the answer to a probkm whose enunciation only differs
from that of the proposed probkm in this, that certain quantities
which were ADDED should have been SUBTRACTED, and vice versa.

164. In what case would the value of the unknown quantity
in Prob. 20, page 94, be negative? Ans. When n>m.

Thus, let m — 20, n = 25, and a = 60 miles;

then x- 6° -6°- 12

~20-25~-5~

To interpret this result, observe that it is impossible that the
second train, which moves the slowest, should overtake the first.
At the time of starting, the distance between them was 60 miles,
and each subsequent hour the distance increases. If, however,
we suppose the two trains to have been moving uniformly along
an endless road, it is obvious that at some former time they must
have been together.

This negative result indicates that the problem is impossible
if understood in its strict arithmetical sense. But if the prob
lem had been stated thus :

Two trains of cars, 60 miles apart, are moving in the same
direction, the forward one 25 miles per hour, the other 20.
How long since they were together?

The problem would have furnished the equation

whenoe x= +12.

118 ALGEBKA.

If we wish to include both of these cases in the same enun
ciation, the question should be, Required the time of their being
together, leaving it uncertain whether the time w as past or future*

EXAMPLES.

1. What number is that whose fourth part exceeds its third;
part by 16? Ans. —192.

How should the enunciation be modified in order that the
result may be positive?

2. The sum of two numbers is 2, and their difference 8-
What are those numbers? Ans. — 3 and +5.

How should the enunciation be modified in order that both
results may be positive ?

3. What fraction is that from the numerator of which if 4
be subtracted the value is one half, but if 7 be subtracted

from the denominator its value is one fifth ? —5

Ans.

-18

How should the enunciation be modified in order that the
problem may be possible in its arithmetical sense?

4. Find two numbers whose difference is 6, such that four
times the less may exceed five times the greater by 12.

Ans. —42 and —36.

Change the enunciation of the problem so that these num
bers, taken with the contrary sign, may be the answers to the
modified problem,

~^

165. 3d. We may obtain for the unknown quantity values of

the form of zero, or -r.
j\.

In what case would the value of the unknown quantity in
Prob. 20, page 94, become zero, and what would this value
signify ?

Ans. This value becomes zero when a = 0, which signifies
that the two trains are together at the outset.

In what case would the value of the unknown quantity in
Prob. 50, page 98, become zero, and what would this value
signify?

DISCUSSION OF PROBLEMS. 119

Ans. When a — rib, which signifies that the age of the father
is now n times that of the son.

In what case would the values of the unknown quantities in
Prob. 38, page 96, become zero, and what would these values
signify ?

When a problem gives zero for the value of the unknown
quantity, this value is sometimes applicable to the problem,
and sometimes it indicates an impossibility in the proposed
question.

166. 4th. We may obtain for the unknown quantity values
j^

of the form of — .

In what case does the value of the unknown quantity in

A

Prob. 20, page 94, reduce to -^-, and how shall we interpret
this result? Ans. When m = n.

On referring to the enunciation of the problem, we see that
it is absolutely impossible to satisfy it ; that is, there can be no
point of meeting; for the two trains, being separated by the dis
tance a, and moving equally fast, will always continue at the

same distance from each other. The result ^ may then be re
garded as indicating an impossibility.

The symbol Q is sometimes employed to represent infinity,
and for the following reason :

•If the denominator of a fraction is made to diminish, while
the numerator remains unchanged, the value of the fraction
must increase.

For example, let m— n = 0.01 ;

then a=— ^-=-^

m — n .01

Let m— n — 0.0001;

then x=,_JL_=_JjL. = i0

m^-n .0001

Hence, if the difference in the rates of motion is not zero, the

120 ALGEBRA.

two trains must meet, and the time will become greater and
greater as this difference is diminished. If, then, we suppose
this difference to be less than any assignable quantity, the time

represented by - - will be greater than any assignable quantity.

rffL — Tif

Hence we infer that every expression of the form — found

for the unknown quantity indicates the impossibility of satis-
fying the problem, at least in finite numbers.

In what case would the value of the unknown quantity in

Prob. 10, page 92, reduce to the form -^, and how shall we in
terpret this result?

167. The s^ymbol 0, called zero, is sometimes used to denote
the absence of value, and sometimes to denote a quantity less
than any assignable value.

The symbol oo , called infinity, is used to denote a quantity
greater than any assignable value. A line produced beyond any
assignable limit is said to be of infinite length ; and time ex
tended beyond any assignable limit is called infinite duration.

We have seen that when the denominator of the fraction

^2~ becomes less than any assignable quantity, the value of
the fraction becomes greater than akf assignable quantity.
Hence we conclude that ^ — oo;

that is, a finite quantity divided by tero is an expression for in
finity.

Also, if the denominator of a fraction be made to increase
while the numerator remains unchanged, the value of the frac
tion must diminish; and when the denominator becomes greater
than any assignable quantity, the value of the fraction must be
come less than any assignable quantity. Hence we conclude that

-=0;

00

that is, a finite quantity divided ly infinity is an expression for
zero

DISCUSSION OF PEOBLEMS.

121

168. 5th. We may obtain for the unknown quantity values
of the form of ^.

In what case does the value of the unknown quantity in
Prob. 20, page 94, reduce to 77, and how shall we interpret this
lesult? Ans. When a = 0, and m — n.

To interpret this result, let us recur to the enunciation, and
observe that, since a is zero, both trains start from the same
point; and since they both travel at the same rate, they will
always remain together ; and, therefore, the required point of
meeting will be any where in the road traveled over. The
problem, then, is entirely indeterminate, or admits of an infinite

number of solutions ; and the expression ~ may represent any
finite quantity.
We infer, therefore, that an expression of the form of Q found

for the unknown quantity generally indicates that it may have
any value whatever. In some cases, however, this value is
subject to limitations.

In what case would the values of the unknown quantities

in Prob. 44, page 97, reduce to ^, and how would they satisfy
the conditions of the problem? Ans. When a = 6 = c,

which indicates that the coins are all of the same value. B
might therefore be paid in either kind of coin ; but there is a
limitation, viz., that the value of the coins must be one dollar.
In what case do the values of the unknown quantities in

Prob. 38, page 96, reduce to -r, and how shall we interpret
his result?

169. The expression - may be conceived to result from a

fraction whose numerator and denominator both diminish si
multaneously, but in such a manner as to preserve the same
relative value. If both numerator and denominator of a frac
tion are divided by the same quantity, its value remains ur>

K

122 ALGEBRA.

changed. Hence, if ~ represent any fraction, we may conceive

both numerator and denominator to be divided by 10, 100,
1000, etc., until each becomes less than any assignable quanti
ty, or 0. The fraction then reduces to the form of ^ but the
ralue of the fraction has throughout remained unchanged.

For example, we may suppose the numerator to represent
the circumference of a circle, and the denominator to represent
its diameter. The value of the fraction in this case is known
to be 3.1416. If now we suppose the circle to diminish until
it becomes a mere point, the circumference and diameter both
become zero, but the value of the fraction has throughout re
mained the same. Hence, in this case, we have

Again, suppose the numerator to represent the area of a cir
cle, and the denominator the area of the circumscribed square;
then the value of the fraction becomes .7854C But this value
remains unchanged, although the circle may be supposed to
diminish until it becomes a mere point. Hence, in this case,
we have 0

Hence we conclude that the symbol -^ may represent any
finite quantity.

So, also, we may conceive both numerator and denominator
of a fraction to be multiplied by 10, 100, 1000, etc., until each*
becomes greater than any assignable quantity ; the fraction

then reduces to the form of — . Hence we conclude that the

oo °°

symbol — may also represent any finite quantity.

INEQUALITIES.

170. An inequality is an expression denoting that one quan
tity »is greater or less than another. Thus Sx > 2a5 denotes
that three times the quantity x is greater than twice the prod-
act of the quantities a and b.

INEQUALITIES. 123

171. In treating of inequalities, the terms greater and less
must be understood in their algebraic sense ; that is, a negative
quantity standing alone is regarded as less than zero; and of
two negative quantities, that which is numerically the greatest
is considered as the least; for if from the same number we sub
tract successively numbers larger and larger, the remainders
must continually diminish. Take any number, 5 for example,
and from it subtract successively 1, 2, 3, 4, 5, 6, 7, 8, 9, etc.,
we obtain

5-1, 5-2, 5-3, 5-4, 5-5, 5-6, 5-7, 5-8, 5-9, etc.,
or, reducing, we have

4, 3, 2, 1, 0, -1, -2, -3, -4, etc.

Hence we see that — 1 should be regarded as less than zero ;
—2 less than —1; —3 less than —2, etc.

172. Tw® inequalities are said to subsist in the same sense
when the greater quantity stands at the left in both, or at the
right in both ; and in a contrary sense when the greater quanti
ty stands at the right in one and at the left in the other. Thus
9>7 and 7>6, or 5<8 and 3<4, are inequalities which sub
sist in the same sense; but the inequalities 10>6 and 3<7
subsist in a contrary sense.

173. Properties of Inequalities. — 1st. If the same quantity be
added to or subtracted from each member of an inequality, the re
suiting inequality will ahuays subsist in the same sense.

Thus, 8>3.

Adding 5 to each member, we have

and subtracting 5 from each member, we have

8_5>3-5.
Again, take the inequality

-3<-2.
Adding 6 to each member, we have

_3 + 6<-2 + 6, or 3<4;
and subtracting 6 from each member^

_3-6<-2-6, or -9<-&

124 ALGEBRA.

174. Hence we conclude that we may transpose a term from
one member of an inequality to the other, provided we change
its sign.

Thus, suppose a2 + 62>3£2-2a2.

Adding 2a2 to each member of the inequality, it becomes

Subtracting b2 from each member, we have

a24-2a2>362-&2,
or 3a2>262.

175. 2d. If we add together the corresponding members of two or
more inequalities which subsist in the same sense, the resulting in
equality will always subsist in the same sense.

Thus, S>4

4>2

7>3
Adding, we obtain 16 > 9.

176. 3d. If one inequality be subtracted from another which sub
sists in the same sense, the result will not always be an inequality
subsisting in the same sense.

Take the two inequalities 4<7
and 2<3

Subtracting, we have 4— 2<7— 3, or 2<4,

where the result is an inequality subsisting in the same sense.

But take 9<10

and 6< 8

Subtracting, we have 9 — 6>10— 8, or 3>2,

where the result is an inequality subsisting in the contrary sense.
We should therefore avoid as much as possible the use of
this transformation, or, when we employ it, determine in what
sense the resulting inequality subsists.

177. 4th. If we multiply or divide each member of an inequality
"by the same positive quantity, the resulting inequality will subsist
in the same sense.

INEQUALITIES. 126

Thus, if a<b,

then ma<mb,

, a b

and — <— .

m m

Also, if — a>— b,

then —ma>—mb.

, a b

and > .

m m

Hence an inequality may be cleared of fractions. Thus, sup
pose we have a2 — b2 c2—d2
~2d~>~3a~'

Multiplying each member by 6ad, it becomes
3a(a2-b2)>2d(c2-d2).

178. 5i\i.\Tf we multiply or divide each member of an inequality
by the same negative number, the resulting inequality will subsist
in the contrary sense.

Take, for example, 8>7.

Multiplying each member by —3, we have the opposite in
equality — 24 < — 21.

So, also, 15>12.

Dividing each member by — 3, we have
-5<-4,

Therefore, if we multiply or divide the two members of an
inequality by an algebraic quantity, it is necessary to ascertain
whether the multiplier or divisor is negative, for in this case
the resulting inequality subsists in a contrary sense.

179. 6th. If the signs of all the terms of an inequality be changed,
the sign of inequality must be reversed.

For to change all the signs is equivalent to multiplying each
member of the inequality by —1.

180. Reduction of Inequalities. — The principles now establish
ed enable us to reduce an inequality so that the unknown quan
tity may stand alone as one member of the inequality. The

126 ALGEBRA.

other member will then denote one limit of the unknown
quantity.

EXAMPLES.

1. Find a limit of x in the inequality

7 5rc 95 0

6~T<12~
Multiplying each member by 12, we have

14-15x<95-24x.
Transposing, 9x<81.

Dividing, • . x<9.

2.

3. S,-2>-

_ x x—

5. - --- ^-<x -- - — . Ans.

D 1O

6. Given

to find the limits of x.

Aiis.

7. A man, being asked how many dollars he gave for his
watch, replied, If you multiply the price by 4, and to the prod
uct add 60, the sum will exceed 256; but if you multiply the
price by 3, and from the product subtract 40, the remainder
will be less than 113. Eequired the price of the watch.

8. What number is that whose half and third part added
together are less than 105 ; but its half diminished by its fifth
part is greater than 33?

9. The double of a number diminished by 6 is greater than
22, and triple the number diminished by 6 is less than double
the number increased by 10. Eequired the number.

INVOLUTION. 127

CHAPTER XL

INVOLUTION.

181. A power of a quantity is the product obtained by tak
ing that quantity any number of times as a factor.

Thus the first power of 3 is 3 ;

the second power of 3 is 3 x 3, or 9 ;

the fourth power of 3 is 3 x 3 x 3 x 3, or 81, etc.
Involution is the process of raising a quantity to any power.

182. A power is indicated by means of an exponent. The
exponent i^a number or letter written a little above a quantity
to the right, and shows how many times that quantity is taken
as a factor.

Thus the first power of a is a1, where the exponent is 1,
which, however, is commonly omitted.

The second power of a is a x a, or a2, where the exponent 2
denotes that a is taken twice as a factor to produce the power
aa.

The third power of a is a x a x a, or a3, where the exponent
3 denotes that a is taken three times as a factor to produce the
power aaa.

The fourth power of a is a x a x a X a, or a4.

Also the rcth power of a is a x a x a x a, etc., or a repeated
as a factor n times, and is written an.

The second power is commonly called the square, and the
third power the cube.

183. Exponents may be applied to polynomials as well as
to monomials.

Thus (a+£ + c)3 is the same as

or the third power of the entire expression a~\-l>

128 ALGEBRA.

Powers of Monomials.

184. Let it be required to find the third power or cube of
2a3b2.

According to the rule for multiplication, we have

(2aW)3 = 2M2 x 2a3b2 x 2a3b2=2 x 2 x 2a3a3aW62=8a966.

In a similar manner any monomial may be raised to any
power.

Hence, to raise a monomial to any power, we have the fol
lowing

RULE.

Raise the numerical coefficient to the required power, and multi
ply the exponent of each of the letters by the exponent of the re
quired power.

185. Sign of the Power. — With respect to the signs, it is ob
vious from the rules for multiplication that if the given mono
mial be positive, all of its powers are positive ; but if the mo
nomial be negative, its square is positive, its cube negative, its
fourth power positive, and so on.

Thus —ax— a=-fa2,

— ax —ax — a—— a3,
— ax— ax— ax— a=+a4,
— ax — ax — ax — ax — a = — a5, etc.

In general, any even power of a negative quantity is positive,
and every odd power negative ; but all powers of a positive quan
tity are positive.

EXAMPLES.

1. Find the square of lla3M2. Ans. 121a66W4.

2. Find the square of — lSx2yz3.
8. Find the cube of 7ab2x2.

4. Find the cube of -8xy2z3.

5. Find the fourth power of 4a52c3.

6. Find the fourth power of —5a3l2x.

7. Find the fifth power of 2aZ>3x2.

8. Find the fifth power of -3ab2x*.

9. Find the sixth power of 3ab2x*.

INVOLUTION.

129

10. Find the sixth power of — 2azb3x*.

11. Find the seventh power of 2a2x3y.

12. Find the rath power of abV.

186. Powers of Fractions.— Let it be required to find the

., . -, f 2ab2

third power of -— .

From the rule for the multiplication of fractions, we have

33 3c ~27c3'

In a similar manner any fraction may be raised to any
power. Hence, to raise a fraction to any power, we have the
following

EULE.

Raise both numerator and denominator to the required power.
\

EXAMPLES.

- 3a&2c3
1. Find the square of . Ans.

2. Find the square of — -jr

_ fiox2

3. Find the cube of -— — -

oran

o

*

4. Find the cube of —

A /y nr^y/

5, Find the fourth power of — jr~-.

— 36r7i

6. Find the fourth power of — p-r

7. Find the fifth power of -

\C1 A ^

8. Find the fifth power of

9. Find the sixth power of

F2

130 ALGEBRA.

187. Negative Exponents. — The rule of Art. 184, for raising
a monomial to any power, holds true when the exponents of any
of the letters are negative, and also when the exponent of the re
quired power is negative.

Let it be required to find the square of a~3. This expres
sion may be written -3, which, raised to the second power, be
comes -g, or a~6, the same result as would be obtained by
multiplying the exponent —3 by 2.

Also, let it be required to find that power of 2am2 whose
exponent is —3.

The expression (2am2)-3 may be written 77- — r-^, which
r (2am2)3

equals 3 6. Transferring the factors to the numerator, we
have 2-3GT37ft-6, or -|a~3m~6.

EXAMPLES.

Find the -value of each of the following expressions.
1. (3a26~4)2. Ans.

2/fj 273 4 \? A

, ^/a 0' c x] . ^Lns.

3. (3ab'2x~ly~z)~2. Ans.

4. (-4a2x-y)-2. Ans.

5. ( — 6ab~5x~2)3'
Q f 3a2x~3^2)~3.

8. (-

9. (-
10. (-

188. Powers of Polynomials. — A polynomial may be raised
to any power by the process of continued multiplication. If
the quantity be multiplied by itself, the product will be the
second power; if the second power be multiplied by the orig
inal quantity, the product will be the third power, and so on.
Hence we have the following

INVOLUTION. 131

KULE.

Multiply the uanlity l>y itself until it has been taken as a facto
as many times as there are units in the exponent of the required
power.

EXAMPLES.

1. Find the square of 2a+362. Ans. 4a2+12a&2+9&*.

2. Find the square of a+m—n.

3. Find the cube of 2a2+3a-l.

4. Find the cube of

5. Find the cube of

6. Find the fourth power of a— b.

7. Find the fourth power of 2a — 3

8. Find the fourth power of a3+&3.

9. Find the fifth power of a— b.

1A .p,. j . ~—

10. Find the square of - =^. Ans. -=-

2

a^f-Zabxy + bW

11. Find the cube of — '• .

m— n

12. Find the cube of ^-^-9.

a— o2

189. Square of a Polynomial. — We have seen, Art. 66, that
the square of a binomial may be formed without the labor of
actual multiplication. The same principle may be extended
to polynomials of any number of terms. By actual multipli
cation, we find the square of a -\-b-\-c to be

that is, the square of a trinomial consists of the square of each
term, together with twice the product of all the terms multiplied to
gether two and two.

In the same manner we find the square of a-\-b-\-c-{-d to be

that is, the square of any polynomial consists of the square of each
term, together with twice the sum of the products of all Hie terms
multiplied together two and two.

132 ALGEBRA.

EXAMPLES.

1. Find the square of a-\-b-\-c+d+x.

2. Find the square of a — b + c.

3. Find the square of l + 2x+3x2.

4. Find the square of 1 — x+x2— x3.

5. Find the square of a— 2^+3a6— m.

6. Find the square of 1 — 3x+3x2— x3.

7. Find the square of a— 26 + 3c— 4d.

In Chapter XVIII. will be given a method by which any
power of a binomial may be obtained without the labor of
multiplication. "^~

EVOLUTION. 133

CHAPTER XII.

EVOLUTION.

190. A root of a quantity is one of the equal factors which,
multiplied together, will produce that quantity.

If a quantity be resolved into two equal factors, one of them
is called the square root.

If a 'quantity be resolved into three equal factors, one of
them is called the cube root.

If a quantity be resolved into four equal factors, one of
them is called the fourth root, and so on.
t .

191. Evolution is the process of extracting any root of a
given quantity.

Evolution is indicated by the radical sign y~.
Thus, <\fa denotes the square root of a.

YCL denotes the cube root of a.

\/a denotes the nib. root of a.

192. Surds. — When a root of an algebraic quantity which
is required can not be exactly obtained, it is called an irration
al or surd quantity.

Thus, Va? is called a surd, -v/3 is also a surd, because the
square root of 8 can not be expressed in numbers with perfect
exactness.

A rational quantity is one which can be expressed in finite
terms, and without any radical sign ; as, a, 5a2, etc.

193. An imaginary root is one which can not be extracted
on account of the sign of the given quantity. Thus the square
root of — 4 is impossible, because no quantity raised to an even
power can produce a negative result.

A root which is not imaginary is said to be real.

134 ALGEBRA.

Roots of Monomials.

194. According to Art. 184, in order to raise a monomial tc
any power, we raise the numerical coefficient to the required
power, and multiply the exponent of each of the letters by the
exponent of the power required. Hence, conversely, to ex
tract any root of a monomial, we extract the root of the nu
merical coefficient, and divide the exponent of each letter by
the index of the required root.

Thus the cube root of 64a663 is 4a26.

195. Sign of the Root. — We have seen, Art. 185, that all pow
ers of a positive quantity are positive ; but the even powers
of a negative quantity are positive, while the odd powers are
negative.

Thus -fa, when raised to different powers in succession,
will give +ttj _|_a2? _|_a3? +a4? +a5? _j_a6j +a7^ etc>

and — a, in like manner, will give

—a, + a2, —a3, -fa4, —a5, -fa6, —a7, etc.

Hence it appears that if the root to be extracted be express
ed by an odd number, the sign of the root will be the same as
the sign of the proposed quantity. Thus, \/— a3= —a; and

If the root to be extracted be expressed by an even number,
and the quantity proposed be positive, the root may be either
positive or negative. Thus, -\/~a?= ±a.

If the root proposed to be extracted be expressed by an even
number, and the sign of the proposed quantity be negative, the
root can not be extracted, because no quantity raised to an
even power can produce a negative result.

196. Hence, to extract any r#ot of a monomial, we have the
following

RULE.

1st. Extract the required root of the numerical coefficient.
Id. Divide the exponent of each literal factor ly the index of the
required root.

EVOLUTION. 135

3c7. Every even root of a positive quantity must have the double
sign ±, and every odd root of any quantity must have the same
sign as that quantity.

From Art. 186, it is obvious that to extract any root of a
fraction, we must divide the root of the numerator by the root
of the denominator. Thus,

/ ^

a

"I'

EXAMPLES.

1. Find the square root of 64a6£4. Am. ±8a3&2.

2. Find the square root of 196a2&4c6x8. Ans.

3. Find the square root of 225a2m68x6.

4. Find the cube root of 64a3&6x9. Ans.

5. Find the cube root of — 125a3x6?/9. Ans. —5ax2y3.

6. Find the cube root of -343a6W.

7. Find the fourth root of 81a4&8. Ans.

8. Find the fourth root of 256a4612x16.

9. Find the fifth root of -32a56]0x15. Ans. -

10. Find the square root of — — g— . Ans. ±- — — .

11. Find the square root of

97/737,6
12. Find the cube root of ~

2mx3'

13. Find the cube root of — --_

216dV *

14. Find the fourth root of 8 16.

15. Find the square root of 64a~2&~4x4. Ans. ±8a~lb~z

16. Find the cube root of — 5l2a~3b~6x3.

17. Find the fourth root of 256a~4Z>-8x4.

18. Find the fifth root of -32a~wb~l5x5. Ans. -2a~2b-

19. Find the square root of (a — l>fx*. Ans. ±(a

20. Find the cube root of (a+b)3(x+y)6.

*b

136 ALGEBKA.

Square Root of Polynomials.

197. In order to discover a rule for extracting the square
root of a polynomial, let us consider the square of a+&, which
is a2-f-2a& + 62, If we arrange the terms of the square accord
ing to the dimensions of one letter, a, the first term will be the
square of the first term of the root; and since, in the present
case, the first term of the square is a2, the first term of the root
must be a.

Having found the first term of the root, we must consider
the rest of the square, namely, 2a& + 62, to see how we can de
rive from it the second term of the root. Now this remainder
may be put under the form (2a + b)b; whence it appears that
we shall find the second term of the root if we divide the re
mainder by 2a + &. The first part of this divisor, 2a, is double
of the first term already determined ; the second part, &, is yet
unknown, and it is necessary at present to leave its place empty.
Nevertheless, we may commence the division, employing only
the term 2ct; but as soon as the quotient is found, which in
the present case is 6, we must put it in the vacant place, and
thus render the divisor complete.

The whole process, therefore, may be represented as fol
lows:

If the square contained additional terms, we might continue
the process in a similar manner. We may represent the first
two terms of the root, a +5, by a single letter, m, and the re
maining terms by c. The square of m+c will be mz+2mc+c2.
The square of the first two terms has already been subtracted
from the given polynomial. If we divide the remainder by
2m as a partial divisor, we shall obtain c, which we place in
the root, and also at the right of 2m, to complete the divisor.
"We then multiply the complete divisor by c, and subtract the

EVOLUTION. 137

product from the dividend, and thus we continue until all the
terms of the root have been obtained.

198. Hence we derive the following

KULE.

1st. Arrange the terms according to the powers of some one let
ter; take the square root of the first term for the first term of the
required root, and subtract its square from the given polynomial

2d. Divide the first term of the remainder by twice the root al
ready found, and annex the result both to the root and the divisor.
Multiply the divisor thus completed by the last term of the root, and
subtract the product from the last remainder.

3d. Double the entire root already found for a second divisor.
Divide the first term of the last remainder by the first term of the
second divisor for the third term of the root, and annex the result
both to the root and to the second divisor, and proceed as before
until all the terms of the root have been obtained.

If the given polynomial be an exact square, we shall at last
find a remainder equal to zero.

EXAMPLES.

1. Extract the square root of a4— 2a3x+3a2x2— 2a
a4 - 2a3x + BaV - 2«x3 + x4 ( a2 - ax + x*

2a2—

2a2x2—

For verification, multiply the root a2— ax+x2 by itself, and
we shall obtain the original polynomial.

2. Extract the square root of a24-2«& + 2ac-f&2-f 2£c+c2.

3. Extract the square root of

4 Extract the square root of

8ax* 4- 4a2x2 + 4x4 + 1 662x2 -f 1 664 + 1 60 b2x.

Ans

138 ALGEBRA.

5. Extract the square root of

6. Extract the square root of 8a&3 + a4— 4a3Z>+4&4.

7. Extract the square root of 4x4-l-12x3 + 5x'2— 6as+l.

8. Extract the square root of

4z4 - I2ax3 4- 25a2x2 - 24a3x 1 6a4.

5. 2x2—
9. Extract the square root of

-f 49a2x2 - 24a3x + 1 6a4.

10. Extract the square root of x*— x3 + ? +4#— 2 +

4

x

2 x 2
' X -+-

y

199. TFAen a Trinomial is a Perfect Square. — The square of
a-\-~b is a2 + 2afr + &2, and the square of a — b is a2— 2ab-\-b2.
Hence the square root of a2±2ab + b2 is a±b; that is, a trino
mial is a perfect square when two of its terms are squares, and
the third is the double product of the roots of these squares.

Whenever, therefore, we meet with a quantity of this de
scription, we may 'know that its square root is a binomial ; and
the root may be found by extracting the roots of the two terms
which are complete squares, and connecting them by the sigp
of the other term.

EXAMPLES.

1. Find the square root of 4«2+12a/;4-9&2. Ans.

2. Find the square root of 9a2— 2

3. Find the square root of 9a4—

4. Find the square root of 4a2-fl4«&-f 16&2, if possible.

,.w

No algebraic binomial can be a perfect square, for the square
of a monomial is a monomial, and the square of a binomial
necessarily consists of three distinct terms.

EVOLUTION. 139

Square Root of Numbers.

200. The preceding rule is applicable to the extraction of
the square root of numbers; for every number may be re
garded as an algebraic polynomial, or as composed of a cer
tain number of units, tens, hundreds, etc. Thus

529 is equivalent to 500 + 20 + 9;
also, 841 " 800+40 + 1.

If, then, 841 is the square of a number composed of tens
and units, it must contain the square of the tens, plus twice the
product of the tens by the units, plus the square of the units. But
these three terms are blended together in 841, and hence arises
the peculiar difficulty in determining its root. The following
principles will, however, enable us to separate these terms,
and thus detect the root.

201. Is/. For every two figures of the square there will be one
figure in the root, and also one for any odd figure. Thus

the square of 1 is 1

" 9 " 81

99 " 98,01

999 " 99,80,01

the square of 1 is 1

10 " 1,00

100 " 1,00,00

1000 " 1,00,00,00

The smallest number consisting of two figures is 10, arid its
square is the smallest number of three figures. The smallest
number of three figures is 100, and its square is the smallest
number of five figures, and so on. Therefore the square root
of every number composed of one or two figures will contain
one figure ; the square root of every number composed of three
or four figures will contain two figures; of a number from five
to six figures will contain three figures, and so on.

Hence, if we divide the number into periods of two figures,
commencing at the. units' place, the number of periods will
indicate the number of figures in the square root.

202. 2d. The first figure of the root will be the square root of
the greatest square number contained in the first period on the lejl.

140 ALGEBRA.

For the square of tens can give no significant figure in the
first right hand period, the square of hundreds can give no fig-
ure in the first two periods on the right, and the square of
the highest figure in the root can give no figure except in the
first period on the left.

Let it be required to extract the square root of 5329.

This number contains two

periods, indicating that there 53,29 (70 + 3, the root.

will be two places in the 49 00

root. Let a + b denote the 140 + 3)~T29
root, where a is the value of 4 29

the figure in the tens' place,

and I of that in the units' place. Then a must be the great
est multiple of 10, which has its square less than 5300 ; this is
found to be 70. Subtract a2, that is the square of 70, from
the given number, and the remainder is 429, which must be
equal to (2a + b)b. Divide this remainder by 2a, that is by
140, and the quotient is 3, which is the value of b. Com
pleting the divisor, we have 2a + 6 = 143; whence (2a + b)b,
that is 143 X 3, or 429, is the quantity to be subtracted ; and
as there is now no remainder, we conclude that 70 + 3, or 73,
is the required square root.

For the sake of brevity, the ciphers may be omitted, pro
vided we retain the proper local values of the figures.

If the root consists of three places of figures, let a represent
the hundreds, and b the tens; then, having obtained a and b as
before, let the hundreds and tens together be considered as a
new value of a, and find a new value of b for the units.

Required the square root of 568516.

Having found 75, the square root of 56,85,16(754

che greatest square number contained in 49

the first two periods, we bring down the 145 ) 785
last period, and have 6016 for a new div- 725

idend. We then take 2a, or 150, for a 1504) 6016
partial divisor, whence we obtain b = 4: 6016

for the last figure of the root. The en
tire root is therefore 754.

EVOLUTION. 141

203. Hence, for the extraction of the square root of num
bers, we derive the following

RULE.

1st. Separate the given number into periods of two figures each,
beginning from the units1 place.

2d. Find the greatest number whose square is contained in the
left-hand period ; this is the first figure of the required root. Sub
tract its square from the first period, and to the remainder bring
down the second period for a dividend.

3d. Double the root already found for a divisor , and find how
many times it is contained in the dividend, exclusive of its right-
hand figure ; annex the result both to the root and the divisor.

4dh. Multiply the divisor thus increased by the last figure of the
root, subtract the product from the dividend, and to the remainder
bring down the next period for a new dividend.

6th. Double the whole root now found for a new divisor, and
proceed as before, continuing the operation until all the periods are
brought down.

In applying the preceding rule, it may happen that the prod
uct of the complete divisor by the last figure of the root is
greater than the dividend. This indicates that the last figure
of the root was taken too large, and this happens because ^the
divisor is at first incomplete — that is, is too small. In such a
case, we must diminish the last figure of the root by unity until
we obtain a product which is not greater than the dividend.

EXAMPLES.

1. What is the square root of 294849? Ans. 543,

2. What is the square root of 840889 ?

3. What is the square root of 1142761 ?

4. What is the square root of 32239684?

5. What is the square root of 72777961 ?

6. What is the square root of 3518743761 ?

204. Square Root of Fractions. — We have seen that the
root of a fraction is equal to the root of its numerator di-

142 ALGEBRA.

vided by the root of its denominator Hence the square root

The number 5.29 may be written — -, and its square root

O D -j Qfif\f) A

is — , or 2.3. So, also, 18.6624 may be written , and

its square root is — v or 4.32. That is, the square root of a
100

decimal fraction, or of a whole number followed by a decimal
fraction, may be found in the same manner as that of a whole
number, if we divide it into periods commencing with the deci
mal point.

In the extraction of the- square root of an integer, if there is
still a remainder after we have obtained the units' figure of the
root, it indicates that the proposed number has not an exact
square root. We may, if we please, proceed with the approxi
mation to any desired extent by supposing a decimal point at
the end of the proposed number, and annexing any even num
ber of ciphers, and continuing the operation. We thus obtain
a decimal part to be added to the integral part already found.

So, also, if a decimal number has no exact square root, we
may annex ciphers and proceed with the approximation to
any desired extent.

EXAMPLES.

TX7, , . , - 1089 0 33

1. What is the square root of ? Am. — .

^ooy uo

2. What is the square root of

-^
1/0569

3. What is the square root of „,— -.,,.„., ?

67551961

4. What is the square root of 9.878449 ?

5. What is the square root of 58.614336 ?

6. What is the square root of .558009 ?

7. What is the square root of .03478225 ?

Find the square roots of the following numbers to five deci
mal places.

EVOLUTION. 143

8. Of 2. Ans. 1.41421.

9. Of 10. Ans. 3.16227.
10. Of 9.1.

11. Of 4f.

12. Of TV

13. Of A.

Cube Root of a Polynomial.

205. We already know that the cube of a+b is a
-\-3ab2 + b3. If, then, the cube were given, and we were re
quired to find its root, it might be done by the following
method.

When the terms are arranged according to the powers of
one letter, a, we at once know, from the first term, a3, that
a must be one term of the root. If, then, wre subtract its
cube from the proposed polynomial, we obtain the remainder
8a2b + Bab2 <f 63, which must furnish the second term of the root.

Now this remainder may be put under the form

whence it appears that we shall find the second term of the
root if we divide the remainder by 3a2+3ab + b2. But, as this
second term is supposed to be unknown, the divisor can not be
completed. Nevertheless, we know the first term, 3a2, that is
thrice the square of the first term already found, and by means
of this we can find the other part, b; viz., by dividing the first
term of the remainder by 3a2. We then complete the divisor
by adding to it Sab+b2, If this complete divisor be multi
plied by 6, it will give the last three terms of the power.
Let it be required to find the cube root of 8a3

Having found the first term of the root, 2a, and subtracted
its cube, we divide the first term of the remainder, 36«25, by
three times the square of 2a, that is 12a2, and we obtain Sb
for the second term of the root. We then complete the divi
sor by adding to it three times the product of the two terms of

144 ALGEBKA.

the root, which is 18a5, together with the square of the last
term 36, which is 9&2. Multiplying then the complete divisoi
by 3Z>, and subtracting the product from the last remainder,
nothing is left. Hence the required cube root is 2a+35.
This result may be easily verified by multiplication.

206. If the root contains three terms, as a + 6 + c, we may
put a + b = m. Then

(a + b + c)3 = (m + c)3 = m3 + 3ra2c + 3mc2 + c3.
If we proceed as in the last example, we shall find a + &, and
we subtract its cube from the given polynomial. There will
then remain 3m2c+3?wc2 + c3, which may be written

(3ra2 + 3rac+c2)c.

We perceive that 3m2 will be the new trial divisor to obtain
c. We then complete the divisor by adding to it Smc-fc2.
Let it be required to find the cube root of 8a6 — 365a54-
a* - 63£3a3

8a6 - 366rt5 + GGZ*2a4 - 63Z>3a3 + 336*a2 - 966o + 66(2a2 - 3ba + 63
So6

1 2a* — 1 8ba3 + 9/>2a2 ) — 366a5 + GG6~a* — G363«3

— 366a5 + 5462a4 — 2 7i3a3

12a*-

The first term of the root is 2a2, and subtracting its cube, the
first term of the remainder is — 365a5, which, divided by 3
times the square of 2a2, gives — Sba for the second term of the
root. Complete the divisor as in the last example, and multi
ply it by —3ba. Subtracting the product from the last remain
der, the first term of the second remainder is 1262a4.

To form the new trial divisor, we take three times the square
of the part of the root already found, viz., 2cr2— 3ba. Divide
the first term of the remainder by 12a*, and we obtain bz for the
last term of the root. We now complete the divisor by add
ing to it three times the product of the third term by the sum
of the first two terms, and also the square of the last term,
Multiplying the divisor thus completed by 52, we find the prod-

EVOLUTION. 145

uct equal to the last remainder. Hence the required cube root
is2a2—

207. Hence, for extracting the cube root of a polynomial, we
derive the following

RULE.

1st. Arrange the terms according to the powers of some one let
ter ; take the cube root of the first term, and subtract the cube from
the given polynomial.

2d. Divide the first term of the remainder by three times the
square of the root already found ; the quotient will be the second
term of the root.

3d. Complete the divisor by adding to it three times the product
of the two terms of the root and the square of the second term.

4th. Multiply the divisor thus increased by the last term of the
root, and subtract the product from the last remainder.

5th. Take * three times the square of the part of the root already
found for a new trial divisor, and proceed by division to find an
other term of the root.

6th. Complete the divisor by adding to it three- times the product
of the last term by the sum of the first two terms, and also the square
of the last term, with which proceed as before till the entire root has
been obtained.

We may dispense with forming the complete divisor accord
ing to the rule if each time that we find a new term of the
root we raise the entire root already found to the third power,
and subtract the cube from the given polynomial.

EXAMPLES.

1. What is the cube root of a6— 6a5 + 15a4— 20a3 + 15a2—
8a-f 1? Ans. a2-

2. What is the cube root of 6x6-40x3+z6 + 96;r

Ans.

3. What is the cube root of 18x4 + 36x2-f 24cc+ 8 + 32x3-f-
x6 + 6xfl? Ans.

4. What is the cube root of 3£6 + &6 - 563 - 1 + 36 ?

Ans.
G

146 ALGEBRA.

5. What is the cube root of 8x6— 36cc5 + 66x4— 63x3-h33x2-^
9x+l?

6. What is the cube root of 8x6+4;8ax5+ 60a2x4— 80a3z3—

7. What is the cube root of 8x6 — 36ax5 + 102a2x4— 171a3a3
5z + 64a6 ?

./foctf o/" Numbers.

208. The preceding rule is applicable to the extraction of
the cube root of numbers ; but a difficulty in applying it arises
from the fact that the terms of the powers are all blended to
gether in the given number. They may, however, be separated
by attending to the following principles :

1st. For every three figures of the cube there will be one figure in
the root, and also one for any additional figure or figures. Thus,

the cube of 1 is 1

9 " 729

99 " 970,299
999 " 997,002,999

the cube of 1 is 1

10 " 1,000

" 100 " 1,000,000

1000 " 1,000,000,000

Hence we see that the cube root of a number consisting of
from one to three figures will contain one figure ; the cube root
of a number consisting of from four to six figures will contain
two figures ; of a number from seven to nine figures will con
tain three figures, and so on.

Hence, if we divide the number into periods of three figures,
commencing at units' place, the number of periods will indi
cate the number of figures in the cube root

209. 2c?. The first figure of the root will be the cube root of the
greatest cube number contained in the first period on the left.

For the cube of tens can give no significant figure in the first
right-hand period ; the cube of hundreds can give no figure in
the first two periods on the right ; and the cube of the high
est figure in the root can give no figures except in the first
period on the left.

Let it be required to extract the cube root of 438976.

This number contains two periods, indicating that there will

EVOLUTION. 147

be two places in 438,976 ( 70 f 6, the root,

the root. Let a be 343,000

f ,, 70ax3 = H700

the value of the 70x6x3- 1260

95976
95976

figure in the tens' G2— 5?

place, and b of that comPlete divisor> 15996
in the units' place. Then a must be the greatest multiple of
10 which has its cube less than 438000 ; that is, a must be 70.
Subtract the cube of 70 from the given number, and the re
mainder is 95976. This remainder corresponds to 3a2b -\-3ab*
+ &3, which may be written

Divide this remainder by 3a2, that is, by 14700, and the quo
tient is 6, which is the value of b. Complete the divisor by
adding to it 3a5, or 1260, and b2, or 36. The complete divisor
is thus found to be 15996, which, multiplied by 6, gives 95976.
Subtracting, the remainder is zero, and we conclude that 70 + 6,
or 76, is the required cube root.

For the sake of brevity the ciphers may be omitted, provided
we retain the proper local values of the figures.

If the root consists of more than two places of figures, the
method will be substantially the same.

Let it be required to extract the cube root of 279,726,264.
O^Q 7o« OA/I fAKA Having found 65, the cube root of

27y,7ZO,4O4:(o54: Xl , , . , „

the greatest cube contained in the first
two periods, we bring down the last
period, and have 5101264 for a new
dividend. We then take three times
the square of the root already found,
5101 264 or 12675, for a partial divisor, whence

5101 264 we °ktam 4 for the last figure of the

— root. We then complete the divisor

by adding to it three times the product of 4 by 65, and the
square of 4, regard being paid to the proper local values of the
figures. The complete divisor is thus found to be 1275316,
which, multiplied by 4, gives 5101264. Hence 654 is the re«
quired cube root.

108

90

11725
12675

780

__ 16
1275316

148 ALGEBEA.

210. Hence, for the extraction of the cube root of numbers,
we derive the following

RULE.

1st. Separate the given number into periods of three figures each,
"beginning at the units' place.

2d. Find the greatest cube contained in the left-hand period ; its
cube root is the first figure of the required root. Subtract the cube
from the first period, and to the remainder bring down the second
period for a dividend.

3d. Take three hundred times the square of the root already found
for a trial divisor; find how many times it is contained in the
dividend, and write the quotient for the second figure of the root.

4th. Complete the divisor by adding to it thirty times the product
of the two figures of the root, and the square of the second figure.

6th. Multiply the divisor thus increased by the last figure of Hie
root; subtract the product from the dividend, and to the remainder
"bring down the next period for a new dividend.

6th. Take three hundred times the square of the whole root now
found for a new trial divisor, and by division obtain another figure
of the root.

1th. Complete the divisor by adding to it thirty times the product
of the last figure by the former figures, and also the square of the
last figure, with which proceed as before, continuing the operation
until all the periods are brought down.

It will be observed that three times the square of the tens,
when their local value is regarded, is the same as three hund
red times the square of this digit, not regarding its local value.

In applying the preceding rule, it may happen that the prod
uct of the complete divisor by the last figure of the root is
greater than the dividend. This indicates that the last figure
of the root was taken too large, and this happens because the
divisor is at first incomplete, that is, too small. In such a case
we must diminish the last figure of the root by unity, until we
obtain a product which is not greater than the dividend.

EXAMPLES.

1. Find the cube root of 163667323. Ans. 547.

EVOLUTION. 149

2. Find the cube root of 39651821. Ans 341.

3. Find the cube root of 4019679. Ans. 159.

4. Find the cube root of 12895213625.

5. Find the cube root of 183056925752.

6. Find the cube root of 759299343867.

211. Cube Root of Fractions. — The cube root of a fraction is
equal to the root of its numerator divided by the root of its de
nominator. Hence the cube root of -YTT$ is T7^.

The number 12.167 may be written \Voir-5 and its cube root
is f-f, or 2.3. That is, the cube root of a decimal fraction, or
of a whole number followed by a decimal fraction, may be
found in the same manner as that of a whole number, if we
divide it into periods commencing with the decimal point.

In the extraction of the cube root of an integer, if there is
still a remainder after we have obtained the units' figure of the
root, it indicates that the proposed number has not an exact
cube root. We may, if we please, proceed with the approxi
mation to any desired extent, by supposing a decimal point at
the end of the proposed number, and annexing any number
of periods of three ciphers each, and continuing the operation.
We thus obtain a decimal part to be added to the integral part
already found.

So, also, if a decimal number has no exact cube root, we may
annex ciphers, and proceed with the approximation to any de
sired extent.

EXAMPLES.

1. Find the cube root of A8^-. Ans. £f .

2. Find the cube root of 14^-. Ans. 2f

3. Find the cube root of 13.312053.

4. Find the cube root of 1892.819053.

5. Find the cube root of .001879080904.

Find the cube roots of the following numbers to 5 decimal
places :

6. 15.25.

Ans. 2.47984.

10. 11.

7. 3.7.

Ans. 1.54668.

11. f

8. 100.1.

Ans. 4.64314.

12. f.

9. 4.

Ans. 1.58740.

150 ALGEBRA.

CHAPTER XIII.

RADICAL QUANTITIES.

212. A radical quantity is an indicated root of a quantity:
as ^/a, y/a, etc. Radical quantities may be either surd or ra
tional.

Radical quantities are divided into degrees, the degree being
denoted by the index of the root. Thus, V3 is a radical of the
second degree ; v/5 is a radical of the third degree, etc.

213. The coefficient of a radical is the number or letter pre
fixed to it, showing how often the radical is to be taken. Thus,
in the expression 2 -v/^ 2 is the coefficient of the radical.

Similar radicals are those which have the same index and
the same quantity under the radical sign. Thus, B\/a and 5\/a
are similar radicals. Also 7Vb and 10v/5 are similar radicals.

214. Use of fractional Exponents. — We have seen, Art. 196,
that in order to extract any root of a monomial, we must di
vide the exponent of each literal factor by the index of the re
quired root. Thus the square root of a4 is a2, and in the same

manner the square root of a3 may be written a*, that of a5 will

5 1

be a*, and that of a, or a1, is a . Whence we see that
a is equivalent to Va,
a? " Vo5,

a* " Ve?, etc.

2

So, also, the cube root of a2 may be written a ; the cube

4 1

root of a4 is a7; and the cube root of a, or a1, is a . Whence
we see that

RADICAL QUANTITIES.

a is equivalent to V/a,
a* " I/a*,

a? " V/a", etc.

In the same manner, or is equivalent to v/a»
a^ " Va,

a? " I/a5,

a" KU/ .

That isj $e numerator of a fractional exponent denotes the power,
and the denominator the root to be extracted.

Let it be required to extract the cube root of — -. This quan
tity, Art. 187, is equivalent to a~4. Now, to extract the cube
root of a~\ we must divide its exponent by 3, which gives us

-4 1 1

a T. But the cube root of -j may also be represented by —

Hence — is equivalent to a T.

a*

So, also, -T is equivalent to a ,

a

4 « .-«;

a~
1 _=

Thus we see that the principle of Art. 77, that a factor may
be transferred from the numerator to the denominator of a frac
tion, or from the denominator to the numerator by changing
the sign of its exponent, is applicable also to fractional exponents.

"We may therefore entirely reject the radical signs hitherto
employed, and substitute for them fractional exponents, and
many of the difficulties which occur in the reduction of radical
quantities are thus made to disappear.

152 ALGEBRA.

To reduce a Radical to its simplest Form.

215. A radical is in its simplest form when it has under the
radical sign no factor which is a perfect power corresponding
to the degree of the radical.

Radical quantities may frequently be simplified by the ap
plication of the following principle : the nth root of the product
of two or more factors is equal to the product of the nth roots of those
factors ; or, in algebraic language,

For each of these expressions, raised to the nth power, will
give the same quantity.

