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DEPARTMENT Oi' EDUCATION 
LEEAHD STANFORD JUNIOR UNIVEBSITI 



ALGEBRA 



THE MACMILLAN COMPANY 

NEW YORK • BOSTON • CHICAGO • DALLAS 
ATLANTA • SAN FRANCISCO 

MACMILLAN & CO., Limited 

LONDON • BOMBAY • CALCUTTA 
MELBOURNE 

THE MACMILLAN CO. OF CANADA, Ltd. 

TORONTO 



ALGEBRA 



FOR COLLEGES AND SCHOOLS 



BY 

H. S. HALL, M.A., and S. R. KNIGHT, B.A. 



REVISED AND ENLARGED 
FOR THE USE OF AMERICAN SCHOOLS 

BY 

F. L. SEVENOAK, A.M. 

PRINCIPAL OP THE ACADEMIC DEPARTMENT 
STEVENS INSTITUTE OP TECHNOLOGY 



N*fa got* 
THE MACMILLAN COMPANY 

1913 

All rights reserved 



629209 



Ooptbight, 18^1897, 1918, 
By THE MACMILLAN COMPANY. 



First edition, 1895. Second edition, revised and enlarged, 1896. 
Reprinted with corrections, January, 1897; August, October, 1897; 
August, 1898; October, 1898; July, August, 1899; September, 1899; 
January, July, 1900; July, 1901; April, August, October, 1902; 
March, September, December, 1903; February, July, 1904; May, 
1905; January, July, 1906; March, 1907; January, October, 1908; 
March, August, 1909; July, 1910; January, 191 1; March, 19x2; 
February, 19x3. Third edition, September, 1913. 



Nortoootj 3Bre*8 

J. 8. Cuahing Co. — Berwick & Smith Co. 

Norwood, Mass., U.S.A. 



PREFACE. 

Within a comparatively short time, the algebra require- 
ment for admission to many of our Colleges and Schools 
of Science has been much increased in both thoroughness of 
preparation and amount of subject-matter. This increase 
has made necessary the rearrangement and extension of 
elementary algebra, and it is for this reason that the present 
revision of Hall and Knight's Elementary Algebra has been 
undertaken. 

The marked success of the work, and the hearty endorse- 
ment by many of our ablest educators of the treatment of 
the subject as therein presented, warrant the belief that the 
present edition, with its additional subject-matter, will be 
found a desirable arrangement and satisfactory treatment 
of every part of the subject required for admission to any 
of our Colleges or Schools of Technology. 

Many changes in the original chapters have been made, 
among which we would call attention to the following : A 
proof, by mathematical induction, of the binomial theorem 
for positive integral index has been added to Chapter xxxix. ; 
a method of finding a factor that will rationalize any bino- 
mial surd follows the treatment of binomial quadratic surds ; 
Chapter xlh. has been re-written in part, and appears as a 
chapter on equations in quadratic form; and the chapter 



VI PREFACE. 

on logarithms has been enlarged by the addition of a four- 
place table of logarithms with explanation of its use. 

Chapters xxi., xxv., xxx., xxxiii., xxxvm., xlii., xliii., 
xliv., xlv., xlvi., xlvii., and xlviii. treat of portions of 
the subject that have not appeared in former editions. 
A chapter on General Theory of Equations is not usually 
found in an Elementary Algebra, but properly finds here 
a place in accordance with the purpose of the present re- 
vision ; and its introduction makes the work available for 
use in college classes. Carefully selected exercises are 
given with each chapter, and at the end of the work a large 
miscellaneous collection will be found. 

The Higher Algebra of Messrs. Hall and Knight has been 
drawn upon, and the works of Todhunter, Chrystal, and 
DeMorgan consulted in preparing the new chapters. 

I gratefully acknowledge my indebtedness to Prof. J. 

Burkitt Webb of the Stevens Institute of Technology both 

for contributions of subject-matter, and valuable suggestions 

as to methods of treatment. My thanks are also due to 

Prof. W. H. Bristol, of the same institution, for suggestions 

as to the arrangement of the chapter on General Theory of 

Equations. 

FRANK L. SEVENOAK. 

June, 1896. 



PREFACE TO SECOND EDITION. 

The printing of the present edition from entirely new 
plates has enabled us to correct a few typographical errors 
found in the first edition, and give, at the suggestion of 
friends, a somewhat fuller explanation of the more diffi- 
cult parts of the subject. We hope that the addition of 
new material to Chapters in., iv., v., x., xx., xxi., xlii., 
xlviii., and of several sets of Miscellaneous Examples, 
will render the book still more acceptable to those whose 
commendation of the former edition has given us much 
pleasure. 

June, 1896. 



vn 



PREFACE TO THIRD EDITION. 

The present edition contains additional material on 
Graphs. The subject has been treated in a single chapter 
(Chapter xlix.) with the idea that the majority of teachers 
prefer such an arrangement. 

Class-room experience has shown that Arts. 632-645 may- 
be used with profit in connection with a first reading of 
Chapter xvn., and Arts. 646-649, 657-660 in connection 
with Chapter xxvi. 

August, 1913. 



via 



CONTENTS. 



CHAPTER 

I. Definitions. Substitutions 



II. 
IIL 



IV. 



V. 



VI. 



VII. 



Negative Quantities. Addition of Like Terms 

Simple Brackets. Addition. Subtraction 
Commutative Law for Addition and Subtraction 
Associative Law for Addition and Subtraction 
Dimension, Degree, Ascending and Descending Powers 
Miscellaneous Examples I. 



Multiplication 

Commutative Law for Multiplication 
Associative Law for Multiplication . 
Index Law for Multiplication . 
Distributive Law for Multiplication . 
Multiplication of Compound Expressions 

Rule of Signs 

Note on Arithmetical and Symbolical Algebra 
Distributing the Product . 
Products written by Inspection . 
Multiplication by Detached Coefficients 



Division 

Index Law for Division 
Division of Compound Expressions . 
Important Cases in Division 
Horner's Method of Synthetic Division 

Removal and Insertion of Brackets . 
Miscellaneous Examples II. 

Simple Equations .... 

ix 



PAGE 
1 

9 

13 
15 
15 
17 
23 

24 
25 
25 
25 
26 
27 
28 
29 
31 
35 
37 

38 
39 
40 
44 
46 

48 
53 

55 



X CONTENTS. 

CHAPTER PAGE 

VIII. Symbolical Expression 60 

IX. Problems Leading to Simple Equations ... 66 

X. Resolution into Factors • . 70 

Trinomial Expressions 72 

Difference of Two Squares . . . . . .78 

Important Cases 82 

Sum or Difference of Two Cubes 82 

Miscellaneous Cases .83 

Converse Use of Factors 85 

The Factor 'Theorem 87 

XI. Highest Common Factor 92 

Factor Theorem Employed 98 

XII. Lowest Common Multiple 102 

XIII. Fractions 107 

Reduction to Lowest Terms 108 

Multiplication and Division 110 

Addition and Subtraction 114 

Rules for Change of Sign . . ... 121 

Cyclic Order 124 

XIV. Complex Fractions. Mixed Expressions . . . 126 
Miscellaneous Exercise on Fractions .... 132 

XV. Fractional and Literal Equations 136 

XVI. Problems 141 

Miscellaneous Examples III. 144 

XVII. Simultaneous Equations 147 

Elimination by Addition or Subtraction . . . 148 

Elimination by Substitution 149 

Elimination by Comparison 160 

Equations involving Three Unknown Quantities . . 152 

Reciprocal 164 

Literal Simultaneous Equations 167 

XVm. Problems 160 

XIX. Indeterminate and Impossible Problems. Negative Re- 
sults 165 

Meaning of ^, 1, ?, - 167 

6 °° oo 



CONTENTS. 



XI 



CHAPTER 

XX. 



XXI. 



XXII. 



XXIII. 



XXIV. 



XXV. 
XXVL 



PAGE 

Involution 169 

To square a Multinomial 171 

To cube a Multinomial 173 

Application of Binomial Theorem .... 173 

Evolution 176 

Square Root of a Multinomial 177 

Cube Root of a Multinomial . . . . . . 181 

Some Higher Roots 185 

The nth Root of a Multinomial 186 

Square and Cube Root of Numbers .... 187 

The Theory of Indices 191 

Meaning of Fractional Exponent 193 

Meaning of Zero Exponent 193 

Meaning of Negative Exponent 194 

Surds (Radicals) 203 

Reduction of Surds 205 

Addition and Subtraction of Surds .... 207 

Multiplication of Surds 208 

Division of Surds 209 

Compound Surds 211 

Factor which will rationalize any Binomial Surd . . 214 

Properties of Quadratic Surds 216 

Square Root of a Binomial Surd 217 

Square Root of a Binomial Surd by Inspection . . 218 

Equations involving Surds 220 

Imaginary Quantities 222 

Imaginary Unit 223 

Fundamental Algebraic Operations .... 225 

Problems 227 

Miscellaneous Examples IV 232 

Quadratic Equations . . . ... . . 235 

Pure Quadratic Equations 236 

Affected Quadratic Equations 237 

Solution by Formula 239 

Solution by Factoring 241 

Formation of Equations with Given Roots . . . 242 
Values found for the Unknown Quantity which do not 

satisfy the Original Equation 243 



Xll CONTENTS. 

CHAPTER PAGE 

XXVII. Equations in Quadratic Form 245 

XX vIIL Simultaneous Equations involving Quadratics . . 249 

Homogeneous Equations of the Same Degree . . 253 

Symmetrical Equations 254 

Miscellaneous Cases 255 

XXIX. Problems 258 

XXX. Theory of Quadratic Equations .... 264 

Number of the Roots 264 

Character of the Roots 265 

Relation of Roots and Coefficients .... 266 

Formation of Equations with Given Roots . . 267 

Miscellaneous Theorems 272 

Remainder Theorem 272 

Symmetry 273 

XXXI. Indeterminate Equations of the First Degree . . 276 

XXXII. Inequalities 279 

Miscellaneous Examples V 283 

XXXIII. Ratio 286 

Proportion 291 

Transformations that may be made in a Proportion . 293 

Variation 296 

XXXIV. Arithmetical Progression 301 

Geometrical Progression 308 

Harmonical Progression 313 

XXXV. Permutations and Combinations . . . .319 

XXXVI. Probability (Chance) 331 

Miscellaneous Examples VI 334 

XXXVII. Binomial Theorem: Proof for Positive Integral 

Index 333 

The General Term 341 

Simplest Form of the Binomial Theorem . . . 342 

Proof by Mathematical Induction .... 343 

Equal Coefficients 345 

Greatest Coefficient 345 



CONTENTS. Xlll 

CHAPTER PAGE 

Greatest Term 345 

Sum of the Coefficients 347 

Expansion of Multinomials 347 

Application when Index is Negative or Fractional . 347 

XXXVIII. Logarithms 362 

Properties of Logarithms 353 

Characteristic and Mantissa 354 

Advantages of Common Logarithms . . . 356 

Logarithms transformed from Base a to Base b . 367 

Logarithms in Arithmetical Calculation . . . 358 

Four-place Table of Logarithms .... 360 

Use of the Table 362 

Cologarithms 366 

Exponential Equations 366 

XXXIX. Interest and Annuities 367 

XL. Limiting Values and Vanishing Fractions . . 370 

XLI. Convergency and Divergency of Series . . . 377 

Tests for Convergency 377 

Auxiliary Series 383 

XLIL Undetermined Coefficients 386 

Functions of Finite Dimensions .... 386 

Functions of Infinite Dimensions .... 390 

Expansion of Fractions into Series .... 391 

Expansion of Surds into Series .... 392 

Reversion of Series 393 

Partial Fractions . 394 

The General Term 397 

XLIII. Continued Fractions 400 

Formation of Successive Convergents . . . 403 

Limits to Error 407 

Recurring Continued Fractions .... 409 

XLIV. Summation of Series 413 

Scale of Relation 413 

The Sum of n Terms 415 



XIV CONTENTS. 

CHAPTER PAGK 

The General Term ....... 417 

The Method of Differences 418 

Any Required Term 419 

The Sum of n Terms 420 

Piles of Shot and Shells 422 

Interpolation 425 

XLV. Binomial Theorem : Proof for Any Index . . . 426 

XL VI. Exponential and Logarithmic Series .... 430 

XLVH. Determinants 485 

Minors 437 

Vanishing of a Determinant 438 

Multiplication of a Determinant 438 

A Determinant expressed as the Sum of Two Other 

Determinants 430 

Simplification of Determinants 440 

Solution of Simultaneous Equations of the First 

Degree 444 

Signs of the Terms 447 

Determinant of Lower Order 448 

XL VIII. Theory of Equations 460 

Horner's Method of Synthetic Division . . . 461 

Number of the Roots 463 

Depression of Equations 463 

Formation of Equations 454 

Relations between the Roots and the Coefficients . 454 

Fractional Roots 468 

Imaginary Roots 458 

Transformation of Equations ..... 460 

Standard Form of Reciprocal Equations . . . 466 

Descartes' Rule of Signs 469 

Derived Functions 472 

Equal Roots 472 

Location of the Roots 474 

Sturm's Theorem and Method 476 

Graphical Representation of Functions . . . 480 

Solution of Higher Numerical Equations . . . 483 

Newton's Method 483 



CONTENTS. XV 

CHAPTER PAGE 

The Cube Roots of Unity 486 

Cardan's Method 487 

Biquadratic Equations 488 

Incommensurable Roots 491 

Horner's Method of Approximation .... 492 

Any Root of Any Number 495 

Miscellaneous Examples VII 496 

XLIX. Graphical Representation of Functions .... 617 

Application to Simultaneous Equations . . . 524 

Measurement on Different Scales 536 

Practical Applications 545 

Miscellaneous Applications of Linear Graphs . . 559 



ALGEBRA. 



CHAPTER I. 

Definitions. Substitutions. 

1. Algebra treats of quantities as in Arithmetic, but with 
greater generality; for while the quantities used in arith- 
metical processes are denoted by figures, which have a 
single definite value, algebraic quantities are denoted by 
symbols, which may have any value we choose to assign 
to them. 

The symbols of quantity employed axe usually the letters 
of our own alphabet ; and, though there is no restriction as 
to the numerical values a symbol may represent, it is under- 
stood that in the same piece of work it keeps the same value 
throughout. Thus, when we say " let a equal 1," we do not 
mean that a must have the value 1 always, but only in the 
particular example we are. considering. Moreover, we may 
operate with symbols without assigning to them any par- 
ticular numerical value; indeed it is with such operations 
that Algebra is chiefly concerned. 

We begin with the definitions of Algebra^ premising that 
the symbols +,—, x, -*-,(), = will have the same mean- 
ings as in Arithmetic. Also, for the present, it will be 
assumed that all the algebraic symbols employed represent 
integral numbers. 

2. An algebraic expression is a collection of symbols; it 
may consist of one or more terms, which are the parts sepa 

B 1 



2 ALGEBRA. 

rated from each other by the signs + and — . Thus, 
7 a 4- 5 6 — 3c — x+2y is an expression consisting of five 
terms. 

Note. When no sign precedes a term the sign + is understood. 

3. Expressions are either simple or compound. A simple 
expression consists of one term, as 5 a. A compound expres- 
sion consists of two or more terms. Compound expressions 
may be further distinguished. Thus an expression of two 
terms, as 3 a — 2 b, is often called a binomial, and one of 
three terms, as 2 a -f 3 6 -f c, a trinomial. Simple expres- 
sions are frequently spoken of as monomials, and compound 
expressions as multinomials or polynomials. 

4. When two or more quantities are multiplied together 
the result is called the product. One important difference 
between the notation of Arithmetic and Algebra should be 
here remarked. In Arithmetic the product of 2 and 3 is 
written 2x3, whereas in Algebra the product of a and b 
may be written in any of the forms a x b, a • b, or ab. The 
form ab is the most usual. Thus, if a = 2, b = 3, the prod- 
uct ab = axb = 2 x3 = 6; but in Arithmetic 23 means 
"twenty-three," or 2 x 10 -f 3. 

5. Each of the quantities multiplied togetner to form a 
product is called a factor of the product. Thus 5, a, b are 
the factors of the product 5 ab. 

Notb. The beginner should carefully notice the difference be- 
tween term and factor. 

6. When one of the factors of an expression is a numeri- 
cal quantity, it is called the coefficient of the remaining 
factors. Thus, in the expression 5 ab, 5 is the coefficient. 
But the word coefficient is also used in a wider sense, and it 
is sometimes convenient to consider any factor, or factors, 
of a product as the coefficient of the remaining factors. 
Thus, in the product 6 abc, 6 a may be appropriately called 



DEFINITIONS. SUBSTITUTIONS. 8 

the coefficient of be. A coefficient which is not merely 
numerical is sometimes called a literal coefficient. 

Note. When the coefficient is unity it is usually Omitted, and we 
write simply a, instead of 1 a. 

7. A power of a quantity is the product obtained by re- 
peating that quantity any number of times as a factor, and 
is expressed by writing the number of factors to the right 
of the quantity and above it. Thus, 

a x a is called the second power of a, and is written a 2 ; 
a x a x a is called the third power of a, and is written a 3 ; 

and so on. 

The index or exponent is the number which expresses the 
power of any quantity. Thus 2, 5, 7 are respectively the 
indices of a*, a B , a 7 . 

Note, a 2 is usually read **o squared" ; a 8 is read "a cubed" ; 
a* is read " a to the fourth " ; and so on. 

When the index is unity it is omitted, and we write simply a, 
instead of a 1 . Thus a, 1 a, a 1 , and 1 a 1 all have the same meaning. 

8. The beginner must be careful to distinguish between 
coefficient and index. 

Ex. 1. What is the difference in meaning between 3 a and a 8 ? 
By 3 a we mean the product of the quantities 3 and a. 
By a 8 we mean the product of the quantities a, a, a. 
Thus, if a = 4, 

3a = 3xa = 3x4 = 12; 

a 8 =axaxa = 4x4x4 = 04. 

Ex. 2. If b = 6, distinguish between 4 b 2 and 2 b*. 
Here 46 2 = 4x6x6 = 4x6x6 = 100; 
whereas 26* = 2x6x6x6x6 = 2x6x6x6x6 = 1260. 

Ex. 3. If x = 1, find the value of 6 «*. 

Here 6«* = 6xa;xxxa;xa; = 6xlxlxlxl = 6. 

Note. The beginner should observe that every power of 1 is S. 

Ex. 4. If a = 4, x = 1, find the value of 6 x°. 
&x* = 6xsc tt = 6xl 4 = 6xl = 6. 



4 ALGEBRA. 

9. The Sign of Continuation, ..., is read "and so on." 

10. The Sign of Deduction, .*., is read "therefore" or 
"hence." 

11. In arithmetical multiplication the order in which the 
factors of a product are written is immaterial. Thus 

3x4 = 4x3. 

In like manner in Algebra ab and ba each denote the 
product of the two quantities represented by the letters a 
and b, and have therefore the same value. Although it is 
immaterial in what order the factors of a product are written, 
it is usual to arrange them alphabetically. Fractional co- 
efficients which are greater than unity are usually kept in 
the form of improper fractions. 

Ex. If a = 6, x = 7, z = 6, find the value of |# axz. 
Here }$axs = j$x6x7x6 = 278. 

EXAMPLES I. a. 

Ka = 7,6 = 2,c = l,a; = 5,y = 3, find the value of 
1. 14 x. 8. Sax. 5. bby. 7. 3d 2 . 

8. x 8 . 4. a 8 . 6. b 6 . 8. 2 ax. 

Ifc = 8, & = 6, c = 4, x = 1, y = 3, find the value of 
11. 3 c 2 . 14. 9xy. 17. x 8 . 80. 6*. 

18. 7y 8 . 15. 8& 8 . 18. 7 j/*. 81. y*. 

13. bob. 16. Sx 6 . 19. c*. 88. x*. 

If a = 6, b = 1, c = 6, x = 4, find the value of 

86. £x*. 88. fx 8 . 30. 3*. 38. 8\ 

87. f^c 8 . 89. &ax. 31. 2«. 88. 7*. 

12. When several different quantities are multiplied to- 
gether a notation similar to that of Art. 7 is adopted. Thus 
aabbbbcddd is written a?b A cd?. And conversely 7 a*ccP has the 
same meaning as7xaxaxaxcxdxd. 

Ex. 1. If x = 5, y = 3, find the value of 4 x 2 y 8 . 

4xty 8 = 4x6 2 x3 8 = 4x25x27 = 2700. 



9. 


6 c*. 


10. 


4^. 


83. 


y*. 


84. 


a*. 


85. 


b*. 


84. 


■feacx. 


35. 


\bcx. 



DEFINITIONS, SUBSTITUTIONS. 6 

Ex. 2. If a = 4, 6 = 9, x = 6, find the value of ^~ 

27 a 8 

$bx* = $ xQx 6 2 _ 8x9x 36 _ . _ u 
27 a 8 27 x 4 8 27 x 64 * ** 

13. If one factor of a product is equal to 0, the product 
must be equal to 0, whatever values the other factors may have. 
A factor is usually called a zero factor. 

For instance, if x = 0, then a&xy* contains a zero factor. 
Therefore a&'oy 2 = when x = 0, whatever be the values of 
a, 6, y. 

Again, if c = 0, then c 8 = ; therefore a&W = 0, whatever 
values a and b may have. 

Note. Every power of is 0. 

EXAMPLES I. b. 

If a = 7, 5 = 2, c = 0, a; = 6, y = 3, find the value of 

1. 4ac a . 3. 86 2 y. 5. I*) 2 *;. 7. \xy^. 9. a*cy. 

2. a 8 6. 4. 3sy 2 . 6. f&Y*. 8. a*c. 10. 8afy. 

If a — 2, 6 = 3, c = 1, p = 0, g = 4, r = 6, find the value of 

11. 5^r. 14. ±«£ 17. **-. 20. 3«2». 23. *£* 

86 9a 8 9<r 64r« 

18. 8a^. 15. 3 a 26«. 18. 6a>C. 21. 2^a». 24. ^!. 

9^ s 32 

13. *£*. 16. J6<r. 19. ^. 22. c»6*. 25. 5*. 

6 2 * 7r <? 

14. Definition. The square root of any proposed ex- 
pression is that quantity whose square, or second power, is 
equal to the given expression. Thus the square root of 81 
is 9, because 9 2 = 81. 

The square root of a is denoted by -f/a, or more simply ya. 

Similarly the cube, fourth, fifth, etc., root of any expres- 
sion is that quantity whose third, fourth, fifth, etc., power 
is equal to the given expression. 

The roots are denoted by the symbols ;/, ^/, ty, etc. 

Examples : ^7 = 3 ; because 3 8 = 27. 

^32 = 2 ; because 2 8 = 32. 



6 ALGEBRA. 

The symbol -y/ is sometimes called the radical sign. 

Ex. 1. Find the value of 6^/(6 aWc), when a = 3, 5 = 1, c = 8. 

6V(0a 8 &*c)= 6 x y/(fi x 3» x l 4 x 8) 

= 6 x V(6 x 27 x 8) = 6 x V1296 
= 6 x 36 = 180. 

Ex. 2. Find the value of Jf^LY when a = 9, b = 8, * = 6. 



8 |/«&1\ = '// P x 3< \ if/ 9x81 \ _ »// 9x9x9 \ 
\V8^/ \\8x5»/ \\8xl26/ \V 1000 J 



10* 



i. v( 2 «)- •• 2«v( 2a y)- 

2. VC**). 10. 6yV( 4 **). 



* V(^) 



EXAMPLES L o. 
If a = 8, c = 0, A = 9, x = 4, y = 1, find the value Of 

8. ^{2 ax), u. 3cV(*aO. 

4. y/(2 ak*). 12 2xy v /'( 4 y 6 )- 

i tS£ »• v(S)- "• w# 

If a = 4, 6 = 1, o = 2, d = 9, as = 6, y = 8, find the value of 

W- V(8«<0. 85. __! w . \\t**\. 

SO. 6V(46»). V(»«W) >I\j*W 

21. 7V(6<&0. je 1 . jo. 

22. VO^)- V( 6c4a 



m 



28 V(H?)- *• ^/(ra) 

* V(e55)- 



81. V^ c - 

82. vy - - 

1 88. V &d - 

^(8a& 2 c) .84. V<**- 



15. We now proceed to find the numerical value of ex- 
pressions which contain more than one term. In these, 
each term can be dealt with singly by the rules already 



DEFINITIONS. SUBSTITUTIONS. 7 

given, and by combining the terms the numerical value of 
the whole expression is obtained. 

16. We have already, in Art. 8, called attention to the 
importance of carefully distinguishing between coefficient 
and index; confusion between these is such a fruitful source 
of error with beginners that it may not be unnecessary once 
more to dwell on the distinction. 

Ex. 1. When c = 6, find the value of c* - 4c + 2c 8 - 3c*. 

Here c* = 6* = 6x6x5x6 = 625; 

4c = 4 x 5 = 20; 

2c 8 = 2 x 6 8 = 2 x 6 x ) y 6 = 250 ; 

3c 2 = 3x6 2 = 3x 6x6 = 76 
Hence the value of the expression = 626 - 20 + 260 — 76 = 780. 

Ex. 2. When p = 9, r = Q t Jfe = 4, find the Value of 

= JW + V«i-2 

= ix 1+9-2 = 7*. 

17. By Art. 13 any term which contains a zero factor is 
itself zero, and may be called a zero term. 

Ex. 1. If a = 2, b = 0, x = 3, y = 1, find the value of 

4 a 8 - ab* + 2xy* + Sabz. 
The expression = (4 x 2 8 )- +(2 x 3 x 1) + 

= 32 - + 6 + = 38. 
Note. The two zero terms do not affect the result. 

Ex. 2. FincTthe value of J x 2 - cPy + 7 abx - % y 8 , when 

a = 5, 6 = 0, x = 7, y = 1. 
f x 2 - afy + 7 a&z - f y 8 = f x 7 2 - 6 s x 1 + - } x 1* 

= 29J - 26 - 2J = 1 ^. 

Note. The zero term does not affect the result. 

18. In working examples the student should pay atten- 
tion to the following hints: 

1. Too much importance cannot be attached to neatness 
of style and arrangement. The beginner should ^emembei 
that neatness is in itself conducive to accuracy. 



8 ALGEBRA. 

2. The sign = should never be used except to connect 
quantities which are equal. Beginners should be particu- 
larly careful not to employ the sign of equality in any 
vague and inexact sense. 

3. Unless the expressions are very short the signs of 
equality in the steps of the work should be placed one 
under the other. 

4. It should be clearly brought out how each step follows 
from the one before it ; for this purpose it will sometimes 
be advisable to add short verbal explanations; tjie impor- 
tance of this will be 3een later. 

EXAMPLES I. d. 

If a = 2, b = 3, c = 1, d = 0, find the numerical value of 

1. 6a + 66-8c + 9d. 6. 26c + Scd -4da + 5a6. 

2. 8a-46 + 6c + 5<f. 7. Sbcd + &cda-7 dab + abc 

3. Gab-3cd+2da-5cb+2db. 8. a 2 + 6 2 +c* + (P. 
ft. abc + bed + cda + dab. 0. 2a a + 36 8 -4c*. 
5. 3abc-2bcd+2cda-4dab. 10. a* + b*-c*. 

If a = 1, b = 2, c = 3, d = 0, find the numerical value of 

11. a 8 + & 8 + c 8 + (« 8 . 12. J6c 8 -a 8 -5 8 -Ja6«c. 

13. 3a&c-6 2 c-6a 8 . 

14. 2 a 2 + 2 6 2 + 2c 2 + 2d 2 - 2 6c - 2cd - 2 da - 2a*. 

15. a 2 + 2& 2 + 2c 2 + d 2 + 2a& + 2&c ffed. 

16. 2c 2 + 2a 2 + 26 2 -4c& + 6a&cd. 

17. 13a 2 + J 3 L c* + 20a&-16ac-16&c. 

18. 6a.&-iac 2 -2a + io 4 -3d + tc*. 

19. 125 6*c - 9 # + 3 a&c 2 d. 

If a = 8, b = 6, c = 1, & = 9, y = 4, find the value of 

so. §.-♦•> + 1* M . ^j-^ + g 

T 1/* cxy 
easy 2 a 



CHAPTER II. 
Negative Quantities. Addition of Like Terms 

19. In his arithmetical work the student has been accus- 
tomed to deal with numerical quantities connected by the 
signs -f and — ; and in finding the value of an expression 
such as If + 7£ — 3| + 6 — 4£ he understands that the quan- 
tities to which the sign -f is prefixed are additive, and those 
to which the sign — is prefixed are subtractive, while the 
first quantity, If, to which no sign is prefixed, is counted 
among the additive terms. The same notions prevail in 
Algebra ; thus in using the expression 7 a + 3b — 4c — 2d 
we understand the symbols 7 a and 3 b to be additive, while 
4 c and 2 d are subtractive. 

20. But in Arithmetic the sum of the additive terms is 
always greater than the sum of the subtractive terms ; and 
if the reverse were the case, the result would have no arith- 
metical meaning. In Algebra, however, not only may the 
sum of the subtractive terms exceed that of the additive, 
but a subtractive term may stand alone, and yet have a 
meaning quite intelligible. 

Hence all algebraic quantities may be divided into pos- 
itive quantities and negative quantities, according as they 
are expressed with the sign + or the sign — ; and this is 
quite irrespective of any actual process of addition and sub- 
traction. 

This idea may be made clearer by one or two simple illus- 
trations. 

(i) Suppose a man were to gain $ 100 and then lose $ 70, 
his total gain would be $ 30. But if he first gains $ 70 and 
then loses $ 100 the result of his trading is a loss of $ 30. 





10 ALGEBRA. 

The corresponding algebraic statements would be 

$100 -$70 = + $30, 
$70 -$100 = -$30, 

and the negative quantity in the second case is interpreted 
as a debt, thakis, a sum of money opposite in character to 
the positive quantity, or gain, in the first case ; in fact it 
^jnaflMSe said to possess a subtractive quality which would 
produce its effect on other transactions, or perhaps wholly 
counterbalance a sum gained. 

(ii) Suppose a man starting from a given point were to 
walk along a straight road 100 yards forwards and then 
70 yards backwards, his distance from the starting-point 
would be 30 yards. But if he first walks 70 yards forwards 
and then 100 yards backwards his distance from the starting- 
point would be 30 yards, but on the opposite side of it. As 
before we have 

100 yards — 70 yards = 4-30 yards, 
70 yards — 100 yards = — 30 yards. 

In each of these cases the man's absolute distance from the 
starting-point is the same ; but by taking the positive and 
negative signs into account, we see that — 30 is a distance 
from the starting-point equal in magnitude but opposite in 
direction to the distance represented by +30. Thus the 
negative sign may here be taken as indicating a reversal of 
direction. 

Many other illustrations might be chosen ; but it will be 
sufficient here to remind the student that a subtractive 
quantity is always opppsite in character to an additive 
quantity of equal absolute value. 

Note. Absolute value is the value taken independently of the 
signs + and — . 

2L Definition. When any number of quantities are 
connected by the signs + and — , the resulting expression 
is called their algebraic sum. Thus 11 a — 27 a + 13 b — 5 b 
is an algebraic sum. This expression, however, is not, as 
will be shown, in its simplest form. 



ADDITION OF LIKE TEBMS. 11 

22. Addition is the process of finding in simplest form the 
algebraic sum of any number of quantities. 

23. Like terms, or similar terms, do not differ, or differ 
only in their numerical coefficients. Other terms are called 
unlike, or dissimilar. Thus 3a, 7a; 5a% 2a*b\ 3a s & 8 , 
— 4a s 6 2 are pairs of like terms; and 4a, 36; 7a 2 , 9a 2 6 
are pairs of unlike terms. 

ADDITION OF LIKE TERMS. 

Rule I. The sum of a number of like terms is a like term. 

Rule II. If all the terms are positive, add the coefficients. 

Ex. Find the value of 8 a + 5 a. 

Here we have to increase 8 like things by 5 like things of the 
same kind, and the aggregate is 13 of such things ; 
for instance, . 8 lbs. + 6 lbs. = 13 lbs. 

Hence also, 8a + 6 a = 18 a. 

Similarly, 8a45a + a + 2a + 6a = 22a. 

Rule III. If all the terms are negative, add the coefficients 
numerically and prefix the minus sign to the sum. 

Ex. To find the sum of — 3 as, — 6 x y — 7 as, — x. 

Here the word sum indicates the aggregate of 4 subtractive quanti- 
ties of like character. In other words, we have to take away succes- 
sively 3, 6, 7, 1 like things, and the result is the same as taking away 
3 + 5 + 7 + 1 such things in the aggregate. 

Thus the sum of — 3 as, — 6 as, — 7 x, — x, is — 16 x. 

Rule IV. If the terms are not all of the same sign, add 
together separately the coefficients of aU the positive terms and 
the coefficients of all the negative terms; the difference of these 
two results, preceded by the sign of the greater, vriU give the 
coefficient of the sum required. 

Ex. 1. The sum of 17x and — 8 as is as, for the difference of 17 
and 8 is 9, and the greater is positive. 

Ex. 2. To find the sum of 8 a, — 9 a, — a, 3 a, 4 a, — 11a, a. 
The sum of the coefficients of the positive terms is 16. 
The sum of the coefficients of the negative terms is 21. 
The difference of these is 5, and the sign of the greater is negative ; 
hence the required sum is — 6 a. 



12 ALGEBRA. 

We used not, however, adhere strictly to this rule, for the 
terms may be added or subtracted in the order we find most 
convenient. 

This process is called collecting terms. 

Ex. 3. Find the sum of } a, 3 a, — J a, —2 a. 
The sum = 3}a - 2 Ja = lja = Ja. 

Note. The sum of two quantities numerically equal but with 
opposite signs is zero. The sum of 5 a and — 6 a is 0. 

EXAMPLES n. 
Find the sum of 

1. 5a, 7a, 11a, a, 23a. 9. - 116, -56, - 8& f -&. 

2. 4x, x, 3x, 7x, 9 x. 10. 5x, — x, — 3x, 2x, — x. 

8. 76, 106, 116, 96, 26. 11. 26y, -lly,- 15y,y, -3y,2y. 

4. 6c, 8c, 2c, 15c, 19c, 100c, c. 12. 5/ f - 9/, - 3/, 21/, - 30/. 

5. — 3x, — 5x, — 11 x, — 7x. 13. 2«, — 38, $, — *, — 5*, 5 a. 

6. -56,-66,-116,-186. 14. 7y, - lly, 16y, -3y, - 2y. 

7. — 3y, — 7y, — y 9 — 2y, — 4y. 15. 5x, - 7x, — 2x, 7x, 2x, — 5x 

8. - c, - 2c, - 50c, - 13c. 16. 7a6, - 3a6, - 5a6, 2a6, a6. 

Find the value of 

IT. -9x 2 + H« 2 + 3x 9 -4x 2 . 19. 3a 8 -7a 8 -8a 8 +2a 8 -llo 8 . 
18. 3a 2 x- ISatx + cPx. 20. 4X 8 - 5s 8 - 8s 8 - 7s 8 . 

31. 4 a 2 6 2 __ a 2&2 _ 7 a 2 fe 2 + 5 2 6 2 __ a 2&il. 

22. -9x*-4x*-12x*+13x*-7x*. 

23. 7 a&cd*- 11 a&cd- 41 a&ca* + 2 a6ca\ 

24. Jx-^x + x+fx. 25. fa + fa-Ja. 

26. -56 + J6-|6 + 26-i6 + J6. 

27. -}x 2 -2x 2 -f5c 2 + « 2 + lx 2 + ^* 2 - 

28. — a6 — %ab — Ja6 — Ja6 — Ja6 + a6 + ^a6. 

29. |x-|x + |«-2x + Qz-$x + x. 
80. -|x 2 -|x 2 -Jx 2 -}x 2 -x a . 



CHAPTER III. 
SiMPiiE Brackets. Addition. Subtraction. 

24. When a number of arithmetical quantities are con- 
nected by the signs -f- and — , the value of the result is the 
same in whatever order the terms are taken. This also 
holds in the case of algebraic quantities. 

Thus a — b + c is equivalent to a + c — b, for in the first 
of the two expressions b is taken from a, and c added to the 
result ; in the second c is added to a, and b taken from the 
result. Similar reasoning applies to all algebraic expres- 
sions. Hence we may write the terms of an expression in 
any order we please. 

Thus it appears that the expression a - b may be written 
in the equivalent form — b + a. 

To illustrate this we may suppose, as in Art. 20, that a 
represents a gain of a dollars, and — b a loss of b dollars : 
it is clearly immaterial whether the cain precedes the loss, 
or the loss precedes the gain. 

25. Brackets ( ) are used, as in Arithmetic, to indicate 
chat the terms enclosed within them are to be considered 
as one quantity. The full use of brackets will be con- 
sidered in Chap. vi. ; here we shall deal only with the 
simpler cases. 

8 + (13 + 5) means that 13 and 5 are to be added and 
their sum added to 8. It is clear that 13 and 5 may be 
added separately or together without altering the result. 

Thus 8 + (13+5) = 8 + 13 + 5 = 26. • 

Similarly a + (b + c) means that the sum of b and c is to 
be added to a. 

Thus a + (b + c) = a + b + c 

13 



14 ALGEBRA. 

8 + (13 — 6) means that to 8 we are to add the excess of 
13 over 5 ; now if we add 13 to 8 we have added 5 too much, 
and must therefore take 5 from the result. 

Thus 8 + (13 - 6) = 8 + 13 - 5 = 16. 

Similarly a+(b—c) means that to a we are to add b, dimin- 
ished by c 

Thus a + (b — c)=sa + b — o . . . . (1). 

In like manner, 

a + &-^c + (d-e--/) = a + &-c + d-e-/. (2). 
Conversely, 

a+b-c+d-e -/= a + 6-c + (d-e -/). (3). 
Again, a — b+c=a+c— b, [Art. 24.] 

= the sum of a and c — b, 
= the sum of a and — b + c, [Art. 24] 
therefore a — 6 + c = a+(— b + c) . . . . (4). 

By considering the results (1), (2), (3), (4), we are led tc 
the following rule : , 

Rule. When an expression within brackets is preceded by 
the sign +, the brackets can be removed without making any 
change in the expression. 

Conversely : Any part of an expression may be enclosed 
within brackets and the sign + prefixed, the sign oj every term 
within Hie brackets remaining unaltered. 

Thus the expression a — b + e — d + e may be written in 
any of the following ways, 

a + (— 6 + c — d + e) 
a — b + (c — d + e), 
a — b + c + (— d + e). 

26. The expression a— (b + c) means that from a we are 
to take the sum of b and c. The result will be the same 
whether b and c are subtracted separately or in one sum. 

Thus a — (6 + c) = a — 6 — a 



SIMPLE BRACKETS. 15 

Again, a — (b — c) means that from a we are to subtract 
the excess of b over c. If from a we take b we get a — 6 •, 
but by so doing we shall have taken away c too much, and 
must therefore add cto a — b. Thus 

a — (b — c) = a — b + c. 
In like manner, a~b — (c — d — e) = a-— 6 — c + d + e. 

Accordingly the following rule may be enunciated : 

Rule. When an expression within brackets is preceded by 
the sign — , the brackets may be removed if the sign of every 
term within the brackets be changed. 

Conversely : Any part of an expression may be enclosed 
within brackets and the sign — prefixed, provided the sign of 
every term within the brackets be changed. 

Thus the expression a — 5 + c + d — e may be written in 
any of the following ways, 

a — (+ b — c — d + e), 
a — b — (— c — d + e), 
a — b + c — (— d + e). 

We have now established the following results : 
L Additions and subtractions may be made in any order. 
Thus a + b — o + d — e— /=a — c + 6 + d— /— e 

— a — c — /+ d + b — e. 
This is known as the Commutative Law for Addition and 
Subtraction. 

II. The terms of an expression may be grouped in any 
manner. 

Thusa + 6-o + d-c-/=(a + 6)-c+(d-e)-/ 
= a +(b - c) + (d - e)-/= a + b -(c - d)-(e +f). 

This is known as the Associative Law for Addition and 
Subtraction. 

ADDITION OF UNLIKE TERMS. 

27. When two or more like terms are to be added together 
we have seen that they may be collected and the result 



16 ALGEBRA. 

expressed as a single like term. If, however, the terms are 
unlike, they cannot be collected. Thus in finding the sum of 
two unlike quantities a and b, all that can be done is to 
connect them by the sign of addition and leave the result in 
the form a + b. 

Also, by the rules for removing brackets, a-h(— 6)=a— b; 
that is, the algebraic sum of a and — b is written in the 
form a - b. 

28. It will be observed that in Algebra the word sum 
is used in a wider sense than in Arithmetic. Thus, in the 
language of Arithmetic, a — b signifies that b is to be sub- 
tracted from a, and bears that meaning only; but in Algebra 
it also means the sum of the two quantities a and — b 
without any regard to the relative magnitudes of a and b. 

Ex. 1. Find the sum of 3 a -66 + 2c;2a + 36-d*; _4a + 2 6. 

Thesum=(3a-66 + 2c) + (2a + 36-a") + (-4a + 26) 
= 3a-66 + 2c + 2a + 36-d*-4a + 26 
= 3a + 2a-4a-66 + 36 + 26 + 2c-d 
= a + 2 c — d , 

by collecting like terms. 

The addition is more conveniently eifected by the fol- 
lowing rule: 

Rule. Arrange the expressions in lines so that the like 
terms may be in the same vertical columns: then add each 
column, beginning with that on the left. 

ft — 5 ft 4. 2 c ^ ne a lS e ^ ra ^° sum °* tne terms in the first 

„ /? column is a, that of the terms in the second 

a , ~~ column is zero. The single terms in the third 

— 4a + 2o an( j f ourtn columns are brought down without 

a + 2 c - d chang e. 

Ex. 2. Add together - 5 aft + 6 be — 7 ac ; Sab + 3 ac — 2 ad ; — 2 ab 
-f 4 ac + 6 ad ; 6c — 3 aft + 4 ac\ 

-6aft + 66c-7ac 
« 7) j_ q _ 9 ^ Here we first rearrange the expressions 

„ , , A , - , so that like terms are in the same ver- 

— 2 ab -f 4 ac + 5 aa 

— 3 ft -I- 6c 4- 4 ad ^ ca ^ columns, and then add up eact 

— ^ — , . - , r^ — ^ column separately. 

— 2 aft + < 6c + 7 ad ^ 



ADDITION. 17 



EXAMPLES in. a. 
Find the sum of 



2a-36 + & 
2a-6 + 3a 
2x + y — 3 s. 
2z + 3y — z. 



1. a + 26-3c; -3a + 6 + 2c 

2. 3a + 26-c; -a + 36 + 2c 

3. — 3x + 2y + z;x — 3y + 2z 

4. -aj+2y + 3s; 3x-y + 2z 

5. 4a + 36 + 6c; -2a + 36-8c;a-6 + c. 

6. -15a-196-18c;14a+16 6 + 8c;a + 66 + 9a 

7. 25a-156 + c; 13a-10 6 + 4c; a + 206-c. 

8. -16a - 106 + 6c; lOa + 56 + c; 6a + 66-c. 

9. 5 ax — 7 6y + as ; az + 2by — cz; — 3ax + 2 by + 3cz. 

10. 20p + g-r; p-20g+r; p + g-20r. 

Add together the following expressions 

11. — 6 ab + 6 6c — 7 ca ; 8 a& — 4 6c + 3 ca ; — 2 a& - 2 6c + 4 ca. 

12. 15a6-27 6c-6ca; 14 a6 - 18 6c+ 10 ca; -49a6+45 6c-3ca. 
18. 6 a6 + 6c — 3 ca ; a6 — 6c + ca ; — a6 + 6c + 2 ca. 

14. pq + gr — rp ; — jwj + gr + rp ; pq — gr + rp- 

15. x + y + z; 2x + 3y — 2z; 3x — 4y + 3. 

16. 2a-36 + c; 15a-216-8c; 3a + 246 + 7c. 

17.- 4 xy — 9yz + 2 zx ; — 25 xy + 2iyz — zx; 23 ay — 1 5 y« + «x. 

18. 17 a6 — 13 6c + 8ca ; — 6 a6 + 9 6c — 7 ca ; 2 a6 — 7 6c - ca. 

19. 47x-QSy + z; -25x+15y-3s; -22x + 48y + 16s. 

20. 23a-176-2c; -9a+156 + 7c; -13a + 36-4c. 

DIMENSION, DEGREE, ASCENDING AND DESCENDING 

POWERS. 

29. Each of the letters composing a term is called a 
dimension of the term, and the number of letters involved is 
called the degree of the term. Thus the product abc is said 
to be of three dimensions, or of the third degree; and ax* is 
said to be of Jive dimensions, or of the fifth degree, 

A numerical coefficient is not counted. Thus 8a 2 6 5 and 
oW are each of seven dimensions, or of the seventh degree. 

But it is sometimes useful to speak of the dimensions of 
an expression with regard to any one of the letters it in- 



18 ALGEBRA 

volves. For instance, the expression 8a?b 4 c, which is of 
eight dimensions, may be said to be of three dimensions 
in a, of four dimensions in b, and of one dimension in c 

30. A compound expressi n is said to be homogeneous 
when all its terms are of the same degree. Thus 8 o* — a 4 b* 
-f 9 ab 9 is a homogeneous expression of six dimensions, or of 
the sixth degree. 

3L Different powers of the same letter are unlike terms ; 
thus the result of adding together 2 x* and 3 x* cannot be ex- 
pressed by a single term, bu :'; must be left in the form 
2a? + 3a* 

Similarly, tho algebraic sum of 5 a*b* — 3 oft 8 and — b 4 is 
5 dfb 2 — 3 ab* — b\ This expression is in its simplest form 
and cannot be abridged. 

32. In adding together several algebraic expressions con- 
taining terms with different powers of the same letter, it 
will be found convenient to arrange all the expressions in 
descending or ascending powers of that letter. This will be 
made clear by tho following examples. 

Ex. 1. Add together 3x* + 7 + Qx — 6x 2 ; 2x*-8 — 9x; 
4x-2x 8 + 3x 2 ; Sx*-9z-x 2 ; x-x 2 -x 8 + 4. 

In "..riting the first expression we put in 

3x 8 -6x 2 + 6x + 7 the first term the highest power of x, in 

2 x 2 — 9 x — 8 the second term the next highest power, 

— 2x 8 + 3x 2 -f4x and so on till the last term, in which x does 
3 x s — x 2 — 9 x not appear. The other expressions are 

— x 8 — x 2 + x+4 arranged in the same way, so that in each 
3aj8_ 2x 2 — 7 x + 3 column we have like powers of the same 

letter. The result Is in descending powers of x. 

Ex. 2. Add together 

3a6 2 -26 8 + a 8 ; 5a 2 6 -a& 2 - 3c 8 ; 8a 8 + 66 8 ; 9a*b -2a* + ab 2 . 
-2b* + Sab 2 + a 8 

— ab 2 + 6 a 2 b — 3 a 8 Here each expression is arranged 
'6 b* + 8 a 8 according to descending powers of 6, 

ab* + 9 a 2 b — 2 a 8 and ascending powers of a. 

36 8 + 3aft 2 + 14a 2 6 + 4a 8 



ADDITION. 19 

EXAMPLES HI. b. 

find the sum of the following expressions : 

1. 2ab + 3ac + Qabc; —5ab + 2bc — babe; Sab — 2bc- Sac 

2. 2x 2 -2xy + 3y 2 ; 4y 2 + 6xy-2x 2 ; a^-^a^-ey 2 . 

3. 3a 2 -7a&-46 2 ; -6a 2 + 9ab - 36 2 ; 4a 2 + afc + 66 2 . 

4. zt + xy-i/ 2 ; -z 2 + yz + y 2 ; -x* + xz + z*. 