Thus, the nth power of V~ab is ah.
And the nib. power of Va X VI is (Va)n x (\fbT, or db.

Hence, since the same powers of the quantities n]/ab and
v/a X Vb are equal, the quantities themselves must be equal.

Let it be required to reduce V48a3x2 to its simplest form.

This expression may be put under the form Vl6a2#2 x VSa.

But Vl6a2x2 is equal to 4ax.

Hence V^Sa3x2=4ax V3a.

Hence, to reduce a radical to its simplest form, we have the
following

EULE.

Resolve the quantity under the radical sign into two factors, one
of which is the greatest perfect power corresponding in degree to the
radical. Extract the required root of this factor, and prefix it to
the other factor, which must be left under the sign.

EXAMPLES.

Eeduce the following radicals to their simplest forms.

l25a3. Ans. 5a V5a.

. W\/2aa

KADICAL QUANTITIES.

153

6.

216. When the quantity under the radical sign is a fraction,
it is often convenient to multiply both its terms by such a
quantity as will make the denominator a perfect power of the
degree indicated. Then, after simplifying, the factor remain
ing under fohe radical sign will be entire.

14. 3 A/|. Ans. SVI^SVrl^lVlO.

15. 4-Y/1+6V3?. Ans. |V2-f3VT4.

Ans.

IVlo.

. 2 Vab.

Ans. j-
b

23. 23v/i+3v/i.
24.

Ans.

Ans.

217. The following principle can frequently be employed in
simplifying radicals :

TJie mnih root of any quantity is equal to the mth root of the nth

root of that quantity. That is,

G 2

154: ALGEBRA.

For, if we raise each expression to the mth power, it becomes V#.
Thus, the fourth root = the square root of the square root.

the sixth root = the square root of the cube root, or

the cube root of the square root,
the eighth root = the square root of the fourth root, or
the fourth root of the square root,
the ninth root =the cube root of the cube root.
Hence, ivhen the index of a root is the product of two or more
factors, we may obtain the root required by extracting in succession
the roots denoted by those factors.

Ex. 1. Let it be required to extract the sixth root of 64.
The square root of 64 is 8, and the cube root of 8 is 2. Hence
the sixth root of 64 is 2.

Ex. 2. Extract the eighth root of 256. Ans. 2.

Ex. 3. Find the fourth root of 1874161. Ans. 37.

Ex. 4. Find the sixth root of 148035889. Ans. 23.

Ex. 5. Find the ninth root of 387420489. Ans. 9.

Ex. 6. Find the eighth root of 2562890625. Ans. 15.

218. When the index of a root is the product of two or more
factors, and one of the roots can be extracted, while the other
can not, a radical may be simplified by extracting one of the roots.

Thus, t/9=\/3.

Keduce the following radicals to their simplest forms :

Ex. 1. 6v/4o"2. Ans. ]/fa

Ex.2.

Ex.3.

Ex.4.

Ex.5.

Ex.6. \/^4-. Ans.

RADICAL QUANTITIES. 155

To introduce a Factor under the Radical Sign.

219. The square root of the square of a is obviously a, and
the cube root of the cube of a is a, etc. That is,

a=}/cP = Va? = Vdi1 etc.
Whence, also, a \/l — v/o5 X \/b = \/a2b.

Hence, to introduce a factor under the radical sign, we have
the following

RULE.

Raise the factor to a power denoted by the index of the required
root, and write it as a factor under the radical sign.

EXAMPLES.

1. Keduce ax2 to a radical of the second degree.

% Ans.

2. Eeduce 2azbx to a radical of the third degree.

Ans.

3. Keduce 5 + 6 to a radical of the second degree.

Ans.

Transform the following radicals by introducing the coefrV
cients as factors under the radical sign :

4. 6V8J. Ans. Vl26.

5. 4</J+3iV§. Ans. -V/2+V98.

6. a-+a&-iT- Ans.

ab

8. 2V2 + 71/5.

9. r|v/7j|.

^To change the Index of a Radical.
220. From Art. 219, it follows that

Va = W2 = V^3 = v/o*, etc. ;
V/a = v/a5 = I/a5 = Va*, etc. ;
V^=v/a"2 = Va1 = {X?, etc.

156 ALGEBRA.

Hence we see that the index of any radical may be multiplied
by any number, provided we raise the quantity under the radical
sign to a power whose exponent is the same number ; or the index
of any radical may be divided by any number, provided we extract
that root of the quantity under the radical sign whose index is the
same number.

If, instead of the radical sign, we employ fractional expo
nents, we shall have

1234
1234

a*=cP=ci*=aT*, etc.

Hence we see that we may multiply or divide both terms of a
fractional exponent by the same number, without changing the
value of the expression.

EXAMPLES.

Verify the following equations :

1. V2+VS-fVi=Vi-hV5+Vl0.

2. t/^Ts.

8. 2Va*.

4. 2 V^-b = 2 Va3 -3azb+ 3a£2 - b3.

5. SVaic

To reduce Radicals to a Common Index.

221. Let it be required to reduce Va and Va to equivalent
radicals having a common index. Substituting for the radical
signs fractional exponents, the given quantities are

a? and a*

Keducing the exponents to a common denominator, the ex
pressions are a 2

a* and OT,

or Va? and Va\

RADICAL QUANTITIES. 157

which are of the same value as the given quantities, and have
a common index 6. Hence we derive the following

RULE.

Reduce the fractional exponents to a common denominator, raise,
each quantity to the power denoted ly the numerator of its new ex
ponent, and take the root denoted ~by the common denominator.

EXAMPLES.
11 1

1. Reduce a , a , and a? to a common index.

3

TT

Ans. a1"*, a^, and a

1 2

2. Eeduce a?, a2, and b* to a common index.

3 12 4

Ans. a , a , and 8*.
11 i

3. Reduce 2T, 3T, and 5T to a common index.

Ans. V6i, "v/81, and 'v/125.

12 3

4. Reduce 3T, 2T, and 2T to a common index.

. 12v/729, "1/256, and 'v/512.

6. Reduce o , am, and an to a common index.

mn 2»i 2m

J.W5. a2771", a2nm, and a2""*.

6. Reduce y/3, VB, and v/7 to a common index.

7. Reduce V2ab, VSab2, and V&ab3 to a common index.

8. Reduce Va+6, V«— ^ ^n(l V«2— &2 to a common index.

To add Radical Quantities together.

222. When the radical quantities are similar, the common
radical part may be regarded as the unit, and the coefficient
shows how many times this unit is repeated. The sum of the
coefficients of the given radicals will then denote how many
times this unit is to be repeated in the required sum.

If the radicals are not similar they can not be added, because
they have no common unit In such a case, the addition can

158 ALGEBKA.

only be indicated by the algebraic sign. Eadicals which are
apparently dissimilar may become similar when reduced to
their simplest forms. Hence we have the following

RULE.

Reduce each radical to its simplest form. If the resulting radi
cals are similar, add their coefficients, and to their sum annex the
common radical. If they are dissimilar , connect them by the sign
of addition.

EXAMPLES.

1. Find the sum of V27, V58, and V75. Ans. 12 VS.

2. Find the sum of 4Vt±7r 3 A/75, and Vl92.

Ans. 51 VS.

3. Find the sum of V72, VI28, and V162. Ans. 23 V2.

4. Find the sum of A/180, A/405, and A/320. Ans. 23 A/5.

5. Find the sum of 3 A/|, 2 V^V, and 4A/S- ^s- VI5.

6. Find the sum of 'v/500, V/108, and \/256. ^w& 12 V3.

7. Find the sum of V50, 1/135, and v/320. ^^- 9i/5.

8. Find the sum of 2 V|, V60, VIB, and Vf.

9. Find the sum of A/45c3, A/80c3, and A/5a*c.

Ans. (a-f 7c)A/5c.
10. Find the sum of Vl8a5^3 + A/50a363.

^ws. (3a5

^Vrf2

11. Find the sum of

/a2 ac or/\ /7

. T + -T + — JV7?-
\5 d m/ V o

12. Find the sum of -/4a36, -y/25^63, and 5Z> Vo&.

RADICAL QUANTITIES. 159

To find the Difference of Radical Quantities,
223. When the radicals are similar, it is evident that the
subtraction may be performed in the same manner as addition,
except that the signs in the subtrahend are to be changed.
Hence we have the following

RULE.

Reduce each radical to its simplest form. If the resulting radi
cals are similar, find the difference of the coefficients, and to the re
sult annex the common radical part. If they are dissimilar, flit
subtraction can only be indicated.

EXAMPLES.

1. From V448 take Vll2. Ans. 4^7.

2. From 5 V20 take 3 V45. Ans. V5.

3. From 2\/50 take Vl8. Ans. 7 1/2.

4. From VSOa^c take V20a*x3.

5. From 21/720* take Vl62a2.

6. From v/192 take v/24. Ans. 2\/3.

7. From 21/320 take 3v/40.

r> -n i . ,_ ..

8. From X-^- take \-^. J.TW. (3a-l-

2T? multiply Radical Quantities together.

224. We have found, Art. 215, that Va multiplied by Vb is
equal to Vab.

Hence, V% X VS = V$.

If the radicals have coefficients, the product of the coefE*
cients may be taken separately.

Thus, aVxxl)Vy=axbxVxxVy=abVxy;
also, 3-v/8x5V2^15Vl6 = 60.

If the radicals have not a common index, they must first be
reduced to a common index. Hence we have the following

160 ALGEBRA.

RULE.

If necessary, reduce the given radicals to a common index.
tiply the coefficients together for a new coefficient ; also multiply the
quantities under the radical signs together, and place this product
under the common radical sign. Then reduce the result to its sim
plest form.

EXAMPLES.

Find the value of the following expressions :

1. 3V8X2-/6. Ans. 24\/3.

2. 5V8X3V5. Ans. 30VlO.

3. V2 x \/3. Ans. \/72.

4. 5 V3 x 7 VI x V2. Ans. 140.

5. cVaxdVa. Ans. acd.

6. 7v/18x5v/4. Ans. 70v/9.

7. iV6xAl/l7.

8. i 1/18x61/20.

9. 3y/4 x 7 V/6 x J 1/6. ^s. 7 v/15.

225. We have seen, Art. 58, that the exponent of any letter in
a product is equal to the sum of the exponents of this letter in the
multiplicand and multiplier. That is, amxa"=am+n, where m
and n are supposed to be positive whole numbers.

When one or both of the exponents are negative, we must
take the algebraic sum of the exponents. For, suppose n is
negative. Then

am x a~n = am x -, by Art. 76, = ^ = am~n.

The same relation holds true when m and n are fractional ;

p r p+r

that is, a*xas=a« *.

For a* x as= \fa? X t/ar, Art. 214, ^V^ x Va^, Art. 220,

Hence we conclude that the exponent of any letter in a product
is equal to the algebraic sum of the exponents of this letter in the

RADICAL QUANTITIES. 161

multiplicand and multiplier, whether the exponents are positive or
negative, integral or fractional.

EXAMPLES.

1. Multiply 5a* by 3a*. Arts. Ufa*

2. Multiply 2la* by 3a8. Ans. 63a^.

3. Multiply 8x*y* by 4a:*y*

4. Find the product of a2, a , a , and cT^*.

i _.& 4 1

5. Find the product of a , a , a^, and a1"2".

Multiplication of Polynomial Radicals.

226. By combining the preceding rules with that for the
multiplication of polynomials, Art. 61, we may multiply togeth
er radical expressions consisting of any number of terms.

Ex. 1. Let it be required to multiply

-3a -

Ex. 2. Multiply 3 + V5 by 2 — VS. Ans. 1 — V5.

Ex. 3. Multiply 7 + 2 V6 by 9-5 V6. Ans. 3- 17 VS.
Ex. 4. Multiply 9 + 2 VlO by 9-2 VlO. Ans. 41.

Ex. 5. Multiply 3 V45-7 V5 by Vl|+2 V§|. Ans. 34.

Ex. 6. Multiply cVa + dVb by cVa—dVb.

A.ns, ac — ooL .

Ex. 7. Multiply aT— a3 + a¥— a2+a^— a+a¥— 1 by a

.Ans. a4 — 1.

4*

162 ALGEBKA.

To divide one Radical Quantity by another.

227. The division of radical quantities depends upon the fol
lowing principle :

The quotient of the nth roots of two quantities is equal to the nth
root of their quotient ; or,

for the nth power of each of these expressions is ^, Art. 186.

Let it be required to divide 4a2A/6fo/ by 2aVSb.

4a2 /% /—

—\/-^=2aV2y. Ans.

2aVSb
Hence we have the following

EULE.

If necessary, reduce the given radicals to a common index. Di
vide the coefficient of the dividend by that of the divisor for a new
coefficient ; also the quantity under the radical sign in the dividend
by that in the divisor, and place this quotient under the common
radical sign. Then reduce the result to its simplest form.

EXAMPLES.

1. Divide 8VT08 by 2V 6. Ans. 12 1/2.

2. Divide 8 v/512 by 4 i/2. Ans. 8 3v/4.

3. Divide 6 3/54 by 3 v/2. Ans. 6.

4. Divide 4 v/72 by 2 v/18.

5. Divide 4\/6a2?/ by

. _

6. Divide 16(a36)w by 8(ac)m.

7. Divide 4 v/12 by 2 V^. Ans. 2

8. Divide v/64 by 2.

9. Divide Va2bc by \

228. We have seen, J.r£. 72, that /7ie exponent of any letter in

KADICAL QUANTITIES. 163

a quotient is equal to the difference between the exponents of this let
ter in the divisor and dividend.

The same relation holds true whether the exponents are posi
tive or negative, integral or fractional , that is, universally,

For the quotient must be a quantity which, multiplied by
the divisor, shall produce the dividend ; and, according to Art.
225, the exponent of any letter in a product is in all cases
equal to the algebraic sum of the exponents of this letter in
the multiplicand and multiplier. Hence this relation must
hold true universally in division.

EXAMPLES.

1. Divide (ab)3 by (o5)* Ans. (afy*.

2,. 1

2. Divide a¥ by a*.

3. Divide 4:Vab by 2 \/ab. Ans. 2 Vab.

4. Divide 9m2(a— 1$ by Sm(a— bfi. Ans. Sm(a-b)^.

1 1 JL JL

5. Divide a2&3 by aBb5. Ans. a

6. Divide 4* by 2*. Ans, 4=

V2

Division of Polynomial Radicals.

229. By combining the preceding rules with that for the di
vision of polynomials, Art. 80, we may divide one radical ex
pression by another containing any number of terms.

A .2. 1 1 1

Ex. 1. Let it be required to divide a6 — a3 — a2 -far by a^— 1.

$ I ai_i

Ex.2. Divide 8a-6 by 2a~

164 ALGEBKA.

Ex. 3. Divide a-41o^-120 by a^

Ans. a~z —
Ex. 4. Divide a*+64&* by a^+46*.

Ans. a?— 4a^6^"-f 16&*.

Ex. 5. Divide x%— xy^+x^y— y^ by x^—y^.

, Ans. x+y.

To involve a Radical Quantity to any power.

^
230. Let it be required to raise am to the nth power.

Ill l_i_l JL
The square of am is amxam=am m=am.

1 til 1+1+1 A
The cube of am is amxamxam=a™^~m=an.

1111 -+-+- etc H

The nth power of am is am x am x am, etc., =am m m' '' =am.

Hence, to involve a radical quantity to any power, we have
the following

RULE.

Multiply the fractional exponent of the quantity by the exponent
of the required power. If the radical has a coefficient, let this be in
volved separately ; then reduce the result to its simplest form.

If the quantity is under the radical sign, it is generally most
convenient to substitute for this sign the equivalent fractional
exponent ; but if we choose to retain the radical sign, we must
raise the quantity under it to the required power.

EXAMPLES.

1 4

1. Eequired the fourth power of fa*. Ans.

2. Eequired the cube of |V3. Ans. f

3. Eequired the square of 3 v/3.

4. Eequired the cube of 17V21.

5. Eequired the fourth power of iV6. Ans.

6. Eequired the fourth power of 2 v/3o*&.

j

7. Eequired the fourth power of abVab. Ans. a6b*.

RADICAL QUANTITIES. 165

8. Kequired the sixth power of (a+b) *.

Ans. a2 + 2ab + b\

9. Eequired the value of V(l|)7 x V(£f)6. Ans. -&.

10. Kequired the value of V(±ab2)x x V(2a?by.

Ans. (2ab)*.

To Extract any Root of a Radical Quantity.
231. A root of a quantity is a factor which, multiplied by
itself a certain number of times, will produce the given quanti-

-I *
ty. But we have seen that the nth power of am is am. There-

n^ _!_

fore the nth root of am is am. Hence we derive the following

RULE.

Divide the fractional exponent of the quantity by the index of the
required root. If the radical has a coefficient, extract its root sep
arately if possible ; otherwise introduce it under the radical sign.
Then reduce the result to its simplest form.

If the quantity is under the radical sign, and we choose to
retain the sign, we must, if possible, extract the required root
of the quantityunder the radical sign ; otherwise we must mul
tiply the index of the radical by the index of the required root.

EXAMPLES.

1. Find the square root of 9(3)^. Ans. 3 v/3.

2. Find the cube root of $V5. Ans. i 6/2.
8. Find the square root of 103.

4. Find the cube root of

5. Find the fourth root

6. Find the cube root of

7. Find the cube root # | \/| Ans- \/f

8. Find the square root of 3 v/5. Ans. y/135.

9. Find the fourth root of % v/|. Ans. $ \/l2.

166 ALGEBRA.

Operations on Imaginary Quantities.

232. It has been shown, Art. 195, that an even root of a neg
ative quantity is impossible. Thus, V— 4, V — 9, V— 5a are
algebraic symbols representing operations which it is impossi
ble to execute ; for the square of every quantity, whether posi
tive or negative, is necessarily positive. Quantities of this na*
ture are called imaginary or impossible quantities. Neverthe
less, such expressions do frequently occur, and it is necessary
to establish proper rules for operating upon them.

233. The square root of a negative quantity may always be rep
resented by the square root of a positive quantity multiplied by the
square root of —I.

Thus 1/^4 = V±x -1 =

The factor V — 1 is called the imaginary factor, and the other
factor is called its coefficient.

234. When several imaginary factors are to be multiplied
together, it is best to resolve each of them into two factors, of
which one is the square root of a positive quantity, and the
other V — 1. We can then multiply together the coefficients
of the imaginary factor by methods already explained. It
only remains to deduce a rule for multiplying the imaginary
factor into itself; that is, for raising the imaginary factor to a
power whose exponent is equal to the number of factors.

The first power of V— 1 is V— 1.

The second power, by the definition of square root, is — 1.

The third power is the product of the first and second pow

ers, or — Ix V— 1 = — V — 1.

The fourth power is the square of the second, or +1.

The fifth is the product of the first and fourth ; that is, it is
the same as the first; the sixth is the same as the second, and

RADICAL QUANTITIES.

167

so on ; so that all the powers of V— 1 form a repeating cycle
of the following terms :

EXAMPLES.

1. Multiply V — 9 by V— 4.

V^x V^4 = 3V^Tx2V^I=£

2. Multiply 1+V^T by 1— V^T.

3. Multiply Vl8 by V^2.

4. Multiply 5 + 2V^3 by 2-V^1

5. Multiply aV — b by cV—d.

6. Multiply 1-V^l by itself.

7. Multiply 2V3— V— 5 by 4 A/3-

JL «/ »/

J.W5. 1

8. Multiply a+ VbV — 1 by a— VbV^ 1.

9. Multiply aV — a2b3 by V — a4^5.

10. Multiply V — a-\-V~^b by V— a— y^6.

u4ws. — acVbd.
Ans. —2V — 1.

235. Division of Imaginary Quantities. — The quotient of one
imaginary term divided by another is easily found by resolv
ing both terms into factors, as in the preceding article.

Ex. 1. Let it be required to divide V^ab by V — a.

— = y 6, A7Z5.

— a

Ex. 2. Divide Z>V^1 by cV— 1,

Ex. 3. Divide unity by V^T.
Ex. 4. Divide a by 6V— 1.
Ex. 5. Divide a by V«V— 1.

. — V — 1.

168 ALGEBRA.

Ex. 6. Divide V^l2 + V^6+V^9 by
Ex. 7. Divide 2V8-V-10 by --/^2.

Ans

Multipliers which shall cause Surds to become Rational.

236. 1st. When the surd is a monomial.

The quantity Va is rendered rational by multiplying it by

Va.

i i

For -y/a X Va—o^ x a^—a.

1 2

So, also, aT is rendered rational by multiplying it by a7.

i -3 i

Also, a4 is rendered rational by multiplying it by a4 ; and an

by multiplying it by a n.

Hence we deduce the following

RULE.

Multiply the surd l>y the same quantity having such an exponent
as, when added to the exponent of the given surd, shall make unity.

237. 2d, When the surd is a binomial.

If the binomial contains only the square root, multiply the
given binomial by the same terms connected by the opposite sign,
and it will give a rational product.

Thus the expression Va-\-Vb multiplied by Va—Vb gives
for a product a— b.

Also the expression \/a+Vb multiplied by \/a— Vb gives
for a product Va—Vb, which may be rendered rational by
multiplying it by Va+Vb.

In general, m\/a ± Vb may be rendered rational by success
ive multiplications whenever m and n denote any power of 2.
When m and n are not powers of 2, the binomial may still be
rendered rational by multiplication, but the process becomes
more complicated.

Ex. 1. Find a multiplier which shall render 1/6 + VB ration
al, and determine the product.

RADICAL QUANTITIES. 169

Ex. 2. Find a multiplier which shall render A/3 — Vx ration
al, and determine the product.

Ex. 3. Find multipliers which shall render A/3 — v/^ ration
al, and determine the product.

»

238. 3d. When the surd is a trinomial

When a trinomial surd contains only radicals of the second
degree, we may reduce it to a binomial surd by multiplying it
by the same expression, with the sign of one of the terms
changed. Thus, Va+Vb+Vc multiplied by Va+Vb—Vc
gives for a product a + b — c + 2Vabi which may be put under
the form of m-{-2Vab.

Ex.1. Find multipliers that shall make A/5 -f A/3— A/2 ra
tional, and determine the product.

Ex. 2. Find multipliers that shall make 1-f- A/2 + A/3 ration
al, and determine the product.

To transform a Fraction whose Denominator is a Surd in such
a Manner that the Denominator shall be Rational.

239. If we have a radical expression of the form

a Vb+Vc

or 7/7 77=, it may be transformed into an equivalent expres

sion in which the denominator is rational by multiplying both
terms of the fraction by Vb± Vc. Hence the

RULE.

Multiply both numerator and denominator by a factor which
ivill render the denominator rational.

EXAMPLES.

Reduce the following fractions to equivalent fractions having
a rational denominator :

2 2A/3

L

2. _ Ans.

A/5- A/2

IT

170 ALGEBKA.

3. -- -=. Ans.

3-

4.

a+Vb
5.

8. ~1 Ans.')/*.

1+a+Vl — a2 , l + Vl—a2

1+a—Vl — a2' &

240. The utility of the preceding transformations will be seen
if we attempt to compute the numerical value of a fractional
surd.

Ex. 1. Let it be required to find the square root of T ; that is,

Vs

to find the value of the fraction — .

Making the denominator rational, we have ^ , and the

value of the fraction is found to be 0.6546.

7V5

Ex. 2. Compute the value of the fraction — — .

Vll+VS

Ans. 3.1003.

•v/6
Ex. 3. Compute the value of the fraction — — .

V7-h V3

Ans. 0.5595.

1/8

Ex. 4. Compute the value of the fraction — -=— —j=.

2v8+3V5-7v2

Ans. 0.7025.

RADICAL QUANTITIES. 171

94-2A/IO

Ex. 5. Compute the value of the fraction

9-2 VlO

Ans. 5.7278.

Square Root of a Binomial Surd.

241. A binomial surd is a binomial, one or both of whose
terms are surds, as 2-f V3 and 1/5— A/2.

A quadratic surd is the square root of an imperfect square.

If .we square the binomial surd 2+ A/3, we shall obtain
7 + 4A/3. Hence the square root of 7+4V3 is 2+ A/3; that
is, a binomial surd of the form a ± Vl> may sometimes be a perfect
square.

242. The method of extracting the square root of an expres
sion of the- form a± Vb is founded upon the following princi
ples:

1st. The sum or difference of two quadratic surds can not be
equal to a rational quantity.

Let A/a and Vb denote two surd quantities, and, if possible,

where c denotes a rational quantity. By transposing Vb and
squaring both members, we obtain

The second member of the equation contains only rational
quantities, while Vb was supposed to be irrational; that is, we
have an irrational quantity equal to a rational one, which is
impossible. Hence the sum or difference of two quadratic
surds can not be equal to a rational quantity.

243. 2d. In any equation which involves both rational quanti
ties and quadratic surds, the rational parts in the two members are
equal, and also the irrational parts.

Suppose we have

172 ALGEBRA.

Then, if x be not equal to a, suppose it to be equal to a-f m;

then /- /7

y — a-\- Vb,

so that m=Vb—Vy;

that is, a rational quantity is equal to the difference of two
quadratic surds, which, by the last article, is impossible. There
fore x=a, and consequently Vy—Vb.

244. To find an expression for the square root of a± Vb.

Let us assume Va + Vb — Vx + Vy. (1.)

By squaring, a+ Vb—x-\- 2Vxy+y. (2.)

By Art. 243, a=x+y, (3.)

and Vb = 2Vxy. (4.)
Subtracting (4) from (3), we have

a—Vb=x-2Vxy+y. (5.)

By evolution, Va—Vb =Vx—Vy. (6.)
Multiplying (1) by (6), we have

(7.)

Adding (3) and (7), a+V~a^-b = 2x. (8.)

TT- a-hvcfc2— u /n \

Hence, cc = — (9.)

Subtracting (7) from (3),

It is obvious from these equations that x and y will be ra
tional when az — b is a perfect square. If «2 — b be not a perfect
square, the values of ~Jx and Vy will be complex surds.

Hence, to obtain the square root of a binomial surd, we pro
ceed as follows :

Let a represent the rational part, and Vb the radical part,
and find the values of x and y in equations (9) and (10). Then,
if the binomial is of the form a-fVS, its square root will be
Vx+Vy. If the binomial is of the form a— Vb, its square
root will be ^fx— V y. ^

RADICAL QUANTITIES.
EXAMPLES.

1. Required the square root of 4 + 2V3.
Here a =4 and Vb =

Hence

2
Hence Vx+ Vy= V3 + 1, Ans.

Verification. The square of VB + 1 is 3 + 2/84-1=4+21/3.

2. Kequired the square root of 11 + 6V2.
Here a=ll and Vft = 6-v/2; or 6 = 72.

ce=9 and y=2.
Vx+Vy = 3+V%, Ans.

3. Required the square root of 11-2/30. Ans. -v/6— V5.

4. Required the square root of 2+ Vs. Ans. V|+ V^»

5. Required the square root of 7 + 2 VTO. Ans. V5+V2.

6. Required the square root of 18 + 8V5.

Ans. VIO + 2-V/2.

245. This method is applicable even when the binomial con
tains imaginary quantities.

7. Required the square root of 1+4V— 3.
Here a = l and V£=4V^3.

Hence 6= -48 and a2-&=49.

Therefore x=4: and y=— 3.

The required square root is therefore 2 + V — 3,

8. Required the square root of — 2-+|V— 3.

9. Required the square root of 2V — 1 or

Ans.

174 ALGEBRA.

10. Eequired the value of the expression

V6 + 2V5-V6-2V5.
IL Eequired the value of the expression

V4 + 3V^H- V4-3V-20. <4ws. 6.

12. Eequired the square root of — 3+ V — 16.

Ans.

13. Eequired the square root of 8 A/ — 1,

Simple Equations containing Radical Quantities.
246. When the unknown quantity is affected by the radical
sign, we must first render the terms containing the unknown
quantity rational. This may generally be done by successive
involutions. For this purpose we first free the equation from
fractions. If there is but one radical expression, we bring that
to stand alone on one side of the equation, and involve both
members to a power denoted by the index of the radical.

Ex.1. Given x-}- Vx2-3x + 60 = 12 to find x.
Transposing x and squaring each member, we have

whence x =4.

2a
. Given Vx+ va+x=— -p = to find x.

I/ /-» I SY>

Ans. x=%.

Ex.3. Given

Ex, 4. Given V4x2— Ix— 6 = 9 — 2x to find x.

247. If the equation contains two radical expressions com
bined with other terms which are rational, it is generally best
to bring one of the radicals to stand alone on one side of the
equation before involution. One of the radicals will thus be
made to disappear, and, by repeating the operation, the remain
ing radical may be exterminated.

Ex.5. Given Vx-}-19+ Vx+lO — Q to find x.

RADICAL QUANTITIES.

175

By transposition,

Squaring,

Transposing and reducing,

VoH-10=4.

Squaring, x+ 10 = 16 ;

whence x — 6.

Ex.6. Given V36+x=18-h Vx to find x.

Ex.7. Given Vx + ±ab = 2b + Vx to find x.

Ex. 8. Given x— Va2+xVb*+x2 — a?+a to find x.

248. When an equation contains a fraction involving radical
quantities in both numerator and denominator, it is sometimes
best to render the denominator rational by Art. 239 ; but the
best method can only be determined by trial.

Ex.9. Given

to find x.

Multiply both terms of the first fraction by Vx+ Vx—3,
and we have (Vx+ Vx— 3)2 3

or

:x-3'

Extracting the square root,

y^
Clearing of fractions,

whence

Ex. 10. Given

Ex. 11. Given

Ex. 12. Given

176 ALGEBKA.

Ex. 13. Given S^~l =l + £(V&c-l) to find x.

Ex. 14. Given -^ * to find x.

,. a?=/r my* V

\m — n)

Ex.15. Given (VSc-6) (Vie + 25)=(5 + 3VaO (Vx+3) to
find x. Ans. x = 9.

Ex.16. Given V'2x— 3n = 3V?i— V%x to find x.

*iLns. x — An,

^ 17 p. -

Ex.17. Given --= - — ._ - to find x.
Vx+2

— -

Ex.18. Given — = - = — — - to find x.

Ex.19. Given --l^Z to find x.

Ex. 20. Given /4a+x=2 Vb+x— Vx to find x.

(a-
Ans. x=:

: --
2a— 6

EQUATIONS OF THE SECOND DEGREE. 177

CHAPTER XIV.

EQUATIONS OF THE SECOND DEGREE.

249. An equation of the second degree, or a quadratic equation
with one unknown quantity, is one in which the highest power
of the unknown quantity is a square.

250. Every equation of the second degree containing but
one unknown quantity can be reduced to three terms; one con
taining the second power of the unknown quantity, another the
first power, and the third a known quantity ; that is, it can be
reduced to the form _

Suppose we have the equation

Clearing of fractions and expanding, we have

9x2+7x-6 + 4x-8 + 164:=12x2+24x-36.
Transposing and uniting similar terms, we have

-3x2-13x=-186.
Dividing by the coefficient of x2, that is, by —3, we have

which is of the form above given, p in this case being equal to
^, and q being equal to 62.

251. In order to reduce a quadratic equation to three terms, we
must first clear it of fractions, and perform all the operations
indicated. We then transpose all the terms which contain the
unknown quantity to the first member of the equation, and the
known quantities to the second member; unite similar terms,
and divide each term of the resulting equation by the coeffi
cient of x2.

178 ALGEBRA.

252. An equation of the form x2+px=q is called a complete
equation of the second degree, because it contains each class of
terms of which the general equation is susceptible.

253. The coefficient of the first power of the unknown quan
tity may reduce to zero, in which case the equation is said to
be incomplete.

An incomplete equation of the second degree, when reduced,
contains but two terms : one containing the square of the un
known quantity, and the other a known term.

Incomplete Equations of the Second Degree.

254. An incomplete equation of the second degree may be
reduced to the form xz=q.

Extracting the square root of each member, we have
x=±i/q.

If q be a positive number, either integral or fractional, we
can extract its square root, either exactly or approximately, by
the rules of Chap. XII. Hence, to solve an incomplete equa
tion of the second degree, we have the

RULE.

Reduce the equation to the form xz=q, and extract the square
root of each member of the equation.

255. Since the square of both -fm and — m is -\-m2, the
square of -f Vq and that of — Vq are both + q. Hence the
above equation is susceptible of two solutions, or has two roots;
that is, there are two quantities, which, when substituted for x
in the original equation, will render the two members identical.
These are +Vq and — Vq.

Hence, Every incomplete equation of the second degree has two
roots, equal in numerical value, but with opposite signs.

EXAMPLES.

Find the values of x in each of the following equations:
1. 4x2— 7 = 3z2-}-9. Ans. x=±4.

EQUATIONS OF THE SECOND DEGREE. 179

Show that each of these values will satisfy the equation.

Ans. x=±7.

±Vab
Ans. x= — ~ — •

__ 2
* ~ f"~ f

10. xa4-Vx2-17=-

Ans.x=±S.

jZVbfe. Clearing of fractions and transposing, we find in each
member of this equation a binomial factor, which being cancel
ed, the equation is easily solved.

¥

PROBLEMS.

Prob. 1. What two numbers are those whose sum is to the
greater as 10 to 7, and whose sum, multiplied by the less, pro
duces 270?
, Let 10x= their sum.

Then 7x — the greater number,

and 3x = the less.

Whence 30x2 = 270,

and x2 = 9;

therefore a; =±3,

and the numbers are ±21 and ±9.

180 ALGEBRA.

Prob. 2. What two numbers are those whose sum is to the
greater as m to n, and whose sum, multiplied by the less, is
equal to a ?

A / an2 /a(m — n

Ans. ± V —/ x and

v m(m — n)

Prob. 3. What number is that, the third part of whose
square being subtracted from 20, leaves a remainder equal
to 8?

Prob. 4. What number is that, the rath part of whose square
being subtracted from a, leaves a remainder equal to b ?

Ans. ± Vm(a — b).

12 3
Prob. 5. Find three numbers in the ratio of ^, ^ and 7, the

£ O 4:

sum of whose squares is 724.

Prob. 6. Find three numbers in the ratio of m, n, and jo, the
sum of whose squares is equal to a.
Ans.

am? an2

Prob. 7. Divide the number 49 into two such parts that the
quotient of the greater divided by the less may be to the quo
tient of the less divided by the greater as | to -f .

Ans. 21 and 28.

Note. In solving this Problem, it is necessary to assume a
principle employed in Arithmetic, viz., If four quantities are
proportional, the product of the extremes is equal to the product of
the means.

Thus, if a : b ; : c : d,

then ad=bc.

Prob. 8. Divide the number a into two such parts that the
quotient of the greater divided by the less may be to the quo
tient of the less divided by the greater as m to n. ,

aVm , aVn
Ans. — — and — — -.

m -j- Vn
Prob. 9. There are two square grass-plats, a side of one of

EQUATIONS OF THE SECOND DEGREE. 181

which is 10 yards longer than a side of the other, and their
areas are as 25 to 9. What are the lengths of the sides?

Prob. 10. There are two squares whose areas are as m to T?,
and a side of one exceeds a side of the other by a. What are
the lengths of the sides ?

aVm , aVn

Ans. —7= = and — 7=^

Vra — Vn Vm — Vn

Prob. 11. Two travelers, A and B, set out to meet each other,
A leaving Hartford at the same time that B left New York.
On meeting, it appeared that A had traveled 18 miles more
than B, and that A could have gone B's journey in 15-| hours,
but B would have been 28 hours in performing A's journey.
What was the distance between Hartford and New York?

Ans. 126 miles.

Projp. 12. From two places at an unknown distance, two
bodies, A and B, move toward each other, A going a miles more
than B. A would have described B's distance in n hours, and
B would have described A's distance in m hours. What was
the distance of the two places from each other ?

Ans. ax-—=

—

Prob. 13. A vintner draws a certain quantity of wine out of
a full vessel that holds 256 gallons, and then, filling the vessel
with water, draws off the same quantity of liquor as before,
and so on for four draughts, when there were only 81 gallons
of pure wine left. How much wine did he draw each time?

Ans. 64, 48, 36, and 27 gallons.

Note. Suppose - part is drawn each time.

256 256(x _ 1)
Then 256 — - — remains after the first draught.

x x

256(V _ I")2
Similarly, - ' 2 - remains after the second draught, and

so on.

256 (x- 1)4
Hence ~r — -==81.

182 ALGEBRA.

Prob. 14. A number a is diminished by the nth part of it
self, this remainder is diminished by the ?ith part of itself, and
so on to the fourth remainder, which is equal to b. Eequired
the value of n.

A Va

Ans. —~ - -.

Prob. 15. Two workmen, A and B, were engaged to work
for a certain number of days at different rates. At the end of
the time, A, who had played 4 of those days, received 75 shil
lings, but B, who had played 7 of those days, received only
48 shillings. Now had B only played 4 days and A played 7
days, they would have received the same sum. For how
many days were they engaged? Ans. 19 days.

Prob. 16. A person employed two laborers, allowing them
different wages. At the end of a certain number of days, the
first, who had played a days, received m shillings, and the
second, who had played b days, received n shillings. Now if
the second had played a days, and the other b days, they would
both have received the same sum. For how many days were
they engaged?

bVm—aVn ,
Ans. — — -- j=r- clays.
Vm — Vn

Complete Equations of the Second Degree.

256. In order to solve a complete equation of the second de
gree, let the equation be reduced to the form

If we can by any transformation render the first member of
this equation the perfect square of a binomial, we can reduce
the equation to one of the first degree by extracting its square
root.

Now we know that the square of a binomial, x+a, or rr2-f
2ax-\-a2, is composed of the square of the first term, plus twice
the product of the first term by the second, plus the square of
the second term.

Hence, considering x*+px as the first two terms of the

EQUATIONS OF THE SECOND DEGREE. 183

square of a binomial, and consequently px as being twice the
product of the first term of the binomial by the second, it is

evident that the second term of this binomial must be ~.

257. In order, therefore, that the expression x2+px may be
rendered a perfect square, we must add to it the square of this

f)

second term ^ ; and in order that the equality of the two mem
bers may not be destroyed, we must add the same quantity to
the second member of the equation. We shall then have

p2 p*

x2+px + ^-=q+^-.

Taking the square root of each member, we have

whence, by transposition, x= — ^-± y q-\- ^-.

Thus the equation has two roots: one corresponding to the
plus sign of the radical, and the other to the minus sign. These
two roots are

258. Hence, for solving a complete equation of the second
degree, we have the following

RULE.

1st. Reduce the given equation to the form ofx'z-{-px=q.

2d. Add to each member of the equation the square of half the co
efficient of the first power of x.

3d. Extract the square root of loth members, and the equation
will be reduced to one of the first degree, which may he solved in the
usual manner.

259. When the equation has been reduced to the form x2 +
pz=q, its two roots will be equal to half the coefficient of the sec-

184 ALGEBRA.

ond term, taken with a contrary sign, plus or minus the square
root of the second member, increased by the square of half the coeffi
cient of the second term.

Ex. 1. Let it be required to solve the equation
x2-10x=-16.

Completing the square by adding to each member the square
of half the coefficient of the second term, we have

Extracting the root, x— 5~ ±3.

Whence x=5±3= 8 or 2, Ans.

To verify these values of x, substitute them in the original
equation, and we shall have

82-10x8 = 64-80=:-16.
Also, 22-10x2 = 4-20=-16.

Ex. 2. Solve the equation 2x2 + 8x— 20 = 70.

Ans. x=+5 or —9.
Ex. 3. Solve the equation 3o;2—

Eed u ci n g, x2 — x = — f .

Completing the square, x2— x+^=^— |=^V-
Hence #=J±i = -for-J, Ans.

Second Method of completing the Square.

260. The preceding method of completing the square is al
ways applicable ; nevertheless, it sometimes gives rise to incon
venient fractions. In such cases the following method may be
preferred. Let the equation be reduced to the form

ax2 -{-bx = c,

in which a and b are whole numbers, and prime to each other,
but c may be either entire or fractional.

Multiply each member of this equation by 4a, and it becomes

Adding b2 to each member, we have

4a2x2 4- 4abx + b* = 4ac -f b2,

where the first member is a complete square, and its terms are
entire.

Extracting the square root, we have
2ax+ b=

EQUATIONS OF THE SECOND DEGREE. 185

Transposing b, and dividing by 2a,

which is the same result as would be obtained by the former
rule ; but by this method we have avoided the introduction of
fractions in completing the square.

If b is an even number, - will be an entire number ; and it

&

would have been sufficient to multiply each member by a, and

£2

add — to each member. Hence we have the following

RULE.

1st. Reduce the equation to the form ax2 + bx=c, where a and b
are prime to each other.

Zd.^Ifb is an odd number, multiply the equation by four times
the coefficient ofx2, and add to each member the square of the coeffi
cient ofx.

3d. If b is an even number, multiply the equation by the coeffi
cient ofx2, and add to each member the square of half the coefficient
ofx.

Ex. 4. Solve the equation 6x2— ISx-— 6.

Multiplying by 4x6, and adding IS2 to each member, we
have 144:c2-312x-f 169 = 169-144=25.

Extracting the root, I2x — 13 = ± 5.
Whence l2x=lS or 8,

and x=% or J-.

Ex.5. Solve the equation llOx2— 21a= — 1.

Multiplying by 440, and adding 212 to each member, we have

48400x2- 9240x+ 441 = 1.
Extracting the root, 220x- 21 = ± 1.
Whence x=^ or -f^.

Ex.6. Solve the equation 7£2-3£=160.

Ans. x=5 or — £*..

261. Modification of the preceding Method. — The preceding
method sometimes gives rise to numbers which are unneces-

186 ALGEBRA.

sarily large. When the equation has been reduced to the form
ax2+bx=c, it is sufficient to multiply it by any number which
will render the first term a perfect square. Let the resulting
equation be

m2x2-{-nx=q.
The first member will become a complete square by the ad-

(n \2
- — 1 , and the equation will then be
2m/

4m2 * 4m2'
This method is expressed in the following

RULE.

Having reduced the equation to the form ax2-\-bx=c, multiply
the equation by any quantity (the least possible) which will render
the first term a perfect square. Divide the coefficient of x in this
new equation by twice the square root of the coefficient of x2, and
add the square of this result to both members.
Ex.7. Solve the equation 8x24-9x=99.

3249

16
9 , 57

33

x=S or —-5-, Ans.
o

Ex. 8. Solve the equation 16x2— 15^=34.

15\2 2401

8/ 64

17
te=8 or -T,

x=2 or — r— , Ans.
ID

Ex. 9. Solve the equation 12x2

Solve the following equations :

Ex.10, frc2— &o5 + 20J=42t. Ans. x = 7 or -6J.

Ex. 11. x2— £-40 = 170. Ans. x=15 or -14.

Ex.12. 3x2-h2x-9 = 76.

EQUATION'S OF THE SECOND DEGREE. 187

Ex.13. x2

,-, 0 - - 0

Ex. 14. &-— —--—=2.

This equation reduces to x2— 15 %x =—46.

Ans. x— 11 J or 4.

Ex. 15. ^-1^=^=2. Ans. x=3 or 6.

x 2x2

x x+L x+2
Ex.17. x2-xV3=x-%VS.

V3 + 3 1/3-1

Ans. x= — — or — — •

„ ' x— 1 cc—3 2

Ex.18. - -z -- -—— ~. Ans. x=5 or 1.

a:— 2 x— 4 3

,-,- _,„ x a-\-x 5

Ex. 19. --- — — =-. ^n». cc=a or — 2«,

x 2

Ex. 20. x2— (a + l)x + ab = Q. Ans. x = a or b.

Ex.21. (3x-25)(7cc4-29) = 0. Ans. a=8£ or -4f

3x_2 2x-5 10 13 1

Ex. 23. x-lx-

Ans. &=8 or .

™ 01 170 170 51 ,12

Ex.24. -- = pr. Ans. x-^ or — If.

x x+l x+2

-r, a2 -4- ax -\- x2 a2 — ax4-x2 ab

Ex. 25.

a+x a — x 3a—

c=— 3a or 3a

Equations which may be solved like Quadratics.

262. There are many equations of a higher degree than the
second, which may be solved by methods similar to those em
ployed for quadratics. To this class belong all equations which
contain only two powers of the unknown quantity, and in which

188 ALGEBRA.

ike greater exponent is double the less. Such equations are of the
form x*n+px* = q,

where n may be either integral or fractional.

For if we assume y=xn, then 2/2=ic2n, and this equation
becomes y*+py=<i;

whence

Extracting the nth root of each member, we have

Ex.1. Solve the equation x*— I3x2=— 36.
Assuming x2=y, the above becomes

whence 2/=9 or 4.

But, since xz=y, x—±Vy.

Therefore x= ± V$ or ± VI.

Thus x has four values, viz., +3, -3, + 2, -2.
To verify these values:

1st value, ( + 3)4-13(+3)*=-36, i.e., 81-117^-36,
2d value, (-3)4- 13(-3)2=-36, i.e., 81-117 = -36.
3d value, (+2)4-13( + 2)2= _36, i.e., 16- 52= -36.
4th value, (_2)4-13(-2)2= -36, i.e., 16- 52= -36.

Ex. 2. Solve the equation x6-35x3=— 216.
Assuming x3=y, the above becomes

whence y—^ or 8.

Hence x=Vy=3 or 2.

This equation has four other roots which can not be de
termined by this process.

Ex. 3. Solve the equation cc-f 4A/# — 21.

Assuming Vx=y, we have

EQUATIONS OF THE SECOND DEGKEE. 189

whence y=3 or — 7.

Therefore a =9 or 49.

Although the square toot of 9 is generally ambiguous, and
may be either +3 or —8, still, in verifying the preceding val
ues, T/X can not be taken equal to —3, because 9 was obtained
by multiplying +3 by itself. For a like reason, ^/x can not
be taken equal to + 7. A similar remark is applicable to sev
eral of the following examples.

263. The same method of solution may often be extended
to equations in which any algebraic expression occurs with two
exponents, one of which is double the other.

Ex.4. Solve the equation (x2+ x)z-26(x2+x)=-I20.
Assuming x*-\-x=y, this equation becomes

2/2-26y=-120;

whence 2/=20 or 6.

We have now the two equations,

the first of which gives x= — 5 or +4,
and the second gives x=— 3 or + 2.

Thus the equation has four roots,

-5, +4, -3, +2,

and any one of these four values will satisfy the given equa
tion.

Ex.5. Solve the equation Va+12+ tAc+12 = 6.

We find z-f- 12 = 16 or 81.

Hence x=4 or 69.

1 Ex.6. Solve the equation 2cc2+ V2a;2+l = ll.

This equation may be written

Hence 2x2 + l = 9 or 16;

therefore x =+2, -2, + V7|-, -

190 ALGEBRA.

264. Equations of the Fourth Degree. — An equation of the
fourth degree may often be reduced to an equation contain
ing the first and second powers of some compound quantity, "
with known coefficients, in the following manner* Transpose
all the terms to the first member; then extract the square
root to two terms, and see if the remainder (with or without^
the absolute term) is a multiple of the root already obtained.:

Ex.7. Solve the equation x4-12x3 + 44cc2-48x

We may proceed as follows:

x4 - 12z3 + 44x2 - 48x - 9009 = 0 ( x2 - fa

_
- Qx ) -I2x3 4- 44x2

8x2-48z-9009,
or 8(cc2-6x)-9009.

Hence the given equation may be expressed as follows:

Ans. a? =13 or — 7, or 3±3V— 10.

Ex. 8. Solve the equation x4-2x3+x=132.

Ans. x=4: or —3, or J±|V— 43.
Ex. 9. Solve the equation x4-f 4x2 = 12.

Ans.. x=±V% or ± V — 6.

Ex. 10. Solve the equation x6— 8^3=

Ans. x=3 or -v/19.

6. 3.

Ex.11. Solve the equation x5+x5-756.

Ans. cc=243 or — v/28*.
Ex.12. Solve the equation -Jcc6— ^x3~ — •£$.

Ans. a?=

Ex.13. Solve the equation 2xTr+3orT=2.

Ans. x=^ or —8.

Ex.14. Solve the equation %x—$Vx = 22£. ^

Ans. x =49 or — — .

y
Ex.15. Solve the equation vTO+^-v/10+^=2.

Ans. x = 6 or —9.

EQUATIONS OF THE SECOND DEGKEE. 191

Ex. 16. Solve the equation x6H-20x3-10=59.

Ans. x=V& or V~^&.
Ex. 17. Solve the equation 3x2n-2x«4- 3 = 11.

,/ Ans. x=V2 or V— £•

Ex.18. Solve the equation Vl+a;-x2-2(l-ha;-cc2) = -J.

or

Ex. 19. Solve the equation Va5-hVi=20.

4.7ZS. z=256 or 625.

Ex.20. Solve the equation x4—

Ans. x=B or — 1, or l±-/^5.
Ex. 21. Solve the equation cc2+5x+4=5v^2+5x+28.

4ws. x=4: or —9, or — f±-i V-51.
Ex. 22. Solve the equation x2 + 5=2Vx2— 2

Ans. x— 1.

Ex.23. Solve the equation (x+ Vx)*-(x+ Vx)2 = 20592.

Ans. x — ^ or Ifr
Ex. 24. Solve the equation x+ V25+cc=157.

Ans. ccml44 or 171.
Ex.25. Solve the equation Vx—l=x—l.

Ans. x=I or 2.

265. We have seen that every equation of the second de
gree has two roots ; that is, there are two quantities which ,
when substituted for x in the original equation, will render
the two members identical. In like manner, we shall find
that every equation of the third degree has three roots, an equa-
tion of the fourth degree has four roots, and, in general, an
equation of the rath degree has m roots.

Before determining the degree of an equation, it should be
freed from fractions, from negative exponents, and from the
radical signs which affect its unknown quantities. Several of
the preceding examples are thus found to furnish equations

192 ALGEBRA.

of the fourth degree, while others furnish equations of the
second degree.

The above method of solving the equation x2n+pxn-=q will
not always give us all of the roots, and we must have recourse
to different processes to discover the remaining roots. The
subject will be more fully treated in Chapter XXL

Problems producing Equations of the Second Degree.

Prob. 1. It is required to find two numbers such that their
difference shall be 8 and their product 240.
Let x=the least number.
Then will 0:4-8 = the greater.
And by the question, x(cc+8)=x'2-\-8x= 240.
Therefore x =12, the less number,

x +8 = 20, the greater.
Proof. 20 — 12 = 8, the first condition.

20x12 = 240, the second condition.

Prob. 2. The Keceiving Eeservoir at Yorkville is a rectan
gle, 60 rods longer than it is broad, and its area is 5500 square
rods. Eequired its length and breadth.

Prob. 3. What two numbers are those whose difference is

2a, and product If

Ans. a±Va2 + bj and — a± yet2 4-6.

Prob. 4. It is required to divide the number 60 into two
such parts that their product shall be 864.

Let x = one of the parts.

Then will 60 - x = the other part.

And by the question, #(60— a?) = 60#-o?2=864.

The parts are 36 and 24, Ans.

Prob. 5. In a parcel which contains 52 coins of silver and
copper, each silver coin is worth as many cents as there are
copper coins, and each copper coin is worth as many cents as
there are silver coins, and the whole are worth two dollars.
How many are there of each ?

Prob. 6. What two numbers are those whose sum is 2a and
Vroduct b? Ans. a+Va2 — b and a-Va*-b.

EQUATIONS OF THE SECOND DEGEEE. 193

Prob. 7. There is a number consisting of two digits whose
sum is 10, and the sum of their squares is 58. Kequired the
number.

Let x=the first digit.

Then will 10— x=the second digit.

And x2+(10-*)2 = 2*2-2

that is x2 — I0x= — 21

= 7 or 3.

Hence the number is 73 or 37.

The two values of x are the required digits whose sum Is
10. It will be observed that we put x to represent the first
digit, whereas we find it may equal the second as well as the
first. The reason is, that we have here imposed a condition
which does not enter into the equation. If x represent either
of the required digits, then 10— a? will represent the other, and
hence the values of x found by solving the equation should
give both digits. Beginners are very apt thus, in the state
ment of a problem, to impose conditions which do not appear
in the equation.

The preceding example, and all others of the same class,
may be solved without completing the square. Thus,

Let x represent the half difference of the two digits.

Then, according to the principle on page 89, 5+ x will rep
resent the greater of the two digits, and 5— # the less.

The square of 5+o? is 25 + 103?+ a?2,
" 5-o? " 25-10J7+ x2.

The sum is 50 +2o?2, which, according to

the problem, =58.

Hence 2x2= 8,

or oc2= 4,

and x =±2.

Therefore, 5+# =7, the greater digit,

and 5—o? =3, the less digit.

Prob. 8. Find two numbers such that the product of their
sum and difference may be 5, and the product of the sum of
their squares and the difference of their squares may be 65.

194 ALGEBRA.

Prob. 9. Find two numbers such that the product of their
sum and difference may be a, and the product of the sum of
their squares and the difference of their squares may be ma.

Am.

Prob. 10. A laborer dug two trenches, whose united length
was 26 yards, for 356 shillings, and the digging of each of
them cost as many shillings per yard as there were yards in
its length. What was the length of each ?

Ans. 10, or 16 yards.

Prob. 11. What two numbers are those whose sum is 2a,
and the sum of their squares is 2b?

Ans. a+Vb — a'z, and a—Vb — a2.

Prob. 12. A farmer bought a number of sheep for 80 dollars,
and if he had bought four more for the same money, he would
have paid one dollar less for each. How many did he buy ?

Let x represent the number of sheep.

80

Then will — be the price of each.
x

80
And — - would be the price of each if he had bought four

more for the same money.
But by the question we have
80 _ 80
x -SC4.4+1-

Solving this equation, we obtain #=16, Ans.

Prob. 13. A person bought a number of articles for a dol
lars. If he had bought 2b more for the same money, he would
have paid c dollars less for each. How many did he buy ?

Ans. -

Prob. 14. It is required to find three numbers such that the
product of the first and second may be 15, the product of the
first and third 21, and the sum of the squares of the second and
third 74. Ans. 3, 5, and 7.

EQUATIONS OF THE SECOND DEGREE. 195

Prob. 15. It is required to find three numbers such that the
product of the first and second may be a, the product of the
first and third b, and the sum of the squares of the second and

third c. , _ . _ , _

/a2 + b* / c 7 / c

Ans, v/__; a\/^^- &V££&5-

Prob. 16. The sum of two numbers is 16, and the sum of
their cubes 1072. What are the numbers? Ans. 7 and 9.

Prob. 17. The sum of two numbers is 2a, and the sum of
their cubes is 2b. What are the numbers ?

Ans. a -f-

Prob. 18. Two magnets, whose powers of attraction are as
4 to 9, are placed at a distance of 20 inches from each other.
It as required to find, on the line which joins their centres, the
point where a needle would be equally attracted by both, ad
mitting, that the intensity of magnetic attraction varies inverse
ly as the square of the distance.

A \ & inches from the weakest magnet,

( or —40 inches from the weakest magnet.

Prob. 19. Two magnets, whose powers are as m to ??, are
placed at a distance of a feet from each other. It is required
to find, on the line which joins their centres, the point which
is equally attracted by both.

The distance from the magnet m is '

Ans.

The distance from the magnet n is

—.

Vm ± V n

Prob. 20. A set out from C toward D, and traveled 6 miles
an hour. After he had gone 45 miles, B set out from D to
ward C, and went every hour -^V of the entire distance ; and
after he had traveled as many hours as he went miles in one
hour, he met A. Required the distance between the places C
and D. Ans. Either 100 miles, or 180 miles.

196 ALGEBRA.

Prob. 21. A set oat from C toward D, and traveled a miles
per hour. After he had gone b miles, B set out from D toward
C, and went every hour |th of the entire distance ; and after
he had traveled as many hours as he went miles in one hour,
he met A. Kequired the distance between the places C and D.

Prob. 22. By selling my horse for 24 dollars, I lose as much
per cent, as the horse cost me. What was the first cost of the
horse? Ans. 40 or 60 dollars.

Prob. 23. A fruit-dealer receives an order to buy 18 melons
provided they can be bought at 18 cents a piece ; but if they
should be dearer or cheaper than 18 cents, he is to buy as
many less or more than 18 as each costs more or less than 18
cents. He paid in all $3.15. How many melons did he buy?

Ans. Either 15 or 21.

Prob. 24. A line of given length (a) is bisected and pro
duced ; find the length of the produced part, so that the rect
angle contained by half the line, and the line made up of the
half and the produced part, may be equal to the square on the
produced part.

Equations of the Second Degree containing Two Unknown
Quantities.

266. An equation containing two unknown quantities is
said to be of the second degree when the highest sum of the expo
nents of the unknown quantities in any term is two. Thus

a? + ^ = 13, (1.)

and x+xy+y = ll, (2.)

are equations of the second degree.

267. The solution of two equations of the second degree
containing two unknown quantities generally involves the so
lution of an equation of the fourth degree containing one un-

EQUATIONS OF THE SECOND DEGREE. 197

known quantity. Thus, from equation (2), we find

11-aj

y=~z+r-

Substituting this value for y in equation (1) and reducing,
we have

an equation which can not be solved by the preceding methods.
Yet there are particular cases in which simultaneous equations
of a degree higher than the first may be solved by the rules for
quadratic equations. The following are the principal cases of
this kind :

268. 1st. When one of the equations is of the first degree and the
other of the second. — We find an expression for the value of one
of the unknown quantities in the former equation, and substi
tute this value for its equal in the other equation.

(x2 + 3xy-y2 = 23{ .
Ex. 1. Given j • 9 — 7 I x y'

From the second equation we find

Substituting this value for x in the first equation, we have
49-2%+42/2 + 21y-6/-2/2=:23?

which may be solved in the usual manner.

(03=3 or Q.

Ans. 3 .

=2 or --

,, 0 n. - <_ . , ,

Ex. 2. Given < 01 f to ^nd x and y-

[ ZX — o?/=:l

( x = 5 or —
Ans. J

= or —

(x + y_ j

Ex. 3. Given •] ~S~ ~~Xy V to find x and y.
(9?/-9x=18)

9?/

Ans.

( x=2 or —4-.

1

(y=4 or |.

For equations of this class there are in general two sets of
values of x and y.

198 ALGEBRA.

269. 2d. When both of the equations are. of the second degree^
and homogeneous. — Substitute for one of the unknown quanti
ties the product of the other by a third unknown quantity.

i/v»2 l npji — 1 0 /
, \ to find x and y.
xy—2y2 = L )

If we assume x=vy, we shall have

v2y2+vy2=~L2, whence y1 — -

12
Therefore -^—

— 0.

v2-hv v— 2

From which we obtain v=S or 3.

Substituting either of these values in one of the preceding
expressions for ?/2, we shall obtain the values of y ; and since
x=vy, we may easily obtain the values of x.

Ans. _l

y = ± 1 or ± — r=.

y 6

Ex. 5. Given

—=

Assuming x—vy^ we find v= j or — .

«^^-«. x « ,

Ex. 6. Given | 5J2+^1:41 [ to find x and y.

A 13

Assuming x=vy, we find v = « or

o

— ±1 or db

=:db8 Or ±

EQUATIONS OF THE SECOND DEGREE. 199

For equations of this class there are in general four sets of
values of x and y. It should be borne in mind that to any
one of the four values of x there corresponds only one of the
four values of y. Thus, when x in the 6th example is -f 1, y
must be -f 3, and can not be one of the other three values
given above.

270. 3d. When the unknown quantities enter each equation sym
metrically. — Substitute for the unknown quantities the sum and
difference of two other quantities, or the sum and product of
two other quantities.

/ r^l nil \

_j_^.-18
Ex. 7. Given I y x V to find x and y.

(
Let us assume

Then x-\-y=2z-l2 or z = 6.

That is, x—Q-\-v and y—Q—v.

But from the first equation we find

Substituting the preceding values of x and y in this equation,
and reducing, we have

whence v=±2.

Therefore x=4 or 8, and y— 8 or 4.

T? , n- ,

Ex. 8. Given ] y 0 > to find x and y.
( x+y=8 )

lx=3 or 5.
Ans. \ K

\y=5 or 3.

Ex. 9. Given | f t^f BQQA \ ^ find x and y.
22=330)

( x=5 or 6.
Ans. 1 a

( y—6 or 5.

271. 4th. When the same algebraic expression is involved to
different powers, it is sometimes best to regard this expression
as the unknown quantity.

200 ALGEBRA.

Ex.10. Given j -2+2-/+2/*+2*=120-22n to find , and y,

( xy — y = 8 j

The first equation may be written

Eegarding x+y as a single quantity, we find its value to be
either 10 or —12.

Proceeding now as in Art. 268, we find
x=6 or 9, or — 9 + ^5;
y—4: or 1, or — 3±V6.

Ex. 11. Given - to find x and

( x+z/=6 j

Eegarding xy as the unknown quantity, its value from the
first equation is found to be either

8 or -12.

(x=2 or 4, or 3± V%1.
Ans. \ , _

( y=4 or 2, or 3 + V21.
<x* 4x_85)

Ex. 12. Given •] y2 y ~ 9 [to find x and y.
( x-y=2 )

Eegarding - as the unknown quantity, we find its value to
be either 5 17

"

For several of these examples there are other roots, some
of which can not be obtained by the processes heretofore ex
plained. The roots of two simultaneous equations are some
times infinite, as in the equations a?2— xy — 8, and a?2—?/2 = 12,
since two quantities that are infinitely great may differ \)j a
finite quantity.

Solve the following groups of simultaneous equations :

Ex.13.

-
Ans. < *

EQUATIONS OF THE SECOND DEGREE.

Ex.15. =84> ; An ,

=U)

201

=8or2.
y=2or8.

„
Ex'

o:=6or30.

V* oo (*5-2/5 = 3093|
Ex'22' < s-t^S

Ex 23

j

Ex.24. ]

(

. Put

0^—5 or —2.
y=2 or —5.

J.ns. )

2.

PROBLEMS.

1. Divide the number 100 into two such parts that the sum
of their square roots may be 14. Ans. 64 and 36.

2. Divide the number a into two such parts that the sum
of their square roots may be b. ,

A ^ I _ /o T%

3. The sum of two numbers is 8, and the sum of their fourth

powers is 706. What are the numbers ? Ans. 3 and 5.

12

202 ALGEBRA.

4. The sum of two numbers is 2a, and the sum of their
fourth powers is 26. What are the numbers?

Ans. a± V—Saz±V8a*+b.

5. The sum of two numbers is 6, and the sum of their fifth
powers is 1056. What are the numbers? Ans. 2 and 4.

6. The sum of two numbers is 2a, and the sum of their fifth
powers is b. What are the numbers ?

7. What two numbers are those whose product is 120 ; and
if the greater be increased by 8 and the less by 5, the product
of the two numbers thus obtained shall be 300 ?

Ans. 12 and 10, or 16 and 7.5.

8. What two numbers are those whose product is a; and
if the greater be increased by b and the less by c, the product
of the two numbers thus obtained shall be d?

Ans. Tr±V^i --- ,

, d—a — bc

where m= - — „
c

9. Find two numbers such that their sum, their product,
and the difference of their squares may be all equal to one
another.

that is, 2.618, and 1.618, nearly.

10. Divide the number 100 into two such parts that their
product may be equal to the difference of their squares.

Ans. 38.197, and 61.803.

11. Divide the number a into two such parts that theii
product may be equal to the difference of their squares.

3a±aV5 , — a + aVE
Ans. - - and - - •

EQUATIONS OF THE SECOND DEGREE. 203

12. The sum of two numbers is a, and the sum of their re
ciprocals is b. Kequired the numbers.

General Properties of Equations of the Second Degree.

272. Every equation of the second degree containing but one
unknown quantity has two roots, and only two.

We have seen, Art. 250, that every equation of the second
degree containing but one unknown quantity can be reduced
to the form x2+px = q. We have also found, Art. 257, that
this equation has two roots, viz.,

This equation can not have more than two roots ; for, if pos
sible, suppose it to have three roots, and represent these roots
by a/, x", and x'". Then, since a root of an equation is such
a number as, substituted for the unknown quantity, will satisfy
the equation, we must have

q, (1.)

q, (2.)

= q. (3.)

Subtracting (2) from (1), we have

Dividing by x'—x", we have

(x'+x")+^=0- (4.)

In the same manner, we find

(a/+a/")+*=0. (5.)

Subtracting (5) from (4), we have
o;"-a;'"=:0;

that is, the third supposed root is identical with the second ;
hence there can not be three different roots to a quadratic
equation.

273. The algebraic sum of the two roots is equal to the coeffi
cient of the second term of the equation taken with the contrary sign.

204 ALGEBRA.

If we add together the two values of x in the general equa
tion, the radical parts having opposite signs disappear, and we
obtain p p

_ ±_ _ -*• - . rf\

2 2~ p'

Thus, in the equation x2— I0x=— 16, the two roots are 8
and 2, whose sum is + 10, the coefficient of x taken with the
contrary sign.

If the two roots are equal numerically, but have opposite
signs, their sum is zero, and the second term of the equation
vanishes. Thus the two roots of the equation xz—I6 are +4
and —4, whose sum is zero. This equation may be written

274. The product of the two roots is equal to the second member
of the equation taken with the contrary sign.

If we multiply together the two values of x (observing that
the product of the sum and difference of two quantities is
equal to the difference of their squares), we obtain

Thus, in the equation x"2— 10x= — 16, the product of the
two roots 8 and 2 is +16, which is equal to the second mem
ber of the equation taken with the contrary sign.

275. The last two principles enable us to form an equation
whose roots shall be any given quantities.

Ex. 1. Find the equation whose roots are 3 and 5.

According to Art. 273, the coefficient of the second term of
the equation must be — 8 ; and, according to Art. 274, the sec
ond member of the equation must be —15. Hence the equa
tion is a;2— 8x=— 15.

Ex. 2. Find the equation whose roots are —4 and — 7.

Ex. 3. Find the equation whose roots are 5 and —9.

Ex.4. Find the equation whose roots are —6 and +11.

Ex. 5. Find the equation whose roots are 1 and —2.

Ex. 6. Find the equation whose roots are — -J and

Ex. 7. Find the equation whose roots are — -J- and

EQUATIONS OF THE SECOND DEGREE. 205

Ex. 8. Find the equation whose roots are 1 ± \/5.
Ex. 9. Find the equation whose roots are 1 ± V^5.

276. Every equation of the second degree whose roots are a and
b,may be reduced to the form (x—a)(x—b) — 0.
Take the general equation

and write it x2+px— q — 0.

Then, by Art. 273, p=-(a + b);
and by Art. 274, q= —ab.

Hence, by substitution,

or, resolving into factors,

(x — a)(x-b) = Q.

Thus the equation x2— 10x=— 16, whose roots are 8 and 2,
may be resolved into the factors x — 8 = 0 and x— 2 = 0.

It is also obvious that if a is a root of an equation of the
second degree, the equation must be divisible by x— a. Thus
the preceding equation is divisible by x— 8, giving the quo
tient x—2.

Ex. 1. The roots of the equation x2-f 6x4-8 = 0 are —2 and
— 4. Eesolve the first member into its factors.

Ex. 2. The roots of the equation x24- 6x— 27 = 0 are +3
and —9. Resolve the first member into its factors.

Ex. 3. The roots of the equation x2 — 2x— 24 = 0 are +6
and —4. Resolve the first member into its factors.

Ex.4. Resolve the equation x24-73x+7SO = 0 into simple
factors. Ans. (x + 60) (x + 13) = 0.

Ex.5. Resolve the equation x2— 88x4-1612 = 0 into simple
factors. Ans. (x-62)(x-26) = 0.

Ex.6. Resolve the equation 2x2+x— 6 = 0 into simple fac
tors. Ans. 2(x+2)0-!) = 0.

Ex.7. Resolve the equation 3x2— Wx— 25 = 0 into simple
Actors. A ns. 3 (x — 5) (x + f ) = 0.

206 ALGEBRA.

Discussion of the General Equation of the Second Degree.

277. In the general equation of the second degree x2-{-px—q.
the coefficient of x, as well as the absolute term, may be either
positive or negative. We may therefore have the four fol
lowing forms:

First form, x2 -\-px- q.

Second form, x2—px — q.

Third form, x2 -\-px- — q.

Fourth form, x2 —px ——q.

From these equations we obtain

4

We will now consider what conditions will render these
roots positive or negative, equal or unequal, real or imaginary.

278. Positive and negative roots.

Since "A"^ *s &reater tnan ~n

y 4- +2 must be greater than If-
For the same reason, y ~ — q must be less than ^.

Therefore, in the first and second forms, the sign of the roots
will correspond to the sign of the radicals; but in the third
and fourth forms the sign of the roots will correspond to the
sign of the rational parts. Hence, in the first form, one root is
positive and the other negative, and the negative root is numerical
ly the greatest; as in the equation x2+x=6, whose roots are
-4-2 and —3.

EQUATIONS OF THE SECOND DEGEEE. 207

In the second form one root is positive and the other negative,
and the positive root is numerically the greatest, as in the equa
tion x2— x — Q, whose roots are —2 and +3.

In the third form both roots are negative, as in the equation
xz-\-6x=— 6, whose roots are —2 and —3.

In the fourth form both roots are positive, as in the equation.
<c2— 5x=— 6, whose roots are +2 and +3.

279. Equal and unequal roots.

In the first and second forms the two roots are always unequal;
but in the third and fourth forms, when q is numerically equal

to — , the radical part of both values of x becomes zero, and
4

the two roots are then said to be equal In this case

7) T)

the third form gives x= — ~±0= — £
and the fourth form gives

Thus, in the equation x2 + 6x= — 9, the two roots are —3
and —3. We say that in this case the equation has two
roots, because it is the product of the two factors 03+8 = 0
and x + 3 = 0.

So, also, in the equation x2— 6x= — 9, the two roots are -f 3
and +3.

280. Real and imaginary roots.

rpT.
Since ^-, being a square, is positive for all real values of p,

p2

it follows that the expression ±- -{-q can only be rendered neg
ative by the sign of q ; that is, the quantity under the radical
sign can only be negative when q is negative and numerically

TJ2

greater than -—. Hence, in the first and second forms, both roots
are always real; but in the third and fourth forms both roots arc
imaginary when q is numerically greater than — .

208 ALGEBRA.

Thus, in the equation £2+4x= — 6, the two roots are
and in the equation x2— 4x= — 6, the two roots are

It will be observed that when one of the roots is imaginary,
the other is imaginary also.

281. Imaginary roots indicate impossible conditions in the pro
posed question which furnished the equation.

The demonstration of this principle depends upon the fol
lowing proposition : the greatest product which can be obtained
by dividing a number into two parts and multiplying them togeth'
er is the square of half that number.

Let p represent the given number, and d the difference of
the parts. ,

Then, from page 89, |?+- = the greater part,

T P d , ,

and ^-—- = the less part,

and ^ — — = the product of the parts.

Now, since p is a given quantity, it is plain that the prod
uct will be the greatest possible when c?=0; that is, the great-

JD

est product is the square of ^-, half the given number.

For example, let 10 be the number to be divided.
We have 10 = 1 + 9; and 9x1= 9.

10 = 2 + 8; and 8x2 = 16.
10 = 3 + 7; and 7x3 = 21.
10=4 + 6; and 6x4 = 24.
10=5 + 5; and 5x5 = 25.

Thus we see that the smaller the difference of the two parts,
the greater is their product ; and this product is greatest when
the two parts are equal.

Now, in the equation x2—px = — j, p is the sum of the two
roots, and q is their product. Therefore q can never be great
er than •£-.

EQUATIONS OF THE SECOND DEGREE. 209

If, then, any problem furnishes an equation in which q is

p2
negative, and numerically greater than ^j-, we infer that the

conditions of the question are incompatible with each other.

Suppose it is required to divide 6 into two parts such that
their product shall be 10.

Let x represent one of the parts, and 6 — x the other part

Then, by the conditions,

x(Q-x) = W;

whence ' x2— 6x=— 10,

and x = 3±V^l.

The imaginary value of x indicates that it is impossible to
find two numbers whose sum is 6 and product 10. From the
preceding proposition, it appears that 9 is the greatest product
which can be obtained by dividing 6 into two parts and multi
plying them together.

Discussion of Particular Problems.

282. In discussing particular problems which involve equa
tions of the second degree, we meet with all the different cases
which are presented by equations of the first degree, and some
peculiarities besides. All the different cases enumerated in
Chapter X. are presented by Prob. 19, page 195, when we
make different suppositions upon the values of a, m, and n;
but we need not dwell upon them.

The peculiarities exhibited by equations of the second de
gree are double values of x, and imaginary values.

283. Double Values of the Unknown Quantity. — We have seen
that every equation of the second degree has two roots. Some
times both of these values are applicable to the problem which
furnishes the equation. Thus, in Prob. 20, page 195, we obtain
either 100 or 180 miles for the distance between the places C

and D.

C E D

Let E represent the position of A when B sets out on his

210 ALGEBRA.

journey. Then, if we suppose CD equals 100 miles, ED will
equal 55 miles, of which A will travel 30 miles (being 6 miles
an hour for 5 hours), and B will travel 25 miles (being 5 miles
an hour for 5 hours).

If we suppose CD equals 180 miles, ED will equal 135 miles,
of which A will travel 54 miles (being 6 miles an hour for 9
hours), and B will travel 81 miles (being 9 miles an hour for
9 hours).

This problem, therefore, admits of two positive answers, both
equally applicable to the question. Problems 22 and 23, page
196, are of the same kind.

In Problem 18, page 195, one of the values of x is positive
and the other negative.

C' A C B

Let the weaker magnet be placed at A, and the stronger at
B ; then C will represent the position of a needle equally at
tracted by both magnets. According to the first value, the
distance AC =8 inches, and CB = 12 inches. Now, at the dis
tance of 8 inches, the attraction of the weaker magnet will be

4 *
represented by —2 ; and at the distance of 12 inches, the at-

9

traction of the other magnet will be represented by ir^, and

these two powers are equal ; for

1-JL

S2~122'

But there is another point, C', which equally satisfies the
conditions of the question ; and this point is 40 inches to the
left of A, and therefore 60 inches to the left of B ; for

A JL

402~602'

284. Imaginary Values of the Unknown Quantity. — We have
seen that an imaginary root indicates impossible conditions in
the proposed question which furnished the equation. In sev
eral of the preceding problems the values of x become imag
inary in particular cases.

EQUATIONS OF THE SECOND DEGREE. 211

When will the values of x in Prob. 6, page 192, be imag
inary ? Ans. When b > a2.

What is the impossibility involved in this supposition ?

Ans. It is impossible that the product of two numbers can
be greater than the square of half their sum.

When will the values of x in Prob. 11, page 194, be imag
inary ? Ans. When a2 > b ; or (2a)2 > 46.

What is the impossibility involved in this supposition?

Ans. The square of the sum of two numbers can not be
greater than twice the sum of their squares.

When will the values of x in Prob. 17, page 195, be imag
inary ? Ans. When a3 > b ; or (2a)3 > 85.

What is the impossibility of this supposition ?

Ans. The cube of the sum of two numbers can not be great
er than four times the sum of their cubes.

When will the values of x in Prob. 4, page 180, be imag
inary, and what is the impossibility of this supposition?

285. Geometrical Construction of Equations of the Second De
gree. — The roots of an equation of the second degree may be
represented by a simple geometrical figure. This may be done
for each of the four forms :

First form. — The first form gives for x the two values
TJ f~r? v

Xz=~f + V T+?> a *=-f--Y.f
Draw the line AB, and make it equal to -^fq. From B draw
BO perpendicular to AB, and make it equal

to ~r. Join A and C ; then AC will repre-

j-i . -

sent the value of ¥-+. For

B AB2+BC2 (Geom., Prop. 11, Bk. IY.).

With C as a centre, and CB as a radius, describe a circle
cutting AC in D, and AC produced in E. For the first value
of x the radical is positive, and is set off from A toward C ;

then — |J is set off from C to D ; and AD. estimated from A
2i

to D, represents the first value of x.

212 ALGEBRA.

For the second value of x we begin at E, and set off EC

/y-\

equal to — |j ; we then set off the minus radical from C to A j

and EA, estimated from E to A, represents the second value
of x.

Second form. — The second form gives for x the two values

The first value of x is represented by AE estimated from A
to E. The second value is -f DC — CA, the latter being esti
mated from C to A. Hence the second value is represented
by DA estimated from D to A.

TJiirdform. — The third form gives for x the two values

*T-g, and »=~|.

Draw an indefinite line FA, and from any point, as A, set
0 off a distance AB=— J?. We set off its

value to the left, because £ is negative.
JF D 'E A At B draw BC perpendicular to FA, and
make it equal to Vq. With C as a centre, and a radius equal

to -^, describe an arc of a circle cutting FA in D and E. Now
A ,

the value of y ^- — q will be BD or BE. The first value of

x will be represented by — AB + BE, which is equal to — AE.
The second will be represented by — AB — BD, which is equal
to — AD; so that both of the roots are negative, and are esti
mated in the same direction, from A toward the left.

Fourth form. — The fourth form gives for x the two values

T~?' and x=~y 4 ~q-

Construct the radical part of the value of x as in the last

79

~

a

79

case. Then, since ~ is positive, we set off its value AB from

EQUATIONS OF THE SECOND DEGREE. 213

A toward the right. To AB we add BD,
which gives AD for the first value of x;
and from AB we subtract BE, which leaves
AE for the second value of x. Both val
ues are positive, and are estimated in the
same direction, from A toward the right.

Equal Roots.— If the radius CE be taken equal to CB, that

is, if Vq is equal to -f, the arc described with the centre C will

A

be tangent to AF, the two points D and E will unite, and the
two values of x become equal to each other. In this case the
radical part of the value of x becomes zero.

Imaginary Boots. — If the radius of the circle described with
the centre C be taken less than CB, it will not

/ meet the line AF. In this case q is numerical-

2

ly greater than ±L, and the radical part of the

value of x is imaginary.

214 ALGEBRA.

CHAPTER XV.

RATIO AND PROPORTION.

286. Ratio is the relation which one quantity bears to an
other with respect to magnitude. Ratio is denoted by two
points like the colon (:) placed between the quantities com
pared. Thus the ratio of a to b is written a : b.

The first quantity is called the antecedent of the ratio, and
the second the consequent. The two quantities compared are
called the terms of the ratio, and together they form a couplet.
The quantities compared may be polynomials; nevertheless,
each quantity is called one term of the ratio.

287. A ratio is measured by the fraction whose numerator
is the antecedent and whose denominator is the consequent of

the ratio. Thus the ratio of a to b is measured by j.

288. A compound ratio is the ratio arising from multiplying
together the corresponding terms of two or more simple ratios.
Thus the ratio of a to b compounded with the ratio of c to d
becomes ac to bd.

The ratio compounded of the ratios 3 to 5 and 7 to 9 is
21 to 45.

289. The duplicate ratio of two quantities is the ratio of
their squares. Thus the duplicate ratio of 2 to 3 is 4 to 9 ;
the duplicate ratio of a to b is a2 to b2.

290. The triplicate ratio of two quantities is the ratio of their
cubes. Thus the triplicate ratio of a to b is a3 to b3.

291. If the terms of a ratio are both multiplied or both divided

RATIO AND PROPORTION. 215

by the same quantity, the value of the ratio remains unchanged.
The ratio of a to b is represented by the fraction j, and the

value of a fraction is not changed if we multiply or divide
both numerator and denominator by the same quantity. Thus

a ma

b=mb>

7 7 a b

or a : b=.ma : mb— - : -.

.;./:. n n

PROPORTION.

292. Proportion is an equality of ratios. Thus, if a, b, c, d
are four quantities such that a when divided by b gives the
same quotient as c when divided by d, these four quantities
are called proportionals. This proportion may be written thus,

a : b : : c : d,
or a:b=.c:d,

°r f=>

b d
In either case the proportion is read a is to b as c is to d.

293. The terms of a proportion are the four quantities which
are compared. The first and fourth terms are called the ex
tremes, the second and third the means. The first term is called
the first antecedent, the second term the first consequent, the third
term the second antecedent, and the fourth term the second con
sequent.

294. When the first of a series of quantities has the same
ratio to the second that the second has to the third, or the
third to the fourth, and so on, these quantities are said to be in
continued proportion, and any one of them is a mean propor
tional between the two adjacent ones. Thus, if

a : b : : b : c : : c : d : : d : e,

then a, b, c, d, and e are in continued proportion, and b is a
mean proportional between a and c, c is a mean proportional
between b and d, and so on.

216 ALGEBRA.

295. Alternation is when antecedent is compared with ante
cedent and consequent with consequent. Thus, if

a : b : : c : d,
then, by alternation, a : c : : b : d. See Art. 301.

296. Inversion is when antecedents are made consequents,
and consequents are made antecedents. Thus, if

a:b::c:d)
then, inversely, b: a:: d: c. See Art. 302.

297. Composition is when the sum of antecedent and conse
quent is compared with either antecedent or consequent.
Thus, if a b::c:d,

then, by composition, a-+ b : a : : c+d : c,

and a-\-b:b:: c+d : d. See Art. 304.

298. Division is when the difference of antecedent and con
sequent is compared with either antecedent or consequent.
Thus, if a:b::c:d,

then, by division, a- -b : a : : c— d : c,

and a— b :b:: c—d : d. See Art. 305.

299. If four quantities are in proportion, the product of the ex
tremes is equal to the product of the means.

Let a:b :: c: d.

Then | = J, Art. 292.

Multiplying each of these equals by bd, we have ad—lc.

300. Conversely, if the product of tiuo quantities is equal to the
product of two other quantities, the first two may be made the ex
tremes, and the other two the means of a proportion.

Let ad—lc.

Dividing each of these equals by bd, we have

a_ £

b~d?
or a:b::c:d, Art. 292.

RATIO AND PROPORTION. 217

EXAMPLES.

1. Given the first three terms of a proportion, 24, 15, and
40, to find the fourth term.

2. Given the first three terms of a proportion, 3a&3, 4a262,
and 9a3b, to find the fourth term.

3. Given the last three terms of a proportion, 4a365, 3a363,
and 2a5£, to find the first term.

4. Given the first, second, and fourth terms of a proportion,
5/, 7x2#3, and 2lxGy, to find the third term.

5. Given the first, third, and fourth terms of a proportion,
a +5, a2— &2, and (a — &)2, to find the second term.

Which of the following proportions are correct, and which
are incorrect?

6. 3a+46: 9a + 86:: a-2b: Ba— 4b.

7. 9a2-4£2 : I5a2-25ab+8b2 : : 15a2+25a6+852: 25a2-1662.
8.

9.

301. If four quantities are in proportion, they will be in pro
portion when taken alternately.

Let a : b :: c: d;

then

Multiplying by 6,

Dividing bye, ~c=\>

or a:c::b:d.

302. If four quantities are in proportion, they ivill be in pro
portion wJien taken inversely.

Let a:b::c:d;

then £ = £

6 a

Divide unity by each of these equal quantities, and we have

b_d
a~ c'

or b : a : : d : c.

K

218 ALGEBRA.

303. fiatios tiial are equal to the same ratio are equal to each
other.

If albumin, (1.)

and c : d : : m : «, (2.)

then a : b : : c : d

From proportion (1), y = — .

From proportion (2)^ -^ = — .

TT a c

Hence 1 =d>

or aibiicid.

304. //" /cwr quantities are proportional, they will be propor
tional by composition.

Let aibiicid;

, a c

then j- = -3.

Add unity to each of these equals, and we have

. , . a+b c+d
that is, = ,

or a + b i bn c-frf: d.

305. If four quantities are proportional, they will be propor
tional by division.

L : a ib nc id;

then | = |.

Subtract unity from each of these equals, and we have

a 1 — c 1
b~~ ~d~

, . a — ^ c— c?

that is, — T— = — -i— ,

or a — b-b:ic—did.

RATIO AND PROPORTION. 219

306. If four quantities are proportional, the sum of the first
and second is to vieir difference^as the sum of the third and fourth
is to their difference.

L -.-: a:b::c:d.

By composition. Art. 304^

a+b:b::c+d:d.
By alternation, Art 301,

a+b:c+d::b:d.
Also by division, Art. 305,

a — b: b:: c—d'.d;

by alternation, a— b : c—d ::b:d.

By equality of ratios, Art 303,

a+bi c+d'.i a—b: c — d,
or a-r-b: a—b:: c+d: c— d.

307. If four quantities are in proportion, any equimultiples of
the first couplet will be proportional to any equimultiples of the
second coupkL

Let a:b::c:d;

*** f-3- :'"..•

Multiply both terms of the first fraction by w», and both
terms of the second fraction by w, and we have

: _ :
. :
or maimb-.-ncind,

303. If four quantities are in proportion, any equimultiples of
the antecedents wiO. be proportional to any equimultiples of the con-

L I a:b::c:d;

then a _ c

b~d'

Multiply each of these equals by m, and we have
ma me

220 ALGEBRA.

Divide each of these equals by w,
ma me
nb ~~ nd>
or ma :nb::mc: rid.

309. If any number of quantities are proportional, any one an
tecedent is to its consequent as the sum of all the antecedents is to
the sum of all the consequents.

Let a : b : : c : d : : e : /;

then, since a : b : : c : d,

ad = bc; (1.)

and, since a : b : : e : /,

«/=**/ (2-)

also ab = ba. (3.)

Adding (1), (2), and (3),

ab+ad+af = ba+bc+be;
that is, a (b + d+f) = b(a + c + e).

Hence, Art. 300, a : b :: a+c + e : b + d+f

310, If there are two sets of proportional quantities, the prod
ucts of the corresponding terms will be proportional.

Let a : b : : c : d,

and e:f::g:h;

then ae : bf : : eg : dh.

-

Multiplying together these equal quantities, we have

ae_cg_
bf ~ dh1
or ae :bf:: eg : dh.

311. If four quantities are in proportion, like powers or roots
of these quantities will also be in proportion.

RATIO AND PKOPOKTION.

221

a:b::c:d;

a c

T — -j-
b d

Eaising each of these equals to the nth power, we obtain

Let
then

that is, an:bn :: cn: dn.

In the same manner, we find

i i. i i
an : bn :: cn : dn.

312. If three quantities are in continued proportion, the product
of the extremes is equal to the square of the mean.

If a:b::b:c,

then, by Art. 299, ac=bb=b2.

313. If three quantities are in continued proportion, the first is
to the third in the duplicate ratio of the first to the second.

Let a:b::b:c;

a _ b
b~ c
Multiply each of these equals by y, and we have

a a a b
bXb=bXc;

a_a?

c 62'
a : c : : a2 : b2.

then

that is,
or

314. If four quantities are in continued proportion, the first is
to the fourth in the triplicate ratio of the first to the second.
Let a:b ::b : c:: c: d ;

a b

be

a _ c

b~d'

a _ a

b=b'

then

and

also

(2.)
(3.)

222 ALGEBRA.

Multiplying together (1), (2), and (3), we have
a3 dbc a

nence a : d : : a3 : b3.

VARIATION.

315. Proportions are often expressed in an abridged form.
Thus, if A and B represent two sums of money put out for
one year at the same rate of interest, then

A : B : : interest of A : interest of B.

This is briefly expressed by saying that the interest varies as
the principal. A peculiar character ( a ) is used to denote this
relation. 1 hus interest oc principal

denotes that the interest varies as the principal.

316. One quantity is said to vary directly as another when
the two quantities increase or decrease together in the same
ratio. Thus, in the above example, A varies directly as the
interest of A. In such a case, either quantity is equal to the
other multiplied by some constant number.

Thus, if the interest varies as the principal, then the interest
equals the product of the principal by some constant number,
which is the rate of interest.

If A a B, then A = raB.

If the space (S) described by a falling body varies as the
square of the time (T), then

S = mT2,
where m represents a constant multiplier.

317. One quantity may vary directly as the product of sev
eral others.

Thus, if a body moves with uniform velocity, the space de
scribed is measured by the product of the time by the velocity.
If we put S to represent the space described, T the time of
motion, and V the uniform velocity, then we shall have

S a T x V.

RATIO AND PROPORTION. 223

Also the area of a rectangular figure varies as the product
of its length and breadth.

The weight of a stick of timber varies as the product of its
length x its breadth x its depth x its density.

318. One quantity is said to vary inversely as another when
the first varies as the reciprocal of the second. Thus, if the
area of a triangle be invariable, the altitude varies inversely as
the base.

If the product of two quantities is constant, then one varies
inversely as the other.

In uniform motion the space described is measured by the
product of the time by the velocity ; that is,

SocTxV;

S
whence T a =.

If the space be supposed to remain constant, then

Toe — •

y

that is, the time required to travel a given distance varies in
versely as the velocity.

Conversely, if one quantity varies inversely as another, the
product of the two quantities is constant

Thus, if T a i

then the product of T by Y is equal to a constant quantity.

319. One quantity is said to vary directly as a second, and
inversely as a third, when it varies as the product of the second
by the reciprocal of the third. Thus, according to the New
tonian law of gravitation, the attraction (G) of any heavenly
body varies directly as the quantity of matter (Q), and inverse
ly as the square of the distance (D).

That is, Gr a .

224 ALGEBRA.

EXAMPLES.

~. | x+y:x:: 5:3) ,

1. Given < „ f to find x and y.

( xy=6 }

Since x+y: x :: 5 : 3,

by division, Art. 305, y : x : : 2 : 3.

2x
Hence 3z/=2x, and y=-~--

Substituting this value of y in the second equation, we ob
tain %X2

T-:

Therefore x=±S,

and y = ±2.

~. ( x + y : x— y :: 3 : 1 } « , ,

2. Given i *, » k/i f to find x and y.

( ^_ ?/3 — 56 )

From the first equation, Art. 306,

2x : 2y : : 4 : 2 ;
whence x = 2y.

Substituting this value of x in the second equation, we find
y=2, and x=4.

8. .Given \ ^+^ : <*-g : : 64 : X j to find c, and y.

I a??/=63 j

By JLr^. 311, ic+y : x— y : : 8 : 1.

By Art. 306, 2x : 2y : : 9 : 7.

9y
Hence x = -^-.

Substituting this value of x in the second equation, we find
y=±7, and x=±9.

(x3-y3:(x-y)3::Ql:l[ 4
4. Given a and

From the first equation, by division, Art. 305,

Sxy(x— y) : (x—yf : : 60 : 1.
Hence 960 : (x-yf : : 60 : 1,

or 16 : (x-y? ::!:!.

Therefore

Hence
and

By addition,

Hence

Therefore
and

RATIO AND PROPORTION.
X — ?/=±4.

5. Given

>. Given -j

( V^-

4xy=l2SO.
+ 2xy +?/2=

x=+20 or ±16,
y=±16 or ±20.

x-\ry-

225

\ to find # and y.
=o )

J.W5.

=4 or 2.

=2 or 4.

— Va— x=V?/— x , ,,

"y J- to find x and y.

y—x+ Va—x : Va—x : : 5 : 2

Ans.

O

7. Given x+Vx : x—Vx :: 3Vx+6 :%Vx to find x.

J.W5. x=9 or 4.

8. What number is that to which if 1, 5, and 13 be severally
added, the first sum shall be to the second as the second to the
third? Ans. 8.

9. What number is that to which if a, 5, and c be severally
added, the first sum shall be to the second as the second to the

third? . l2 — ac

Ans. - .

a — 2b-}-c

10. What two numbers are as 2 to 3, to each of which if 4
be added, the sums will be as 5 to 7 ?

y 11. What two numbers are as m to w, to each of which if a
'be added, the sums will be as p to q ?

Ans. am(P—9) . an (P-V)
mq — np ' mq — np '

12. What two numbers are those whose difference, sum, and
product are as the numbers 2, 3, and 5 respectively ?

Ans. 2 and 10.

K 2

226 ALGEBRA.

13. What two numbers are those whose difference, sum, and
product are as the numbers m, n, and p ?

IP A %P
Ans. — ^— and — ±— .

n+m n — m

14. Find two numbers, the greater of which shall be to the
less as their sum to 42, and as their difference to 6.

Ans. 32 and 24.

15. Find two numbers, the greater of which shall be to the
less as their sum to a and their difference to b.

2

16. There are two numbers which are in the ratio of 3 to 2,
the difference of whose fourth powers is to the sum of their
cubes as 26 to 7. Kequired the numbers. Ans. 6 and 4.

17. What two numbers are in the ratio of m to w, the differ
ence of whose fourth powers is to the sum of their cubes as

P to qf o o *

* A mp rnr+n9 n np m34-wa

Ans. - - x — — , and — x — — , .
q m* — n* q m*~n*

18. Two circular metallic plates, each an inch thick, whose
diameters are 6 and 8 inches respectively, are melted and form
ed into a single circular plate 1 inch thick. Find its diameter,
admitting that the area of a circle varies as the square of its
diameter.

19. Find the radius of a sphere whose volume is equal to
the sum of the volumes of three spheres whose radii are 3, 4,
and 5 inches, admitting that the volume of a sphere varies as
the cube of its radius.

20. Find the radius of a sphere whose volume is equal to
the sum of the volumes of three spheres whose radii are r, r',
and r".

PROGRESSIONS. 227

CHAPTER XYI.

PROGRESSIONS.
ARITHMETICAL PROGRESSION.

320. An arithmetical progression is a series of quantities which
increase or decrease by a common difference. Thus the fol
lowing series are in arithmetical progression :

1,3,5,7,9,...

20, 17, 14, 11, 8, ...

a, a+c?, a-\-2d, a + 3c?, . . .

a, a— d, a—2d, a— 3c?, . . .

In the first example the common difference is -f2, and the
series forms an increasing arithmetical progression ; in the sec
ond example the common difference is —3, and the series forms
a decreasing arithmetical progression. In the third example the
common difference is +c?, and in the fourth example it is — d.