5. -xS-^y + Sy 2 ; 3x 2 + 4xy-6;/ 2 ; x^xy + y 2 . 

6. gt-fleP + z-1; 2x 2 ~2aj + 2; -ixS+Sx + l. 

7. 2x 8 -x 2 -x; 4x 8 + 8x 2 + 7x; -fla^-exP + x. 

8. 9x 2 -7x+6; -14x 2 + 16x-6; 20x 2 -40x-17. 

9. 10x« + 6x + 8; 3x 8 -4x 9 -6; 2x 8 -2x-3. 

10. a 8 — aft + be ; ab + 6 8 — ac ; ac — be + c 8 . 

11. S^-S^ + d 8 ; 6 8 ~2a 8 .+ 3d 8 ; 4c 8 - 2a 8 -3<J" 

12. 6x 8 -2x+l; 2x 8 + x + 6; x 2 -7x 8 + 2x-4. 

13. a 8 -a 2 + 8a; 3a 8 + 4a 2 + 8a; 5a 8 -6a 2 -11a. 

14. x 2 + ^-2xy; 2« 2 -3y 2 -4y«; 2x 2 -2s 2 -3x*. 

15. x*-2y* + x; y 8 ~2x 8 + y; x 2 + 2j/ 2 -x + y 8 . 

16. xS + Sx^ + Sxy 2 ; - Sx*y -6xy 2 - x 8 ; 3x 2 y + 4xy 2 . 

17. a 8 + 6a& 2 + 6 8 ; 6 8 -10a6 2 -a 8 ; 5a& 2 - 26 8 + 2a 2 6. 

18. x 6 -4x 4 y-6x 8 y 8 ; 3x*y + 2x*y 8 - 6xy* ; 3xV + 6xy*-y* 

19. a 8 -4a 2 6 + 6a&c; a 2 6 - 10 abc+ c 8 ; & 8 + 3a 2 6 + a&c. 

90. x 8 -4x 2 y + 6xy 2 ; 2x 2 y-3xy 2 +2y 8 ; y 8 + 3x 2 y + 4xy 2 . 

Add together the following expressions s 

91. ia-i&; -a + $6; fa-6. 

29. -|a-J&; -Ja + f&; -2a-&. 

93. -2a + fc; - Ja-26; }6-3c. 

94. -Vo-^c;2o-3&;^6-c. 

96. fx^ixy-Jy 2 ; -x 2 - f xy + 2y*; f a^-xy- Jy» 
26. 3a 2 -|a6-i& 2 ; - fa 2 + 2a6 - |6 2 ; -fa 2 -aft + 6 2 . 

27. f^-Jxy + Ay 2 ; -|« 2 + i**y-y 2 ; l* 2 -** + iy 2 . 

98. -ixS + Saa^-fa^; x 8 -Yax 2 + ia 2 x; - Ja^+ia 2 ^ 

99. fx 2 -fxy-7y 2 ; fxy + ^y 2 ; -|x 2 + 4y 2 . 

Sa Ja 8 -2a 2 6-J6 8 ; fa 2 & - f a& 2 + 26 8 ; - fa 8 * afc 2 + J6 8 . 



20 ALGEBRA. 

SUBTRACTION. 

33. Subtraction is the inverse of Addition. The simplest 
cases have been considered under the head of addition of 
like terms, of which some are negative. [Art. 23.] 

Thus 5a — 3a = 2a, 

3a — 7a = — 4 a, 
— 3a — 6a = — 9a. 

Also, by the rule for removing brackets [Art. 26], 

3a-(-8a)=3a + 8a 
= 11 a, 
and — 3a— (— 8a)= — 3a + 8a 

= 5 a, 

SUBTRACTION OF UNLIKE TERMS. 

34. The method is shown in the following example : 

Ex. Subtract 3 a — 2 b — c from 4a — 36 + 6 c. 

The result of subtraction = 4 a — 36 + 6c — (3 a — 26 — c) 

= 4a-3& + 5c-3a + 26 + c 
= 4o-3a-36 + 26 + 6c + c 
= a — b + 6 a 

It is, however, more convenient to arrange the work as follows, the 
signs of all the terms in the lowe: line being changed, 

4a-36+6c 
-8a + 26+ c 

by addition a -6 + 6c 

Rule. Change the sign of every term in the expression to 
be subtracted, and add it to the other expression. 

Note. It is not necessary that in the expression to be subtracted 
the signs should be actually changed ; the operation of changing signs 
ought to be performed mentally. 



SUBTRACTION. 21 

Ex.1. From 5x 2 + xy-3y 2 take 2x a + 8xy- 70«. 

6x 2 + xy-3y* 
2x 2 + 8xy-7y 2 

3x 2 -7xy + 4y 2 . 

Ex. 2. Subtract 3a; 2 - 2x from 1 - z*. 

Terms containing different powers of the same letter being unlike 
must stand in different columns. 

_ %9 _j_ i The rearrangement of terras in the first 

3 x 2 — 2 x ^ ne * s not nece88ar t/i DU * ft i 8 convenient, 

because it gives the result of subtraction in 

^x 8 — 3x 2 + 2x + l. descending powers of x. 

BXAMPI<ES III. o. 
Subtract 

1. 4a — 36 + c from 2a- 36 -c. 

2. a -36 + 6c from 4a-86 + c. 

3. 2x-8y + s from 16x + 10y -18s. 

4. 16a-276 + 8c from 10a + 36 + 4c. 

5. — 10 x — 14 y + 15 3 from x — y — «. 

6. - lla& + 6cd from - 106c + a6-4ca\ 

7. 4a -36 + 15c from 26a -166 — 18c. 

8. — 16x- 18y-16* from -6x+ 8y + 1 z. 

9. a& + cd — ac — 6a* from ab + cd f- ac + 6a*. 

10. — a6 + cd — ac + 6<f from a6 — cd + ac - 6d. 

From 

11. 3a6 + 5cd-4ac-66(2 take 3a6 + 6co* - 3ac- 66A 

12. yz — xz + xy take — xy + ys — xz. 

13. -2x»-x 2 -3x + 2 take x*-x + l. 

14. - 8x 2 y + 15 xy* + lOxyz take 4^ - 6xy 2 - 6xy*. 

15. £ a — 6 + J c take J a + J 6 — J c. 

16. }x + y — z take Jx — £y — \z. 

17. -a-36 take fa + J6-}c. 

18. Jx-fy + ^s take - Jx + fy-^*. 

19. -£x-$y-6s take fx-$y--^s. 
«0. -Jx + fy- J take ix-fy-i. 



22 ALGEBRA. 

EXAMPLES m. d. 
From 

1. 3xy — 6yz + 8x« take — 4xy + 2ys — 10 xz. 

2. -8xV + 16x 8 y + 13x^ take 4xV + 7* 8 y--8xy 8 . 

3. -8 + 6a6 + o 2 6 2 take 4-3o6-6a 2 6 a . 

4. a 2 &c + b*ca + c 2 a& take Sa 2 bc - SJftja - 4c 2 a&. 

5. -7a 2 & + 8o6 2 + cd take 5o 2 6-7a& 2 + 6cd. 

6. -Sxty + Sx^-x 2 ? 2 take 8 x 2 ? - 6 xy 2 + x 2 ? 2 - 

7. 10 a 2 6 2 + 15 aft 2 + 8 a 2 6 take - 10 a 2 &* + 16 a& 2 - 8 a 2 6. 
.8. 4x 2 -3x + 2 take -6x 2 + 6x-7. 

9. x 8 + llx 2 + 4 take 8x 2 -5x-3. 

10. -Sa^+dx^+lS take QaV- 8x*-&. 

Subtract 

11. x 8 - x 2 + x + 1 from x 8 + x 2 - x + 1. 

12. Sx^-Sxty + x 8 -? 8 from z? + S&y + Szy* + y*. 

13. 6 8 + c 8 -2a6c from a 8 + 6 8 -3a&c. 

14. 7xy 2 -y 8 -3x 2 y + 6x 8 from 8X 8 + 7xty - 3xy 2 - y 8 . 

15. x^ + S + x-Sx 8 from 6x*-8x 8 -2x 2 + 7. 

16. a 8 + & 8 + c 8 -3a6c from 7a6c-3o 8 + 6& 8 -c 8 . 

17. 1— x + x 6 — x 4 — x 8 from x* — 1 + x — x 2 . 

18. 7a*-8o 2 + 3a 6 + a from a 2 - 6a 8 - 7 + 7a 8 . 

19. 10a 2 & + 8a& 2 -8a 8 & 8 -&* from 5 a 2 6 - 6 aft 2 - 7 aW. 

20. a 8 - 6 8 + 8a6 2 - 7 a 2 & from - 8a6 2 + 16a 2 6 + ft 8 . 

From 

21. Jx 2 - Jxy-fy 2 take -tx 2 + xy-i^. 

22. | a 2 - ia - 1 take - |a 2 + a - J. 

23. ix 2 -ix + i take Jx-l + Jx 2 . 

24. fx 2 -f ax take J- Jx 2 - Jox. 

25. fa^-Jx^-y 2 take Jx^-f^-Jxy 8 . 

26. Ja 8 -2ox 2 -£a 2 x take Ja 2 x + Ja 8 - Jax 2 . 

35. We shall close this chapter with an exercise contain* 
ing miscellaneous examples of Addition and Subtraction. 



SUBTRACTION. * 23 



MISCELLANEOUS EXAMPLES I. 

1. To the sum of 2a — 36 — 2 c and 2 b — a + 7 c add the sum of 
a — 4 c + 7 6 and c — 6 6. 

2. From 6a 8 -f 3x — 1 take the sum of 2x — 6 + 7x 2 and 3x 2 + 

4 •— 2x 8 + x. 

3. Subtract 3 a - 7 a 8 + 6 a 2 from the sum of 2 + 8 a 8 — a 8 and 

2o 8 -3o 2 + a-2. 

4. Subtract 6x 2 + 3x — 1 from 2 as 8 , and add the result to 
3a; 2 -|- 3x - 1. 

5. Add the sum of 2 y — 3 y 2 and 1 — 6 y 8 to the remainder when 
1 — 2 y 2 -f y is subtracted from 5 y 8 . 

6. Take a 2 — y 2 from 3 xy — 4 y 2 , and add the remainder to the 
sum of 4xy - x 2 - 3y 2 and 2x 2 + 6y 2 . 

7. Find the sum of 6 a — 7 & -f c and 3 b — 9 a, and subtract the 
result from c — 4 6. 

8. Add together 3x 2 - 7 x + 5 and 2X 8 + 6x - 3, and diminish 
the result by 3 x 2 + 2. 

9. What expression must be added to 5 x 2 — 7 x + 2 to produce 
7x 2 -l? 

10. What expression must be added to 4 x 8 — 3 x 2 + 2 to produce 
4x 8 + 7x-6? 

11. What expression must be subtracted from Sa — 6 6 + c so as 
to leave 2a — 46 + c? 

12. What expression must be subtracted from 9x 2 +llx — 6 so 
as to leave 6x 2 - 17x + 3 ? 

13. From what expression must 11 a 2 — bab — 7 be be subtracted 
so as to give for remainder 5 a 2 + 7 a& + 7 &c ? 

14. From what expression must 3a& + 6 6c — 6 ca be subtracted so 
as to leave a remainder 6ca — 5 be? 

15. To what expression must 7 X s —_ 6x 2 — 6x be added so as to 
make9x 8 -6x-7x 2 ? 

16. To what expression must 6 ab — 11 be — 7 ca be added so as to 
produce zero ? 

17. If 3x 2 — 7x + 2 be subtracted from zero, what will be the 
result? 

18. Subtract 3 x 8 — 7 x + 1 from 2 x 2 — 5 x — 3, then subtract the 
difference from zero, and add this last result to 2 x 2 — 2 x 8 — 4. 

19. Subtract 3x 2 — 6 x + 1 from unity, and add 6x 2 — 6x to the 
result. 



CHAPTER IV. 
Multiplication. 

36. Multiplication in its primary sense signifies repeated 
addition. 

Thus 3x4 = 3 taken 4 times 

= 3 + 3 + 3 + 3. 

Here the multiplier contains 4 units, and the number of 
times we take 3 is the same as the number of units in 4. 

Again a xb = a taken b times 

= a + a + a+ •••, the number of terms being b. 

Also 3x4 = 4x3; and so long as a and b denote pos- 
itive whole numbers, it is easy to show that a xb = b x a. 

37. When the two quantities to be multiplied together are 
not positive whole numbers, we may define multiplication as 
an operation performed on one quantity which when performed 
on unity produces the other. For example, to multiply 4 by ^, 
we perform on £ that operation which when performed on 
unity gives ^ ; that is, we must divide £ into 7 equal parts 

and take 3 of them. Now each part will be equal to , 

5x7 

4x3 

and the result of taking 3 of such parts is expressed by - — -• 

5x7 

„ 4 3 4x3 

Hence ;X- = - — -■ 

5 7 5x7 

24 



MULTIPLICATION. 26 

Also, by the last article, 

4x3 ^ 3x4 ^3 4 
5x7 7x5 7 5 

4 3_3 4 

• • ^ X — _ X • 

5 7 7 5 

The reasoning is clearly general, and we may now say 
that a x b = b x a, where a and b are any positive quanti- 
ties, integral or fractional. 

The same is true for any number of quantities, hence the 
factors of a product may be taken in any order. This is the 
Commutative Law for Multiplication. 



38. Again, the factors of a product may be grouped in any 

way we please. 

Thus abed = a xb x c xd 

= (ab) x (cd) = a x (be) x d = a x (bed). 

This is the Associative Law for Multiplication. 

39. Since, by definition, a s = aaa, and a* = aaaaa, 
.\ a 3 x a 5 = aaa x aaaaa = aaaaaaaa = a 8 = a 8 * 5 ; 

that is, the index of a letter in the product is the sum of its 
indices in the factors of the product. This is the Index Law 
for Multiplication. 

Again, 5a* = 5aa, and 7a 8 = 7 aaa. 

.-. 5 a 2 x 7 a 8 = 5 x 7 x aaaaa = 35 a 5 . 

When the expressions to be multiplied together contain 
powers of different letters, a similar method is used. 

Ex. 6 a 8 6 2 x 8 a a &x* = 5 aaabb x 8 aabxxx = 40 cfibhfi. 

Note. The beginner must be careful to observe that in this proc- 
ess of multiplication the indices of one letter cannot combine in any 
way with those of another. Thus the expression 40 a b b*& admits of 
no further simplification. 



26 



ALGEBRA. 



40. Rule. To multiply two simple expressions together, 
multiply the coefficients together and prefix their product to the 
product of the different letters, giving to each letter an index 
equal to the sum of the indices that letter has in the separate 
factors. 

The rule may be extended to cases where more than two 
expressions are to be multiplied together. 

Ex. 1. Find the product of x% ac 8 , and x 8 . 

The product = x 2 x s 8 x s 8 = x 2+ * x x* = x*+*+* = z w . 

The product of three or more expressions is called the 
continued product. 

Ex. 2. Find the continued product of 6&V, 8y 2 sfi, and 3xz*. 
The product = 5a?V x Sy 2 * 6 x 3 as** = 120 x*y*afi. 



MULTIPLICATION OF A COMPOUND EXPRESSION BY 

A SIMPLE EXPRESSION. 

41. By definition, 

(a + b)m— wi + m + w+- taken a + b times 
= (ra + m + m + ••• taken a times), 

together with (m + m+m + "' taken b times) 

= am -{-bm (1). 

Also (a — b)m= m + m-\- m + ••• taken a — b times 

= (m + ra + ra -f ••• taken a times), 

diminished by (m -f m + m -f- ••• taken b times) 

= am — bm (2). 

Similarly, (a — b + c)m 

= am — bm -f cm. 

Hence the product of a compound expression by a single 
factor is the algebraic sum of the partial products of each term 
of the compound expression by that factor. This is known as 
the Distributive Law for Multiplication. 



MULTIPLICATION. 27 

Ex. 3(2a + 3&-4c)=6a + 96-12c. 

(4s 2 - 7 y - 8 s 8 ) x 3xy 2 = 12afy 2 - 21xy 8 _- 24 sjfls 8 . 

Note. It should be observed that for the present a, 6, c, m de- 
note positive whole numbers, and that a is supposed to be greater 
than b, 

EXAMPLES IV. a. 
Find value of 

1. 5* 2 x7aA 6. 2a6cx3oc 8 . 11. «yx6aV. 

2. 4 a 8 x 5 a 8 . 7. 2 a 8 6 8 x 2 a 8 6 8 . 12. abc x «y«. 

8. 7 aft x 8 aW. 8. 5 a 2 6 x 2 a. 13. 3 oWs 8 x 5 a 8 6x. 

4. 6»y 2 x 5X 8 . 9. 4a 2 6 8 x 7 a 6 . 14. 4 a 8 6x x 7 d 2 **. 

5. Scflbxlfi. 10. SaWxa^y*. 15. Sa^xxSca. 

Multiply 

16. 6«V by Qahfi. 21. 5s + 3y by 2a& 

17. 2aty by afy 7 . 82. a 2 + ft 2 - d 2 by a 8 6. 

18. Sctoty 1 by a 2 ^. 28. &c + ca-a& by abc 

19. aft + be by a 8 6. 24. 6 a 2 + 3 ft 2 - 2 c 2 by 4 a*bc\ 

20. 6a6-76x by 4a 2 foe 8 . 25. bx*y + xy*-TriW by 3s 8 . 

MULTIPLICATION OF COMPOUND EXPRESSIONS. 

42. If in Art. 41 we write c + d for m in (1), we have 
(a + &)(c + d)= a(c 4- d)+ 6(c + d) 

= (c + d)a + (c + d)6 [Art. 37.] 

= ac + ad + be + bd. 
Again, from (2) 

(a — b)(c 4- d)= a(c 4- d) — b(e 4- d) 

= (c + d)a — (c + d)b 
= ac + ad —(be -f bd) 
= ac + ad —be — bd. 

Similarly, by writing c — d f or m in (1) 

(a + 6)(c - d)= a(c - d)+ b(c - d) 

= (c — d)a +(c — d)b 
*= ac — ad + be — bd. 



28 ALGEBRA. 

Also, from (2) 

(a — b)(c — d)=a(c — d)— b(c — d) 

= (c — d)a—(c — d)b 
= ac — ad —(be — bd) 
= ac — ad — bc + bd. 

If we consider each term on the right-hand side of this 
last result, and the way in which it arises, we find that 

(+a)x(+c)= + ac, 

(_&)x(-d)= + H 
(-&)x(+c) = -6c, 
(+ a) x (— d) = — ad. 

These results enable us to state what is known as the 
Rule of Signs in multiplication. 

Rule of Signs. The product of two terms with like signs is 
positive; the product of two terms with unlike signs is neg- 
ative. 

43. The rule of signs, and especially the use of the nega- 
tive multiplier, will probably present some difficulty to the 
beginner. Perhaps the following numerical instances may 
be useful in illustrating the interpretation that may be 
given to multiplication by a negative quantity. 

To multiply 3 by — 4 we must do to 3 what is done to 
unity to obtain — 4. Now — 4 means that unity is taken 
4 times and the result made negative; therefore 3 x(— 4) 
implies that 3 is to be taken 4 times and the product made 
negative. 

But 3 taken 4 times gives + 12. 

... 3 x (-4) = -12. 

Similarly, — 3 x — 4 indicates that — 3 is to be taken 4 
times, and the sign changed ; the first operation gives — 12, 
and the second + 12. 

Thus (-3)x(-4)= + 12. 



MULTIPLICATION. 29 

Hence, multiplication by a negative quantity indicates that we 
are to proceed just as if the multiplier were positive, and then 
change the sign of the product. 

44. Note on Arithmetical and Symbolical Algebra. 
Arithmetical Algebra is that part of the science which deals 

solely with symbols and operations arithmetically intelli- 
gible. Starting from purely arithmetical definitions, we are 
enabled to prove certain fundamental laws. 

Symbolical Algebra assumes these laws to be true in every 
case, and thence finds what meaning must be attached to 
symbols and operations which under unrestricted conditions 
no longer bear an arithmetical meaning. Thus the results 
of Arts. 41 and 42 were proved from arithmetical definitions 
which require the symbols to be positive whole numbers, 
such that a is greater than b and c is greater than d. By 
the principles of Symbolical Algebra we assume these re- 
sults to be universally true when all restrictions are removed, 
and accept the interpretation to which we are led thereby. 

Henceforth we are able to apply the Law of Distribution 
and the Rule of Signs without any restriction as to the 
symbols used. 

45. To familiarize the beginner with the principles we 
have just explained we add a few examples in substitutions 
where some of the symbols denote negative quantities. 

Ex. 1. If a = — 4, find the value of a 8 . 

Here a 8 = (- 4)* =(- 4>x(-4)x (-4) = - 64. 

Ex.2. If a = -l, 6 = 3, c=-2, find the value of -ZatbcP, 
Here - 3o*6c 8 = - 3 x (- l) 4 x 3 x (- 2)» 

= -3xl x3x(-8)=72. 

EXAMPLES IV. to. 
If a = —2, 6 = 3, c = — 1, x = — 5, y = 4, find the value of 



1. 


3 a 2 6. 


6. 


3 a 2 c. 


11. 


- 4 a*c*. 


16. 


4c 5 « 8 . 


2. 


8 a6c 2 . 


7. 


- 6 2 c 2 . 


12. 


Sc*z*. 


17. 


- 6 a 2 6 2 c 2 . 


3. 


-6c 8 . 


8. 


3 a*c*. 


13. 


6a 2 z 2 . 


18. 


-7a 8 c 8 . 


4. 


6 a?c 2 . 


9. 


- 7 a 8 6c. 


14. 


— 7 c*zy. 


19. 


8c*a^. 


5. 


4c 8 y. 


10. 


- 2 a*6s. 


15. 


— Sax 9 . 


20. 


7a*c*. 



80 ALGEBRA. 

H a = — 4, 6 = — 3, c = — 1, /= 0, as = 4, y=sl t find the value oi 
21. 3a 2 +&c-4cy. 34. 3 aty* - 6 b*x - 2c 8 . 

83. 2 a& 2 - 3 be 2 + 2/s. 26. 2a 8 -S& 8 + 7ey*. 

23. /a 3 -263-cx». 26. 36V -4& 2 /-6c*«. ' 

27. 2 v /(oc)-8V(a^) + V(6 a c 4 )- 

23. 3 V(acx) - 2 vtW - 6 V(<fy)« 

29. 7 V(a^) - 3 V(6 4 c a ) + 6 V(/**). 

30. 3 c V(3 6c) - 6 V(4 c 2 ? 8 ) - 2 cy V( 3 W) • 

46. The following examples further illustrate the rule of 
signs and the law of indices. 

Ex. 1. Multiply 4 a by - 3 6. 

By the rule of signs the product is negative ; also 4 a x 3 b = 12 ab. 

.-. 4ax(-36) = -12a6. 

Ex. 2. Multiply — 6 a& 8 x by — ab*x. 

Here the absolute value of the product is 6 a s &*a; 2 9 and by {he rule 
of signs the product is positive. 

.-. (-&ab*x)x(-a&x)=6a*lfix*. 
Ex. 8. Find the continued product of 3 a 2 6, — 2 a 8 &V — aft 4 . 

o or / o rmn a jsm Tnis result, however, may be 

3a 2 6 x(— 2a 8 6 2 ) = — Qafib 8 ; , V . 

v written down at once ; for 

(-6a 6 6 8 )x(-a& 4 ) = + 6a 6 & 7 . 9 _ . t . t _ 4 A R . T 

v y v y 8 a 2 6 x 2 a 8 6 2 x aft 4 = 6 a 6 6 T , 

Thus the complete product is , , A , , . , A , 

fl 6 7 and by the rule of signs the re- 

quired product is positive. 

Ex. 4. Multiply 6 a 8 - 6 a?b - 4 ab 2 by - 3 aft 2 . 

The product is the algebraic sum of 'the partial products formed 
according to the rule enunciated in Art. 40 ; 

thus (6 a 8 - 6 a 2 6 - 4 aft 2 ) x (- 3 ab 2 ) = - 18 a 4 ^ 2 + 16 a*b» + 12 a 2 6 4 . 

EXAMPLES IV. O. 

Multiply together 

1. ax and — 3 ax. 5. — abed and — 3 a 2 6 8 c*<P. 

2.-2 abx and — 7 a&x. 6. xys and — 6x*y*z. 

3. a 2 6 and — ab 2 . 7. 3 xy + 4 y« and — 12 xyz. 

4. 6 x*y and — 10 ay. 8. ab — 6c and a 2 bc*. 



MULTIPLICATION. 81 

9. -x-y-*and -Sac. 18. - 2 a 2 b - 4 ad 2 and - 7a 2 6 2 . 

10. a 2 - ft 2 + c 2 and a&c. 13. 6x 2 y - 6xy 2 + 8xV» and 3xy. 

11. —ab + bG — ca and — a&c. 14. — 7 xfy— 6 xy 8 and — &xPjp. 

15. — 5 xy' 2 z + 3 xy« 2 — 8 x*yz and xys. 

16. 4 xVs 2 — 8 xyz and — 12 xfys 8 . 18. 8 xyz — 10 x*yz* and — xy«. 

17. - 13 xy 2 - 15 x*y and - 7 x 8 ? 8 . 19. abc - a 2 6c - ab*c and - a&c. 

20. — a 2 bc + 6 2 ca — c*db and — a&. 

Find the product of 

21. 2a — 35 + 4c and — fa. 28. }o — J6 — c and fox. 

22. 3x-2y-4and -fx. 24. $ a 5 ^ 2 - $ ax 8 and - J a 8 *. 

25. -$a*x 2 and -}a 2 +ax-fx 2 . 
26. -Jxyand -3x 2 +fxy. 27. - Jxfy 2 and - Jx 2 + 2y*. 

47. The results of Art. 41 may be extended to the case 
where one or both of the expressions to be multiplied 
together contain more than two terms. For instance 

(a — & -f c) ra = am — bm + cm\ 

replacing m by x — y, we have 

(a - b + c)(x — y) = a(x — y)- b(x - y)+ c(x - y) 

= (ax — ay) — (bx — by)+(cx — cy) 
= aa? — ay — foe + &y -f «c — cy. 

48. These results enable us to state the general rule for 
multiplying together any two compound expressions. 

Rule. Multiply each term of the first expression by each 
term of the second. When the terms multiplied together have 
like signs, prefix to the product the sign +, when unlike prefix 
— ; the algebraic sum of the partial products so formed 
gives the complete product. 

This process is called Distributing the Product 
Ex. 1. Multiply x + 8 by x + 7. 

x + 8 
x + 7 
x 2 + 8x 
+ 7x + 56 
by addition x 2 -f!5x + 66 



82 



ALGEBRA. 



Note. We begin on the left and work to the right, placing the 
second result one place to the right, so that like terms may stand in 
the same vertical column. 

Ex. 3. Multiply 2x - Sy by 4x - 7 y. 

2x -8y 
4x — 7y 



by addition 



8x a -12xy 

— 14xy4- 21^ 

8x*-26xy 4-210* 



EXAMPLES IV. CL 



Find the product of 

1. x + 6 and x + 10. 

2. x 4- 6 and x — 6. 
8. x — 7 and x — 10. 

4. a; — 7 and x + 10. 

5. x 4- 7 and x — 10. 

6. x 4- 7 and x 4- 10. 

7. x + 6 and x — 6. 

8. x 4- 8 and x — 4. 

9. x — 12 and x — 1. 

10. x + 12 and x — 1. 

11. x — 16 and x4- 16. 

12. x — 16 and — x 4- 3. 

13. — x — 2 and — x — 3. 

14. — x 4- 7 and x — 7. 

15. — x + 6 and — x — 6. 

16. x - 13 and x 4- 14. 

17. x - 17 and x 4- 18. 

18. x + 19 and x - 20. 

19. _ x - 16 and - x 4- 16. 

20. - x + 21 and x - 21. 



21. 2x — 3 and x + 8. 

22. 2x + 3 and x - 8. 

23. x — 6 and 2 x — 1. 

24. 2 x — 6 and x — 1. 

25. 3x -5 and 2x4-7. 

26. 3x + 6 and 2x-7. 

27. 6x-6 and 2x4-3. 

28. ox + 6 and 2x-3. 

29. 3 x — 6 y and 3 x 4- 6 y, 

30. 3x — 6 y and 3 x — 6 y. 
81. a — 2b and a + 3 6. 

32. a — 7 6 and a 4- 8 6. 

33. 3a -66 and a -86. 

34. a — 9 b and a 4- 6 6. 

35. x 4- a and x — b. 
86. x — a and x 4- 6. 

37. x — 2 a and x 4- 3 6. 

38. ax — by and ax + by. 
39 xy — a& and xy 4- a&. 

, 40. 2pq — 3r and 2pg + 3 r. 



MULTIPLICATION. 33 

49. We shall now give a few examples of greater 
difficulty. 

Ex. 1. Find the product of Sx 2 — 2x — 6 and 2 x — 5. 

3 x 2 — 2 x — 5 Each term of the first expression is mul- 

2 x _ 5 tiplied by 2 jc, the first term of the second 

a ^ __ 4 2 _ i o x expression; then each term of the first 

— lfi 2 4- 10 rofi expression is multiplied by — 6 ; like terms 

-— - — <<% ^-— • are placed in the same columns and the 

6*»-19** +26 resu ite added. 

Ex. 2. Multiply a-b + Scby a + 2b 

a - 6 + 8c 
a + 26 

a* — a& + 3 oc 

2ab -26 2 + 66c 

a 2 + a6 + 3ac-26 2 + 66c 

When the coefficients are fractional, we use the ordinary 
process of Multiplication, combining the fractional coeffi- 
cients by the rules of Arithmetic. 

Ex. 8. Multiply Ja 2 - £ ab + f ft 2 by \ a + \b. 

i« 2 - lab +|& 2 

ja + jb 

Jo 8 - \a 2 b + \ab 2 

+ itfb-laV + jb* 

ia 8 -&a 2 & + £a& 2 + J&» 

50. If the expressions are not arranged according to 
powers ascending or descending of some common letter, a 
rearrangement will be found convenient. 

Ex. Multiply 2xz-z 2 + 2x 2 -3yz + xy by x-y + 2z. 

2» 2 + xy + 2xz-3yz-z 2 
x — y + 2z 

2x*+ x 2 y+2x 2 z-3xyz- xz 2 

— 2x 2 y -2xyz - xy* + 3y 2 z + yz 2 
4x 2 z + 2xyz + 4xz 2 -6y3 2 -2* 8 

2s» - x*y + 6x 2 z - Zxyz + 3xz 2 - xy 2 + 3y 2 z- byz 2 -2z* 
•o 



84 ALGEBBA 

EXAMPLES IV. e 
Multiply together 

1. a + b + c and a + b — e. 

2. a — 2b + c and a + 2 6 — & 

3. a 2 - aft + b 2 and a* + aft + 6*. 

4. x 2 + 3y 2 and x + 4y. 

5. x 8 -2x 2 + 8 and 3 + 2. 

6. x* - xty 2 + y* and x 2 + y». 

7. x 2 + xy + y 2 and x — y. 

8. a 2 - 2 ox + 4x 2 and a 2 + 2 ox + 4«& 

9. 16a 2 +12a& + 96 2 and 4a-Sft. 

10. cPx — ax 2 + x 8 — a 8 and x + a. 

11. x 2 + x - 2 and x 2 + x - 6. 

12. 2x 8 -3x 2 + 2x and 2x 2 + 8x + 8. 

13. - a 5 + a 4 6 - a 8 ^ 2 and - a - 6. 

14. X8-7X + 5 and x 2 -2x + 8. 

15. a 8 + 2a 2 6 + 2a& 2 and a 2 -2a&+26*. 

16. 4x 2 + 6xy + 9y 2 and 2x-3y. 

17. x 2 — 3xy — y 2 and — x 2 + xy + y 8 . 

18. ^-a^ + o 8 and a 8 + a 2 ft 2 + ft 8 . 

19. x 2 - 2xy + y 2 and x 2 + 2xy + y*. 

20. ab + cd + ac+ bd and a& + cd — ac — bd. 

21. - 3 a 2 6 2 + 4 a& 8 + 16 a*b and 6 a 2 b 2 + aft 8 - 3 8*. 

22. 27x 8 -86ax 2 + 48a 2 x-64a 8 and 3x + 4o. 

23. a 2 - 5a& - b 2 and a 2 + 5a& + ft 2 . 

24. x 2 — xy + x + y 2 + y + 1 and x + y — 1. 

25. a 2 + 6 2 + c 2 -6c-ca-aft and a + 8 + c 

26. — x*y + y 4 + « 2 y 2 + x* — xy 8 and x + y. 

27. x 12 - x 9 y 2 + x«y* - x*y« + y 8 and x 8 + y a . 

28. 8a 2 + 2a + 2a 8 + l + a 4 and a 2 -2a + l. 

29. — ax 2 + 3 axy 2 — 9 ay* and — ox — 8 ay 2 . 

30. -2x 8 y + y 4 + 3x 2 y 2 + x*-2xy 8 and x 2 + 2xy+.y i . 

31. §a 2 + Ja + J and }a - J. 
82. ix 2 -2x+J and |x + J. 



MULTIPLICATION. 86 

38. fa 2 + xy + j y a and \x - Jy. 

84. |a& 2 -as-}a 2 and }a? - }ax+ fco 9 . 

35. J* 2 — fas-f and J a: 2 + }a?-f. 

86. i<xx + ix 2 + Ja 2 and f a 2 + fas 2 - Jo*. 

Note. Examples involving literal, fractional, and negative expo- 
nents will be found in the chapter on the Theory of Indices. 

51. Products Written by Inspection. Although the result 
of multiplying together two binomial factors, such as a? + 8 
and x — 7, can always be obtained by the methods already 
explained, it is of the utmost importance that the student 
should learn to write the product rapidly by inspection. 

This is done by observing in what way the coefficients of 
the terms in the product arise, and noticing that they result 
from the combination of the numerical coefficients in the 
two binomials which are multiplied together ; thus 

(a? + 8) (a? + 7) = a? + 8a? + 7a? + 56 

= 3?* + 15 a? + 56. 

(a?- 8) (a? - 7)= x* - 8a? - 7a? + 56 

= a? 2 -15 a? + 56. 

(a? + 8) (a? - 7)= a?* + 8a? - 7a? -56 

= X* •{• X — 56. 

(a?- 8) (a? + 7)= a? - 8a? + 7a?- 56 

ssx 2 — x — 56. 

In each of these results we notice that : 

1. The product consists of three terms. 

2. The first term is the product of the first terms of the 
two binomial expressions. 

3. The third term is the product of the second terms of 
the two binomial expressions. 

4. The middle term has for its coefficient the sum of the 
numerical quantities (taken with their proper signs) in the 
second terms of the twt binomial expressions. 



86 ALGEBRA. 

The intermediate step in the work may be omitted, and 
the products written at once, as in the following examples : 

(a? + 2) (x + 3) = x 9 + 5 x + 6. 
(x - 3) (x + 4) = x 9 + x - 12. 
(x + 6) (x -9) = a 8 - 3a? -54. 
(a? — 4 y) (a? — 10 y) = a? 2 — 14 xy + 40 y 1 . 
(a;-6y)(aj + 4y)=a?-2a^-24y > . 

By an easy extension of these principles we may write 
che product of any two binomials. 

Thus (2x + 3y) (x - y)= 2a? + Sxy -2xy-3tf 

= 2a? + xy-3y*. 
(3a?-4y)(2a? + y)=6a?-8ajy + 3ajy-4^ 

= 6a? — 5 xy — 4y*. 

EXAMPLES IV. f. 

Write the values of the following products : 

1. (a + 8)(a;-6). 15. (a -6)(a + 18). 

8. (s + 6)(as-l). 16. (a + 3)(a + 3). 

8. (3 - 8)(s + 10). 17. (a - ll)(a + 11). 

4. (x-l)(x + 5). 18. (a-8)(a-8). 

5. (s-f 7)(s-9). 19. (3-3a)(a; + 2a). 

6. (*-10)(a;-8). 30. (x + Ga)(x - 5 a). 

7. (3 -4) (a +11). 21. (aj + 3a)(aj-8a). 

8. (x-2)(aj + 4). 82. (« + 4y)(aj-2y). 

9. (x+2)(aj-2). 28. (z + 7y)(a>- 7y). 

10. (a-l)(a + l). 24. (a- 3y)(x - 3y) 

11. (a + 9)(a-6). 25. (3 x - 1) (a + 1). 

12. (a-3)(a-f 12). 28. (2 x -f 5) (2 x - 1). 

13. (a-8)(a + 4). 27. (S a; + 7) (2 x - 3). 

14. (a-8)(a + 8).. 28. (4x- 3)(2a; + 3). 



MULTIPLICATION. 87 

(8* + 8)(8a>-8). 88. (2x + 7y)(2s- 6y). 

80. (2aj-5)(2*-5). 84. (5a + 3a)(5a;-3a). 

81. (3s-2y)(3x + y). 85. (2s- 5a)(x + 5a). 
88. (3x + 2y)(3x + 2y). 86. (2z + a)(2z + a). 

MULTIPLICATION BY DETACHED COEFFICIENTS. 

52. In the following cases we lessen the labor of multipli- 
cation by using the Method of Detached Coefficients: 

(i.) When two compound expressions contain but one 
letter. 

(ii.) When two compound expressions are homogeneous 
and contain but two letters. 

Ex.1. Multiply 2s 8 -4s 2 + 5s -5 by 3s a + 4s-2. 
Writing coefficients only, 



2- 


4 + 


5- 


5 




3 + 


4- 


2 






6- 


12 + 15 - 


15 




+ 


8- 


16 + 20 - 


-20 




— 


4 + 


8- 


-10 + 10 



6- 4- 5 + 13-30+10 

Inserting the literal factors according to the law of their formation, 
which is readily seen, we have for the complete product, 

6 x 6 - 4x* - 5x* + 13s 2 - 80s + 10. 

Ex. 8. Multiply 3a* + 2o«6 + 4a6 8 + 2 6* by 2a a - 6 2 . 

34-24-0-4-44-2 ^ tne ^ rst ex P reS8 i° n the term con- 

2 4-0 — 1 taining a 2 6 2 is missing, so we write a 

H ■ 4 .a I a 1.4 zero i* 1 tne corresponding term in the 

+ ]_oJ]q_o — 4 — 2 ** ne °* coen * c i ent ?. I n the second ex- 

- — ~ — ~~ ~~ A ~ Q pression we write a zero for the coeffi- 

6 + 4-3+6 + 4-4-2 cient of the missing term ah% 

The law of formation of literal factors is readily seen, and we have 
for the complete product, 

Ga*+ Aa*b - 3a 4 6 2 + 6a*b* + 4a 2 &* _4a6 5 -26* 

EXAMPLES IV. fir. 

1. Multiply s 6 + x* + x 2 + 2x+l by a 8 + s - 2. 

2. Multiply a 8 + 6 a 2 6 + 12 ab 2 + 8 6 8 by 3 a 8 + 2 6 8 . 
8. Multiply 2a*-3a 2 + 4a + 4 by 2 a 2 - 3 a - 2. 

4. Multiply 3 s 6 + 2s*y - xh/ 1 + xy* by x 2 + 4 ay - 6^. 



CHAPTER V. 



Division, 



53. When a quantity a is divided by the quantity b, the 
Quotient is defined to be that which when multiplied by b 

produces a. The operation is denoted by a -f- b, % or a/6; 

in each of these modes of expression a is called the dividend, 
and b the divisor. 
Division is thus the inverse of multiplication, and 

(a -*- b) x b = a. 

54. The Rale of Signs holds for division. 

Thus a&H-a = 2$ = ^i=& 

a a 

a a 

a»+(-a)— «*.«, (-«)*(-») fc 

— a — a 

-ad-K-a) = ^ = (-")xft = fc 

— a —a 

Hence in division as well as multiplication 

like signs produce +, 
unlike signs produce — . 

55. Since Division is the inverse of Multiplication, it fol- 
lows that the Laws of Commutation, Association, and Distri- 
bution, which have been established for Multiplication, hold 
for Division. 

88 



DIVISION. 89 



DIVISION OF SIMPLE EXPRESSIONS. 

56. The method is shown in the following examples : 

Ex. 1. Since the product of 4 and x is 4z, it follows that when 4x 
is divided by x the quotient is 4, 
or otherwise, 4 x ■+• x = 4. 

Ex.3. Divide 27 a 6 by 9 a 8 . _- . .. ... . 

97 «6 97 ™ e ^mov© * rom the divisor ana 

The quotient = ^ = ±L£22«1 dividend the factors common to 

9 a Vaaa ^^ M ^ Arifchmetic# 
= 3 aa = 3 a a . 
Therefore 27 a 5 + 9 a 8 = 3 a 2 . 

Ex. 3. Divide 35 a*b*cfi by 7 aW*. 

The quotient = 36 aaa ' bb ' «* = Sqq.c = 6a«c. 

1 a.bb.cc 

We see, in each case, that $e indeo; o/ any Jefter in ^e 

quotient is the difference of the indices of that letter in the 

dividend and divisor. This is called the Index Law for 
Division. 

We can now state the complete rule : 

Rule. The index of each letter in the quotient is obtained 
by subtracting the index of that letter in the divisor from that 
in the dividend. 

To the result so obtained prefix with its proper sign the quo- 
tient of the coefficient of the dividend by that of the divisor. 

Ex. 4. Divide 46 a*b 2 x* by -9a 8 6x a . 

The quotient = (- 5) x a 6 " 8 6 a ^a*-* 
= - 6 a 8 6x 2 . 

Ex.5. -21a 2 6 8 -i-(-7a 2 & 2 ) = 3&. 

Note. If we apply the rule to divide any power of a letter by the 
same power of the letter, we are led to a curious conclusion. 

Thus, by the rule a* + a* = a 8 " 8 = a ; 

but also a 8 -*- a 8 = — = 1.* 

a 8 

.\ a° = l. 

This result will appear somewhat strange to the beginner, but its 
full significance will be explained in the chapter on the Theory of 
Indices. 



40 ALGEBRA. 

DIVISION OF A COMPOUND EXPRESSION BY A SIMPLE 

EXPRESSION. 

57. Rule. To divide a compound expression by a single 
factor, divide each term separately by that factor, and take the 
algebraic sum of the partial quotients so obtained. 

This follows at once from Art. 40. 

Ex. 1. (9x- 12y + 3s)-*--3=-3x + 4y-0. 

Ex. 2. (36 a 8 & 2 - 24 aW - 20 a*ft 2 ) + 4 a 2 6 = 9 ab - 6 6* - 6 a*6. 

Ex. 3. (2x 2 -5a^ + fa;V)^-}* = -4*+10y-8xy». 

EXAMPLES V. a. 
Divide 

1. 3a* by x 2 . 15. - SOy 8 ** by -6x»y. 

2. 27 x* by -Ox 8 . 16. x 8 - 3x 2 + x by x. 

3. -35x«by7x 8 . 17. x« - 7x6 + 4x* by x 2 . 

4. xfy 8 by x 2 y. 18. 10x 7 - 8x« + 3x* by x 8 . 

5. a 4 x 8 by - aH*- 19. 16x 6 -26x* by -Sx 8 . 

6. 12 a 5 & 6 c* by - 3 <fibH. 20. -24x« - 32 x* by - 8X 8 . 

7. - a 6 c 9 by - acfi. 21. 34 xV - 61 xV by 17 xy. 

8. 15 xfy 7 * 4 by SxtyV- 23- a*-ab- ac by - a. 

9. - 16x 8 y 2 by - 4xy*. 23. a 8 - a 2 6 - a 2 6 2 by a 2 . 

10. -48a 9 by -8a 8 . 24. Sx 9 - 9x 2 y- 12xy 2 by -3x. 

11. 63 tfbW by 9 aWc*. 25. 4x*y 4 -8x 8 y 2 +6xy 8 by -2xy. 

12. 7 a 2 6c by - 7 a?bc. 26. Jxfy 2 - 3xV by - Jxfy 2 . 

13. 28 a 4 6 8 by - 4 a 8 6. 27. - £ x 2 + Jxy + -^x by - f x. 
14 16 tftyx 2 by - 2 xy. 28. - 2 a 5 x 8 + J a 4 x* by $a*x. 

DIVISION OF COMPOUND EXPRESSIONS. 

58. We employ the following rule : 

Rule. 1. Arrange divisor and dividend according to ascend, 
ing or descending powers of some common letter. 

2. Divide the term on the left of the dividend by the term 
on the left of the divisor, and put the result in the quotient 



DIVISION. 41 

3. Multiply the whole divisor by this quotient, and put the 
product under the dividend. 

4. Subtract and bring down from the dividend as many 
terms as may be necessary. 

Repeat these operations till all the terms from the dividend 
are brought down. 

Ex. 1. Divide x 2 + 11 x + 30 by x + 6. 

Arrange the work thus : 

a? + 6)x 2 +lla; + 30( 

divide a 2 , the first term of the dividend, by », the first term of the 
divisor ; the quotient is x. Multiply the whole divisor by x, and put 
the product x 2 + 6 x under the dividend. We then have 

* + 6)x 2 +lla; + 30(aj 

x 2 + 6x 

by subtraction 6 x + 30 

On repeating the process above explained we find that the next 
term in the quotient is + 5. 

The entire operation is more compactly written as follows : 

* + Q)x* + 11 x + 30(a? + 5 

x*+ 6a? 

6x + 30 
5z + 30 

The reason for the rule is this: the dividend may be 
divided into as many parts as may be convenient, and the 
complete quotient is found by taking the sum of all the par- 
tial quotients. Thus x 2 -+ 11 x -+ 30 is divided by the above 
process into two parts, namely, x 2 + 6 x, and 5 x + 30, and 
each of these is divided by x + 6; thus ■ we obtain the 
complete quotient x + 5. 

Ex. 2. Divide 24x 2 - 65 xy + 21 y* by Bx - Sy. 

Sx -Sy}24x 2 - 65 sy + 21y 2 (Sx - 7y. 
24a 2 - 9xy 

-56sey + 21y 2 
-66xy + 21y 2 



42 ALGEBRA. 

EXAMPLES V. U, 
Divide 

L x* + 3x + 2byx+L a 4x3 + 23* + 15 by 4x + S 

t. a^-Tx + W by *-a 10. 6x*-7x-3 by 2x-3. 

3. a 3 -lla + 30 by a-6. 11. 3x« + x - 14 by x - 2. 

4. fl*-49a + 600 by a-26. IS. 3**-*- 14 by x + 2. 

5. 3x* + 10x + 3 by x + 3. 13. 6x»- 31x + 36 by 2x- 7. 

6. 2x 2 + llx + 5 by 2x + l. 14. 12a 2 -7ax-12x*by 3a-4x, 

7. 6x 2 + llx + 2 by x + 2. 16. 16a*+17fltt-4s* by 3a+4x 

8. 2x 2 + 17x + 21 by 2x + 8. 16. 12a 2 -llac-36<*by4a-9c 

17. -4xy-16y 2 + 96x 2 by 12x-6y. 

18. 7x 8 + 96x 2 -28x by 7x-2. 20. 27x 8 +9x 2 -3x-lDby 3x-2. 

19. 100x 8 -3x-13x 2 by3+25x. 21. 16a 8 -40a 2 +39a-9by8a-3. 

59. The process of Art. 58 is applicable to cases in which 
the divisor consists of more than two terms. 

Ex.1. Divide 6x*-x* + 4x 8 -6x 2 -x- 15 by 2x*-x + 3. 