321. In an arithmetical progression having a finite number
of terms, there are five quantities to be considered, viz., the first
term, the last term, the number of terms, the common differ
ence, and the sum of the terms. When any three of them are
given, the other two may be found. We will denote

the first term by a,

the last term by Z,

the number of terms by n,

the common difference by c?,

and the sum of the terms by s.

The first term and the last term are called the extremes, and
all the other terms are called arithmetical means.

322. In an arithmetical progression the last term is equal to the
first term plus the product of the common difference ~by the number
of terms less one.

228 ALGEBRA.

Let the terms of the series be represented by
a, a+d, a + 2d, a + 8t/, a + 4cZ, etc.

Since the coefficient of d in the second term is 1, in the third
term 2, in the fourth term 3, and so on, the nth term of the
series will be

or l=a+(n — l)c?,

in which d is positive or negative according as the series is an

increasing or a decreasing one.

323. The sum of any number of terms in arithmetical pr ogres*
sion is equal to one half the sum of the two extremes multiplied by
the number of terms.

The term preceding the last will be I— d, the term preceding
that I— 2d, and so on. If the terms of the series be written in
the reverse order, the sum will be the same as when written in
the direct order. Hence we have

— &Q+ ----
Adding these equations term by term, we have

Here a+l is taken n times ; hence

n
or s—-(a+T).

£k

324. Tn an arithmetical progression the sum of the extremes is
equal to the sum of any tivo terms equidistant from the extremes.

This principle follows from the preceding demonstration.
It may also be shown independently as follows :
The rath term from the beginning is a + (m—I)d.
The rath, term from the end is I— (m— l)d.

And the sum of these terms is a-\-l.

325. To insert any number of arithmetical means betiveen two
given terms.

The whole number of terms in the series consists of the two

PROGRESSIONS.

229

extremes and all the intermediate terms. If, ihen, m repre
sents the number of means, m + 2 will be the whole number
of terms. •

Substituting m-f 2 for n in the formula, Art. 322, we have

or

d= -- -=the common difference.

whence the required means are easily obtained by addition.
326. The two equations

contain five quantities, a, Z, TI, d, 5, of which any three being
given, the other two can be found. We may therefore have
ten different cases, each requiring the determination of two dif
ferent formulae. These formulae are exhibited in the following
table, and should be verified by the student.

No.
1.

2.
3.

4.
5.

7.
8.
9.
10.

Given.

a, cZ, ?i,

?, 4 71,

a, Z, ft,
a, 71, 5,

71, d, 5,
Z, 71, 5,

a, cZ, Z,
a, Z, 5,
a, d, 5,

z, (U

Re
quired.

a, 5,

cZ,Z,
a, Z,

<7, C?,
71,5,
71, CZ,
Z, 71,

rf ^a-

Formulae.
^ -1-71 T2Z (T? 1W1

?*-!'

5 0-l)rZ

5 — .^^(^-l-Zj.

7 25
Z = a.

71

5 (V?_l")r7

71 2 '

25 _

a— I •

»' 2 •
2»^_2s

71

Z-a

71 — - 1 . 1 •

tt — 1 -1 \'

n(n — 1)
(Z+a)(Z — a-hcZ)

d

25

cZ P~a"

<=-*d±y2ds+^

dY. n-d 2a±V(2a d)> + 8ds

2d

2Z+d±y(2l+df $ds

a ^c?±'y/(/4-^c/^)2 — 2o!!s

n ±— i .

230 ALGEBKA.

EXAMPLES.

1. The first term of an arithmetical progression is 2, and the
common difference is 4 ; what is the 10th term ?

.4ns. 38.

2. The first term is 40, and the common difference —3 ; what
is the 10th term?

3. The first term is 1, and the common difference f ; what
is the 10th term ?

4. The first term is 1, and the common difference — J ; what
is the 10th term ?

5. The first term is 5, the common difference is 10, and the
number of terms is 60 ; what is their sum ?

Ans. 18000.

6. The first term is 116, the common difference is —4, and
the number of terms is 25 ; what is their sum ?

7. The first term is 1, the common difference is f , and the
number of terms is 12 ; what is their sum ?

8. The first term is 1-f , the common difference is — f ; and
the number of terms is 10 ; what is their sum ?

9. Required the number of terms of a progression whose sum
is 442, whose first term is 2, and common difference 3.

Ans. 17.

10. Required the first term of a progression whose sum is
99, whose last term is 19, and common difference 2.

11. The sum of a progression is 1455, the first term 5, and
the last term 92 ; what is the common difference ?

12. Required the sum of 101 terms of the series

1, 3, 5, 7, 9, etc. Ans. 10201.

13. Find the nth term of the series

1, 3, 5, 7, 9, etc. Ans. 2n-l.

14. Find the sum of n terms of the series

1, 3, 5, 7, 9, etc. Ans. n\

15. Find the sum of n terms of the series of numbers

1, 2, 3, 4, 5, etc.

Ans.

PROGRESSIONS. 231

16. Find the sum of n terms of the series

2, 4, 6,8, etc. Ans. n(n + l).

17. Find 6 arithmetical means between 1 and 50.

18. Find 7 arithmetical means between -J- and 3.

19. A body falls 16 feet during the first second, and in each
succeeding second 32 feet more than in the one immediately
preceding ; if it continue falling for 20 seconds, how many feet
will it pass over in the last second, and how many in the whole
time?

Ans. 624 feet in the last second, and
6400 feet in the whole time.

20. One hundred stones being placed on the ground in a
straight line at the distance of two yards from each other, how
far will a person travel who shall bring them one by one to a
basket which is placed two yards from the first stone?

Ans. 20200 yards.

PROBLEMS.

327. When of the five quantities a, /, w, c?, s, no three are
directly given, it may be necessary to represent the series by
the use of two unknown quantities. The form of the series
which will be found most convenient will depend upon the
conditions of the problem. If x denote the first term and y
the common difference, then

x, x+y, x+Zy, x+3y, etc.,
•will represent a series in arithmetical progression.

It will, however, generally be found most convenient to rep
resent the series in such a manner that the common difference
may disappear in taking the sum of the terms. Thus a pro
gression of three terms may be represented by

x-y, x, x + y;

one of four terms by x— 3y, %—y, %+y, x+Sy;
one of five terms by x — 2y, x—y, x, x+y, x + 2y.

Prob. 1. A number consisting of three digits which are in
arithmetical progression, being divided by the sum of its digits,
gives a quotient 26 ; and if 198 be added to it, the digits will
be inverted ; required the number. Ans. 234.

232 ALGEBRA.

Prob. 2. Find three numbers in arithmetical progression the
sum of whose squares shall be 1232, and the square of the
mean greater than the product of the two extremes by 16.

Ans. 16, 20, and 24.

Prob. 3. Find three numbers in arithmetical progression the
sum of whose squares shall be a, and the square of the mean
greater than the product of the two extremes by b.

/a-'2b /T /a-2b
Ans. Y — g Vb; y ^ ; and y

Prob. 4. Find four numbers in arithmetical progression
whose sum is 28, and continued product 585.

Ans. 1, 5, 9, 13.

Prob. 5. A sets out for a certain place, and travels 1 mile
the first day, 2 the second, 3 the third, and so on. In five days
afterward B sets out, and travels 12 miles a day. How long
will A travel before he is overtaken by B ?

Ans. 8 or 15 days.

This is another example of an equation of the second de
gree, in which the two roots are both positive. The following
diagram exhibits the daily progress of each traveler. The di
visions above the horizontal line represent the distances trav
eled each day by A ; those below the line the distances trav
eled by B.

A. 12345 6 7 8 9 10 11 12 13 14 15

UNI

1 1

1

1

1

1

1

1

1

1

2

3

4

1
5

1
6

7

1
8

1
9 1

B.

It is readily seen from the figure that A is in advance of B
until the end of his 8th day, when B overtakes and passes him.
After the 12th day, A gains upon B, and passes him on the
15th day, after which he is continually gaining upon B, and
could not be again overtaken.

Prob. 6. A goes 1 mile the first day, 2 the second, and so
on. B starts a days later, and travels b miles per day. How
long will A travel before he is overtaken by B ?

_--
Ans. - - ' — - days.

PROGRESSIONS. 233

In what case would B never overtake A?

For instance, in the preceding example, if B had started one
day later, he could never have overtaken A.

Prob. 7. A traveler set out from a certain place and went 1
mile the first day, 3 the second, 5 the third, and so on. After
he had been gone three days, a second traveler sets out, and
goes 12 miles the first day, 13 the second, and so on. After
how many days will they be together ?

Ans. In 2 or 9 days.

Let the student illustrate this example by a diagram like the
preceding.

Prob. 8. A and B, 165 miles distant from each other, set out
with a design to meet. A travels 1 mile the first day, 2 the
second, 3 the third, and so on. B travels 20 miles the first
day, 18 the second, 16 the third, and so on. In how many
days will they meet? Ans. 10 or 33 days.

GEOMETEICAL PROGRESSION.

328. A geometrical progression is a series of quantities each of
which is equal to the product of the preceding one by a constant
factor.

The constant factor is called the ratio of the series.

329. When the first term is positive, and the ratio greater
than unity, the series forms an increasing geometrical progres
sion, as

2, 4, 8, 16, 32, etc.,
in which the ratio is 2.

When the ratio is less than unity, the series forms a decreas
ing geometrical progression, as

81, 27, 9, 3, etc.,
in which the ratio is -J.

330. In a geometrical progression having a finite number
of terms, there ^XQ five quantities to be considered, viz., the first

234 ALGEBRA.

term, the last term, the number of terms, the ratio, and the sum

of the terms. When any three of these are given, the other

two may be found. We will denote

the first term by a,

the last term by lt

the number of terms by n,
the ratio by r,

and the sum of the terms by s.
The first term and the last term are called the extremes, and

all the other terms are called geometrical means.

331. In a geometrical progression, the last term Is equal to the
product of the first term by that power of the ratio whose exponent
is one less than the number of terms.

According to the definition, the second term is equal to the
first multiplied by r, that is, it is equal to ar / the third term
is equal to the second multiplied by r, that is, it is equal to
ar2 ; the fourth term is equal to the third multiplied by r, that
is, it is equal to ar3; and so on. Hence the nth term of the
series will be equal to arn~l ; hence we shall have

l=arn~l.

332. To find the sum of any number of terms in geometrical
progression, multiply the last term by the ratio, subtract the first
term, and divide the remainder by the ratio less one.

From the definition, we have

Multiplying this equation by r, we have

rs = ar-\-ar2-\- .... -f arl~* + arn
Subtracting the first equation from the second, member from
member, we have rs—s=arn—a.

arn — a

Hence s = - — ;

r — 1

or, substituting the value of I already found, we have

rl—a

PROGRESSIONS. 235

If we had subtracted the second equation from the first, we
should have found

• a — rl

S=T^?

which is the most convenient formula when r is less than
unity, and the series is, therefore, a decreasing one.

333. To find the sum of a decreasing geometrical series
when the number of terms is infinite, divide the first term by
unity diminished by the ratio.

The sum of the terms of a decreasing series may be repre
sented by the formula

a— rl

=T~^'

Now, in a decreasing series, each term is less than the pre
ceding, and the greater the number of terms, the smaller will
be the last term of the series. If the number of terms be in
finite, the last term of the series will be less than any assigna
ble number, and rl may be neglected in comparison with a.
In this case the formula reduces to

s=

l-r

334. To find any number of geometrical means between two given
lerms.

In order to solve this problem, it is necessary to know the
ratio. If m represent the number of means, m + 2 will be the
whole number of terms. Hence, putting w + 2 for n in the
formula, Art. 331, we have

whence we obtain

r=

That is, to find the ratio, divide the last term by the first term,
and extract the root which is denoted by the number of means plus
one. Having found the ratio, the required means may be ob
tained by continued multiplication.

236

ALGEBRA.

335. The two equations

5 =

arn— a
7^T~'

contain five quantities, a, Z, ??, r, 5, of which any three being
given, the other two can be found. We may therefore have
ten different cases, each requiring the determination of two quan
tities, thus giving rise to twenty different formulae. The first
four of the following cases are readily solved. The fifth and
sixth cases involve the solution of equations of a higher degree
than the second. When n is not large, the value of the un
known quantity can generally be found by a few trials. The
four remaining cases, when n is the quantity sought, involve
the solution of an exponential equation. See Art. 416. These
different cases are all exhibited in the following table for con
venient reference.

No.

Given.

Re
quired.

Formulas.

1.

a, r, 7i,

u

7 j arn — a

• r-l '

2.
3.

4.
6.

7

a,Z, n,
a, 7i, s,

a, 5,

Z Zr»-Z

1 n n

T> ( — 1 * <?

W ' IE J_'

6.

Z, 71,5,

a,r,

a(s _ a)n-1 — Z (5 — 1 )n~ 1 ; (s — Z)rn — s?<n "1 = — I.

8.
9.
10.

a,r,Z,
a, Z, 5,
a, r, 5,

5,71,

r,n,

Z,71,

a, n,

Ir—a lo«;. Z— log. a t 1

r — 1 log. r
s — a log. Z — log. a ^ ^

r — , , 7i — i / N i / 7N 1 i.
s — Z log. (s — a) — log. (,9 — Z)

r log. r

a — lr— (r— 1)5, TI_ • • T5 1 1

PROGRESSIONS. 237

EXAMPLES.

1. Find the 12th term of the series 1, 3, 9, 27, etc.
We have 1= arn~~L = 3u = 177147, Ans.

2. Given the first term 2, the ratio 3, and the number of
terms 10 ; to find the last term. Ans. 39366.

3. Find the sum of 14 terms of the series 1, 2, 4, 8, 16, etc.

firfU f*

5 = ^_f? = 2"-l = 16383, A^
T—L

4. Find the sum of 12 terms of the series 1, 3, 9, 27, etc.

Ans. 265,720.

5. Given the first term 1, the last term 512, and the sum of
the terms 1023 ; to find the ratio.

6. Given the last term 2048, the number of terms 12, and
the ratio 2 ; to find the first term.

7. Find the sum of 6 terms of the series 6, 4J, 3f , etc.

Am. 19||f

8. Find the sum of 15 terms of the series 8, 4, 2, 1, etc.

Anc 1 ^2 04 T

Ans. iCM

9. Find three geometrical means between 2 and 162.

10. Find two geometrical means between 4 and 256.

11. Find three geometrical means between a and b.

Ans.

12. Find the value of 1 + |-+J+-J-+, etc., to infinity.

a 1

13. Find the value ofl-f-J-f-|--f--g1f+, etc., to infinity.

Ans. f.

14. Find the value ofl-j-J-j-^-f-^-f-, etc., to infinity,

15. Find the ratio of an infinite progression whose first term
is 1, and the sum of the series f. Ans. -J.

16. Find the first term of an infinite progression whose ratio
is ^V, and the sum f . Ans. %.

17. Find the first term of an infinite progression of which

the ratio is -, and the sum T.

18. Find the value of the series 3-f 2+|+, etc., to infinity.

238 ALGEBRA.

19. Find the value of the series f+l+|+, etc., to infinity.

20. A gentleman, being asked to dispose of his horse, said
he would sell him on condition of receiving one cent for the
first nail in his shoes, two cents for the second, and so on,
doubling the price of every nail to 32, the number of nails in
his four shoes. What would the horse cost at that rate ?

Ans. $42,949,672.95.

PEOBLEMS.

Prob. 1. Find three numbers in geometrical progression such
that their sum shall be 21, and the sum of their squares 189.
Denote the first term by x and the ratio by y ; then
* + 0^+0^=21, (1.)

Transposing xy in Eq. (1), squaring, and reducing, we have

(3.)

/*
Comparing (2) and (3), xy=§, or #=-.

Substituting this value of x in Eq. (1), and reducing, we have

Whence y=2 or J, and x=3 or 12.

The terms are therefore 3, 6, and 12, or 12, 6, and 3.

Prob. 2. Find four numbers in geometrical progression such
that the sum of the first and second shall be 15, and the sum
of the third and fourth 60.

By the conditions, £c+a^ = 15, (1.)

Multiplying Eq. (1) by 7/2, we have

ay2 + ay* = 150* = 60.

Therefore 2/2=4, and y=±2.

Also cc±2x = 15;

therefore x=5 or —15.

Taking the first value of x and the corresponding value of
?/, we obtain the series 5, 10, 20, and 40.

Taking the second value of x and the corresponding value
of y, we obtain the series —15, +30, —60, and +120;

PROGRESSIONS. 239

which numbers also perfectly satisfy the problem understood
algebraically. If, however, it is required that the terms of the
progression be positive, the last value of x would be inapplica
ble to the problem, though satisfying the algebraic equation.

Several of the following problems also have two solutions, if
we admit negative values.

Prob. 3. Find three numbers in geometrical progression such
that their sum shall be 210, and the last shall exceed the first
by 90. Ans. 30, 60, and 120.

Prob. 4. Find three numbers in geometrical progression such
that their sum shall be 42, and the sum of the first and last
shall be 34. Ans. 2, 8, and 32.

Prob. 5. Find three numbers in geometrical progression such
that their continued product may be 64, and the sum of their
cubes 584. Ans. 2, 4, and 8.

Prob. 6. Find four numbers in geometrical progression such
that the difference between the first and second may be 4, and
the difference between the third and fourth 36.

Ans. 2, 6, 18, and 54.

Prob. 7. Find four numbers in geometrical progression such
that the sum of the first and third may be a, and the sum of
the second and fourth may be b.

a3 tfl abz b3

and

Prob. 8. Find four numbers in geometrical progression such
that the fourth shall exceed the second by 24, and the sum of
the extremes shall be to the sum of the means as 7 to 3.

Ans. 1, 3, 9, and 27.

Prob. 9. The sum of $700 was divided among four persons,
whose shares were in geometrical progression, and the differ
ence between the greatest and least was to the difference be
tween the means as 37 to 12. What were their respective
shares ? Ans. 108, 144, 192, and 256.

^ Prob. 10. Find six numbers in geometrical progression such
that their sum shall be 1365, and the sum of the third and
fourth shall be 80. Ans. 1, 4, 16, 64, 256, and 1024.

240 ALGEBKA.

CHAPTER XVII.

CONTINUED FRACTIONS. — PERMUTATIONS AND COMBINATIONS, \

336. A continued fraction is one whose numerator is unity,
and its denominator an integer plus a fraction, whose numera
tor is likewise unity, and its denominator an integer plus a
fraction, and so on.

The general form of a continued fraction is

c+1

d+j etc.

When the number of terms a, &, c, etc., \sfinite, the continued
fraction is said to be terminating ; such a continued fraction
may be reduced to an ordinary fraction by performing the op
erations indicated.

337. To convert any given fraction into a continued fraction.

Let - be the given fraction ; divide m by n ; let A be the
n

quotient, and p the remainder : thus,

n n n

P
"Divide n by^>; let a be the quotient, and q the remainder:

thus,

n a .1

— =a+--=a-\ — .

P P P_
2

Similarly, ^ = 6+-=6+-«,

r

CONTINUED FRACTIONS. 241

and so on, so that we have

Z> + , etc.

We see, then, that to convert a given fraction into a contin
ued fraction, we proceed as if we were finding the greatest com
mon divisor of the numerator and denominator ; and we must,
therefore, at last arrive at a point where the remainder is zero,
and the operation terminates ; hence every rational fraction can
"be converted into a terminating continued fraction.

Ex. 1. Transform - into a continued fraction.

22 + 1

Ex. 2. Transform |44 into a continued fraction.

Ans.

1 + 1

1 + 1

Ex. 3. Transform |4? into a continued fraction.

1

Ans.

2 + 1

_

2 + 1

Ex. 4. Transform |4r into a continued fraction.
Ex. 5. Transform -f-g-J- into a continued fraction.
Ex. 6. Transform -J|-- into a continued fraction.

338. To find the value of a terminating continued fraction,
Ex. 1. Find the value of the continued fraction

_

s+i-

Beginning with the last fraction, we have

L

242 ALGEBRA.

Hence

Therefore

and

Ex. 2. Find the value of the continued fraction

2 + 1

3 + 1

2 + 1

Ex- 3. Find the value of the continued fraction
1

3 + 1

2 + 1

2+i
Ex. 4. Find the value of the continued fraction

l+

2 + 1

1 + 1

1+1

1+A-

339. To find the value of an infinite continued fraction.

Let the fraction be

1

a + 1

c+, etc.

An approximate value of this fraction is obtained by omit
ting all its terms beyond any assumed fraction, and obtaining
the value of the resulting fraction, as in the previous article.

CONTINUED FRACTIONS. 243

Thus we obtain
1st approximate value, - ;

2d approximate value, o+l=-

6
1_
3d approximate value, a+1 , 1

7. , -i OC-\-L

(bc + l)d+b

4th approximate value. -7-^ — q^— ; — £-= , — r

' (ab+I)cd+ad+ab+l, etc.

340. The fractions formed by taking one, two, three, etc.,
of the quotients of the continued fraction are called converging
fractions, or conver gents.

The convergents, taken in order, are alternately less and great
er than the continued fraction.

The first convergent - is too great, because the- denominator
a j

a is too small ; the second convergent —r — -- is too small, be
cause a-f-y is too great, and so on.

341. When a fraction has been transformed into a continued
fraction, its approximate value may be found by taking a few
of the first terms of the continued fraction.

Thus an approximate value of -J4r is -J, which is the first
term of its continued fraction.

By taking two terms, we obtain |4, which is a nearer ap
proximation ; and three terms would give a still more accu
rate value.

Ex. 1. Find approximate values of the fraction 1513/3.

Ans. i, I, ||.

Ex. 2. Find approximate values of the fraction

Ex. 3. Find approximate values of the fraction

244 ALGEBKA.

342. By the preceding method we are enabled to discover
the approximate value of a fraction expressed in large num
bers, and this principle has some important applications, par
ticularly in Astronomy.

Ex. 4. The ratio of the circumference of a circle to its diam-
ter is 3.1415926. Find approximate values for this ratio.

22 333 355

. -y,

Ex. 5. The length of the tropical year is 365d 5h. 48m. 48s.
Find approximate values for the ratio of 5 A. 48 ra. 485. to 24
hours.

Ans. J, A, A, T3TTT-

Ex. 6. In 87969 years the earth makes 277287 conjunctions
with Mercury. Find approximate values for the ratio of 87969
to 277287.

Ex. 7. In 57551 years the earth makes 36000 conjunctions
with Venus. Find approximate values for the r^,tio of 57551
to 36000.

Ans. |, fjf

Ex. 8. In 295306 years the moon makes 3652422 synodical
revolutions. Find an approximate value for the ratio of 295306
to 3652422. Ans. -j^.

Ex. 9. One French metre is equal to 3.2809 English feet.
Find approximate values for the ratio of a metre to a foot.

Ex. 10. One French kilogramme is equal to 2.2046 pounds
avoirdupois. Find approximate values for the ratio of a kilo
gramme to a pound.

Ex. 11. One French litre is equal to 0.2201 English gallons.
Find approximate values for the ratio of a litre to a gallon.

PERMUTATIONS AND COMBINATIONS.

343. The different orders in which things can be arranged
are called their permutations. In forming permutations, all of
the things or a part only may be taken at a time.

Thus the permutations of the three letters a, 6, c, taken all
together, are

abc, acb. bac, bca, cab, cba.

PEKMUTATIONS AND COMBINATIONS. 245

The permutations of the same letters taken two at a time are

ab, ac, ba, be, ca, cb.

The permutations of the same letters taken one at a time are

a, b, c.

344. The number of permutations of n things taken m at a time
is equal to the continued product of the natural series of numbers
from n down to n—m + l.

Suppose the things to be n letters, a, 5, c, d

The number of permutations of n letters, taken singly or
one at a time, is evidently equal to the number of letters, or
to n.

If we wish to form all the permutations of n letters taken
two at a time, we must write after each letter each of the
Ti—1 remaining letters. We shall thus obtain n(n—l) permu
tations.

If we wish to form all the permutations of n letters taken
three at a time, we must write after each of the permutations
of n letters taken two at a time each of the n— 2 remaining let
ters. We shall thus obtain n(n— l)(n— 2) permutations.

In the same manner we shall find that the number of permu
tations of n letters taken four at a time is
n(n-l)(n-2)(n-3).

Hence we may conclude that the number of permutations
of n letters taken m at a time is

n(n-l)(n-2)(n—S) (n-m + 1).

345. The number of permutations of n things taken all together
is equal to the continued product of the natural series of numbers
from 1 to n.

If we suppose that each permutation comprehends all the n
letters; that is, if m — n, the preceding formula becomes

n(n-l)(n-2) 3x2x1;

or, inverting the order of the factors,

1.2.3.4. . . . (7i-l)w,

which expresses the number of permutations of n things taken
all together.

246 ALGEBRA.

For the sake of brevity, 1.2.3.4.... (n— l)n is often de-
noted by \n; that is, \n denotes the product of the natural num
bers from 1 to n inclusive.

346. The combinations of things are the different collections
which can be formed out of them without regarding the order
in which the things are placed.

Thus the three letters a, 6, c, taken all together, form but one
combination, abc.

Taken two and two, they form three combinations, ab, ac, be.

347. The number of combinations of n things, taken m at a time,
is equal to the continued product of the natural series of numbers
from n down to n — m-\-\ divided by the continued product of the
natural series of numbers from 1 to m.

The number of combinations of n letters taken separately,
or one at a time, is evidently n.

The number of combinations of n letters taken two at a

. n(n — l)
time is • '.

L ,&

For the number of permutations of n letters taken two at a
time is n(n — 1), and there are two permutations (ab, bd) corre
sponding to one combination of two letters ; therefore the num
ber of combinations will be found by dividing the number of
permutations by 2.

The number of combinations of n letters taken three at a

. n(n-l)(n-2)
time is -— - — yv '-.

For the number of permutations of n letters taken three at
a time is n(n — ~L)(n — 2), and there are 1.2.3 permutations for
one combination of these letters ; therefore the number of com
binations will be found by dividing the number of permuta
tions by 1.2.3.

In the same manner we shall find the number of combina
tions of n letters taken m at a time to be

77(71— l)(n — 2) . . . . (n — m-
1.2.3 . m

PERMUTATIONS AND COMBINATIONS. 247

EXAMPLES.

1. How many different permutations may be formed of 8
letters taken 5 at a time ? Arts. 8.7.6.5.4= 6720.

2. How many different permutations may be formed of the
26 letters of the alphabet taken 4 at a time ?

Ans. 358800.

3. How many different permutations may be formed of 12
letters taken 6 at a time ? Ans. 665280.

4. How many different permutations may be formed of 8
things taken all together?

Ans. 1.2.3.4.5.6.7.8=40320.

5. How many different permutations may be made of the
letters in the word Roma taken all together?

6. How many different permutations may be made of the
letters in the word virtue taken all together?

7. What is the number of different arrangements which can
be formed of 12 persons at a dinner- table ?

Ans. 479001600.

8. How many different combinations may be formed of 6
letters taken 3 at a time ?

6.5.4 on

Ans- ro=2a

9. How many different combinations may be formed of 8
letters taken 4 at a time ? Ans. 70.

10. How many different combinations may be formed of 10
letters taken 6 at a time? Ans. 210.

11. A telegraph has m arms, and each arm is capable of n
distinct positions; find the total number of signals which can
be made with the telegraph.

12. How many different numbers can be formed with the
digits 1, 2, 3, 4, 5, 6, 7, 8, 9, each of these digits occurring once,
and only once, in each number ?

248 ALGEBKA.

CHAPTER XVIII.

BINOMIAL THEOREM.

348. The binomial theorem, or binomial formula, is a formula
discovered by Newton, by means of which we may obtain any
power of a binomial x+a, without obtaining the preceding
powers.

349. By actual multiplication, we find the successive powers
of x+a to be as follows:

(x + a)2 — x2 -f 2ax + a2,
(x + a)3 = x3 + 3aa?2 + 3a2# + a3,
(x + a)4 = x4 + 4ax3 + 6a2x2 + 4a3x + a4,
(SB + a)5 = ^5 + 5acc4 + lOaV + 10«3x2 + 5a*x + a6.
The powers of a? — a, found in the same manner, are as fol
lows:

On comparing the powers of x-\-a with those of x— a, we
perceive that they only differ in the signs of certain terms. In
the powers of x+a, all the terms are positive. In the powrers
of x — a, the terms containing the odd powers of a have the sign
minus, while the terms containing the even powers have the
sign plus. The reason of this is obvious; for, since —a is the
only negative term of the root, the terms of the power can only
be rendered negative by a. A term which contains the factor
—a an even number of times will therefore be positive; if it
contain it an odd number of times it must be negative. Hence
it appears that it is only necessary to seek for a method of ob
taining the powers of x-\-a, for these will become the powers
of x— a by simply changing the signs of the alternate terms.

BINOMIAL THEOREM.

249

350. Law of the Exponents. — The exponents of x and of a in
the different powers follow a simple law. In the first term of
each power, x is raised to the required power of the binomial ;
and in the following terms the exponents of x continually de
crease by unity to zero, while the exponents of a increase by
unity from zero up to the required power of the binomial.

351. Law of the Coefficients. — The coefficient of the first term
is unity ; that of the second term is the exponent of the power ;
and the coefficients of terms equidistant from the extremes are
equal to each other; but after the first two terms it is not ob
vious how to obtain the coefficients of the fourth and higher
powers.

In order to discover the law of the coefficients, we will form
the product of several binomial factors whose second terms are
all different; thus,

(x+a)(x+b)=x*+a x+ab.

(x+a)(x+b)(x+c)(x+d)=x*+a

+ ad
+bd
+ cd

x -\-dbc.

x2-{-abc
+ abd

-\-bcd

x -\-dbcd.

In each of these products the exponent of x in the first term
is equal to the number of binomial factors, and in the follow-
ng terms continually decreases by one. The coefficient of the
first term is unity ; the coefficient of the second term is the sum of
the second terms of the binomial factors ; the coefficient of the third
term is the sum of all their products taken two and two, and so on.
The last term is the product of the second terms of the binomial
factors.

250 ALGEBRA.

352 We will now prove that if the laws of formation jusi
stated are true for any power, they will also hold true for the forma
tion of the next higher power.

Suppose that we have found the product of ra binomials
a? + a, a? +5, .... x+L Let Pj denote the sum of the second
terms of the binomials, P2 the sum of the different products of
these second terms taken two and two, P3 the sum of their
products taken three and three, and so on ; and let Pm denote
the product of all these second terms. The product of the
given binomials will then be

Multiplying this polynomial by a new binomial, a?+?, we ob
tain the following product :

-{-I

— 1

+ /P,

+P1

l-l

The law of the exponents of x remains the same. The co
efficient of the first term is still equal to unity, and that of the
second term is the sum of the second terms of the ra + 1 bino
mials. The coefficient of the third term consists of the sum
of the products of the second terms of the m binomial factors
taken two and two, increased by the sum of the same second
terms multiplied by Z, which is equivalent to the sum of the
products of the second terms of the ra+1 binomials taken two
and two. The coefficient of the fourth term consists of the sum
of the products of the second terms of the m factors of the first
product taken three and three, increased by the sum of the
products of their second terms taken two and two multiplied
by 7, which is equivalent to the sum of the products of the sec
ond terms of the ra + 1 binomials taken three and three, and
so on. The last term is equal to the product of the second
terms of the m binomial factors multiplied by ?, which is equiv
alent to the product of the second terms of the ra + 1 binomials.

Hence the law which was supposed true for m factors is true
for ra + 1 factors; and therefore, since it has been verified for
two factors, it is true for three ; being true for three factors, it
is also true for four, and so on ; therefore the law is general.

BINOMIAL THEOKEM. 251

353. Powers of a Binomial. — If now, in the preceding bino
mial factors, we suppose the second terms to be all equal to a,
the product of these binomials will become the rath power of

The coefficient of the second term of the product becomes
equal to a multiplied by the number of factors; that is, it is
equal to ma.

The coefficient of the third term reduces to a2 repeated as
many times as there are different combinations of m letters

taken two and two ; that is, to — ^— — - a2.

1. Z

The coefficient of the fourth term reduces to a3 repeated as
many times as there are different combinations of m letters

. . m(m — l)(m — 2) _
taken three and three; that is -- TOO - a » anc* so on-

L . Zt.O

The last term will be am.

Hence the mth power of oc+a may be expressed as follows:

(x 4- a)m =m m~l m~ )

1.2.3

354. We perceive that if the. coefficient of any terra be multi
plied by the exponent of x in that term, and the product be divided
by -the exponent of a in that term increased by unity, it will give the
coefficient of the succeeding term.

Forming thus the seventh power of x+a, we obtain

We have thus deduced

Sir Isaac Newton's Binomial Theorem.

355. In any power of a binomial x-\-a, the exponent of x begins
in the first term with the exponent of the power, and in the follow
ing terms continually decreases by one. The exponent of a com
mences with one in the second term of the power, and continually
increases by one.

The coefficient of the first term is one, that of the second is the ex

252 ALGEBKA.

ponent of the power ; and if the coefficient of any term be multi
plied by the exponent of x in that term, and divided by the exponent
of a increased by one, it will give the coefficient of the succeeding term.

356. The coefficient of the nth term from the beginning is equal
to the coefficient of the nth term from the end.

If we change the places of x and a, we shall have, by the law
of formation,

(a + x)™ = a™ + mxa™~^ + "- xiam-<i +

•L.Z

m(m-l)(m-2)

1.2.3

The second member of this equation is the same as the sec
ond member of the equation in Art. 353, but taken in a reverse
order. Comparing the two, we see that the coefficient of the
second term from the beginning is equal to the coefficient of
the second term from the end ; the coefficient of the third from
the beginning is equal to that of the third from the end, and
so on. Hence, in forming any power of a binomial, it is only
necessary to compute the coefficients for half the terms ; we
then repeat the same numbers in a reverse order.

357. The mth power ofx-\-a contains m + ~L terms. This ap
pears from the law of formation of the powers of a binomial
developed in Art. 352. Thus the fourth power of x-}-a con
tains five terms ; the sixth power contains seven terms, etc.'

358. The sum of the coefficients of the terms in the nth power of
x-\-a is equal to the nth power of 2.

For, suppose cc=l and a=l, then each term of the formula
without the coefficients reduces to unity, and the sum of the
terms is simply the sum of the coefficients. In this case
(x-\-a)m becomes (l + l)m, or 2m.
Thus the coefficients of the

second power are 1+2 + 1= 4=22,

third " 1 + 3 + 3 + 1= 8=23,

fourth " l+4+6+4+l = 16 = 24, etc.

BINOMIAL THEOREM. 253

359. To obtain the development of (x — a)m, it is sufficient to
change +a into —a in the development of (x + d)m. In con
sequence of this substitution, the terms which contain the odd
powers of a will have the minus sign, while the signs of the re
maining terms will be unchanged. We shall therefore have

1.2.3

EXAMPLES.

1. Find the sixth power of a -}-b.
The terms without the coefficients are

a6, a56, a462, a363, a2£4, ab5, b6.
The coefficients are

1 « 6x5 15x4 20x3 15x2 6x1
' ' 2 ' 3 ' ~T~' ~T~'~1T
that is, 1, 6, 15, 20, 15, 6, 1.
Prefixing the coefficients, we obtain

2. Find the ninth power of a— b.
The terms without the coefficients are

, ,
The coefficients are

9x8 36x7 84x6 126x5 126x4 84x3 36x2 9x1
2 ' 3 ' 4 ' 5 ' 6 ' ~T~' ~~8~' ~9~
that is,

1, 9, 36, 84, 126, 126, 84, 36, 9, 1.
Prefixing the coefficients, we obtain

It should be remembered that it is only necessary to com
pute the coefficients of half the terms independently.

3. Find the seventh power of a — x.

4. Find the third term of (a + b)15.

5. Find the forty-ninth term of (a— a;)50.

6. Find the middle term of (a -fa;)10.

254: ALGEBRA.

360. A Binomial with Coefficients. — If the terms of the given
binomial have coefficients or exponents, we ma;y obtain any
power of it by means of the binomial formula. For this pur
pose, each term must be raised to its proper power denoted by
the exponents in the binomial formula.

7. Find the fourth power of 2x+3a.

For convenience, let us substitute y for 2x and b for 3a.
Then (y + fe)4 =y + ty*b + 6y2b2 + tyb3 4- fe4.

Eestoring the values of?/ and fe,
the first term will be (2ce)4=16£e*,

the second term will be 4(2x)3x3a =
the third term will be 6(2x)2 x (3«)2
the fourth term will be 4(2x)x(3a)3 =
the fifth term will be (3a)* = 81a*.

Therefore (2x+3a)4=:16x4 + 96x3a+216x2tt2

It is recommended to write the three factors of each term in
a vertical column, and then perform the multiplication as indi
cated below :

Coefficients, 1+4+6 +4+1

Powers of 2x, 16x4 + 8x3 + 4x2 + 2x + 1

Powers of 3«, 1 + 3a + 9a2 + 27 a3

8. Find the fifth power of 2ax— 3b.

Coefficients, 1 +5 +10 +10 +5 +1

Powers of 2aar, 32a5.r5+ 16a4a:4 + Sa5x3 + 4a2ar2 + 2nx + 1
Powers of -36, 1 36 + 962 27b3 + 81 &* -2436s

(2ax - 36)5 =32a!'x6 - 240a4x46 + 720a3a:362 - 1080a2a:263 + 8 Waxb* - 24366.

9, Find the fourth power of 2x+5a2.

Ans. 1 6x4 + 1 60x3a2 + 600x2a4 + lOOOxa6 + 625a8.

10. Find the fourth power of x3 + 4?/2.

11, Find the sixth power of a3 + Safe.

12. Find the seventh power of 2a — Sfe.

Ans. 128a7 -

13. Find the fifth power of 5a2 —

BINOMIAL THEOREM. 255

14. Find the sixth power of a2x + by2.

15. Find the fifth power of ax—1.

16. Find the fifth term of (a2-62)12.

17. Find the fifth term of (Sx^-ty^)9.

18. Find the sixth power of 5 — ^.

Powers and Roots of Polynomials.

361- If it is required to raise & polynomial to any power, we
iiiay, by substituting other letters, reduce it to the form of a
binomial. We obtain the power of this binomial by the gen
eral formula ; then, restoring the original letters, and perform
ing the operations indicated, we obtain the required power of
the proposed polynomial.

Ex. 1. Let it be required to raise a + b-\-c to the third power.

If we put b-\-c=m, we shall have

(a + 1 4- c)3 = (a 4- m)3 = a3 + 3a2m + 3am2 + m3,
or a3 + Sa2(b + c) + 3a(6 + c)2 + (ft + c)3.

Developing the powers of the binomial b-\-c, and performing
the operations indicated, we obtain

Ex. 2. Find the fifth power of

Ex. 3. Find the fourth power of a2 —

Ans. a8 - 4«7£ + 1 Oa6£2 - 1 6a5ft3 + 1 9a4&4

- 16a3// + 10a2^6-4(/&7 + 1*.
Ex. 4. Find the fifth power of 1+2^-f 3x2.
Ex. 5. Find the sixth power of a
Ans. a6-f 6a5& + 6a5c +

60a36c2 + 20a3c3

+ 15a2c* + 6«Z>5
4 4- 6ac5 -f Z>6 4- Qb

362. The binomial theorem will inform us how to extract
any root of a polynomial. We know that the mih power of

256 ALGEBRA.

x + a is xm+ max™-^ + other terms. The first term of the root
is, therefore, the mth root of the first term of the polynomial.
Also the second term of the root may be found by dividing the
second term of the polynomial by mxm~l ; that is, the first term
of the root raised to the next inferior power, and multiplied
by the exponent of the given power. Hence, for extracting
any root of a polynomial, we have the following

RULE.

Arrange the terms according to the powers of one of the letters,
and take the mth root of the first term for the first term of the re
quired root.

Subtract the mth power of this term of the root from the given
polynomial, and divide the first term of the remainder ly m times
the (m — 1) power of this root ; the quotient will be the second term
of the root.

Subtract the mth power of the terms already found from the given
polynomial, and, using the same divisor, proceed in like manner to
find the remaining terms of the root.

Ex. 1. Find the fourth root of

32a3) —

Here we take the fourth root of 16a4, which is 2a, for the
first term of the required root, subtract its fourth power, and
bring down the first term of the remainder, — 96a3£. For a
divisor, we raise the first term of the root to the third power
and multiply it by 4, making 32a3. Dividing, we obtain — Sx
for the second term of the root. The quantity 2a — 3x, being
raised to the fourth power, is found to be equal to the proposed
polynomial.

Ex. 2. Find the fifth root of

-40a2f lOa-1.

Ans. 2x—~L

BINOMIAL THEOREM. 257

Ex. 3. Find the fourth root of

336o;5+81z8-216x7--56x4+16-224x3 + 64x.

Am. 3x2— 2z— 2.

363. To extract any Root of a Number. — The preceding meth
od may be applied to the extraction of any root of a number.
Let n be the index of the root, n being any whole number.
For a reason similar to that given for the square and cube roots,
we must first divide the number into periods of n figures each,
beginning at the right. The left-hand period may contain less
than n figures. Then the first figure of the required root will
be the nth root of the greatest nth power contained in the first
period on the left. If we subtract the nth power of this root
from the given number, and divide the remainder by n times
the (n— l)th power of the first figure, regarding its local value,
the quotient will be the second figure of the root, or possibly
a figure too large. The result may be tested by raising the
whole root now found to the nth power ; and if there are other
figures they may be found in the same manner.

In the extraction of the nth root of an integer, if there is
still a remainder after we have obtained the units' figure of the
root, it indicates that the proposed number has not an exact nth
root. We may, if we please, proceed with the approximation
to any desired extent by annexing any number of periods of
n ciphers each, and continuing the operation. We thus ob
tain a decimal part to be added to the integral part already
found.

So, also, if a decimal number has no exact nth root, we may
annex ciphers, and proceed with the approximation to any de
sired extent, dividing the number into periods commencing
with the decimal point.

Ex, 1. Find the fifth root of 33554432.

335.54432(32
243

6.3* =405) 925
325= 33554432.

Ex. 2. Find the fifth root of 4984209207.

258 ALGEBRA.

Ex. 3. Find the fifth root of 10.

Ex. 4. Find the fifth root off. Am. .922.

364. When the index of the required root is composed of
two factors, we may obtain the root required by the successive
extraction of simpler roots, Art. 217. For the mnih root of
any number is equal to the mth root of the nth root of that
number.

Thus we may obtain the fourth root by extracting the square
root of the square root.

We may obtain the sixth root by extracting the cube root
of the square root, or the square root of the cube root. It is,
however, best to extract the roots of the lowest degrees first,
because the operation is less laborious.

We may obtain the eighth root by extracting the square root
three times successively. We may obtain the ninth root by ex
tracting the cube root twice successively.

Ex. 1. Find the fourth root of

6a262 + a4 - 4a36 - 4a £3 + 64.

Ex. 2. Find the sixth root of

Ex. 3. Find the eighth root of

SERIES. 259

CHAPTER XIX.

SERIES.

365. A series is a succession of terms each of which is de°
rived from one or more of the preceding ones by a fixed law.
This law is called the law of the series. The number of terms
of the series is generally unlimited. Arithmetical and geomet
rical progressions afford examples of series.

366. A converging series is one in which the sum of the first
n terms can not numerically exceed some finite quantity, how
ever great n may be.

Thus, 1, J, i, -J, TV? etc-j is a converging series.

367. A diverging series is one in which n can be taken so
large that the sum of the first n terms is numerically greater
than any finite quantity.

Thus, 1, 2, 3, 4, 5, 6, etc., is a diverging series.

368. When a certain number of terms are given, and the
law of the series is known, we may find any term of the series,
or the sum of any number of terms. This may generally be
done by the method of differences.

369. To find the several orders of differences for arty series:
Subtract the first term from the second, the second from the

third, the third from the fourth, etc. ; we shall thus form a new
series, which is called the first order of differences.

Subtract the first term of this new series from the second, the
second from the third, etc. ; we shall thus form a third series,
called the second order of differences.

Proceed in like manner for the third, fourth, etc., orders of
differences, and so on till they terminate, or are carried as far
as may be thought necessary.

260

ALGEBRA.

Ex. 1. Find the several orders of differences of trie series oi
square 'numbers 1, 4, 9, 16, etc.

Squares.
1

4
9

16
25

Ex. 2. Find the several orders of differences of the series of
cube numbers 1, 8, 27, etc.

1st Diff.

2d Diff.

3dD

3

5

2

0

7

2

0

9

2

0

2

Cubes.

1st Diff.

2d Diff.

3d Diff.

4th Diff.

1

8

7

12

27

19

18

6

0

64

37

24

6

0

125

61

30

6

0

216

91

36

6

0

Ex. 3. Find the several orders of differences of the series of
fourth powers 1, 16, 81, 256, 625, 1296, etc.

Ex. 4. Find the several orders of differences of the series of
fifth powers 1, 32, 243, 1024, 3125, 7776, 16807, etc.

Ex. 5. Find the several orders of differences of the series of
numbers 1, 3, 6, 10, 15, 21, etc.

370. To find the nth term of any series:

Let a, 6, c, d, e, etc., represent the proposed series. If we
subtract each term from the next succeeding one, we shall ob
tain the first order of differences ; if we subtract each term of
this new series from the succeeding term, we shall obtain the
second order of differences, and so on, as exhibited in the fol
lowing table :

4th Order of Differences.

e_4c?-j-6c—

Series.

1st Diff.

2d Differences.

3d Order of Differences.

a

7

b-a

0

c
d

c-b
d-c

d-2c+b
e—2d+c

d—Bc+Sb— a
e—3d+Sc—b

e —d

e

SERIES. 261

Let D', D", D'", D"", etc., represent the first terms of the
several orders of differences. Then we shall have
T>'—b— a, whence b=a+~D'.

D"=c-26+a, " c

etc., etc.

The coefficients of the value of c, the third term of the pro
posed series, are 1, 2, 1, which are the coefficients of the second
power of a binomial ; the coefficients of the value of d, the
fourth term, are 1, 3, 3, 1, which are the coefficients of the third
power of a binomial, and so on. Hence we infer that the co
efficients of the nth term of the series are the coefficients of the
(tt— l)th power of a binomial. If we denote the nth term of
the series by Tn, we shall have

9

L Z.o

+ , etc.

Ex. 1. Find the 12th term of the series 2, 6, 12, 20, 30, etc.
The first order of differences, 4, 6, 8, 10, etc.
" second order of differences, 2, 2, 2, etc.
" third order of differences, 0, 0.
Here D'=4, D"=2, and D'" = 0. Also a = 2 and n= 12.
Hence T13 = 2 + 11D' + 55D" = 2 +44 + 110 = 156, Ans.
Ex. 2. Find the twentieth term of the series

1, 3, 6, 10, 15, 21, etc.
Here D'=2, D/r = l, a=l, and n = 20.
Therefore T20=1-|-19D/+171D'/ = 1 + 38 + 171=210, Ant.
Ex. 3. Find the thirteenth term of the series

1, 5, 14, 30, 55, 91, etc.
Ex. 4. Find the fifteenth term of the series

1, 4, 9, 16, 25, 36, etc.
Ex. 5. Find the twentieth term of the series

1, 8, 27, 64, 125, etc.
Ex. 6. Find the ftth term of the series 1, 3, 6, 10, 15, 21, etc.

rc
Ans. -

262 ALGEBRA.

Ex. 7. Find the nth term of the series 1, 4, 10, 20, 35, etc.

Ans. *

Ex. 8. Find the nth term of the series 1, 5, 15, 35, 70, 126, etc,

ttfoi+l)(n + 2)(tt + 3)
24

?/r

371. To find the sum of n terms of any series:
Let us assume the series

0, a, a-f-&, a+6-fc, a-J-6+c+c?, etc. (1.)
Subtracting each term from the next succeeding, we obtain
the first order of differences,

a, bj c, d, etc. (2.)

Now it is clear that the sum of n terms of the series (2) is
equal to the (n + l)th term of series (1); and the nth order of
differences in series (2) is the (n+l)th order in series (1). If,
then, we denote the sum of n terms of series (2) by S, which is
the same as the (n-j-l)th term of (1), we may obtain the value
of S from the formula of the preceding article by substituting

0 for a,
tt + 1 for n,
a for D',
V for D", etc.
Hence

etc.

When any one of the successive orders of differences be
comes zero, this formula gives the exact sum of the terms.
When no order of differences becomes zero, the formula may
still give approximate results, which will, in general, be nearer
the truth the greater the number of terms employed.

EXAMPLES.

1. Find the sum of 15 terms of the series

1, 3, 6, 10, 15, 21, etc.
Here o=l, D' = 2, D" = l, D'" = 0.
Therefore 8=15-1-15.14+5.7.13=680, Ana.

SERIES. 263

2. Find the sum of 20 terms of the series

1, 4, 10, 20, 35, etc.

3. Find the sum of n terms of the series

1, 2, 3, 4, 5, 6, etc.

A
Ans.

4. Find the sum of n terms of the series

I2, 22, 32, 42, 52, etc.

6

5. Find the sum of n terms of the series
I3, 23, 33, 43, 53, etc.

(

6. Find the sum of n terms of the series

1, 3, 6, 10, 15, etc.

Ans.

7. Find the sum of n terms of the series

1.2, 2.3, 3.4, 4.5, 5.6, etc.

An8.

8. Find the sum of n terms of the series
1, 4, 10, 20, 35, etc.

2.3.4

Interpolation.

372. Interpolation is the process by which, when we have
given a certain number of terms of a series, we compute inter
mediate terms which conform to the law of the series.

Interpolation may, in most cases, be effected by the use of
the formula of Art. 370. If in this formula we substitute w + 1
for n, we shall have

Tn+1=a+MD'+^D"+"(1\-2V+, etc,

which expresses the value of that term of the series which has
n terms before it. When n is a fraction less than unity, Trt+i

264 ALGEBKA.

stands for a term between the first and second of the given
terms. When n is greater than 1 and less than 2, the inter
mediate term will lie between the second and third of the given
terms, and so on. In general, the preceding formula will give
the value of such intermediate terms.

EXAMPLES.

1. Given the cube root of 60 equal to 3.914868,

" " " 61 " 3.936497,

" " 62 " 3.957891,

" " " 63 " 3.979057,

" " " 64 " 4.000000,

to find the cube root of 60.25.
Here D' = + .021629, D"=-. 000235, D//;=+. 000007, etc

a=3.914868, and w = .25.
Substituting the value of n in the formula, we have

Tn+i^a+iD'-A^'+T^'''-, etc.
The value of the 1st term is +3.914868,

" " 2d " + .005407,

" " 3d « + .000022,

" " 4th " + .000000.

Hence the cube root of 60.25 is 3.920297.

2. Find the cube root of 60.5. Ans. 3.925712.

3. Find the cube root of 60.75. Ans. 3.931112.

4. Find the cube root of 60.6. Ans. 3.927874.

5. Find the cube root of 60.33. Ans. 3.922031.

6. Given the square root of 30 equal to 5.477226,

" " " 31 " 5.567764,

« " " 32 " 5.656854,

" " " 33 " 5.744563,

« " " 34 " 5.830952,

to find the square root of 30.3 Ans. 5.504544,

7. Find the square root of 30.4. Ans. 5.513619.

8. Find the square root of 30.5. Ans. 5.522681.

9. Find the square root of 30.6. Ans. 5.531727.
10. Find the square root of 30.8. Ans. 5.549775.

SERIES. 265

Development of Algebraic Expressions into Series.

373. An irreducible fraction may be converted into an in
finite series by dividing the numerator by the denominator, ac
cording to the usual method of division.

Ex. 1. Expand - - into an infinite series.
1 — x

x3+x* + 1 etc.

i-rflS

X3

Hence - -- = 1+ x-f x2+#3+x4+x5-f , etc., to infinity.
J. — x

Suppose x=^, we shall then have

, etc.

Suppose x—%, we shall then have

, etc.

Ex. 2. Convert - - into an infinite series.
1-fo?

Ans. 1— x-\-x2— x*-\-x*— x54-, etc.
Suppose 03=4, we shall then have

T|-|=i=l-i+i-H-rV-A+, etc.

Ex. 3. Convert - into an infinite series.
a+x

_. XX2 X3 X4

Ans. 1 --- \- — — -H — j— , etc.
a a2 a3 a*

Ex. 4. Convert - into an infinite series.
a—x

_, X X2 X3 X*

Ans. 1+- +— +_4-_4- etc.
•- a a2 a3 «4

266 ALGEBRA.

1+x

Ex. 5. Convert into an infinite series.

1— x

Ex. 6. Convert into an infinite series.

a — x

, 2x 2x2 , 2x3 2x* ,
Ans. 1 + — H — 2-+—+—+, etc-
a a2 a3 a4

Ex. 7. Convert into an infinite series.

1— x+x2

Ans. 1+x— x3— x4+x6+x7— , etc.

1 — x

Ex. 8. Convert into an infinite series.

1— x+x2

Ans. 1— x2— x3+x5+x6— x8— , etc.

~1 i /Y*

Ex. 9. Convert = ^ into an infinite series.

1— x — x2

374. An algebraic expression which is not a perfect square
may be developed into an infinite series by extracting its square
root according to the method of Art. 198.

Ex. 1. Develop the square root of 1 -fx into an infinite series.
x x2 x3 5x4
2~8 +16-128+' 6tC-

x

x*
4

X2 X3 . X4

x3 x6

5x4 x5 x6

SERIES. 267

Hence the square root of 1 -f x is equal to

Suppose a5=l, we shall have

V2 = l+i— J + Vo~ rfff + , etc.
Ex. 2. Develop the square root of a2 + x into an infinite series.

Ex. 3. Develop the square root of a4 +x into an infinite series.
Ex. 4. Develop the square root of a*— x into an infinite series.
Ex.5. Develop the square root of «2 + x2 into an infinite
series. \

Method of Undetermined Coefficients.

375. One of the most useful methods of developing algebraic
expressions into series is the method of undetermined coefficients.
It consists in assuming the required development in the form
of a series with unknown coefficients, and afterward finding the
value of these coefficients. This method is founded upon the
properties of identical equations.

376. An identical equation is one in which the two members
are identical, or may be reduced to identity by performing the
operations indicated in them. As

ax 4- b = ax -f 5,

a — x

a ax

a—- =- .

1+x l+x

377. It follows from the definition that an identical equation
is satisfied by each and every value which may be assigned to a let
ter which it contains, provided that value is the same in both
members of the equation.

Every identical equation containing but one unknown quan
tity can be reduced to the form of

3-K etc.

268 ALGEBRA.

378. If an equation of the form

A + Bx + Cx2 + , etc. =
must be satisfied for each and every value given to x, then the co
efficients of the like powers of x in the two members are equal each
to each.

For, since this equation must be satisfied for every value of
ic, it must be satisfied when x = 0. But upon this supposition
all the terms vanish except two, and we have

A = A'.
Suppressing these two equal terms, we have

Bx + Cx2 + , etc.=B/x + C'x2 + J etc.
Dividing each term by x, we obtain

B + Cx + , etc. = B/ + C'x + , etc.

Since this equation must be satisfied for every value of x, it
must be satisfied when x=0. But upon this supposition

B = B'.

In the same manner we can prove that

0=0',
D=D', etc.

379. Whenever we have an equation of the form

M + Nx + Px2 + Qx3 + , etc. = 0,

which is true for every value of x, all the coefficients ofx are equal
to zero.

For, if we transpose all the terms of the equation in the last
article to the left-hand member we shall have

But it has been shown that A = A', B = B', etc.- whence
A— A' = 0, B— B' = 0, etc. If we substitute M for A — A', and
N for B — B', etc., the equation will be

M + Nx + Px24-Qx3 + , etc. = 0.
whence M-0 N = 0 P = 0 etc.

-

Ex. 1. Expand the fraction ^-^ — into an infinite series.

1 — 3x

It is plain that this development is possible, for we may divide
the numerator by the denominator, as explained in Art 373.
Let us, then, assume the identical equation

SERIES. 269

etc.,

where the coefficients A, B, C, D are supposed to be independ
ent of x, but dependent on the known terms of the fraction.

In order to obtain the values of these coefficients, let us clear
this equation of fractions, and we shall have

(E-3D)x4+, etc.

Now, since this is supposed to be an identical equation, the
coefficients of the like powers of x in the two members are
equal each to each.
Therefore A = l.

B-3A = 2, whence B=5;
C-3B = 0, " 0 = 15;
D-3C = 0, " D=45;
E-3D = 0, " E = 135, etc.

Substituting these values of the coefficients in the assumed
series, we obtain

where the coefficient of each term after the second is three
times the coefficient of the preceding term.

380. The method thus exemplified is expressed in the fol
lowing

RULE.

Assume the proposed expression equal to a series of the form
A + Bx+C^2 + , etc. ; clear the equation of fractions, or raise it to
its proper power, and place the coefficients of the like powers of x in
the two members equal each to each. Then find from these equa
tions the values of A, B, C, etc., and substitute these values in the,
assumed development.

Ex. 2. Expand the fraction - into an infinite series.

1 — 2X+X2

Assume - — — - - = A + Bx+Cx2 + Dcc3+Ex4+, etc.

JL • — ^u^L* ~Y~ OC

Clearing of fractions, we have

270 ALGEBRA.

, etc.

Therefore we must have
A«lf
B-2A = 0, whence B = 2

:, z=-=5, etc.

Therefore _ =l + 2as+3s8 + 4a? + 5a4 + , etc.

Ex. 3. Expand the fraction + X- into an infinite series,

l — x— x2

where the coefficient of each term is equal to the sum of the
coefficients of the two preceding terms.

Ex. 4. Expand - — ~X 2 into an infinite series.

J. ~~~" £OC -~~ Ot£

What is the law of the coefficients in this series?

Ex. 5. Expand Vl — x into an infinite series.

x x2 x3 5cc4
'2~8"~l6~~128~~256

x x2 x3 5cc4 7x5
Ans. 1- - -~ etc.

1 — x

Ex. 6. Expand •= - 5 into an infinite series.
1xx2

Ans. 1 — 2x+x2+x3 — 2x4+x5-j-o;6— , etc.
Ex. 7. Expand Va2— x2 into an infinite series.

381. Proper Form of the assumed Series. — In applying the
method of undetermined coefficients to develop algebraic ex
pressions into series, we should determine what power of the
variable will be contained in the first term of the development,
and assume a corresponding series of terms. Generally the
first term of the development is constant, or contains x0' but
the first term of the series may contain x with any exponent
either positive or negative. If the assumed development com
mences with a power of x lower than is necessary, no error will

SEKIES. 271

result, for the coefficients of the redundant terms will reduce
to zero. But if the assumed development commences with a
power of x higher than it should, the fact will be indicated by
an absurdity in one of the resulting equations.

The form of the series which should be adopted in each case
may be determined by putting x=0, and observing the nature
of the result. If in this case the proposed expression becomes
equal to a finite quantity, the first term of the series will not
contain x. If the expression reduces to zero, the first term will

A

contain x ; and if the expression reduces to the form — , then

the first term of the development must contain x with a nega
tive exponent.

Let it be required to develop — 5 into a series.
Assume — - -=A+~Bx+

OX — X

Clearing of fractions, we have

etc.,

whence, according to Art. 378, we obtain 1 = 0, which is ab
surd, and shows that the assumed form is not applicable in the
present case.

Let us, however, assume

etc.

Clearing of fractions, we have

l = 3A+(3B-A)x+(3C-B)x2+(3D-C)#3+, etc.
Therefore 3A = 1, whence A = £;

Substituting these values, we find

/•yi — 1 /y& /-v» /y»2

I i * ' \ ' \ rt-f />

o^ ^2 ~""~Q~"t~'q"i~97r~' ftT"1"' eic*

OX — X O t/ ^i I OX

272 ALGEBRA.

To Resolve, a Fraction into Simpler Fractions.

382. When the denominator of a fraction can be resolved
into factors, the principles now developed enable us to resolve
the fraction itself into two or more simpler fractions, having these
factors for denominators. In such a case, the given fraction is
the sum of the partial fractions.

Ex. 1. Eesolve the fraction -= — into partial fractions.

x2— 5x + 6

We perceive that x2-5x+6 = (x-2)(cc-3).

. 5x-12 A B
Assume -= — = — —^ = -+•

x2—5x+6 x—2 x— 3'
in which the values of A and B are to be determined.
Clearing of fractions, we have

By the principle of Art. 378, A + B=5,
and 3A + 2B = 12.

From which we obtain A =2 and B = 3.

Substituting in the assumed equation, we have

5a?-12 2 3
#2_5x+6~x— 2 a?— 3'

5x-i- 1

Ex. 2. Eesolve — — T into partial fractions.
x2— 1

2
An*.

07+1 X—
£)£, _ ^Q

Ex. 3. Eesolve — — ^ - ^ ^nto Partia^ fractions.

3 2
Ans. -- -H

.

x — o x — o

3x2— 1
Ex. 4. Eesolve — ^ -- into partial fractions.

00 — — OC

Ans. — — r-H

— — - ,

cc-f-1 x— 1 x

K ~ , 2x2— 6x-f 6 ...... .

Ex. 5. Eesolve 7 - —7 — — ^ into partial fractions.

(x-l)(x-'2)(x-3)

1 2 3

Ans. -

-2 x-3°

SERIES.

278

into partial fractions.

Ex. 6. Resolve

Ex. 7. Eesolve

Reversion of Series.

383. ^7ie reversion of a series is the finding the value of the
unknown quantity contained in an infinite series by means of
another series involving the powers of some other quantity.

This may be accomplished by the method of undetermined
coefficients in a mode similar to that employed in Art. 379.

Ex. 1. Given the series y—x+x2-\-x*+, etc., to find the value
of x in terms of y.

Assume x=Ay+Vy* + Cy3 + Dy* + . etc.

Find, by involution, the values of x2, x3, x4, and x5, carrying
each result only to the term containing y5. Then, substituting
these values for x, cc2, x3, etc., in the given equation, we shall have

+2AB
+ A3

__J-2AC

HB

+ 3^

+ A4

+ 2AD

+ 2BC

+ 3AB2
+4A3B

Since this is an identical equation, we place the coefficients
of the like powers of y in the two members equal to each other,
and we obtain

A=+l, B=-l, C=+l, D=-l, E=+l, etc.

Hence we have x=y—y2 + y* — if-}-y5 — ) etc., Ans.

Ex. 2. Given the series y=x— — -f- — -5- + , etc., to find the

A <± O

value of x in terms of y. 2 ?,s ? 4

^4.7Z5. x=7/+^-+4-+^4-, etc.

M ?

274 ALGEBRA.

/y>2 ,y>3 ^4 /Y"

Ex. 3. Given the series y=oc— — -{-— ——-}-— — ^ etc., to find

,4 o 4 o

the value of x in terms of y.

Ans. x=y + 1l + j^ + ^+^f^> + , etc.

Ex. 4. Given the series i/—x-\-x3-\-x5-\-x1 -\-x*-\-, etc., to find
the value of x in terms of y.

Ans. x — y — ?/3 + 2?/5 — 5?/7-fl4?/9 — , etc.

Ex. 5. Given the series 2/=^ + 3x2+5o;3 + 7a?4H-9x5 + , etc.. to
find the value of x in terms of y.

Ans. x=y— 3/+13?/3— 67/4-381?/5— , etc.

384. When the sum of a series is known, we may sometimes
obtain the approximate value of the unknown quantity by re
verting the series.

Ex.1. Given ^+ixH^3 + ^ir^4 + ;r«V<j^5 + , etc.=l, to
find the value of x.

If we call s the sum of the series, and proceed as in the last
article, we shall have

X=2s-s2+fs3--i-s44-fs5-, etc.

Substituting the value of s, we find

03 = 4— -rV+ y-V— imr+Wg-o — , etc., or x = 0.446354 nearly.

Ex. 2. Given 2x + 3x3 + 4x5 + 5x7 + , etc.= J, to find the value

ofx- 5 353 1955 152s7

or a=J-Tf* + T^-TBy»rT+, etc.=:0.2300 nearly.

^3 O" o^

Ex.3. Given .T— — +——— + , etc. =^, to find the value of x.
o o <

).,
or a?=t+A-+Tnrer+ asUo6 + , etc. = .34625 nearly.

/yi2 /y>3 O^ 'T1

Ex. 4. Given x-\- — -\- — -}- — + — -{-^ etc.=-J-, to find the value

Jj O 4: O

ofa;-

,
=s__ + ___ + ___ + i etc.,

Or 33 = "5 - 15 7T ~T~ 7 5 0 '1 A 0001~375000 - 1125000o~i~7

or a = 0.1812692 nearly.

SERIES. 275

Binomial Theorem.

385. In Art. 353 the binomial theorem was demonstrated
for the case in which m is a positive whole number. By means
of the method of undetermined coefficients we can prove that
this formula is true, whether m is positive or negative, entire or
fractional. The demonstration of this theorem depends upon
the following proposition :

(yn Jn

386. The value of - — — , when a— &, is in all cases nan~l,

whether n is positive or negative, integral or fractional.

first. It was shown in Art. 83 that when n is a positive
whole number, an—bn is exactly divisible by a — b, and the

quotient is an-l + an-*b + an-*bz+ +6n~1. The number of

terms in this quotient is equal to n; for b is contained in all the
terms except the first, and the exponents of b are 1, 2, 3, etc.,
to n— 1, so that the number of terms containing b is n— 1, and
the whole number of terms is equal to n. Now, when a—b,
each term of the above quotient becomes a71"1, and, since there
are n terms in the quotient, this quotient reduces to nan~l.

Second. Suppose n to be a positive fraction, or n=—, where

p and q are positive whole numbers.

1 P

Let ai=x, whence aq=xp, and a=xq.

i p

Also, let &=y, whence b^—yP, and b = yQ.

Then, substituting, we have

P P xP-yP

an—bn aq — bi xp—yp x—y
a_& ~~ a— b ~ x<i—yi~ afl—y*

x-y

But p and q are positive integers; therefore, when a =b, and,
consequently, x=y, according to case first, the numerator of
the last fraction becomes pxp~l, and the denominator becomes
qx*-1 ; that is, the fraction reduces to

pxP~l p

r, or —xP-<i.

qx<i-1 q

276 ALGEBRA.

1

Substituting for x its value afl, the fraction reduces to

p £Z« p *_.!

— a v .or -a« .or «a*~*.
2 2

Third. Suppose n to be negative, and either integral or frac«
tional ; or let n——m. Then we shall have

I __ l_

an—bn_a-m—b-m_a™~'b™__ 1 bm—am I am—bm

a—b ~ a—b ' a—b ~ ambm' a—-b ~ ~a"'bm' a — b '
Now, when a—b^ the first factor of the last expression re

duces to -- ^, or — a~2m, and the second factor (by one of the
a

preceding cases) reduces to mam~l. Hence the expression be
comes —ar*mXmaP-'\ or — ma-™-1, QTnan~l.

387. It is required to obtain a general formula expressing the
value of(x + a)m, whether m be positive or negative, integral or

fractional.

(a\ / a\m

1 + -); therefore (x+a)m=xm(l-\--) .

(a\m
1+-) , we have

only to multiply it by xm to obtain that of (x+a)m.

Let -=z; then, to develop (l + z)"1, assume

(l + 2)™^A + Bs+O;2+TV+, etc., (1.)

in which A, B, C, D, etc., are coefficients independent of z, and
we are to determine their values,

Now this equation must be true for any value of z; it must
therefore be true when 2 = 0, in which case A = l.
Substituting this value of A in Eq. (1), it becomes

(1 + 2)*» = 1 + Bz + Cz2 + Dz2 + , etc. (2.)

Since Eq. (2) is to be true for all values of z, let z=n • then
(2) becomes

(l + ?2)m=1 + B7^^-(>^2 + D?^3 + , etc. (3.)

Subtracting (3) from (2), member from member, we have

SERIES. 277

Dividing the first member of (4) by (1 + z) — (1 + n)7 and the
second by its equal z— n, we have

z—n z — n

But when z=n, or 1 + 2 = 1 +n, the first member of equation
(5) becomes m(l + z)711-1.

g2 _ 72,2

Also, — =« + n, when s=w, becomes 2z.

2 — ?1
Z3 — ft3

— — z2+zn-\-n2, when 3=n, becomes 3z2, etc,

£ —"• Tc-

These values substituted in (5) give

ra(l-|-s)m-1=:B+2Cs+3Ds2 + 4Es3 + , etc< (6>j

Multiplying both members of Eq. (6) by 1 + z, we have
m(l + 2)-=B + (2C + B)s + (3D + 2C>2+(4:E + 3D>3 + 3etc. (7.)
If we multiply Eq. (2) by m, we have

m(l + s)m = m + ?7^Bz + mCs2 + mD23^-, etc. (8.)

The first members of Eq. (7) and (8) are equal; hence their
second members are also equal, and we have

-, etc.—

etc. (9.)

This equation is an identical equation ; that is, it is true for
all values of z. Therefore the coefficients of the like powers
of z in the two members are equal each to each, and we have

B=fli.

2C+B=wB, whence C = m^"^;

2t

^-l)(!!!=2)t etc.

^.O

Substituting these values in (2), we have
(l+^ = l + m.+^-^^ + ro(i"-y'-2^ + ,etC. (10.)

If in this equation we restore the value of z, which is ^, we

have

a m(m — l) a2 m(m -!)(???, -2) a3

a;

- . _. _ - + etc.;

xj x 2 ,x2 2.3 3

278 ALGEBRA.

and multiplying both members by xm, we obtain

o-io I <rv) _ ~I 1

(x + a)m = x

-'a' + ,etc, (11.)

which is the general formula for the development of any bino
mial (x + a)m, whatever be the values of x and a, and whether
m be positive or negative, integral or fractional ; and this for
mula is known as the Binomial Theorem of Sir Isaac Newton.

388. When the Series is Finite. — The preceding development
is a series of an infinite number of terms ; but when m is a pos
itive integer, the series will terminate at the (??z + l)th term,
and all the succeeding terms will become zero. For the second
term of Eq. (11) contains the factor m, the third term the factor
m— 1, the fourth term the factor m — 2, and the (m + 2)d term
contains the factor m — m, or 0, which reduces that term to 0;
and since all the succeeding terms also contain the same factor,
they also become 0. There will therefore remain only m-f-1
terms.

When m is not a positive integer, it is evident that no one
of the factors m, m— 1, m — 2, m— 3, etc., can be equal to 0,
so that in that case the development will be an infinite series.

389. Expansion of Binomials with negative integral Exponents
This is effected by substitution in formula (11).

Ex. 1. Expand - - or (a + b}~1 into an infinite series.
a + b

In (11) let ra= — 1, and we find
the coefficient of the second term is — 1,

" " third " is •~1*~2=:+1,

A

" " fourth " is ~

,

o

" fifth " is ~1*~"4=+1, etc.

SERIES. 279

Hence we have

, etc.,
1 _1 IV* &_3

M^~«~«2 a3~«4

which is an infinite series, and the law of the series is obvious.
We might have obtained the same result by the ordinary meth
od of division.

Ex. 2. Expand -f - =— or (a + 6)~2 into an infinite series.
(a+Vf

Ans. a-2-2a-364-3a-462-4a-5£34-5a-664-, etc.,

or -5 -- H — r -- r + ~6~~; e^c->

a2 a3 a4 a5 a6

where the law of the series is obvious.

Ex. 3. Expand -- j or (a—b)-1 into an infinite series.

Ans. a-1 + a-*b + a-*bz a-4b3 + etc.

Ex. 4. Expand -, r— or (a—b)~2 into an infinite series.

(a — b)2

Ex. 5. Expand (a-f&)~3 into an infinite series.

Ans. a~3 — Sa~~4b + 6a~5b2 — Wa~Gb3 + 15a~7^4 — , etc.
Ex. 6. Expand (a — b)-4 into an infinite series.

Ex. 7. Expand (i + 2x)~5 into an infinite series.

Ans. 1 — 10x-}-60x2-280x3+, etc.

390. Expansion of Binomials with positive Fractional Exponents.

Ex. 1. Expand Va-\-b or (a+6)2 into an infinite series.
Represent the coefficients of the different terms by A, B, Q
D, etc. ; then

A= +1,

B= n =+*.

2.4'

3 2.4.6'

?7-3 1.3.5

280 ALGEBRA.

Hence we have

JLJL1JL 1 3 10 5 1 Q P; 7

(7 \ 2 2 -1- ~~ 2 7 -*" *?70 -L . O ~1F7 o i»O»U -- 5~74

a+^a'+- 2*-« 2

+ , etc.

The factors which form the coefficients are kept distinct, in
order to show more clearly the law of the series. The numer-
ators of the coefficients contain the series of odd numbers, 1, 3,
5, 7, etc., while the denominators contain the even numbers,
2, 4, 6, 8, etc.

Ex. 2. Expand (x— a)2 into an infinite series.

Ex. 3. Expand (a2 + x)3 into an infinite series.

, a-^x a-V 3«-5x3 3.5a~V
An*. a+— __-4.____2T_nr_ + , etc.

x x2 Sx3 3.5x4

a + " + -ete-

Ex. 4. Expand (a-f-5)* into an infinite series.

b W 2.563 2.5.8i4

Ex. 5. Expand (a3— 63)8 into an infinite series.

2.5/>9

,

AnS. a 1 1 — — -—

fl 6o g n Q>

3a3 3.6ci6 3.6.9a9

_2_

Ex. 6. Expand (a+^c)8 into an infinite series.
Ex. 7. Expand (a—b)^ into an infinite series.

ij-i A 3^2 3.7£3 3.7.1164

~

2 4.8.12a3~4.8.12.16a4 '
Ex. 8. Expand (1— x}5 into an infinite series.

x 4x2 4.9x3 4..
~5~630~5.10.15~5.10.15.20~' C

J|L

>F
391. Expansion of Binomials ivith negative Fractional Expo

nents.

Ex. 1. Expand - —^ or (a + b) a into an infinite series.

SERIES. 281

The terms without the coefficients are

a~* a~h, a~h*, a~~h3, a~V, etc.
Kepresent the coefficients by A, B, C, D, etc. ; then
A- +1,

B- n = -

n-cl_ 1.3.5
~1T~~274~6'

Hence we obtain

-4 -i 1 -|7 , 1.3 -|79 1.3.5 -$73 , 1.3.5.7 -S
=a -ga b + -a ^-jf^ ^+0^ ~

— , etc.

ft 8&a 3.5Z>3
+

2.4.6a32.4.6.8a4 '*

Ex. 2. Expand (a2— #)~2 into an infinite series.

1 , a; , 1.3a;2 , 1.3. ox3 , 1.3.6.7a^

S- a+2a3 + 2.4a5+2.4.6a7+^4r6.8a9 + ' 6

97?

Ex. 3. Expand . into an infinite series.

Vx2+a4

A m j-t a4 . l-3«8 1.3.5ft12 1.3.5.7a16
?' 1 + +

Ex.4. Expand (a-fx) * into an infinite series.

_i 1 _| 1.4 -f 2 1 . 4 . 7 -L° 3 1.4.7.10 -J.J3 4
~3a 3.6a ~3 . 6 . 9a X 3. 679712° 'C •'

Ex. 5. Expand (a2— x2) * into an infinite series.

1 J-, , x2 , 1.5x4 1.5.9x6 )

AnS' V* \ 1+4^+4^+OTl2^ + ' 6ta f

Ex. 6. Expand (1+x) 5 into an infinite series.
x 6x2 6.11x3 G.ll.i

282 ALGEBKA.

392. Extraction of any Root of a Surd Number. — The approx
imate value of a surd root may be found by the binomial
theorem by dividing the number into two parts, and consider
ing it as a binomial.

Ex. 1. Find the square root of 10.

second "

+ .1666667

third

- .0046296

fourth "

4- .0002572

fifth

- .0000179

sixth "

+ .0000014

seventh "

- .0000001

If, in Ex. 3, Art. 390, we make a2 = 9 and as=l, we shall have

1 -I o QfC Q £i 7

27~3~2.4.33 2.4.6.3b~2. 4 . 6 . 8 .37+2.4.6.8.10. 39~' 6tC*

The value of the first term is 3.0000000

u
u

u
tt
u

Their sum is 3.1622777,

which is the square root of 10 correct to seven decimal places.
Ex. 2. Find the square root of 99.

-/99= VlOO— 1 = (100— 1) .
Substituting in Ex. 3, Art. 390, we have

V99 = 10 — • — — t 5— , etc.

The value of the first term is 10.0000000

" " second " - .0500000

" third " - .0001250

" " fourth " - .0000006

Their sum is 9.9498744,

which is the square root of 99 correct to seven decimal places.

393. The method here exemplified for finding the nth root
of any number is expressed in the following

RULE.

Find, by trial, the nearest integral root (a), and divide the given
number into two parts, one of which is the nth power of (a). Con-

SERIES. 283

sider these two parts as the terms of a binomial, and develop it into
a series by the binomial theorem.

Ex. 3. Find the cube root of 9 to seven decimal places.

1 2 2.5 2.5.8

+ ^~+"«~ll"h' e

= 2.0800838.

Ex. 4. Find the cube root of 31 to seven decimal places.
A qli 4 2.42 2.5.43 2.5.S.44 1

r + 3.27~3.6.272 + 3.6.9.273~3.6.9.12.274+' ')

= 3.1413806.
Ex. 5. Find the fifth root of 30 to seven decimal places.

2 2.4 2.4.9

5.16 5.10.162 5.10.15.163~'e:
= 1.9743506.

^

234 ALGEBKA.

CHAPTEE XX.

LOGARITHMS.

394. The logarithm of a number is the exponent of the power
to which a constant number must be raised in order to be equal
to the proposed number. The constant number is called the
base of the system.

Thus, if a denote any positive number except unity, and
a2=m, then 2 is the exponent of the power to which a must
be raised to equal m; that is, 2 is the logarithm of m in the
system whose base is a. If a*=ra, then x is the logarithm of
m in the system whose base is a.

395. If we suppose a to remain constant while m assumes in
succession every value from zero to infinity, the corresponding
values of x will constitute a system of logarithms.

Since an indefinite number of different values may be attrib
uted to a, it follows that there may be an indefinite number of sys
tems of logarithms. Only two systems, however, have come into
general use, viz., that system whose base is 10, called Briggs's
system, or the common system of logarithms ; and that system
whose base is 2.718 + , called the Naperian system, or hyperbolic
system of logarithms.

Properties of Logarithms in general.

396. TJie logarithm of the product of two or more numbers is
equal to the sum of the logarithms of those numbers.

Let a denote the base of the system ; also, let m and n be
any two numbers, and x and y their logarithms. Then, by the
definition of logarithms, we have

a*=m, (1.)

a.v=n. (2.)

Multiplying together equations (1) and (2) member by mem
ber, we have a*+y=mn.

LOGARITHMS. 285

Therefore, according to the definition of logarithms, x+y is
the logarithm of mn, since it is the exponent of that power of
the base which is equal to mn.

For convenience, we will use log. to denote logarithm, and
we have

x + y = log. mn = log. m + log. n.

Hence we see that if it is required to multiply two or more
numbers together, we have only to take their logarithms from
a table and add them together; then find the number corre
sponding to the resulting logarithm, and it will be the product
required.

397. The logarithm of the quotient of two numbers is equal to
the logarithm of the dividend diminished by that of the divisor.
If we divide Eq. (1) by Eq. (2), member by member, we shall

m

have ax~v=—.

n

Therefore, according to the definition, x — y is the logarithm

777

of — , since it is the exponent of that power of the base a which
is equal to — . That is,

n
x

— y = log. f~-l=~]pg. m— log. n.

Hence we see that if we wish to divide one number by an
other, -we have only to take their logarithms from the table and
subtract the logarithm of the divisor from that of the dividend ;
then find the number corresponding to the resulting logarithm,
and it will be the quotient required.

398. The logarithm of any power of a number is equal to the
logarithm of that number multiplied by the exponent of the power.

If we raise both members of Eq. (1) to any power denoted
by Pi we have apx=mp.

Therefore, according to the definition, px is the logarithm of
mp, since it is the exponent of that power of the base which is
equal to mP. That is,

—p log. m.

286 ALGEBRA.

Therefore, to involve a given number to any power, we
multiply the logarithm of the number by the exponent of the
power; the product is the logarithm of the required power.

399. The logarithm of any root of a number is equal to the log
arithm of that number divided by the index of the root

If we extract the rth root of both members of Eq. (1), we

X

shall have ar= y/ra.

Therefore, according to the definition, - is the logarithm of

l/ra. That is

x , r i — loo;, m
-—log. \/m=— — .

r r

Therefore, to extract any root of a number, we divide the
logarithm of the number by the index of the root ; the quotient
is the logarithm of the required root.

400. The following examples will show the application of
the preceding principles :

Ex. 1. log. (abcd)=log. a + log. &-|-log. c-f-log. d.

Ex. 2. log. f^-J =log. a + log. 6 + log. c— log. d— log. e.

Ex. 3. log. (ambncp) — m log. a + n log. b+p log. c.

(a?n£n\
— — J =ra log. a + w log. l—p log. c.

Ex. 5. log. \/a6=i(log. a+log. b).

Ex. 6. log. y • ^-=£{log. a + 2 log. Z>+4 log. c-5 log. d}.

Ex. 7. log. (a3 \/^) = log. (a^) =^ log. a,
Ex. 8. Iog.(a2-x2)^log.{((z+x)(a-cc)} =log.
Ex. 9. log. Va2 — xz= i log. (a+£c)+ J log. (a

Ex. 10. log. r=i %• 8+* lo& 4-i loS- 6~T log. 2.

6 x v/2

LOGARITHMS. 287

401. In all systems of logarithms, the logarithm of unity is zero.
For in the equation ax—n,

if we make n = l, the corresponding value of x will be 0, since
a°= 1, Art. 75 ; that is, log. 1 = 0.

402. In all systems of logarithms, the logarithm of the base is
unity.

For a*=a;

that is, log. a = l.

Common Logarithms.

403. Since the base of the common system of logarithms is
10, all numbers in this system are to be regarded as powers of
10. Thus, since

10°=1, we have log. 1 = 0;

10! = 10, " log. 10 = 1;

102=100, " log. 100 = 2;

103=1000, " log. 1000 = 3, etc.

From this it appears that in Briggs's system the logarithm
of any number between 1 and 10 is some number between 0
and 1 ; that is, it is a fraction less than unity, and is generally
expressed as a decimal. The logarithm of any number between
10 and 100 is some number between 1 and 2 ; that is, it is equal
to 1 plus a decimal. The logarithm of any number between
100 and 1000 is some number between 2 and 3 ; that is, it is
equal to 2 plus a decimal ; and so on.

404. The same principle may be extended to fractions by
means of negative exponents. Thus, since

10-^-A- or 0.1, we have log. 0.1 = -1;

10-^yio- or 0.01, " log. 0.01 = -2;

10-3=T<nnr or 0.001, " log. 0.001 = -3 ;

10-4=Toiin7 or 0.0001, " log. 0.0001 = -4, etc.
Hence it appears that the logarithm of every number be
tween 1 and 0.1 is some number between 0 and —1, or may be
represented by —1 plus a decimal. The logarithm of every
number between 0.1 and 0.01 is some number between —1 and

288 ALGEBRA.

— 2, or may be represented by —2 plus a decimal. The loga
rithm of every number between 0.01 and 0.001 is some number
between —2 and —3, or may be represented by —3 plus a
decimal, and so on.

405. Hence we see that the logarithms of most numbers must
consist of two parts, an integral part and a decimal part. The
former part is called the characteristic or index of the logarithm.
The characteristic may always be determined by the following

RULE.

The characteristic of the logarithm of any number is equal to
the number of places by which the first significant figure of that
number is removed from the unifs place, and is positive when this
figure is to the left, negative ivhen it is to the right, and zero when
it is in the unit's place.

Thus the characteristic of the logarithm of 397 is +2, and
that of 5673 is +3, while the characteristic of the logarithm
of 0.0046 is -3.

406. The same decimal part is common to the logarithms of
all numbers composed of the same significant figures.

For, since the logarithm of 10 is 1, it follows from Art. 397
that if a number be divided by 10, its logarithm will be dimin
ished by 1, the decimal part remaining unchanged. Thus, if
we denote the decimal part of the logarithm of 3456 by m, we
shall have

log. 3456 =3 + m.
log. 345.6 = 2 + m.
log. 34.56 = 1 + m.
log. 3.456 = 0+w.

log. .3456=— 1+m.
log. .03456= -2 + m.
log. .003456 =-3 + m.
log. .0003456=- 4+ m.

Table of Logarithms.

40?. The table on pages 290, 291, contains the decimal part
of the common logarithm of the series of natural numbers from
100 to 999, carried to four decimal places. Since these num
bers are all decimals, the decimal point is omitted, and the char
acteristic is to be supplied according to the rule in Art. 405,

LOGARITHMS. 289

408. To find the logarithm of any number consisting of not
more than three figures. — Look on one of the pages of the ta
ble, along the left-hand column marked No., for the two left-
hand figures, and the third figure at the head of one of the
other columns. Opposite to the first two figures, and in the
column under the third figure, will be found the decimal part
of its logarithm. To this must be prefixed the characteristic,
according to the rule in Art. 405. Thus

the logarithm of 347 is 2.5403;
871 is 2.9400.

The logarithm of 63, or 63.0, is 1.7993;
" 5, or 5.00, is 0.6900;

" 0.235 is 1.3711.

The minus sign is here placed over the characteristic, to show
that that alone is negative, while the decimal part of the loga
rithm is positive.

409. To find the logarithm of any number containing more
than three figures. — By inspecting the table, we shall find that
within certain limits the differences of logarithms are propor
tional to the differences of their corresponding numbers. Thus

the logarithm of 216 is 2.3345;
" 217 is 2.3365;

" 218 is 2.3385.

Here the difference between the successive logarithms, called
the tabular difference, is constantly 20, corresponding to a differ
ence of unity in the natural numbers. If, then, we suppose the
logarithms to increase at the same rate as their corresponding
numbers (as they do nearly), a difference of 0.1 in the numbers
should correspond to a difference of 2 in the logarithms; a dif
ference of 0.2 in the numbers should correspond to a differ
ence of 4 in the logarithms, etc. Hence

the logarithm of 216.1 must be 2.3347;

216.2 " 2.3349, etc.

In order to facilitate the computation, there is given, on the
right margin of each page, the proportional part for the fourth
figure of the natural number, corresponding to tabular differ-

N

290

ABLE OF COMMON LOGARITHMS.

•'NO.

10
1 1

12

i3
i4
i5
16

17

18

X9

20
21
22
23
24
25
26
27

0

i

2

3

4 | 5

6

7

8

9

PE

I
2
3

4
5
6

7

8

9

i

2

3
4
5
6

7
8

9
i

2

3
4
5
6

8
9

i

2

3

4
5
6

8
9

OPO

43
~

17

22
26

3o
34
39

38

~4
8
ii
i5
J9

23

27
3o
34

33

RTIO

42

T

8
i3
'7

21
25
29

34
38

_37_

4

7
ii
i5
19

22
26

3o
33

32

*ML

4i

PA]

4o

ITS

3(

oooo
o4i4
0792
1 1 39
i46i
1761

2o4l

23o4
2553

oo43
o453
0828
n73
1492
1790
2068
233o
2577

0086
0492

o864
1206
i523
1818
2og5
2355
2601

0128
o53i
o899

I239

i553

i847

2122
2380
2625

OI7O

o569
o934

I27I

1 584
i875
2148
24o5
2648

0212

o6o7
0969
i3o3
i6i4
1903

2I75
2430
2672

0253
o645
ioo4
i335
1 644
i93i

2201

2455
2695

O294
0682

io38
i367
i673
i959

2227
2480
27l8

o334
o7i9

IO72

i399
I7o3
i987

2253

25o4

2742

o374
o755
1 106
i43o

I732
20l4
2279

2529

2765

4

8

12

z6

21
25
29

33

37

36

"7

7
1 1

i4
18

22
25
29
32

3i

4
8

12

16

20

24

28

36

35

~

7
1 1

i4

18

21

25

28

32

3o

i

\
i:
ij

2C
2!
2r

3:

3J

3z
K

u

i-
24

2/

3;

24

2788
3oio

3222

3424

36i7

38o2
3979
4i5o
43:4

2810
3o32
3243
3444
3636
3820
3997
4i66
433o

2833
3o54
3263
3464
3655
3838
4oi4
4i83
4346

2856
3o75
3284
3483
3674
3856
4o3i
4200
4362

2878
3096
33o4
35o2
3692
3874
4o48
4216
4378

2900

3n8
3324

3522

37n

3892
4o65

4232

4393

2923

SiSg

3345
354i
3729
39o9
4082
4249
44o9

2945
3 1 60
3365
356o
3747
3927
4o99
4265
44s5

2967

3i8i
3385
3579
3766
3945
4n6
4281
444o

2989

3201

34o4
3598
3784
3962
4i33
4298
4456

28
29

3o
3i

32

33
34
35
36

37
38
39
4o

4i

42

43
44
45

46"
47
48

r49
5o
5i

5 2

53

54

4472
4624
4771
4914
5o5i
5i85
53i5
544 1
5563

4487
4639
4786
4928
5o65
5i98
5328
5453
5575

45o2
4654
48oo
4g42
5o79

521 I

534o

5465
5587

45i8
4669
48i4
4955
5092

5224

5353
5478
5599

4533
4683
4829
4969
5io5
5237
5366
5490
56n

4548
4698
4843
4983
5i 19
525o
5378
55o2
5623

4564
47i3

4857

4997
5 1 32
5263
5391
55i4
5635

4579
4728
487i
5oi i
5i45
5276
54o3
5527
5647

4594
4742
4886
5024
5i59
5289
54i6
5539
5658

46o9

4?5?
49oo

5o38
5172
53o2
5428
555i
5670

3

7
10

i3

J7
20

23

26
3o

28

3
6
10
i3
16
*9

22
26
29

27

3
6
9

12

16
19

22

25

28
26

3
6
9

12

i5

18

21

24
27

25

(
(
i:
I<
!•

2(
2v
2(

•2i

5682
5798
59n
6021
6128
6232
6335
6435
6532

5694
58o9
5922
6o3i
6i38
6243
6345
6444
6542

57o5
582i
5933
6042
6i49
6253
6355
6454
655i

57i7
5832
5944
6o53
6160
6263"
6365
6464
656i

5729
5843
5955
6064
6i7o
6274
6375
64 74
65 7 1

574o
5855
5966
6o75
6180
6284
6385
6484
658o

5752
5866
5977
6o85
6191
6294
6395
6493
659o

5763
5877
5988
6096
6201
63o4
64o5
65o3
6599

5775

5888

5999
6io7

6212
63i4
64i5
65i3
6609

5786
5899
6010
6117
6222
6325
6425

6522

6618

3
6
8
ii
i4

*7

20

22
25

3
5
8
1 1
i4
16

*9

24

3
5
8

10

i3
16
18

21

23

3
5

8

10

i3

i5
1 8*

20
23,

(

1C
It

\L

l"t

<5

H

6628
6721
6812

6902

699o
7076
7160
7243
7324

6637
673o
6821
691 1
6998
7o84
7i68
y25 1
7332

6646
6739
683o
6920
7oo7

7°93

7177
7259
734o

6656
6749
68 °9
6928
7oi6

7IOI

7i85

7267

7348

6665
6758
6848
6937

7024
7I 10

7I93

7275

7356

6675
6767
68 5 7
6946
7o33
7118
7202
7284
7364

6684
6776
6866
6955
7042
7126
7210

7292
7372

6693
6785
6875
6964
7o5o
7i35

72l8

73oo
738o

6702

6794
6884
69-72
7o59
7i43

7226

73o8
7388

6712
68o3
6893
698i
7067
7i52
7235
73i6
7396

TABLE OF COMMON LOGARITHMS.

No.

o

i

2

3

4

5

6

7

8

9

PR

I

2

3

4
5
6

8
9

i

2

3

4
5
6

8
9

i

2

3
4
5
6

8
9

i

2

3

4
5
6

7
8

Q

OPOl

23
2

5
7
9

i4
16
18

21

18

2

4
5

7

9
ii
i3
i4
16

i3

mo

22
2

4
7
9
1 1
i3
i5
18
20

'7

2

3
5
7
9
10

12
14

i5

12

NAL

21

2

4
6
8
1 1
i3
i5
i?
19

2L

2

3
5
6

8

10

ii
i3
i4

ii

PAI

20
2

4
0
8
10

12

i4
16

18

2L

2

3
5
6
8

9
ii

12

i4

10

tTS

12

2
i

6
8
10
ii
i3
i5
17

2*

i

3
4
6

8
10
1 1
i3

9

55
56

57

58
59
60
61
62
63

?4o4
7482
7559
7634
7709
7782
7853
7924
7993

74l2

749o
7566
•7642
77i6
7789
7860
793i
8000

74i9
7497
7574
764g
7723

7796

7868

7938
8oo7

7427

75o5
7582
7657
773i
78o3
7875
7945
8oi4

7435
75i3
7589
7664
7738
78io
7882

7952

8021

7443

7520

7597

7672

7745
78i8
7889
7959
8028

745i
7528
76o4
7679
7752
7825
7896
7966
8o35

7459
7536

76l2

7686
776o
7832
79°3
7973
8o4i

7466
7543

7619
7694
7767
7839
79io

798°
8o48

7474
755i
7627
7701
7774
7846
79i7

7987
8o55

64
65
66
67
68
69
7°
71
72

73

74
75
76

77

78

79
80
81

"sT

83
84
85
86
87
'88

•89

J?l

91
92
93

94
95
96

97

98

99

8062
8129
8i95
8261
8325
8388
845 1
85i3
8573

"8633"
8692

875i
8808
8865

892I

8976
9o3i
9o85

8o69
8i36
8202
8267
833i
8395
8457
85i9
8579

8o75
8142
8209
8274
8338
84oi
8463
8525
8585

8082
8i49
82i5
8280
8344
84o7
847o
853i
859i

8o89
8i56
8222
8287
835i
84i4
8476
8537
8597

8o96
8162
8228
8293
8357
8420
8482
8543
86o3

8102
8i69
8235
8299
8363
8426
8488
8549
86o9

8io9
8176
8241
83o6
8370
8432
8494
8555
86i5

8116
8182
8248
83i2
8376
8439
85oo
856i
8621

8122

8i89
8254
83i9
8382
8445
85o6
8567
8627

8639
8698
8756
88i4
887i
8927
8982
9o36
9o9o

8645
87o4
8762
8820
8876
8932
8987
9042
9096

865i
87io

8768
8825
8882
8938
8993
9o47

9IOI

8657
87i6
8774
883i
8887
8943
8998
9o53
9io6

8663
8722

8779
8837
8893
8949
9oo4
9o58

9I 12

8669
8727
8785
8842
8899
8954
9oo9
9o63
9n7

8675
8733
879i
8848
89o4
896o
9oi5
9o69

9I22

8681
8739

8797
8854
89io
8965

9020

9o74

9I28

8686
8745
8802
8859
89i5
897i

9025

9°79
9i33

i
3
4
5

7

8

9

10
12

8

I

2

4
5
6

7

8

10

ii

7

i

2

3
4
6

7

8

9

10

C

I
2

3
4
5
6

7
8

9

5

i

2

3
4
5
5
6

8

4

o
i
i

2
2

2

3
3

4

9i38
9i9i
9243
9294
9345
9395
9445
9494
9542

9i43
9i96

9248

9299
935o
94oo
945o
9499
9547

9149
9201
9253
93o4
9355
94o5
9455
95o4
9552

9i54
92o6
9258

9^9
936o

94io
946o

9<?9
9557

9i59

92I2
9263

93i5
9365
94i5
9465
95i3
9562

9i65

92I7

9269

9320

937o

9420

9469
95i8
9566

9i7o

9222

9274

9325
9375
94s5
9474
9533
957i

9i75

9227
9279

933o
938o
943o

9479
9528
9576

9i8o

9232

9284

9335
9385
9435
9484
9533
958i

9i86
9238
9289
934o
939o
944o
9489
9538
9586

i

2
2
3

4
5
6
6
7

i
i

2

3
4
4

5
6
6

i
i

2
2

?