2x 2 -x + 3)6x 5 - x* + 4x 8 -6x 2 -x-16(3x 8 + x 2 -2x-6 
6x*-3x* + 9x 8 

2x*-5x 8 - 6x* 
2x*- x»+ 3x* 

-4x«- 8x 2 - x 
~4x»+ 2x 2 -6x 

-10x 2 + 5x-15 
-10x 2 + 5x-15 

Ex. 2. Divide a 8 + ft 8 + c 8 - 8 ale by a + b'+ e. 

a + 6 + c) a 8 - 3 a&c + & 8 + c 8 (a 2 -a&-<w + &*-&o + cP 
a 8 + a 2 b + a^ 

- a 2 b - a 2 c - 3 abc 

- g*b-ab*- abc 

-a 2 c + a6 2 -2a6c 

— o 2 c — qfrc — qc 2 

a& 2 - a&c + ac 2 + 6 8 

q&j +5 8 + ft*o 

- a&c + ac 2 - 6 2 c 

- a&c - & 2 c - 6c 2 

ac 2 + 6c 2 + c 8 

Note. The result of this division will be referred to later. 



DIVISION. 48 

60. Sometimes it will be found convenient to arrange the 
expressions in ascending powers of some common letter. 

Ex. Divide 2o* + 10 - 16a-80a a -f 16a* by 2-4a-6a 2 . 

2 - 4a - 608)10 - 16a - 30a 2 + 2 a* -f 16a*(6 + 2a - So* 
10 -20a -26a 2 

4a -14a 2 + 2a 8 
4 q - 8a 2 -10a* 

- 6<*+ 12a* -ft-lfta* 

- 6a* + 12q»+16a* 

63. When the coefficients are fractional, the ordinary 
process may still be employed. 

Ex. Divide Jaj» + ^a^ + ^V 8 byi* + Jy. 

iaJ + iirH^ + A^ + A^a^-iair + ilf 1 

In the examples given hitherto the divisor has been exactly 
contained in the dividend. When the division is not exact, 
the work should be carried on until the remainder is of lower 
dimensions [Art. 29] than the divisor. 

examples v. a 
Divide 

L ^-*~9£-12by«* + 3* + 8. 
8. 2y»-8y*-6y-lby2y»-6y-l. 
8. 6m 8 -m 2 - 14m + 3 by Sm 2 + 4m- 1. 
4. 6a 5 -13a* + 4a* + 3a 2 by 3a*-2a a -a 
ft. x* + x»+7* 2 -6a; + 8bya; 2 + 2a; + 8. 
a o 4 -a 8 -8a a + 12a-9by a 2 + 2a-3. 

7. a* + 6a* + I3a 2 + 12a + 4bya 2 + 3a + 2. 

8. 2&*-& 8 + 4a; 2 + 7a; + 1 by x 2 -s+3. 

a x B -6a5* + 9«*-6a; 2 -x + 2byx 2 -8a; + 2. 
10. «»-4« 1 -l-8«« + 8« 2 -8a5 + 2byfl^-.*«-SL 



44 



n. 

12. 
13. 
14. 
15. 
16. 
17. 
18. 
10. 
20. 
21. 
23. 
24. 
25. 
27. 
28. 



ALGEBRA. 

30 ac* + 11 x* - 82 a;* - bz + 3 by 2x - 4 + 3a?. 
30y + 9 - 71 y» -f 28 y* - 35 y* by 4y 2 - 13 y + 6. 
# _ 15^4 + 4 #i + 7^2 _ 7 j. + 2 by 3 k* - jfc + 1. 

R J. Q«nL 31 m X Qm2 i l m 8 i ~,K V~ O o _ 



6 



0/^-10^ + 4^ + 7^-7^ + 2 by 3*8-* + !. 

15 + 2 m* - 31 m 4- 9»» 2 + 4 ro* + m* by 3 - 2m - m*. 

2 a* - 8a; + z* + 12 - 7 a; 2 by a; 2 + 2 - 3x. 

a 6 - 2 a* - 4** + 19a; 2 by a* - 7a; + 6. 

192 - a* + 128a; + 4a; 2 - 8a? by 16 - a; 2 . 

14x* + 45 a% + 78 x 2 y« + 45 ay 8 + 14 y* by 2a; 2 + 5a;y + 7 y». 

x 6 -x*y + x*y 2 -a* + a; 2 -y*bya*-a;-y. 

x 6 + x*y - x*y 2 + X s - 2a;y 2 + y* by a; 2 + xy - y\ 

a 9 - b 9 by a* - &». 22. a; 9 - y 9 by a; 2 + xy + y*. 

»7 _ 2y" - 7 a£y* - Txy 12 + 14x*y8 by x - 2y 2 . 

a 8 + 3a 2 & + 6 8 - 1 + 8a6 2 by a + b - 1. 

a^-ySbya^ + aj^ + aj^ + yS. 26. a 12 - ft* 2 by a* - P. 

a w + 2a 6 6« + 6 12 by a* + 2a 2 6 2 + 6*. 

1 — a 8 - 8a£ - 6 aa; by 1 — a — 2 «. 



Find the quotient of 

29. Ja 8 - Ja 2 a; + y aa; 2 - 27a* by £a- 8a. 

30. Aa>-Atf» + Aa- &t>yi«-i. 

31. la^ + T^a^byi^ + iac. 

32. A* 5 -if*aa* by fa-fa;. 

33. , 9 lT a*- 8 f a»-{tf + *a + J£ly!tf-J-a 

34. 36a; 2 + iy 2 + J-4a^-6a;-r iyby6a;-Jy 



-f 



62. Important Cases in Division. The following examples 
in division may be easily verified ; they are of great im- 
portance and should be carefully noticed. 



I. 



- = x + y, 

x-y 

Qp — iP 

2- = x* + xy + V s , 

x-y 

~^ =x s + x?y + xy 2 + !f, 



DIVISION. 



45 



and so on ; the divisor being x — y, the terms in the quotient 
all positive, and the index in the dividend either odd or even. 



n. 



* + y 
rf+jf 

x + y 



X*- 


-ay+aft 




X*- 


-x*y + xSp- 


-a^ + Sft 


X*- 


-x*y + xY- 


-xY+xY 



-a^ + S 8 , 



and so on; the divisor being x + y, the terms in the quotient 
alternately positive and negative, and the index in the divi-' 
dend always odd. 

x + y 

x + y 

— =-^L = x 5 - x*y + xY - x*y* + xt? - if, 
x + y 

and so on ; the divisor being x + y, the terms in the quotient 
alternately positive and negative, and the index in the divi- 
dend always even. 

IV. The expressions vP + ip, x* + y*, x*+if ••• (where 
the index is even, and the terms both positive), are never 
divisible by x + y or x — y. 

All these different cases may be more concisely stated 
as follows: 

(1) jr* — /* is divisible by x — / if n be any whole number. 

(2) x n +y n is divisible by x +/ if n be any odd whole number. 

(3) x n — /* is divisible by x -f / if n be any e/e/i whole number. 

(4) jr* +jr» is never divisible by * +/ or jr — / when /i is an e/eii 
whole number. 



Note. General proofs of these statements will be found in Art. 10& 



48 



ALGEBRA. 



DIVISION BY DETACHED COEFFICIENTS. 

63. In Art. 52 we considered certain cases of compound 
expressions in which the work of multiplication could be 
shortened by using the Method of Detached Coefficients. In 
the same cases the labor of division can be considerably 
abridged by using detached coefficients, and employing an 
arrangement of terms known as Horner's Method of Synthetic 
Division. The following examples illustrate the method : 

Ex. 1. Divide 3a*-8x*-5x 8 +26a*-28a;+24 by x»-2aj*-4*+8. 



1 

2 

4 

-8 



3-8- 6 + 26-28 + 24 Dividend 
6 + 12-24 
_ 4_ 8 + 16 

+ 6 + 12-24 



Quotient 3-2+ 3+ 0+ 0+ 

Inserting the literal factors in the quotient according to the law 
of their formation, which is readily seen, we have for a complete 
quotient, 3 x 2 — 2 x + 3. 

Explanation. The column of figures to the left of the vertical line 
consists of the coefficients of the divisor, the sign of each after the first 
being changed, which enables us to replace the process of subtraction 
by that of addition at each successive stage of the work. Dividing the 
first term of the dividend by the first term of the divisor, we obtain 3, 
the first term of the quotient. Multiplying 2, 4, and — 8, the remain- 
ing terms of the divisor, by this first term of the quotient gives the 
second horizontal line. We then add the terms in the second column 
to the right of the vertical line and obtain — 2, which is the coefficient 
of the second term of the quotient. With this coefficient as a multi- 
plier, and using 2, 4, and — 8 again as a multiplicand, we form the third 
horizontal line. Adding the terms in the third column gives 3, which 
is the third term of the quotient. With this coefficient as a multiplier 
and the same multiplicand as before, we form the fourth horizontal 
line. As only zeros now appear in the quotient, the division is exact. 

Ex. 2. Divide 2 a 1 + 7 cfib + 12 a b b 2 + 10 a*6 8 - 4 a*b* by 2 a« + 
8 a 2 b — 6 8 to four terms in the quotient. 



2 

-8 



1 



2 + 7 + 12 + 10 

-3+ 0+ 1 

-6+0 

- 9 



Quotient 1 + 2+ 3+ 1 



-^aW + Oa^ + Oofi* 



+ 2 
+ 
-3 



+ 3 

+ 



+ 1 



-6 



+ 3 +1 



DIVISION. 47 

Inserting literal factors, a* + 2 a*b + 8 a 2 6 2 + aft 8 is the complete 
quotient, and — 6a?b* + 8a 2 & 5 + aft 5 is the remainder. 

Explanation. The term ab l in the divisor is missing, so we write 
for the coefficient of this term in the column of figures on the left 
of the vertical line. We add the columns as in Ex. 1, but as the 
first term of the divisor is 2, we divide each, sum by 2 before placing 
the result in the line of quotients. We then use these quotients as 
multipliers, the multiplicand being in each case — 3, 0, and 1, and 
form the horizontal lines as in Ex. 1. Having obtained the required 
number of terms in the quotient, the remainder is found by adding 
the rest of the columns and setting down the results without dividing 
by 2. By continuing the first horizontal line (dividend), as shown 
in this example, we at once see what literal factors the remainder 
must contain. 

EXAMPLES V. <L 
Divide : 

1. a 4 - 4 a* + 2 fl* -f 4 a + 1 by a 2 - 2 a - 1. • 

8. a* - 4 a 8 6 + 6 a 2 6 2 + 6* - 4 a&* by a 2 + 6 2 - 2 ah. 

3. a* - 10 a*6 + 16 a 8 6 2 - 12 a 2 b* + ab* + 2 b* by (a - ft) 2 . 

4. a 8 - 2 6*x* + 6 8 by a 8 + bx 2 + b*x + 6 8 . 

5. x 6 — 3 x*y* + 8 xy* — 6 y 6 by x 2 — 4 xy + y* to four terms in the 
quotient. 



CHAPTER VI. 
Removal and Insertion op Brackets. 

64. We frequently find it necessary to enclose within 
brackets part of an expression already enclosed within 
brackets. For this purpose it is usual to employ brackets 
of different forms. The brackets in common use are ( ), 
\ I) C ]• Sometimes a line called a vinculum is drawn 
over the symbols to be connected ; thus a — b + c is used 
withp the same meaning as a — (b 4- c), and hence 

a — b + c = a — 6 — c. 

65. To remove brackets it is usually best to begin with 
the inside pair, and in dealing with each pair in succession 
we apply the rules already given in Arts. 25, 26. 

Ex. 1. Simplify, by removing brackets, the expression 



a-26-[4a-66-{3a-c+(5a-26- 3a -c 4-26)}]. 

Removing the brackets one by one, we have 

a _ 2 6 -[4 a - 6 b - {3 a - c + (5a - 2 b - 3a + c - 2 6)}] 
= a-26-[4a-66-{3a-c + 6a -26 - 3a + c -26}] 
= a-26-[4a-66- 3a + c- 5a ;f26-H 3a - c + 26] 
= a-26- 4a + 66 + 3a-c+ 5a -26 -8a + c -26 
= 2 a, by collecting like terms. 

Ex. 2. Simplify the expression 

-[-2a;-{3y-(2a-3y) + (3aj-2y)} + 2aj]. 

The expression = -[—2x-{3y-2x + Sy + 3x-2y} + 2z] 

= -[-2x-3y + 2x-3y-3x + 2y+.2x'] 
= 2x+3y-2x + 3y + 33-2y-2x 

= x -f 4 y. 

48 



REMOVAL AND INSERTION OF BRACKETS. 49 



EXAMPLES VI. ft. 

Simplify by removing brackets : 

1. a-(b-c)+a + (b-c)+b-(c + a). 

2. a -[& + {a -(& + a)}]. 3. a -[2a -{86 -(4c -2 a)}] 

4. { a -(6-c)} + {6-(c-a)}-{c-(a-6)}. 

5. 2a-(6&+[3c-a])-(5a-[& + c]). 

6. _{_[-.(a-&^)]}. 7. -(-(-(_x)))-(-(-y)). 
8. _[a_{&_(c-a)}]-[6-{c -(a -&)}]. 

»• -[-{-(& + c-a)}] + [-{-(c + a-&)}]. 

10. -6x-[3y-{2x-(2y-x)}]. 

11. _(_(_<,))_(_(_(_*))). 

12. 8a-[a-f &-{a + & + c-(a + 6 + c + d)}]. 

13. -2a-[3x + {3c-(4y + 3x + 2a)}]. 

14. 3x-[6y-{6s-(4x-7y)}]. 

15. _[6x-(llsr-3x)]-[5y-(3a;-6y)]. 

16. -[15a>-{14y-(15s+12y)-(10x-15s)}]. 

17. 8x-{16y-[3x-(12y-x)-8y]+x}. 

18. — [x — {« + (a — z) — (z — x) — «} — x]. 

19. — [a + {a — (a — x) — (a-f a)— a} — a]. 

20. — [a — {a + (x — a)— (x — a) — a} — 2 a]. 

66. A coefficient placed before any bracket indicates that 
every term of the expression within the bracket is to be 
multiplied by that coefficient. 

Note. The line between the numerator and denominator of a 
fraction is a kind of vinculum. Thus x ~~ is equivalent to \(x — 6). 

Again, an expression of the form ^/(x 4- y) is often written Vx + y, 
the line above being regarded as a vinculum indicating the square root 
of the compound expression x + y taken as a whole. 

Thus V25 + 144 = y/W9 = 13, 

whereas y/2b + V 144 = 5 + 12 = 17. 

67. Sometimes it is advisable to simplify in the course of 
the work. 



60 ALGEBRA. 



Ex. Find the value of 



84 - 7 [ - 11* - 4{- 17* + 3 (8 - - 5*)}]. 

The expression = 84-7[-ll*-4{-17* + 3(8-0 + 5*)}] 

= 84-7[-ll*-4{-17* + 8(6*-l)}] 
= 84 - 7 [- 11* - 4{ - 17* + 16*- 8}] 
= 84 - 7 [- 11 * - 4{ - 2* - 3}] 
= 84-7[-ll* + 8*+12] 
-84 -7 [-3* +12] 
= 84 + 21*-84 
= 21*. 

When the beginner has had a little practice, the number oi 
steps may be considerably diminished. 

EXAMPLES VI. bt 

Simplify by removing brackets : 

1. a -[2 6 + {3 c - 3a -(a + &)} + 2 a -(ft + 8c)]. 

2. a + b-(c + a -[& + c-(jn + b-{c+a -(ft + c - a)})]). 
8. a-(6-c)-[a-6-c-2{6 + c-3(c-a)-d}]. 

4. 2x-(3y-4*)-{2*-(3y + 4*)}-{3y-(4* + 2*)}. 

5. b + c -(a + b -[c + a -(b + c - {a + b ~(c + a - 6)})]). 

6. a-(&-c)-[a-&-c-2{ft-f c}]. 

7. 3a3-[ea a -{8&2_(9c?-2a a )}]. 

8. 6 -(c - a) -[& - a - c - 2{c + a - 3 (a - 6)- d}]. 

9. - 20 (a - d) + 3 (6 - c) - 2 [6 + c + a* - 8 {c + a* - 4 (d - a)}]. 

10. - 4 (a + d) + 24 (b - c) - 2 [c + d + a - 8{d + a - 4 (b + c)}] 

11. - 10 (a + 6) - [c + a + b - 3 {a + 2 6 - (c + a - ft)}] + 4 a 

12. a-2(&-c)-[-{-(4a-ft-c- 2 {a + ft + «})}]. 

13. 2(3&-6a)-7[a-6{2-6(a-6)}]. 

14. 6 {a - 2 [6 - 3 (c + d)]} - 4{a - 3 [6 -4(c - d)]}> 

15. 6{a - 2 [a - 2 (a + *)]} - 4{a - 2 [a - 2 (a + *)]}. 

16. -10{a-6[a-(ft-c)]} + 60{6-(c + a)}. 
IT. _8{-2[-4(-a)]} + 6{-2[-2(-a)]}> 
1* -2{-[-(*-y)]} + {-2[-(*-i0]}. 



REMOVAL AND INSERTION OF BRACKETS. 61 

* «■-• '-^-HK'-lhit-K-V)]}- 

81. Hi ( a -b)-S(b-e)}-{^- c -^}-l{<>-*-K*-V} 

INSERTION OF BRACKETS. 

6& The converse operation of inserting brackets is im- 
portant. The rules for doing this have been enunciated in 
Arts. 25, 26 ; for convenience we repeat them. 

Rule I. Any part of an expression may be enclosed within 
brackets and the sign -f prefixed, the' sign of every term within 
the brackets remaining unaltered, 

Ex. a — 6 + c — d — e = a — 6+(c — d — e). 

Rule II. Any part of an expression may be evdosed within 
brackets and the sign — prefixed, provided the sign of every 
term within the brackets be changed. 

Ex. a — 6 + c — d — e = a — (6 — c) — (d + e). 

G9. The terms of an expression can be bracketed in various 
ways. 

Ex. The expression ax — bx + ex — ay + by — cy 
nay be written (ax — bx) + (ex — ay) + (by — cy), 
or (ax — bx + ex) — (ay — by + cy) f 

or (<kc — ay) — (&« — by) + (ex — cy). 

70. A factor, common to every term within a bracket, 
may be removed and placed outside as a multiplier of the 
expression within the bracket. 

Ex. 1. In the expression 

ax 9 — cx + 7 — dx* + bx — c — dafi + bx 2 — 2 s 

bracket together the powers of a; so as to have the sign + before each 
bracket. 



52 ALGEBRA. 

The expression = (owe 8 - die 8 ) + (6x 2 - dx 2 ) + (6* - ex - 2«) + (7- e t 

= 3fl(a - d)+ **(& - d) + z(b - c - 2) + (7 - c) 
= (a - d)&* +(6 - d)x 2 +(& - c - 2)x + 7 - c. 

In this last result the compound expressions a — d, 6— d, 6 — e — 2 
are regarded as the coefficients of x* f x 2 , and x respectively. 

Ex. 2. In the expression — a*x — 7a + a*y + S — 2 x — ab bracket 
together the powers of a so as to have the sign — before each bracket. 
The expression = — (a*x — a*y) — (7 a + ab) — (2 x — 3) 

= -a 2 (x-j0-a(7 + &)-(2x-3) 
= -(* - V)<* 2 -(7 + 6)a -(2* - 8). 

EXAMPLES VI. O. 

In the following expressions bracket the powers of x so that the 
signs before all the brackets shall be positive : 

1. ax*-f &x 2 + 5 + 2&x~6x 2 + 2x*-3x. 

2. Sbx 2 - 7-2» + a6 + 5ax 8 + ca;-4a; 2 -6a! 8 . 

3. 2-7x 8 +6ax 2 -2cx + 9ax 8 + 7x-3x 2 . 

In the following expressions bracket the powers of x so that the 
signs before all the brackets shall be negative ; 

4. ax 2 + 6x*-a 2 x*-2bx*-Sx*-bx*. 

5. 7x 8 -3c 2 x dbx 6 + 5 ax + 7 x 5 - abcx*. 

6. 3 6 2 x* - foe - ax* - ex* - 6c*x - 7 x*. 

Simplify the following expressions, and in each result regroup the 
terms according to powers of x : 

7. ox 8 - 2 ex -[6x 2 - {ex - dx -(6x 8 + 3cx 2 )} -(ex 2 - bx)]. 

8. 5 ox 8 - 7(6x - ex 2 )- {6 6x 2 -(3 ax 2 + 2 ax)-4cx*}. 

9. ax 2 -3{-ax 8 + 36x-4[icx 8 -f(ax-6x 2 )]}. 

10. x 6 -46x*-iri2ax-4'|36x*-9^-6x 6 ]-|ax*}l. 

71. In certain cases of addition, multiplication, etc., of 
expressions which involve literal coefficients, the results 
may be more conveniently written by grouping the terms 
according to powers of some common letter. 

Ex. 1. Add together ax 8 -2 6x 2 +3, 6x-cx 8 -x 2 , and x 8 — ox 2 + ox. 

The sum = ax 8 - 2 6x 2 + 3 + bx - ex 8 - x 2 4- x 8 - ax 2 + ex 
= ax 8 - ex 8 + x 8 - ax 2 - 2 6x 2 - x 2 + bx + ex + 3 
=(a - c -r l)xs - (a + 2 6 + l)x 2 + (6 + c)x + 8. 



MISCELLANEOUS EXAMPLES II. 53 

Ex. 8. Multiply ax 2 — 2 bx + 3 c by px — q. 

The product = (ax 2 — 2 foe + 3 c) ( px - g) 

= apx 8 — 2 bpx 2 -f 3 cpx — aqx 2 + 2 6gx — 3 cq 
= apx 8 -(2 6p + aq)& + (3cp + 2 5g)x - 3 eg. 

EXAMPLES VI. d. 

Add together the following expressions, and in each case arrange 
the result according to powers of x-: 

1. ax 8 — 2 ex, foe 2 — ex 8 , and ex 2 — x. 

2. x 2 - x - 1, ax 2 - tie 8 , 6s + x 8 . 

3. a'¥-6a;, 2ox 2 -5ax 8 , 2x 8 -foe 2 -ax. 

4. ax 2 + foe — c, gx — r — px 2 , x 2 + 2 x + 3. 

5. px 8 — gx, gx 2 — px, g — x 8 , px 2 + gx 8 . 

Multiply together the following expressions, and in each case 
arrange the result according to powers of x: 

6. ax 2 + bx -f 1 and ex -f 2. 9. 2 x 2 — 3x — 1 and bx + c. 

7. ex 2 — 2 x + 3 and ax — b. 10. ax' 2 — 2 foe -f 3 c and x — 1. 

8. ax 2 — bx — c and px + g. 11. px 2 — 2 x — g and ax — 3. 

12. x 8 + ox 2 — foe — c and x 8 — ax 2 — bx -f c. 

13. ox 8 - x 2 + 3x - 6 and ox 8 + x 2 + 3 x + fo 

14. x* — ax 8 — foe 2 + ex + d and x* + ox 8 — foe 2 — ex -f & 

MISCELLANEOUS EXAMPLES II. 

1. Find value of (a - 6) 2 +(& - c) 2 -f (a - 6) + 2 c 2 when = 1, 
6 = 2, c = - 8. 

2. Find the sum of 2x, 8x 2 , 6, -Sx 8 , -4, x, -6x 2 , 8X 8 , 
arranging result in descending powers. 

3. Diminish the sum of ft 8 + 7 6* -6 and 4 ft 2 -36 + 7 by 
11 b 2 + 2. 

4. Show that (l+x) 2 (l+y 2 )-(l+x 2 )(l+y) 2 =2 (x-y)(l-xy). 

5. Simplify (a + 6) (a -f c) - (a - 6) (a - c). 

6. Subtract the sum of 3 wi 8 — 4 m + 1 and m 2 -3m from 
4m 8 + 2 m 2 -7 m. 

7. What expression must be taken from the sum of a + 3 &, 
4a 2 — i> a, ft 2 4- 2 a, 2 a — 3 b 2 in order to produce a 2 — 6 2 ? 

8. Find value of a 2 +(c+d)( J(— \+d\ when a=2, c=9, d=a 



64 ALGEBRA. 

9. Multiply a 2 + (b - c) 8 by b + e + 1. 

10. Divide 343x 8 + 612y 8 by 7x + 8y. 

11. If a = 1, b = 2, c = 3, d = 4, find the value of 

12. What number must be added to 2 x 2 — Sxy 2 + to produce 
7x 2 + }xy a -x«y + 6? 

13. Simplify (x- y)_{3x-(x'+ y)} + {(2x- 3y)-(x-2y)}. 

14. Show that a(a - l)(a - 2)(a - 3) = (a 2 - 3 a + l) 2 - 1. 
18. x4-lO« a + 0-r-x 2 -2x~3. 

16. Simplify 9a-(2&-c) + 2d-(6a + 3&)+4c-2(f, and 
find its value when a = 7, 6 = — 3, c = — 4. 

17. Multiply 3 a 2 6 - 4 a& 8 c + 2 cWc 8 by -6aW, and divide 
the result by 3 ab 2 c 2 . 

18. If a = - 1, 6 = 2, c = 0, d = 1, find the value of 

ad + ac - a 2 - cd + c 2 - a + 2 c + a 2 6 + 2 a 8 . 

19. Find the sum of 3 a + 2 6, - 5 c - 2 a*, 3 e + 6/, 6 - a + 2 a\ 
-2a-36 + 5c-2/. 

20. Subtract ax 2 — 4 from zero, and add the difference to the sum 
of 2 x 8 — 6 x and unity. 

21. Multiply ix 2 + J«y-iy a by Ax 2 -ixy + f^. 

22. Divide 6a«-a 8 &+2 a 2 6 2 +13a6 8 +4 6* by 2a 2 -3o6+4 6 2 . 

23. Simplify 5x* -8X 8 - (2x 2 - 7) - (x* + 5) + (Sx 8 -x), and 
subtract the result from 4 x* — x + 2. 



24. Simplify by removing brackets 6 [x - 4 {x - 3(2 x - 3 x f 2)}] . 

25. If a = 1, b = 2, c = 3, and d = 4, find value of 



26. Express by means of symbols 

(1) 6's excess over c is greater than a by 7. 

(2) Three times the sum of a and 2 6 is less by 5 than the product 
of b and c 

27. Simplify 

8a a -(4 a-6 2 )-{2 a 2 - (3 6-a 2 )-2 6-3 a}-{6 6-7 a-O^-ft 2 )}. 

Find the continued product of 

x 2 + xy + y 2 , x 2 - xy + y 2 , x* - x 2 ^ 2 + y*. 



CHAPTER VII. 

Simple Equations. 

72. An equation asserts that two expressions are equal, 
but we do not usually employ the word equation in so wide 
a sense. 

Thus the statement x + 3-\-x = 2x + 3, which is always 
true whatever value x may have, is called an identical equa- 
tion, or an identity. The sign of identity frequently used 
is =. 

The parts of an equation to the right and left of the sign 
of equality are called members or sides of the equation, and 
are distinguished as the right side and left side. 

73. Certain equations are only true for particular values 
of the symbols employed. Thus 3 x = 6 is only true when 
2 = 2, and is called an equation of condition, or more usually 
an equation. Consequently an identity is an equation which 
is always true whatever bo the valuer of :he symbols in- 
volved; whereas an equation, in the ordinary use of the 
word, is only true for particular values of the symbols. In 
the above example 3 x = 6, the value 2 is said to satisfy the 
equation. The object of the present chapter is to explain 
how to treat an equation of the simplest kind in order to 
discover the value which satisfies it. 

74. The letter whose value it is required to find is called 
the unknown quantity. The process of finding its value is 
called solving the equation. The value so found is called 
the root or the solution of the equation. 

75. The solution of equations, and the operations sub- 
sidiary to it, form an extremely important part of Mathe- 

66 



50 ALGEBRA. 

matics. All sorts of mathematical problems consist in the 
indirect determination of some quantity by means of its 
relations to other quantities which are known, and these 
relations are all expressed by means of equations. The 
operation in general of solving a problem in Mathematics, 
other than a transformation, is first, to express the con- 
ditions of the problem by means of one or more equations, 
and secondly, to solve these equations. For example, the 
problem which is expressed by the equation above given 
is the very simple question, "What is the number such 
that if multiplied by 3, the product is 6 ? " In the present 
chapter, it is the second of these two operations, the solu- 
tion of an equation, that is considered. 

76. An equation which involves the unknown quantity 
in the first degree is called a simple equation. 

The process of solving a simple equation depends upon 
the following axioms : 

1. If to equals we add equals, the sums are equal. 

2. If 'from equals we take equals, the remainders are 
equal. 

3. If equals are multiplied by equals, the products are 
equal. 

4. If equals are divided by equals, the quotients are 
equal. 

77. Consider the equation 7 x = 14. 

It is required to find what numerical value x must have 
consistent with this statement. 
Dividing both sides by 7, we get 

x = 2 (Axiom 4). 

Similarly, if - = — 6, 

multiplying both sides by 2, we get 

x = — 12 . . . . (Axiom 3). 
Again, in the equation Ix — 2 a? — x = 23 + 15 — 10, by 
collecting terms, we have 4 x = 28. 

,\ x= 7. 



8IMPLE EQUATIONS. 6? 



TRANSPOSITION OF TERMS. 

7a To solve 3a>-8 = a> + 12. 

Here the unknown quantity occurs on both sides of the 
equation. We can, however, transpose any term from one 
side to the other by simply changing its sign. This we pro- 
ceed to show. 

Subtract x from both sides of the equation, and we get 

Sx — x — 8 = 12 . . . (Axiom 2). 
Adding 8 to both sides, we have 

3a? -x = 12 + 8 . . (Axiom 1). 

Thus we see that +- x has been removed from one side, 
and appears as — x on the other ; and — 8 has been re- 
moved from one side and appears as + 8 on the other. 

It is evident that similar steps may be employed in all 
cases. Hence we may enunciate the following rule : 

Rule. Any term may be transposed from one side of the 
equation to the other by changing its sign. 

79. We may change the sign of every term in an equation; 
for this is equivalent to multiplying both sides by — 1, 
which does not destroy the equality (Axiom 3). 

Ex. Take the equation — 3 x — 12 = x — 24. 

Multiplying both sides by - 1, 3 x + 12 = - x + 24, 
which is the original equation with the sign of every term changed. 

80. We can now give a general rule for solving a simple 
equation with one unknown quantity. 

Rule. Transpose all the terms containing the unknown 
quantity to one side of the equation, and the known quan- 
tities to the other. Collect the terms on each side ; divide 
both sides by the coefficient of the unknown quantity, and 
the value required is obtained. 

Ex. 1. Solve 5(x-3)-7(6-a0+3 = 24-3(8-a;). 

Removing brackets, 5x - 15 - 42 + Ix + 3 = 24 - 24 -f 3s; 
transposing, bx+7x-3x = 24 - 24 + 16 + 42-3; 

collecting terms, 9 x = 64. 

.-. jc = 6. 



58 ALGEBBA. 

Ex.8. Solve 5ft-(4*-7)(3«-5) = 6-8(4ft-0)(«-ly 
Simplifying, we have 

6ft -(12 a*- 41 x + 36) = 6 -8(4** - 13* + 9), 

and by removing brackets, 

6 x - 12 x* + 41 x - 35 = 6 - lfrx 2 + 89 ft - 27. 

Erase the term — 12 x 2 on each side and transpose; 

thus 6ft + 41a; - 39ft = 6-27 + 36; 

collecting terms, 7 x = 14. 

.\ ft = 2. 

Note. Since the — sign before a bracket affects every term within 
it, in the first line of work of Ex. 2, we do not remove the brackets 
until we' have formed the products. 

Ex. 3. Solve 7 ft - 6[x - {7 - 6(ft - 3)}] = 3 ft + 1. 
Removing brackets, we have 

7 X _ 6[ft - {7 - 6a; + 18}]= 3ft + 1, 
7ft-6[ft-26 + 6ft]=3ft + l, 
Ix - 5ft + 125- 30ft = 3ft + 1; 
transposing, 7 ft — 5x — 30ft — 3 ft = 1 — 125 ; 

collecting terms, — $1 x = — 124 ; 

.-. ft = 4. 

81. It is extremely useful for the beginner to acquire 
the habit of occasionally verifying, that is, proving the truth 
of his results. Proofs of this kind are interesting and con- 
vincing; and the habit of applying such tests tends to 
make the student self-reliant and confident in his own 
accuracy. 

In the case of simple equations we have only to show 
that when we substitute the value of x in the two sides of 
the equation we obtain the same result. 

Ex. To show that x = 2 satisfies the equation 

6x _(4 X _ 7)(3 X - 6) = 6 - 3(4a - 9)(a; - 1). Ex. 2, Art. 80. 

When x=2, the left side 6ft-(4ft-7)(3ft-6)=10-(8-7)(6-6) 

= 10-1=9. 

The right side6 - 3(4ft - 9)(x - 1) = 6 - 3(8 - 9)(2 - 1) 

= 6-3(-l)=9. 

Thus, since these two results are the same, x = 2 satisfies the 
equation. 



SIMPLE EQUATIONS. 6$ 

EXAMPLES Vn. 

Solve the following equations : 

1. 3x + 16 = x + 26. 2. 206-8 = 8* -7. 

8. 3x + 4 = 6(x-2). 4. 2x + 3 = 16 -(2*- 8) 

6. 8(x-l)+17(x-3)=4(4x-9)+4. 

6. 15(x-l)+4(x + 3) = 2(7 + x). 

7. 6x-6(x-6) = 2(x + 6)+6(x-4). 

8. 8(x-3)-(6-2x)=2(x + 2)-6(6-«). 

9. 7(25-x)-2x = 2(3x-26). 

10. 3(169-x)-(78 + x) = 29x. 

11. 6x - 17 + 3x -6 = 8a;-7 -8* + 116. 

13. 7a: -39- 10x + 15 = 100 -33* + 26. 
18. 118-66x-123 = 16x + 35-120ae. 

14. 157 - 21(05 + 3) = 163 - 16(2x - 6). 

15. 179 - 18(3 - 10) = 168 - 3(a5 - 17). 

16. 97 - 6(05 + 20) = 111 - 8(05 + 8). 

17. 05 -[3 + {a; -(8 + 05)}]= 5. 

18. 6x-(3x-7)-{4-2x-(6x-8)} = 10. 

19. 14x-(6x-9)~{4-3x-(2x-3)} = 30. 

20. 25x-19-[3-{4x-6}]=3x-(6x-6). 

21. (x+l)(2x+l) = (x + 3)(2x + 3)-14. 

22. (x + l) 2 -(x a - 1)= x(2x + 1)- 2(x + 2)(x+ 1)+ 20. 
28. 2(x+l)(x + 3)+8=(2x+l)(x + 6). 

24. 6(x 2 - 3x + 2)- 2(x a - 1)= 4(x + l)(x + 2)- 24. 

26. 2(x - 4)-(x 2 + x - 20)= 4x a -(5x + 3)(x - 4)- 64. 

26. (x + 16) (x - 3) - (x a - 6 x + 9) = 30 - 15(x - 1). 

27. 2x-6{3x-7(4x-9)} = 66. 

28. 20(2 - x)+ 3(x - 7)- 2[x + 9 - 3{9 - 4(2 - «)}]= 22. 

29. x + 2-[x-8-2{8-3(5-x)-x}] = 0. 

80. 3(6-6x)-6[x-6{l-8(x-6)}]=23. 

81. (x+l)(2x + 3)=2(x + l) 2 + 8; 
32. 3(x-l) 2 -3(x 2 -l) = x-16. 
88. (3x+l)(2x-7) = 6(x-3) 2 + 7. 

84. x 2 -8x + 25 = x(x-4)-26(x-5)-16L 

35. x (x + 1) + (x + l)(x + 2) = (x + 2)(x + 8)+ x(x + 4)- ft 

86. 2(x + 2)(x-4) = x(2x + l)-21. 

37. (x+l) 2 + 2(x + 3) 2 = 3x(x + 2)+36. 

38. 4(x + 6)« - (2 x 4- 1) 2 = 3(x - 5)+ 180. 

39. 84 + (x + 4)(x - 3)(x + 5) = (x + l)(x+ 2)(x + 3). 

40. (x + l)(x + 2)(* + 6)=x 8 + 9x 2 + 4(7x-l). 



CHAPTER Vni. 

Symbolical Expression. 

82. In solving algebraic problems the chief difficulty of 
the beginner is to express the conditions of the question 
by means of symbols. A question proposed in algebraic 
symbols will frequently be found puzzling, when a similar 
arithmetical question would present no difficulty. Thus, 
the answer to the question "find a number greater than x 
by a " may not be self-evident to the beginner, who would 
of course readily answer an analogous arithmetical question, 
"find a number greater than 50 by 6." The process of 
addition which gives the answer in the second case supplies 
the necessary hint ; and, just as the number which is greater 
than 50 by 6 is 50 -f 6, so the number which is greater than 
x by a is x -f- a. 

83. The following examples will perhaps be the best intro- 
duction to the subject of this chapter. After the first we 
leave to the student the choice of arithmetical instances, 
should he find them necessary. 

Ex. 1. By how much does x exceed 17 ? 

Take a numerical instance ; " by how much does 27 exceed 17 ? ** 

The answer obviously is 10, wbich is equal to 27 — 17. 

Hence the excess of x over 17 is x — 17. 

Similarly the defect of x from 17 is 17 — x. 

Ex. 2. If x is one part of 45 the other part is 46 — x. 

Ex. 3. How far can a man walk in a hours at the rate of 4 miles 
an hour ? 

In 1 hour he walks 4 miles. 

In a hours he walks a times as far, that is, 4 a miles. 

60 



SYMBOLICAL EXPRESSION. 61 

Ex. 4. A and B are playing for money ; A begins with 8 p and B 
with q dimes: after B has won $z, how many dimes has each ? 
What B has won A has lost, 

.•• A has 10 ( p — x) dimes, 
B has q + 10 x dimes. 

EXAMPLES VUL a. 

1. What must be added to x to make y f 

2. By what must 3 be multiplied to make a ? 

3. What dividend gives b as the quotient when 5 is the divisor ? 

4. What is the defect of 2 c from 3 d. 

5. By how much does 3 k exceed k ? 

6. If 100 be divided into two parts, and one part be x, what is 
the other ? 

7. What number is less than 20 by c ? 

8. What is the price in cents of a oranges at ten cents a dozen ? 

9. If the difference of two numbers be 11, and if the smaller be se, 
what is the greater ? 

10. If the sum of two numbers be c, and one of them is 20, what is 
the other ? 

11. What is the excess of 90 over x ? 
18. By how much does x exceed 30 ? 

18. If 100 contains x 5 times, what is the value of x . 

14. What is the cost in dollars of 40 books at x dimes each ? 

15. In x years a man will be 36 years old, what is his present age ? 

16. How old will a man be in a years if his present age is x years ? 

17. If x men take 5 days to reap a field, how long will one man 
take? 

18. What value of x will make 5x equal to 20 ? 

19. What is the price in dimes of 120 apples, when the cost of two 
dozen is x cents ? 

80. How many hours will it take to walk x miles at 4 miles an 
hour? 

81. How far can I walk in x hours at the rate of y miles an hour ? 

88. How many miles is it between two places, if a train travelling 
p miles an hour takes 5 hours to perform the journey ? 



62 ALGEBRA. 

28. A man has a dollars and b dimes, how many cents has he ? 

24. If I spend x half-dollars out of a sum of $20, how many half* 
dollars have I left ? 

25. Out of a purse containing fa and b dimes a man spends c 
cents ; express in cents the sum left. 

26. By how much does 2 x — 6 exceed x + 1 ? 

27. What number must be taken from a — 2 b to leave a — 8 b ? 

28. If a bill is shared equally amongst x persons, and each pays 
four dimes, how many cents does the bill amount to ? 

29. If I give away c dimes out of a purse containing a dollars and 
b half-dollars, how many dimes have I left ? 

80. If I spend x quarters a week, how many dollars do I save out 
of a yearly income of $y ? 

81. A bookshelf contains x Latin, y Greek, and z English books ; 
if there are 100 books, how many are there in other languages ? 

82. I have x dollars in my purse, y dimes in one pocket, and z 
cents in another ; if I give away a half-dollar, how many cents have I 
left ? 

38. In a class of x boys, y work at Classics, z at Mathematics, and 
the rest are idle ; what is the excess of workers over idlers ? 

84. We add a few harder examples worked out in full. 

Ex. 1. What is the present age of a man who x years hence will 
be m times as old as his son now aged y years ? 

In x years the son's age will be y + x years ; hence the father's age 
will be m(y + x) years ; therefore now the father's age is m(y + ») — 
x years. 

Ex. 2. Find the simple interest on $k in n years at /per cent. 

Interest on $ 100 for 1 year is $/, 

•* * X ••••'•' $iwp 

$k $-^, 

v v 100 

nkf 
. \ Interest on $ k f or n years is $ ^. 

Ex. 8. A room is x yards long, y feet broad, and a feet high ; find 
how many square yards of carpet will be required for the floor, and 
how many square yards of paper for the walls. 



SYMBOLICAL EXPRESSION. 63 

(1) The area of the floor is 3 xy square feet ; 

3 set/ xtt 
• \ the number of square yards of carpet required is -^ = -^ 

(2) The perimeter of the room is 2(3 x + y) feet ; 

. *.' the area of the walls is 2 a (3 x + y) square feet ; 

2a(Sx + y) 
. \ number of square yards of paper required is — ^— g — — 

Ex. 4. The digits of a number beginning from the left are a t b> c\ 
what is the number ? 

Here e is the digit in the units' place ; b standing in the tens' place 
represents b tens ; similarly a represents a hundreds. 

The number is therefore equal to a hundreds + b tens + e units 

= 100a+10& + e. 

If the digits of the number are inverted, a new number is formed 
which is symbolically expressed by 

100c +106 + a. 

Ex. 6. What is (1) the sum, (2) the product of three consecutive 
numbers of which the least is n ? 

The numbers consecutive to n are n + 1, n + 2. 

. \ the sum = n + (n + 1) + (n + 2) 
= 3n + 3. 

And the product = n(n + l)(n + 2). 

We may remark here that any even number may be de- 
noted by 2n, where n is any positive whole number; for 
this expression is exactly divisible by 2. 

Similarly, any odd number may be denoted by 2 n +- 1 ; 
for this expression when divided by 2 leaves remainder 1. 

EXAMPLES VIII. b. 

1. Write four consecutive numbers of which x is the least. 

2. Write three consecutive numbers of which y is the greatest. 

3. Write five consecutive numbers of which x is the middle one. 

4. What is the next even number after 2 n ? 

5. What is the odd number next before 2 x + 1 ? 

8. Find the sum of three consecutive odd numbers of which the 
middle one is 2 n + 1. 



64 ALGEBRA. 

7. A man makes a journey of x miles. He travels a miles by coach, ' 
b by train, and finishes the journey by boat. How far does the boat 

carry him ? • 

8. A horse eats a bushels and a donkey b bushels of corn in a j 
week ; how many bushels will they together consume in n weeks ? 

9. If a man was x years old 5 years ago, how old will he be y 
years hence ? 

10. A boy is x years old, and five years hence his age will be half 
that of his father. How old is the father now ? 

11. What is the age of a man who y years ago was m times as old 
as a child then aged x years ? 

12. A's age is double B's, B's is three times C's, and C is x years 
old : find A's age. 

13. What is the interest on $1000 in b years at c per cent. ? 

14. What is the interest on $ a; in a years at 5 per cent. ? 

15. What is the interest on $50 a in a years at a per cent. ? 

16. What is the interest on $24 xy in x months at y per cent, per 
annum? 

17. A room is x yards in length, and y feet in breadth ; how many 
square feet are there in the area of the floor ? 

18. A square room measures x feet each way ; how many square 
yards of carpet will be required to cover it ? 

19. A room is p feet long and x yards in width ; how many yards 
of carpet two feet wide will be required for the floor ? 

90. What is cost in dollars of carpeting a room a yards long, b feet 
broad, with carpet costing c dimes a square yard ? 

21. A room is a yards long and b yards broad ; in the middle there 
is a carpet c feet square ; how many square yards of oil-cloth will be 
required to cover the rest of the floor ? 

22. How long will it take a person to walk b miles if he walks 20 ' 
miles in c hours ? 

23. A train is running with a velocity of x feet per second ; how 
many miles will it travel in y hours ? 

24. How many men will be required to do in x hours what y men 
do in xz hours ? 



CHAPTER IX. 
Problems Leading to Simple Equations. 

85. The principles of the last chapter may now be em 
ployed to solve various problems. 

The method of procedure is as follows : 

Represent the unknown quantity by a symbol, as x, and 
express in symbolical language the conditions of the ques- 
tion ; we thus obtain a simple equation which can be solved 
by the methods already given in Chapter vii. 

Note. Unknown quantities are usually represented by the last 
letters of the alphabet. 

Ex. 1. Find two numbers whose sum is 28, and whose difference is 4. 
Let x represent the smaller number, then x + 4 represents the 
greater. 

Their sum is x + (as + 4), which is to be equal to 28. 

Hence x + x + 4 = 28 ; 

2s = 24; 

.-. 3 = 12, 

and x + 4 = 16, 

so that the numbers are 12 and 16. 

The beginner is advised to test his solution by proving 
that it satisfies the conditions of the question. 

Ex. 2. Divide 60 into two parts, so that three times the greater 
may exceed 100 by as much as 8 times the less falls short of 200. 
Let x represent the greater part, then 60 — x represents the less 
Three times the greater part is 3 x, and its excess over 100 is 

Zx - 100. 

Eight times the less is 8(60 - a), and its defect from 200 is 

200 - 8(60 - x). 

w 65 



66 ALGEBRA. 

Whence the symbolical statement of the question is 

3x - 100 = 200 - 8(00 - x); 
Sx - 100 = 200 - 480 + 8a?, 
480 - 100 - 200 = Sx - 3aj, 
6x = 180; 
. •. x = 86, the greater part* 
and 60 — x = 24, the less. 

Ex. 3. Divide $47 between A, B, C, so that A may have $10 more 
than B, and B $ 8 more than C. 

Suppose that C has x dollars ; then B has x + 8 dollars, and A has 
x -f 8 + 10 dollars. 

Hence x + (x + 8) + (a; + 8 + 10) = 47 ; 

x + x + 8 + x + 8 + 10 = 47, 

3a; = 21; 

• • x ^— i , 
so that C has $7, B $ 16, A $25. 

Ex. 4. A person spent $ 112.80 in buying geese and ducks ; if each 
goose cost 14 dimes, and each duck 6 dimes, and if the total number 
of birds bought was 108, how many of each did he buy ? 

In questions of this kind it is of essential importance to 
have all quantities expressed in the same denomination ; in 
the present instance it will be convenient to express the 
money in dimes. 

Let x represent the number of geese, then 108 — x represents the 
number of ducks. 

Since each goose cost 14 dimes, x geese cost 14 x dimes. 

And since each duck cost 6 dimes, 108 — x ducks cost 6(108 — x 
dimes. 

Therefore the amount spent is 

14b + 6(108 - a) dimes ; 

but the question states that the amount is also 9112.80, that is, 1128 
dimes. 

Hence 14 x + 6(108 - x) = 1128 ; 

dividing by 2, 7 x + 324 - 3 x = 664, 

4s = 240; 
.•. x = 60, the number of geese ; 
and 108 — x = 48, the number of ducks. 



PROBLEMS LEADING TO SIMPLE EQUATIONS. 67 

Ex. 5. A is twice as old as B ; ten years ago he was four times as 
old ; what are their present ages ? 

Let x represent B's age in years, then 2 x represents A's age. 
Ten years ago their ages were respectively, x — 10 and 2 x — 10 
years ; thus we have 2 x — 10 = 4(x — 10) ; 

2s-10 = 4a;-40, 
2s = 30; 
. \ x = 15, 
so that B is 16 years old, A 30 years. 

Note. In the above examples the unknown quantity x represents 
a number of dollars, ducks, years, etc. ; and the student must be care- 
ful to avoid beginning a solution with a supposition of the kind, u let 
x = A's share," or u let x = the ducks," or any statement so vague 
and inexact. 