4
4
5
5

i
i

2

2

3
3
4
4
S

959o
9638
9685
973i

9777
9823
9868
9912
9956

9595
9643
9689
9736
9782
9827
9872
9917
9961

96oo
9647
9694
974i
9786
9832
9877

992I

9965

96o5
9652

9699
9745
979i
9836
988i
9926
9969

96o9
9657
9703
975o
9795
984i
9886
993o
9974

96 1/'
966i
97o8
9754
98oo
9845
989o
9934
9978

96i9
9666
97i3
9759
98o5
985o
9894
9939
9983

9624
967i
97i7
9763
98o9
9854

9899
9943

9987

9628
9675

9722

9768
98i4
9859
99o3
9948
999 '

9633
968o

9727
9773
98i8
9863
99o8
9952
99q6

292 ALGEBRA.

ences from 43 to 4. Thus, on page 291, near the top, we see
that when the tabular difference is 20, the corrections for .1,
.2, .3, etc., are 2, 4, 6, etc.

It is obvious that the correction for a figure in the fifth place
of the natural number must be one tenth of the correction for
the same figure if it stood in the fourth place. Such a correc
tion would, however, generally be inappreciable in logarithms
which extend only to four decimal places.

EXAMPLES.

Find the logarithm of 4576. Ans. 3.6605.

13.78. Ans. 1.1392.

" " 1.682. Ans. 0.2258.

.03211. 4ws._2.5066.

" " .4735. Ans. 1.6753.

15983. Ans. 4.2036.

The logarithms here given are only approximate. We can
obtain the exact logarithm of very few numbers ; but by taking
a sufficient number of decimals we can approach as nearly as
we please to the true logarithm.

410. To find the natural number corresponding to any loga
rithm. — Look in the table for the decimal part of the loga
rithm, neglecting the characteristic; and if the decimal is ex
actly found, the first two figures of the corresponding natural
number will be found opposite to it in the column headed No.,
and the third figure will be found at the top of the page. This
number must be made to correspond with the characteristic by
pointing off decimals or annexing ciphers. Thus
the natural number belonging to the logarithm 3.3692 is 2340;
" " " " 1.5378 is 34.5.

If the decimal part of the logarithm is not exactly contained
in the table, look for the nearest less logarithm, and take out
the three figures of the corresponding natural number as be
fore. The additional figure or figures may be obtained by
means of the proportional parts on the margin of the pnge.

Find the number corresponding to the logarithm 3.3685.

LOGARITHMS. 293

The next less logarithm in the table is .3674, and the three
corresponding figures of the natural number are 233. Their
logarithm is less than the one proposed by 11, and the tabular
difference is 18. By referring to the margin of page 291, we
find that, with a difference of 18, the figure corresponding to
the proportional part 11 is 6. Hence, since the characteristic
of the proposed logarithm is 3, the required natural number is
'2336.

EXAMPLES.

1. Find the number corresponding to the logarithm 2.5386.

Ans. 345.6.

2. Find the number corresponding to the logarithm 0.2345.

Ans. 1.716.

3. Find the number corresponding to the logarithm 1.9946.

Ans. 9^.76.

4. Find the number corresponding to the logarithm 1.6478.

Ans. 0.4444.

411. Multiplication ~by Logarithms. — According to Art. 396,
to find the product of two numbers we have the following

RULE.

Add the logarithms of the factors ; the sum will be the logarithm
of the product.

The word sum is here to be understood in its algebraic sense.
The decimal part of a logarithm is invariably positive ; but the
characteristic may be either positive or negative.

Ex. 1. Find the product of 57.98 by 3.12.

The logarithm of 57.98 is 1.7633.

3.12 is 0.4942.

The log. of the product 180.9 is 2.2575.

Ex. 2. Find the product of 0.00563 by 172.5.

The logarithm of 0.00563 is "3.7505.
" 172.5 is 2.2368.

The log. of the product 0.971 is L9873.

Ex. 3. Find the product of 54.32 by 6.543.

Ex. 4. Find the product of 3.854 by 0.5761.

294 ALGEBRA.

412. Division by Logarithms. — According to Art. 397, to find
the quotient of two numbers we have the following

KULE.

From the logarithm of the dividend subtract the logarithm of
the divisor ; the difference will be the logarithm of the quotient.

The word difference is here to be understood in its algebraic
sense ; the decimal part of the logarithm being invariably pos
itive, while the characteristic may be either positive or nega
tive.

Ex. 1. Find the quotient of 888.7 divided by 42.24.

The logarithm of 888.7 is 2.9488.

" 42.24 is 1.6257.

The quotient is 21.04, whose log. is 1.3231.

Ex. 2. Find the quotient of 0.8692 divided by 42.32.

The logarithm of 0.8692 is T.9391.

" 42.32 is 1.6265.

The quotient is 0.02054, whose log. is 2^3126.

Ex. 3. Find the quotient of 380.7 divided by 13.75.
Ex. 4. Find the quotient of 24.93 divided by .0785.

413. Involution by Logarithms. — According to Art. 398, to
involve a number to any power we have the following

RULE.

Multiply the logarithm of the number by the exponent of the
power required.

It should be remembered that what is carried from the deci
mal part of the logarithm is positive, whether the characteris
tic be positive or negative.

Ex. 1. Find the fifth power of 2.846.

The logarithm of 2.846 is 0.4542.

5

The fifth power is 186.65, whose log. is 2.2710.

Ex. 2. Find the cube of .07654.

LOGARITHMS. 295

The logarithm of .07654 is 2^8839.

3

The cube is 0.0004484, whose log. is 46517,

Ex. 3. Find the 20th power of 1.06.
Ex. 4. Find the seventh power of 0.8952.

414. Evolution by Logarithms. — According to Art. 399, to
extract any root of a number we have the following

RULE.

Divide the logarithm of the number by the index of tJie root re
quired.

Ex. 1. Find the cube root of 482.4.

The logarithm of 482.4 is 2.6834.

Dividing by 3, we have 0.8945, which corresponds to 7.843,
which is therefore the root required.

Ex. 2. Find the 100th root of 365. Ans. 1.061.

When the characteristic of the logarithm is negative, and is
not divisible by the given divisor, we may increase the charac
teristic by any number which will make it exactly divisible,
provided we prefix an equal positive number to the decimal
part of the logarithm.

Ex. 3. Find the seventh root of 0.005846.

The logarithm of 0.005846 is "3.7669, which may be written

7+4.7669.

Dividing by 7, we have 1.6810, which is the logarithm </
.4797, which is therefore the root required.

Ex. 4. Find the 10th root of 0.007815.

415. Proportion by Logarithms. — The fourth term of a pro
portion is found by multiplying together the second and third
terms and dividing by the first. Hence, to find the fourth term
of a proportion by logarithms, we have the following

296 ALGEBRA.

RULE.

Add the logarithms of the second and third terms, and from
sum subtract the logarithm of the first term.

Ex. 1. Find a fourth proportional to 72.34, 2.519, and 357.5.

Ans. 12.45.

Ex. 2. Find a fourth proportional to 43.17, 275, and 5.762.
Ex. 3. Find a fourth proportional to 5.745, 781.2, and 54.27.

Exponential Equations.

416. An exponential equation is one in which the unknown
quantity occurs as an exponent. Thus,

ax = b

is an exponential equation, from which, when a and ~b are
known, the value of x may be found. If a — 2 and Z> = 8, the
equation becomes 2^ — 8,

in which the value of x is evidently 3, since 23 — 8.
If a = 16 and Z> = 2, the equation becomes

16*=2,

in which the value of x is evidently J, since 16—2.

417. Solution by Logarithms. — When b is not an exact power
or root of a, the equation is most readily solved by means of
logarithms. Taking the logarithm of each member of the
equation ax — b, we have

x log. a=log. &,

, log. 1)

whence X=Y~ —.

log. a

Ex. 1. Solve the equation 3* =20.

log. 20 1.3010

x = -r- — r~ A^~, =2.727 nearly,
log. 3 .47v 1

Ex. 2. Solve the equation 5* = 12.

(2V
-j =J.

Ex. 4. Solve the equation 10* =7.

i

Ex. 5. Solve the equation 12* =3.

3

Ex. 6. Solve the equation 12* = To

LOGARITHMS. 297

Compound Interest.

418. Interest is money paid for the use of money. When the
interest, as soon as it becomes due, is added to the principal, and
interest is charged upon the whole, it is called compound interest.

419. To find the amount of a given sum in any time at com
pound interest. It is evident that $1.00 at 5 per cent, interest
becomes at the end of the year a principal of $1.05 ; and, since
the amount at the end of each year must be proportioned to
the principal at the beginning of the year, the amount at the
end of two years will be given by the proportion

1.00: 1.05:: 1.05 : (1.05)2.

The sum (1.05)2 must now be considered as the principal,
and the amount at the end of three years will be given by the
proportion

1.00: 1.05 ::(1.05)2:(1.05)3.

In the same manner, we find that the amount of $1.00 for
n years at 5 per cent, compound interest is (1.05)n.

For the same reason, the amount for n years at 6 per cent, is
(1.06)n. It is also evident that the amount of P dollars for a
given time must be P times the amount of one dollar.
Hence, if we put

P to represent the principal,
r the interest of one dollar for one year,
n the number of years for which interest is taken,
A the amount of the given principal for n years,
we shall have A = P (1 + r)n.

This equation contains four quantities, A, P, n, r, any three
of which being given, the fourth may be found. The compu
tation is most readily performed by means of logarithms. Tak
ing the logarithms of both members of the preceding equation
and reducing, we find

log. A=log. P + nxlog. (1 + r),
log. P = log. A — n X log. ( 1 -f r),

, /-, , N log. A — log. P
log.(l+r)=- n - ,

_log. A -log. P

298 ALGEBRA.

Ex. 1. How much would 500 dollars amount to in five years
at 6 per cent, compound interest ?

The log. of 1.06 is 0.0253

6

0.1265

The log. of 500 is 2.6990

The amount is $669.10, whose log. is 2.8255.

Ex. 2. What principal at 6 per cent, compound interest will
amount to 500 dollars in seven years? Ans. $332.60.

Ex. 3. At what rate per cent, must 500 dollars be put out at
compound interest so that it may amount to $680.30 in seven
years ? Ans. 4% per cent.

Ex. 4. In what time will 500 dollars amount to 900 dollars
at 6 per cent, compound interest ? Ans. 10TTT years.

Ex. 5. How much would 400 dollars amount to in nine years
at 5 per cent, compound interest?

Ex. 6. What principal at 5 per cent, compound interest will
amount to 400 dollars in eight years?

Ex. 7. At what rate per cent, must 400 dollars be put out
at compound interest so that it may amount to $620.70 in nine
years ?

Ex. 8. In what time will a sum of money double at 6 per
cent, compound interest?

Ex. 9. In what time will a sum of money double at 5 per
cent, compound interest?

Annuities.

420. An annuity is a sum of money stipulated to be paid an
nually, and to continue for a given number of years, for life,
or forever.

421. To find the amount of an annuity left unpaid for any num*
her of years, allowing compound interest.

Let a denote the annuity, n the number of -years, r the in
terest of one dollar for one year, and A the required amount.
The amount due at the end of the first year is a.
At the end of the second year the amount of the first an-

LOGARITHMS. 299

unity is a(l-f r), and a second payment becomes due; hence
the whole sum due at the end of the second year is a + a(l-f r).

At the end of the third year a third payment a becomes due,
together with the interest on a -{-a (I -\-r) ; hence the whole sum
due at the end of the third year is a+a(l-|-r)-|-a(l-f r)2, or
a{l + (l + r)-f (1 + r)2}, and so on.

Hence the amount due at the end of n years is

These terms form a geometrical progression in which the ratio
is 1+r. Hence, by Art. 332, the sum of the series is

422. To find the present value of an annuity, to continue for a
certain number of years, allowing compound interest.

The present value of the annuity must be such a sum as,
if put out to interest for n years at the rate r, would amount
to the same as the amount of the annuity at the end of that
period.

If P denote the present value of the annuity, then the amount
of the annuity will be P(l-fr)n, which must be equal to

a.

Therefore p=g-(l+r)'- 1

r

Ex. 1. How much will an annuity of 500 dollars amount to
in 15 years at four per cent, compound interest ?

(l+r)» =1.7987

(l+r)n— 1 = .7987, whose log. is L9024

the log. of .04 is 2".6Q21

L3003

the log. of 500 is 2.6990
The amount is $9983, whose log. is 3.9993.

Ex. 2. What is the present value of an annuity of 500 dol-

300 ALGEBKA.

lars to continue for 20 years, interest being allowed at the rate
of four per cent, per annum ?

(l+r)n =2.188

(l;+r)*— 1=1.188, whose log. is 0.0748
the log. of (l+r)n is 0.3400

1.7348
-=12500, whose log. is 4.0969

The present value is $6787, whose log. is 3.8317.

Ex. 3. How much will an annuity of 600 dollars amount to
in 12 years at three per cent, compound interest?

Ex. 4. What is the present value of an annuity of 600 dollars
to continue for 12 years at three per cent, compound interest ?

Ex. 5. In what time will an annuity of 500 dollars amount
to 5000 dollars at 4 per cent, compound interest?

Ans. In 8-f years,

Increase of Population.

423. The natural increase of population in a country is some
times computed in the same way as compound interest. Know
ing the population at two different dates, we compute the rate
of increase by Art. 419, and from this we may compute the pop
ulation at any future time on the supposition of a uniform rate
of increase. Such computations, however, are not very reliable,
for in some countries the population is stationary, and in others
it is decreasing.

Ex. 1. The number of the inhabitants of the United States
in 1790 was 3,930,000, and in 1860 it was 31,445,000. What
was the average increase for every ten years ?

Ans. 34§ per cent.

Ex. 2. Suppose the rate of increase to remain the same for
the next ten years, what would be the number of inhabitants
in 1870? Ans. 42,330,000.

Ex. 3. At the same rate, in what time would the number in
1860 be doubled ? Ans. 23£ years.

Ex. 4. At the same rate, in what time would the number in
1860 be tripled?

LOGARITHMS. 301

To find the Logarithm of any given Number.
424. If m and n denote any two numbers, and x and y their

/y» 1 si i

logarithms, then — — — will be the logarithm of Vmn. For, ac
cording to Art. 396, ax+y—mn, and, taking the square root of

x+y .

each member, we have a 2 —Vmn. Therefore, ^ is the

logarithm of Vmn, since it is the exponent of that power of
the base which is equal to Vmn.

Now, in Briggs's system, the logarithm of 10 is 1, of 100 is

14-2
2, etc. Hence the logarithm of VlO x 100 is ~ - ; that is, the

Zi

logarithm of 31.6228 is 1.5.

1 + 1.5

So, also, the logarithm of VlO x 31.6228 is - - ; that is,

2

the logarithm of 17.7828 is 1.25, and so on for any number of
logarithms.

In this manner were the first logarithmic tables computed ;
but more expeditious methods have since been discovered. It
is found more convenient to express the logarithm of a number
in the form of a series.

425. Logarithms computed ly Series. — The computation of log
arithms by series requires the solution of the equation

in which a is the base of the system, n any number, and x is
the logarithm of that number. In order that a and n may be
expanded into a series by the binomial theorem, we will con
vert them into binomials, and assume a = l-\-b and n = l + m;
then we shall have

where x is the logarithm of 1 + m, to the base 1-f &, or a.
Involving each member to a power denoted by ?/, we have

Expanding both members by the binomial theorem, we have

302 ALGEBKA.

xy(xy— 1)79 xy(xy—

i eto.

. o

Canceling unity from both members and dividing by y, we
have

t eto.)=

i etc.

This equation is true for all values of y ; it will therefore be
true when y=0. Upon this supposition, the equation becomes

,, b2 b3 . . w2 m3

35 (&— g +-3-' etc.)=m — — +- — , etc.,

m2 m3
m- — + -g — ,etc.

whence x=log. (l+m)= -- - — — — .

If we put M=

the last equation becomes

(yyj2 *yyi 3

cc^log. w-log. (l + m) = M(m— — +— — , etc.). (1.)

'We have thus obtained an expression for the logarithm of
the number 1-f-m or n. This expression consists of two fac
tors, viz., the quantity M, which is constant, since it depends
simply upon the base of the system ; and the quantity within
the parenthesis, which depends upon the proposed number.
The constant factor M is called the modulus of the system.

426. To determine the Base of Napier's System. — In Napier's
system of logarithms the modulus is assumed equal to unity.
From this condition the base may be determined. Equation
(1), Art. 425, in this case becomes

m3 m4

x=m-—+— -— +, etc.

LOGARITHMS. 303

Reverting this series, Art. 883, Ex. 3, we obtain

X2 X3 X*

w=a:+¥+0+0^+' etc'

But, by hypothesis, ax — n — l + m; therefore

X2 X3 CC4

If x be taken equal to unity, we have

By taking nine terms of this series, we find

a=2.718282,
which is the base of Napier's system.

427. The logarithm of a number in any system is equal to tJw
modulus of that system multiplied by the Naperian logarithm of the
number.

If we designate Naperian logarithms by Nap. log., and log
arithms in any other system by log., then, since the modulus
of Napier's system is unity, we have

m? m3
log. (l+m) = M(m- +-, etc.),

Nap. log.(l + 7n)=m— -+ — , etc.

/i O

Hence log. (1 + m) = M x Nap. log. (1 -f m),

or M

Nap. log.
where 1-f m may designate any number whatever.

428. To render the Logarithmic Series converging. — The for
mula of Art. 425,

fm /YD

m--+— , etc.), (1.)

can not be employed for the computation of logarithms when
m is greater than unity, because the series does not converge.
This series may, however, be transformed into a converging
series in the following manner :

304 ALGEBRA.

Substitute — m for m, and we shall have

Subtracting Eq. (2) from Eq. (1), observing that log. (l + m]
— log. (1— m) = log. - — , we shall have

J. Tfi

1 + ra ^r/ . m3 . m5 .

Now, since this is true for every value of m, put

1
m — — — r, whence

'

— — ,
1— m p

and the preceding series, by substitution, becomes
log. 2^=Iog.(p+1)_log./)=2M^+__L_+^:_+, et,);

429. This series converges rapidly, and may be employed
for the computation of logarithms in the Naperian or the com
mon systems. It is only necessary to compute the logarithms
of prime numbers directly, since the logarithm of any other
number may be obtained by adding the logarithms of its sev
eral factors. Making £> = !, 2, 4, 6, etc., successively, we obtain
the following

Naperian or Hyperbolic Logarithms.

=°-693147

=1.098612

log. 4 = 2 log. 2 =1.886294

log. 6=log. 3-f log. 2 =1.791769

log. 8 = 3 log. 2 =2.079442

log. 9 = 2 log. 3 =2.197225

log. 10= log. 5 -flog. 2 =2.302585
etc., etc., etc.

LOGARITHMS. 305

430. To construct a Table of Common Logarithms. — In order
to compute logarithms of the common system, we must first
determine the value of the modulus. In Art 427, we found

M= 1°g-(1+m)
Nap. log. (1 + ra)'

If ~L-{-m=a, the base of the system, then log. a=l, and we
have

M=Af— n - 5
Nap. log. a

that is, the modulus of any system is the reciprocal of the Naperian
logarithm of the base of the system.

The base of the common system is 10, whose Naperian log
arithm is 2.302585. Hence

_

which is the modulus of the common system.

We can now compute the common logarithms by multiply
ing the corresponding Naperian logarithms by .434294, Art.
427. In this manner was the table on pages 290-1 computed.

431. Results. — The base of Briggs's system is 10.

" Napier's " 2.71828.

The modulus of Briggs's system is 0.43429.

" Napier's " 1.

Since, in Briggs's system, all numbers are to be regarded as
powers of 10, we have

10°-602=4, etc.

In Napier's system, all numbers are to be regarded as powers
.71828. Thus,

2.7180.603^2,
2.7 IS1-098^ 3,
2.7181-886=4, etc.

306 ALGEBKA.

CHAPTER XXL

GENERAL THEOliY OF EQUATIONS.

432. A cubic equation with one unknown quantity is an
equation in which the highest power of this quantity is of the
third degree, as, for example, x3— -6x2+8x— 15 = 0. All equa
tions of the third degree with one unknown quantity may be
reduced to the form

A biquadratic equation with one unknown quantity is an
equation in which the highest power of this quantity is of the
fourth degree, as, for example, x4 — 6x34-7cc2 + 5x— 4 = 0. Every
equation of the fourth degree with one unknown quantity may
be reduced to the form

The general form of an equation of the fifth degree with one
unknown quantity is

x5 + ax* + bx3 + ex2 + dx -f e = 0 ;

and the general form of an equation of the nth degree with one
unknown quantity is

xn + Axn~l + 'Bxn-*+Cxn-5-{- .... +Tx+Y=±;0. (1.)

This equation will be frequently referred to hereafter by the
name of the general equation of the nth degree, or simply as
Equation (1).

An equation not given in this form may be reduced to it by
transposing all the terms to the first member, arranging them
according to the descending powers of the unknown quantity,
and dividing by the coefficient of the first term. In this equa
tion n is a positive whole number, but the coefficients A, B, C,
etc., may be either positive or negative, entire or fractional,
rational or irrational, real or imaginary. The term "V may be
regarded as the coefficient of a?°, and is called the absolute term
of the equation.

It is obvious that if we could solve this equation we should

GENERAL THEORY OF EQUATIONS. 307

have the solution of every equation that could be proposed.
Unfortunately, no general solution has ever been discovered ;
yet many important properties are known which enable us to
solve any numerical equation.

433. Any expression, either numerical or algebraic, real or
imaginary, which, being substituted for x in Equation (1), will
satisfy it, that is, make the two members equal, is called a root
of the equation.

It is assumed that Eq. (1) has at least one root ; for, since the
first member is equal to zero, it will be so for some value of
,T, either real or imaginary, and this value of x is by definition
a root.

434. If a is a root of the general equation of the nth degree, its
first member can be exactly divided by x — a.

For we may divide the first member by x— a, according to
the usual rule for division, and continue the operation until a
remainder is found which does not contain x. Let Q denote
the quotient, and R the remainder, if there be one. Then we
shall have

^+A:rn-1+Bxw-2_|_.... +To3+V = Q(a?-a) + R (2.)

Now, if a is a root of the proposed equation, it will reduce
the first member of (2) to 0; it will also reduce Q(#— a) to 0;
hence R is also equal to 0. But, by hypothesis, R does no1
contain x; it is therefore equal to 0, whatever value be attrib
uted to x, and, consequently, the first member is exactly di
visible by x— a.

435. If the first member of the general equation of the nth de
gree is exactly divisible by x— a, then a is a root of the equation.

For suppose the division performed, and let Q denote the
quotient; then we shall have

xn+Axn~l + 'Bxr*-2+ .... +Tx+ Y=Q(x— a).

If, in this equation, we make cc = a, the second member re«
duces to 0; consequently the first member reduces to 0; and,
therefore, a is a root of the equation.

308 ALGEBRA.

EXAMPLES.

1. Prove that 1 is a root of the equation

The first member is divisible by as— • 1, and gives x2~-5x-\-6 = Q.

2. Prove that 2 is a root of the equation

x3-x-Q = 0.
The first member is divisible by x— 2, and gives x2 + 2x+ 3 = 0,

3. Prove that 2 is a root of the equation

4. Prove that 4 is a root of the equation

5. Prove that — 1 is a root of the equation

^_38^3 + 210x2+ 538x+ 289=0.

6. Prove that —5 is a root of the equation

x5 + 6x4_10a?3-112x2-207x-110:=0.

7. Prove that 3 is a root of the equation

436. Every equation of the nth degree containing but one un
known quantity has n roots and no more.

Since the equation has at least one root, denote that root by
a; then will the first member be divisible by x— a, and the quo
tient will be of the form

05"-! + A/a5n-2+B'o5w-3+ .... -]_T'ff+ V,
and the given equation may be written under the form

(x—a)(xn-* + A'xn-* + .... + T'o; + Y') = 0. (3.)

Now equation (3) may be satisfied by supposing either of its
factors equal to zero. If the second factor equals zero, we
shall have

xn~l+ AV>-2-j-B'xn-3+ .... + T'o;+ V' = 0. (4.)

Kow equation (4) has at least one root ; denote that root by
I; then will the first member be divisible by x— 6, and equation
(4) can be written under the form

which reduces Eq. (3) to the form of

GENERAL THEORY OF EQUATIONS. 309

By continuing this process, it may be shown that the first
member will ultimately be resolved into n binomial factors of
the form x — a, x—b, x—c, etc. Hence equation (1) may b^
written under the form

(x-a)(x-b)(x-c)(x-d).. . . (x-k)(x-l) = 0. (5.)

This equation may be satisfied by any one of the n values,
x=a, x=bj x=Cj etc., and, consequently, these values are the
roots of the equation.

The equation has no more than n roots, because if we ascribe
to x a value which is not one of the n values a, 6, c, etc., this
value will not cause any one of the factors of Eq. (5) to be zero,
and the product of several factors can not be zero when neither
of the factors is zero.

If both members of Eq. (5) be divided by either of the fac
tors x — a, x—b, etc., it will be reduced to an equation of the
next inferior degree; and if we can depress any equation to a
quadratic, its roots can be determined by methods already ex
plained.

Ex. 1. One root of the equation

is 1 ; what are the other roots ?
Ex. 2. Two roots of the equation

are 1 and 3 ; what are the other roots?
Ex. 3. Two roots of the equation

are 3 and 5 ; what are the other roots? Ans. 2± V3.

Ex. 4. Two roots of the equation

are 2 and 3 ; what are the other roots? — 3± A/5

Ans. - - — • — .

4

Ex. 5. Two roots of the equation

x4-6x3 + 24£-16 = 0
are 2 and —2 ; what are the other roots? Ans. 3± VS.

437. The n r6ots of an equation of the nth degree are not
necessarily all different from each other. Any number, and, in-

310 ALGEBRA. ,

deed, all of them, may be equal. When we s&y that an equation
of the nth degree has n roots, we simply mean that its first
member can be resolved into n binomial factors, equal or un
equal, and each factor contains one root.

Thus the equation x3— 6x2+12x— 8 = 0 can be resolved into
the factors (x-2)(x-2)(x-2) = 0, or (x-2)3 = 0; whence it
appears that the three roots of this equation are 2, 2, 2. But,
in general, the several roots of an equation differ from each
other numerically.

The equation x3 = 8 has apparently but one root, viz., 2, but
by the method of the preceding article we can discover two
other roots. Dividing x3 — 8 by x — 2, we obtain x2-±-2x-\-4: = 0.
Solving this equation, we find x= — 1± V— 3. Thus, the three
roots of the equation x3 — 8 are

2; _l + -v/^3; -1-V^3.

The student should verify the last two values by actual mul
tiplication.

Ex. 1. Find the four roots of the equation x4— 81 = 0.
Ex. 2. Find the six roots of the equation x6— 64=0.

438. The coefficient of the second term in the equation of the nth
degree is equal to the algebraic sum of the roots with their signs
changed.

The coefficient of the third term is equal to the algebraic sum of
the products of all the roots, taken in sets of two.

The coefficient of the fourth term is equal to the algebraic sum of
the products of all the roots, taken in sets of three, with their signs
changed.

The last term is equal to the continued product of all the roots
with their signs changed.

Let a,b,c,d,.... I, represent the roots of an equation of the
nih degree. This equation will accordingly contain the factors
x—a, x—b, etc.; that is, we shall have

(x-a)(x-b}(x-c)(x-d) .... (£c-Z) = 0.

If we perform the multiplication as in Art. 351, we shall
have

GENERAL THEORY OF EQUATIONS.

311

xn— a

-b

— c

-d

etc.

+ bd

etc.

— abd

— acd
-bed

etc.

±(abcd.... l)=

which results are seen to conform to the laws above stated. By
the method employed in Art. 352 it may be proved that if these
laws hold true for the product of n binomial factors, they will
also hold true for the product of ?i + l binomial factors. But
we have found by actual multiplication that these laws are true
for the product of four factors, hence they are true for the
product of five factors. Being true for five, they must be true
for six, and so on for any number of factors.

It will be perceived that these properties include those of
quadratic equations mentioned on pages 203-5.

If the roots are all negative, the signs of all the terms of the
equation will be positive, because all the signs of the factors
of which the equation is composed are positive.

If the roots are all positive, the signs of the terms will be
alternately positive and negative.

If the sum of the positive roots is numerically equal to the
sum of the negative roots, their algebraic sum will be zero;
consequently the coefficient of the second term of the equation
will be zero, and that term will disappear from the equation.
Conversely, if the second term of the equation is wanting, the
sum of the positive roots is numerically equal to the sum of
the negative roots.

Ex. 1. Form the equation whose roots are 1, 2, and 3.

For this purpose we must multiply together the factors
Ta;—l? x — 2, x— 3, and we obtain x3 — 6x2 + llx— 6 = 0.

This example conforms to the rules above given for the co
efficients. Thus the coefficient of the second term is equal to
the sum of all the roots, 1 + 2 + 3, with their signs changed.

The coefficient of the third term is the sum of the products
of the roots taken two and two; thus,

1x2 + 1x3 + 2x3 = 11.

312 ALGEBRA.

The last term is the product of all the roots, 1x2x3, with
their signs changed.

Ex. 2. Form the equation whose roots are 2, 3, 5, and — 6.

Ans. cc4-4x3-29;r2 + lt>6x-180 = 0.

Show how these coefficients conform to the laws above given.
Ex. 3. Form the equation whose roots are

1, 1, 1, -1, and -2.

Ex. 4. Form the equation whose roots are
1, 3, 5, -2, -4, and -6.
Ans. £6 + 3£5-41x4-87x3+400
Ex. 5. Form the equation whose roots are

l±v^2 and 2±V^3.
Ex. 6. Form the equation whose roots are
^l and 2±V3.

439. Since the last term is the continued product of all the
roots of an equation, it must be exactly divisible by each of them.

For example, take the equation x3 — x— 6 = 0. Its roots must
all be divisors of the last term, 6 ; hence, if the equation has a
rational root, it must be one of the numbers 1, 2, 3, or 6, either
positive or negative ; and, by trial, we can easily ascertain
whether either of these numbers will satisfy the equation. We
thus find that +2 is one of the roots, and, by the method of
Art. 436, we find the remaining roots to be — 1± V — 2.

If the last term of an equation vanishes, as in the example
^4 + 2x3 + 3x2 + 6x = 0, the equation is divisible by x— 0, and
consequently 0 is one of its roots. If the last two terms van
ish, then two of its roots are equal to zero.

440. If the coefficients of an equation are whole numbers, and
ike coefficient of its first term unity, the equation can not have a
root which is a rational fraction.

Suppose, if possible, that T is a root of the general equation

of the nth. degree, where j- represents a rational fraction ex

pressed in its lowest terms. Substitute this value for x in the
given equation, and we have

GENERAL THEORY OF EQUATIONS. 313

Multiplying each term by bn~\ and transposing, we obtain

Now, by supposition, a, 6, A, B, C, etc., are whole numbers;
hence the right-hand member of the equation is a whole num
ber.

But, by hypothesis, | is an irreducible fraction ; that is, a
and b contain no common factor. Consequently, an and b will
contain no common factor; that is, ~ is a fraction in its low
est terms. Hence the supposition that the irreducible fraction
7 is a root of the equation leads to this absurdity, that an irre

ducible fraction is equal to a whole number.

This proposition only asserts that every commensurable root
must be an integer. The roots can not be of the form of -|, |-,
•f, etc. The equation may have other roots which are incom
mensurable or imaginary, as 2±V3, 1 ± V — 2.

.

441. Any equation having fractional coefficients can be trans*
formed into another which has alt its coefficients integers, and the
coefficient of its first term unity.

Reduce the equation to the form

Axn + 'Bzn-l + Cxn-*+ ____ -fTcc-fV=0,
in which A> B, C, etc., are all integers, either positive or neg
ative.

Substitute fur x the value x=-r-,

A.

and the equation becomes

which, multiplied by A71"1, becomes

2/n-f By'1-1 + AC?/'1-2 + .... -f A'l-2T?/4- A'^VrrrO.

O

814 ALGEBKA.

in which the coefficients are all integers, and that of the first
term unity.

The substitution of ~ for x is not always the one which leads

to the most simple result ; but when A contains two or more
equal factors, each factor need scarcely ever be repeated more
than once.

^x2 5x 2
Ex. 1. Transform the equation x3 — —-\ — - — -= 0 into an-

A 4 y

other whose coefficients are integers, and that of the first term
unity.

Clearing of fractions, we have

?y
Substituting ^ for x, the transformed equation is

y3 9?/2 45y R _0
6~-~6""~6~- =°J

or ?/3-9?/2+45?/-48 = 0.

Transform the following equations into others whose coeffi
cients are integers, and that of the first term unity.

Ex.2. x3 + 2x2 + |-^0. Ans. ?/3-hl2?/2-f9?/-24 = 0.
Ex.3. x3+°?—+2 = 0.

Ex.4.
Ex.5.
Ex.6.

.

o

442. If in any complete equation involving but one unknown
quantity the signs of the alternate terms be changed, the signs of all
the roots will be changed.

Take the general equation of the nth degree,

xn + Ax"-1 + Bxn-2 + (V*-3+ .... =o. (1.)
in which the signs may follow each other in any order whatever.

GENERAL THEORY OF EQUATIONS. 315

If we change the signs of the alternate terms, we shall have

xn — Axn~l + ~Bxn~* — Gxn~s 4- ____ =0. (2 .)
or, changing the sign of every term of the last equation,

— xn 4- Axn~l — Bxn~2 + Cxn~* — ____ =0. (3. )

Now, substituting +a for x in equation (1) will give the
same result as substituting —a in equation (2), if n be an even
number; or substituting —a in equation (3), if n be an odd
number. If, then, a is a root of equation (1), —a will be a root
of equation (2), and, of course, a root of equation (3), which is
identical with it.

Hence we see that the positive roots may be changed into
negative roots, and the reverse, by simply changing the signs
of the alternate terms ; so that the finding the real roots of any
equation is reduced to finding positive roots only.

This rule assumes that the proposed equation is complete;
that is, that it has all the terms which can occur in an equation
of its degree. If the equation be incomplete, we must intro
duce any missing term with zero for its coefficient.

Ex. 1. The roots of the equation x3 — 2x2—5x-\-Q—0 are 1,
3, and —2 ; what are the roots of the equation

Ex. 2. The roots of the equation x3 — 6x24 lice— 6 = 0 are 1,
2} and 3 ; what are the roots of the equation

Ex. 3. The roots of the equation x4 — 6x3 + 5x2 -f2#— 10 = 0
are —1, +5, 1 + V— 1, and 1— V — 1 ; what are the roots of
the equation x4 -f 6x3 -f 5x2 — 2x — 10 = 0 ?

443. If an equation whose coefficients are all real contains im
aginary roots, the number of these roots must be even.

If an equation whose coefficients are all real has a root of
the form a + bV — 1, then will a — bV — 1 be also a root of the
equation. For, let a + bV — 1 be substituted for x in the equa
tion, the result will consist of a series of terms, of which those
involving only the powers of a and the even powers of bV — I
will be real, and those which involve the odd powers o
will be imaginary.

316 ALGEBKA.

If we denote the sum of the real terras by P, and the sum
of the imaginary terms by Q\/ — 1, the equation becomes
P+QV^7=0.

But, according to Art. 243, this equation can only be true
when we have separately P — 0 and Q = 0.

If we substitute a — bV — l for x in the proposed equation,
the result will differ from the preceding only in the signs of
the odd powers of b V— 1, so that the result will be P— QV^l.

But we have found that P = 0 and Q=0 ; hence P-Q\/37[ ^0.
Therefore a— bV — 1, when substituted for x, satisfies the equa
tion, and, consequently, it is a root of the equation.

It may be proved in a similar manner that if an equation
whose coefficients are all rational, has a root of the form a-\- Vb,
then will a— Vb be also a root of the equation.

Ex.1. One root of the equation x3— 2x-j-4 = 0 is 1 + V — 1 ;
what are the other roots?

Ex. 2. One root of the equation x3 — x2 — 7x -f 15 = 0 is
2+ V— 1 ; what are the other roots?

Ex. 3. One root of the equation x3 — xz + 3x + 5 = 0 is

1 + 2V — 1 , what are the other roots?

Ex. 4. One root of the equation x4 — 4x3 + 4x — 1 — 0 is

2 + A/3 ; what are the other roots ?

Ex. 5. Two roots of the equation

are _l_|_-y/_i and 1— V— 3; what are the other six roots?

444. Any equation involving but one unknown quantity may be
transformed into another whose roots differ from those of the pro
posed equation by any given quantity.

Let it be required to transform the general equation of the
nth degree into another whose roots shall be less than those of
the proposed equation by a constant difference h.

Assume y—x—li, whence x = y+h.

Substituting y-\-h for x in the proposed equation, we have

GENERAL THEORY OF EQUATIONS, 317

Developing the different powers of y + h by the binomial
formula, and arranging according to the powers of y, we have
if+nhyn~l

+ (n-I)Ah
+ B

+ (w-2)B//

+ C

which equation satisfies the proposed condition, since y is less
than x by h. If we assume y = x-{-h, or x=y—h, we shall ob
tain in the same manner an equation whose roots are greater
than those of the given equation by h.

Ex. 1. Find the equation whose roots are greater by 1 than
those of the equation x3-\-Sxz— 4x+l = 0.
We must here substitute y—l in place of x.

Ans. ?/3_7?/-(-7 — 0.

Ex. 2. Find the equation whose roots are less by 1 than those
of the equation x3— 2o?2+3as— 4=0.

Ans. y3 + y2 + 2y—2 = Q.

Ex. 3. Find the equation whose roots are greater by 3 than
those of the equation 'x4 + 9

Ans.

Ex. 4. Find the equation whose roots are less by 2 than those
of the equation 5x4 — 12x3 + 3x2 + 4^-5 = 0.

Ans. 5?/4 + 28?/3 + 51?/2+32?/-l = 0.

Ex. 5. Find the equation whose roots are greater by 2 than
those of the equation x5 + 10x4H-4

Ans.

445. Any complete equation may be transformed into another
wliose second term is wanting.

Since h in the preceding article may be assumed of any
value, we may put nh + A = 0, which will cause the second term

of the general development to disappear. Hence li— - and

A w'

x = y — -. Hence, to transform an equation into another which

?&

wants the second term, substitute for the unknown quantity a new
unknown quantity minus the coefficient of the second term divided
ly the highest exponent of the unknown quantity.

318 ALGEBRA.

Ex.l. Transform the equation x3— 6x2 + 8x — 2 = 0 into an
other whose second term is wanting.

Put #=2/4-2. Ans. y3—4y—2 = 0.

Ex.2. Transform the equation x4—l6x3—6x-}-l6~0 into
another whose second term is wanting.

Put sc=2/+4. Ans. y4-

Ex. 3. Transform the equation

into another whose second term is wanting.

Ans. ?/5-78?/3+412?/2-757?/-h401=0.

Ex.4. Transform the equation x4— 8x3-f5 = 0 into another
whose second term is wanting.

According to Art. 438, when the second term of an equation
is wanting, the sum of the positive roots is numerically equal
to the sum of the negative roots.

446. If two numbers, substituted for tht unknown quantity in an
equation, give results with contrary signs, there must be at least one
real root included between those numbers.

Let us denote the real roots of the general equation of the
nth degree by a, I, c, etc., and suppose them arranged in the
order of their magnitude, a being algebraically the smallest,
that is, nearest to — a ; b the next smallest, and so on. The
equation may be written under the following form,
(x—a)(x— b)(x— c)(x-d) ..... =0.

Now let us suppose x to increase from — a toward 4- a ,
assuming, in succession, every possible value. As long as x is
less than a, every factor of the above expression will be nega
tive, and the entire product will be positive or negative accord
ing as the number of factors is even or odd. When x becomes
equal to a, the whole product becomes equal to 0. But if x be
greater than a and less than 5, the factor x— a will be positive,
while all the other factors will be negative. Hence, when x
changes from a value less than a to a value greater than a and
less than 6, the sign of the whole product changes from -h to —
or from — to -f-. When x becomes equal to b, the product
again becomes zero ; and as x increases from b to c, the factor

GENERAL THEORY OF EQUATIONS. 319

#_& becomes positive, and the sign of the product changes
again from — to -f or from + to — ; and, in general, the prod
uct changes its sign as often as the value of x passes over a real
root of the equation.

Hence, if two numbers substituted for x in an equation give
results with contrary signs, there must be some intermediate
number which reduces the first member to 0, and this number
is a root of the equation.

If the two numbers which give results with contrary signs
differ from each other only by unity, it is plain that we have
found the integral part of a root.

If two numbers, substituted for x in an equation, give results
with like signs, then between these numbers there will either
be no root, or some even number of roots. The last case may
include imaginary roots.

Forifa + 6'/— 1 be a root of the equation, then will a — IV — 1
be also a root. Now

a result which is always positive; that is, the quadratic factor
corresponding to a pair of imaginary roots of an equation whose
coefficients are real, is always positive.

Ex. 1. Find the first figure of one of the roots of the equa
tion xt+xz+x— 100 = 0.

When x=41 the first member of the equation reduces to
— 16; and when 05 =5, it reduces to -f 55. Hence there must
be a root between 4 and 5 ; that is, 4 is the first figure of one
of the roots.

Ex. 2. Find the first figure of one of the roots of the equa
tion cc3-6x2 + 9x-10 = 0.

Ex. 3. Find the first figure of each of the roots of the equa
tion x3— ±x2—

447. In a series of terms, two successive signs constitute a
permanence when the signs are alike, and & variation when they
are unlike. Thus, in the equation x3 — 2x2 — 5#+6 = 0, the signs
of the first two terms constitute a variation, the signs of the
second and third constitute a permanence, and those of the
third and fourth also a variation.

320 ALGEBRA.

Descartes' s Rule of Signs.

448. Every equation must have as many variations of sign as
it has positive roots, and as many permanences of sign as it has
negative ?*oots.

According to Art. 436, the first member of tbe general equa
tion of the Tith degree may be regarded as the prcrduct of n bi
nomial factors of the form x—a,x — b, etc. The above theorem
will then be demonstrated if we prove that the multiplication
of a polynomial by a new factor, x — a, corresponding to a pos
itive root, will introduce at least one variation, and that the mul
tiplication by a factor, x+a, will introduce at least one perma*
nence.

Suppose, for example, that the signs of the terms in the

original polynomial are -M --\ 1 1- , and we have

to multiply the polynomial by a binomial in which the signs
of the terms are -| — . If we write down simply the signs
which occur in the process and in the result, we have

± -

"We perceive that the signs in the upper line of the partial
products must all be the same as in the given polynomial ; but
those in the lower line are all contrary to those of the given
polynomial, and advanced one term toward the right. When
the corresponding terms of the two partial products have dif
ferent signs, the sign of that term in the result will depend
upon the relative magnitude of the two terms, and may be
either -f or — . Such terms have been indicated by the double
sign ± ; and it will be observed that the permanences in the
given polynomial are changed into signs of ambiguity. Hence,
take the ambiguous sign as you will, the permanences in the
final product are not increased by the introduction of the posi
tive root -fa, but the number of signs is increased by one, ancl

GENERAL THEORY OF EQUATIONS.

321

therefore the number of variations must be increased by one.
Hence each factor corresponding to a positive root must intro
duce at least one new variation, so that there must be as many
variations as there are positive roots.

In the same manner we may prove that the multiplication
by a factor, x + a, corresponding to a negative root, must intn>
duce at least one new permanence ; so that there must be as
many permanences as there are negative roots.

If all the roots of an equation are real, the number of posi
tive roots is equal to the number of variations, and the number
of negative roots is equal to the number of permanences. If
the equation is incomplete, we must supply the place of any
deficient term with ±0 before applying the preceding rule.

Ex.1. The equation x5-3x4-5x3 + 15^2+4x-12 = 0 has
five real roots; how many of them are positive?

Ex.2. The equation cc4— 3x3— 15x2 +49^-12 = 0 has four
real roots ; how many of them are negative ?

Ex. 3. The equation

has six real roots; how many of them are positive?

Derived Polynomials.

449. If we take the general equation of the nth degree, and
substitute y + h in place of x, it becomes

Developing the powers of the binomial y+h, and arranging
in the order of the powers of h, we have

A2

y

-i

1.2

The part of this development which is independent of h is
of the same form as the original polynomial, and we will des

O 2

322 ALGEBRA.

ignate it by X. We will denote the coefficient of h by Xu the

iiz
coefficient of -^-^ by X2, etc- The preceding development may

then be written

450. The polynomials X1? X2, etc., are called derived poly
nomials, or simply derivatives. XT is called the first derivative
of X, X2 the second derivative, and so on. X is called the prim
itive polynomial. Each derived polynomial is deduced from the
preceding by multiplying each term by the exponent of the leading
letter in that term, and then diminishing the exponent of the lead
ing letter by unity.