EXAMPLES IX. 

1. One number exceeds another by 5, and their sum is 20 ; find 
them. 

2. The difference between two numbers is 8 ; if 2 be added to the 
greater the result will be three times the smaller ; find the numbers. 

3. Find a number such that its excess over 50 may be greater by 
11 than its defect from 89. 

4. What number is that which exceeds 8 by as much as its double 
exceeds 20 ? 

5. Find the number which multiplied by 4 exceeds 40 as much as 
40 exceeds the original number. ' 

8. A man walks 10 miles, then travels a certain distance by train, 
and then twice as far by coach. If the whole journey is 70 miles, how 
far tioes he travel by train ? 

7. What two numbers are those whose sum is 68, and difference 28 ? 

8. If 288 be added to a certain number, the result will be equal to 
three times the excess of the number over 12 ; find the number. 

9. Twenty-three times a certain number is as much above 14 as 
16 is above seven times the number ; find it. 

10. Divide 105 into two parts, one of which diminished by 20 shall 
be equal to the other diminished by 15. 

11. Divide 128 into two parts, one of which is three times as large 
as the other. 

18. Find three consecutive numbers whose sum shall equal 84. 



68 ALGEBRA. 

13. The difference of the squares of two consecutive numbers is 
35 ; find them. 

14. The sum of two numbers is 8, and one of them with 22 added 
to it is five times the other ; find the numbers. 

15. Find two numbers differing by 10 whose sum is equal to twice 
their difference. 

16. A and B begin to play each with $60. If they play till A's 
money is double B's, what does A win ? 

17. Find a number such that if 5, 15, and 35 are added to it, the 
product of the first and third results may be equal to the square of 
the second. 

18. The difference between the squares of two consecutive num- 
bers is 121 ; find the numbers. 

19. The difference of two numbers is 3, and the difference of their 
squares is 27 ; find the numbers. 

SO. Divide $380 between A, B, and C, so that B may have $30 
more than A, and C may have $20 more than B. 

21. A sum of $7 is made up of 46 coins which are either quarters 
or dimes ; how many are there of each ? 

22. If silk costs five times a& much as linen, and I spend $48 in 
buying 22 yards of silk and 50 yards of linen, find the cost of each 
per yard. 

23. A father is four times as old as his son ; in 24 years he will 
only be twice as old ; find their ages. 

24. A is 25 years older than B, and A's age is as much above 20 
as B's is below 85 ; find their ages. 

25. A's age is three times B's, and in 18 years A will be twice as 
old as B ; find their ages. 

26. A is four times as old as B, and in 20 years will be twice as 
old as C, who is 5 years older than B ; find their ages. 

27. A's age is six times B's, and fifteen years hence A will be three 
times as old as B ; find their ages.. 

28. A sum of $16 was paid in dollars, half-dollars, and dimes. 
The number of half-dollars used was four times the number of dollars 
and twice the number of dimes ; how many were there of each ? 

29. The sum of the ages of A and B is 30 years, and five years 
hence A will be three times as old as B ; find their present ages. 

30. I spend $69.30 in buying 20 yards of calico and 30 yards of 
silk ; the silk costs as many quarters per yard as the calico costs cents 
per yard ; find the price of each. 



PROBLEMS LEADING TO SIMPLE EQUATIONS. 69 

81. I purchase 127 bushels of grain. If the number of bushels of 
wheat be double that of the corn, and seven more than five times the 
number of bushels of corn equals the number of bushels of oats, find 
the number of bushels of each. 

82. The length of a room exceeds its breadth by 3 feet ; if the 
length had been increased by 3 feet, and the breadth diminished by 2 
feet, the area would not have been altered ; find the dimensions. 

33. The length of a room exceeds its breadth by 8 feet ; if each 
had been increased by 2 feet, the area would have been increased by 
60 feet ; find the original dimensions of the room. 

84. A and B start from the same place walking at different rates ; 
when A has walked 15 miles B doubles his pace, and 6 hours later 
passes A ; if A walks at the rate of 5 miles an hour, what is B's rate 
at first? 

35. A sum of money is divided among A, B, and C, so that A and 
B have together $20, A and C $30, and B and C $40 ; find the share 
of each. 

86. A man sold two pieces of cloth, losing $ 6 more on the one than 
on the other. If his entire loss was $ 4 less than four times the smaller 
loss, find the amount lost on each piece. 

37. Two men received the same sum for their labor; but if one 
had received $ 10 more, and the other $ 8 less, then one would have 
had three times as much as the other. What did each receive ? 

38. In a certain examination the number of successful candidates 
was four times the number of those who failed. If there had been 
14 more candidates and 6 less had failed, the number of those who 
passed would have been five times the number of those who failed. 
Find the number of candidates. 

39. A purse contains 14 coins, some of which are quarters and the 
rest dimes. If the coins are worth $2 altogether, how many are there 
of each kind ? 

40. An estate was divided among three persons in such a way that 
the share of the first was three times that of the second, and the share 
of the second twice that of the third. The first received $ 900 more 
than the third. How much did each receive ? 



CHAPTER X. 

Resolution into Factobs. 

86. Definition. When an algebraic expression is the 
product of two or more expressions, each of these latter 
quantities is called a factor of it, and the determination of 
these quantities is called the resolution of the expression 
into its factors. 

87. Rational expressions do not contain square or other 
roots (Art. 14) in any term. 

88. Integral expressions do not contain a letter in the 
denominator of any term. Thus, a 2 -f 3 xy -f- 2 y 2 , and 
$x?-\-$xy — $y* are integral expressions. 

89. In this chapter we shall explain the principal rules 
by which the resolution of rational and integral expressions 
into their component factors, which are rational and integral 
expressions, may be effected. 

WHEN EACH OF THE TERMS IS DIVISIBLE BY A 

COMMON FACTOR. 

90. The expression may be simplified by dividing each 
term separately by this factor, and enclosing the quotient 
within brackets; the common factor being placed outside 
as a coefficient. 

Ex. 1. The terms of the expression 3 a 2 — 6 ab have a common 

factor 3 a. 

.-. 3a 2 -6a& = 3a(a-26). 

Ex. 2. 5 a^x 3 - 16 abx* - 20 6 8 x 2 = 5 bxHatx - 3 a - 4 6 a ). 

70 



RESOLUTION INTO FACTORS. 71 

EXAMPLES X. a. 
Resolve into factors : 

1. a?- ax. 5. &x-2x*. 9. 15a a ~226o*. 

2. x*-x 2 . 6. bax-ha*x*. 10. 64-81x. 

3. 2a-2a 2 . 7> 15 + 26x 2 . 11. 10s 8 -26zty. 

4. 7p 2 +p. 8. 16a + tex*y. 12. Sx*-x 2 + x. 

13. 3o*-3a 8 6 + 6o 2 6 2 . 16. 6 a* - 10 a 2 a; 8 - 16 a 8 x 8 . 

14. 2 ay - 6 xhp- + 2 ay 8 . 17. 7 a - 7 a 8 + 14 a*. 

15. 6 se 8 - 9 a 2 2/ + 12 xy 2 . 18. 38 a 8 ** + 67 a*x 2 . 

WHEN THE TERMS CAN BE GROUPED SO AS TO 
CONTAIN A COMMON FACTOR. 

91. Ex. 1. Resolve into factors x 2 — ax + bx — ab. 

Noticing that the first two terms contain a factor x, and the last two 
terms a factor 6, we enclose the first two terms in one bracket, and 
the last two in another. Thus, 

x 2 — ax'+ bx — ab = (x 2 — ax) + (bx — ab) 

= x(x — a) + 6(x — a) . . . (1) 
= (z-a)(a; + 6), 
since each bracket of (1) contains the same factor x — a. 
Ex. 2. Resolve into factors 6 x 2 — 9 gkc + 4 6se — 6 ab. 

6 a; 2 - 9 ax + 4 6x - 6 a6 = (6 a 2 - 9 ax) + (4 bx - 6 a&) 

= 3 s(2 a; - 3 a) + 2 6(2 a; - 3 a) 
= (2s-3a)(3se + 2&). 

Ex. 3. Resolve into factors 12 a 2 — 4 ab — 3 ax 2 + 6a; 2 . 

12 a 2 - 4 a& - 3 ax 2 + bx 2 = (12 a 2 - 4 a6)-(3 ax 2 - bx 2 ) 

= 4a(3a-6)-a; 2 (3a-&) 
= (3a-6)(4a-a; 2 ). 

Note. In the first line of work it is usually sufficient to see that 
each pair contains some common factor. Any suitably chosen pairs 
will bring out the same result. Thus, In the last example, by a 
different arrangement, we have 

12 a 2 - 4 ab - 3 ax 2 + bx 2 = (12 a 2 - 3 ax 2 )-(4 ab - bx 2 ) 

= 3 a(4 a - x 2 ) - 6(4 a - x 2 ) 
= (4a-s 2 )(3a-6). 

The same result as before, for the order in which the factors of a 
product are written is of course immaterial. 



72 



ALGEBRA. 



Resolve into factors : 

1. a 2 + ab + ac + be. 

8. a 2 — ac + ab — be. 

3. a*c* + acd + a&c + fta". 

4. a a + 8a + ac + 8c. 
ft. 2as + cx + 2c + C*. 

6. a; 2 — ox + 5 x — 6 a. 

7. 6a + a& + 6& + &*. 

8. a& — 6y - ay + y 2 . 

9. rose — my — ns + ny. 

10. was — ma + nx — na. 

11. 2 ax + ay + 2 bx + &y. 



X. b. 

18. Sax — foe — 8ay + 5y. 
18. 6x2 + 307-200; -ay. 

14. mx — 2my-nz + 2ny. 

15. ax 2 — 3 foey — axy + 3 6y*. 

16. x 2 + mxy — 4 xy — 4 my 8 . 

17. 2s*-* 8 + 4* -2. 

18. 8s* + 6a£ + 3a;+ 6. 

19. x* + x* + 2a; + 2. 

*>• 1^-2^ + y-i. 

21. oxy + bcxy — az — bee. 
88. Z^ 2 + g*x* -a(p-af*. 



TRINOMIAL EXPRESSIONS. 

92. When the Coefficient of the Highest Power is Unity, 

Before proceeding to the next case of resolution into factors 
the student is advised to refer to Chap. iv. Art. 51. Atten- 
tion has there been drawn to the way in which, in forming 
the product of two binomials, the coefficients of the different 
terms combine so as to give a trinomial result Thus, by 
Art. 51, 

(x + 5)(x + 3)=x* + 8x + 15 .... (1), 

(x-5)(x-3)=x*-8x + 15 .... (2), 
(a? + 5)(a?-3)=oj a + 2a:-15 .... (3), 
ix-5)(x + 3)=x*-2x-15 .... (4), 

We now propose to consider the converse problem : namely, 
the resolution of a trinomial expression, similar to those which 
occur on the right-hand side of the above identities, into its 
component binomial factors. 

By examining the above results, we notice that : 

1. The first term of both the factors is x. 

2. The product of the second terms of the two factors is 
equal to the third term of the trinomial ; thus in (2) above 
we see that 15 is the product of — 5 and — 3 ; while in 
(3) — 15 is the product of + 5 and — 3. 



RESOLUTION INTO FACTORS. 78 

3. The algebraic sum of the second terms of the two factors 
is equal to the coefficient ofxin the trinomial ; thus in (4) the 
sum of — 5 and +- 3 gives — 2, the coefficient of x in the tri- 
nomial. 

In showing the application of these laws we will first con- 
sider a case where the third term of the trinomial is positive. 

Ex. 1. Resolve into factors x 2 + 11 * 4- 24. 

The second terms of the factors must be such that their product is 
4- 24, and their sum + 11. It is clear that they must be + 8 and + 3. 

.-. x 2 + llx + 24=(x+8)(x + 3). 

Ex. 8. Resolve into factors x 2 — 10 x + 24. 

The second terms of the factors must be such that their product is 
+ 24, and their sum — 10. Hence they must both be negative, and it 
is easy to see that they must be — 6 and — 4. 

* .-. a; 2 - 10 a; + 24 = (x - 6)(x - 4). 

Ex. 3. x 2 - 18s + 81 =(a$ - 9)(x - 9) = (x - 9) 2 . 

Ex. 4. x* + 10x 2 + 26 =(x 2 + 5)(x 2 + 5) = (x 2 + 5)«. 

Ex. 5. Resolve into factors x 2 — 11 ax + 10 a 2 . 
The second terms of the factors must be such that their product is 
+ 10 a 2 , and their sum — 11 a. Hence they must be — 10 a and — a. 

.«. x 2 - 11 ax + 10 a 2 = (x - 10 a) (x - a). 

Note. In examples of this kind the student should always verify 
his results, by forming the product (mentally, as explained in Chapter 
iv.) of the factors he has chosen. 

EXAMPLES X. C. 
Resolve into factors : 

1. fl 2 + 3a + 2. 7. x 2 -21x + 108. 13. x 2 + 20x + 96. 

2. a 2 + 2 a + 1. 8. x 2 - 21 x + 80. 14. x 2 - 26x + 165. 

3. a 2 + 7 a + 12. 9. x 2 + 21x + 90. 15. x 2 - 21 x + 104. 

4. a 2 -7a + 12. 10. x 2 -19x + 84. 16. a 2 + 30a + 225. 

5. x 2 -llx + 30. 11. x 2 -19x + 78. 17. a 2 + 54a + 729. 
a x 2 -15x + 56. 12. x 2 -18x + 45. 18. a 2 -38a + 361- 

19. a 2 - 14 ab + 49 6 2 . 23. x 2 :- 23 xy + 132 y 2 . 

20. a 2 + 6a6 + 66 2 . 24. x 2 - 26xy + 169^. 

21. ro 2 -13w» + 40n 2 . 25. x* + 8x 2 + 7. 

22. ro 2 -22row + 106w 2 . 26. x* + 9 x 2 y 2 J- 14 y*. 



74 ALGEBRA, 

27. sfi + 49^ + 6000*. 88. 20 + 9* + ofc 

28. x*y 2 + 34st/ + 289. 34. 132- 23a; + a& 

29. a*&* + 37 a 2 6 2 + 300. 35. 88 + 19a; + a£. 

30. a 2 -29a& + 546 2 . 36. 130 + 31 xy + x 2 y 9 . 

31. ** + 162 a; 2 +6661. 37. 204 - 29 x 2 + a*. 

32. 12-7a; + * 2 . 88. 216 + 35a; + S 2 . 

93. Next consider a case where the third term of the tri- 
nomial is negative. 

Ex. 1. Resolve into factors a; 2 + 2 as — 86. 

The second terms of the factors must be such that their product is 

— 35, and their algebraic sum + 2. Hence they must have opposite 
signs, and the greater of them must be positive in order to give its 
sign to their sum. 

The required terms are therefore + 7 and — 6. 

.-. a; 2 + 2 x - 36 = (x + 7)(a; - 6). 

Ex. 2. Resolve into factors x 2 — 3 x — 64. 

The second terms of the factors must be such that their product is 

— 64, and their algebraic sum — 3. Hence they must have opposite 
signs, and the greater of them must be negative in order to give its 
sign to their sum. 

The required terms are therefore — 9 and + 6. 

.-. a; 2 -3 a; -64 = (a;- 9)(a; + 6). 

Remembering that in these cases the numerical quantities must have 
opposite signs, if preferred, the following method may be adopted. 

Ex. 3. Resolve into factors a; 2 !/ 2 + 23 xy — 420. 

Find two numbers whose product is 420, and whose difference is 23. 
These are 35 and 12 ; hence inserting the signs so that the positive 
may predominate, we have 

x*y 2 + 23 xy - 420 = (xy + 35) (xy - 12). 

EXAMPLES X. d. 

Resolve into factors : 

1. a; 2 - x - 2. 6. a; 2 + 2 x - 3. 9. a a + a - 20. 

2. x 2 + x - 2. 6. x 2 + x - 56. 10. a 2 - 4 a - 117. 
8. a; 2 - x - 6. 7. x 2 - 4 x - 12. 11. a; 2 + 9 x - 36. 
4. x 2 - 2 a; - 3. 8. a 2 - a - 20. 12. a; 2 + x - 166. 



RESOLUTION INTO FACTORS. 75 

18. a? + »-110. 18. x*j/* - 6 xy - 24. 88. a 2 - 11 a - 26. 

14. & - 9s - 90. 19. a 2 + ox - 42 a 2 . 84. aty* +14 ay-240. 

15. a 3 - * - 240. 80. s 2 -32 sy-105 y 2 . 85. a* - a 2 6 2 - 56 6*. 
18. a 2 - 12 a - 85. 81. a 2 + 18s - 116. 86. x* - 14a: 2 - 61. 
17. a 2 - 11 a - 152. 88. a 2 + 16 a* - 260. 87. y*+6 x*tf-Vl **. 

88. a 2 + 12 abx - 28 6*s 2 . 81. «* - a 2 * 2 - 132 a*. 

89. a 2 - 18 axy - 243 sefy 2 . 88. x* - a% 2 - 462 a 4 . 
80. sc* + 13 a*s 2 - 300 a*. - 88. x« + x 8 - 870. 

84. 2 + a - x 2 . 86. 110 - * - a* 2 . 38. 120 - 7 a* - a 2 * 8 . 

85. 6 + « - a; 2 . 87. 380 - x - aft 

94. When the Coefficient of the Highest Power is not Unity. 

Again, referring to Chap. iv. Art. 51, we may write the 
following results: 

(3aj + 2)(a5 + 4)=3a* + 14a> + 8 . . . (1), 

(3aj-2)(a?-4)=3^-14a?4-8 . . . (2), 

(3a? + 2)(a;-4)=3a5 2 -10a>-8 . . . (3), 
(3s-2)(aj + 4)=3ar 8 + 10a>-8 # . , (4). 

The converse problem presents more difficulty than the 
cases we have yet considered. 

Consider the result 3 x* - 14 x + 8 = (3 x - 2) (x - 4). 

The first term 3 x 2 is the product of 3 x and x. 

The third term -f 8 is the product of — 2 and — 4. 

The middle term — 14 x is the result of adding together 
the two products 3 x x — 4 and x x — 2. 

Again, consider the result 3 a? 2 — 10 a?— 8= (3 aj+2)(a?— 4). 

The first term 3 x 2 is the product of 3 a; and a?. 

The third term — 8 is the product of -f- 2 and — 4. 

The middle term — 10 x is the result of adding together 
the two products 3 x x — 4 and x x 2 ; and its sign is nega- 
tive because the greater of these two products is negative. 

Considering in a similar manner results (1) and (4), we 
see that : 

1. If the third term of the trinomial is positive, then the 
second terms of its factors have both the same sign, and this 
sign is the same as that of the middle term of the trinomial 



76 ALGEBRA. 

2. If the third term of the trinomial is negative, then the 
second terms of its factors have opposite signs. 

95. The beginner will frequently find that it is not easy 
to select the proper factors at the first trial. Practice alone 
will enable him to detect at a glance whether any pair he 
has chosen will combine so as to give the correct coefficients 
of the expression to be resolved. 

Ex. Resolve into factors 7 x 2 — 19 x — 6. 

Write (7 x 3) (x 2) for a first trial, noticing that 8 and 2 must 
have opposite signs. These factors give 7 x 2 and — for the first and 
third terms. But since 7x2— 3x1 = 11, the combination fails to 
give the correct coefficient of the middle term. 

Next try (7 x 2) (x 3). 

Since 7x3 — 2x1 = 19, these factors will be correct if we insert 
the signs so that the negative shall predominate. 

Thus 7X 2 - 19x-6=(7x + 2)(x-3). 

[Verify by mental multiplication.] 

96. In actual work it will not be necessary to put down 
all these steps at length. The student will soon find that 
the different cases may be rapidly reviewed, and the unsuit- 
able combinations rejected at once. 

Ex.1. Resolve into factors 14 x 2 + 29 x- 15 (1), 

14x 2 -29x-15 (2). 

In each case we may write (7 x 3) (2 x 5) as a first trial, noticing 
that 3 and 5 must have opposite signs. 

And since 7 x 6 — 3 x 2 = 29, we have only now to insert the proper 
signs in each factor. 

In (1) the positive sign must predominate. 

In (2) the negative sign must predominate. 

Therefore 14x 2 + 29x - 15 =(7x - 3)(2x + 6). 
14x 2 - 29x - 15 =(7x + 3)(2x - 6). 

Ex. 2. Resolve into factors 5 x 2 + 17 x + 6 (1), 

5x 2 -17x + 6 (2). 

In (1) we notice that the factors which give 6 are both positive. 
In (2) we notice that the factors which give 6 are both negative. 






RESOLUTION INTO FACTORS. 77 

And therefore for (1) we may write (5x + )(* + ). 

(2) we may write (6a; — )(as — ). 
And, since 6 x 3 + 1 x 2 = 17, we see that n 

5 a; 2 + 17 x + 6 = (5 x + 2) (x + 3). 
6a; 2 - 17 x + 6 =(5 x - 2)(x - 3). 

In each expression the third term 6 also admits of factors 6 and 1, 
but this is one of the cases referred to above which the student would 
reject at once as unsuitable. 

Ex. 3. 9a; 2 - 48 xy + 64 y 2 = (3x - 8y)(3aj - 8y) 

= (3x-8y) 2 . 

Ex. 4. 6 + 7a; - 6a; 2 =(3 + 6x)(2 - a;). 

Note. In Chapter xxyi. a method of obtaining the factors of any 
trinomial in the form ax 2 + bx + c is given. 

EXAMPLES X. 6. 
Resolve into factors : 

1. 2x 2 + 3x+l. 14. 2x 2 -x-l. 27. 16a; 2 - 77a + 10. 

% 3x 2 + 5x-f2. 15. 3x 2 + 7x-6. 28. 12 a; 2 - 31a;- 15. 

3. 2x 2 + 5x + 2. 16. 2x 2 + x-28. 29. 24x 2 + 22x-21. 

4. 3a; 2 + 10a; + 3. 17. 3 a: 2 + 13 x - 30. 30. 72x 2 -145x+72. 
6. 2x 2 + 9x + 4. 18. 6x 2 + 7x-3. 81. 24x 2 -29xy-4y*. 

6. 3x 2 + 8x + 4. 19. 2a; 2 -x -15. 32. 2-3x-2x 2 . 

7. 2a; 2 + 11a; + 6. 20. 3a; 2 + 19a;- 14. 33. 6 + 5x-6x 2 . 

8. 3x 2 + llx + 6. 21. 6x 2 -31x + 36. 34. 4-5x-6x 2 . 

9. 5x 2 + llx + 2. 22. 4x 2 + a;-14. 35. 5 + 32x-21x 2 . 

10. 3x 2 + x - 2. 23. 3x 2 - 13x + 14. 36. 18 - 33x + 5 x 2 . 

11. 4x 2 + llx-3. 24. 4x 2 + 23x+15. 37. 8 + 6x-5x 2 . 

12. Sa^+Hx-S. 25. 2x 2 -6xy-3y 2 . 38. 20-9x-20x 2 . 

13. 2x 2 +16x-8. 26. 8x 2 -38x+35. 39. 10-6x-15x*. 

97. We add an exercise containing miscellaneous exam- 
ples on the preceding cases. 

EXAMPLES X. f. 
Resolve into factors : 

1. x 2 + 13x + 42. 3. 2 x 2 -f 7 x + 6. 

2. 143 - 24 ax + a 2 * 2 . 4. a 2 b ? - 3 abc - 10 c 2 . 



78 



ALGEBRA. 



5. a^ + s-fl. 

6. 2ax 2 -h3axp-2bxy'-Sb^. 

7. x 2 + 7x0-600*. 

8. a a -o0-21O0 a . 

9. x a - 21a; + 110. 

10. 24 + 37*-72x*. 

11. 98- 7a*-* 8 . 

12. 3x 2 + 23x+14. , 

13. 2x 2 +3x-2. 

14. x 2 - 20x0 -960*. 
1ft. a*-20a&x + 75&***. 
18. a 2 - 24a + 95. 

17. 7 + 10* + 3* 2 . 

18. a 2 -4a -21. 

19. ^ + 43x^4-3900*. 



90. x 2 + 23x + 102. 

2L amaP+bmzg—anxff—bnjj*. 

22. 6x 2 -7x-8. 

88. S + llx-4x*. 

24. 12x 2 - 23x0 + 100*. 

2ft. 3x 2 +7x + 4. 

98. a 2 -32a + 256. 

97. 3**- 19*- 14. 

. x 2 - 19s + 90. 

. x 2 + 8x-40. 

80. xV-16xj/ + 39, 

81. 204-6*-**. 

82. 15x* + 224*-16. 
88. 3x 2 + 41x + 26. 
84. 65 + 8x0- 



WHEN AN EXPRESSION IS THE DIFFERENCE OF TWC 

SQUARES. 

98. By multiplying a +- b by a — 6 we obtain the identity 

(a + &)(a-6)=a a -6 2 , 
a result which may be verbally expressed as follows: 

The product of the sum and the difference of any two quan- 
tities is equal to the difference of their squares. 

Conversely, the difference of the squares of any two quan- 
tities is equal to the product of the sum and the difference oj 
the two quantities. 

Ex. 1. Resolve into factors 25 x* — 16 f*. 

26 x 2 - I62/ 2 =(5x) 2 -(4y)«. 

Therefore the first factor is the sum of 5x and 40, and the second 
factor is the difference of 5 x and 4 y. 

.-. 25x 2 - 16i/ 2 = (6x + 4y)(5x-4y). 

The intermediate steps may usually be omitted. 
Ex. 2. 1 - 49c« =(1 + 7c 8 )(l - 7 c*). 



RESOLUTION INTO FACTORS. 79 

The difference of the squares of two numerical quantities 
is sometimes conveniently found by the aid of the formula 

a 2 -& 2 = (a+&)(a-6). 

Ex. 8. (829)2 _ (171)2 = (329 + 171) (329 - 171) 

s 600 x 168 = 79000. 

EXAMPLES X. ff. 
Resolve into factors : 

1. a?-4. 4. c*-144. 7. 400 - a 2 . 10. 86a*-266* 

2. a 2 -81. 6. 49-c 2 . 8. «*-9a 2 . 11. 9x 2 -l. 

8. y 2 -100. 6. 121 -a; 2 . 9. y 2 -26» 2 . 12. 86p 2 -49g 2 . 

18. 4jfc*-l. 24. 1-860*. 85. ^-26^. 

14. 49 -100 A 2 . 26. 9s* -a 2 . 86. 25-64**. 

15. 1-260?. 28. 81a*-25a 2 . 87. 121a*-81a*. 

16. 9a!*-y 2 . 27. art* 2 -49. 88. p 2 g 2 -64o*. 

17. tf*q 2 -S6. 28. a 2 -64 3°. 39. 64s 2 -26*«. 

18. aW-l&cP. 29. a 2 6*-9a* 40. 49a* -16y*. 

19. a* -9. 80. afy«-4. 41. 81^ -26 6 2 . 

20. 9 a* -121. 81. l-a 2 6*. 42. 16a> u -9y«. 

21. 26a: 2 -64. 82. 9 -4a 2 . . 48. 36s»-49a M . 

22. 81a* -49a*. 83. 9a* -256*. 44. l-100a 6 6 4 c*. 

23. a*-25. 84. **-166*. 45. 2bz™-16a*. * 

Find, by resolving Into factors, the value of : 

46. (576) 2 - (425) 2 . 49. (339) 2 - (319)*. 52. ;i723)« - (277)*. 

47. (121)* - (120)*. 50. (753) 2 -(263) 2 . 53. (1639) 2 - (739)* 

48. (760)* -(260) 2 . 51. (101)*- (99) 2 . 54. (1811) 2 -(689) a . 

99. When One or Both of the Squares is a Compound Expres- 
sion. We employ the method of the preceding articles, as 
is shown in the following examples : 

Ex. 1. Resolve into factors (a + 2 6) 2 - 16 ai*. 
The sum of a + 2 b and 4x is a + 2 6 + 4a\ and their difference is 
« + 2ft-4z. 

.-. (a + 2 6) 2 - 16x 2 =(a + 26 + 4x)(a + 26 - 4*). 



80 ALGEBRA. 

Ex. a. Resolve into factors x 2 - (2 6 - 8 c) 2 . 

The sum of x and 2 6 — 3 c is x -f 2 6 — 3 c, and their difference is 

3 -(2 6 - 3c) = x - 26 + 8c. 

.•. x 2 -(26 - 3c) 2 =(x + 26 -3c)(x- 26 + Sc). 

If the factors contain like terms, they should be collected 
so as to give the result in its simplest form. 

Ex. 8. (3x + 7|0 2 -(2x-3y) 2 

= {(3x+7y) + (2x-3|0}{(3x + 7|0-(2x-8i0} 
= (3x:f-7y + 2x-8y)(3x+7y-2x + 3y) 

= (5x + 4y)(x+10j0. 

EXAMPLES X. h. 
Resolve into factors : 

1. (a + 6) 2 -c 2 . 4. (x + 2y) 2 -a 2 . 7. (x+6c)*-l. 

2. ( a -6) 2 -c 2 . 5. fa + 3 6) 2 -16x 2 . 8. (a-2x) 2 -6 2 . 

3. ( a + y )a_4^. 6 . (x + 5a) 2 ~9y*. 9. (2x-3a) 2 -9c* 

10. 9x 2 -(2a-36) 2 . 18. (6 - c) 2 - (a - x) 2 . 

11. l-( a -6) 2 . 19. (4a + x) 2 -(6 + y) 2 . 

12. c 2 -(5a-86) 2 . 80. (a + 26) 2 -(3x + 4y) 2 . 

13. (a + 6) 2 -(c + d) 2 . 21. 1 - (7 a - 3 6) 2 . 

14. (a-6) 2 -(x + y) 2 . 22. (a - 6) 2 - (x - y) 2 . 

15. (7x + y) 2 -l. 28. (a-Sx^-lOy 2 . 

16. (a + 6) 2 - (m - n) 2 . 24. (2a-5x) 2 -l. 

17. (a - w) 2 - (6 + ro) 2 . 25. (a + 6 - c) 2 -(x-y + *)*. 

Resolve into factors and simplify: 

28. (x + y) 2 -x 2 . 27. x 2 -(y-x) 2 . 28. (x + 8y)«-4y 2 . 

29. (24x + y) 2 -(23x-y) 2 . 84. 10 a 2 - (3 a + l) 2 . 

30. (6x + 2y) 2 -(3x-y) 2 . 85. (2a + 6 -c) 2 -(a - 6 + c) 2 . 

31. 9x 2 -(3x-6y) 2 . 86. (x - 7y + s) 2 -(7y -s) 2 . 

32. (7x+3) 2 -(5x-4) 2 . 87. (x + y - 8) 2 -(x-8) 2 . 

33. (3a + l) 2 -(2a-l) 2 . 38. (2x + a - 3) 2 -(3 -2x) 2 . 

100. Compound Expressions Arranged as the Difference oi 
two Squares. By suitably grouping the terms, compound 
expressions can often be expressed as the difference of two 
squares, and so be resolved into factors. 



BESOLUTION INTO FACTORS. 



81 



Ex. 1. Resolve into factors a 2 — 2 ax + x 8 — 46*. 

a 2 - 2 ax + x 2 - 46 2 = (a 2 - 2 ax + a 2 ) - 46* 

= (a - a) 2 - (2 6)» 
= (a-x + 26)(a-x-26). 

Ex. 8. Resolve into factors 9 a 2 - c 2 + 4 ex - 4 x*. 

9a 2 -c 2 + 4cx-4x 2 = 9a 2 -(c 2 -4cx + 4x«) 

= (3a) 2 -(c-2x) 2 
= (3a + c - 2x)(8a - c + 2x). 
Ex. 8. Resolve into factors 12 xy + 26 - 4x 2 - 9 y 2 . 
12xy + 26 - 4x 2 - 9y 2 = 26 -(4x 2 - 12xy + 9y 2 ) 

= (5) 2 -(2x-3y) 2 
= (6 + 2x-3|0(6-2x + 3y). 

Ex. 4. Resolve into factors 26a* - a 2 - c 2 + 6 2 + d 2 + 2ac. 

Here the terms 2bd and 2oc suggest the proper preliminary ai 
rangement of the expression. Thus 

26d - a 2 - c 3 + 6 2 + # + 2ac = ft 2 + 2 6d + tf» - a 2 + 2 ac - c 2 

= 6 2 + 2 bd + d 2 - (a 2 - 2 ac + c 2 ) 
= (& + d) 2 -(a-c) 2 
=(6 + d + a - c)(6 + <l - a + e). 



EXAMPLES X. 
Resolve into factors : 

1. x 2 +2xy+y 2 -o 2 . 

2. o 2 -2a6+6 2 -x 2 . 
8. x 2 -6ax+9a 2 -166 a . 

4. 4a 2 +4a6+6 2 -9c 2 . 

5. x 2 +a 2 +2ax-y a . 

6. 2aH-a 2 +y a -x 2 . 

7. x a -a 2 -2a6-6 2 



k. 



8. y 2 -c 2 +2cx-*». 

9. l-* 2 -2xy-i 



10. c 2 -x 2 -y 2 +2xy. 

11. x 2 +y 2 +2xy-4xy. 
18. a 2 -4a&+46 2 -9a 2 c a . 
18. x 2 +2xy+y 2 -a 2 -2a6-W 
14. a 2 -2a&+6 2 -c 2 -2cd-d* 



16. x 2 -4ax + 4a 2 -& 2 -t-26y-y*. 
18. y 2 + 2 by + ft 2 - a 2 - 6ax - 9a 2 . 

17. x 2 -2x + l -a 2 -4a&-46 2 . 

18. 9a 2 -6a + l-x 2 -8dx-l0d 2 . 

19. x*-a 2 + f/ 2 -b*-2xy + 2ab. 
90. a 2 + & 2 -2a&-c 2 — d*-2cd. 

21. 4 x 2 - 12ax - c 2 - k* - 2 ck + 9a 2 . 

22. a 2 + 66x-96 2 x 2 -10a6- l+266». 
28. a*-26x 8 + 8a 2 x 2 -9 + 30x 8 + 16x*. 



82 ALGEBRA. 

101. Important Cases. By a slight modification some ex- 
pressions admit of being written in the form of the difference 
of two squares, and may then be resolved into factors by the 

method of Art. 98. 

« 

Ex. 1. Resolve into factors x 4 + x 2 ^ 8 + y*. 

x 4 + x*y 2 + y* = (x 4 + 2 x V + y 4 ) - «V 

= 0» 2 + V 8 ) 2 ~ (ay) 2 

= (x 2 + y 2 + xy)(x 2 + y»-a^) 

= (x 2 + xy + ^(x 2 - xy + y 2 ) 

Ex. 2. Resolve into factors x 4 — 15 x 2 y 2 -f 9 y 4 . 

x* - 16 xV + 9y* = (x* - 6xV + 9y*)- 9x*y* 

= (*P-8yi)*-(3fly)t 
= (x 2 - 3y 2 + 3x20(* 2 - 8^ - 8xy;. 

EXAMPLES X. L 
Resolve into factors •• 

1. x* + 16x a + 266. 6. 4x 4 + 9y 4 -93xfy 8 . 

8. 81 a 4 + 9 a 2 6 2 + M. 7. 4 ro 4 + 9 n* - 24 ro 2 n 2 . 

8. x* + y*-7xV- 8. 9x 4 + 4y 4 + llx 2 y 2 . 

4. m 4 + n 4 -18ro 2 n 2 . 9. x 4 - ^x^ 2 * 26^. 

ft. x 4 - 6 x 2 y 2 + y 4 . 10. 16 a 4 + 6* - 28 a 2 6*. 

WHEN AN EXPRESSION IS THE SUM OR DIFFERENCE 

OF TWO CUBES. 

102. If we divide a 8 + b* by a + 5 the quotient is 
a* — a& 4- b 2 ; and if we divide a 3 — 6 8 by a — 6 the quotient 
is a 2 + ab + 6*. 

We have therefore the following identities : 

a 8 + b 8 =(a + &)(a 2 - a& 4- &*); 
a 8 - 6 8 = (a - 6) (a 2 + a& + ft 2 ). 

These results are very important, and enable us to resolve 
into factors any expression which can be written as the sun- 
or the difference of two cubes. 

Ex. 1. 8x 8 -27y 8 =(2x)»^(3y)« 

= (2x-3y)(4x 2 + 6xy + 9y 2 ). 
Note. The middle term 6 xy is the product of 2 x and 3 y. 



RESOLUTION INTO FACTOR8. 8S 

Ex. 2. 64a*+ 1 = (4a) 8 + (1) 8 

=(4a + l)(16a 2 -4a + l). 

We may usually omit the intermediate step and write the 
factors at once. 

Ex.8. 343a 8 -27a*==(7a 2 -8aO(49a* + 21a 2 x + 9a0). 
Ex. 4. 8a? + 729 = (2a; 8 + 9)(4 & - 18a* + 81). 

EXAMPLES X. m. 
Resolve into factors : 

1. a*-? 8 . 5. 8«8-y8. 9. a 8 & 8 -c». 18. 126 + a«. 

2. a^ + y 8 . 6. ^+8^. 10. 8a* + 27 y 8 . 14. 216 - a 8 . 
8. & - 1. 7. 27 ofi + 1. 11. 1 - 343 s 8 . 15. a 8 6 8 + 612. 
4. 1-fa 8 . 8. l-Sy 8 . 12. 64 + y 8 . 18. lOOOy 8 -!. 

17. « 8 + 64y 8 . 25. xV + s 8 . 33. 8a*-««. 

18. 27 - 1000 x 8 . 26. a*b*& - 1. 84. 216 ofi - ft 8 . 

19. a 8 6 8 + 216 c 8 . 27. 343 3* + 1000 y 8 . 85. a 8 + 343 & 8 . 

20. 343-8S 8 . 28. 729 a 8 - 64 6 8 . 36. a« + 7296 8 . 

21. a 8 + 27 6 8 . 29. 8 a 8 6 8 + 126 s 8 . 37. 8 a 8 - 729 y 8 . 

22. 27 a 8 - 64 y 8 . 30. afy 8 - 216 s 8 . 38. p 8 g 8 - 27 a 8 . 

23. 126 x 9 - 1. 31. afi - 27 y 8 . 89. s 8 - 64 y 8 . 

24. 216 p 8 - 343. 32. 64 a 6 + 125 y 8 . 40. x*y* - 612. 

103. Miscellaneous Cases of Resolution into Factors. • 
Ex. 1. Resolve into factors 16 a 4 — 81 6 4 . 

16 a* - 81 6 4 =(4 a 2 + 9 6 2 )(4 a 2 - 9 ft 2 ) 

= (4 a 2 + 9 ft 2 ) (2 a + 3 6) (2 a - 3 6). 

Ex. 2. Resolve into factors x* — y 6 . 

a* — y 6 = (se 8 4- y 8 ) (re 8 — y 8 ) 

= (x + y)(x 2 - gy + &)(x - y)(a5 2 + a# + y 2 ). 

Note. When an expression can be arranged either as the difference 
of two squares, or as the difference of two cubes, each of the methods 
explained in Arts. 98, 102 will be applicable. It will, however, be 
found simplest to first use the rule for resolving into factors the differ- 
ence of two squares. 

In all cases where an expression to be resolved contains a simple 
factor common to each of its terms, this should be first taken outside 
a bracket as explained in Art. 90. 



84 ALGEBRA. 

Ex. 8. Resolve into factors 28 x*y -f 64 x«y - 60 x*y. 

28 x*y + 64 x*y - 60 a*y =r 4 x*y (7 x 2 + 16 x - 15) 

= 4x2y(7a;-6)(flB + 3). 

Ex. 4. Resolve into factors a^p 2 - 8 j^p 2 - 4 xV + 32 y*tf. 

The expression = p^x 8 - 8 y 8 ) - 4 g 2 (s* - 8 y 8 ) 

= (x*-8y*)(p*-4g2) 
= (x - 2 y)(x 2 + 2xy + 4y»)Q> + 2g)(p - 2$). 

Ex. 5. Resolve into factors 4 x 2 — 26 y 2 + 2 a; + 6 y. 

4 x 2 - 26 y 2 + 2 x + 5 y =(2 x + 5 y)(2 x - 6 y) + 2 x + 6 y 

= (2 x + 6 y) (2 x - 5 y + 1). 

EXAMPLES X. o. 
Resolve into two or more factors : 

1. a 2 -y 2 -2 yz-z*. S3. a 2 x - 6 2 x + a*y - 6*y. 

2. x 6 - y 6 ^ 6 . 24. x 4 + 4 xV* 2 + 4 y 4 * 4 . 

3. 6 x 2 - x - 77. 25. a 8 6» + 512. 

4. 729y*-64x«. 26. 2x 2 +17x + 36. 

5. x* - 4096. 27. . 600 x 2 y - 20 y 8 . 

6. 2win+2xy+m 2 +n 2 -x 2 -y 2 . 28. a 6 -8a 2 6 8 . 

7. S3x 2 -16x-66. 29. a*x* - 16 xfy*- 

g. a*+b*-c i -d*+2a?b*-2c*d?. 80. 6 2 + c 2 - a 2 -2 6c. 

9. m*x + m8y - n 8 x - n 8 y. 81. 6x* - 15X 8 - 90x 2 . 

10. (a + & + c) 2 -(a-6-c) 2 . 32. 14 a?x* - 36 a 8 *? + 14 a*x. 

11. 4 + 4 x + 2 ay + x 2 - a 2 - y 2 . 33. x 8 - 1. 

12. x 2 - lOx - 119. 84. 1 -(ro 2 + n 2 )+ 2 win. 

18. a*-b*-c 2 +cP-2(ad-bc). 35. 75x 4 -48a 4 . 36. 5a*b*-6ab 

14. x 2 - a 2 + y 2 - 2 xy. 87. 8 x*y + 62 xy + 60 y. 

15. a 2 +x 2 -(y 2 +3 2 )-2(y*-ax). 38. 3 xfy 2 + 26 axy + 36 a 2 . 

16. 21 x 2 + 82 x - 39. 39. 729 a 7 6 - a& 7 . 

17. 1 - a*x 2 - b*y* + 2 a6xy. 40. a*x* - 64 ay. 

18. ctd* - c 2 - aW* + a 2 . 41. a 12 - 6 12 . 

19. a^ - aV - *>*& + W- 42- 24x 2 y 2 - SOxy 8 -36y*. 

20. x 2 - 6x - 247. 43. (a + 6) 4 - 1. 

21. a*x*-#x*-aty + <W' 44- ^-(fc + c) 4 . 

22. acx* - &cx + adx - 6d. 45. (c + a") 8 - 1. 



RESOLUTION INTO PAOTOES. 86 

46. 1 -(x - y)«. 58. (a+6) 2 + o + 6. 

47. 260(a - 6)« + 2. 53. a 8 + & 8 + a + b. 

48. (c + d)*+ (c - d)«. 54. a a -96 2 + a + 35. 

49. 8 (x + y)*- (2 x - y)«. 55. 4 (x - y)»- (x - y). 

50. x 2 - 4 y 2 + x - 2 y. 56. x*y - x 2 y 3 - xfy 2 + sy*. 

51. a- - ft 2 + a - 6. 57. 4 a 2 - 9 6 2 + 2 a - 3 &. 

58. Resolve x 16 — y 1 * into five factors. 



CONVERSE USE OF FACTORS. 

104. The actual processes of multiplication and division 
can often be partially or wholly avoided by a skilful use of 
factors. 

It should be observed that the formulae which the student 
has seen exemplified in the preceding pages are just as use- 
ful in their converse as in their direct application. Thus 
the formula for resolving into factors the difference of two 
squares is equally useful as enabling us to write at once the 
product of the sum and the difference of two quantities. 

Ex. 1. Multiply 2a+3&-cby2a-35 + c. 
These expressions may be arranged thus : 

2a+(36-c) and 2a -(36 -c). 
Hence the product = {2 a + (3 b - c)}{2 a - (8 b - c)} 

= (2 a) 2 - (3 b - c) 2 [Art 98.] 

= 4a 2 -(9& 2 -6&c + c 8 ) 

= 4a 2 -96 2 + 66c-c 2 . 

Ex. 8. Divide the product of 2x 2 + x - 6, and Ox 2 - 5x + 1 by 
3x 2 + 5x-2. 

Denoting the division by writing the divisor under the dividend 
(Art. 53), with a horizontal line between thein, the required quotient 

_ (2x- 2 + x-6)(6x 2 -6x+l) 
~~ 3x 2 + 5x-2 

_ (2x - 3) ( x + 2)(3x - l)(2x - 1) 
"~ (3x-l)(x + 2) 

c(2x-3)(2x-l). 



86 ALGEBRA. t 

Ex. 8. Prove the identity 

17(6* + 3a) 2 - 2(40x + 27 a)(6 x + 3a)= 25 x* - 9a*. 

Since each term of the first expression contains the factor 6 x + 3 a, 
the first side 

=(6* + 3a){17(6x + 3a)- 2(40 a: + 27 a)} 

= (6x + 3a)(85x + 61a-80x-64a) 

= (6x+3a)(6x-8a) 

= 26x 2 -9a 2 . 

Ex. 4. Show that (2x + 3y~*) 8 + (Sx + 7y + *) 8 is divisible by 
6(x + 2 y). 

The given expression is of the form A* -f JB 8 , and therefore has a 
divisor of the form A + B . 

Therefore (2x + 3y-«) 8 +(3x + 7y + *) 8 

is divisible by (2x + 3y - s) + (3x + 7y + z) 9 

that is, by 6x+10y, 

or by 6(x + 2y). 

EXAMPLES X. p. 

Find the product of 

1. 2x-7y + 3sand2x-f 7y-8s. 

5. 3x 2 -4xy + 7y 2 and3x 2 + 4xy + 7y 2 . 
.8. 6x 2 + 6xy-9y 2 and6x 2 -5xy-9y 2 . 

4. 7x 2 -8xy + 3y 2 and7x 2 + 8xy-8y 2 . 

6. x 8 + 2x 2 y + 2xy 2 + y 8 and x 8 - 2xty + 2xy 2 - y 8 . 

6. (x + y) 2 + 2(x + y) + 4 and (x + y) 2 - 2(x + y) + 4. 

7. Multiply the square of a + 3 b by a 2 - 6 ab + 9 6 2 . 

8. Divide(4x + 3y - 2s) 2 -(3x - 2y -f 3s) 2 byx + 6y- 60. 

9. Divide x 8 + 16a 4 x* + 266a 8 by x 2 + 2ax + 4a 2 . 

10. Divide (x 2 + 7x + 10)(x + 3) by x 2 + 6x + 6. 

11. Divide (3x -f 4y - 2s) 2 -(2x + 3y - 4s) 2 by x + y + 2*. 

Prove the following identities ; 

12. (a + &) 8 -(a - 6) 2 (a + &) = 4a&(a + 6). 
18. c*-d*-(c-d") 8 (c + d")=2cd(c 2 -eP). 
14. (x + y) 4 -3xK* + y) 2 =(x + y)(x 8 + y 8 ). 



RESOLUTION INTO FACTORS. 87 

15. Show that the square of x + 1 exactly divides 

(x 8 + x 2 + 4) 8 -(x 8 - 2x + 8)». 

la Show that (3x 2 -7x+2) 8 -(x 2 -8x+8) 8 is divisible by 2x-3, 
and by x + 2. 

105. The Factor Theorem. If any rational and integral 
expression containing x becomes equal to when a is written 
for x, it is exactly divisible by x — a. 

Let P stand for the expression. Divide P by as — a until 
the remainder no longer contains x. Let R denote this 
remainder, and Q the quotient obtained. Then 

P=Q(a-a) + 2*. 

Since this equation is true for all values of x, we will 
assume that x equals a. By hypothesis, the substitution of 
a for x makes P equal to ; thus, 

0=Q(0) + i*. 

A i* = 0. 