Ex. 1. What are the successive derivatives of

(1st. 3x2-
Ans. \ 2d. 6a-

( 3d. 6.
Ex. 2. What are the successive derivatives of

Ex. 3. What are the successive derivatives of

x5-f3x4 + 2x3-3x2-2x-2?
Ex. 4. What is the first derivative of

+ ____ +Tx+V?

Equal Roots.

451. We have seen, Art. 436, that if a, I, c, etc., are the roots
of the general equation of the nth degree, the equation may be
written

X = (x-a)(x-l)(x-c). . . . (x-k)(x-l) = 0.

When the equation has two roots equal to a, there will be
two factors equal to x—a; that is, the first member will be di
visible by (x—a)2; when there are three roots equal to a, the
first member will be divisible by (x—a)3; and if there are n
roots equal to a, the first member will contain the factor (x— o)n-
The first derivative will contain the factor n(x — a)71"1; that is,

GENERAL THEORY OF EQUATIONS. 323

x— a occurs (n— 1) times as a factor in the first derivative.
The greatest common divisor of the primitive polynomial, and
its first derivative, must therefore contain the factor x—a, re
peated once less than in the primitive polynomial.

Hence, to determine whether an equation has equal roots, we
have the following

RULE.

Find the greatest common divisor between the given polynomial
and its first derivative. If there is no common divisor the equation
has no equal roots. If there is a common divisor, place this equal
to zero, and solve the resulting equation.

Ex. 1. Find the equal roots of the equation

The first derivative is 8xz— 16x-f-21.

The greatest common divisor between this and the given
polynomial is x— 3.

Hence the equation has two roots, each equal to 3.
Ex. 2. Find the equal roots of the equation

Ans. Two roots equal to 5.
Ex. 3. Find the equal roots of the equation

Ans. Two roots equal to 2.
Ex. 4. Find the equal roots of the equation
x*-6xz— 8x-3 = 0.

Ans. Three roots equal to —1.
Ex. 5. Find the equal roots of the equation

Ex. 6. Find the equal roots of the equation

-j- 20x4-16 = 0.

/Sturm's Theorem.

452. The object of Sturm's theorem is to determine the number
of the real roots of an equation, and likewise the situation of
these roots, or their initial figures when the roots are irrational

According to Art. 446, if we suppose x to assume in succes
sion every possible value from — a to + a, and determine the

824 ALGEBRA.

number of times that the first member of the equation changes
its sign, we shall have the number of real roots, and, conse*
quently, the number of imaginary roots in the equation, since
the real and imaginary roots are together equal in number to
the degree of the equation. Sturm's theorem enables us easily
to determine the number of such changes of sign.

453. Sturm's Functions. — Let the first member of the general
equation of the nth degree, after having been freed from its
equal roots, be denoted by X, and let its first derivative be de
noted by Xr We now apply to X and Xx the process of find
ing their greatest common divisor, with this modification, that
we change the sign of each remainder before taking it as a di
visor; that is, divide X by X1? and denote the remainder with
its sign changed by E; also, divide Xa by E, and denote the
remainder with its sign changed by Ep and so on to En, which
will be a numerical remainder independent of x, since, by hy
pothesis, the equation X=0 has no equal roots.

We thus obtain the series of quantities

X, X1? E, Er E2, .... En,

each of which is of a lower degree with respect to x than the
preceding; and the last is altogether independent of x, that is,
does not contain x.

We now substitute for x in the above functions any two
numbers, p and q, of which p is less than q. The substitution
of p will give results either positive or negative. If we only
take account of the signs of the results, we shall obtain a cer
tain number of variations and a certain number of perma
nences.

The substitution of q for x will give a second series of signs,
presenting a certain number of variations and permanences.
The following, then, is the Theorem of Sturm.

454. Jfj in the series of functions X, X15 E, E15 .... En, we
substitute in place of x any two numbers, p and q, either positive
or negative, and note the signs of the results, the difference between
the number of variations of sign when x—p and when x = q is equal

GENERAL THEORY OF EQUATIONS. 325'

to the number of real roots of the equation X = 0 comprised between
p and q.

Let Q, Qj, Q2, . . . . Qn, denote the quotients in the success
ive divisions. Now, since the dividend is equal to the product
of the divisor and quotient plus the remainder, or minus the
remainder with its sign changed, we must have the following
aquations :

X=X1Q-R, (1.)

X1 = RQ, -R>, (2.)

R=R1Q2-RZ, (3.)

Rn_2 = Rn-lQn — Rn- (n — 1).

From these equations we deduce the following conclusions:

455. If, in the series of functions X, Xr R, etc., any number be
substituted for x, two consecutive functions can not reduce to zero at
the same lime.

For, if possible, suppose X1 = 0 and R = 0; then, by Eq. (2),
we shall have R^O. Also, since R=0 and E1 = 0, by Eq. (3)
we must have R2 = 0; and from the next equation R3 — 0, and
so on to the last equation, which will give Rn=0, which is im
possible, since it was shown that this final remainder is inde
pendent of a:, and must therefore remain unchanged for every
value of x.

456. When, by the substitution of any number for x. any one of
these functions becomes zero, the two adjacent functions must have
contrary signs for the same value of x.

For, suppose Rj in Eq. (3) becomes equal to zero, then this
equation will reduce to R= — R2; that is, R and R2 have con
trary signs.

457. If a is a root of the equation X = 0, the signs ofK. and X,
will constitute a variation for a value of x which is a little less than
a, and a permanence for a value of x which is a little greater than a.

Let h denote a positive quantity as small as we please, and
let us substitute a+h for x in the equation X = 0. According

u2<> ALGEBRA.

to Art. 449, the development will be of the form

A2
X + XjA-fX.^ — + other terms involving higher powers of h<

Now, if a is a root of the proposed equation, it must reduce
the polynomial X to zero, and the development becomes

h2

— -j-other terms involving higher powers of A,

2

or 7z(X1 + Xa + , etc.).

Zi

Also, if we substitute a -\- h for x in the first derived polyno
mial, the development will be of the form

Xj + X27i-|- other terms involving higher powers of h.

Now a value may be assigned to h so small that the first
term of each of these developments shall be greater than the
sum of all the subsequent terms. For if h be made indefinitely
small, then will X2/i be indefinitely small in comparison with
Xj, which is finite; and, since the following terms contain
higher powers of h than the first, each will be indefinitely small
in comparison with the preceding term ; and, since the number
of terms is finite, the first term must be greater than the sum
of the subsequent terms. Hence, when li is taken indefinitely
small, the sum of the terms of the two developments must have
the same sign as their first terms,

X^ and Xr

When h is positive, these terms must both have the same
sign ; and when li is negative, they must have contrary signs ;
that is, the signs of the two functions X and Xj constitute a
variation when x — a—li^ and a permanence when x=a+h.

458. Demonstration of Sturm's Tlieorem. — Suppose all the real
roots of the equations

X = 0, X1 = 0, K=:0, R^O, etc.,

to be arranged in a series in the order of magnitude, beginning
with the least. Let^> be less than the least of these roots, and
let it increase continually until it becomes equal to q, which we
suppose to be greater than the greatest of these roots. Now,
so long as p is less than any of the roots, no change of sign will

GENERAL THEORY OF EQUATIONS. 327

occur from the substitution of p for x in any of these functions,
Art. 446. But suppose j> to pass from a number a little small
er to a number a little greater than a root of the equation X = 0,
the sign of X will be changed from + to — or from — to +,
Art. 446. The signs of X and Xx constitute a variation before
the change, and a permanence after the change, Art. 457 ; that
is, there is a variation lost or changed into a permanence.

Again, while p increases from a number a little smaller to a
number a little greater than another root of the equation X=0,
a second variation will be changed into a permanence, and so
on for the other roots of the given equation.

But when p arrives at a root of any of the other functions
X1} R, E1} its substitution for x reduces that polynomial to
zero, and neither the preceding nor succeeding functions can
vanish for the same value of x, Art. 455 ; and these two adja
cent functions have contrary signs, Art. 456. Hence the en
tire number of variations of sign is not affected by the vanish
ing of any function intermediate between X and En, for the
three adjacent functions must reduce to +0— or — 0 + .

Here is one variation, and there will also be one variation
if we supply the place of the 0 with either + or — ; thus,
+ ± — or — ± +.

Thus we have proved that during all the changes of p,
Sturm's functions never lose a variation except when p passes
through a root of the equation X = 0, and they never gain a
variation. Hence the number of variations lost while x in
creases from p to q is equal to the number of the roots of the
equation. X — 0, which lie between _p and q.

Now, since all the real roots must be comprised within the
limits — oc and -f oc , if we substitute these values for x in the
series of functions X, X15 etc., the number of variations lost
will indicate the whole number of real roots. A third suppo
sition that x — 0 will show how many of these roots are posi
tive and how many negative ; and if we wish to determine
smaller limits of the roots, we must try other numbers. It is
generally best in the first instance to make trial of such num
bers as are most convenient in computation, as 1, 2, 10, etc.

328 ALGEBKA.

EXAMPLES.

1. Find the number and situation of the real roots of the
equation x3 — 3x2 — 4x + 13 = 0

Here we have X=x3— 3x2-4^+13, and X1=:3x2-6a7~4.
Dividing X by X17 we find for a remainder — 14x-f 35. Ee-
jecting the factor 7, and changing the sign of the result, we
have E=2x— 5. Multiplying X: by 4, and dividing by E, we
find for a remainder —1. Changing the sign, we have Ex= +1.

Hence we have

If we substitute — oc for x in the polynomial X, the sign of
the result is — ; if we substitute — oc for x in the polynomial
X1? the sign of the result is + ; if we substitute — oc for x in
the expression 2x— 5, the sign of the result is — ; and E1? being
independent of x, will remain + for every value of x, so that,
by supposing x= — oc, we obtain the series of signs

+ +•

Proceeding in the same manner for other assumed values of
a?, we shall obtain the following results :

Assumed Values of a:. Resulting Signs. Variations.

— oc + — + giving 3 variations.

-3 — + — + "3 "

- 2 + + + "2

0 + - - + " 2 "

1 + • + " 2 "

2 + — + " 2 "
2J - 0 + " 1

3 + + + + " 0
+ oc + + + + "0

We perceive that no change of sign in either function occurs
by the substitution for x of any number less than — 3 ; but, in
passing from —3 to —2, the function X changes its sign from
— to +, by which one variation is lost. In passing from 2

GENERAL THEORY OF EQUATIONS. 329

to 2J, the function X again changes its sign, and a second va
riation of sign is lost. Also, in passing from 2-J to 3, the func
tion X again changes its sign, and a third variation is lost; and
there are no further changes of sign arising from the substitu
tion of any number between 3 and + oc.

Hence the given equation has 3 real roots j one situated be
tween —2 and —3, one between 2 and 2-J-, and a third between
2j- and 3. The initial figures of the roots are therefore —2,
+ 2, and +2.

There are three changes of sign of the primitive function,
two of the first derived function, and one of the second derived
function ; but no variation is lost by the change of sign of
either of the derived functions; while every change of sign of
the primitive function occasions a loss of one variation.
• 2. Find the number and situation of the real roots of the
equation x3— 5x2 + 8x— 1 = 0.

Here we have

^=-2295.

When x= — cc, the signs are — -\ , giving 2 variations,

x=+(xj il + + H "1 "

Hence this equation has but one real root, and, consequently,
must have two imaginary roots. Moreover, it is easily proved
that the real root lies between 0 and +1.

3. Find the number and situation of the real roots of the
equation ^4 °^3

Here we have

, or 2x3-3x2-7x+5,
K=17x2-23x-45,

R2=+ 524535.
When x= — oc, the signs are + — + — +, giving 4 variations,

Hence the four roots of this equation are real.

33 0 ALGEBRA.

Substituting different values for x, we find that

when x= — 3, the signs are H ---- 1 ---- h, giving 4 variations,

x=-2, _ + +- + , » 3

05= -1, _ + + __+, « 3

a?= 0, + + - - + , " 2 "

#= + 1, " + - --+, " 2

x=+2, " H ----- h, " 2 "

a=+3, + + + + +, " 0

Hence this equation has one negative root between —2 and
—3, one negative root between 0 and — 1, one positive root
between 2 and 2-J, and another positive root between 2J and 3.

4. Find the number and situation of the real roots of the
equation x3— 7x4-7 = 0.

Ans. Three; viz., one between —3 and. —4, one be
tween 1 and 1-J, and the other between 1-J and 2.

5. Find the number and situation of the real roots of the
equation 2x4 — 20x + 19 = 0.

Ans. Two; viz., one between 1 and 1-J, the other
between \\ and 2.

6. Find the number and situation of the real roots of the
equation x5 + 2x4 + 3^3 + 4x2 + 5x— 20 = 0.

Ans. One, situated between 1 and 2.

7. Find the number and situation of the real roots of the
equation x3 + 3x2 + 5x— 178 = 0.

Ans. One, situated between 4 and 5.

8. Find the number and situation of the real roots of the
equation x4-12x2 + 12x-3 = 0.

Ans. Four; viz., one between —3 and —4, one be
tween 0 and -J, one between % and 1, and the
other between 2 and 3.

9. Find the number and situation of the real roots of the
equation x4-8x3 + 14x2 + 4x-8 = 0.

Ans. Four; viz., one between —1 and 0, one between
0 and +1, one between 2 and 3, and the other
between 5 and 6.

GENERAL THEORY OF EQUATIONS. 331

Solution of Simultaneous Equations of any Degree.

459. One of the most general methods for the elimination
of unknown quantities from a system of equations, depends
upon the principle of the greatest common divisor.

Suppose we have two equations involving x and y. We
first transpose all the terms to one member, so that the equa
tions will be of the form

A=0, B = 0.

We arrange the terms in the order of the powers of a?, and
we will suppose that the polynomial B is not of a higher de
gree than A. We divide A by B, as in the method of finding
the greatest common divisor, Art. 95, and continue the opera
tion as far as possible without introducing fractional quotients
having x in the denominator. Let Q represent the quotient,
and E the remainder ; we shall then have

But, since A and B are each equal to zero, it follows that R
must be equal to zero. If, then, there are certain values of x
and y which render A and B equal to zero, these values should
be the roots of the equations

B = 0, R = 0.

We now divide B by K, and continue the operation as far as
possible without introducing fractional quotients having x in
the denominator. Let R/ denote the remainder after this divi
sion. For the same reason as before, R' must equal zero, and
we thus obtain the two equations

R = 0, R' = 0,

whose roots must satisfy the equations A = 0, B — 0. If we con
tinue to divide each remainder by the succeeding, and suppose
that each remainder is of a lower degree with respect to x than
the divisor, we shall at last obtain a remainder which does
not contain x. Let R" denote this remainder. The equation
R7/ = 0 will furnish the values of ?/, and the equation R' = 0 will
furnish the corresponding values of x.

If we have three equations involving three unknown quan
tities, we commence by reducing them to two equations with

332 ALGEBRA.

two unknown quantities, and subsequently to a single final
equation by a process similar to that above explained.

f 2 I 2 ~l Q i\

Ex. 1. Solve the two equations j ~

Divide the first polynomial by the second, as follows :
x + y— 5

-(y-6)ag-

2?/2— 1%+12, the remainder.
This remainder must be equal to zero ; that is,

whence 2/ — 2 or 3.

When y=2, x = 3;

Ex. 2. Solve the equations | '

[xy+xy2— 12 = 0.

Multiply the first polynomial by y, to make its first term di
visible, and proceed as follows :

12— 3 0# -hi 2 ?/2, the remainder.

Hence 12-30?/ + 12?/2 = 0;

therefore 2/=z2 or -J-.

When = 2x = 2

Ex. 3. Solve the equations | ^y-

( cc3?/ — x3 + a:— 3 = 0.

The first remainder is 3?/ — 3, which, being placed equal to
0, gives y = l, whence a? =3.
Ex. 4. Solve the equations

(3?/2— ?y + l)-?/3 + ?/2— 2?/ = 0,

The remainder after the first division is x— 2y, and after the
second division y2 — y. Hence we conclude
x — 2y = 0, and y2—y — 0.

GENERAL THEORY OF EQUATIONS. 833

Whence we have y=l or 0,

x = 2 or 0.

v - r o.i *!. *•

Ex.5. Solve the equations

The remainder after the first division is

Hence we have 12 (y— l)(a?— 1) = 0, which equation may be
satisfied by supposing y— 1 = 0 or x— 1 = 0.
When a? = l, y—— 1 or 0,

y = l, ar= 2 or 3.

IT fl G i *i r

Ex.6. Solve the equations

The first division gives a remainder cc(y— 2)+2/2— 4, whence
we have

^
and we may have either

y— 2 = 0, or ic

If we divide the first member of the second equation by
jc+^-{-2, we obtain the remainder ?/2— 5?/+6, which also equals
zero; whence y=2 or 3.

When y = 2, x= —4 or 0,

y=3, o;=-5.

334 ALGEBKA.

CHAPTEE XXII.

SOLUTION OF NUMERICAL EQUATIONS OF HIGHER DEGREES.

461. Equations of the third and fourth degrees can some«
times be solved by direct methods ; but these methods are com
plicated, and are of limited application. No general solution
of an equation higher than the fourth degree has yet been dis
covered. To obtain the roots of numerical equations of degrees
higher than the second, we must generally employ tentative
methods, or methods which involve approximation.

462. Commensurable Roots of an Equation. — Any equation
having fractional coefficients can be transformed into another
which has all its coefficients integers, and the coefficient of its
first term unity, Art. 441, and such an equation can not have
a root which is a rational fraction, Art. 440 ; that is, every com
mensurable root of this equation must be an integer. Every
integral root of this equation is a divisor of the last term, Art.
439. Hence, to find the commensurable roots of an equation,
we need only make trial of the integral divisors of the last term.

463. Method of finding the Roots. — In order to discover a con
venient method of finding the roots, we will form the equation
whose roots are 2, 3, 4, and 5. This equation, Art. 436, may
be expressed thus,

(a- 2) (a- 3) (a- 4) (a- 5) = 0.

If we perform the multiplication here indicated, we shall ob
tain £4-14x3 + 71x2-154;r+120 = 0.

"We know that this equation is divisible by x— 5, and we
will perform the operation by an abridged method. Since the
coefficients of the quotient depend simply upon the coefficients
of the divisor and dividend, and not upon the literal parts of
the terms, we may obtain the coefficients of the quotient by op
erating upon the coefficients of the divisor and dividend by the

NUMERICAL EQUATIONS OF HIGHER DEGREES. 335

usual method. To the coefficients thus found the proper let
ters may afterward be annexed. The operation may then be
exhibited as follows :

A B C D ^ r

1-14 + 71-154+120
1- 5

1—5, divisor,

1 — 9 + 26—24, quotient.

- 9 + 71

- 9+45

+ 26-154
+ 26-130

- 24 + 120
- 24 + 120.

Supplying the powers of x, we obtain for a quotient

In applying this method of division, care should be taken to
arrange the terms in the order of the powers of x ; and if the
series of powers of x in the dividend is incomplete, we must
supply the place of the deficient term by a cipher.

The preceding operation may be still further abridged by
performing the successive subtractions mentally, and simply
writing the results. Eepresent the root 5 by r, and the coeffi
cients of the given equation by A, B, C, D, . . . . V.

We first multiply — r by A, and subtract the product from
B; the remainder, — 9, we multiply by — r, and subtract the
product from C; the remainder, +26, we multiply by —r, and
subtract the product from D ; the remainder, —24, we multi
ply by — r, and, subtracting from V, nothing remains. If we
take the root r with a positive sign, we may substitute in the
above process addition for subtraction ; and if we set down only
the successive remainders, the work will be as follows :

A B C D V r

1-14+71-154+120(5

1_ 9 + 26- 24,
and the rule will be

Multiply A by r, and add the product to B ; set down the sum,
multiply it ~by r, and add the product to C ; set down the sum, mul-

336 ALGEBRA.

tiply it by r, and add the product to D, and so on. The final prod
uct should be equal to the last term V, taken with a contrary sign.
The coefficients above obtained are the coefficients of a cubic
equation whose roots are 2, 3, and 4. The polynomial may
therefore be divided by x — 4, and the operation will be as
follows :

1-5+ 6.

These, again, are the coefficients of a quadratic equation
whose roots are 2 and 3. Dividing again by x— 3, we have

1-5 + 6(3
1-2,
which are the coefficients of the binomial factor x — 2.

These three operations of division may be exhibited together
as follows:

1-14+71-154 + 120
1_ 9 + 26- 24
1- 5+ 6
1- 2

5, first divisor.
4, second divisor.
3, third divisor.

464. How to find all the Integral Roots. — The method here ex
plained will enable us to find all the integral roots of an equa
tion. For this purpose, we make trial of different numbers in
succession, all of which must be divisors of the last term of the
equation. If any division leaves a remainder, we reject this di
visor ; if the division leaves no remainder, the divisor employed
is a root of the equation. Thus, by a few trials, all the integral
roots may be easily found.

The labor will often be diminished by first finding positive
and negative limits of the roots, for no number need be tried
which does not fall within these limits.

Ex. 2. Find the seven roots of the equation

We take the coefficients separately, as in the last example,
and try in succession all the divisors of 36, both positive and
negative, rejecting such as leave a remainder. The operation
is as follows:

NUMERICAL EQUATIONS OF HIGHER DEGREES.

337

1 + 1-14-14+49+49-36-36

14-2-12-26 + 23 +

l + 4_ 4-34-45-18

1 + 74.17 + 17+ 6

1 + 6 + 11+ 6

1+5+ 6

1 + 3
Hence the seven roots are

1, 2, 3, -1, -1, -2, -3.
Ex. 3. Find the six roots of the equation

1, first divisor.

2, second divisor.

3, third divisor.
— 1, fourth divisor.
— 1, fifth divisor.
—2, sixth divisor.
—3, seventh divisor,

— 1584=0.

1+ 5-81- 85 + 964+ 780-1584

1+ 6-75-160 + 804 + 1584

1 + 10-35-300-396

1 + 16 + 61+ 66

1 + 14+33

1 + 11
The six roots, therefore, are

1, 4, 6, -2, -3, -11.
Ex. 4. Find the five roots of the equation

1.

4.

6.

- 2.

- 3.

-IL

1 + 6-10-112-207-110

1.

1 + 5-15- 97-110 -2.

1 + 3-21- 55 -5.

1-2-11
Three of the roots, therefore, are

-1, -2, -5.

The two remainining roots may be found by the ordinary
method of quadratic equations. Supplying the letters to the
last coefficients, we have

xz-2x-ll = 0.

Hence cc~l±VT2.

Ex. 5. Find the four roots of the equation

Ans. 1, 2, 3, and 6.

Ex- 6. Find the four roots of the equation

338 ALGEBRA.

Ex. ?t, Find the four roots of tbe equation
x4_55x2-30z+ 504 = 0.
Ex. 8. Find the four roots of the equation

Ex. 9. Find the four roots of the equation
x4-z3-z2-fl9z-42 = 0.
Ex. 10. Find the five roots of the equation

465. Incommensurable Roots. — If a high numerical equation
is found to contain no commensurable roots, or, if after remov
ing the commensurable roots, the depressed equation is still
of a higher degree than the second, we must proceed by ap
proximation to find the incommensurable roots. Different
methods may be employed for this purpose ; but the following
method, which is substantially the same as published by Horner
in 1819, is generally to be preferred.

Find, by Sturm's Theorem, or by trial, Art. 446, the integral
part of a root, and transform the given equation into another
whose roots shall be less than those of the preceding by the
number just found, Art. 444. Find, by Art. 446, the first fig
ure of the root of this equation, which will be the first decimal
figure of the root of the original equation. Transform the last
equation into another whose roots shall be less than those of
the preceding by the figure last found. Find, as before, the
first figure of the root of this equation, which will be the sec
ond decimal figure of the root of the original equation. By
proceeding in this manner from one transformation to another,
we may discover the successive figures of the root, and may
carry the approximation to any degree of accuracy required.

Ex. 1. Find an approximate root of the equation

We have found, page 330, that this equation has but one real
root, and that it lies between 4 and 5. The first figure of the
root therefore is 4. Transform this equation into anotner whose
roots shall be less than those of the proposed equation by 4,
which is done by substituting 2/4-4 for x. "We thus obtain

'

NUMERICAL EQUATIONS OF HIGHER DEGREES. 339

The first figure of the root of this equation is .5. Transform
the last equation into another whose roots shall be less by .5,
which is done by substituting z-f.5 for y. We thus obtain
z3-fl6.5224-92.75z=3.625.

The first figure of the root of this equation is .03. Transform
the last equation into another whose roots shall be less by .03,
which is done by substituting v-\-.OB for z. We thus obtain

The first figure of the root of this equation is .008. Trans
form the last equation into another whose roots shall be less
by .008, and thus proceed for any number of figures required.

486. How the Operation may be abridged. — This method would
be very tedious if we were obliged to deduce the successive
equations from each other by the ordinary method of substitu
tion ; but they may be derived from each other by a simple
law. Thus, let

Ax34-Bx2+Cx=V (1.)

be any cubic equation, and let the first figure of its root be de
noted by r, the second by r', the third by r", and so on.

If we substitute r for x in equation (1), we shall have
Ar3 + Br2-f O = V, nearly,

Whence r=~ _T . ,. (2.)

C + Br-f A?-2

If we put y for the sum of all the figures of the root except
the first, we shall have x=r-\-y ; and, substituting this value
for x in equation (1), we obtain

4- Br2+2Bn/ -fB?/2 V =V;

+ Cr + Cy j
or, arranging according to the powers of y, we have

Let us put Br for the coefficient of y2, C' for the coefficient of
?/, and V7 for the right member of the equation, and we have
A^+By + C'y=V. (3.)

This equation is of the same form as equation (1) ; and, pro
ceeding in the same manner, we shall find

840 ALGEBRA.

V7

where r' is the first figure of the root of equation (3), or the
second figure of the root of equation (1).

Putting z for the sum of all the remaining figures, we have
y=r' + z; and, substituting this value in equation (3), we shall
obtain a new equation of the same form, which may be written

and in the same manner we may proceed with the remaining
figures.

Equation (2) furnishes the value of the first figure of the root ;
equation (4) the second figure, and similar equations would fur
nish the remaining figures. Each of these expressions involves
the unknown quantity which is sought, and might therefore
appear to be useless in practice. When, however, the root has
been found to several decimal places, the value of the terms

Br and Ar2 will be very small compared with C, and r will be

y
very nearly equal to — . We may therefore employ C as an

0

approximate divisor, which will probably furnish a new figure
of the root. Thus, in the last example, all the figures of the
root after the first are found by division.

46-77 =.5,
3.62 -92.75 = .03,

If we multiply the first coefficient A by r, the first figure
of the root, and add the product to the second coefficient, we
shall have

B + Ar. (6.)

If we multiply expression (6) by r, and add the product to
the third coefficient, we shall have

C + Br+Ar2. (7.)

If we multiply expression (7) by r, and subtract the product
from V, we shall have

Y_O-Br2-Ar3,
which is the quantity represented by Vx in equation (3).

NUMERICAL EQUATIONS OF HIGHER DEGREES. 341

If we multiply the first coefficient A by r, and add the prod
uct to expression (6), we shall have

B + 2Ar. (8.)

If we multiply expression (8) by r, and add the product to
expression (7), we shall have

C + 2Br+3Ar2,
which is the coefficient of y in equation (3).

If we multiply the first coefficient A by r, and add the prod
uct to expression (8), we shall have

which is the coefficient of y2 in equation (3).

We have thus obtained the coefficients of the first transformed
equation ; and, by operating in the same manner upon these
coefficients, we shall obtain the coefficients of the second trans
formed equation, and so on ; and the successive figures of the
root are indicated by dividing V by C, Y' by C', Y" by C", etc.

467. The results of the preceding discussion are expressed
in the following

RULE.

Represent the coefficients of the different terms by A, B, C, and
the right-hand member of the equation by Y. Having found r, the
first figure of the root, multiply A by r, and add the product to B.
Set down the sum under B ; multiply this sum by r, and add the
product to C. Set down the sum under C ; multiply it by r, and
subtract the product from Y; the remainder will be the FIRST DIV
IDEND.

Again, multiply A by r, and add the product to the last number
under B. Multiply this sum by r, and add the product to the last
number under C ; this result will be the FIRST TRIAL DIVISOR.

Again, multiply A by r, and add the product to the last number
under B.

Find the second figure of the root by dividing the first dividend
by the first trial divisor, and proceed with this second figure pre
cisely as was done with the first figure, carefully regarding the local
value of the figures.

The second figure of the root obtained by division will fre-

342

ALGEBRA.

qently furnish a result too large to be subtracted from the re
mainder V7, in which case we must assume a different figure.
After the second figure of the root has been obtained, there will
seldom be any further uncertainty of this kind.

It may happen that one of the trial divisors becomes zero.
In this case equation (2) becomes

whence

= ™ Or r — \/ -=:

that is, the next figure of the root will be indicated by dividing
the last dividend by the last number under B, and extracting
the square root of the quotient.

The entire operation for finding a root of the equation

may be exhibited as follows :

A B

1 +3
4
7
4
11
4

C Y r

+5 =178 (4.5388=x.
28 132

33 46= 1st dividend.
44 42.375

77 = 1st divisor.

7.75

3.625 =2d dividend.
2.797377

15.5
.5

84.75
8.00
92.75 =2d divisor.
.4959

.827623 = 3d dividend.
.751003872

16.0
.5

.076619128= 4th dividend

16.53
3
16.56
3

93.2459
.4968

93.7427 = 3d divisor.
.132784

16.598
8

93.875484
.132848

16.606 94.008332 =4th divisor.
Having found one root, we may depress the equation

NUMERICAL EQUATIONS OF HIGHER DEGREES. 343

to a quadratic by dividing it by x— 4.5388. We thus obtain

where x is evidently imaginary, because q is negative and
greater than £ See Art. 280.

After thus obtaining the root to five or six decimal places,
several more figures will be correctly obtained by simply di
viding the last dividend by the last divisor.

Ex. 2. Find all the roots of the equation

The first figure of one of the roots we readily find to be 3.
We then proceed, according to the Rule, to obtain the root to
four decimal places, after which two more will be obtained
correctly by division.
ABC Y r

I + 11 -102 =-181 (3.21312=x.

_3 42 _180

14—60 -1 = 1st dividend.

_3 51 -.992

17 -^9 = 1st divisor. - .008" = 2d dividend.
3 4.04 -.006739

20.2 -4.96 -.001261 = 3d dividend.

_2 4.08 -.001217403

20.4 - 0.88 = 2d divisor. -.000043597 = 4th dividend.
2 .2061

20.61 -.6739
_ 1 .2062

20.62 -.4677 = 3d divisor.
_ 1_ .061899

2O633 -.405801
_ 8 .061908
20.636 — .343893 = 4th divisor.

The two remaining roots may be found in the same way, or
by depressing the original equation to a quadratic. Those

roots are,

3.22952
—17.44265.

344 ALGEBKA.

When a power of x is wanting in the proposed equation, we
must supply its place with a cipher.

Ex. 3. Find all the roots of the cubic equation
33-70;=:-. 7.

The work of the following example is exhibited in an ab
breviated form. Thus, when we multiply A by r, and add the
product to B, we set down simply this result. We do the same
in the next column, thus dispensing with half the number of
lines employed in the preceding example. Moreover, we may
omit the ciphers on the left of the successive dividends, if we
pay proper attention to the local value of the figures. Thus
it will be seen that in the operation for finding each successive
figure of the root, the decimals under B increase one place,
those under C increase two places, and those under Y increase
three places.
1 + 0 .-7 = -7 (1.356895867=*.

1 -6 -6

2 —4= 1st div'r. — 1= 1st dividend.
3.3 -3.01 -.903

3.6 -1.93=2d div'r. -97= 2d dividend.
3.95 -1.7325 86625

4.00 r. 1.5325= 3d div'r. 10375= 3d dividend.
4.056 -1.508164 9048984

4.062 -1.483792 =4th div'r. 1326016= 4th dividend.
4.0688 -1.48053696 1184429568

4.0696 -1.47728128 = 5th div. 141586432 = 5th div'd.
4.07049 -1.4769149359 132922344231

4.07058 - 1,4765485837 = 6th div. 8664087769 = 6th div'd
Having proceeded thus far, four more figures of the root,

5867, are found by dividing the sixth dividend by the sixth di

visor.

We may find the two remaining roots by the same process ;

or, after having obtained one root, we may depress the equation

to a quadratic equation by dividing by #—1.356895867, and
we shall obtain

x2 4- 1.35689586705-5.158833606 = 0.

NUMERICAL EQUATIONS OF HIGHER DEGREES. 345

Solving this equation, we obtain _
x=-. 678447933 ±V5. 619125204.

( -3.048917,

Hence the three roots are ...... < 1.356896,

^r ( 1.692021.

Ex. 4. Find a root of the equation 2#3 + 3x2 = 850.
2+3 +0 =850 (7.0502562208

17 119 833

31 336 = 1st divisor. 17 = 1st dividend.

45.10 338.2550 16.912750

45.20 340.5150= 2d divisor. 87250=2d dividend.

45.3004 340.52406008 68104812016

45.3008 340.533 12024= 3d div. 19145187984= 3d div'd.

45.30130 340.5353853050 17026769265250

45.30140 340.5376503750 =4th d. 2118418718750=4th div.

Dividing the fourth dividend by the fourth divisor, we ob
tain the figures 62208, which make the root correct to the
tenth decimal place.

The two remaining values of x may be easily shown to be
imaginary.

When a negative root is to be found, we change the signs
of the alternate terms of the equation, Art. 442, and proceed
as for a positive root.

Ex. 5. Find a root of the equation 5x3— 6a?2 + 3ic=— 85.

Changing the signs of the alternate terms, it becomes

+6 +3 +85 (2.16139.

16 35 70

26 87 = 1st divisor. 15 = 1st dividend.

36.5 90.65 9.065

37.0 94.35 = 2d divisor. 5.935 = 2d dividend.
37.80 96.6180 6.797080

.38.10 98.9040= 3d divisor. 137920= 3d dividend.
38.405 98.942405 98942405

38.410 98.980815 = 4th divisor. 38977595 = 4th div'd.
38.4165 98.99233995 29697701985

38.4180 99.00386535= 5th div'r. 9279893015 =5th div'd.

P 2

346 ALGEBRA.

Hence one root of the equation

=— 85
is -2.16139.

The same method is applicable to the extraction of the cube
root of numbers.

7': Ex. 6. Let it be required to extract the cube root of 9 ; in
Dther words, it is required to find a root of the equation

#3 = 9.

1+0 +0 =9 (2.0800838.

24 8

4 12 = 1st divisor. 1 = 1st dividend.

6.08 12.4864 .998912

6.16 12.9792 = 2d divisor. 1088 = 2d dividend.
6.24008 12.9796992064 1038375936512

6.24016 12.9801984192 = 3dd. 49624063488= 3d div.
6.240243 12.980217139929 38940651419787

6.240246 12.980235860667 =4thd. 10683412068213 =4thd.

Ex. 7. Find all the roots of the equation

( 1.02804.
Ans. \ 6.57653.
( 7.39543.
Ex. 8. Find all the roots of the equation

( -1.12061.
Ans. •] —3.34730.
( -4.53209.
Ex. 9. Extract the cube root of 48228544.

Ans. 364.

Ex. 10. There are two numbers whose difference is 2, and
whose product, multiplied by their sum, makes 100. What
are those numbers ?

Ex. 11. Find two numbers whose difference is 6, and such
that their sum, multiplied by the difference of their cubes, may
produce 5000.

Ex. 12. There are two numbers whose difference is 4 ; and

NUMERICAL EQUATIONS OF HIGHER DEGREES. 347

the product of this difference, by the sum of their cubes, is
3400. What are the numbers ?

Ex. 13. Several persons form a partnership, and establish a
certain capital, to which each contributes ten times as many
dollars as there are persons in company. They gain 6 plus
the number of partners per cent., and the whole profit is $392.
How many partners were there ?

Ex. 14. There is a number consisting of three digits such
that the sum of the first and second is 9 ; the sum of the first
and third is 12 ; and if the product of the three digits be in
creased by 38 times the first digit, the sum will be 336. Ke=
quired the number.

( 636,

Ans. 1 or 725,

( or 814.

Ex. 15. A company of merchants have a common stock of
$4775, and each contributes to it twenty-five times as many
dollars as there are partners, with which they gain as much
per cent, as there are partners. Now, on dividing the profit,
it is found, after each has received six times as many dollars
as there are persons in the company, that there still remains
$126. Kequired the number of merchants.

^ Ans. 7, 8, or 9.

EQUATIONS OF THE FOURTH AND HIGHER DEGREES.

468. It may be easily shown that the method here employed
for cubic equations is applicable to equations of every degree.
For the fourth degree we shall have one more column of prod
ucts, but the operations are all conducted in the same manner,
as will be seen from the following example.

Ex. 1. Find the four roots of the equation

By Sturm's Theorem, we have found that these roots are all
real; three positive, and one negative.
We then proceed as follows :

348

ALGEBRA.

1-8
-3
+ 2
7
12.2
12.4
12.6
12.83
12.86
12.89

+ 14 +4 =8 (5.2360679.
- 1 - I -5
-f 9 + 44 = 1st divisor. 13 = 1st dividend.
44 53.288 10.6576

46.44 63.072 = 2d div'r. 2.3424 =2d dividend.
48.92 64.626747 1.93880241

51.44 66.193068 = 3ddiv. .40359759 = 3d div'd.
51.8249 66.509117736 .399054706416

52.2107 66.825633024=4thd. 4542883584= 4th d

52.5974

12.926 52.674956
12.932 52.752548
and by division we obtain the four figures 0679.

The other three roots may be found in the same manner.

C- .7320508,

Hence the four roots are

5.2360679.
Ex. 2. Find a root of the equation

x5 + 2^4 + 3x3 + 4x2 + 5x = 20.

"We have found, by Sturm's Theorem, that this equation has
a real root between 1 and 2.
"We then proceed as follows :

1+2

+ 3

+4

+5

+20 (1.125789.

3

6

10

15

15

4

10

20

35= 1st divisor.

~5 = 1st dividend.

5

15

35

38.7171

3.87171

6

21

37.171

42.6585 =2d divisor.

1.12829 =2d dividend.

7.1

21.71

39.414

43.5027

2016

.87005

44032

7.2

22.43

41.730

44.3566

2080, 3d div'r.

.25823

55968, 3d div'd.

7.3

23.16

42.211

008

44.5731

44750625

.22286

5723753125

7.4

23.90

42.695

032

44.7902

83203125, 4th

d. 3536

9873046875,

4th d.

7.5

2

24.05

04

43.182

080

7.5

4

24.20

12

43.304

790125

7.5

f>

24.35

24

43.427

690500

7.5

8

24.50

40

7.6

0,5

24.54

2025

7.6

10

24.58

0075

Dividing the fourth dividend by the fourth divisor, we ob
tain the figures 789,

NUMERICAL EQUATIONS OF HIGHER DEGREES. 349

When we wish to obtain a root correct to a limited number
of places, we may save much of the labor of the operation by
cutting off all figures beyond a certain decimal. Thus if, in the
example above, we cut off all beyond five decimal places in the
successive dividends, and all beyond four decimal places in the
divisors, it will not affect the first six decimal places in the root.
Ex. 3. Find the roots of the equation x4— 12x2 + 12x=3.

f -3.907378,

, j + .443277,

AUS' 1 -f .606018,

1+2.858083.

Ex. 4. Find the roots of the equation

x4-l6x3 + 79x2— U0x=— 58.

f +0.58579,

j +3.35425,

M +3.41421,

[ +8.64575.

Ex. 5. Find the roots of the equation

+ 0.934685,
+ 3.308424,
Ans A +3.824325,
+4.879508,
I +7.053058.
Ex. 6. Required the fourth root of 18339659776.

Ans. 368.
Ex. 7. Required the fifth root of 26286674882643.

Ans. 483.

Ex. 8. There is a number consisting of four digits such that
the sum of the first and second is 9 ; the sum of the first and
third is 10 ; the sum of the first and fourth is 11 ; and if the
product of the four digits be increased by 36 times the product
of the first and third, the sum will be equal to 3024 diminished
by 300 times the first digit. Required the number.

-4 f 6345,

or 7234,

or 9012.

350 ALGEBRA.

469. Newton's Method of Approximation.

Let x3 + ~Bxz + Cx='V be an equation to be solved.
by trial, a number, r, nearly equal to the root sought, and let
r-\-h denote the exact value of the root, so that h is a small
fraction which is to be determined. Substitute r-\-h for x in
the given equation, and there will result a new equation con
taining only h and known quantities. Now, since h is sup
posed to be a small fraction, h2 and h3 will be small compared
with h; and if we reject the terms which contain the second
and third powers of /*, we shall have, approximately,

-3r2-2Br-CT

This correction applied to the assumed root gives a closer
approximation to the value of x. Eepeat the operation with
this corrected value of r, and a second correction will be ob
tained which will give a nearer value of the root ; and, by suc
cessive repetitions, the value of the root may be obtained to
any required degree of accuracy.

The value of h may, however, be found more briefly by ob
serving that the numerator is the first member of the equation
after V has been transposed and x changed to r; and the de
nominator is the first derived function of the numerator with a
negative sign, Art. 450.

EXAMPLES.

1. Find a root of the equation x3+2xz+3x=50.
For the numerator of the value of A, we have

-- 50

Hence h= — =-= — -. -- — .

—8r2—4:r— 3

We find, by trial, that x is nearly equal to 3. If we substi
tute 3 for r, we shall have

Hence ce=2.9 nearly. If we substitute this new value of r,
we shall find the value of A to be 4-. 00228.

NUMERICAL EQUATIONS OF HIGHER DEGREES. 351

Hence x= 2. 90228. If we repeat the operation with this
last value of r, we shall find the value of h to be +.0000034
Hence #=2.9022834.

2. Find a root of the equation x5— 6#=10.

Here h=

Assume r~2, and we obtain

A=-^, or -0.14.

Hence #=1.86 nearly. If we assume r=1.86, we shall find
the value of h to be —.021.

Hence #=1.839 nearly. If we assume r— 1.839, we shall
find the value of A to be +.00001266.

Therefore x= 1.83901266.

3. Find a root of the equation x3— 9#= 10.

Ans. #=3.4494897.

4. Find a root of the equation #3 + 9#2+ 4^=80.

Ans. #=2.4721359.

470. Approximation by Double Position. — Find, by trial, two
numbers, r and r', as near as possible to the true value of x ;
substitute them successively for x in the given equation, and
let E and E' represent the errors which result from these sub
stitutions. We assume that the errors of the results are pro
portional to the errors of the assumed numbers. This suppo
sition is not entirely correct ; but if we employ numbers near
to the true values, the error of this supposition is generally not
very great, and the error becomes less and less the further we
carry the approximation. We have then
E:E'::#-r:#-V.

Whence, Art. 305, E-E' : r'—ri: E : x-r;
that is, As the difference of the errors is to the difference of the two
assumed numbers, so is either error to the correction required in the
corresponding assumed number.

This correction, being added to the assumed number when
it is too small, or subtracted when too great, will give a near
approximation to the true root. This result, and some other

352 ALGEBRA.

number, may now be used as new values of r and r' for obtain
ing a still nearer approximation, and so on.

It is generally most convenient to assume two numbers which
differ only by unity in the last figure on the right, or one of
the values of r already used, together with the approximate
root, may be employed for the two assumed numbers.

This method of approximation is applicable to many equa
tions which can not be solved by either of the preceding
methods.

EXAMPLES.

1. Find one root of the equation x3+x2+x— 100 = 0.
When 4 and 5 are substituted for x in this equation, the re
sults are —16 and +55.

Hence 55 + 16:5—4:: 16:. 22.

Therefore x^=4.22 nearly.

We now assume the two values 4.2 and 4.3, and, substi
tuting them for x in the given equation, we obtain the results
-4.072 and +2.297.

Hence 4.072 + 2.297 : 4.3-4.2 :: 2.297: .036.
Therefore x =4.264 nearly.

Assuming again the two values 4.264 and 4.265, and sub
stituting them for x, we obtain the results —.027552 and
+.036535.

Hence .064087 : .001 : : .027552 : .0004299.
Therefore #=4.2644299 very nearly.

2. Find one root of the equation x3 + 2xz— 23x-70 = 0.

Ans. x=5.13458.

3. Find one root of the equation x4—3x2— 75^—10000 = 0.

Ans. 33=10.2610.

4. Find one root of the equation

x5 + 3x4 + 2o?3-3x2-2x-2 = 0.

Ans. 05=1.059109.

471. The different Roots of Unity. — The equation xn — a would
appear to have but one root, that is, x — \/a; but, by Art. 436,
it must have n roots; that is, the nth root of a must have n dif
ferent values. Unity must therefore have two square roots,

NUMERICAL EQUATIONS OF HIGHER DEGREES. 353

three cube roots, four fourth roots, five fifth roots, six sixth
roots, and so on.

Ex. 1. Find the two roots of the equation x2 — \.
Extracting the square root, we find x— -f-1 or —1.
Ex. 2. Find the three roots of the equation x3 = 1.
Since one root of this equation is x=~L, the proposed equa
tion must be divisible by x— 1 ; and dividing, we obtain

cc2+x-f 1 = 0.
Now the roots of this equation are

x=—
Hence the required roots are

+ 1, i(-l + v^3), and i(-l_Z
which are -the cube roots of unity ; and these results may be
easily verified.

Ex. 3. Find the four roots of the equation x* — I.
The square root of this equation is

x2=-f 1, or =-1.
Hence the required roots are

+ 1, -1, +V^1, -V^I.
Ex. 4. Find the five roots of the equation cc5 = l.
Since one root of this equation is a? = l, the proposed equa
tion must be divisible by x— 1 ; and dividing, we obtain

Dividing again by #2, we have

Now put v=x-\ — , (2.)

Ju

whence vz—x'2-\-2-\ —

x2

which, being substituted in equation (1), gives

v* + v-l = Q.
This equation, solved by the usual method, gives

354 ALGEBKA.

Now equation (2) gives

a?2— vx=— 1.

Whence x=%[v+ Vv2— 4], and x=^[v—Vv2—^]9
from which, by substituting the value of v, we obtain

and

Hence the fifth roots of unity are
1.

Ex. 5. Find the six roots of the equation cc6 = l.
These are found by taking the square roots of the cube roots
Hence we have

+ 1, -1, i±iV^3, _i±4/^3.
Ex. 6. Find the four roots of the equation x*=— 1, or

The first member may be made a complete square by adding
2z2; that is, x4 + 2^2 + 1 = 2z2,

whence #2 + 1 = ± x \/2 .

By transposition and completing the square,

Hence
that is,
or

These four values, together with the four values found in
Ex. 3, are the eight roots of the equation

EXAMPLES FOR PRACTICE. 355

EXAMPLES FOR PEACTICE.

EQUATIONS OF THE FIRST DEGREE WITH ONE UNKNOWN-
QUANTITY.

Ex.1. Given 12i+3z-6-y=^-5f, to find the value
Of x. Ans. x=I39%.