As the remainder is 0, the expression is exactly divisible 
by x—a. 

The following examples illustrate the application of this 
principle : 

Ex. 1. Resolve into factors x* + 3x* - 13 x - 16. 
By trial we find that this expression becomes when x = 3 ; hence, 
x — 3 is a factor. Dividing by x — 3, we obtain the quotient 

x 2 + 6x + 6. 

The factors of this expression are easily seen to be x + 1 and x + 6 ; 
hence 

x' + Sx 8 - 13x-16=(x-3)(x+l)(x+6). . 

Ex. 3. Resolve into factors x 8 + 6x 2 + 11 x + 6. 

It is evident that substituting a positive number for x will not make 
the expression equal to 0. By substituting — 1, however, for x v the 
expression becomes — 1+6 — 11 + 6, or 0; hence, 

x 8 + 6x 2 + llx + 6 

is divisible by x+ 1. Dividing by x+ 1, we obtain the quotient 
x 2 + 6 x + 6, and factoring this expression we have 

x 8 + 6x 2 + llx + 6 =(x + l)(x + 2)(x + 8). 



88 ALGEBRA. 

Note. The student should notice that the only numerical values 
that need be substituted for x are the factors of the last term of the 
expression, and that we change the sign of the factor substituted 
before connecting it with x. Thus, in Ex. 1, the factor 3 gives a 
divisor x — 3, and in Ex. 2, the factor — 1 gives a divisor x + 1. 

Ex. 3. Without actual division, show that bx 5 — 6 x*+l is divisible 
by x — 1. 

If the expression is divisible by x— 1, it will become 0, or "vanish," 
when 1 is substituted for x. Making this substitution, we obtain ; 
hence the division is possible. 

EXAMPLES X. r. 

Without actual division, show that x — 2 is a factor of each of the 
following expressions : 

1. a* -6a: +2. 8. a* -7s 2 + 16*- 12. 

3. a* + a; 2 -4a:-4. 4. 3»-8a0 + 17x- 10. 

Determine by inspection whether x -f 3 is a factor of any of the 
following expressions : 

5. x*~7x + 6. -7. a* + 6a; + 6. 

6. a* + 6x 2 + 11 a; + 6. 8. a* + 3a5 2 + x + 3. 
9. Show that 32 x 10 - 33 a* + 1 is divisible by x - 1. 

Resolve into factors . 

10. 2a* + 4x 2 -2a;-4. 11. 3a* -6a* ~3x + 6. 

106. We shall employ the Factor Theorem in giving 
general proofs of the statements made in Art. 62. 

We suppose n to be a positive integer. 

(I.) af — #" is always divisible by x — y. 

By Art. 105, x n —y n is exactly divisible by x—y if the sub- 
stitution of y for x in the expression x* — #* gives zero as 
a result. Making this substitution we have 

a; n — y n = y n — y* = 0. 

Therefore #* — y n is always divisible by x — y. 

(II.) a n — y n is divisible by x-\-y when n is even. 

If this be true, the substitution of — y for x in the ex- 
pression a*" — y n gives zero as a result. Making this sub- 
stitution we have 

af — 2f = (— #)" — #". 



RESOLUTION INTO FACTORS. 

When n is even, this expression becomes #* — #*, or zero. 
Therefore af — y* is divisible by x + y when n is even. 

(III.) af + #* is never divisible by x — y. 

Here the substitution of y for x in the expression af +#• 
gives 

a*" + y* = #" +jT = 2y*. 

As this expression is not zero, x* + #* is never divisible 
by a? — #. 

(IV.) af -f- #" is divisible by a? + y when /i is odd. 

Here the substitution of — y for x in the expression 
a*" + y* gives 

af + 3T = (-y) ,, + 3r. 

When w is odd, this expression becomes — y" + #", or zero. 
Therefore af + y" is divisible by a? + y when w is odd. 

The results of the present article may be conveniently 
stated as follows : 

(i.) For all positive integral values of n, 
a?" — y" = (a; — y) (x"~ l + x*~*y + aT^y 2 H hy*" 1 - 

(ii.) When w is odd, 

x* + y« = (x + y)(x*- l -x^*y + x*^tf -*** + #*-*). 

(iii.) When n is even, 

af — y* = (a? + y) (a?*" 1 — a^ _, y + af 1 " 8 ^ 1 — ••• — y*" 1 ). 

107. We shall now discuss some cases of greater difficulty, 
and also show how certain expressions of frequent occur- 
rence, which are not integral, may be separated into factors. 
The student may omit this portion of the chapter until 
reading the subject a second time. 

Ex. 1. Resolve a 9 - 64 a 8 - a 6 + 64 into six factors. 

The expression 

= a* (a* - 64) - (o« - 64) 

= (a 8 -64)(a 8 -l) 

= (a 8 + 8)(a 8 -8)(a 8 -l) 

= (a + 2)(a 2 -2a + 4)(a - 2)(a* + 2a + 4)(a - l)(a« + a + 1). 



dO ALGEBRA. 

Ex. 3. a(a-l)a?-(a-&-l)ay-&(& + l)¥* 
= {ax - (6 + 1) y} {(a - 1) « + &srt. 

108. From Ex. 2, Art. 59, we see that the quotient of 

a 8 + 6 8 + c? — 3abc by a + 6 + c is a 2 + 6 2 + c 2 — 6c — ca — ab. 
Thus a 8 + 6 8 -f-c 8 -3a6c 

= (a + 6 + c)(a 8 + 6 2 + c 8 -6c-ca-a6) . . (1). 

This result is important and should be carefully remem- 
bered. We may note that the expression on the left consists 
of the sum of the cubes of three quantities a, 6, c, diminished 
by three times the product dbc. Whenever an expression 
admits of a similar arrangement, the above formula will 
enable us to resolve it into factors. 

Ex. 1. Resolve into factors a 8 — 6 8 -f c 8 + 3 abc. 

a 8 - 6 s + c 8 + Sabc = a 8 + (— ft) 8 + c 8 - 3 a (- b) c 

= (<*-& + c)(a 2 + & 2 + c 2 + 6c-ca + a&), 
— & taking the place of b in formula (1). 

Ex. 2. 

a* _ 8 y 8 - 27 - 18&y = a* + (- 2y) 8 + (- 3) 8 - Sx (- 2y)(- 3) 

= (x - 2y - S)(x* + 4^ + 9 - 6 y + Sx + 2xy). 

109. Expressions which can be put into the form a? 8 ± — 

may be separated into factors by the rules for resolving the 
sum or the difference of two cubes. [Art. 102.] 

Ex. 1. \ - 27 & 8 = (? Y- (3 62)8 

. " -e-")(j+ ! J !+ "'> 

Ex. 2. Kesolve cfte 8 -—-*» + 4 i* 1 * £our factors. 
cto? - ^—<# + ^ = « , (o a - l)-4(« a - !) 






RESOLUTION INTO FACTORS. 91 

EXAMPLES X. 8. 

Resolve into two or more factors : 

1. x*y + 3acy 2 -3s«-y 8 . 5. a 8 + (a + b)ax + bx 2 . 

2. 4 mn 2 - 20 n 8 + 46 nm 2 - 9m 8 . 6. pn(rn 2 + 1) - m(& + n 2 ). 

3. ab(x 2 +l)+x(a 2 + V). 7. 66x(a 2 + 1)- a(4x 2 + 9 ft 2 ). 

4. *%?(** -1) +&?(**-«*). 8. (2 a 2 + 3 y 2 )x + (2 x* + 3 a 2 )y 

9. (2aj 2 -3a 3 )y + (2a 2 -3y 2 )a;. 

10. a(a - l)x 2 + (2 a 2 - l)x + a(a + 1). 

11. 3x 2 -(4a + 2&)x + a 2 + 2a&. 

12. 2 a 2 x a - 2(3 & - 4 c)(& - c)y 2 + a&xy. 

13. (a 2 - 3a + 2)x 2 + (2 a 2 - 4a + l)x + a{a - 1). 

14. a(a+l)x a + (a + &)xy-&(&-l)y 2 . 

15. ft 8 + c 8 - 1 + 3 &c. 18. a 8 - 27 & 8 + c 8 + 9 abc. 

16. a 8 + 8 c 8 + 1 - 6 ac. 19. a 8 - & 8 - c 8 - 3 a&c. 

17. a 8 + 6 8 +8c 8 -6a&c. 20. 8 a 8 + 27 & 8 + c 8 - 18 abc. 

"■Si- 1 - 28 -S +y8 - ' 86 -!l +1000 - 

22.216a 8 -! 8 . 24. ^-1. *>• £-^' 

8 729 612 x 8 

27. Resolve x 8 + 81 x 4 + 6561 into three factors. 

28. Resolve (a* - 2 a 2 b 2 - & 4 ) 2 - 4 a*& 4 into four factors. 

29. Resolve 4(a& + cd) 2 - (a 2 + b 2 - c 2 - d 2 ) 2 into four factors. 

80. Resolve x 8 into four factors. 

266 

81. Resolve x w — y 18 into six factors. 
Resolve into four factors : 

86. ^-.^£--J- + £ 
72 32 9x 2 4 

87. x 6 -26x 2 + 6J-ix*. 

4 



32. 


«!-8x-a 8 + 8x 8 . 

X 2 


33. 


x 9 + x 3 y-8x 8 ^ 8 -8i^. 


34. 


x 9 + x 6 + 64x 8 + 64. 


86. 





Resolve into five factors : 

88. xf + x^-iex^-ie. 89. lexf-si^-ie^H-si. 



CHAPTER XI. 
Highest Common Factor. 

110. Definition. The Highest Common Factor of two 
or more algebraic expressions is the expression of highest 
dimensions (Art. 29) which divides each of them without 
remainder. 

The abbreviation H. C. F. is sometimes used instead of 
the words highest common factor. 

SIMPLE EXPRESSIONS. 

111. The H. C. F. can be written by inspection. 

Ex. 1. The highest common factor of a 4 , a 8 , a 2 , a 6 is a 2 . 

Ex. 2. The highest common factor of a 8 6 4 , «& 6 c 2 , a 2 b 7 c is ab* ; for 
a is the highest power of a that will divide a 8 , a, a 2 ; b 4 is the highest 
power of b that will divide 6 4 , & 6 , b 1 ; and c is not a common factor. 

112. If the expressions have numerical coefficients, find 
by Arithmetic their greatest common measure, and prefix 
it as a coefficient to the algebraic highest common factor. 

Ex. The highest common factor of 21 a 4 ofiy t S&cfixty, 28 cfixy 4 is 
7 ofay ; for it consists of the product of 

(1) the numerical greatest common measure of the coefficients ; 

(2) the highest power of each letter which divides every one of the 
given expressions. 

EXAMPLES XI. a. 
Find the highest common factor of 

1. 4a6 2 , 2a 2 &. 3. 6xy% 8&Vs 2 . 5. 6a 8 & 8 , 15 abc*. 

2. Sx 2 ^ 2 , x 8 y 2 . 4. a6c, 2a6 2 c. 6. Ox 2 ? 2 * 2 , 12xy«s. 

02 



HIGHEST COMMON FACTOR. 



93 



7. 4a 2 6 8 c 2 , 6a*b*c*. 

8. 7 a^c 6 , 14 abW. 

9. 49 as*, 63 ay 2 , 66 az\ 

10. 17 a&*c, 34 a*bc, 61a&c 2 . 

11. fl^xfy 2 , Wxy 2 , c*x*y. 



IS. 25x^3, 100x 2 ys, 126 xy. 

13. a 2 bpxy, b' 2 qxy, a*bxt*. 

14. 16a 6 W, 60 a 8 & 7 c«, 26 aWc*. 

15. 36 aW, 42 a 8 c& 2 , 30ac 2 6». 

16. 24a 8 6 2 c 8 , 16 a*b*<fl, 40 a'W. 



COMPOUND EXPRESSIONS. 

113. H. C. F. of Compound Expressions which can be fac- 
tored by Inspection. The method employed is similar to 
that of the preceding article. 

Ex. 1. Find the highest common factor of 

4 ex* and 2cx 8 + 4c 2 x 2 . 

It will be easy to pick out the common factors if the expressions 
are arranged as follows : 

4 ex 8 = 4cx 8 , 

2 ex* + 4 c*x 2 = 2 ex 2 (x + 2 c) ; 

therefore the H. C. F. is 2 ex 2 . 

Ex. 2. Find the highest common factor of 

3a 2 +9a&, a 8 -9a& 2 , a 8 + 6 a 2 6 + 9 afl». 

Resolving each expression into its factors, we have 

3a 2 + 9a& = 3a(a + 36), 
a 8 - 9a& 2 = a(a + 3 &)(a - 36), 
a 8 + 6 a 2 & + 9 ab* = a(a + 3 &) (a + 3 6); 
therefore the H. C. F. is a(a + 3 &). 

114. When there are two or more expressions containing 
different powers of the same compound factor, the student 
should be careful to notice that the highest common factor 
must contain the highest power of the compound factor 
which is common to all the given expressions. 

Ex. 1. The highest common factor of 

x(a — x) 2 , a (a — x) 8 , and 2 ax(a - x) 6 is (a — x) 2 . 

Ex. 2. Find the highest common factor of 

ax 2 + 2 a*x + a 8 , 2 ax 2 - 4 aH - 6 a 8 , 3 (ax + a 1 ) 8 . 



84 ALGEBRA. 

Resolving the expressions into factors, we have 

(W 2 + 2fl^ + o 8 = a(& + 2 ax + a*) 

= a(x + a) 2 (1% 

2ax 2 ~4a 2 x-6o 8 = 2a(x 2 -2ax-8a a ) 

= 2a(x + a)(x-3a) . . .. (2), 
3 (ax + a 2 )« = 3 a 2 (x + a) 2 (3). 

Therefore from (1), (2), (3), by inspection, the highest common 
factor is a(x + a). 

EXAMPLES XL b. 
Find the highest common factor of 

1. a 2 + ab, a 2 -ft 2 . 13. 6 6x + 4 &y, 9 ex + 6 cy. 

3. (x + y) 2 , x 2 - y 2 . 13. x 2 + x, (x + l) 2 , x 8 + 1. 
8. 2 x 2 — 2 xy, x 8 — xty. 14. xy — y t x*y — xy. 

4. 6x 2 -9xy, 4x 2 -9y 2 . 15. x 2 - 2 xy + y 2 , (x - y)K 

5. x 8 + x*y, x 8 + y 8 . 16. x 8 + a^ x 4 - a*. 

6. a 8 6 - ab 9 , aPV* - a 2 6 6 . 17. x 8 + 8 y 8 , x 2 + xy - 2 y 2 . 

7. a 8 - a 2 x, a 8 - ax 2 , a 4 - ax 8 . 18. x 4 - 27 a^, (x - 3 a) 2 . 

8. a 2 -4x 2 , a 2 + 2 ax. 19. x 2 + 3x+ 2, x 2 -4. 

9. a 2 - x 2 , a 2 - ax, a^x- ax 2 . 20. x 2 - x - 20, x 2 - 9x + 20 

10. 4x 2 + 2xy, 12x^-3^. 21. x 2 - 18x + 45, x 2 -9. 

11. 20x-4, 50x 2 -2. 22. 2x 2 -7x+3, 3x 2 -7x-6. 
28. x 6 - xy 2 , x 8 + x*y + xy + y 2 . 

24. a 8 x - a 2 &x - 6 a&^x, a 2 6x 2 - 4 aft 2 * 8 + 3 b*x*. 

25. 2x 2 + 9x + 4, 2x 2 + llx+6, 2x 2 -3x-2. 

26. 3x* + 8x 8 + 4x 2 , 3x 6 + llx* + 6x 8 , 3x* - 16X 8 - 12x*. 

27. 2x*+5x 8 +3x 2 , Ox^ + lSa^ + ex 2 , 2x* - 7X 8 - 15 x 2 . 

28. 12x + 6x 2 + 6, 6x + 3x 2 + 3, 18x + 3x 2 + 15. 

29. x 4 + 4x 2 + 3, x* + 5x 2 + 6, 3x* + llx 2 + 6. 

80. 2 a 2 + 7 ad + 6 d 2 , 2 a 2 + 9 ad + 9 d 2 , 6 a 2 + 11 atf + 8 d 3 . 

81. 2x 2 + 8xy + 6y 2 , 4x 2 + 14xy + 6y 2 , 2x 2 + lOxy + 12y*. 

IIS. H. C. F. of Compound Expressions which cannot be 
factored by Inspection. To find the highest common factor 
in such cases, we adopt a method analogous to that used in 
Arithmetic for finding the greatest common measure of two 
or more numbers. 



HIGHEST COMMON FACTOB. 



95 



Note. The term greatest common measure is sometimes used 
instead of highest common factor ; but, strictly speaking, the term 
greatest common measure ought to be confined to arithmetical quanti- 
ties; for the highest common factor is not necessarily the greatest 
common measure in all cases, as will appear later. (Art. 121.) 

116. We begin by working out examples illustrative of 
the algebraic process of finding the highest common factor, 
postponing for the present the complete proof of the rules 
we use. But we may conveniently enunciate two principles, 
which the student should bear in mind in reading the 
examples which follow. 

I. If an expression contains a certain factor, any multiple 
of the expression is divisible by that factor. 

II. If two expressions have a common factor, it will divide 
their sum and their difference ; and also the sum and the 
difference of any multiples of them. 

Ex. Find the highest common factor of 

4s*-3ai 2 -24 x-9 and Sx*- 2x* -53x-89. 



2x 



8 



4s»- 
4s» 


-3x 2 - 
-6a 2 - 


24 s- 
21s 


9 




2a; 2 - 
2s 2 - 


3s- 
Qx 


9 




3»- 
3*- 


9 
9 



8s* -2z 2 -53a; -39 

8s»-6s 2 -48a;-18 

4 X 2_ 6&-21 

4s 2 - 6a;-18 

x- 3 



2 



2 



Therefore the H. C. F. is x — 3. 

Explanation. First arrange the given expressions according to 
descending or ascending powers of x. The expressions so arranged 
having their first terms of the same order, we take for divisor that 
whose highest power has the smaller coefficient. Arrange the work 
in parallel columns as above. When the first remainder 4 b 2 — 6 s— 21 
is made the divisor we put the quotient x to the left of the dividend. 
Again, when the second remainder 2 x 2 — 3 x — 9 is in turn made the 
divisor, the quotient 2 is placed to the right ; and so on. As in Arith 
metic, the last divisor x — 3 is the highest common factor required. 

117. This method is only useful to determine the com' 
pound factor of the highest common factor. Simple factors 
of the given expressions must be first removed from them, 



96 



ALGEBRA. 



and the highest common factor of these, if any, must be 
observed and multiplied into the compound factor given by 
the rule. 

Ex. Find the highest common factor of 

24 a* - 2a; 8 - 60 x 2 - 32x and 18x* - 6x* - 89 z* - 18 x. 
We have 24 a* - 2x 8 - 60 x 2 - 32x = 2x(12x 8 - x 2 - 30x - 16), 
and 18x* - 6X 8 - 39x 2 - 18x = 3x(6a* - 2X 2 - 13x - 6). 

Also 2 x and 3 x have the common factor x. Removing the simple 
factors 2 x and 3 x, and reserving their common factor x, we continue 
as in Art. 116. 



2x 



6x«-2x 2 -13x-6 
6x»-8x 2 - 8x 


12x* 
12 x 8 


- x 2 -30x-16 
-4x 2 -26x-12 


6x 2 - 5x-6 

6x 2 - 8x-8 

3x + 2 




Sx*- 4x- 4 
3x 2 + 2x 

- 6x- 4 

- 6x- 4 



2 



x 



-2 



Therefore the H. C. F. is x(3 x + 2). 

118. So far the process of Arithmetic has been found 
exactly applicable to the algebraic expressions we have con- 
sidered. But in many cases certain modifications of the 
arithmetical method will be found necessary. These will 
be more clearly understood if it is remembered that, at 
every stage of the work, the remainder must contain as a 
factor of itself the highest common factor we are seeking. 
[See Art. 116, 1 & II.] 

Ex. 1. Find the highest common factor of * 

Sx 8 - 13x 2 + 23x - 21 and 6X 8 + x 3 - 44x + 21. 



Sx 8 -13x 2 + 23x-21 



6x*+ x 2 -44x + 21 

6x 8 -26x 2 + 46x-42 

27x 2 -90x + 63 



2 



Here on making 27 x 2 — 90 x -f 63 a divisor, we find that it is not 
contained in 3X 8 — 13 x* 2 + 23 x — 21 with an integral quotient. But 
noticing that 27 x 2 -90x+63 may be written in the form9(3x 2 -10x+7), 
and also bearing in mind that every remainder in the course of the 
work contains the H. 0. F., we conclude that the H. C. F. we are 
Beeking is contained in 9(3 x 2 — 10 x + 7). But the two original ex- 



HIGHEST COMMON FACTOR. 



97 



pressions have no simple factors, therefore their H. C. F. can hare 
none. We may therefore reject the factor 9 and go on with divisor 
8 a* — 10 x + 7. Besoming the work, we have 



-1 



3a*-13 a* + 28a;-21 
3s 8 -10a* + 7a; 

- 3a* + 16a; -21 

- 3a*+10a;- 7 

2)6 x - 14 



3a*-10a; + 7 
3 a*- 7 a; 

- 3a; + 7 

- 3a;+7 



-1 



Sx- 7 

Therefore the highest common factor is 3 x — 7, 
The factor 2 has been removed on the same grounds as the factor 
9 above. 

Ex. S. Find the highest common factor of 

2a* + a*-x-2 (1), 

and 3 a* -2 a* -ha; -2 (2). 

As the expressions stand we cannot begin to divide one by the other 
without using a fractional quotient. The difficulty may be obviated 
by introducing a suitable factor, just as in the last case we found it 
useful to remove a factor when we could no longer proceed, with the 
division in the ordinary way. The given expressions have no common 
simple factor, hence their H. C. F. cannot be affected if we multiply 
either of them by any simple factor. 

Multiply (2) by 2, and use (1) as a divisor : 



-2a> 



17 a; 



14 



2a* + a; 2 - x- 
7 


2 

14 


6a*- 4a* + 2a;- 4 
6a* + 3a; 2 - 3a;- 6 


14a* + 7a; 2 - 7a;- 
14a* -10a; 2 - 4a; 


- 7a* + 6*+ 2 
17 


17a*- 3a;- 
17 a; 2 -17 a; 


14 

14 
14 


-119 a* -f 86a; + 84 
-119 a* + 21 x + 98 


14a;- 
. 14a;- 


64)64 a; -64 
x- 1 



-7 



Therefore the H. C.F. is x - 1. 

After the first division the factor 7 is introduced because the first 
remainder -7x 2 + 5a;+2 will not divide 2 a* + a* - x - 2. 

At the next stage the factor 17 is introduced for a similar reason, 
and finally the factor 64 is removed as explained in Ex. 1. 

From these examples it appears that we may multiply or divide 
either of the given expressions, or any of the remainders which occur 
in the course of the work, by any factor which does not divide both 
of the given expressions. 



98 ALGEBRA. 

t 
Note. If, in Ex. 2, the expressions had been arranged in ascend 
ing powers of x, it would have been found unnecessary to introduce 
a numerical factor in the course of the work. 

119. The use of the Factor Theorem (Art! 105) often 
lessens, in a very marked degree, the work of finding the 
highest common factor. Thus in Ex. 2 of the preceding 
article it is easily seen that both expressions become equal 
to when 1 is substituted for x. t hence x — \ is a factor. 
Dividing the first of the given expressions by as — 1, we 
obtain a quotient 2 x 2 + 3 x +- 2. It is evident that this 
will not divide the second expression, hence x — 1 is the 
H. C. F. 

120. When the Method of Division by Detached Co- 
efficients (Art. 63) is employed in finding the H. C. F., 
the following is a convenient arrangement. 

Ex. Find the H.C.F. of 

x* + 3x*+12a;-16 and x 8 - 13 x + 12. 

We write the literal factors of the dividend until we reach a term 
of the same degree as the first term of the divisor. 

x* 


+ 13 
-12 

a; +3; 13 + 39-62 

The addition of the terms in the third column gives 13 » 2 , which is 
of lower degree than the first term of the divisor, hence we can 
proceed no further with the division and have for a remainder 
13 x 2 + 39 x — 52. Removing from this remainder the factor 13, as 
it is not a factor of the given expressions, we have for a second divi- 
sor x 2 + 3 x — 4. The first divisor, as written . before the signs werv 
changed, forms the second dividend : 



flc* + 3« 8 + 


0+12- 


-16 





+ 13-12 








+ 39- 


-36 



x 2 


x8 + 0z 2 -13 + 12 


-3 


-3+4 


+ 4 


+ 9-12 



a; -3; 

since there is no remainder, the last divisor, as written before the signs 
were changed, is the H. C. F. Thus x 2 + 3 x — 4 is the H. C. F. 



HIGHEST COMMON FACTOR. 99 

Let the two expressions in Ex. 2, Art. 118, be 
written in the form 

2x t + x*-x-2=(x-l)(2x* + 3x + 2), 

3x*-2x 2 + x-2=(x-l)(3x* + x + 2). 

Then their highest common factor is x — 1, and therefore 
2x* + 3x + 2 and 3x* + x + 2 have no crtgebrafo common 
divisor. If, however, we put a? =6, then 

2aj» + s 2 -a;-2 = 460, 
and 3a 8 -2 a 2 -!- a; -2 = 580; 

and the greatest common measure of 460 and 580 is 20; 
whereas 5 is the numerical value x — 1, the algebraic highest 
common factor. Thus the numerical values of the algebraic 
highest common factor and of the arithmetical greatest com- 
mon measure do not in this case agree. 

The reason may be explained as follows : when x = 6, the 
expressions 2 x 2 4- 3 x + 2 and 3 x 2 + x -f- 2 become equal to 
92 and 116 respectively, and have a common arithmetical 
factor 4; whereas the expressions have no algebraic com- 
mon factor. 

It will thus often happen that the highest common factor 
of two expressions and their numerical greatest common 
measure, when the letters have particular values, are not 
the same ; for this reason the term greatest common meas- 
ure is inappropriate when applied to algebraic quantities. 

EXAMPLES XI. O. 

Find the highest common factor of the following expressions: 

1. x 8 + 2x 2 -13x + 10, x* + x 2 -10x + 8. 

2. x 8 - 6x 2 - 99a; + 40, a* - 6x 2 - 86 x + 36. 

3. x* + 2x 2 -Sx-16, x* + 3x 2 - 8x - 24. 

4. a4_a.2_5a._3, 38 - 4x 2 - llx-6. 

5. x« + 3x 2 -8x-24, x« + 3x 2 -.3x-9. 

6. a 8 - 5a 2 x + 7 af - 3 x 8 , a 8 - 3 ax 2 + 2 x 8 . 

7. 2x«-6x 2 +llx + 7, 4x 8 -llx 2 + 26x + 7, 
'¥.'2x» + 4x 2 -7x-14, 6x 8 -10« a -21x + 36. 



1 



100 ALGEBRA. 

9. 3z«-3x*-2a?-a;-l, 9x* -3a*- ar- 1. 

10. 3x 8 -3ax8 + 2a 2 x-2a 8 , 3 x 8 + 12 as 2 + 2 cflx + 8 a*. 

11. 2x 8 -9ax 2 + 9a 2 x-7a 8 , 4s 8 -20ax 2 + 20a*x- 16 a 8 . 

12. 10x 8 + 25ax 2 -6a 8 , 4x 8 +9ox a -2a 2 x-o 8 . 

13. 24x*y+72x 8 y 2 -6sV-90xy* > 6x*y 2 +13a% 8 -4* 2 y*-15a^«. 

14. 4 x*a 2 + 10 x 4 a 8 -60x 8 a*+54x 2 a 6 , 24x 5 a 8 +30x 8 a 6 -126x 2 a«. 

15. 4x 6 + 14x* + 20x 8 + 70x 2 , 8x* + 28x* - 8x* - 12x* + 66a* 

16. 72 a*- 12 ax 2 + 72 a 2 x-420 a 8 , 18 a^+42 ax 2 - 282 0^+270 a 8 . 

17. x 6 - x 8 - x + 1, x 1 + a; 6 + a* - 1. 

18. l + x + a^-x*, l-a^-xP + x 7 . 

122. The statements of Art. 116 may be proved as fol 
lows: 

I. If F divides A it will also divide mA. 
For suppose A = aF, then mA = maF. 
Thus F is a factor of mA 

II. If F divides A and 5, then it will divide mA ± nB. 
For suppose A = aF, B = fc-F, 

then mA ± nB = maF ± nbF 

= F(ma ± nb). 

Thus F divides mA ± nB. 



We may now enunciate and prove the rule for find- 
ing the highest common factor of any two compound alge- 
braic expressions. 

We suppose that any simple factors are first removed. 
(See Example, Art. 117.) 

Let A and B be the two expressions after the simple 
factors have been removed. Let them be arranged in de- 
scending or ascending powers of some common letter ; also 
let the highest power of that letter in B be not less than 
the highest power in A. 

Divide B by A ; let p be the quotient, and C the remain- 
der. Suppose G to have a simple factor m. Remove this 
factor, and so obtain a new divisor D. Further, suppose 
that in order to make A divisible by D it is necessary to 



HIGHEST COMMON FACTOR. 101 

multiply A by a simple factor w. Let q be the next quotient 
and E the remainder. Finally, divide 2) by E\ let r be the 
quotient, and suppose that there is no remainder. Then 
E will be the H. C. F. required. 
. The work will stand thus : 

A)B(p 

pA 

m)C_ 

D)nA(q 
qD 

E)D(r 

rE 

First, to show that E is a common factor of A and B. 

By examining the steps of the work, it is clear that E 
divides 2), therefore also qD ; therefore qD + E, therefore 
nA ; therefore A, since n is a simple factor. 

Again E divides D, therefore mD, that is, C. And since 
E divides A and C, it also divides pA + C, that is, B. 
Hence E divides both A and B. 

Secondly, to show that E is the highest common factor. 

If not, let there be a factor X of higher dimensions 
than E. 

Then X divides A and B, therefore B — pA, that is, (7; 
therefore D (since m is a simple factor) ; therefore nA — qD, 
that is, E. 

Thus X divides E\ which is impossible, since by hypoth- 
esis, X is of higher dimensions than E. 

Therefore E is the highest common factor. 

124. The highest common factor of three expressions 
A, B, C may be obtained as follows : 

First determine F the highest common factor of A and B; 
next find G the highest common factor of F and C ; then 
G will be the required highest common factor of A, B, C. 

For F contains every factor which is common to A and B, 
and G is the highest common factor of F and C. Therefore 
G is the highest common factor of A, B, G. 



CHAPTER XII. 

Lowest Common Multiple. 

Definition. The Lowest Common Multiple of two 
or more algebraic expressions is the expression of lowest 
dimensions which is divisible by each of them without 
remainder. 

The abbreviation L. C. M. is sometimes used instead of the 
words lowest common multiple. 

SIMPLE EXPRESSIONS. 

126. The L. C. M. can be written by inspection. 

Ex. 1. The lowest common multiple of a 4 , a 3 , a 2 , cfi is a 6 . 

Ex. 2. The lowest common multiple of a?b*, ab 5 , a 2 b 7 is a 8 b 7 ; for 
a 8 is the lowest power of a that is divisible by each of the quantities 
a 8 , a, a 2 ; and b 1 is the lowest power of 6 that is divisible by each of 
the quantities b 4 , b 5 , 6 7 . 

• 127. If the expressions have numerical coefficients, find 
by Arithmetic their least common multiple, and prefix it as 
a coefficient to the algebraic lowest common multiple. 

Ex. The lowest common multiple of 21 a*x*y, 35 afhty, 28 a*xy* is 
420 a 4 * 4 !/ 4 ; for it consists of the product of 

(1) the numerical least common multiple of the coefficients ; 

(2) the lowest power of each letter which is divisible by every power 
of that letter occurring in the given expressions. 

EXAMPLES XII. a. 

Find the lowest common multiple of 

1. z*y 2 ,xyz. 4. 12ab,Sxy. 7. 2x, Sy, 4 s. 

2. 3x 2 yz, 4 sty 8 . 5. ac, be, ab. 8. 3x 2 , 4y 2 , 3 s 2 . 

3. 6 a 2 bc*, 4 ab 2 c 6. a 2 c, be 2 , cb 2 . 9. 7 a 2 , 2 ab, 3 6*. 

102 



LOWEST COMMON MULTIPLE. 103 

10. a*bc, 6 2 ca, c 2 ab. 13. 35 a 2 c 8 6, 42 aW, 30 ac 2 6 8 . 

11. 6 a 2 c, 6 c6 2 , 3 6c 2 . 14. 66 a*6 2 c 8 , 44 a 8 6*c 2 , 24 a 2 6 8 c*. 
18. 2 xy, 3 ay, 4&y. 15. 7 a 2 6, 4 ac 2 , 6 act, 21 6c. 

COMPOUND EXPRESSIONS. 

128. L. C. M. of Compound Expressions which can be fac- 
tored by Inspection. The method employed is similar to 
that of the preceding article. 

Ex. 1. The lowest common multiple of 6x 2 (a — x) 2 , 8a 8 (a — x) s 
and 12 ax(a — as) 6 is 24 ahP(a — x) 6 '. 
For it consists of the product of 

(1) the numerical L. C. M. of the coefficients ; 

(2) the lowest power of each factor which is divisible by every 
power of that factor occurring in the given expressions. 

Ex. 2. Find the lowest common multiple of 

3a 2 + 9a6, 2a 8 -18a6 2 , a 8 + 6 a 2 b + 9 ab 2 . 
3a 2 + 9a6 = 3a(a + 36), 
2a 8 - 18a6 2 = 2a(a + 3 6)(a-36), 
fjfi + 6a 2 6 + 9a6 2 = a(a + 3 6)(a + 36) 

= a(a + 3 6) 2 . 
Therefore the L. C. M. is 6 a(a + 3 6) 2 (a - 3 6). 

EXAMPLES XII. b. 

Find the lowest common multiple of 
1. x 2 , x 2 -3s. 2. 21s 8 , 7x 2 (x + l). 3. a 2 + ab, ab + 6 2 . 

4. 4afy-y, 2x 2 + x. 10. (a - x) 2 , a 2 - x 2 . 

5. 6a 2 -2a, 9s 2 -3s. IX, (1+x) 8 , 1 + x 8 . 

6. x 2 + 2 a, re 2 + 3.x + 2. 12. x 2 + 4 & + 4, x 2 + 6 x + 6. 

7. x 2 - 3 x + 2, x 2 - 1. 13. x 2 - 5 x + 4, x 2 - 6 & + 8. 

8. (a + a) 8 , a 8 + x 8 . 14. 1 - x 8 , (1 - x) 8 . 

9. a 2 + x 2 , (a + x) 2 . 15. (a - x) 8 , a 8 - x*. 

16. a 2 + * - 20, x 2 - 10 x + 24, x 2 - x - 30. 

17. x 2 + * - 42, x 2 - 11 x + 30, x 2 + 2x - 35. 

18. 2 x 2 + 3 x + 1, 2 x 2 + 6 x + 2, x 2 + 3 x + 2. 

19. a 2 -x 2 , (a-*) 2 , a 8 -a 8 . 



104 ALGEBRA. 

90. 3x 2 + llx + 6, 3x* + 8x + 4, x 2 + 6x + 6. 

21. 6x 2 +llx + 2, 6x 2 + 16x + 3, x 2 + 6x + 6. 

82. 1 + x 2 , (1 + x) 2 , 1 + x 8 . 

23. 2x 2 + 3x-2, 2x 2 + 16x-8, x 9 + 10x + 16. 

24. 3x 2 -x-14, 3aJ 2 -13x+ 14, x 2 -4. 

26. 12x 2 + 3x-42, 12«»H-30x 2 + 12 a, 32 x 2 - 40 x - 28. 

28. 3x 4 + 26x» + 35x 2 , 6 a 2 + 38 X - 28, 27x 8 + 27x 2 - 30x. 

27. eOxi + Sa^-Sx 2 , 60x 2 y + 32 xy + 4y, 40 xfy - 2 xty - 2 xy. 
23. 8x 2 -38xy + 35jft 4x 2 -xy-6y 2 , 2x 2 -6xy-7y 2 . 

29. 12x 2 -23x2/ + 10y 2 , 4x 2 -9xy + 5y 2 , 3x 2 -5xy + 2y 2 . 

30. 6 ax*+7 a^-S ate, 3 a 2 x 2 +14 a«x - 6 a 4 , 6 x 2 +39 ax+45 a 2 . 

31. 4ax 2 y 2 +H axy 2 -3 ay 2 , 3x^+7 xV^-Oxy 8 , 24ax 2 -22ax+4a. 

129. L. C. M. of Compound Expressions which cannot be 
factored by Inspection. When the given expressions are 
such that their factors cannot be determined by inspection, 
they must be resolved by finding the highest common factor. 

Ex. Find the lowest common multiple of 

2x* + x 8 -20x 2 -7x + 24 and 2x* + 3x 8 --13x 2 -7x+ 16. 

The highest common factor is x 2 + 2 x — 3. 
By division, we obtain 

2 x* + x 8 - 20 x 2 - 7 x + 24 = (x 2 + 2 x - 3) (2 x 2 - 3 x - 8). 

2x* + 3x 8 -13x 2 -7x+15=(x 2 + 2x-3)(2x 2 -x-6). 

Therefore the L. C. M. is (x 2 +2 x-3)(2 x 2 - 3 x-8)(2 x 2 -x-6). 

130. We may now give the proof of the rule for finding 
the lowest common multiple of two compound algebraic 
expressions. 

Let A and B be the two expressions, and F their highest 
common factor. Also suppose that a and b are the respec- 
tive quotients when A and B are divided by F\ then 
A = aF, B = bF. Therefore, since a and b have no com- 
mon factor, the lowest common multiple of A and B is 
abF, by inspection. 



LOWEST COMMON MULTIPLE. 105 

131. There is an important relation between the highest 
common factor and the lowest common multiple of two ex- 
pressions which it is desirable to notice. 

Let F be the highest common factor, and X the lowest 
common multiple of A and B. Then, as in the preceding 
article, 

A = aF, B = bF, 

and X = abF. 

Therefore the product AB = aF x bF 

= Fx abF 
= FX (1). 

Hence the product of two expressions is equal to the product 
of their highest common factor and lowest common multiple. 

Again, from (1) X =«| X J5 = | x A; 

hence the lowest common multiple of two expressions may be 
found by dividing their product by their highest common fac- 
tor; or by dividing either of them by their highest common 
factor, and multiplying the quotient by the other. 



The lowest common multiple of three expressions 
A, B, C may be obtained as follows : 

First find X, the L. C. M. of A and B. Next find F, the 
L. C. M. of X and (7; then Y will be the required L. C. M. 

of A, b, a 

For T is the expression of lowest dimensions which is 
divisible by X and C, and X is the expression of lowest 
dimensions divisible by A and B. Therefore T is the ex- 
pression of lowest dimensions divisible by all three. 

EXAMPLES XII. o. 

1. Find the highest common factor and the lowest common multi« 
pie of x 2 - 5 x + 6, x 2 - 4, x 8 - 3z - 2. 

2. Find the lowest common multiple of 

ab(x 2 + 1) + x(a 2 + 6«) and ab(x 2 - 1) + x(a 2 - ft 2 ). 

3. Find the lowest common multiple of xy — bx, xy — ay, 

lP — 3 by + 2 b 2 y xy — 2 bx — ay + 2 ab, xy — bx — ay + ab. 



106 ALGEBRA. 

4. Find the highest common factor and the lowest common multl 
pie of x* + 2x 2 -3x, 2x? + 5a; 2 - Sz. 

5. Find the lowest common multiple of 

1 - x, (1 - x 2 ) 2 , (1 + ^ 

6. Find the lowest common multiple of 

x 2 - 10* + 24, x 2 - 8a; + 12, x 2 - 6x + 8. 

7. Find the highest common factor and the lowest common molt: 
pie of 6a? + x 2 -6a;-2, 6x* + 5a; 2 - 3x- 2. 

8. Find the lowest common multiple of 

(6c 2 - a&c) 2 , b\ac 2 - a 8 ), a 2 c 2 + 2 ac 8 + «*. 

9. Find the lowest common multiple of 

x 3 - y\ xhf - j/ 4 , y\x - y) 2 , x 2 + xy + j/ 2 . . 

Also find the highest common factor of the first three expressions. 

,10. Find the highest common factor of 

6x 2 -13x + 6, 2x 2 + 6x-12, 6x 2 -x-12. 

Also show that the lowest common multiple is the product of the 
three quantities divided by the square of the highest common factor. 

11. Find the lowest common multiple of 

x* + ax* + a 8 x + a 4 , x* + a^x 2 + a*. 

12. Find the highest common factor and the lowest common multi- 
ple of 3X 8 - 7x 2 y + 5xy' 2 - y*, x 2 y + Sxy 2 - 3X 8 - y 8 , 

3x»+ Sx^ + xi/ 2 -^ 8 . 

13. Find the highest common factor of 

4x 8 - 10x 2 + 4x+2, 3x* - 2x 8 -3x + 2. 

14. Find the lowest common multiple of 

a 2 - b 2 , a 8 - & 8 , a 8 - a 2 b - ab* - 2 &*. 

15. Find the highest common factor and the lowest common multi- 
ple of (2 x 2 - 3 a 2 )y + (2 a 2 - 3 y 2 )x, (2 a 2 -f 3 y 2 )x + (2 x 2 + 3a 2 )y. 

16. Find the highest common factor and the lowest common multi- 
ple of x 8 -9x 2 + 26x-24, x 8 -12x 2 -f 47x-60. 

17. Find the highest common factor of 

x 8 -15ax 2 + 48a 2 x+ 64 a 8 , x 2 - 10 ax + 16 a 2 . 

18. Find the lowest common multiple of 

21 x(xy - y*\ 2 , 35(xV* -xV) y 15|/(x 2 + xy) 2 . 



CHAPTER XIII. 



Fractions. 



I. Definition. If a quantity x be divided into b 
equal parts, and a of these parts be taken, the result is 

called the fraction ^ of x. 

b 

If a? be the unit, the fraction ^ of a? is called simply "the 

b 

fraction - " ; so that the fraction - represents a equal parts, 

b of which make up the unit 

The quantity above the horizontal line is spoken of as 
the numerator, and that below the line as the denominator 
of the fraction. 

134. A Simple Fraction is one of which the numerator 
and denominator are whole numbers. 



REDUCTION OF FRACTIONS. 

To reduce a fraction is to change its form without 
changing its value. 

136. To prove that ? = ^, where a, 6, m are positive 
integers. 

By - we mean a equal parts, b of which make up the 
unit, (1); 

by — - we mean ma equal parts, mb of which make up 
** the unit, (2). 

107 



108 ALGEBRA. 

But b parts in (1) = mb parts in (2) ; 

•\ 1 part in (1) = m parts in (2) ; 
.\ a parts in (1) = ma parts in (2) j 

that is, t = — -• 

b mb 

Conversely, 2? = i 

Hence, The value of a fraction is not altered if we multiply 
or divide the numerator and denominator by tlie same quantity. 

I37. Redaction of a Fraction to its Lowest Terms. As 
shown in the preceding article an algebraic fraction may 
be changed into an equivalent fraction by dividing numera- 
tor and denominator by any common factor ; if this factor 
be the highest common factor, the resulting fraction is said 
to be reduced to its lowest terms. 

Ex. 1. Reduce to lowest terms — ^—r--> 

36 a 6 * 2 

24 oWs 8 __ 2 8 x 3 a*cW __ 2 (fte 

36a 6 x 2 2 2 x3 2 a 6 z 2 3o 2 ' 

24 a?c*x* 
Ex. 2. Reduce to lowest terms - — — -^ — " a • 

18 a 8 x 2 - 12 a** 8 
24 a 8 ^ 2 _ 24 a*c*x* _ 4a<fl 
ISaW - 12 a 2 xfi 6 a% 2 (3 a - 2 x) 3a -2» 

Ex. 3. Reduce to lowest terms """ ' » 

9ajy-12y8 

6s 2 -8sy = 2g(3s-4y) = 2x # 
9ajy-12y 2 3^(3* -4y) 3y # 

Note. The beginner should be careful not to begin cancelling 
until he has expressed both numerator and denominator in the most 
convenient form, by resolution into factors where necessary. 

EXAMPLES XIII. a. 

Reduce to lowest terms : 

1 12 mrt^ 3 ax ft abx + bx* 

15 mPnp* ' a 2 x 2 — ax * acx + ex 2 

46s 8 y 4 s * 4 3a 2 -6q& 6 15 o 2 6 2 c 



69 «V** ' 2a 2 6-4a6 2 100(a 8 -a 2 6) 



FRACTIONS. 109 

7 4s a -0y« 9 x(2tf-Bax ) t u (xy-3^) a 

4x 2 + 6xy" * a(4a?x-9&) " x 8 y 2 -27y 6 

g 20(x* -yi) , 1Q x«-2xy a . 12 x a -6s 

Sx^ + Sxy + Sy 2 g*-4aEV + 4y* x 2 -4x-6 

13 3s a + 6s 15 x 8 y + 2x a y + 4xy 
x 2 + 4x + 4 # ' a 8 - 8 

14 5 q 8 6 + 10 qg&a .- 3 q* + 9 q 8 6 + 6 oW 
3a 2 6 2 + 6a& 8 ' a* + a 8 6-2a 2 6 2 

17 a*- 14a; 2 -51 _- 2x a + 17x + 21 21 3x a -f-23x-f-14 
x*-2x 2 -16 # * 3x 2 + 26x + 35* ' 3x 2 + 41x + 26 

18 ^ + ay-tf . ao a^-Ma 2 23 27q + q* 

x 8 -? 8 ox 2 +9ax + 20a 18a-6a 2 + 2a 8 

138. When the factors of the numerator and denominator 
cannot be determined by inspection, we find the highest com- 
mon factor, by the rules given in Chapter xi. 

Ex. Reduce to lowest terms 8 ** ~ 13x ' + 28ag - 21 . 

15x 8 -38x a -2x + 21 

First Method. The H. C. F. of numerator and denominator is 
3a; -7. 

Dividing numerator and denominator by 3x — 7, we obtain as re- 
spective quotients x a — 2 x + 3 and 5x 2 — x — 3. 

3x 8 -13x 2 4-23x-21 _ (3x-7)(x 2 -2x-f-3) _ x 2 -2x + 3 
15x 8 --38x a -2x + 21"~(3x--7)(5x 2 -x--3)~5x 2 -x--3' 

This is the simplest solution for the beginner ; but in this 
and similar cases we may often effect the reduction without 
actually going through the process of finding the highest 
common factor. 

Second Method. By Art. 116, the H. C. F. of numerator and 
denominator must be a factor of their sum 18 x 8 — 51 x 2 + 21 x, that 
is, of 3x(3x — 7)(2x — 1). If there be a common divisor it must 
clearly be 3 x — 7 ; hence arranging numerator and denominator so as 
to show 3 x — 7 as a factor, 

<u * ♦• x 2 (3x-7)-2x(3x-7)+3(3x-7) 

the fraction = , \, n ^ ^ ={ — ^ =£ 

5x 2 (3x — 7) - x(3 x - 7) - 3(3 x - 7) 

_ (3x-7)(x a -2x-f-3) _ x 2 -2x + 3 
""(3x-7)(5x a -x-3) 5x a -x-8 



110 



ALGEBRA. 



139. If either numerator or denominator can readily be 
resolved into factors we may use the following method. 

Ex. Reduce to lowest terms — g 8 -t-3x 2 -4a? — # 

7x 8 -18x 2 + 6x + 5 

The numerator = x(x* + 3x — 4) = x(x + 4) (x — 1). 

Of these factors the only one which can be a common divisor is 
x— 1. Hence, arranging the denominator, 



the fraction = 



s(x + 4)(x-l) 

7x 2 (x - 1)- lla&(3 * - 1)- 6(* - 1) 



__ s(x + 4)(s-l) __ s(g + 4) 
"" (x - 1 ) (7 x 2 - 1 1 x - 6) " 7x*-ll*-5 * 



EXAMPLES XIIL b. 



Reduce to lowest terms : 

x q 8 - qgft - q&2 - 2 b* 

q 8 + 3q 2 6 + 3q& 2 + 26 8 ' 

2 g 8 -5g 2 +7a?-3 

x 8 -3x + 2 

3 q 8 + 2q 2 -13q+10 

q» + q 2 -10q + 8 

4 2xS-f 5x 2 y-30sy 2 -f 27 

4x 8 + 6xy 2 -21y 8 

5 4q 8 -f-12a 2 &-«& 2 -15 6 8 
6q8 + 13q^-4a6 2 -156 8 ' 

l + 2x 2 + x 8 + 2x* 



6. 



7. 



8. 



l + 3x 2 + 2x 8 + 3x* 
a?2 - 2 x + 1 
3x 8 + 7x-10* 

3 a 8 - 3 q 2 5 + aft 2 - b* 
4q 2 -5a& + 6 2 



9 4s»-f Sax* + cP | 

x 4 + qx 8 -f q 8 * + q* 
10 *s 8 - 10s 2 + 4s + 2 
3x*-2s 8 -3x + 2* 

n 16 s* - 72 x 2 q 2 + 81 q* 
4x 2 +12qx + 9q a 

12 6g 8 + a; 2 -5x-2 
6x 8 + 6x 2 -3x-2 

13 5x 8 + 2x 2 -15x-6 
7x 8 -4x 2 -21x+12* 

14 4x* + llx 2 + 25 
4x*-9x 2 + 30x-26 

15 3 x 8 - 27 qx 2 + 78 q^ - 72 a* 
2x3 + 10ttx 2 - 4q^ - 48a 8 

le qx 8 - 5q 2 s 2 - 99q 8 x + 40 a* 
x 4 - 6 qx 8 - 86 q 2 x 2 + 36 a 8 *" 



MULTIPLICATION AND DIVISION OF FRACTIONS. 

140. Rale I. To multiply a fraction by an integer. Multi- 
ply the numerator by that integer; or, if the denominator be 
divisible by the integer, divide the denominator by it. 



FRACTIONS. Ill 

The proof is as follows : 

(1) — represents a equal parts, b of which make up the 

unit ; --- represents ac equal parts, b of which make up the 
b 

unit; and the number of parts taken in the second fraction 

is c times the number taken in the first ; 

that is, Jxca^. 

o 

Hence - x b = — = a ; that is, the fraction - is the quantity 

which must be multiplied by & in order to obtain a. Now 
the quantity which must be multiplied by b in order to 
obtain a is the quotient resulting from the division of a by b 
[Art. 53]; therefore we may define a fraction thus: the 

fraction 5 is the quotient of a divided by b. 
o 



141. Rule II. To divide a fraction by an integer. Divide 

the numerator, if it be divisible, by the integer; or, if the numer- 
ator be not divisible, multiply the denominator by thai integer. 

The proof is as follows : 

(1) ^ represents ac equal parts, 6 of which make up the 

unit ; % represents a equal parts, b of which make up the 
o 

unit. 

The number of parts taken in the first fraction is c times 
the number taken in the second. Therefore the second frao 
tion is the quotient of the first fraction divided by e; 

thatia, ^ + c = r 

o ft 



112 ALGEBRA. 

(2) But if the numerator be not divisible by c, we have 

5 — 25. 

6 ~~bc' 

.% 2 -|. c = f? -i- c = 2- f by the preceding case. 

142. Rule III. To multiply together two or more fractions. 
Multiply the numerators for a new numerator, and the denomi- 
nators/or a new denominator. 

To find the value of ? x 1- 

6 d 

Let x = ? x |. 

6 d 

Multiplying each side by 6 x d, we have 

xx 6 xd = ?x- t x5 xd 
6 (2 

= ^x6x^X(J [Art. 37.] 

6 d J 

= a x a [Art. 140.] 

.\ aj6d = ac. 

Dividing each side by bd, we have 

Similarly, ?x^x^, = ^,; and so for any number of 
b d f bdf 

fractions. 

143. Rule IV. To divide one fraction by another. Invert 
the divisor, and proceed as in multiplication. 

Since division is the inverse of multiplication, we may 

ft (* 

define the quotient x, when - is divided by -, to be such that 

6 d 

v c a 



FRACTIONS. 113 

Multiplying by -, we have ajx-x- = ?x*| 

c d e b e 

ad 
be 

Hence ?-^ = ^ = 2 x £ [Art 142.] 

O a DC 
Ex.1. Simplify p* x ^ x 6bc 



The expression = 



8 6 4q 8 6q&c 
2x8x6xq&c 8 6d» 



i)x4x6xa*6 2 c 12a»& 



Ex. 8. Simplify ™±*« x 4q^6«. 

4a» 12 a 4-18 

2q 2 4- 3q 4 q 2 - 6 q _ q(2q 4- 3) 2q(2q-8) __ 2q -8 
4a» 12a + 18 4a 8 6(2a + 3) 12a ' 

by cancelling those factors which are common to both numerator and 
denominator. 

Ex. 3. Simplify — * w x — = — !=--: -»- * ^ « 

ax-a 2 9x 2 -4q 2 3ax + 2a« 

The expression 

_ 6x 2 -ax-2a* v , x-a x 3qg-f-2q a 
ax-a* 9x 2 -4q 2 2x4-a 

(8 as - 2q)(2g + q) a; - q «(3s + 2q) _- 

"" q(«-q) (3x + 2q)(3«- 2q; 2* 4- a "^ 

since all the factors cancel each other. 



EXAMPLES XIII. O. 



Simplify 
1. 



7q 2 ft* x 18ac g c > 5 14s 2 -7s % 2g-l 

Ooa;^ 16qc*' " 12x 8 + 24« 2 * a + 2* 

^ 21 A; 2 ?) 8 . 28 j> 2 A« e q 2 fr 2 4-8qft t db 4- 3 

13 wm 2 "*" 39 tn*n 8 ' " 4q 2 -l 2q + l' 

8 2^ x hzH . 21s 2 y 8 g 2 T ^ s 2 -4q 2 x 2q 
Syz Txs/ 2 ' iOxtftz " qx 4- 2 q a z - 2 q* 

4. *» 2 x 36 j) 8 ? 2 , 15 mpz* 8 q 2 -121 t q + 11 

8n 81 mn 27 n 2 *^' ' q 2 - 4 ' q 4- 2 ' 

I 



114 ALGEB&A. 

9 16s 2 - Pa 2 x x-2 12 x« + 3x + 2 x x 2 + 7x + 12 

x 2 -4 4* -3a" " x 2 + 9x + 20 x 2 + 6x + 6" # 

10 25a 2 -& 2 x s(3q+2) lg 2x 2 + 5x + 2 x x 2 + 4x 

9a 2 x 2 -4x 2 6a + b ' x 2 -4 2x 2 + 9x + 4 

u a^ + 5g + 6 x 2 -2x-3 M 6* -27 ft 4ft 2 -25 

x a_i X 2_9 * " 26 9 + 66 26 s -116 + 16* 

15 2x 2 -f 13x+15 . 2x a + llx + 6, 

4x 2 -9 **" 4x 2 -l 
le 3 a 2 + 3 ax 2 a 2 + ax - 8 a? 
(a-x) 2 a 2 -x 2 

a-a.Ha-^is t x a -12x-46 
x 2 -4x-45 x 2 - 6x -27* 
2 + 5x-f-3x 2 t 4a* t 
(1 + x) 8 1 + x 8 " 



17. 
18. 



1©. 2x 2 -x-l ^ 4x 2 + x-14 



20 



22. 



2x 2 + 6x + 2 16x 2 -49 
a 2_2q&-36 2 , a 2 -4afr + 3ft» 
(a + 6) 8 a 8 + 6* 

21 s*-fla* + 36as , s 4 + 216x < 
x 2 -49 x 2 -x-42 

64 p 2 ? 2 - g* (x - 2J 2 x 2 -4 

x 2 - 4 X Spq + s 2 + (x + 2)*' 
rf-2q-8 tf-q-12 . 

(a + 3) 2 a 2 + 9 

24. x 2 -x-20 x x 2 -x-2 t x + 1 

x 2 -26 x 2 + 2x-8 x 2 + 5* 

25 s 2 -18s + 80 x x 2 -6x-7 ^ g-j-6 
x 2 -6x-50 x 2 -16x + 66 «-l 
at 1-s 2 l-3x 2 + 2x 8 

■* r^ a-x) 8 

m x 2 -5x-14 x 2 -llx + 18 s 2 -16s + 63 r 
(x-2) 2 x 2 -4 ax 8 -4ax« + 4« 

ADDITION AND SUBTRACTION OF FRACTIONS 

*ma *% ff.G ad + be 

144. To prove 7 + 0= — S — 

b d bd 

Webave r^S^^ir [Artl86.1 

0a a bd, 



FRACTIONS. US 

Thus in each case we divide the unit into bd equal parts, 
and we take first ad of these parts, and then be of them; 
that is, we take ad -f- be of the bd parts of the unit ; and this 

is expressed by the fraction — i — . 

bd 

a . c ad + bc 



Similarly, 



b d bd 
b d~~ bd ' 



145. Here the fractions have been both expressed with a 
common denominator bd. But if b and d have a common 
factor, the prqduct bd is not the lowest common denominator, 

and the fraction — — — will not be in its lowest terms. To 

bd 

avoid working with fractions which are not in their lowest 
terms, we take the lowest common denominator, which is the 
lowest common multiple of the denominators of the given 
fractions. 

■ 

Rule I. To reduce fractions to their lowest common denomi- 
nator. Find the L. C. M. of the given denominators, and take 
it for the common denominator; divide it by the denominator 
of the first fraction, and multiply the numerator of this fraction 
by the quotient so obtained; and do the same with all the other 
given fractions. 

Ex. 1. Express with lowest common denominator 

a b c 

• , •, ' • •• 

Sxy 6 xyz 2 yz 

The lowest common multiple of the denominators is 6 xyz. Dividing 

this by each of the denominators in turn, and multiplying the corn - 

sponding numerators by the respective quotients, we have the equiva 

lent fractions 

2az b Sex 

6 xyz bxyz 6 xyz 
Sz. 8. Express with lowest common denominator 

5 * and 4a 



8a(x-a) 3«(x 2 -an 



116 ALGEBRA. 

The lowest common denominator is 6 ax(x — a)(x + a). 
We must, therefore, multiply the numerators by 8 x(x + a) and 2 a 
respectively. 

Hence the equivalent fractions are 

16 * 2 (* + «) and 8g * 

6 ox(x — a)(x 4- a) 6 ax(x - a)(x + a) 

146. We may now enunciate the rule for the addition or 
subtraction of fractions. 

Rule II. To add or subtract fractions. Beduce them to the 
lowest common denominator ; add or subtract the numerators, 
and retain the common denominator. 

Ex. 1. Find the value of ^p^ + 5a? " 4a - 

3a 9a 

The lowest common denominator is 9 a. 

Therefore the expression = «(»* + «)+ 5*-*« 

9a 

— 6g + 3q-f5s — 4a __ llx — g | 
9a 9a 

Ex. 8. Find the value of »=*! + **=« _ 8 *- 2q . 

xy ay ax 

The lowest common denominator is axy. 

Thus the expression = < x ~ »») + »(«*-«)-»(«»-»«) 

axy 

_ as — 2 ay 4- 3 xy — as — 8xy 4-2 ay 

~~ axy ~~ * 

since the terms in the numerator destroy each other. 



EXAMPLES XIII. <t 
Find the value of 

- 2x-l.x-6.x-4 A 2x4-5 *4-3 27 

1. — — — A \- • 4. ■ ■ — — — — • 

3 6 4 x 2x 8x» 

„ 2x — 3 x4-2, 6x4-8 . a—b,b—c.c—a 

9 612 ab bo ea- 
rn x-7x-9 X4-3 ^ g-26 q -5& . a4-7 6, 
' 16 26 46 ' 2a 4a 8a 



FRACTIONS. 117 

T g-g+qj-g-* 8 -* 10. L-*f-*+* + f . 

x a 2ax xy xy* x*i/* 

a 2a 2 -6 2 6 2 -c 2 c 2 -^ 2 1t 2x-3y , 3x-2« , 6 

•• — — • — t— • 11. £ H H -• 

a 2 b 2 C 2 xy xz x 

9. s-3 3^-9 8-x» ^ a 2 - 6c ac - 6 2 aft - <g 

6x 10 x 2 16 x 8 fcc ac ab 

Ex. 8. Simplify 2s-3a _2x zl a. 

x — 2a x — a 

The lowest common denominator is (x-2o)(x- a). 

-. , 4 . , (2x-3a)(x-a)--(2x-a)(x-2a) 

Therefore the expression = - ^ ^^-9 r-^ ^ 

^ (x — 2 a) (x — a) 

_ 2 s 2 - 5 ax + 3 a 2 - (2 x* - 5 ax + 2 a 2 ) 
"" (x — 2a)(x — a) 

2X 2 - 6 ax + 3 a 2 - 2x 2 + 6 ax - 2 a 2 



(x — 2a)(x — a) 



a 2 



(as — 2 a) (x — a) 

Note. In finding the value of such an expression as 

-(2as-a)(x-2a), 

the beginner should first express the product in brackets, and then 
remove the brackets, as we have done. After a little practice he will 
be able to take both steps together. 

The work will sometimes be shortened by first reducing 
the fractions to their lowest terms. 

Ex.4. Simplify x * + 6 *9 " * * -%- 

x 2 -16y 2 2x 2 + 8xy 

llie expression = ** + ***-*# _ _JL_ 

a^-iey 2 x + 4y 

x 2 + 5 xy - 4 y 2 - y (x - 4 y) 
*~ x 2 -16y 2 

x 2 + 5xy - 4y 2 - xy + 4y» 
x 2 - 16 y 2 

_ x* + 4xy _. 



a^-lOy 2 x-4y 



118 ALGEBRA. 



Find the value of 



EXAMPLES XIII. e. 



I 1 . 1 , ^ _a b_^ - *_+2__?_=_?, 

x + 2 x + 3 x + a x+b " x - 2 x + 2* 

q 2 1 - a , b . jc-4 jb-7 

x + 8 x + 4 x—a x—b x—2 x — b 

8b __3 1__ # - a + x q — 3 9 a ° 2 



* — 6 x + 2 a — x a + x x-a rf — tf 

10. -£- + _*_. 18. — I— + l 



x*- 4 (x - 2)* x(x - y) y(x + jr) 

11. *+_ȣ-. 19. 2 2 



s-8 j*-9 a(x«-a 2 ) x(x + a)« 

12 1 ' x + V 20 _££_ + _ £L— . 

2x-Sy 4x 2 -9y 2 ' * 26«P-y* 6x + y 

18 __1 1 21 y ' l g 

1-x 8 (1-x) 8 ' x^ - y 2 ) y(x 2 + y 2 ) 

14 * + <i _ s 2 + 2 a 2 go x + a x — a 



x-2a x 2 -4a 2 (x 2 + a 2 ) (x + a) 2 

15 4a 2 +ft 2 2a -6 ' g8 x 2 -4a 2 x + 4a 

' 4 a 2 - ft 2 2 a + b x 2 - 2 ax x + 2 a 

16. -J* *£L. ^ x 2 + xy + y 2 | x 2 -xy + y 2 

x 2 — y 2 % + y x + y * — v 

17. -* *_ 26. — ! (« + »*)' . 

1-x 2 1+x 2 a-2x a*-8x* 



a 2 - ab + ft 2 a 2 + ab + ft 2 

147. Some modification of the foregoing general methods 
may sometimes be used with advantage. The most useful 
artifices are explained in the examples which follow, but no 
general rules can be given which will apply to all cases. 

Ex. 1. Simplify 2+|_a±*_ «- 

a — 4 a — 3 a* — lo 

Taking the first two fractions together, we have the expression 

= q 2_9_( g 2_i6) 8 7 8 

(a-4)(a-3) a 2 - 16 ( a -4)(a-8) ( a + 4)(a-4) 

7(q + 4)-8(q-3) 52 - a 

= (a + 4)(a - 4) (a - 3) ~~ (a + 4)(a - 4)(a - 3)' 



FRACTIONS. 119 



Ex. 8. Simplify 



The expression = — 



2a; 2 + a;-l 3x 2 + 4a; + l 
1 1 



Ex. 8. Simplify 



(2s - l)(s + 1) (3a; + l)(x + 1) 
3s-M + 2s-l 
= (2x-l)(a; + l)(Sa; + l) 

_ 5a; m 

= (2x-l)(a;+l)(3a;+l) 

1 1 2x 4a* 



a — z a + x a 2 + z 2 a 4 4- &* 

Here it should be evident that the first two denominators give 
L. CM. a 2 — x 2 , which readily combines with a 2 -f z 2 to give L. C. M. 
a 4 — s 4 , which again combines with a* + x* to give L. C. M. a 8 — z* 
Hence it will be convenient to proceed as follows : 

The expression = a + x "~ ^ a ~~ x ' — — — ••• 

a 2 -z 2 

2x 2x 

a 2 - x 2 a 2 + ai* 

4a* 4a* 8* 7 



a 4 — x* a*+rc* a 8 — afi 

EXAMPLES XIH. | . 
Find the value of 

1.-1 L.+ *» 8. l 



x + y x-y x 2 -ip 2(a-6) 2(a+&) a 2 -&* 

8. 1 , 1 3s 9 1 .2. 3 

2z + y 2z-y 4z 2 -y* 1 -» r I /■-' (1+a;) 2 

8 5 3a; 4-13a; 1Q 2q 5 4(3q-f2) 

l + 2a; l-2x l-4aj 2 ' " 2a~3 Qa+9 3(4a 2 -9)' 

A 2a ,36 86 2 ,, 3 , 2 ,5a; 



2a+36 2ft-?>b Aa 2 -9b 2 z-2 Sx + 6 a; 2 -4 

5. -JL_ + i i is. -* y_+&y + xy*. 

a + x (a + x) 2 a 2 — z*. xfi + y 8 a* — y 8 z* — y 6 

6. A-* _ 1_ 13. . } _.+ J 



9_a* 3 + a 8-a a?-9x+20 ae 2 -lla; + 30 



6(x*-l) 2(*-l) S(x+1) x8-7a; + 12 a*-5a; + 



120 ALGEBRA. 

4 8 



l5 ' 2a?-s-l 2x 2 + x-8 7 ' 4-7a-2a« 3-a-10a* 

16. I =-* 18 5 2 



19. 



2s 2 -a;-l 6x 2 -x-2 5 + x-18a; 2 2 + 6x + 2x* 

1 1 1 



z+1 (x+l)(s-f2) (x + l)(x + 2)(x + 3) 

^ 5s 15 (a; -1) 9 (x + 8) 

2(z+l)(x-3) 16(x-3)(x-2) 16(x + l)(x-2) 

21 fl + 36 q + 26 a + b 

" 4(a + 6)(a + 2&) (a + 6)(a + 36) 4(a + 26)(a + 36) 

22. , 2 + 2 X 



23. 



x 2 -3x + 2 z*-x-2 x*-l 

x 15 12 



<b 2 + 5x + 6 «a + 9x + 14 x 2 -f!0x + 21 



84.^ + ^4- 4 * + 2 



x 2 -l 2x+l 2x 2 + 3x + l 

25 5(2s-S) 7 s 12(3s+l) 

' ll(6s 2 + s-l) 6s 2 + 7s-3 ll(4aa + 8a; + 8) 

26. *^-?-*Il2 + _I_. 28. 1+2fl Mfl 8o 



s + 2 s + 3 s-1 l-2a l + 2a (l-2a)» 

27. £=i-*±i *_. 89. X * ** 



x _4 3 + 3 x a -16 1 + x 1 + a* (1-x) 8 

80 24 x 3 + 2 s . 3 - 2 s 

9-12x + 4s 2 3-2x 3 + 2x* 

31. *---! -*«-. 82. -JL- + 1 4 « 



3-x S + x 9 + x 2 2a + 3 2a -3 4 a 2 + 9 

8 & _l +_J_ T+ I 



4(l + x) 4(1-*) 2(1 + **) 
84. 3 ,+ 1 — 



8(a-x) 8(a + x) 4(a 4 + x«) 
86. -l^=+-i -i— 86. 6 



4 + x» 2-* 2 + x 3-6x 3 + 6x 2 + 8x* 

87. 1—-- . .« . + J 



2o-8x 3 a 2 + 48 x 2 2a + 8x 
88. * + -* _«— . 89. 2x 4 * 2 * 



6a 2 + 64 3a-9 3a 2 -27 2 + x (2 + x) 2 4-x« 

40. -1 1— + * * 



8-8x 8 + 8x 4 + 4 X s 2 + 2se« 



FRACTIONS. 121 

148. We have thus far assumed both numerator and 
denominator to be positive integers, and have shown in 
Art. 140 that a fraction itself is the quotient resulting from 
the division of numerator by denominator. But in alge- 
bra division is a process not restricted to positive integers, 
and we shall now extend this definition as follows: The 

algebraic fraction - is the quotient resulting from the division 
of a by by where a and b may have any values whatever. 

149. By the preceding article ^^ is the quotient result- 
ing from the division of — a by — b ; and this is obtained by 
dividing a by b, and, by the rule of signs, prefixing -f . 

Therefore -=i? = + ? = !? (1). 

— b b b 

Again, ^— is the quotient resulting from the division of 

the rule of signs, prefixing — . 

Therefore ^? = -2 (2). 

b o 

Similarly, -^- = -7 (3). 

— o o 

These results may be enunciated as follows: 

(1) If the signs of both numerator and denominator of a 
fraction be changed, the sign of the whole fraction will be un- 
changed. 

(2) If the sign of either numerator or denominator alone 
be changed, the sign of the whole fraction will be changed. 

The principles here involved are so useful in certain cases 
of reduction of fractions that we quote them in another 
form, which will sometimes be found more easy of application. 

1. We may change the sign of every term in the numerator 
and denominator of a fraction without altering its value. 

2. We may change the sign of a fraction by simply changing 
the sign of every terrn in either the numerator or denominator 



122 ALGEBRA. 

Note. The student should keep clearly in his mind the distinction 
between term and factor. The rule governing change of sign for 
factors will be given in Art. 150. 

Ex L b — a __ — b + a __ a — b t 

y — x — y + x x — y 

Ex. 2 x-x 2 __ _ -x + x 2 = __ ad» — g 

2y 2y 2y 

•p a 3x 3 a; _ 3x 



4-x 2 -4 + x 2 x*-4 
The intermediate step may usually be omitted. 

Ex.4. Simplify -*- + -**- + a(Sx " a X 

x + a x — a a 2 — x 2 

Here it is evident that the lowest common denominator of the first 
two fractions is x 2 — a 2 , therefore it will be convenient to alter the sign 
of the denominator in the third fraction.. 

Thus the expression = -g- + — -. a ( Sx " a ) 

x + a x — a x 2 — a 3 

__ q(x — o) -f 2 s(x + a) — q(3x — a) 

x 2 -a 2 

^ qx-fl^-f 2x 2 + 2qx-3qx + q a _ 2s» 

x 2 - a 2 x* - a a# 

Ex.5. Simplify „ 5 o + l*g^i + 



3x-3 1- x 2 2x + 2 
The expression = ^ 3 f ~" 1 + * 



Simplify 



3(x - 1) x 2 - 1 2(x + 1) 

__ 10(x + 1) - 6(3x - 1)4- 3(x - 1) 
6(x 2 - 1) 

_ 10x + 10-18x + 6 + 3x-3 _ 13 -5s 
OCx 2 - 1) 6(x 2 - 1)" 

EXAMPLES XTTI. gr. 



x _1 1. 1 8 x+2q . 2(q 2 -4qs) ^Sa 

4x — 4 5x + 6 1— x 2 " x+a a 2 — x 2 x— a 

a __3 2 5q - x— a . q 2 +3qx . x+a t 

* 1 + a 1-a a 2 -l" ' x+a a 2 -x 2 x-a 



FRACTIONS. 123 



5. -^--+r-^+r i ^ 10. :r JL ^+ * 



2*+l 2*-l \-4& 6a + 6 6-6a 8a 4 -8 

6. -1* ? 2_. u _£_ + _*%L. 

1 — x 2 a — ,1 ac + 1 x 8 — y 8 y 6 — x 6 

7 2-5x 3+x 2x(2x-ll) , w x 2 - y 2 gy - y 8 , 

* x+3 3-x x 2 -9 * ' xy xy-x 2 ' 

8 3-2x 2x + 3 12 18 x^ + y 2 ■ x . y 
2x + 3 3-2x 4x 2 -9* ' x 2 -^ x + y y-x' 

g _6 3 11 M x 2 + 2x + 4 s 2 -2sc + 4 

* 2ft + 2 46-4 6-66*' ' x + 2 " 2-x 

15 X I 3g I 1 

* 2a+5ft 25ft 2 -4a* 2a-6ft 

16. 2 6~q 3x(a-ft) 6-2a 
x-b ft 2 -*; 2 ft + x ' 



w ax 2 -f b . 2(ftx + ax' 2 ) ax 2 - ft T 
" 2x-l l-4x 2 2x + l 

18. a + c + &+c 



19. 



(a - &)(« - a) (6 - a)(x - 6) 
a — c b — c 



21. 



(a - ft)(x - a) (ft - a)(ft - x) 

2 

(» - a){a - ft) ~*~ {x - ft)(ft -a) (x - a)(x -ft) 

1.1 1 



2Q *a + y | a + b + y x + y-a 



(a 2 - ft 2 ) (a 2 + ft 2 ) (ft 2 - a 2 ) (x 2 + a 2 ) (a 2 + a 2 ) (x 2 + ft 2 ) 
1 + _±a_ + 1 2a 



x + a x 2 — a 2 a — x x 2 + a 
88. _» W-?-+ * 



x+a x + 3a a — x x — 3a 
84. .- 1 ,- * .+ 1 * 



4a»(o + x) 4a»(a;-a) 2a 2 (a 2 + :e 2 ) o 8 -^ 

25 * g_., x8 + y S 4. xy 

' x*-y 2 x 2 + y 2 y*-x* (x + y)(x 2 + y 2 ) 

86. -_$ + 2 + « 4 + 64 * - 



a(a 2 - ft 2 ) ft (a 2 + ft 2 ) aft (6* - a*) ft 8 - a 8 



124 ALGEBRA. 

150. From Art. 149 it follows that: 

(1) Changing the signs of an odd number of factors of 
numerator or denominator changes the sign before the 
fraction. 

(2) Changing the signs of an even number of factors of 
numerator or denominator does not change the sign before 
the fraction. 

Consider the expression 



(a-b)(a-c) (b-c)(b-a) (c-a)(c-b) ' 

By changing the sign of the second factor of each denomi- 
nator, we obtain 

111 



± * m 

(o-6)(c-o) (&-c)(a-6) (c-a)(b-c) ' w ' 

Now it is readily seen that the L. C. M. of the denomi- 
nators is (a — b)(b — c)(c — a), and the expression 

__ — (b — c) — (c — a) — (a — b) 
"~ (a — 6)(6 — c)(c — a) 

(a — b)(b — c)(c — a) 



There is a peculiarity in the arrangement of this 
example which it is desirable to notice. In the expression 

(1) the letters occur in what is known as 
* Cyclic Order ; that is, b follows a, a follows 
c, c follows b. Thus, if a, 6, c are arranged 
round the circumference of a circle, as in 
the annexed diagram, if we start from any 
letter and move round in the direction of 
the arrows, the other letters follow in cyclic 
order, namely abc, bca, cab. 

The observance of this principle is especially important 
in a large class of examples in which the differences of 
three letters are involved. Thus we are observing cyclic 
order when we write b — c, c — a, a — b; whereas we are 




FRACTIONS. 125 

violating cyclic order by the use of arrangements such as 
b — c, a — c, a — b, or a — c, b — a, b — c. It will always 
be found that the work is rendered shorter and easier by 
following cyclic order from the beginning, and adhering 
to it throughout the question. 



EXAMPLES xm. ta. 
Find the value of 



1. 



2. 



8. 



(a-b)(a-c) (6-c)(6-a) (c-a)(c-6) 
b , c , a 



(o-6)(a-c) (6-c)(6-a) (c-a)(c-b) 

z . x . y 



(x-y)(x-z) (y-z)(y-x) (*-x)(s- 



4. v±* + • + * + x + y 

(x-y)(x-z) (y-z)(y-x) 0-x)(s-jr) 

6 b — c , c — a , a — b 

• -; ttz t+t; m r + 



(a-6)(a-c) (&-c)(6-a) (c-a)(c-b) 

g g2 y g . y 2 zx . s 2 sy 

(* -*)(*-*) (F-«)(F-«) («-*)(*- 

7. 1 + a | 1+ft , _l+_o_ 



( a -6)(o_c) (&-c)(6-a) (c-a)(c-6) 

8. p-a t g-q | r-a 

(p-q)(p-r) (q-r)(q-p) (r-p)(r-q) 

e p + q-r j gH-r-p , r+g-g. , 

(P -q)(P~r) (g-r)(g-p) (r-j>)(r-g) 

10. _ «\ + „ * + <* 



(a 2 - &*) (a 2 - c 2 ) (6 2 - c 2 ) (6 2 - a 2 ) (c 2 - a 2 ) (c 2 - &*) 

ii x +y + X +V + ^H-y 

(P-q)(P-r) (Q-r)(q-p) (r-j>)(r-g) 

It, g + r j r + jp t jp + g 

(*-iO(a-*) (y-«)(y-«) («-«)(«-y) 



CHAPTER XIV. 
Complex Fractions. Mixed Expressions. 



We now propose to consider some miscellaneous 
questions involving fractions of a more complicated kind 
than those already discussed. 

In the previous chapter, the numerator and denominator 
have been regarded as integers ; but cases frequently occur 
in which the numerator or denominator of a fraction is itself 
fractional. 

153. A Complex Fraction is one that has a fraction in the 
numerator, or in the denominator, or in both. 

a a 

ti h K 

Thus -, -, - are Complex- Fractions. 

O {C c 

c d 

In the last of these types, the outside quantities, a and d, 
are sometimes referred to as the extremes, while the two 
middle quantities, b and c, are called the means. 

a 

154. By definition (Art. 148) - is the quotient resulting 

c 

d 

from the division of - by - ; and this by Art. 143 is — • 

b d be 

a 

b __ad 

c ~~ be 

d 

126 



. . 



COMPLEX FRACTIONS. 127 

From the preceding article we deduce an easy method 
of writing down the simplified form of a complex fraction. 

Multiply the extremes for a new numerator, and the means 
for a new denominator. 

a + x 
-p b _ abja + x) _ q 

a*-x* " 6(a 2 - x 2 ) a^x 
ab 
by cancelling common factors in numerator and denominator. 



\. The student should especially notice the following 
cases, and should be able to write off the results readily. 

1 ' a H b b 

-=Ur=lX- = -, 

a or a a 



r= a -s- r = a x b = ab. 
1 o 

b 

1 

a = l^l = l b_b 

l~~a ' b~~ a l~~a 



157. We now proceed to show how complex fraction! 
oan be reduced by the rules already given. 



Ex.1. 



a , c 

b d (a , c\ (a c\ ad+bc ad — be 



Ex.* 



a 


c 


'Kb^dj^Kb" 


"*;- 


' bd 


1 bd 


b 


d 


' 












_ ad+bc bd 


r 


ad + 6c g 








bd ' ad — 


be 


ad — bo 






a* 










34 


' X 

a 4 " 
"a 8 


■(■♦SM- 


a*\ 


x* + a* 

X 


as* — o* 






_s 2 + a 2 x 


a* 


X* 





x* - a* ai 2 - a 2 



128 ALGEBRA. 

Ex. 8. Simplify - ^— . 

a — 6 a + b 

Thenumerator = (« 2 + ^-(a 2 - 5^ = 4a 2 » 2 

(a 2 + 6 s ) (a 2 - b*) (a 2 + ft 2 ) (a 2 - &} 

4ab 



Similarly, the denominator = 
Hence the fraction = 



4a 2 6 2 4ab 



(a 2 + ft 2 ) (a 2 - 6 2 ) (a + 6) (a - 6) 

_ 4a 2 6 2 * (a-f6)(o>-&) _ q& 

- ( fl 2 + 6 2) ( a 2_ 52) x 4 a6 - a 2 + £2* 

Notb. To ensure accuracy and neatness, when the numerator and 
denominator are somewhat complicated, the beginner is .advised to 
simplify each separately as in the above example. 

In the case of complex fractions like the following, called 
Continued Fractions, we begin from the lowest fraction; and 
simplify step by step. 

_ . ». ... 9x 2 -64 



1. 



X 1 






1 x 






4 + x 




_ 9s 2 -64 9x*- 


•64 

I + x 
4 




r. 1 r. 1 - 




4+^-x 




4 + x 






9x 2 -64 9x*-64 






~~4x-4 -(4 + «) ^ 3x-8 




4 4 






= 4(9* 2 -64) 

3x-8 v ' 






EXAMPLES XIV. a. 






Find the value of 






»_i. 1 + 1 « + $ 

n m n x y c d 


4. 


i+5 

X 


m n x y d 




X 



COMPLEX FRACTIONS. 129 

2+*« i_2? « + c 

5* . 7* » 9* ■« XI* • 

8 x* d n + j> 

8~~ 1 X ~" 

8. =-r« 8. r« 10. 12. - 

8a c n x 

,o * „ x x* x* „ 2x2.fl.-6 

* 776T8- 14 ' -9 "• 4 t 

1+- + -5 - — » -5""" 1 

xx 2 x x 2 

le 2 / 1 1\ -- / g8-68 g 8 + 6 8 \ 4qft 

' 1-x 3 V1-* l+«/" ' \a-6 a + b)' a 2 -6« 

18 f q2 - ax 4- x 2 a 2 + qx + x 2 \ . x 8 
* \ a — x a + x / a 2 — 

\l+x x / \l4-x x J 



x* 



80. 



a + b a — b 



* 

21. 
28. 


a — 6 a + b 


i a 2 + & 2 
(a + &)* 

a , x 
x 2 a 2 


I-JL+I 
a* ax x* 

1 1 


28. 

24 
85. 


8x-2 3x + 2 


1 
9-± 

' X 2 

1 l x 


1 2 a: 2 
l + xH-^- 

1 — X 
1 




a 2 -l 
1 



26. 



4x + - Ax 



1 + 2(x + y) 



6-x 

27. 2 

x + -™ 



y + 2 



28. 



i l + z 

x-1 

X 



29 *-Zl 



x-2-- * 



x — — — 
x-2 



80. 



x--JW * + l 



x + i * - i 

a — 1 x x 



130 ALGEBRA. 



Sometimes it is convenient to express a single frac- 
tion as a group of fractions. 

n y bx*y- 103y a + 15y» _ bx*y lOay 8 15y» 

~2y 3~ l '2*a 



MIXED EXPRESSIONS. 

159. We may often express a fraction in an equivalent 
form, partly integral and partly fractional. It is then called 
a Mixed Expression. 

5 



3 + 2 3 + 2 3 + 2 

Ex.3. 8g-2 = 8(3 + 5)-15-2 = 8(g+5)-17 _ s 17 

3 + 6 3 + 6 3 + 6 « + 6 

In some cases actual division may be advisable. 
Ex.3. Showthat 2a?2 ~ 7g "" 1 = 2g- 1- 4 



3-3 3-3 

Performing the indicated division, we obtain a quotient 2 as — 1, and 
a remainder — 4. 

Therefore 2x *~ 7x ~ X = 2x - 1 - 4 



«-3 s-8 

160. If the numerator be of lower dimensions (Art. 29) 
than the denominator, we may still perform the division, 
and express the result in a form which is partly integral 
and partly fractional. 

Ex. Provethat _l£— = 2»- 6x* + 18a? 5 - M * 



1 + 33 2 1 + S«* 

By division l + 8« 2 )2a; <2a;-6x 8 + 18»* 

23 + 63* 

-638-183 5 



183 6 

183* + 543* 

- 64x* 

Whence the result follows. 



MIXED EXPRESSIONS. 181 

Here the division may be carried on to any number of terms in 
the quotient, and we can stop at any term we please by taking for our 
remainder the fraction whose numerator is the remainder last found, 
and whose denominator is the divisor. 

Thus, if we carried on the quotient to four terms, we should have 

2x = 2*-6*»+18**-64*7 + 102x * 



l + 3* a 1 + 8*2 

The terms in the quotient may be fractional ; thus if x 2 
is divided by x* — a 8 , the first four terms of the quotient are 

- + ^ + % + ^z> and the remainder is 5L. 
x or z 7 a, a? 

161. Miscellaneous examples in multiplication and divis- 
ion occur which can be dealt with by the preceding rules 
for the reduction of fractions. 

Ex. Multiply *+2q- a * ft by2*-a- 2a * 



2* + 3q * + a 

Theproduct= f*+2a- g * \x(2x- a--2*?-\ 
* \ 2x + 3q/ V * + «/ 

__ 2** + 7q* + 6q* - qg 2* a + ax - q a - 2qg 
2* + 8q * + q 

_ 2* a + 7q* + 5q a x 2* 2 + q*-3q a 
2* + 3q * + q 

* - _ (2* + 5q)(* + q) (2*+3q)C*-q) 
2* + 3q * + a 

s=(2* + 6q)(*-a> 

EXAMPLES XIV. b. 

Express each of the following fractions as a group of simple frac 
tions in lowest terms : 

« Sz*y + xy*-y* a «+.&+■« 

9xy aoc 

« Sa*x-4a*x* + 6ax* fi 6c + ca + q& T 

12 q* * ' " qftc 

. q»-3q 2 6 + 3q& 2 + & 8 a c fibc- 3q6 8 c + 2q6c 
o. * o. — • 

2 qfr ,6 abc 



132 ALGEBRA. 

Perform the following divisions, giving the remainder after four 
terms in the quotient : 

7. s-Kl+as). 9. (l+*) + (l-aj). 11. x*+(x~r 3). 

g. a _s_( a _&). io. l-*-(l-x + x 2 ). 12. 1+Cl-xV*. 

13. Show that ^ini!L = a + 26+ 36 * 



49 



(a - 6) a a - 6 

14. Show that a*- xy +y»--^- = ^ult 

x+ y x + y 

15. Show that mx * - U*-**+ 1 = l2x-25 + 

6^ + 9s-2 x + 2 

16. Showthatl-f fla+62 " ca =^ + 6 + C >^- f6 - C > 

2a6 2a6 

17. Divide x + 16 * ~ 2 ] by x - 1 + 13 



x 2 - 16 x + 4 

18. Multiply a* - 2 ax + 4x' — ££- by 3- **i a + 4 *\ - 

rj a + 2x a 2 + 2ax + 4x* 

19. Divide 6 2 + 36 - 2 li- by 36 + 6 - 26 * 



6-3 " 6-3 

20. Divide a 2 + 96 2 +-^~ by a + 36+ 136 * 



a 2 -96 2 " a -36 

21. Multiply 4a , + U* + fcfI by 1-_£±S-- 

162. We add an exercise in which most of the processes 
connected with fractions will be illustrated. 



EXAMPLES XIV. O. 

Simplify the following fractions : 

4 a(a 2 - x 2 ) f a 2 - ax a 2 + 2 ax + x* l 
3 6(c 2 -x 2 )* 4 ~L6c+&x X c 2 -2cx + x 2 _r 

x(x + a) (x -f 2 a) x(x + a)(2x+a) 
* 3a 6a 

s. iM l —) ? 4 /«±iy_(^iiy 

6\a-6 a + 2 6/ a 2 + a6-26 2 \«-W \s + y' 

r 2 , 2 4x 

0. + 



x-1 x + 1 x 2 - x + 1 

u-** 1-W \ « / 



FRACTIONS. 1&& 



i.l 1 !_+_ .* 

a? (as + 1) 2 « + l l + s + a* 
8 1 + a? 8 e 2aG 8 -9a 2 + 27 

1 + 2& + 2«P + *P' ' Sac 8 - 81a:+ 162 

io a (qa "~ &a)a; fl(fl2 ~ &a)a * 



it f ag* — g* . aff + qac ) x afi — <&Kp . /a? q\ 
l«a_2ox + a 2 ' a;-a'J a£ + a 8 \<* */" 

' a 2 + oa; + a; 2 ' a 8 — a 8 

18 s*-2a? 2 + l 1C a 8 + fl(l + <t)V + tf 

16. »+ 2 8 « 



a a + 1 a + 2 j + 1 



a 



16. o o .^v — t-t; + stX 



2ac a + 9*+9 ' 2 2*-3 ^ 9 
17. 2 2 



« 8 + a; 2 + a:+l aj 8 -x 2 + *-i 

18 l-o 2 .1/1 . 1 V 

" (1 + <kb) 2 - (a + *) 2 ' 2\l-« l + «/ 

19 2a«-ai 2 -2a;+l «0 a; 2 -fls + 8 

a* -3a; + 2 " 4a£-21a; 2 +15a; + 20 

21 2q x — a .• 2 



(as -2 a)* » 2 -6ax + 6a 2 «-3a 

2\a 2 -aJ 2 / 2 a-* \a + */ 

23 «/_l 1 \ x g 2 -y 2 . 1 , 

2\aj-y a; + y/ x*y + xy 2 x + y 

* + y L2\a; + y x-yJ x*y + xy*A 
36. (.,_.-|)(., + ._8) + („-i). 

la a + ac a — z> \a — z a + xf 



134 



ALGEBRA. 



2a- 



87. 



2x 



2a-l 4fc*-.l 
a-b 



6 + 



l + q& 



q — 



28 

q-ft 
l-q& 



• ( 6+ A)( 6 -«^)(£tI} 



1 + a& 1 — a6 



** + 



30. 



y 



— « 






a^ + y 8 ' 






a + 



as. 



l+««6 



. i-sy 

1 + ab \ —xy 



a + b . q — ft 
M q — 6 q-f-ft aft 8 — q 8 5 



q + b q — 6 



84. 



(l-a^Cl-s 8 ) x? + aJ 8 



x(l+x)(l-x) 2 



x 2 



* {*-M-3M-i> 



i+i i 

36. — - — x 



1_ 

m 



i i ' l 

± m* + ±- m-l+± 

m m m 

?+?_! i+V 1+ g 
87. 4-^X * *• 



* 2 . a; 



7* + v +1 



x — y x A 



38. 
39. 
40. 
41. 



y x 
2 1 



a 2 _ 2 a* - 1 a* + 1 q 2 + 2 
1.1 3 



+ 
+ 



6m-2n 3m + 2n 6m + 2n 
3 .3 , . 1 



+ 



+ 



s-1 



4(1 - a;) 2 8(1 - as) 8(1 + x) 4(1 + x*) 

4 5 1 1__ 

9(*-2) + 9(x+l) 3(s + l) 2 a . + 2+ I 

x 



45. 
46. 

47. 



52. 



54. 



FRACTIONS. 135 



:+ ' 



42 ( X \*\( l \ ?L_ 

' \v x/W-rfJ tf + xy^xy-i/* 

X 2 -(y-z) 2 »■-(«-«)! g 2 -(s-y)« 
* (a + *)» - *■ "^ (a; + y)« - * + (y + *)« - rf # 

a;2-(y-2g)g ya-.(2g-a?) a 4z*-( X -y)* 
*^ (2* + afl«-y« + (as + y)*--4* + (y + 2*)*-»** 



(« - y)(y - «) + (y - *)(* - oc) + (g- s)(s - y) 
«(« - as) + y(x - y) + z(y - z) 

g — ft — c . 6 — c — o c — g — & 

(a - &)(a - c) (6 - c)(& - a) (c - a)(c - 6)' 

c + q . a + b . fr + c 

(a - &)(a - c) (6 - c)(& - a) (c - a)(c - b) 



gg-(2y-3g)g 4y 2 -(3g-a;) 2 9g»-(s-2y)« 
(3s + a;) 2 -4y 2+ (a; + 2y) 2 -9s 2 + (2 y + 3 *)* - a; 2 ' 

9y 2 -(4s-2aQ 2 16g 2 -(2a;-3y) 2 4s 2 -(3y-4g) 2 
*' (2x + 3y) 2 -16s 2 + (3y + 4*)* -4s* + (4« + 2a;) 2 - 9y* 



1 x + a 1 x — a 

<tx x x 2 + a 2 , x x 2 + a 2 
00. z ■ H 



1 a + x 1 a — x 



a at + x 2 a a 2 + x* 
M (a;+q)(a;+6)-(y+q)(y+&) (x-a)(y-b)-(x-b)(y-a) 



x — y (a — b) 

I_a + x a-x \ . ( a 2 + x 2 q 2 -a? 2 \ 

\a 2 - ax + x 2 a 2 + ax + a 2 / ' U 8 - a 8 a* + x*)' 



a; — 1 . z-1 a; + 3 a? + 3 

3 s-2 7 s + 4 
g + 2 . x + 2 a?-2 a?-2 

4 *-3 3 aj-1 



/3x4 L o^\ 2 _ 1 9 33 -a; 2 

\1 -h 3a; 2 / . x 2 3a; 2 + 1 

3a?-l . x ~*~ 3 2(a; 2 + 3) ' 

«»~3a; x 2 (x*-x)* 



CHAPTER XV. 



Fbactional and Litbbal Equations. 

163. In this chapter we propose to give a miscellaneous 
collection of equations. Some of these will serve as a 
useful exercise for revision of the methods already explained 
in previous chapters; but we also add others presenting 
more difficulty, the solution of which will often be facili- 
tated by some special artifice. 

The following examples worked in full will sufficiently 
illustrate the most useful methods. 

Ex.1. Solve 4 -*^li = iL--l. 

8 22 2 

Multiply by 88, which is the least common multiple of the denomi- 
nators, and we get 

352-ll(x-9)=4a;-44; 
removing brackets, 852 — Ilje-f99 = 4a5 — 44; 
transposing, — 11 x — 4 a; = — 44 — 352 — 09 ; 

collecting terms and changing signs, 16 * = 495 ; 

.-. a5 = 33. 

Note. In this equation — is regarded as a single term with 

the minus sign before it. In fact it is equivalent to — } (as — 9), the 
line between the numerator and denominator having the same effect 
as a bracket. 

Ex.* Solve 8x±2?_|*+l = 2xj 1 8_ 1> 

20 3 x + 4 5 

Multiply by 20, and we have 

8 g + 23- 20 ( 6a;4 " 2 ) -=8s+12-20. 
3aj-f4 

By transposition, 31 = 20 ( 5a; + 2 ) . 

136 



FRACTIONAL AND LITERAL EQUATIONS. 137 

Multiplying across, 98 x + 124 = 20(5 x + 2), 

84 = 7x; 

.-. 3 = 12. 

Ex.3. Solve ir±V*ni = «=i + *zi 

a; -10 x-6 x-7 x-9 

This equation might be solved by clearing of fractions, but the 
work would be very laborious. The solution will be much simpli- 
fied by proceeding as follows : 

Transposing, -^-^ & — o _ £Lz_I _ x ~ 4 . 

* *'• x-10 x-7 x-9 x-6 

Simplifying each side separately, we have 

(s - 8)(x - 7)-(x - 5) (as - 10) _ (x - 7)(x - 6) -(x - 4)(x - 9) . 
(x - 10)(x - 7) (a; - 9) (as - 6) ' 

. x*-15x+56-(g 2 -15x+50) _ x 2 -13x+42--(x 2 -13x+36) . 
(x - 10) {x - 7) (x-9)(x-6) 5 

" (a; - 10) (x - 7) (x-9)(x-6) 

Hence, since the numerators are equal, the denominators must 
be equal ; that is, 

(x - 10)(x - 7) = (x - 9)(x - 6), 

x 2 - 17 x + 70 = x 2 - 15 x + 64, 

16 = 2x; 

.«. x = 8. 