Ex.2. Given a(2x+l$b-10a)=b(x + 7b), to find tbe valu8
of x. Ans. x—6a—7b.

Ex. 3. Given 2-^^=1-^^, to find the value of a.

Ans. x—4:^.

/y\ fY*

Ex. 4. Given m-\ — =n—p—T, to find the value of x.

a

T
o

(n—p—m)ab
Ans. x=±— -j-2 — .
-f a + b

2^-3 4*-9 8x-27 16x-81 9 ,

Ex.6. Given ^g -- ^0 -- 3^" "2T~ "40' U
the value of x. Ans. x—Q.

a4 — 54

Ex. 6. Given a3 + ^ + ^2H-^3 = - — , to find tne value of ^

cc

Ans. x — a—b.

Ex.7. Given z^l_^+l3=26+^, to find the value

. x — a—l.

Ex. 8. Given ^-^+^-^=-15, to find the value of x.
5 10 4 o

Ans. x=

Ex.9. Given llix=^+66|-5a?-91, to find the value

o

of x.

Ex 10 Given {5a+^-5 = -, to find the value of x.
x x

3a-6

Ans. x = — - — .
4

356 ALGEBRA.

Ex. 11. Given c=a+^~ ^, to find the value of x.

a(m— 3c+3a)
Ans. x——^— —'-.

c—a-\-m

Ex. 12. Given — - --- - — = — — - to find the value of x.
o o 44

Ans. as=-f.

Ex. 13. Given (m— x)(n—x) — (p-\- x)(x—q), to find the val
ue of x.

mn-4- pq

Ans. x=- - -^-^ — .
m+n+p— q

fio"_99

Ex.14. Given 8x-28=(4x-f 21)^ £f, to find the value

ox -f- 14

Of #. .

Ex. 15. Given x=a+-|+^, to find the value of #.

Ans. g=

.

—cf

/Y» /y^y»

Ex.16. Given--l- i+3«J = 0, to find the value of a:.

c— art

Ex. 17. Given (8-3x)2+(4-4x)2^(9-5x)2, to find the val
ue of x. Ans. x=TV

„ 16x+7 05-16 2^ + 1

Ex.18. Given — — -- \- = — ^ — , to find tbe value

£i~t Lit - \)X O

of x. Ans. a3=17.

Ex.19. Given ^^-^^=11-1^, to find the value

Of 2. ^.725.05=7.

-, OA r. l-2x 5-605 8 l-3x2

Ex. 20. Gwen --__— . ^ to find the

value of x. -^ Ans. x— f.

EXAMPLES FOR PRACTICE. 357

PROBLEMS INVOLVING EQUATIONS OF THE FIRST DEGREE WITH
ONE UNKNOWN QUANTITY.

Prob. 1. Said an old miser, For 50 years I have saved 200
dollars annually ; and for many years, each of my four sons has
saved annually the same sum, viz., the oldest for 27 years past,
the second since 24 years, the third since 19, and the fourth
since 16 years. How long since the savings of the four sons
amounted in the aggregate to as much as those of the father?

Ans. 12 years.

Prob. 2. From four towns, A, B, C, D, lying along the same
road, four persons start in the stage-coach for the same place,
E. The distance from A to B is 19 miles, from B to C 3 miles,
and from C to D 5 miles. It subsequently appeared that the
person who started from A paid as much fare as the three oth
er persons together ; and the fare per mile was the same for
each. It is required to determine the distance from D to E.

Ans. 7 miles.

Prob. 3. Five towns, A, B, C, D, E, are situated along the
same highway. The distance from A to B is 37 miles, from B
to D 34, and from D to E 14 miles. A merchant at C, situ
ated between A and D, receives at one time 8 tons of goods
from A, and 6 tons from B. At another time he receives 11
tons from D, and 9 from E, and in the latter case he paid the
same amount for freight as in the former, the rate of transporta
tion being the same in both cases. It is required to compute
the distance from B to C. "t"~ Ans. 15 miles.

Prob. 4. If 20 quarts of water flow into a reservoir every 3
minutes, after a certain time it will still lack 40 quarts of being
full. But if 52 quarts flow into it every 5 minutes during the
same period, 72 quarts of water will have overflown. What is
the capacity of the reservoir, and how many quarts of water
must flow into it every minute in order that it may be just filled
in the time before mentioned?

Ans. The capacity of the reservoir is 240 quarts, and
8 quarts must flow into it every minute.

Prob. 5. A mason, by working 10 hours daily, could com-

358 ALGEBRA.

plete in a week as much over 888 cubic feet of wall as at pres
ent he completes less than 888 cubic feet, working only 8-J
hours daily. How many cubic feet of wall does he now com
plete weekly ? Ans. 816 cubic feet.

Prob. 6. After a certain time I have $670 to pay, and 4J
months later I have $980 to pay. I settle both bills at once,
at 4f per cent, discount, for $1594.41. When did the first sum
become due? Ans. After 5£ months.

Prob. 7. A merchant gains 8 per cent, when he sells a hogs
head of oil at 36 dollars. How much per cent, does he gain or
lose when he sells a hogshead at 32 dollars?

Ans. He loses 4 per cent.

Prob. 8. A merchant loses 2J per cent, when he sells a bag
of coffee for 39 dollars. How much per cent, does he gain or
lose when he sells a bag of coffee for 41 J- dollars ?

Ans. He gains 3f per cent.

Prob. 9. A merchant owes $2007, to be paid after 5 months,
$3395 after 7 months, and $6740 after 13 months. When
should the entire sum of $12,142 be paid, so that neither party
may sustain any loss? Ans. After 10 months.

Prob. 10. A merchant has three sums of money to pay, viz.,
$1013 after 3J months, $431 four months later, and the third
sum still four months later. How large is the third sum, sup
posing he could pay the three bills together in 6J months with
out loss or gain ? V- Ans. $428.

Prob. 11. A merchant has two kinds of tobacco ; the one cost
40 cents per pound, the other 24 cents. He wishes to mix the
two kinds together, so that he may sell it at 34 cents per pound
without loss or gain. How much must he take of each sort in
order to have 64 pounds of the mixture?

Ans. 40 pounds of the better sort, and 24 pounds of
the poorer.

Prob. 12. A vinegar dealer wishes to dilute his vinegar with
water. At present he sells his vinegar at 6 dollars per hogs
head (120 quarts). How much water must he add to 29^
hogsheads in order to be able to sell the mixture at 4 cents per
quart? Ans. 7f hogsheads.

EXAMPLES FOR PRACTICE. 359

Prob. 13. A metallic compound consists of 4 parts copper
and 3 parts silver. How much copper must be added to 94J
pounds of the compound, in order that the proportions may be
7 parts of copper to 2 parts of silver ?± Ans. 87f pounds.

Prob. 14. In 255 pounds of spirit of wine, water and pure al
cohol are combined by weight in the ratio of 2 to 3. How
much water must be extracted by distillation, in order that the
ratio of the water to the alcohol may be 3 to 17 by weight?

Ans. 75 pounds.

Prob. 15. It is required to diminish each of the factors of the
two unequal products, 52x45 and 66x37, by the same num
ber, so that the new products may be equal to each other.
What is that number? Ans. 17.

Prob. 16. The square of a certain number is 1188 greater than
the square of a number smaller by 6 than the former. What
is that number? Ans. 102.

Prob. 17. I have a certain number of dollars in my posses
sion, which I undertook to arrange in the form of a square, and
found that I wanted 25 dollars to complete the square ; but if
I diminish each side of the square by 2, there remain 31 dollars
over. How many dollars have I? Ans. 200.

Prob. 18. A vine-tiller has a rectangular garden, whose
length is to its breadth as 7 to 5, which he wishes to plant with
vines. If he sets the plants at a certain uniform distance from
each other, he finds that he has 2832 plants remaining. But
if he places them nearer together, so as to make 14 more on
each longer side, and 10 more on each shorter side, he has only
172 plants remaining. How many plants has he?

^ Ans. 14,172.

Prob. 19. In the composition of a certain quantity of gun
powder, the nitre was ten pounds more than two thirds of tho
whole ; the sulphur was four and a half pounds less than one
sixth of the whole ; and the charcoal was two pounds less than
one seventh of the nitre. How many pounds of gunpowder
were there ? Ans. 69 pounds.

Prob. 20. There are three numbers in the ratio of 3, 4, and
5. Five times the first number, together with four times the

362

ALGEBRA.

Ex. 11.

-. .

~~ ' T—

a o

X Z

-+-=4
a c

(5x— 6^ + 42-15
Ex.12. ^7x + 4y-32-19

fcc

-21—

Ex.13. ^ * = 9-y

Ans. =2&.

— 4.

Ex.14, 1 Sx— 5?/ + 72— 75V
( 9x — 112 + 10 — 0 )

Ans. \y=—5.
(2=6.

( 3x — 5?/ + 42 — 5 }
Ex.15, -j lx + 2y— 32 — 2 t

^J«-2'

(4x + 3?/— 2 — 7 )

(z=S.

Ex.16. •< 2x + 3^—42—20 V

An,.\y=i

v ox — ^y + O2^^^o /

( 2 — 2.

f t-r o ~f ^

f 7^-%=n

Ex.l7.-| £l-ll-lt

j y—^-

Ans'\ 2-16.

[t*=25.

J~

fx-12

T? 10 j 2x + 3y— 39 f

iilX. lO. -< ,; f- -.-if

5x — 72— 11

Ans-\ Si

[4?/+32— 41 J

[w=4.

T

Ex.19. j!;^2x-30 f

1 |

^|s

1 w-9.

EXAMPLES FOR PRACTICE. 363

=105] fa?=30.

20 =317 An*

'20' u -741 f ^

J

- 2.

Ex.21, -j lOy— 3a + 3w-2v = 2 Ans. z= 3.

2y-2x=3 u=—l.

PROBLEMS INVOLVING EQUATIONS OF THE FIRST DEGREE WITH
SEVERAL UNKNOWN QUANTITIES.

Prob. 1. Two sums of money, which were put out at inter
est, the one at 5 per cent, the other at 4J per cent., yielded in
one year $284.40 interest. If the former sum had been put out
at 4f per cent., and the latter at 5 per cent., they would have
yielded $4.50 less interest. What were the two sums of money?
Ans. One was $3420, the other $2520.

Prob. 2. There is a number consisting of two digits; the
number is equal to three times the sum of its digits, and if it be
multiplied by three, the result will be equal to the square of the
sum of its digits. Find the number. ^ Ans. 27.

Prob. 3. A merchant sold two bales of goods for the sum of
$987-|, the first at a loss of 8f per cent, the second at a loss of
11J per cent If he had sold the first at a loss of 11 J per cent,
and the second at a loss of 8f- per cent., he would have received
the sum of $992f. How much did each bale cost?

Ans. The first $455, the second $645.

Prob. 4. Two messengers, A and B, from two towns distant
57-J miles from each other, set out to meet each other. If A
starts 5f hours earlier than B, they will meet in 6J- hours after
B starts; but if B starts 5f hours earlier than A, they will meet
in 5f hours after A starts. How many miles does each travel
in an hour? -K Ans. A 3 miles, and B 3^ miles.

Prob. 5. A jeweler has two masses of gold of different de
grees of fineness. If he mixes 10 ounces of the one with 5
ounces of the other, he obtains gold 11 carats fine; but if he

364 ALGEBRA.

mixes 7-J- ounces of the former with 1J- ounces of the latter, he
obtains a mixture 10 carats fine. What was the fineness of
each mass ? Ans. The one 9 carats, the other 15 carats.

Prob. 6. A farmer has a certain number of oxen, and proven
der for a certain number of days. If he sells 75 oxen, his prov
ender will last 20 days longer ; but if he buys 100 more oxen,
his provender will be exhausted 15 days sooner. How many
oxen has he, and how many days will the provender last ?

Ans. 300 oxen, and the provender will last 60 days.

Prob. 7. A certain number of laborers remove a pile of stones
in 6 hours from one place to another. If there had been 2 more
laborers, and if each laborer had each time carried 4 pounds
more, the pile would have been removed in 5 hours ; but if there
had been 3 less laborers, and if each laborer had each time car
ried 5 pounds less, it would have required 8 hours to remove
the pile. How many laborers were there, and how much did
each carry at one time?

Ans. There were 18 laborers, and each carried 50 pounds.

Prob. 8. A heavy wagon requires a certain time to travel
from A to B. A second wagon, which every 4 hours travels 5
miles less than the first, requires 4 hours more than the first to
go from A to B. A third wagon, which every 3 hours travels
8f miles more than the second, requires 7 hours less than the
second to make the same journey. How far is A from B, and
what time does each wagon require to travel this distance?
Ans. From A to B is 60 miles; the first wagon
requires 12 hours, the second 16, and the third
9 hours.

Prob. 9. I have two equal sums to pay, one after 9, and the
other after 15 months. If I settle them both at once, at the
same rate of discount, I must pay for the first sum $1208, and
for the second $1160. How much was each sum, and at what
per cent was the discount reckoned ?

Ans. $1280, and the discount was 7-J per cent.

Prob. 10. A small square lies with one angle in the angle of a
larger square. The excess of the side of the larger square above
that of the smaller is 118 feet; the excess of the square itself

EXAMPLES FOR PRACTICE. 365

is 26,432 square feet. What are the contents of each of the
two squares ?

Ans. The one 29,241, the other 2809 square feet.
Prob. 11. It is required to find two numbers whose sum, dif
ference, and product are in the ratio of the numbers 5, 1, and 18.

Ans. 9 and 6.

Prob. 12. Two numbers are in the ratio of 7 to 3, and their
difference is to their product as 1 to 21. What are the num
bers? -dm 28 and 12.

Prob. 13. Three towns, A, B, and C, lie at the angles of a
triangle. From A by B to C is 164 miles ; from B bj C to A
is 194 miles ; and from C by A to B is 178 miles. How far
are A, B, and C from each other ?

Ans. From A to B 74 miles, from B to C 90, and from

C to A 104 miles.

Prob. 14. A railway train, after traveling for one hour, meets
with an accident which delays it one hour, after which it pro
ceeds at three fifths of its former rate, and arrives at the termi
nus three hours behind time ; had the accident occurred 50
miles further on, the train would have arrived 1 hour and 20
minutes sooner. Required the length of the line.

Ans. 100 miles ; original rate 25 miles per hour.
Prob. 15. A railway train, running from New York to Al
bany, meets on the way with an accident, which causes it to

diminish its speed to -th of what it was before, and it is in con
sequence a hours late. If the accident had happened b miles
nearer Albany, the train would have been c hours late. Find
the rate of the train before the accident occurred.

b(n—I) .,

Ans. — '- miles per hour.

a—c

Prob. 16. Three boys are playing with marbles. Said A to
B, Give me 5 marbles, and I shall have twice as many as you
will have left. Said B to C, Give me 13 marbles, and I shall
have three times as many as you will have left. Said C to A,
Give me 3 marbles, and I shall have six times as many as you
will have left. How many marbles had each boy ?

Ans. A had 7, B 11, and C 21 marbles.

366 ALGEBRA.

Prob. 17. It is required to divide the number 232 into three
parts such that, if to the first we add half the sum of the oth
er two, to the second we add one third the sum of the other
two, and to the third we add one fourth the sum of the other
two, the three results thus obtained shall be equal. What are
the parts ?

Ans. The first 40, the second 88, and the third 104

Prob. 18. Four towns, A, B, C, and D, are situated at the
angles of a quadrilateral figure. When I travel from A by B
and C to D, I pay $6.10 passage-money; when I travel from
A by D and C to B, I pay $5.50. From A by B to C, I pay
the same as from A by D to C ; but from B by A to D, I pay
40 cents less than from B by C to D. What are the distances
of the four towns from each other, supposing I paid in each case
10 cents per mile ?

Ans. From A to B 21, from B to C 17, from C to D
23, and from D to A 15 miles.

Prob. 19. Four players, A, B, C, and D, play four games at
cards. At the first game A, B, and C win, and each of them
doubles his money ; at the second game A, B, and D win, each
of them doubling the money he had at the commencement of
that game; at the third game A, C, and D win; and at the
fourth game B, C, and D win ; and at each game each winner
won as much money as he had at the commencement of that
game. They now count their money, and find that each has
$64. How much had each before commencing play?
'/ Ans. A had $20, B had $36, C had $68, and D had $132.

Prob. 20. A and B start together from the foot of a mountain
to go to the summit. A would reach the summit half an hour
before B, but, missing his way, goes a mile and back again need
lessly, during which he walks at twice his former pace, and
jreaches the top six minutes before B. C starts twenty minutea
after A and B, and, walking at the rate of two and one seventh
miles per hour, arrives at the summit ten minutes after B.
Find the rates of walking of A and B, and the distance from
the foot to the summit of the mountain.

Ans. 2^, 2 ; distance 5 miles.

EXAMPLES FOR PRACTICE. 367

Prob. 21. Find three numbers such that if six be subtracted
from the first and second, the remainders will be in the ratio
of 2 : 3 ; if thirty be added to the first and third, the sums will
be in the ratio of 3 : 4 ; but if ten be subtracted from the sec
ond and third, the remainders will be as 4 : 5.

-f- Ans. 30, 42, 50.

Prob. 22. A and B engage to reap a field of wheat in twelve
days. The times in which they could severally reap an acre
are as 2:3. After some days, finding themselves unable to
finish it in the stipulated time, they call in C to help them,
whose rate of working was such that, if he had wrought with
them from the beginning, it would have been finished in nine
days. Also, the times in which he could have reaped the field
with A alone, and with B alone, are in the ratio of 7 : 8. When
was C called in? Ans. After six days.

EQUATIONS OF THE SECOND DEGREE WITH ONE UNKNOWN

QUANTITY.

A. — INCOMPLETE EQUATIONS OP THE SECOND DEGREE.

-r, _, ri. X + I8 X— 18 5 , f

Ex. 1. Given — H —= -, to find the values of x.

T>_u9 T 9 V

i// "l™ £i & "™~- £i O

Ans. x= ±14.

Ex.2. Given \/4+49—\/4~49 = 7, to find the values
v x2 v x2

of x. Ans. x— ±f.

Ex.3. Given - + -=- + - to find the values of a?.
• 5 x 2 x

Ans. x=db VlO.
Ex. 4. Given x-\- -\/a + x2 — ; to find the values of x.

Ans. x = ± -J (a — 1).

Ex.5. Giveny^ + m2-3 = m + l-y^— 2, to find the
values of x. Ans. x= ±m.

Ex.6. Given y^+29-\/^-34=7, to find the val
ues of x. Ans. x =

868

Ex. 7. Given
ues of x.

ALGEBRA.

=— IT, to find the val

fVl-X2

Ex.8. Given 27(7-x)2-43 = 77-3(7-x)2, to find the val
ues of x.

Remark. Put 7—x=y; first find the value of #, and thence the value of x.

Ans. a?=5 or 9.
Ex. 9. Given - , ~" =5, to find the values of x.

a+Va2— x

.Ans. x = ±

Ex. 10. Given

Vx—Vx—a x—a

find

Ans' -ra

Ex.11. Given

Ex.12. Given

V

X

V.27

=\y, to find the values of 07.

+X

, to find the values of x.

Vl + x 1 — VI— x

j\ Ans. x=

B. — COMPLETE EQUATIONS OF THE SECOND DEGREE.

Ex. 13. Given 557x=5801i+8x2, to find the values of x.

Ans. x=56£ or 12 J.
Ex. 14. Given (7x)2— 7x=l, to find the values of x.

Ans. x=0.2311477 or —0.0882905.

Ex. 15. Given 12x2=2H- Jx, to find the values of x.

Ans. x = l^ or — 1TV

Ex.16. Given 57x—18x2+ 145 = 0, to find the values of x.

Ans. x=4£ or —If.

Ex.17. Given ^(x+l)-l(2x*+x-I) = ^-(x+l), to find
^51 / oO

the values of x. Ans. x— — 1 or f .

EXAMPLES FOK PRACTICE. 369

Ex. 18. Given — ^r=s, to find the values of x-

x—I2 x—Q 6

Ans. x—24: or ^-.

X-4-4: X 4 10

Ex. 19. Given --\ r = -s~, to find the values of x.

X — 4: X + 4 3

Ans. x=8 or —8.
id the values of x.
Ans. x=o or — |-.

£ *W 1 ft T

TT1 ^^^ /^ • a ~T~ ^^^ " — c'c*y -L^-' — •** . r* j AT_ i

Ex. 21. Given _3__^_lg = __lsg, to find the val-
ues of x. Ans. 0^=8 or — 2-j-y^--

Ex. 22. Given - — --f - — =-, to find the values of x.
ox — o Zx — o A ,

Ans. x=.^ or 1.

Ex. 23. Given --\ T=— - — ~r. to find the values of x.

x—I x—l x—3

Ex. 20. Given - - H -- -= - -, to find the values of x.
x+l x+2 ce+3'

^?i5. x= or —

Ex.24. Given (7-4V3)^2+(2- V~3)x=2, to find the val
ues of x. Ans. x=2+V$, or —2(2 + -v/3).

^/r 91 _ 1/^r

Ex.25. Given — ^— + x- =2|, to find the values of x.
21 — Vx Vx

Ans. #=:49 or 196.

Ex. 26. Given \/x+ Vx=20, to find the values of x.
Ans. cc=+44=256 or -54 =

Ex.27. Given < - - - =_+_+- to find the values ofx.
a-fo-j-cc a o x

Ans. x=—a or —b.

Ex. 28. Given -^^J^± to find the values of x.
a+x a—b

Ans. g==

.

2

Ex. 29. Given = find

a + x

Q2

370 ALGEBRA.

Ex.30. Given x*— 4x3 + 7x2 — 60; = 18, to find the values of 9
by a quadratic equation. Ans. x = 3 or — 1, or 1± V— 5.

PROBLEMS INVOLVING EQUATIONS OF THE SECOND DEGREE
WITH ONE UNKNOWN QUANTITY.

Prob. 1. It is required to find three numbers which are in the
ratio of J, -J, and J, and the sum of whose squares is 10,309.

Ans. 78, 52, 39.

Prob. 2. A gentleman buys a certain number of pounds of
salt, four times as much sugar, and eight times as much coffee,
and for each pound of the three articles he paid as many cents
as he bought pounds of that article. For the whole he paid
$3.24. How many pounds of coffee did he buy?

Aits. 16 pounds.

Prob. 3. A rectangular garden was 37 feet broad and 259 feet
long. Its breadth was increased by a certain number of feet,
and its length diminished by seven times that number, by which
means its area was diminished 63 square feet. By how many
feet was the breadth increased ? Ans. 3 feet.

Prob. 4. Find that number whose square added to its cube
is nine times the next higher number. Ans. 3.

Prob. 5. A sets out from New York to Chicago, and B at the
same time from Chicago to New York, and they travel uniform
ly ; A reaches Chicago 16 hours, and B reaches New York 36
hours after they have met on the road. Find in what time
each has performed the journey.

Ans. A 40 hours, B 60 hours.

Prob. 6. A square vineyard, in which the vines are set in
squares so as to be uniformly four feet apart, is to be replanted
so that the vines may be uniformly 3J feet apart. Supposing
8640 more vines are required for this change, what must be the
length of each side of the vineyard? Ans. 672 feet.

Prob. 7. A glass mirror, 33 inches high and 22 inches wide,
is to be set in a frame of uniform breadth, such that the surface
of the frame shall be just equal to that of the glass. What
must be the breadth of the frame? Ans. 5J- inches.

Prob. 8. Required the solution of the preceding problem, if

EXAMPLES FOR PRACTICE. 371

we represent the height of the mirror by a and its breadth by 6,
and it is required that the surface of the frame shall be p times
that of the mirror.

•
4:

Prob. 9. Sixty pounds of a certain quality of sugar cost $2.40
less than sixty pounds of another quality. If I buy sugar of
each quality to the amount of $5.04, 1 obtain of the first kind
8 pounds more than of the second. What was the price of a
pound of each kind ?

Ans. One 14 cents, the other 18 cents.

Prob. 10. A gentleman bought a horse for a certain sum.
He afterward sold him for $144, and gained as much per cent,
as the horse cost him. How much did he pay for the horse ?

Ans. 80 dollars.

Prob. 11. A merchant buys a certain number of barrels of
flour for $216. At another time he expended the same sum of
money for flour, but obtained three barrels less, the price of
flour having risen one dollar per barrel. How many barrels
did he buy in the first case? Ans. 27 barrels.

Prob. 12. A and B contribute together $3400 in trade, A for
12 and B for 16 months. In the distribution, A received $2070,
capital and profits, and B received $1920. What was each one's
capital? Ans. A contributed $1800, and B $1600.

Prob. 13. Supposing the mass of the earth to be 80 times that
of the moon, their distance 240,000 miles, and the force of at
traction to vary directly as the quantity of matter, and inverse
ly as the square of the distance, at what point between them
will a third body be equally attracted by the earth and moon ?

Ans. 24,134 miles from the moon.

Prob. 14. A wall was completed in 5-J days by two masons,
one of whom commenced work 1-J- days later than the other.
In order to complete the wall alone, the first would have re
quired 3 days less than the second. In how many days could
each alone complete the wall ?

Ans. The first in 8, the second in 11 days.

Prob. 15. A courier goes from a place, A, to a place, B, io

372 ALGEBRA.

14 hours. At the same time, another courier starts for B from
a place 10 miles further distant, and expects to reach B at the
same time with the first, by gaining half an hour in every 20
miles. What is the distance from A to B ? Ans. 70 miles.

Prob. 16. From two towns, A and B, which are 104 miles
distant from each other, two wagons start at the same time,
and meet after 10-J hours. One requires for every 8 miles a
quarter of an hour more than the other. How much time does
each require to travel one mile ?

Ans. The one -£%, the other -^ of an hour.

Prob. 17. Two messengers start at the same time from two
towns, A and B, the first toward B, the other toward A, and,
upon meeting, it appeared that the first had traveled 12 miles
more than the second ; also, that if each should continue on at
his former rate, the first would arrive at B in 9 hours, and the
latter at A in 16 hours. What is the distance from A to B?

Ans. 84 miles.

Prob. 18. Two messengers start from the two towns, A and
B, to travel toward each other, but one started two hours ear
lier than the other. They meet each other 2^ hours after the
starting of the second messenger, and they reach the towns A
and B at the same instant. In how many hours did each mes
senger perform the journey ?

Ans. The one in 7, the other in 5 hours.

Prob. 19. Two travelers start from two towns, A and B,
whose distance from each other is 910 miles, and travel uni
formly toward each other. If the first starts 56 hours before
the second, they will meet halfway between A and B. If both
start at the same instant, at the end of 20 hours they will still
be 550 miles from each other. How many hours does each
traveler require to accomplish the distance from A to B ?

Ans. One 182 hours, the other 70 hours.

Prob. 20. A grocer has a cask containing 20 gallons of bran
dy, from which he draws off a certain quantity into another
cask of equal size, and, having filled the last with water, the
first cask was filled with the mixture. It now appears that if
6| gallons of the mixture are drawn off from the first into the

EXAMPLES FOR PRACTICE. 373

second cask, there will be equal quantities of brandy in each,
Eequired the quantity of brandy first drawn off.

Ans. 10 gallons.

Prob. 21. Two merchants sold the same kind of cloth. The
second sold three yards more of it than the first, and together
they received $35. The first said to the second, I should have
received $24 for your cloth ; the other replied, I should have
received $12 j- for yours. How many yards did each of them
sell?

Ans. The first merchant 5 or 15 yards, the second
merchant 8 or 18 yards.

Prob. 22. A and B traveled on the same road, and at the
same rate, from Cumberland to Baltimore. At the 50th mile
stone from Baltimore A overtook a drove of geese, which were
proceeding at the rate of three miles in two hours, and two
hours afterward met a wagon, which was moving at the rate
of nine miles in four hours. B overtook the same drove of
geese at the 45th milestone, and met the same wagon 40 min
utes before he came to the 31st milestone. Where was B when
A reached Baltimore? Ans. 25 miles from Baltimore.

w

EQUATIONS OF THE SECOND DEGREE WITH SEVERAL UNKNOWN

QUANTITIES.

Ex.1. Given (13x)2 + 2?/2 = 177, ) to find the values of x
(2y)2-I3x* = 3, | and y.

Ans. ce= ±1, y — ±2.

Ex.2. Given x^ + fix2— f-.\ 25:7, ) to find the values of
xy — 4:S, f x and y.

Ans. x= ±8, y=±6.

Ex. 3. Given 2(cc+4)2— 5(y— 7)*= 75, ) to find the values
7(>+4)2 + 15(?/-7)2:=1075, j- ofx and ^

Remark. Put x + ±=z, and y — 7—v. First determine z and v, and thence x
and y.

Ans. x= +6 or —14, y = 12 or 2.

Ex.4. Given (x + y)* — 2x2 - 49,) to find the values of x

= 372, j and y.

Ans. x— ±4, y= ±5, or ±13.

374 ALGEBRA.

Ex.5. Given 2cc-l-3?/ = 37J

11 _14 > to find the values of x and
ay ~45' )

. x =5 or -3- = 9 or -7

Ex.6. Given - + =9,

y x V to find the values of x and y.

. x=4: or 2, y=2 or 4.

Ex. 7. Given x2 + 7/2 = 10000,

'

vJ J- to find the values of x and ?/
y - 124, )

^ns. x=96 or 28, y = 28 or 96.

Ex. 8. Given 12 : x : : y : 3, ) . f

* I to find the values of x and y.
yx-\- vy = 5, )

Ans. x=9 or 4, y=4: or 9.

Ex.9. Given (3x+4?/)(7x— 2?/) + 3x+4?/=.44, ) to find the
(3x+4?/)(7x — 2y) — 7cc + 2?/ = 30, f values of x

and y. J.?w. x = l or 1T7T, y = 2 or — iV-

Ex.10. Given — x2+6xz/-9?/24-4x— 12y= 4,) to find the
x2— 2xy f By2— 4x + 5y =53, f values of x
and y. Ans. x — ll or — 7-J-, y = 3 or — 3J.

Ex. 11. Given 2(x2 + 3/2)(x-f?/) =15a#, ) to find the values
4(x4 — y4)(x2 — 2/2)=45x2y3, j ofccand?/.
^dw5. x— 2 or 1, y—~L or 2.

Ex.12. Given (x2 — y*)(x—y) = 16a?y, ) to find the values

Ans. x=9 or 3, ^/^=3 or 9.
Ex.13. Given jc(a5+2M-z) = 27,

Ex.14. Given xy=z, ~\

/ - st

to find the vf ues of x,
and,.

Ans. x = 3, ?/ = 2, 2=4.

}- to find the values of x, y, z, and v.
— ct,

yv = o

Va 3/- /- /-3/-

EXAMPLES FOK PRACTICE. 375

Ex.15. Given a^z = 105,

xyv — 135, to find the values of #, y, z,
xzv — 189, and v.

Ans. #=3, y=§i z = 7, v = 9.

Ex.16. Given x2 + -9+y2=84:J]

y2 ' I to find the values of as

x2 and y.

*+- + y=14.

U J

Ans. cc = 4, y = 2 or 8.

Ex. 17. Given Vy— Va—x = Vy—x, 1 to find the val-
2Vy—x + 2Va—x = oVa—x,j uesofxandy.

Ans. x — a =

Ex.18. Given x3

= ay.} „ , . _

^ v to find the values of x and y.
= bx, j

Ex.19. GivenA/5-v/x + 5\/?/4- Vx+ Vy= 10, / to find the
V^5+ Vy~5 =275, ) values of x

and y.

Remark. Put z*=-\/x + -\fy. Then, from Eq. 1, z=y~5; that is, -\/lc + Vy = 5.
Next put *\/x=^+v, and ~Vy—^— v. Substituting these values in Eq. 2, we
find u2=i, or v=±J.

-13

. a? = U or 4, or —

=4 or 9, or

Several of the following examples have imaginary or incom
mensurable roots which are not here given.

Ex.20. Given (x2 + y2)xy = 13090, ) to find the values of as
x+y — 18, ) and y.

Ans. x~l or 11, y = ll or 7.

Ex. 21. Given 5(x2+?/2)+4x?/ =356, ) to find the values of

'= 62, ) x and y.

Ans. x = 4 or 6, y — 6 or 4.

376 ALGEBRA.

Ex.22. Given (x2^y*)xy=SOO, ) to find the values of on
x* + y* r=337, ) and y.

Ans. x— ±4, y— ±3.

Ex. 23. Given (x*+y2)(x3+y3)=4:55, ) to find the values of
x+y — 5, ) x and y.

Ans. oc — 3 or 2, y = 2 or 3.

Ex.24. Given _,..,,

to nna the values of x

and y.

'-y

Ans. cc=12, y = 6.

Ex.25. Given (tf—xy+y^^+y*) = 91, ) to find the
(x* — xy+y2)(x2+xy-{-y2) = 133, ) values of x
and?/. J-ws. cc= ±3 or ±2, ?/= ±2 or ±3.

Ex. 26. Given (x+y)xy — 30, ) to find the values of x

!=468, ) and y.

Ans. x—2 or 3, y = 3 or 2.
/x y ^2 1

Ex.27. Given cc—v+V -= »1 to nn(^ tne values

v - ' - x-|-?/ > „

' ^ 1 of a? and y.

:41, j

:=±5, y=±4.

Ex.28. Given (x+y)3+x+y=30, } to find the values of x
x—y= 1, ) and ?/.

Remark. Multiply Eq. 1 by ar+y, and we have

Add to each member 9(a;+#)2 + 25, and the square root of each member of the
equation may be extracted.

Ans. x=2, y=l.

Ex.29. Given (x +y)(xy +1)= ISxy ) to find the vat
(x2+7/2)(xY+l) = 208xy ) ues of a? and y.

Remark. Divide Eq. 1 by xy, and Eq. 2 by a;2//2, and we have

x +y +- +-
x y

Put x+-=z, and «+-
x

- +- = 18.

Then z+v = 18, and z2+ua=212.

Whence 2=14 or 4, and t?=4 or 14 ; and hence x and y are easily found

Ans. x-2± -v/3, y=7

EXAMPLES FOE PKACTICE. 377

PROBLEMS INVOLVING- EQUATIONS OF THE SECOND DEGREE
WITH SEVERAL UNKNOWN QUANTITIES.

Prob. 1. If I increase the numerator of a certain fraction by
2, and diminish the denominator by 2, 1 obtain the reciprocal
of the first fraction ; also, if I diminish the numerator by 2, and
increase the denominator by 2, the resulting fraction, increased
by 1^, is equal to the reciprocal of the first fraction. What
is the fraction ? Ans. y.

Prob. 2. It is required to divide the number 102 into three
parts, such that the product of the first and third shall be equal
to 102 times the second part, and the third part shall be 1-J
times the first.

Ans. The first part is 34, the second 17, and the third 51.

Prob. 3. A certain number consists of two digits. If I in
vert the digits, and multiply this new number by the first, I ob
tain for a product 5092 ; but if I divide the first digit by the
second, I obtain 1 for a quotient with 1 for a remainder. What
is the number? Ans. 76.

Prob. 4. The fore wheel of a carriage makes 165 more rev
olutions than the hind wheel in going 5775 feet ; but if the
circumference of each wheel be increased 2-J- feet, the fore
wheel will make only 112 revolutions more than the hind
wheel in the same space. Required the circumference of each
wheel.

Ans. The fore wheel 10 feet, the hind wheel 14 feet.

Prob. 5. A piece of cloth, by being wet in water, shrinks one
eighth in its length and one sixteenth in its breadth. If the
perimeter of the piece is diminished 4J feet, and the surface 5f
square feet, by wetting, what were the length and breadth of
the piece?

Ans. 16 feet long and 2 feet wide.

Prob. 6. A certain number of laborers in 8 hours transport a
pile of stones from one place to another. If there were 8 more
laborers, and if each carried each time 5 pounds less, the pile
would be removed in 7 hours ; but if there were 8 less labor
ers, and if each carried each time 11 pounds more, it would re-

'378 ALGEBKA.

quire 9 hours to remove the pile. How many laborers were
there employed, and how many pounds did each carry ?

Ans. 28 laborers, and each carried 45 pounds ; or 36

laborers, and each carried 77 pounds. -f£.
Prob. 7. A certain capital yields yearly $123-J- interest ; a
second capital, $700 larger, and loaned at J per cent, less, yields
yearly $11-1 more interest than the first. How large was the
first capital, and at what per cent, was it loaned ?

Ans. The capital was $3800, loaned at 3J per cent.
Prob. 8. A person bought a number of $20 railway shares
when they were at a certain rate per cent, discount for $1500 ;
and afterward, when they were at the same rate per cent, pre
mium, sold them all but 60 for $1000. How many did he buy,
and what did he give for each of them ?

Ans. 100 shares at $15 each.

Prob. 9. A rectangular lot is 119 feet long and 19 feet broad.
How much must be added to the breadth, and how much taken
from the length, in order that the perimeter may be increased
by 24 feet, and the contents of the lot remain the same ?

Ans. The length must be diminished 102 feet, and the

breadth increased 114 feet.

Prob. 10. There are two numbers such that their sum and
product together amount to 47 ; also, the sum of their squares
exceeds the sum of the numbers themselves by 62. What are
the numbers?

Ans. 5 and 7.

Prob. 11. The sum of two numbers is a, and the sum of their
reciprocals is b. Kequired the numbers.

Prob. 12. A and B engage to reap a field for $24 ; and as A
alone could reap it in nine days, they promise to complete it in
five days. They found, however, that they were obliged to
call in C to assist them for the last two days, in consequence
of which B received one dollar less than he otherwise would
have done. In what time could B or C alone reap the field?

Ans. B in 15 and C in 18 days.

EXAMPLES FOR PEACTICE. 379

Prob. 13. The sum of the cubes of two numbers is 35, and
the sum of their ninth powers is 20,195. Kequired the numbers.

Ans. 2 and 3.

Prob. 14. There are two numbers whose product is 300 ; and
the difference of their cubes is thirty-seven times the cube ol
their difference. What are the numbers ?

Ans. 20 and 15.

Prob. 15. A merchant had $26,000, which he divided into two
parts, and placed them at interest in such a manner that the in
comes from them were equal. If he had put out the first por
tion at the same rate as the second, he would have drawn for
this part $720 interest ; and if he had placed the second out at
the same rate as the first, he would have drawn for it $980 in
terest. What were the two rates of interest?

Ans. 6 per cent, for the larger sum, and 7 for the smaller.

Prob. 16. A miner bought two cubical masses of ore for $820.
Each of them cost as many dollars per cubic foot as there were
feet in a side of the other; and the base of the greater contain
ed a square yard more than the base of the less. What was
the price of each? Ans. 500 and 320 dollars.

Prob. 17. A gentleman bought a rectangular lot of land at
the rate of ten dollars for every foot in the perimeter. If the
same quantity had been in a square form, and he had bought it
at the same rate, it would have cost him $330 less ; but if he
had bought a square piece of the same perimeter, he would have
had 12J rods more. What were the dimensions of the lot ?

AJIS. 9 by 16 rods.

Prob. 18. A and B put out at interest sums amounting to
$2400. A's rate of interest was one per cent, more than B's;
his yearly interest was five sixths of B's ; and at the end of ten
years his principal and simple interest amounted to five sev
enths of B's. What sum was put at interest by each, and at
what rate?

^ir Ans. A $960 at 5 per cent, B $1440 at 4 per cent.

Prob. 19. A person bought a quantity of cloth of two sorts
for $63. For every yard of the best piece he gave as many
dollars as he had yards in all ; and for every yard of the poor-

380 ALGEBRA.

er, as many dollars as there were yards of the better piece more
than of the poorer. Also, the whole cost of the best piece was
six times that of the poorer. How many yards had he of each ?

Ans. 6 yards of the better and 3 of the poorer.
Prob. 20. A commences a piece of work alone, and labors
for two thirds of the time that B would have required to per;
form the entire work. B then completes the job. Had both
labored together, it would have been completed two days soon
er, and A would have performed only half what he left for B.
Eequired the time in which they would have performed the
work separately.

Remark. Suppose A would have performed the work in x days, and B in y days.

A labors -J- days, and performs — part of the work.
o ox

O0I O~» O*»

B performs 1 —-^-= — - — - part of the work.
ox ox

3x—2y

— - — -xy= time B labored.

• — — +— =whole time consumed.

ox o

-+-~part both did in one day— -.

x y xy

— — =the days of work if both labored together.

xy ,9_3xy—2yz2y
x+y 3x ~3 '

Also, ix-^-= iof°~

*"*+i 2 3x '

Ans. A in 6 days and B in 3 days.

EXAMPLES FOR PRACTICE. 381

PROGRESSIONS.

Ex. 1. What is the sum of the natural series of numbers
1, 2, 3, etc., up to 1000?

Ans. 500,500.

Ex. 2. What is the sum of an arithmetical progression whose
first term is 6, the last term 2833, and the number of terms 38?

Ans. 53,941.

Ex. 3. What is the first term and the sum of the terms of an
arithmetical progression, when the last term is 24, the common
difference -f-, and the number of terms 22 ?

Ans. First term 9, and sum of terms 363.
Ex. 4. Kequired the number and the sum of the terms of an
arithmetical progression, when the first term is — f , the com
mon difference — |, and the last term —2 If.

Ans. Number of terms 25, and sum of terms — 281 J.
Ex. 5. The first term of an arithmetical progression is 5, the
last term 23, and the sum of the terms 392. What is the com
mon difference and the number of terms ?

Ans. Common difference •§, and number of terms 28.
Ex. 6. Between 7 and 13 it is required to interpolate 8 terms
which shall form an arithmetical progression.

Ans. 7f, 8t, 9, 9|, IQfr, 11, llf, 12f

Ex. 7. In an arithmetical progression, the sum of the 19th,
the 43d, and the 57th terms is 827; the sum of the 27th, the
58th, the 69th, and the 73d terms is 1581. What is the first
term and the common difference ?

-4.W.9. The first term is 5, and the common difference 7
Ex. 8. In boring an artesian well 500 feet deep, $3.24 is paid
for the first foot, and 5 cents more for each subsequent foot.
How much was paid for the last foot, and how much for the
whole well ?

Ans. For the last foot $28.19, and for the entire well

$7857i

Ex. 9. According to natural philosophy, a body falling in a
vacuum describes in the first second of its fall 16^ feet, and

382 ALGEBRA.

in each succeeding second 32^ feet more than in the second
immediately preceding. If a body has fallen 20 seconds, how
many feet will it fall in the last second, and how many in the
whole time?

Ans. 627J feet in the last second, and 6433^ feet in
the whole time.

^ Ex. 10. Divide unity into four parts in arithmetical progres
sion, of which the sum of the cubes shall be VT-.

Ex. 11. A ship, with a crew of 175 men, set sail with a sup
ply of water sufficient to last to the end of the voyage ; but in
30 days the scurvy made its appearance, and carried off three
men- every day; and at the same time a storm arose, which
protracted the voyage three weeks. They were, however,
just enabled to arrive in port without any diminution in each
man's daily allowance of water. Eequired the time of the
passage, and the number of men alive when the vessel reached
the harbor.

Ans. The voyage lasted 79 days, and the number of
men alive was 28.

Ifamark. Put x = days the voyage was expected to last.

a;-f-21=days the voyage lasted.

a; + 21-30=a;-9=the days after 30.
On the 31st day the number of men was 172, etc.

Last term = 172-3(o;-10).

Sum of series =(344- 3(o: - 10)) x^?.

Then (344 -3* + 30)^5 = 175 (x -30).

Whence x =58. Also, x +21 = 79 days the voyage lasted.

Ex. 12. The number of deaths in a besieged garrison amount
ed to 6 daily ; and, allowing for this diminution, their stock of
provisions was sufficient to last 8 days. But on the evening
of the sixth day 100 men were killed in a sally, and afterward
the mortality increased to 10 daily. Supposing the stock of
provisions unconsumed at the end of the sixth day to support 6
men for 61 days, it is required to find how long it would sup-

EXAMPLES FOR PRACTICE. 383

port the garrison, and the number of men alive when the pro
visions were exhausted.

Ans. The provisions last 6 days, and 26 men survive.

Remark. Put a:— number of men at first.

x— 42=number expected at end of 8 days.

9r_ 49

— -x8 = 8x — 168 = number of days' provisions.

2

___X6 = 6:E — 90=days' provisions exhausted at end of 6th day.
2

2X— 78 =366 =the remainder.
Whence #=222.

222 — 136 = 86— number of men after the sally.
Put y=number of days the provisions lasted afterward.

Ex. 13. The first term of a geometrical progression is 1, the
ratio 2, arid the number of terms 13. What is the last term
and the sum of the terms ?

Ans. The last term is 4096, and the sum of the terms

8191.

Ex. 14. The first term of a geometrical progression is 7, the
ratio 3, and the number of terms 11. What is the last term
and the sum of the terms ?

Ans. The last term is 413,343, and the sum of the

terms 620,011.

Ex. 15. The sum of the terms of a geometrical progression
is 411,771, the ratio 7, and the number of terms 7. Eequired
the first and last terms.

Ans. First term 3, and last term 352,947.
Ex. 16. Between 1 and £ it is required to interpolate 11
terms forming a continued geometrical progression. What are
the terms?

Ans. 0.9439, 0.8909, 0.8409, 0.7937, 0.7492, 0.7071,

0.6674, 0.6300, 0.5946: 0.5612, 0.5297.

Ex. 17. What will $1200 amount to in 36 years at 4 per
cent, compound interest? Ans. $4924.70.

Ex. 18. A farmer sowed a peck of wheat, and used the whole
crop for seed the following year; the produce of the second
year he uaed for seed the third year, and so on. If in the 10th

884 ALGEBRA.

year the crop was 1,048,576 pecks, by how many times must the
seed have increased each harvest^ supposing the increase to
have been always the same ? Ans. Four times.

Ex. 19. There are three numbers in geometrical progression^
the difference of whose differences is six, and their sum is forty-
two. Kequired the numbers. Ans. 6, 12, and 24.

Ex. 20. There are three numbers in geometrical progression,
the greatest of which exceeds the least by 24 ; and the differ
ence of the squares of the greatest and the least is to the sum
of the squares of all the three numbers as 5 : 7. What are the
numbers? Ans. 8, 16, and 32.

Ex. 21. There are three numbers in geometrical progression
whose continued product is 216, and the sum of their cubes is
1971. Kequired the numbers. Ans. 8, 6, and 12.

Ex. 22. There are four numbers in geometrical progression
whose sum is 850 ; and the difference between the extremes is
to the difference of the means as 87 : 12. What are the num
bers? Ans. 54, 72, 96, 128.

Ex. 23. There are four numbers, the first three of which are
in geometrical progression, and the last three in arithmetical
progression ; the sum of the first and last is 14, and that of the
second and third 12 ; find the numbers.

25 15 9 3
Ans. 2, 4, 8, 12 ; or — — , -, -.

Ex. 24. Three numbers whose sum is 15 are in arithmetical
progression ; if 1, 4, and 19 be added to them respectively, they
are in geometrical progression. Determine the numbers.

Ans. 2, 5, 8.
qfc.

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