Ex.4. Solve 6*:^64_2s-ll = 4x-55__x-6. 

x-13 x-6 x-14 x-7 

Wehave 6 + — i— -(2 + — L_\ = 4 + — ^—--( 1 + -±-\; 
x-13 \ x-Gj x-14 V x-7/' 



x-13 x-6 x-14 x-7 
Simplifying each side separately, we have 



(x - 13)(x - 6) (x - 14) (as - 7) 

(x - 13)(x - 6) = (x - 14) (x - 7), 

x 2 - 19x + 78 = x 2 - 21 x + 98, 

2 x = 20 ; 

.\ x = 10. 



138 ALGEBRA. 

164. To solve equations whose coefficients are decimals^ 
we may express the decimals as common fractions, and pro- 
ceed as before ; but. it is often found more simple to work 
entirely in decimals. 

Ex. 1. Solve .376 x - 1.875 = .12z + 1.185. 

Transposing, 376 * - .12 * = 1. 186 + 1.876 ; 

collecting terms, (.376 - . 12)* = 8.06 9 
that is, .266 s = 3.06; 

... * = ?£? = 12. 
.265 

Ex. a. Solve .(5a; + .26-$* = 1.8 -.76*- J. 
Expressing the decimals as common fractions, we have 

clearing of fractions, 24s + 9 - 4* = 68 - 27* - 12 ; 
transposing, 24* - 4* + 27* = 68 - 12 - 9, 

47* = 47; 

V .\ * =S 1. 

EXAMPLES XV. a. 
- 4(* + 2) _ n , 5* A *-8 , *-8 , 5 _ A 

, * + 20 ,8*_ A a *,* * ,*_75 

8. -_ + T =6. 6. - +§ _ i + g = 7f 

4 2\ Z) 6 8\ 1) 
». _ (a ._ 8 ) + _ r - + - r =7 — . 

10. a ! -^3*-^- 5 -U|(2 a! -67)-|. 

,, 2ac-6 x-3 _ 4s-8 , , ia 4(s+3') _ 8;e+87 7s-29 
' 6 2x-15 10 "' 9 18 6*-12 



FRACTIONAL AND LITERAL EQUATIONS. 189 

18 (2a?-l)(3a; + 8) _ 1 _ ^ 30+6* 60+8* ^ 48_ 

6*(* + 4) ' " x +l " r s+3 *+] 

14. L*±l>- 2x +* = o. ^ * *+l_*-8 *-9 

6* + 3 6* + 2 »• ^Z2-iZI-i36"iT7- 

W- — 7T rr=« — -r— ' • ot *+5 *-6 a;— 4 a;— 15 

*+S *+l 2*+6 2*+2 «. _-_ = _--_ 

"• i^^ol^O""^*-^*^ 22. 5=2- *=±=*zi?--£zll 

17 3 , 30 ^ 3 . 5 *~ 9 aJ - U "- 16 *~ 17 

4-2* 8(1-*) 2-* 2-2* 33, *+3 *+6_*+2 *+5 | 
- " *+6 *+9 *+6 *+8* 

lft. 3 16*+4£ 23 ^ *+2 *-7 g+S^E-6 

* + l 3*+2 + *+l* " * *-6 *+l *-4 # 

o- 4*-17.10*-18 8*-80,6*-4 
*-4 2*-3 2*-7 *-i 

5*-8.6*-44 10*-8 * — 8 
H = • 



* — 2 * — 7 * — 1 * — 6 

o- 2* — 3 .4* — .6 Qfl*— 2 * — 4 M 

.3* -.4 .06* -.07 # 06 .0626 

89. .083(* - .626) = .09(* - .69876). 
30. (2* + 1.6)(3* - 2.26) =s (2* - 1.125)(8* + 1.26). 
81 .3*- 1 _ .6+1.2* g^ 1-1.4* __ .7(*- 1) 

.6* — .4 2* — .1 .2 + * .1 — .6* 

LITERAL EQUATIONS 

165. In the equations we have discussed hitherto the 
coefficients have been numerical quantities, but equations 
often involve literal coefficients. [Art. 6.] These are sup 
posed to be known, and will appear in the solution. 

Ex. 1. Solve (* + a)(* + b) — c (a + c) = (* — c)(* + c) + aft. 
Multiplying out, we have 

«* + a* + 6* + a& — ae — c 2 = «* — d 9 + aft ; 
whence a* + 6* = ac^ 

(a + b)z = aci 

..* = r» 

a + b 



140 ALGEBRA. 

Ex. 8. Solve -2 ^ = «^. 

x — a x — b as — c 

Simplifying the left side, we have 

a (x — b) — b (as — a) __ a — b 
(a? — a)(x — b) x — e* 
(a — 5) a; a — 6 . 
(as — a)(x — b) x — e' 
x = 1 

# " (x — a)(x — 6) x — c 

Multiplying across, a; 2 — ex = x 2 — ox — 6a; + a6, 
ax -f 6x — ex = aft, 
Ca + b — c) x = ab ; 

aft 



• • 5C> ^— 



a + 6 — e 



EXAMPLES XV. b. 

1. ax-26 = 6&x-3a. u 1_1_1_2. 

2. a 2 (x-a)+6 2 (x-&) =a&x * a x x T 

8. x 2 + a 2 = (6-x) 2 . 12# ?/x +1 \3/x_ 1 \ 

4. (x-a)(x + 6) = (x-a + 6) 2 . ' 3\a / *\a / 

5. a(x-2)+2x = 6 + a. w a = c(a _ 6) + &. 

6. ro 2 (m — x) — wmx = w 2 (n + x). * x 

7. (a + x)(6 + x) = x(x-c). 14 &«_3x = 4&_2x # 

8. (a-6)(x-a) = (a-c)(x-6), 6 b a a 

2x4- 3a __2(3x + 2a) lft x— a ^ x— 6 

x + a 3x+a ' b — x a — x 

10. 2 fo- 6 ) = 2x+6 ^ x - a _ (x - &)» 

3x — c 8(x — c) 2 2x — a 

18. (a + 6)x 2 - a(6x + a 2 )= 6x(x - a) + ax(x - 6). 

19. 5(a + x)-(a + x)(b-x) = x 2 + ^-. 

20. 6(«-*)-f(& + a; ) a + a6 (f + 1 ) a = - 

21. a*+a(2a-x)-^ = (x-|V + a*. 

22. (2x-.a)(x+^) = 4x(|-x)-|(a-4x)(2a + 3x). 



CHAPTER XVI. 

• Problems leading to Fb actional and Literal 

Equations. 

166. We here give some problems which lead to equations 
with fractional and literal coefficients. 

Ex. 1. Find two numbers which differ by 4, and such that one-half 
of the greater exceeds one-sixth of the less by 8. 

Let x represent the smaller number, then x + 4 represents the greater. 

One-half of the greater is represented by \{x + 4), and one-sixth of 
the less by \ as. 

Hence J(x + 4) — \x = 8 ; 

multiplying by 6 t 3* + 12 - x = 48 ; 

2 x = 36 ; 
. *. x = 18, the less number, 
and x + 4 = 22, the greater. 

Ex. 2. A has $ 180, and B has $ 84 ; after B has won from A a 
certain sum, A has then five-sixths of what B has ; how much did B 
win? 

Suppose that B wins x dollars, A has then 180 — x dollars, and B 
has 84 + x dollars. 

Hence 180 -a> = J(84 + x); 

1080 -6a; = 420 + 6x, 
lla; = 660; 
.\ x = 60. 
Therefore B wins 960. 

EXAMPLES XVL 

1. Find a number such that the sum of its sixth and ninth parts 
may be equal to 15. 

3. What is the number whose eighth, sixth, and fourth parts 
together make up 13 ? 

3. There is a number whose fifth part is less than its fourth part 
by 3: find it. 

141 



142 ALGEBRA. 

4. Find a number such that sir-sevenths of it shall exceed four- 
fifths of it by 2. 

5. The fifth, fifteenth, and twenty-fifth parts of a number together 
make up 23 : find the number. 

6. Two consecutive numbers are such that one-fourth of the less 
exceeds one-fifth of the greater by 1 : find the numbers. 

7. Two numbers differ by 28, and one is eight-ninths of the other : 
find them. 

8. There are two consecutive numbers such that one-fifth of the 
greater exceeds one-seventh of the less by 3 : find them. 

9. Find three consecutive numbers such that if they be divided 
by 10, 17, and 26, respectively, the sum of the quotients will be 10. 

10. A and B begin to play with equal sums, and when B has lost 
five-elevenths of what he had to begin with, A has gained $ 6 more 
than half of what B has left : what had they at first ? 

11. From a certain number 3 is taken, and the remainder is divided 
by 4 ; the quotient is then increased by 4 and divided by 6, and the 
result is 2 : find the number. 

12. In a cellar one-fifth of the wine is port and one-third claret : 
besides this it contains 15 dozen of sherry and 30 bottles of hock : 
how much port and claret does it contain ? 

13. Two-fifths of A's money is equal to B's, and seven-ninths of 
B's is equal to C's, in all they have $ 770 : what have they each ? 

14. A, B, and C have $1286 among them: A's share is greater 
than five-sixths of B's by $25, and C's is four-fifteenths of B's: find 
the share of each. 

15. A man sold a horse for $35 and half as much as he gave for it, 
and gained thereby $ 10 : what did he pay for the horse ? 

16. The width of a room is two-thirds of its length. If the width 
had been 3 feet more, and the length 3 feet less, the room would have 
been square : find its dimensions. 

17. What is the property of a person whose income is $ 430, when 
he has two-thirds of it invested at 4 per cent, one-fourth at 3 per cent, 
and the remainder at 2 per cent ? 

18. I bought a certain number of apples at three for a cent, and 
five-sixths of that number at four for a cent : by selling them at six- 
teen for six cents I gain 3£ cents : how many apples did I buy ? 

19. Find two numbers such that the one may be n times as great 
as the other, and their sum equal to 6. 



PROBLEMS. 143 

90. A man agreed to work a days on these conditions : for each 
day he worked he was to receive c cents, and for each day he was idle 
he was to forfeit d cents. At the end of a days he received m cents. 
How many days was he idle ? 

21. A sum of money is divided among three persons: the first 
receives a dollars more than a third of the whole sum ; the second 
receives b dollars more than a half of what remains ; and the third 
receives c dollars, the amount which is left. Find the original sum. 

22. Out of a certain sum a man paid $96 ; he loaned half of the 
remainder, and then spent one-fifth of what he had left. After these 
deductions he still had one-tenth of the original sum. How much had 
he at first ? 

28. A man moves 12 miles in an hour and a half, rowing with the 
tide, and requires 4 hours to return, rowing against a tide one-quarter 
as strong : find the velocity of the stronger tide. 

24. A man moves a miles in b hours, rowing with the tide, but 
requires c hours to return, rowing against a tide d times as strong as 
the first: find the velocity of the stronger tide. 

25. A has a certain sum of money from which he gives to B $ 4 
and one-sixth of what remains ; he then gives to C $ 6 and one-fifth 
of what remains, and finds that he has given away half of his money. 
How many dollars had A, and how many dollars did B receive ? 

26. The fore-wheel of a carriage is a feet, and the hind- wheel is b 
feet in circumference. What is the distance passed over when the 
fore-wheel has made c revolutions more than the hind-wheel ? 

27. In a certain weight of gunpowder the nitre composed 10 pounds 
more than two-thirds of the weight, the sulphur 4J pounds less than 
one-sixth, and the charcoal 5} pounds less than one-fifth of the nitre. 
What was the weight of the gunpowder ? 

28. Two-thirds of A's money is equal to B's, and three-fourths of 
B's is equal to C's; together they have $650. What amount has 
each? 

29. A dealer spends $1450 in buying horses at $100 each and 
cows at $ 30 each ; through disease he loses 10 per cent of the horses 
and 20 per cent of the cows. By selling the remainder at the price 
he gave for them he receives $1260: find how many of each kind 
he bought. 

80. A, B, C start from the same place at the rates of c, c -f d, 
c 4- 2 d miles an hour respectively : B starts k hours after A ; how 
long after B must C start in order that they may overtake A at 
the same instant, and how far will they then have walked? 



144 ALGEBRA. 

MISCELLANEOUS EXAMPLES m. 

1. Subtract p 8 — 4p? + 8 from unity, and Sjj 2 — p — 7 from zero* 
and add the results. 

2. Simplify (x - y) 2 + (x - s) 2 + 2 {(x - y) (* - x) + y«}. 

3. Solye 6{x - 2 [a; - S(x - 1)]} = 70. 

4. Divide x* + s 8 - 24 x 2 - 35x + 67 by x 2 + 2x - 8. 

5. Find the factors of 

(i.) (a + 6; 2 — 121 ; (iL)a*-M; (in.) * 2 - 6*- 14. 

6. Find the H. C. F. of a* - 2a 2 + 1 and 2o»+ a 2 + 4a - 7. 

7. A man being asked his age said : " Ten years ago I was five 
times as old as my son, but 20 years hence, I shall be only twice as 
old as he." How old was he ? 

s-1 s-2 8-x 



8. Solve (i.) 6- 



2 3 4 J 



/.. N 4a;-9 x — 3 6a; — 8 x + 6 
v J 27 4 6 2 

9. By how much does y* - 3y 2 + 8y + 9 exceed y-4y«-f 6-y 8 ? 

10. Show that 

(ox + 6y) 2 + {ay - 6x) 2 + c^x 2 + c 2 ^ = (a; 2 + y«) (a 2 + 6 2 + c 2 ). 

11. Solve (i.) ^_3x^ = ^_7a^. 

(ii.) (3a? - l) 2 +(4x - 2) 2 = (6x - 8)*. 

IS. If x = 1, y = 2, s = 3, find the value of 

(x - y)[(x + z) + (y - «)]- a; 2 + y(y + *)--**. 

18. Find the factors of (i.) a 2 +17 a&+60 ft 2 ; (ii.) 10a 2 + 79a - 8. 

14. Simplify (i.) o aC A Q -f — 5^_ ; 
* v y a 2 -4y a ac+ 2cy' 

••• \ a & i a 



a ft + &2 a 2 _ a & a « _ 52 

16. Find the L. C. M. of a* + 6a; 2 + 11 x + 6 and x* - 7x + 6. 

16. The difference between the numerator and the denominator of 
a proper fraction is 8, and if each be increased by 17, the fraction 
becomes equal to f : find it. 

17. Solve (i.) 20(7x + 4)-18(8x + 4)-6 = 26(x + 6); 

(ii.) j(» + l)+J(x + 3)=i(« + 4)+16. 

18. Divide (i.) 2 x (a; 2 - 1) (x + 2) by a; 2 + x - 2 ; 

Oi.) 6x(x-ll)(x 2 -x-166) by x 8 + x 2 - 132 a; 



MISCELLANEOUS EXAMPLES III. 145 

19. A boy is one-sixth the age of his father, and five years older 
than his sister; the united ages of all three being 51, how old is 
each? 

20. Find the continued product of x 2 + ax + a 2 , x 2 — ax + a 2 , 
as* - a*x* + a 4 . 

21. Show without actual division that x — 3 is a factor of the ex- 
pression x 8 — 2x 2 — 5se + 6. 

. Simplify (i.) £ + ?+ 4 2aj - 



9 3 s-6 3(3-6)' 

(IL) _L___i_ + l. 

a — x 2a — x x 

23. Find the H. C. F. of 2 a* 4- a 2 - a -2 and a 6 - a 8 - 2a 3 + 2 a 
by the usual method. Is the work shortened by proceeding as in 
Art. 119 ? Show that the square of the H. C. F. is contained in the 
second expression. 

24. Divide x* + 19 X s - 216 by (a 2 - 3 x + 9)(x - 2). 

25. Simplify (i.) t±l *- - £zJ£ . (ii .) « + _^_. 

* * w 2 p + 2 2p 2 -8 v y « 2 -4 (x-2) 2 

26. Find the factors of (i.) 14 a 2 - 11 a-16 ; (ii.) a*+5 a 2 6 2 +9 6*. 

27. Of a party 5 more than one-third are Americans, 7 less than 
one-half are Englishmen, and the remainder, 8 in number, are Ger- 
mans : find the number in the party. 

28. Solve (i.) 2(5 x - 2) - 3(6 x - 8) = 5(a + 1) - (2 x - 11); 

v ; 2? 18 4 * 

29. Simplify (i.) ; (ii) 1 + a? + a?a + *-& . 

— ° * m a(x 2 +a 2 ) x(x+a) 2 ' K } 1-s 8 (1-s) 8 

80. Find three numbers whose sum is 21, and of which the greatest 
exceeds the least by 4, and the middle one is half the sum of the 
other two. 

81. Employ the Factor Theorem in finding the H. C. F. of 

a 8 - 2 a 2 + 1 and 2 a 8 + a 2 + 4 a - 7. 

32. Show that 3(^x-2)J(^,-2) _8, = 4^ 

x 2 - x - 2 x 2 + x - 2 x 2 - 4 as* - 1 

83. Two trains go from P to Q by different routes, one of which 
is 15 miles longer than the other. A train on the shorter route takes 
6 hours, and a train on the longer, travelling 10 miles less per hour, 
takes 8} hours. Find length of each route. 



146 ALGEBRA. 

84. Find the factors of 

(L)4a«-4xy-15y»; (ii.) 9a?*- 82*^ + 9y*. 

35. Solve !Lzi!£-LLi5 aB !i±l5 + 5-«* 

5 3 3 

36. The number of months in the age of a man on his birthday In 
1875 was exactly half of the number denoting the year in which he 
was born. In what year was he born ? 

37. Simplify (L) 2g ~ 7 _- 2 ( g + 2 ); (ii.) — I — + L_ — 

38. Divide a^+^+^+aV+y 8 by &-&y+x*iP-xy*+y* and 
find the value of the quotient when as = and y = 1. 



89. Simplify (i.) 



1(1-2) 

1 — x _ x\x I m 






n , v x l + s + a£ 

K } 1 t x *l + Sx + Sx* + 2afi l 

1 + & 

40. A regiment has sufficient food for m days; but if it were 
reinforced by p men, would have food enough for n days only. Find 
the number of men in the regiment. 

* 1 *. 1 - * 

41. Solve (i.) «-=-! + —=-£ + 2 = 0; 

a o c 

baa 



42. Simplify (1 



+ a) a +fl+ 2—- 1 

i *-+r+W- 



CHAPTER XVII. 
Simultaneous Equations. 

167. Consider the equation 2 x + 5 y = 23, which contains 
two unknown quantities. 

From this we get y = 23 7 2a? (1). 

5 

Now for every value we give to x there will be one cor- 
responding value of y. Thus we shall be able to find as 
many pairs of values as we please which satisfy the given 
equation. Such an equation is called indetei-minate. 

For instance, if x = 1, then from (1) y = ^. 

Again, if x =a — 2, then y = ^- ; and so on. 

But if also we have a second equation of the same kind 
expressing a different relation between x and y, such as 

3a? + 4y = 24, 

we have from this y = 24 ~" 3a? (2). 

4 

If now we seek values of x and y which satisfy both 
equations, the values of y in (1) and (2) must be identical. 

Therefore 23-2* = 24-3* 

5 4 

Multiplying across, 92 — 8 x = 120 — 15 x\ 

7a> = 28; 
.•. x = 4. 

Substituting this value in equation (1), we have 

= 23-2a? == 23-8 ==3 
y 5 5 

Thus, if both equations are to be satisfied by the same 
values of x and y, there is only one solution possible. 

147 



148 ALGEBRA. 

• 168. Definition. When two or more equations are 
satisfied by the same values of the unknown quantities, they 
are called simultaneous equations. 

169. In the example already worked, we have used the 
method of solution which best illustrates the meaning of 
the term simultaneous equations; but in practice it will be 
found that this is rarely the readiest mode of solution. 
It must be borne in mind that since the two equations are 
simultaneously true, any equation formed by combining 
them will be satisfied by the values of x and y which sat- 
isfy the original equations. Our object will always be to 
obtain an equation which involves one only of the unknown 
quantities. 

170. The process by which we cause either of the un- 
known quantities to disappear is called elimination. It 
may be effected in different ways, but three methods are 
in general use : (1) by Addition or Subtraction ; (2) by Sub 
etitution ; and (3) by Comparison. 

ELIMINATION BY ADDITION OR SUBTRACTION. 

171. Ex.1. Solve 7aj + 2y = 47 (1), 

6s-4y = l (2). 

Here it will be more convenient to eliminate p. 

Multiplying (1) by 2, 14 x -f 4 y = 94, 
and from (2) 6 a? — 4 y = 1; 

adding, 19 a; = 95; 

,\ x = 5. 

To find y, substitute this value of z in either of the given equations. 

Thus from (1) 85 + 2y = 47; 

and x = 6. i 

In this solution we eliminated y by addition. 

Ex. 2. Solve 8x+7y = 27 (1), 

6as + 2y = ld. (2> 



SIMULTANEOUS EQUATIONS. 149 

To eliminate x we multiply (1) by 5 and (2) by 8, so as to make 
the coefficients of x in both equations equal. This gives 

15s + 35y = 135, 

16* + 6y = 48; 

subtracting, 29 y = 87 ; 

.-. if = 3. 

To find x, substitute this value of y in either of the given equations. 
Thus from (1) 3*-*- 21 = 27; 

and y = 8. * 

In this solution we eliminated x by subtraction. 



Rule. Multiply, when necessary, in such a manner as to 
make the coefficients of the unknown quantity to be eliminated 
equal in both equations. Add the resulting equations if these 
coefficients are unlike in sign; subtract if like in sign* 

ELIMINATION BY SUBSTITUTION. 

172. Ex. Solve 2s-5y = l (1), 

7s + 8y = 24 (2). 

Transposing — by in (l\ and dividing by 2, we obtain 

x = * — • 
2 

Substituting this value of x in (2) gives 

7 (^t2)+3 y = 24. 

Whence 86y-f 7 + 6y = 48, 

and 41 y = 41 ; 

.*. |f=l. 
This value substituted in either (1) or (2) gives 

« = 8. 

Rule. Prom one of the equations, find the value of the 
unknown quantity to be eliminated in terms of the other and 
known quantities; then substitute this value for that quantity 
in the other equation, and reduce. 



160 ALGEBRA. 



ELIMINATION BT COMPARISON. 

173. Ex. Solve x + 16y = 5S (1) 

y + 3s = 27 (2). 

From (1) as = 53 -16?, 

■"■» •- S 4 X 

Placing these values of as equal to each other, we have 

* 3 
Whence 169 -45y = 27 -y, 

and 44y = 132; 

.% y = 3. 
Substituting this value in either (1) or (2) gives 

x = S. 

Rule. .FVom eocft equation find the value of the unknown 
quantity to be eliminated in terms of the other and known quan- 
tities; then form an equation with these values, and reduce* 

EXAMPLES XVII. a. 

Solve the equations : 

1. 8s + 4y = 10, 8. 15s + 7y = 29, 15. 89*-8y = 99, 

4a; + y = 9. 9s+16y = 89. 52«- 15y = 80. 

2. 3 + 2y = 13, 9. 14s-3y = 39, 16. 5as=7y-21, 
Sx + y = 14. 6s + 17y = 35. 21s-9y = 75. 

3. 4s + 7j/ = 29, 10. 28s-23y = 33, 17. 6y-6x=18, 

x + Sy = ll. 63x-25y = 101. 12«-9y = 0. 

4. 2s-y = 9, 11. 35«+17y = 86, 18. 8aj = 6y, 
3s-7y = 19. 66aj-13y = 17. 13jc = 8y + L 

5. 6s + 6y=17, 12. 15s + 77y = 92, 19. 3«=7y, 

6s + 6y = 16. 66*-33y = 22. 12y = 5z-l. 

6. 2« + y = 10, 13. 5x-7y = 0, 20. 19a+17y = 0, 
7x-r$y = b3. 7s + 5y = 74. 2a;-;- = 63. 

7. 8je-y = 34, 14. 21x-50y = 60, 21. 93 x + 15 y = 123, 
s + 8y = 53. 28x-27y = 199. 15as + 98y = 201 



SIMULTANEOUS EQUATIONS. 151 

174. We add a few cases in which, before proceeding to 
solve, it will be necessary to simplify the equations. 

Ex.1. Solve 5(x + 2tj)-(8x + lly)=14t (1), 

7«-9y-3(x-4y)=S8 (2). 

From (1) 6* + lOy - 8s - 11 y = 14 ; 

.\ 2s-y = 14 (8). 

From (2) 7*-9y-3a; + 12y = 88; 

.-. 4s+3y = 38 (4). 

From (3) 6z-3y = 42; 

and hence we may find x = 8, and y = 2. 

Ex.8. Solve 3 x - V-y± = ^f^ <*>» 

§£±i-. i( 2s-6) = |, (2). 

Clear of fractions. Thus 
from (1) 42 x - %y + 10 = 28a; - 21 ; 

.-. 14aj-2y=-31 (8). 

From (2) 9y + 12 - 10s + 25 = 15y ; 

.-. 10aj + 6y = 37 (4). 

Eliminating y from (8) and (4), we find that 

Eliminating x from (8) and (4), we find that 

Note. Sometimes, as in the present instance, the value of the 
second unknown is more easily found by elimination than by substi- 
tuting the value of the unknown already found. 

« 

EXAMPLES XVII. b. 

1. ^+» = 16, 8. ^-» = 8, 5. | + |=10, 

» + |=14. *-^=8. 1+0 = 60. 

2 - + v = 6 4. *-y = 6, «. x = Sy, 

' 5 * 1 i-I = 2 - l+v = u 

x — y = 4. 45 o 



152 ALGEBRA* 

4s-y = 20. 7 6 7 5 4 

8. i*-iy = 4, * + ? = 4f 8* + Jy = 17. 
|* + Ay = 3. 8 w 3s-l y = 7 > 

9. 2aj + y = 0, 11. 3x-7y = 0, 2 4 2* 
Jy-3aj = 8. $aj + fy = 7. » + 3y = 9. 

14 f + ? = 3a J -7y-87 = b. 15. 2±I = ^p* = a^jr. 

8 4 10 2 8 



SIMULTANEOUS EQUATIONS INVOLVING THREE 
UNKNOWN QUANTITIES. 

175. In order to solve simultaneous equations which con- 
tain two unknown quantities we have seen that we must have 
two equations. Similarly, we find that in order to solve 
simultaneous equations which contain three unknown quan- 
tities we must have three equations. 

Rule. Eliminate one of the unknowns from any pair of the 
equations, and then eliminate the same unknown from another 
pair. Two equations involving two unknowns are thus obtained, 
which may be solved by the rules already given. The remain- 
ing unknown is then found by substituting in any one of the 
given equations. 

Ex.1. Solve 6je + 2y-5s = 13 (1), 

3x + 3y-2s = 13 (2), 

7a + 6y-3s = 26 (8). 

Choose y as the unknown to be eliminated. 
Multiply (1) by 8 and (2) by 2, 

18a + 6y-15s = 89, 
6x + 6y- 4s=26; 
subtracting, 12 x — 11 z = 13 . • • . • • (4). 

Again, multiply (1) by 6 and (3) by 2, 

30aj+ 10 y- 25s = 65, 

14«+10y- 6z = 52; 

subtracting, 16 a; — 192 = 13 (5), 



SIMULTANEOUS EQUATIONS. 153 

Multiply (4) by 4 and (5) by 3, 

48s -44s = 52, 
48 a -67* = 89; 
subtracting, 13 z = 18 ; 

and from (4) a? = 2, 

from (1) y = S. 4 

Note. After a little practice the student will find that the solution 
may often be considerably shortened by a suitable combination of the 
proposed equations. Thus, in the present instance, by adding (1) and 
(2) and subtracting (3) we obtain 2 x — 4 2 = 0, or x = 2z. Substi- 
tuting in (1) and (2), we have two easy equations in y and z. 

Ex.* Solve f — l=|+l=f + 2, 

2 6 7 

8 2 
From the equation | — 1 = | + 1, 

we have 8s— y = 12 (1). 

Also, from the equation | — 1 = | + 2, 

we have 7s- 2s = 42 (2). 

And, from the equation | + 1 = 18, 

8 2 

we have 2y + 3* = 78 ....... (8). 

Eliminating * from (2) and (3), we have 

21* + 4y = 282'; 

and from (1) 12*-4y = 48; 

whence * = 10, y = 18. Also by substitution in (2) we obtains = 14. 

Ex. 8. Consider the equations 

6x — By— z = 6 (1), 

13s-7y + 3s = 14 (2), 

7s-4y = 8 (SV 

Multiplying (1) by 3 and adding to (2), we have 

28aj- 16y = 32, 
or 7ac-4y = 8. 



154 ALGEBRA, 

Thus the combination of equations (1) and (2) leads us to an 
equation which is identical with (3), and so to find x and y we have 
but a single equation 7 x — 4 y = 8, the solution of which is indeter- 
minate. [Art. 167.] 

In this and similar cases the anomaly arises from the fact that the 
equations are not independent; in other words, one equation is de- 
ductible from the others, and therefore contains no relation between 
the unknown quantities which is not already implied in the other 
equations. 

EXAMPLES XVH. o. 

1. s + 2y + 2*=ll, 9. 8a;-4y = 60-16, 
2*+ y + = 7, 4a;— y— = 5, 
3a; + 4y + = 14. a; = 3y + 2(0-l). 

2. * + 3y + 4s = 14, 1& 6a; + 2y = 14, 
* + 2y + = 7, y-6«=-16. 

2«+ y + 20 = 2. a;+2y + = O. 

8. a; + 4y + 30 = 17, 11. a;-| = 6, 



3a; + 3y + = 16, 



5 



2a; + 2y + = 11. jr-| = 8, 



« - ?,= 10. 



4> 3x-2y + = 2, 
2a; + 8y — = 6, 

«+ y + = 6. y^- z = z + x = x + y^ 

5. 2a;+ y+ = 16, * * 3 2 

s + 2y+ = 9, a;+y + = 27. 

*+ y + 20 = 8. 18 . IL=L? = IL^ == 50-4 a j, 

6. a 5 -2y + 30 = 2, yj# = 2«+l. 
2,-8*+ 0=1, M 2jC + 8y = 6t 

3,- y+2 0=9. • 20 _ J =1 ; 

7. 3x + 2y- = 20, 7x-90 = 8. 

2aj + 8y + 60 = 7O, 16 . £(3 + 0-6)= y-z 

x- y + 60 = 41. =2aj-ll = 9-(* + 20). 

8. 2^ + 3^+40 = 20, , + 20=^+10 
3a; + 4y + 60 = 26, 2 

3a; + 6y + 60 = 81. = 20 + 6 = 110 -(y + 0). 

176. Definition. If the product of two quantities be 

equal to unity, each is said to be the reciprocal of the other. 
Thus if ab = 1, a and b are reciprocals. They are so called 



SIMULTANEOUS EQUATIONS. 156 

1 1 

because a = -, and b = - ; and consequently a is related to b 
b a 

exactly as & is related to a. 

The reciprocals of x and y are - and - respectively, and 

x y 11 

in solving the following equations we consider - and - as 

os y 

the unknown quantities. 

Ex. 1. Solve ?-? = l (1), 

* y 

*M = 7 (8). 

x y 

Multiply (1) by 2 and (2) by 3 ; thus 

16_18 = 2 
* V ' 

*- + l? = 21, 

* y 

adding, ^ = 28; 

x 

multiplying across, 46 = 23 &, 

/. x = 2; 
and by substituting in (1), y = 8. 

Ex.2. Solve — + ^--^-=7 (1), 

2s - 4y 8# 4 





1--L 
x $y 


(2), 




1 1 4- 4 -2A 

— — — — t " — *T5 ••••••••• 

x by z 


(3); 


clearing of fractional coefficients, we obtain 




from (1) 


6,3 4 _« 

— "f" ~" """ — — *5#««#««#rj 

x y z 


<*). 


from (2) 


3 _I-0 


(6), 


from (3) 


15 3 , 60 _ «o 

— — — • — -f- — — OS . •• *•••••• 

x y z 


(«>• 



156 ALGEBRA. 

Multiply (4) by 15 and add the result to (6) ; we have 

105 + 42 = 77 
x y 

dividing by 7, 15 + ? = u r (7). 

x y 

from (6) !§_? = <); 

x y 

adding, ?2=11; 

x 

.*. x = 3, 

from (5} y 

from (4) z 



= 2. J 



EXAMPLES XVII. d. 



1. 5 + 5 = 8, 

x y 


6. 5 + 2 = 30, 
x y 


11. 


-L + J--2 
4x + 3y ' 


15 + § = 4. 

x y 


»-5 = 2. 
x y 




1-JL.l. 
y 2x 


2. $-1 = 8, 

? + li = 8. 
x y 


7. ?-? = 7, 

x y 

efl+'W 

\x y/ 


12. 


2y — x = 4xy, 

4 -? = 9. 
y x 


8. 12-4 = 2, 

x y 


8. 26 + 24 = 1, 
x y 


13. 


l-? + 4 = 0, 
x y 


?-? = 0. 


20f? + ^=7. 
\x yl 




1-1+1 = 0, 

y * 


4. 5 + 15 = 79, 

x y 


9. - + 21 = 42, 
x 2/ 




? + ?=14. 

X 


1?-1 = 44. 
x y 


14__16_ 1 
x y 


14. 


1 + 1 + 1=38, 
x y z 


5. 21 + 1? = 6. 

a; y 


10. 8 + 5 = 1, 

x y 16 




1 + 2-1 = 28, 
x y z 


l_i = -L. 
y x 42' 


9y-22x = 2*£. 
* 26 


1 

X 


+ -L + J- = 20. 
Sy 2z 



SIMULTANEOUS EQUATIONS. 157 



LITERAL SIMULTANEOUS EQUATION& 

177. Ex. 1. Solve ax + by = c (1), 

a'x + b'y = & (2). 

The notation here first used is one that the student will frequently 
meet with in the course of his reading. In the first equation we 
choose certain letters as the coefficients of z and y, and we choose 
corresponding letters with accents to denote corresponding quantities 
in the second equation. There is no necessary connection between the 
values of a and a', read "a and a prime, 11 and they are as different as 
a and b ; but it is often convenient to use the same letter thus slightly 
varied to mark some common meaning of such letters, and thereby 
assist the memory. Thus a and a 1 have a common property as being 
coefficients of x ; b, b' as being coefficients of y. 

Sometimes instead of accents letters are used with a suffix, such as 
«i, a%, as ; b±, b it bz, etc., read " a sub one, a sub two, 1 ' etc. 

To return to the equation ax + by = c (1), 

a f x+b*y = c' (2). 

Multiply (1) by V and (2) by b. Thus 

ab'x + bb f y = 6'c, 
a'bx + bb'y = be 9 ; 
by subtraction, (ab f - a r b)x = b'c - be 9 ; 

As previously explained in Art. 171, we might obtain y 
by substituting this value of x in either of the equations (1) 
or (2) ; but y is more conveniently found by eliminating x, 
as follows : 

Multiplying (1) by a' and (2) by a, we have 

aa'x -f- a*by = a% 
ac-x + db t y = ac l ; 
by subtraction, (a% — ob*) y = a'c — ac f ; 



158 ALGEBRA, 

or, changing signs in the terms of the numerator and denom- 
inator so as to have the same denominator as in (3), 

„ ac* — a'c „ , „ b'c — be? 



ob'-atf ob'-aty 

Ex.8. 8olve «jr« + £zi£ = i (1), 

c — a c — 6 

x + a . y-a _a r2 v 

e a—b e 

From (1) by clearing of fractions, we hare 

x(c - b) - a(c — 6)+ y(c — a) — 6(c - a)=(c — a)(c — 6), 
*(c — 6) + y(c — a) = ac — ab + 6e — a& + c* — ac — 6c 4- a6, 

«(c — 6)4- y( c — a) = c* — a& (3). 

Again, from (2), we have 

x(a — 6)+ a(a — 6) + cy — ca = a(a — 6), 

«(a — 6) + cy = ac (4). 

Multiply (8) by c and (4) by c — a and subtract, 

x{c(c — 6) - (c — a) (a — 6)}= c* — o6c — ac(c — a), 
05(c* - ac + o a - 06) = c(c* -ab-uc + <P); 

.«. » = c; 
and therefore from (4) jr = 6. 



EXAMPLES XVIL 6. 

1. a« + 6y = I, T ? + 2 = _L l & 9»~r6=p(o— y), 
6«4-«y=w»« a b ab q X I y\ 

2. lx + my = n, *-lL = jL. ~a" { ' r=P V + bJ' 
px + qy = r. a' 6' aW % 

8 - «» = *. _ « « A U# m + m' = 1 ' 

6x + ay = c. 8. --* = 0, *__«,- 

4. » + * = * 6s-fay = 4a6. «»' * 

6x -f ay = 6 2 . * 

6. * + «, = «', , S..!!.. »»• *■ + » = * 

6. px-qy = r, £*_!l£ = 3 ^ ( a_6 )*=(«+6)|r, 



SIMULTANEOUS EQUATIONS. 159 

14. (a-6)s+(a+6)y=2a*-26*, ^ _a+ &- = a + &, 
(a + 6)0! — (a — 6)y = 4 aft. " 6« ay 

W. * + 2 = i, * +JL 2 M = <*« + 6». 

a 6 8a 66 3 2 tf 

16. -+2 = 2, ■£ = -£. «0. ay+bz = 2xy, 

a b a ' b ' cy + dx = Sxy. 

*~ x y + x.y a 

IT. — f = If r + = 7* c -t « I 

a 6 6 a 6 SL =Af = — ? — 

« — y m — n 

18. -^- + _1L^ = 2, 

c -f d c — d « + w» = 1 + w . 

«e — aty = c* + d 2 . y+«i J + » 

(x + y)0» 3 + P)= 2(w« + P)+ wl(« + jr). 

28. te + qf = a + 6,axf-^---i- : Wcyf r i J_^ = -l« 

\a-6 a + 6' \6-a 6 + a/ a+b 

94. (a-6)aj+(a + 6)y = 2(a 2 -6 2 ), a&-6y = a* + W. 



CHAPTER XVIII. 
Pboblems leading to Simultaneous Equations, 

178. In the Examples discussed in the last chapter we 
have seen that it is essential to have as many equations as 
there are unknown quantities to determine. Consequently 
in the solution of problems which give rise to simultaneous 
equations, it will always be necessary that the statement 
of the question should contain as many independent con- 
ditions as there are quantities to be determined. 

Ex. 1. Find two numbers whose difference is 11, and one-fifth of 
whose sum is 9. 

Let x represent the greater number, y the less. 

Then *-y = ll (1). 

Also, ^±-* = 9, 

or g + y = 45 . . » (2). 

By addition, 2 x = 56 ; and by subtraction, 2 y s= 84. 
The numbers are therefore 28 and 17. 

Ex. 2. If 15 lbs. of tea and 10 lbs. of coffee together cost $ 15.50, 
and 25 lbs. of tea and 13 lbs. of coffee together cost $24.55, find the 
price of each per pound. 

Suppose a pound of tea to cost x cents and a pound of coffee to 
cost y cents. 

Then from the question, we have 

15a + 10y = 1560 (1), 

25x+13y = 2455 (2) 

Multiplying (1) by 5 and (2) by 3, we have 

75 * + 50 y = 7750, 
75 x + 39 y = 7365. 
Subtracting, llj/ = 385, 

y = 35. 

160 



PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS. 161 

And from (1) 15 x + 350 = 1550. 

Whence 15 x = 1200 ; 

.•. x = 80. 
Therefore the cost of a pound of tea is 80 cents, and the cost of a 
pound of coffee is 35 cents. 

Ex. 3. A person spent $6.80 in buying oranges at the rate of 3 for 
10 cents, and apples at 15 cents a dozen ; if he had bought five times 
as many oranges and a quarter of the number of apples, he would 
have spent $ 25.45. How many of each did he buy ? 

Let x represent the number of oranges and y the number of apples. 

* oranges cost ^ cento, 

O 

y apples cost — ^ cents : 
9 12 

•••^+^=«8o a). 

Again, 5 x oranges cost 5 x x — , or -~? cents, and ^ apples cost 

.-.5|» + ^=2M1S (»>. 

Multiply (1) by 5 and subtract (2) from the result ; 

thett (S-ii>= 8555 

^ = 866, 

.\ y = 144; 
and from (1) x =. 150. 

Thus there were 150 oranges and 144 apples. 

Ex. 4. If the numerator of a fraction is increased by 2 and the 
denominator by 1, it equals f ; and if the numerator and denominator 
are each diminished by 1, it equals £ : find the fraction. 

Let x represent the numerator of the fraction, y the denominator ; 

x 
then the fraction is — 

y 

From the first supposition, *L±_ = 2 (1) 

y + 1 8 w 

V — 1 1 

from the second, - = * (2). 

y- 1 2 w 

These equations give x = 8, y = 16. 

Thus the fraction is ^. 



162 ALGEBRA. 

Ex. 5. The middle digit of a number between 100 and 1000 U 
zero, and the sum of the other digits is 11. If the digits be reversed, 
the number so formed exceeds the original number by 495. Find it. 
Let x represent the digit in the units 1 place ; 
y represent the digit in the hundreds' place ; 
then, since the digit in the tens 1 place is 0, the number will be repre- 
sented oy 100 y + x. [Art. 84, Ex. 4.] 

And if the digits are reversed, the number so formed will be repre- 
sented by 100 x + y. 

.% 100a; + V -(100y + x)= 495, 
or 100 a; + y - 100 y - x = 496; 

/. 99a;-99y = 496, 
that Is, x — y = 6 (1). 

Again, since the sum of the digits is 11, and the middle one is 0, 

we have x + y = 11 (2). 

From (1) and (2) we find x = 8, y = 3. 
Hence the number is 808. 

EXAMPLES XVm. 

1. Find two numbers whose sum is 84, and whose difference is 
10. 

2. The sum of two numbers is 73, and their difference is 37 : find 
the numbers. 

3. One-third of the sum of two numbers is 14, and one-half of 
their difference is 4 : find the numbers. 

4. One-nineteenth of the sum of two numbers is 4, and their 
difference is 30 : find the numbers. 

5. Half the sum of two numbers is 20, and three times their dif- 
ference is 18 : find the numbers. 

6. Six pounds of tea and eleven pounds of sugar cost $ 5.65, and 
eleven pounds of tea and six pounds of sugar cost $ 9.65. Find the 
cost of tea and sugar per pound. 

7. Six horses and seven cows can be bought for $250, and thir- 
teen cows and eleven horses can be bought for $461. What is the 
value of each animal ? 

8. A, B, C, D have $ 290 between them ; A has twice as much as 
C, and B has three times as much as D ; also C and D together have 
$ 50 less than A. Find how much each has. 

9. A, B, C, D have $ 270 between them ; A has three times as 
much as C, and B five times as much as D ; also A and B together 
have $50 less than eight times what C has. Find how much each 
has. 



PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS. 163 

10. Four times B's age exceeds A's age by twenty years, and one- 
third of A's age is less than B's age by two years : find their ages. 

11. One-eleventh of A's age is greater by two years than one- 
seventh of B's, and twice B's age is equal to what A's age was thir- 
teen years ago : find their ages. 

12. In eight hours A walks twelve miles more than B does in seven 
hours ; and in thirteen hours B walks seven miles more than A does 
in nine hours. How many miles does each walk per hour ? 

13. In eleven hours C walks 12£ miles less than D does in twelve 
hours ; and in five hours D walks 3} miles less than does in seven 
hours. How many miles does each walk per hour ? 

14. Find a fraction such that if 1 be added to its denominator it 
reduces to I, and reduces to f on adding 2 to its numerator. 

15. Find a fraction which becomes \ on subtracting 1 from the 
numerator and adding 2 to the denominator, and reduces to J on 
subtracting 7 from the numerator and 2 from the denominator. 

16. If 1 be added to the numerator of a fraction it reduces to J ; 
if 1 be taken from the denominator it reduces to \, Required the 
fraction. 

17. If f be added to the numerator of a certain fraction the frac- 
tion will be increased by ^-, and if J be taken from its denominator 
the fraction becomes § : find it. 

18. The sum of a number of two digits and of the number formed 
by reversing the digits is 110, and the difference of the digits is 6 : 
find the numbers. 

19. The sum of the digits of a number is 13, and the difference 
between the number and that formed by reversing the digits is 27 ; 
find the numbers. 

20. A certain number of two digits is three times the sum of its 
digits, and if 45 be added to it the digits will be reversed : find the 
number. 

21. A certain number between 10 and 100 Is eight times the sum 
of its digits, and if 45 be subtracted from it the digits will be reversed : 
find the number. 

22. A man has a number of silver dollars and dimes, and he ob- 
serves that if the dollars were turned into dimes and the dimes into 
dollars he would gain $ 2.70 ; but if the dollars were turned into half- 
dollars and the dimes into quarters he would lose $ 1.30. How many 
of each had he ? 

28. In a bag containing black and white balls, half the number of 
white is equal to a third of the number of black ; and twice the whole 
number of balls exceeds three times the number of black balls by four. 
How many balls did the bag contain? 



164 ALGEBRA. 

24. A number consists of three digits, the right hand one being 
zero. If the left hand and middle digits be interchanged, the number 
is diminished by 180 ; if the left hand digit be halved and the middle 
and right hand digits be interchanged, the number is diminished by 
464 : find the number. 

26. The wages of 10 men and 8 boys amount to $22.30 ; if 4 men 
together receive $ 3.40 more than 6 boys, what are the wages of each 
man and boy ? 

86. A grocer wishes to mix sugar at 8 cents a pound with another 
sort at 5 cents a pound to make 60 pounds to be sold at 6 cents a 
pound. What quantity of each must he take ? 

27. A traveller walks a certain distance ; had he gone half a mile 
an hour faster, he would have walked it in four-fifths of the time ; 
had he gone half a mile an hour slower, he would have been 2} hours 
longer on the road : find the distance. 

28. A man walks 35 miles partly at the rate of 4 miles an hour, 
and partly at 5; if he had walked at 5 miles an hour when he walked 
at 4, and vice versd, he would have covered 2 miles more in the same 
time : find the time he was walking. 

29. Two persons, 27 miles apart, setting out at the same time are 
together in 9 hours if they walk in the same direction, but in 3 hours 
if they walk in opposite directions : find their rates of walking. 

30. When a certain number of two digits is doubled, and increased 
by 10, the result is the same as if the number had been reversed, and 
doubled, and then diminished by 8; also the number itself exceeds 3 
times the sum of its digits by 18 : find the number. 

31. If I lend a sum of money at 6 per cent, the interest for a cer- 
tain time exceeds the loan by $ 100; but if I lend it at 3 per cent, for 
a fourth of the time, the loan exceeds its interest by $ 425. How 
much do I lend ? 

32. A takes 3 hours longer than 8 to walk 30 miles ; but if he 
doubles his pace he takes 2 hours less time than B : find their rates 
of walking. 



CHAPTER XIX. 

Indbtebminate and Impossible Pboblems. Nega 
tive Results, Meaning of % A % * 

indeterminate and impossible problems. 

179. By reference to Art. 167, it will be seen that a 
single equation involving two unknown quantities is satis- 
fied by an indefinitely great number of sets of values of the 
unknowns involved, and that it is essential to have as many 
equations expressing different, or independent conditions, 
as there are unknown quantities to be determined. If the 
conditions of a problem furnish a less number of independent 
equations than quantities to be determined, the problem is 
said to be indeterminate. If, however, the conditions give 
us a greater number of independent equations than there 
are unknown quantities involved, the problem is impossible. 

Suppose the problem furnishes 

Sx + y = 10, 

2x + y = 5, 

x + y = 3. 

Prom (1) and (2) we obtain x = 5 and y = — 5. From 
(2) and (3) we obtain x = 2 and y = 1. These values can- 
not all be true at the same time, hence the problem is 
impossible. 

NEGATIVE RESULTS. 

180. A is 40 years old, and B's age is three-fifths of A's. 
When wiU A be five times as old as B ? 

165 



166 ALGEBRA. 

Let x represent the number of years that will elapse. 
Then 40 + x = 5 (24 + x) ; 

.\ 40 + a> = 120 + 5x, 
or a = -20. 

According to this analysis, A will be five times as old as 
B in — 20 years. The meaning of this result ought to be 
at once evident to a thoughtful student. Were the result 
any positive number of years, we would simply count that 
number forward from the present time (represented by the 
word " is " in the problem) ; manifestly then the — 20 years 
refers to past time. Hence the problem should read, " A is 
40 years old, and B's age is three-fifths of A's. When was 
A five times as old as B ? " 

Suppose the problem read 

A is 40 years old, and B's age is three-fifths of A's : find 
the time at which A's age is five times that of B. 

Let us assume that x years will elapse. 

Then 40 + x = 5 (24 + x)\ 

,\ a? = — 20. 

Interpreting this result, we see that we should have 
assumed that x years had elapsed. 

The student will notice that the word " will " in the first 
statement suggested that we should assume x as the number 
of years that would elapse, and that the negative result 
showed a fault in the enunciation of the problem ; but that 
the problem, as given in the next discussion, permitted us to 
make one of two possible suppositions as to the nature of 
the unknown quantity, so that the negative result indicates 
simply a wrong choice. 

Hence in the solution of problems involving equations of 
the first degree, negative results indicate 

(1) A fault in the enunciation of the problem, or 

(2) A wrong choice between two possible suppositions, as to 
the nature of the unknown quantity, allowed by the problem. 

Generally it will be easy for the student to make such 
changes as will give an analogous possible problem. 



INDETERMINATE AND IMPOSSIBLE PROBLEMS. 167 

EXAMPLES XIX. 

Make such necessary changes in the statements of the following 
problems as will render them possible arithmetically. 

1. A is 27 years old and B 15 ; in how many years will A be twice 
as old as B ? 

2. What are the two numbers whose difference is 50, and sum 40 ? 

3. If to the sum of twice a certain number and | of the same num- 
ber 10 be added, the result is equal to twice the number. 

4. A man loses $ 400, and then finds that 6 times what he had at 
first is equal to 5 times what he has left. 

5. What fraction is that which becomes f when 1 is subtracted 
from its numerator, and -J when 1 is subtracted from its denominator? 

6. A is to-day 25 years old, and B's age is f of A's : find the date 
when A's age is twice that of B. 

MEANING OF |, |, ?, 2. 
181. Meaning of -• Consider the fraction - in which 

\j x 

the numerator a has a certain fixed value, and the denomi- 
nator a; is a quantity subject to change; then it is clear that 
the smaller x becomes, the larger does the value of the 

fraction - become. For instance, 
x 

J?_=10a, -5L_ = 1000a, a = 100000 a. 

T0~ i looooo 

By making the denominator x sufficiently small, the value 

of the fraction - can be made as large as we please ; that 

x 

is, as the denominator x approaches to the value 0, the fraction 
becomes infinitely great. The symbol oo is used to express 
a quantity infinitely great, or more shortly infinity. The 
full verbal statement, given above, is sometimes written 

o =0 °- 



— • If, in the fraction - 
tor x gradually increases and finally becomes infinitely large, 



Meaning of — If, in the fraction -, the denomina- 

co x 



168 ALGEBRA. 

the fraction - becomes infinitely small ; that is, as the de» 

x 

nominator of a fraction approaches to the value infinity, the 
fraction itself approaches to the value 0. This full verbal 
statement is sometimes written 

5- a 

oo 

183. Meaning of — The symbol - may be indeterminate 

a? 1 — 4 
in form or in fact. Thus the value of when x = 2 

x — 2 

is -, but by putting the fraction in the form \ x + )\ x — } 

... x ~~ 2 

we see that the expression is equivalent to x + 2, which 

q» r.8 

becomes 4 when x = 2. Again, = - when x = a, but 

x — a 

by .putting the fraction in the form (*-«)(«* + «* + <*) 

x — a 

we see that the expression is equivalent to x* + xa + a 2 , 

or 3 a 2 , when x = a. These fractions assumed the form - 



under particular conditions, but it is evident that they 
do not necessarily have the same value. 

On the other hand, the symbol - may show that a value 

is really indeterminate. Thus, solving in the regular way 
the equations 

» + y + 2 = o, 

2a? + 2y + 4 = U, 

we get x = ~~ = --, and we can easily see that x can 

have any value whatever if we give y a value to suit, so 
that the value of x is indeterminate. 

oo 1 

184. Meaning of — Inasmuch as — = 0, what is tiae of 

00 00 

- is equally so of — 
oo 



CHAPTER XX. 

Involution. 

185. Definition. Involution is the general name foi 
repeating an expression as a factor, so as to find its second, 
third, fourth, or any other power. 

Involution may always be effected by actual multipli- 
cation. Here, however, we shall give some rules for writing 
at once 

(1) any power of a monomial ; 

(2) the square and cube of any binomial ; 

(3) the square and cube of any multinomial ; 

(4) any power of a binomial expressed by a positive 
integer. 

186. It is evident from the Rule of Signs that 

(1) no even power of any quantity can be negative; 

(2) any odd power of a quantity will have the same sign 
as the quantity itself. 

Note. It is especially worthy of notice that the square of every 
expression, whether positive or negative, is positive. 

INVOLUTION OF MONOMIALS. 

187. From definition we have, by the rules of multi- 
plication, 

(a*)* = a*-a*-a*=a*+ M = a*. 

(- a 3 )* = (- a 3 ) (- a 8 ) = o*+ 8 = sfi. 

(- a*) 8 =(- a 5 ) (-a*)(- a 6 ) = - a 5+5+5 = - a M . 

(- 3 a*) 4 = (- 3) 4 (a*) 4 = 81 a 12 . 

169 



170 ALGEBRA. 

Hence we obtain the following rule for raising a simple 
expression to any proposed power : 

Rule. (1) Raise the coefficient to the required power by 
Arithmetic, and prefix the proper sign found by Art. 42. 

(2) Multiply the index of every factor of the expression by 
the exponent of the power required. 

Examples. (1) (- 2 a; 2 ) 6 = - 32 a; 10 . 

(2) (- 3 a& 8 ) 6 = 729 cfib^. 

\2x*y) 81 sV* 

It will be seen that In the last case the numerator and the denonv 
nator are operated upon separately. 



EXAMPLES 

Write the square of each of the following expressions : 

1. Sab*. 5. 4xyz*. % __£_. i 3 g2 & 8 x 

2. 5a?V. e. -*<**&». 8a ^ 4c5x * 

„ 11- -2a^. 

3. -2a6c 2 . 2£ 9 _7o&. Sa s 

4. 11 W. * 3y» 3 l». 6a;8 - 

Write the cube of each of the following expressions : 

13. 2a&*. 16. -3a*6. g ^ 19 . 7a 4y. 

14. 3a*. j^ "- -J5 ^ _ Ja6- 

15. -2a 7 c a . * 3y 2 

Write the value of each of the following expressions t 

21. (s a wy. ^ / i y. 26> /**y. M /W 

22 f-flW \ 2a2 / * 8 */ V »W 

a* /3a*\* a* / **V oo / 2a¥\* 

TO SQUARE A BINOMIAL. 

188. By multiplication we have 

(a + b)* = (a + b)(a + b) = a 2 + 2 ab + b* . . (1), 
( a -6) 2 = (a-6)(a-6) = a 2 -2a6 + 6 2 . . (2). 

Rule I. The square of the sum of two quantities is equal 
to the sum of their squares increased by twice their product. 



INVOLUTION. 171 

Rule II. The square of the difference of two quantities is 
ejual to the sum of their squares diminished by twice their 
product. 

Ex.1. (» + 2y)* = 3» + 2.3.2y + (2y)* 

= x 2 + 4^ + 4^. 

Ex.* (2a»-8d*)a=(2a3)»-2.2a 8 .3&2 + (362)* 

= 4a 6 -12o 8 6 a + 96*. 

189. These rules may sometimes be conveniently applied 
to find the squares of numerical quantities. 

Ex. i. The square of 1012 = (1000 + 12)2 

= (1000) 2 + 2 . 1000 . 12 + (12)* 
= 1000000 + 24000 + 144 
= 1024144. 

Ex.2. The square of 98 =(100-2)2 

s(100)«-2.100.2+(2)« 
s 10000 -400 + 4 
= 9604. 



TO SQUARE A MULTINOMIAL. 

190. We may now extend the rules of Art. 188 thus: 

(af6-Hc) 2 =i(a + 6)+ci s 

= (a + b) 2 + 2(a + b)c + c* [Art. 188, Rule 1.] 
= a 2 + 6 2 + c 2 + 2a& + 2ac + 2&c. 

In the same way we may prove 

(a - b + c) 2 = a 2 + b 2 + c 2 - 2 ab + 2ac - 2bc 
(a + b + c + fy 2 = a 2 + b 2 + <? + (& + 2ab + 2ac 

+ 2ad + 2bc + 2bd + 2cd. 

In each instance we observe that the square consists of 

(1) the sum of the squares of the several terms of the 
given expression; 

(2) twice the sum of the products two and two of the 
several terms, taken with their proper signs ; that is, in each 



172 ALGEBRA. 

product the sign is + or — according as the quantities com- 
posing it have like or unlike signs. 

Note. The square terms are always positive. 

The same laws hold whatever be the number of terms in 
the expression to be squared. 

Rule. To find tlte square of any multinomial : to the sum 
of the squares of the several terms add twice the product (with 
tlie proper sign) of each term into each of the terms that follow it. 

Ex.1. (x-2y-Zz)* = x*+iy*+9z*-2.X'2y-2-x-Zz+2-2y-Se 

= x 2 + 4 y 2 + 9« 2 - 4sy -6xz + 12yz. 

Ex.2. (l+2a-3a; 2 ) 2 = l+4a; 2 +9«*+2.1.2a;-2.1.3x a -2.2a;.3a« 

= 1 + 4x 2 -f 9s* + 4x - 6a: 2 - 12a* 
= l + 4z-2a; 2 -12x 8 + 93*. 

EXAMPLES XX. b. 

Write the square of each of the following expressions: 

1. a + 3 6. 3. » — by. 5. Sx-y. 7. 9x — 2y. 

2. a — 36. 4. 2x-\-Sy. 6. Sx+by. 8. 5a6 — c. 
9. a — 6 — c. 14. xy + yz + zx. M a _ 2 ^ , £, 

10. a+6-c. i* <,«_!_,. * 2 4 

10. x —y + a — o. 

11. a + 26 + c. 1fl a « fc 3 

12. 2a-» + 4a 16 ' 2s + 3j, + «-26. » g"^"^ 

13. x 2 -y 2 -z 2 . 17. ro-n-p-g. 20. l^-aj + f. 

TO CUBE A BINOMIAL. 

19L By actual multiplication, we have 

(a + b)* = (a+b)(a + b)(a + b) 

= a 3 + 3a 2 b + 3ab 2 +b* . . . (1), 

(a-b) 3 =a s -3a 2 b + 3ab 2 -b* . . . (2). 

From these results we obtain the following rule : — 

Rule. To find the cube of any binomial : take the cube of 
the first term, three times the square of the first by the second, 
three times the first by the square of the second, and the cube of 
the last. 



INVOLUTION. 173 

If the binomial be the sum of the quantities, all signs will 
be + ; if the difference of two quantities, the signs will be 
alternately + and -, commencing with the first. 

EXAMPLES XX. o. 

Write the cube of each of the following expressions : 

1. x + a. 5. * 2 + 4y*. 9 a _ 26 u x±_^ 

2. x-a. 6. 4s 3 -St/ 2 . 8 8 



3. s-2j,. 7. 2«»-3& 2 . 1Q g +2# ^ a + ^ 

4. 2a6-3c. 8. 5a* -4y*. 8 G 



TO CUBE ANY MULTINOMIAL. 

192. Consider a trinomial : 

= a* + 3a*(b + c) + 3a(ft + c)» + (6 + c) 8 
= a 8 + b* + <? + 3a*(b + c) + 36'(a + c) 
+ 3c 2 (a + b) + 6abc. 

Rule. 2b cute a multinomial : take the cube of each term, 
three times the square of each term by every other term, and 
six times the product of every three different terms. The signs 
are determined by the law of signs for multiplication. 

EXAMPLES XX. d. 
Write the cube of each of the following expressions . 

1. l + * + a& 4. a + bx + z 2 . 7. I + — -* 8 . 

2. 1 + *-*£, 5. 2a + bx-cx*. 8. 1 +*-fs t + a; 1 . 

3. 1- 2a; + 3a 2 . 6. 83 + 2a; 2 -a*. 9. 2 -8a: + a; 2 + 2a* 

TO RAISE A BINOMIAL TO ANY POWER EXPRESSED BY 
A POSITIVE INTEGER [BINOMIAL THEOREM], 

193. By actual multiplication, we obtain the following 
identities : 

(a + &) 8 = o 8 + 3a 2 & + 3a&'+ft 8 ; 

(a + b) 4 = d 4 + 4 a*b + 6 a V + 4 aft 8 -♦ &•. 



174 ALGEBRA. 

In these results, spoken of as expansions, we notice that; 

(1) The number of terms equals the index of tlie binomial 
plus one. 

(2) The exponent of a in the first term is the same as the 
index of the binomial, and decreases by one in each succeeding 
term. 

(3) The quantity b appears for the first time in the second 
term of the expansion with an exponent 1, and its exponent 
increases by one in each succeeding term. 

(4) The coefficient of the first term is 1. 

(5) The coefficient of the second term is the same as the 
index of the binomial. 

(6) The coefficient of any term may be found by multiplying 
the coefficient of the preceding term by the exponent of a in 
that term, and dividing the result by the exponent of b phis 1. 

Ex. 1. Expand (a + 6) 6 . 

(a + by = a? + a* + a 4 + a* + a* + a 

+ 6 + 6 2 + 6 8 + M + &* + 6 8 
Coefficients, 1+6 +15 +20 +15 + 6 + l 
Multiplying, a* + 6 a b b + 15 atb* + 20 a 8 6 8 + 15 a 2 6* + 6 ab 6 + &• 

. Ex. 2. Expand (a - 2 6 2 ) 4 . 

(a-2 6 2 ) 4 =[a+(-2 6 2 )] 4 

= a* + a* + a 2 +a 

+ (_2& 2 ) + (-2& 2 ) 2 +(-2& 2 ) 8 +(-2 6 2 ) 4 

Coefficients, 1 +4 +6 +4 + J. 

Multiplying, a 4 -8a 8 6 2 + 24a 2 & 4 -32a& 6 + 16 ft 8 

Note. The student will observe that in the line of coefficients, 
terms at equal distances from the beginning and the end are equal. 

194. The same method may be used in expanding any 
multinomial. 

Ex. Expand (a + 2 b - c) 8 . 
(a + 2b - c) 8 =[(a + 2 6) + (- c)] 8 

= (a + 2 by + (a + 2 6) 2 +(a + 2 6) 

+ (-<0 +(-<0 2 +(-c)» 
Coefficients, 1 +3 +3 +1 

"Multiplying, (a + 2 6) 8 - 3(a + 2 6) 2 c + 3(a + 2 6)c* - c 8 



INVOLUTION. 175 

Performing the operations indicated, we have 
a 8 + 6 a 2 b - 3 a 2 c + 12 ab 2 - 12 abc + 3 <h? + 8 6 s - 12 6*c 

+ 66c 2 -c 8 . 

Note. A fall discussion of the Binomial Theorem for Positive 
Integral Index is given in Chapter XXXVII. 

EXAMPLES XX. a 
Expand the following expressions : 
1. (s + y) 6 - 9. (i& + ««)*. 14. (o-6 + c)». 

2. (a — ft) 6 . //T27, r 2\6 * 

3. (2« + >)< 10 ' fir-i)" »' (2-|-^. 

4. (3x + 2y)5. u (2abc_2*y t 16> (a + 6 _ 2 c)«. 

5. (2x--3&)». \ 3 5 A 

6. «* + *)•. lg (2a_VV 17 ' (« + » + — «•. 

7. (2a&-c*)«. ' \» *J . 18. (a+2 6+c-2d)» 

8. (3 a 2 6 2 - 2 cd»)6. 13. (a + 26 + 3c)». 19. ( a -26 + c-d)* 

20. Find the middle term of (a + l) 10 . 

21. Find the two middle terms of (x — y) 11 . 

22. Find the term independent of a in ( — — — J • 



CHAPTER XXI. 
Evolution. 

195. The root of any proposed expression is tnat quantity 
which being repeated as a factor the requisite number of 
times produces the given expression. (Art. 14.) 

The operation of finding the root is called Evolution: it 
is the inverse of Involution. 

196. By the Rule of Signs we see that 

(1) any even root of a positive quantity may be either 
positive or negative; 

(2) no negative quantity can have an even root ; 

(3) every odd root of a quantity has the same sign as the 
quantity itself. 

Note. It is especially worthy of notice that every positive quantity 
has two square roots equal in magnitude, but opposite in sign. 

Ex. V( 9 « 2 ^) = ± 3 axS - 

In the present chapter, however, we shall confine our 
attention to the positive root. 

EVOLUTION OF MONOMIALS. 

197. From a consideration of the following examples we 
will be able to deduce a general rule for extracting any 
proposed root of a monomial. 

Examples. (1) V(« 6&4 ) = « 8&2 because (a 8 & 2 ) 2 = cfib*. 

(2) #(- x 9 ) = - x 8 because (- x 8 ) 8 = - x 9 . 

(3) ^(c 20 ) = c* because (c 4 ) 6 = c 20 . 

(4) ^(81 a* 2 ) = 3 x 8 because (3 x 8 )* = 81 x M . 

176 



EVOLUTION. 177 

Rule. (1) Find the root of the coefficient by Arithmetic, 
and prefix the proper sign found by Art. 42. 

(2) Divide the exponent of every factor of the expression 
by the index of the proposed root. 

Examples. (1) <J/(-64a 8 ) = -4aA 

(2) ^(16 a 8 ) =2 a*. 






It will be seen that in the last case we operate separately upon the 
.numerator and the denominator. 

EXAMPLES XXI. a. 

Write the square root of each of the following expressions : 

1. 4a 2 6*. 6. 81a 6 6 8 . 9 324 x 12 u 256 reV 

2. 9 &y\ 6. 100a«. " 1W* 6 " ' 289 j>" 

3. 26sty>. 7. a*>b™c*. 1Q 81 q" M 400 oW» 

4. 16 a***? 8 . 8. aWc* ' «&&*' ' 81«"y»" 

Write the cube root of each of the following expressions: 

13. 27o 6 6 8 c 8 . 16. -343 a 12 © 18 . lg 8s» ^ 27x 2T 

14 -8«W aW " 729 y»" 64 j* 

14. 8a &. n _et 125« 8 &« fll 343** 

16. 64a*U». 1J6 ie - 2165V ~«25* 

Write the value of each of the following expressions : 
22. ^(789flW). gg r/J^ ^ • /«£_. 

\fl68ft66 \627c86 

at 10 / a80aj8(> fifi g ^* 5 

24. ^(-*i° y i8). »*• V~&mo"* \"^T 

EVOLUTION OF MULTINOMIALS. 

196. The Square Root of Any Multinomial. Since the 
square of a + b is a* 4- 2 a& 4- 6 2 , we have to discover a pro- 
cess by which a and 6, the terms of the root, can be found 
when a 2 -f 2ab + b 2 is given. 

The first term, a, is the square root of a*. 

Arrange the terms according to powers of one letter a. 
The first term is a 2 , and its square root is a. Set this 

N 



178 ALGEBRA. 

down as the first term of the required root. Subtract a* 
from the given expression and the remainder is 2 db + W or 
(2a + 6)x&. 

Thus, by the second term of the root, will be the quotient 
when the remainder is divided by 2 a + b. 

This divisor consists of two terms : 

1. The double of a, the term of the root already found. 

2, b } the new ierm itself. 

The work may be arranged as follows : 



a 
a 


> + 2ab + V(a + b 
s 


2a + b 


2od + 6* 
2a& + &* 


Ex. 1. Find the square root of 9x 2 - 42 xy + 49 y* 

9z* - 42xy + 49y 2 (3s - 7y 
: 9x 2 


6*e-7y 


-42xy + 49y 2 
-42sy + 49y 2 



Explanation. The square root of 9 x 2 is 3 as, and this is the first 
term of the root. 

By doubling this we obtain 6z, which is the first term of the 
divisor. Divide — 42 xy, the first term of the remainder, by 6 a; and 
we get — 7 y, the new term in the root, which has to be annexed both 
to the root and divisor. Next multiply the complete divisor by — 7 y 
and subtract the result from the first remainder. There is now no 
remainder and the root has been found. 

The process can be extended so as to find the square root of any 
multinomial. The first two terms of the root will be obtained as 
before. When we have brought down the second remainder, the first 
part of the new divisor is obtained by doubling the terms of the root 
already found. We then divide the first term of the remainder by 
the first term of the new divisor, and set down the result as the next 
term in the root and in the divisor. W i e next multiply the complete 
divisor by the last term of the root and subtract the product from the 
last remainder. If there is now no remainder the root has been 
found ; if there is a remainder we continue the process. 

Ex. 8. Find the square root of 

26x2a 2 - 12xa» + 16 x* + 4a* - 24sta. 



EVOLUTION. 179 

Rearrange in descending powers of x. 

16 a* - 24 aft* + 25« 2 a 2 - 12 xa 8 + 4a*(4x 2 - 3xa + 2 a 2 

16 a* 

8<?-8za -24a£a + 26x 2 a 2 

-24a?a + 9as»a 2 _ 

8a£-6a:a + 2a* 16 x'a 2 - 12 xa* + 4 a* 

16s 2 a 2 -12att 8 + 4a* 



Explanation. When we have obtained two terms in the root, 
4x 2 — 3xa, we have a remainder 

16x 2 a 2 -12xa 8 + 4o*. 

Double the terms of the root already found and place the result, 
8x 2 - 6 a**, as the first part of the divisor. Divide 16x 2 a 2 , the first 
term of the remainder, by 8z 2 , the first term of the divisor ; we get 
+ 2 a 2 which we annex both to the root and divisor. Now multiply 
the complete divisor by 2 a 2 and subtract. There is no remainder and 
the root is found. 

EXAMPLES XXI. b. 
Find the square root of each of the following expressions : 

1. x 2 -10xy + 2by 2 . 6. 4x*-12a*+29x 2 -80aH-26. 

2. 4x 2 -12xy+9y 2 . 7. 9x*-12x*~ 2x* + *as + 1. 

3. $lx*+lSxy + y*. 8. a* -4a* + 6x 2 -4s+ 1. 

4. 26x 2 -30xy + 9y*. 9. 4o* + 4a 8 - 7a 2 -4a + 4. 
6. a*-2a 8 + 3a 2 -2a + l. 10. 1 - 10x + 27a5 2 - lOa^ + s*. 

11. 4x 2 + 9y 2 + 25« 2 +12xy-30y*-20a*. 

12. 16a* + 16x 7 -4x 8 -4x* + x 10 . 

13. x* - 22 x* + 34x 8 + 121 x 2 - 374a; + 289. 

14. 25x 4 -80ax 8 + 49a 2 x 2 -24a 8 x+16a 4 . 
16. 4a* + 4x 2 y 2 -12x 2 * 2 + y 4 -0y 2 3 2 + 93*. 

16. 6 ab 2 c - 4 a?bc + a 2 6 2 + 4 a 2 c 2 + 9 b 2 c 2 - 12 a&A 

17. - 6 6 2 c 2 + 9 c* + 6* - 12 c?a 2 + 4 a 4 + 4 a 2 6 2 . 

18. 4x* + 9y 4 + 13xV-6xy 8 -4x 8 y. 

19. l-4x+10x 2 -20x 8 + 26x*-24x B + 16x 8 . 
80. 6a<^ + 46 2 x*-f a 2 x 10 + 9c 2 -12&cx 2 -4a&x 7 . 

199. When the expression whose root is required contains 
fractional terms, we may proceed as before, the fractional 



180 



ALGEBRA. 



part of the work being performed by the rules explained in 
Chapter xiii. 



200. There is one important point to be observed when 
an expression contains powers of a certain letter and also 
powers of its reciprocal. Thus in the expression 

2a? + 4 + 4 + a 8 +5 + 7s 8 + ^, 
or x or 

the order of descending powers is 

x or or 

i 

and the numerical quantity 4 stands between x and — 

x 

The reason for this arrangement will appear in Chapter 

XXII. 

Ex. Find the square root of 24 + ^ - — + ^ - ^X 

x 2 y y 2 x 

Arrange the expression in descending powers of y. 



16^ 32y , gi _8ag ( x*/4y_ ^g 
x 2 x y jfix x y 

16 y« 



a? 



**-4 



X 
X 






8 + - 



8-** + £ 

y "" 

Ba 
JL 



8-** + ^ 



Here the second term in the root, — 4, arises from division of 
^ by 5J£, and the third term, -, arises from division of 8 by ^ ; 



x 
thus 8 



x 



x &y y 



EVOLUTION. 181 

EXAMPLES XXI. o. 
Find the square root of each of the following expressions 

25 + 6 +f ' 7 ' 64 + 7" + 

2. ?£ + I°* + 25. 8. a5* + 2flc» — a;+i. 

9. -So^^ + a^-Sa + fJa 2 



y 2 v 

x* 2x 



~ by b* 4 x x 2 a 2 



*. £- 



9y* 3y 2 2 16 

e. i^-2+ 25.. ' 18. ^rf+^^aM?* 

26 9a} 2 439 8 

1A 9a 2 6a, 101 4a , 4a* 

14» — r - p -rz — r 



a; 2 6x 26 15a 9a a 
16. 16m* + V*» 2 » + 8m 2 + Jn a + Jn + 1. 

16. 4a* + 32s 2 + 96+^ + ^. 

ar a? 



201. The Cube Root of Any Multinomial. Since the 
cube of a + b is a 3 4- 3 a?b -f 3 aft 2 -f- 6 8 , we have to discover 
a process by which a and 6, the terms of the root, can be 
found when a 8 + 3a 2 6 -f 306* + b s is given. 

The first term a is the cube root of a 8 . 

Arrange the terms according to powers of one letter a ; 
then the first term is a 8 , and its cube root a. Set this down 
as the first term of the required root. Subtract a 8 from the 
given expression, and the remainder is 

3a?b + 3o& 2 + ft 8 or (3a 2 + 3a6 + 6 s ) x b. 

Now the first term of the remainder is the product of 3 a* 
and b. Thus to obtain b we divide the first term of the re- 
mainder by three times the square of the term already found. 

Having found b we can complete the divisor, which con- 
sists of the following three terms : 



] 



182 



ALGEBRA. 



1. Three times the square of a, the term of the root 
already found. 

2. Three times the product of this first term a and the 
new term b. 

3. The square of b. 

The work may be arranged as follows : 

a 8 + 3 a 2 b + 3 a6* + 6* (a + b 



3(a)* = 3a* 

3 x a x b= + 3ab 

(by = +&» 

3a* + 3ab + b 2 



3a 2 b + 3aV 2 + b* 



3a*b + 3dP + & 



Ex. L Find the cube root of 8x* - 36x*y + 64xy 2 - 27 y 8 . 



8x« - S6x*y + 64 xy 2 - 27 y»(2x - 3y 
8x» 



3(2x)« = 12s» 
3x2xx(-8y) = -18xy 
(-3y) a = ± 



12x 2 -18xy + 9y 2 



-86x^ + 64^-27^ 



-SBxfy + 64xy fl -27y» 



Explanation. The cube root of 8 x 8 is 2 as, and this is the first 
term of the root. 

By taking three times the square of this first term we obtain 12 x 2 , 
which is the first term of the divisor, and is called the u trial divisor." 
Divide — 36x 2 y, the first term of the remainder, by 12x 2 and We get 

— 3 y, the new term in the root. To complete the divisor, we first 
annex to the trial divisor three times the product of 2x, the part of 
the root already found, and — 3 y, the new term of the root : this is 

— 18 xy. We then annex the square of — 3y, the new term, and the 
divisor is complete. We next multiply this divisor by the new term, 
and subtract the result from the first remainder. There is now no 
remainder and the root has been found. 

The process can be extended so as to find the cube root of any 
multinomial. The first two terms of the root will be obtained as 
before. When we have brought down the second remainder, we form 
Jie trial divisor by taking three times the square of the two terms of 
the root already found, and proceed as is shown in the following 
example. 



EVOLUTION. 



188 



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184 



ALGEBRA. 



EXAMPLES XXI. d. 
Find the cube root of each of the following expressions : 

1. a 8 + 3 a 2 + 3 a + 1. $. ahfi - 3 ctftffy 2 + 3 axy 4 - |f« 

8. 64 cfi - 144 a*b + 108 ab* - 27 6 8 . 

4. l + 3x + 6x 2 + 7x 8 + 6x* + 8x 8 -f x 8 . 

5. l-6x + 21x 2 -44x 8 + 63x*-64x* + 27x 8 . 

7. 8a«-36a* + 66a4-63a* + 33a 2 -9a+l. 

8. 8x 8 + 12x 6 -30x*-36x 8 + 46x 2 + 27x-27. 

9. 27 x 8 - 54x*a + 117 x*a 2 - 116 x 8 ** 8 + 117 x 2 a* - 64 xa 6 + 27 a 6 . 

10. 27x 8 -27x 8 -18x*+17x 8 + 6x 2 -3x-K 

11. 24xty> + 96 xV - 6xfy + ofi - 96 xy* + 64 y 8 - 66 x 8 ? 8 . 
18. 216 + 342 x 2 + 171 x* + 27 x 6 - 27 x 5 - lOOx 8 - 108 x. 



202. We add some examples of cube root where fractional 
terms occur in the given expressions. 

Ex. Find the cube root of 64 - 27 x 8 + 4 - ^r 

x 8 x 8 

Arrange the expression in ascending powers of x. 

I_§5 + 64-27as»71-S* 
x 8 x 8 \x 2 



x 8 



8 x 



(!)' 



12 
x* 



3x 



2; 
x 2 



x(-3x)= -i? 

x 



(-8x) 2 



+ 9X 2 



l?_l§ + 9x 2 

X* X 



-55 + 64-27X 8 
x 8 



-55+64-27X 8 

x 8 



EXAMPLES XXI. e. 



Find the cube root of each of the following expressions : 

. 27 x 8 



xj^Sx 2 Sx..! 
8 4 2 

8. 8x 8 -4x 2 y 2 + f xj/* 



27 



27 x 2 9x 

64^ 8y* y 

4. ^-6x*+12xV 



8. 

8 



EVOLUTION. 186 



y 2 y x x 2 x* 



27 3 ^ » x 2 z* 

- a* 12a; 2 64s 119 ,108a 48a 2 8a 8 

cP a 2 a x x 2 x 9 

g 64 a* 192 a 2 240 q 160 { 60 a; 12 a? a* 
x 8 a; 2 a a a 2 a* 

9 Gb 6 a , a 8 3 a 2 3 6 2 ft 8 

10. gQa? 4 80s 8 90s 2 . 8a* 108a; 2? 48a* 
y 4 y s y 2 y 6 y y 6 

203. Some Higher Roots. The fourth root of an expres- 
sion is obtained by extracting the square root of the square 
root of the expression. 

Similarly by successive applications of the rule for find- 
ing the square root, we may find the eighth, sixteenth ••• root. 
The sixth root of an expression is found by taking the 
cube root of the square root, or the square root of the cube 
root. 

Similarly by combining the two processes for extraction 
of cube and square roots, other higher roots may be ob- 
tained. 

Ex. 1. Find the fourth root of 

81 a 4 - 216 x*y + 216 x*y 2 - 96 xy* + 16 y 4 . 

Extracting the square root by the rule we obtain 9 x 2 — 12 xy+4y' 2 ; 
and by inspection, the square root of this is 3 x — 2 y, which is the 
required fourth root. 

Ex. 2. Find the sixth root of 

(*-*)"-«H)(*-iM-a' 

By inspection, the square root of this is 

o -I 

which may be written x 8 — 3 a; + - — — ; 

x ac 8 

and the cube root of this is x — -, 

x 
which is the required sixth root. 



186 ALGEBRA, 

We conclude the subject of higher roots by giving a rule, 
which depends upon the Binomial Theorem, for finding the 
nth root of any multinomial. 

(1) Arrange the terms according to the descending powers 
of some letter. 

(2) Take the nth root of the first term, and this wiU be the 
first term of the root. 

(3) When any number op terms of the root have been 
found, subtract from the given multinomial the nth power of 
the part of the root already found, and divide the first terra of 
the remainder by n times the (n — l)th power of the first 
term of the root, and, this will be the next term of the root. 

204. When an expression is not an exact square or cube, 
we may perform the process of evolution, and obtain as 
many terms of the root as we please. 

Ex. To find four terms of the square root of 1 + 2 as — 2ap*. 

l + 2x-2x 2 (l + &-§*» + fa* 



2 + * 

2 + 2«-f« a 



2«-2s 2 

2x + x 2 



2 + 2x-3a£ + f« 8 



-3a* 

-Sx*-3x* + $x* 



Sx*-$x* 

Sx* + Sx*-%x* + jx* 



-^La^ + ja^-fa* 
Thus the required result Is 1 + * — §a£ + fas 8 . 

EXAMPLES XXI. f. 

Find the fourth roots of the following expressions : 
1. a* - 28 x 8 + 294 x* - 1372 x + 2401. 
fl lft 32 ,24 8,1 

a. is 1- — — _ + -_ _. 

m m 2 m 8 m* 
3. a 4 + 8a 8 x+16a* + 32ax 8 + 24a*x 2 . 
"4. 1 + 4x + 2x 2 -8x 8 -5x* + 8x 6 + 2x 8 -4a; 7 + a6 8 . 
5. l + $x + 20x 2 + 8x*-2Gx*-Bx 6 + 2Qx*-&x I + x*. 



EVOLTTTIOtf. 187 

Find the sixth roots of the following expressions : 

a l + 6s + 15a; 2 + 20x 8 + 16&* + 63 6 + & 8 . 

7. sfi - 12 ax? + 240 cfix* - 192 a 6 x + 60 cfix* - 160 a 8 s 8 +64 cfi. 

8. a* - 18a 6 s + 136 a 4 x 2 - 640 a 8 * 8 + \2\bcfix* - 1468 ax 6 + 729a* 

Find the eighth roots of the following expressions : 

9. a*-8xV+28a*y 2 -66aJ 6 y 8 +70^^-66« 8 y 6 +28a; 2 j/ 6 --8a;y 7 +y 8 . 

10. {&* + 2(p - l)a* + (p 2 - 2p - l)aJ2 - 2(p - l)x + 1}*. 

Find to four terms the square root of 

11. o 2 -a;. 13. x 2 + a 2 . 13. a* -8**. 14. 9a 2 + 12as, 

Find to three terms the cube root of 

15. l-6s + 21x 2 . 16. 27 a« - 27 s 8 - 18 s*. 17. 64-48s + 9x 2 . 

16. Find the fifth root of 

a M - 10a 9 + 60a 8 - 160 cfi + 360 cfi - 692 a 8 + 720a* 

- 640 a 8 + 400 a 2 - 160a + 32. 

* 

19. a*> - bcfi + 20a 8 - 60a 7 + 106a 8 - 161a 8 + 210a* - 200 cfi 

+ 160a 2 -80a + 32. 

90. a 10 + 6a* + 6a 8 - 10a T - 16a 8 -f 11 a 6 + 16a* - 10a 8 

-6a 2 + 6a-l. 

205. Square and Cube Root of Numbers. Before leaving 
the subject of Evolution it may be useful to remark that 
the ordinary rules for extracting square and cube roots in 
Arithmetic are based upon the algebraic methods we have 
explained in the present chapter. 

Ex. 1. Find the square root of 6329. 

Since 6329 lies between 4900 and 6400, that is between (70) 2 and 
(80) 2 , its square root consists of two figures and lies between 70 and 
80. Hence, corresponding to a, the first term of the root in the alge- 
braic process of Art. 198, we here have 70. 

The analogy between the algebraic and arithmetical methods will 
be seen by comparing the cases we give below. 

a 2 + 2 ab + 6 2 (a + b 6329(70 + 3 = 73. 

cfi 4900 

2a + 6 " 2'a& + &2 140 + 3 = 143 429 
2 ab + b 2 429 



188 ALGEBRA. 

Ex. 2. Find the square root of 53824. 

Here 53824 lies between 40000 and 00000, that is between (200)* 

and (300) a . 

a b c 
53824(200 + 30 + 2 = 232 
40000 
2a + 6...400 + 30 = 430 



2(a+6)+c ... 460+ 2 = 462 



13824 
12900 



024 
024 



Ex. 3. Find the cube root of 614125. 

Since 614125 lies between 512000 and 720000, that is between (80)* 
and (90) 8 , therefore its cube root consists of two figures and lies 

between 80 and 00. 

a + b 
614125(80 + 5 = 86. 
512000 
8a* = 3x(80) a =19200 
8x0x6 = 3x80x5= 1200 
6 2 = 5x5= 25 



20425 



102125 



102126 



206. We shall now show that in extracting either the 
square or the cube root of any number, when a certain num- 
ber of figures have been obtained by the common rule, that 
number may be nearly doubled by ordinary division. 

207. If the square root of a number consists of 2/1 + 1 
figures, when the first n + 1 of these have been obtained by the 
ordinary method, the remaining n may be obtained by division. 

Let N denote the given number ; a the part of the square 
root already found, that is the first n + 1 figures found by 
the common rule, with n ciphers annexed ; x the remaining 
part of the root. 

Then -y/N=a + X) 

.*. 2T= a 2 + 2ax + a?i 

• *-*=*+£■ a> 



• * 



2a 2a 



EVOLUTION. 189 

Now N—a* is the remainder after n + 1 figures of the 
root, represented by a, have been found; and 2 a is the 
divisor at the same stage of the work. We see from (1) 
that N—a* divided by 2 a gives x, the rest of the quotient 

required, increased by - — We shall show that -— is a 
H ' .2a 2a 

proper fraction, so that by neglecting the remainder arising 

from the division, we obtain x, the rest of the root. 

For x contains n figures, and therefore x* contains 2n 

figures at most ; also a is a number of 2 n -f 1 figures (the 

last n of which are ciphers) and thus 2 a contains 2 n -f 1 

figures at least ; and therefore — is a proper fraction. 

Erom the above investigation, by putting n = 1, we see 
that two at least of the figures of a square root must have 
been obtained in order that the method of division, used 
to obtain the next figure of the square root, may give that 
figure correctly. 

Ex. Find the square root of 200 to five places of decimals. 



200(17.02 
1 



27 
3402 



100 
180 



10000 
6804 



3106 



Here we have obtained four .figures in the square root by the ordi- 
nary method. Three more may be obtained by division only, using 
* X 1702, that is 8404, for divisor, and 3106 as remainder. Thus 

8404)31060(038 
30636 
13240 
10212 



30280 

27232 

8048 



190 ALGEBRA. 

And therefore to five places of decimals V290 = 17.02938. 

It will be noticed that in obtaining the second figure of the root, the 
division of 190 by 20 gives 9 for the next figure ; this is too great, and 
the figure 7 has to be obtained tentatively. 

208. If the cube root of a number consists of 2/? +2 figures, 
when the first n -f- 2 of these have been obtained by the ordi- 
nary method, the remaining n may be obtained by division. 

Let N denote the given number ; a the part of the cube 
root already found, that is, the first n + 2 figures found by 
the common rule, with n ciphers annexed ; x the remaining 
part of the root. 

Then -f/N^a + x; 

N=a? + Sa 2 x + Zax* + &\ 

AT* , rfi /jj* ajS 

3«r a oar 

Now N — a 3 is the remainder after n "4- 2 figures of the 
root, represented by a, have been found; and 3a 2 is the 
divisor at the same stage of the work. We see from (1) 
that jflT— a 8 divided by 3 a 2 gives x, the rest of the quotient 

Qtj Oft 

required, increased by — h^-~z' We shall show that this 

a 3ar 

expression is a proper fraction, so that by neglecting the 

remainder arising from the division, we obtain x, the rest of 

the root. 

By supposition, x is less than 10 w , and a is greater than 

x 2 10 2 * 

10»»-n. therefore — is less than ■— — ; that is, less than 

1 x 3 lO 8 * 1 

— : and — - is less than - — tt^-t^; that is, less than 

10' 3a 2 3xl0 4n+2, ' 

- hence — h-^— s is less than — + - — -r-? — r, and 



3 x 10" +1 ' a Sa 2 10^3 x 10 n+1 ' 

is therefore a proper fraction. 



CHAPTER XXII. 
The Theory of Indices. 

209. Hitherto all the definitions and rules with regard 
to indices have been based upon the supposition that they 
were positive integers ; for instance, 

(1) a u = a*a»a^> to fourteen factors. 

(2) a 14 xa 8 = a 14+8 = a 17 . 

(3) a u + a* = a u -* = a n . 

(4) (a 14 ) 8 = a 14 * 8 = a 42 . 

The object of this chapter is twofold; first, to give 
general proofs which shall establish the laws of combina- 
tion in the case of positive integral indices; secondly, to 
explain how, in strict accordance with these laws, intelli- 
gible meanings may be given to symbols whose indices are 
fractional, zero, or negative. 

We shall begin by proving, directly from the definition 
of a positive integral index, three important propositions. 

210. Definition. When m is a positive integer, a m 
stands for the product of m factors each equal to a. 

2U~ Prop. I. To prove that a m x a n = a™" 1 "*, when m and 
n are positive integers. 

By definition, a m = a • a • a ••• to m factors ; 

a" = a ' a • a ••• to n factors ; 
.*. a m x a n = (a- a • a ••• torn factors) x (a • a • a ••• to n factors) 
= a • a • a ••• to m -f n factors 
__ a »+n^ by definition. 

101 



192 ALGEBRA. 

Cob. If p is also a positive integer, then 

a m x a* x a? = a m+w+ * ; 
and so for any number of factors. 

4 

212. Prop. II. To prove that a m -+- a n = a"-*, when /n 
and /i are positive integers, and /n is greater than /i. 

~» . ~n am a- a 'a*-- to m factors 
a -s- a == — = 

a n a« a • (*••• to n factors 
= a • a • a ••• torn- n factors 
= a w -'\ 

213. Prop. III. To prove that (a m ) n = a mn , when /n and 
n are positive integers. 

(a m ) n = a m • a m • a m • • • to n factors 

= (a*a*a*" torn factors) (a* a 'a*" to m factors) ••• 
the bracket being repeated n times, 

— a'a*a»" to mn factors 



mn 



= a 

214. These are the fundamental laws of combination of 
indices, and they are proved directly from a definition which 
is intelligible only on the supposition that the indices are 
positive and integral. 

But it is found convenient to use fractional and negative 

indices, such as d* 9 a~ 7 ; or, more generally, a q , cr n ; and 

these have at present no intelligible meaning. For the 

definition of a m [Art. 210], upon which we based the three 

propositions just proved, is no longer applicable when m is 

fractional or negative. 

Now it is important that all indices, whether positive or 

negative, integral or fractional, should be governed by the 

same laws. We therefore determine meanings for symbols 
p 

such as a q , a~ n , in the following way : we assume that they 
conform to the fundamental law, a m xo" = a m+n , and accept 



THE THEORY OF INDICES. 193 

the meaning to which this assumption leads us. It will be 
found that the symbols so interpreted will also obey the 
other laws enunciated in Props, n. and in. 

p 

215. To find a meaning for a*, p and q being positive inte- 
gers. 

Since a m x a n = a m+ * is to be true for all values of m and 

n, by replacing each of the indices m and n by -?, we have 

a 9 x a 9 = a 9 9 = a 9 . 

p p p <bp p ?P + P ft? 

Similarly, a 9 x a 9 x a 9 = a 9 x a 9 = a 9 9 = a 9 . 
Proceeding in this way for 4, 5, ••• q factors, we have 

p p p qp 

a 9 x a 9 x a 9 ••• to q factors = a 9 ; 

p 
that is, (a 9 ) 9 = a*. 

Therefore, by taking the gth root, 

p 

a 9 =-Vtf, 
p 
or, in words, a 9 is equal to " the qth root ofaf" 

Examples. (1) x* = y/&. 

(2) J = */a. 

(8) 4* = V* 8 = V64 = 8. 

2. ft 2_i_5 8 

(4) a* xd* = s * = a*. 

a 2 a. 2 3a+4 

(6) ** x &* = A*" 4 "* = k~*-. 

(6) 3 ah* x 4 <A>£ = 12 J + h^ = 12 a*6$. 

216. To find a meaning for a . 

Since a m x a n = a m+n is to be true for all values of m and 
n, by replacing the index m by 0, we have 

. 

9 



a° x a n = a* 4 *" = a n • 



a" 



... a ° = ^-=l. 



a n 



Hence any quantity with zero index is equivalent to 1. 
Ex. x 6c x #-* = x*-c+c-& = jc° = 1. 
o 



1 



194 ALGEBRA. 

217. To find a meaning for a~". 

Since a m x a n = a*"*"" is to be true for all values of m and 
n, by replacing the index m by — w, we have 

or* x a* = a~* +n = a . 
But a°=l. 

Hence a~ n = — t 

a n 

and a n = — • 

a- n 

From this it follows that any factor may be transferred 
from the numerator to the denominator of an expression, or 
vice-versa, by merely changing the sign of the index. 

Examples. (1) x" 8 = — • 

(2) J T =y*=Vy- 

JT 2 
(3) 27-i=-i-=— 1 



27 | ,°/(27)2 y& 32 9 

218. To prove that a m -i-a n =a m ~ n for all values of m and n. 

a m -i-a n = a m x — = a m x a~ n 

a n 

= a m ~" n , by the fundamental law. 
Examples. (1) a s + a 6 = a 8 " 6 = a -2 = —• 

(2) c + c~% = c 1+ * = c^. 

(3) ob»-» -*- #»-« = «■-*- (•-«> = as 8 "*. 

219. The method of finding a meaning for a symbol, as 
explained in the preceding articles, deserves careful atten- 
tion. The usual algebraic process is to make choice of 
symbols, give them meanings, and then prove the rules for 
their combination. Here the process is reversed; the sym- 
bols are given, and the law to which they are to conform, 
and from this the meanings of the symbols are determined. 



THE THEORY OP INDICES. 195 

220. The following examples will illustrate the different 
principles we have established. 

3a~ 2 Sx 



Examples. (1) 
(2) 



5 arty 6 aty 

2q*xq*x6q~£ = 4 fl }+ f-fHH = i a -i = ± 
9a~*xai 3 3 3 « 



(4) 2^a + -~ + a$ = 2a* + 3a* + a* 
a"* 

= 6o 7 + a^ = a*(6 + a 2 ). 

EXAMPLES XXII. a. 
Express with positive indices : 

1. 2 x~K 6. -i-p 9. 2 x* X 3 x-K 14 v * » 

_2 6af* i 4 v^~ 

2. 3 a 3 . 10. U2 a~*. 

3. 4ar 2 a 8 . 7 3ar^ # u . ^ x x -i. 16 . _2_. 

4. 3 + a-*. ^^ 12. «-2 x -i^3x. Vr 

5. J- 8. £Cl 13. J-. 16. -£*. 

4a~ 2 6-« ^«s ^-i 

17. a- 2 af * -?- a- 8 . 18. ^/a" 1 -s- ^a. 19. ^<r» + tya\ 

Express with radical signs and positive indices : 
20. ob*. 26. — . 28. a~~* x 2 a'h •» a~K 



rf 



32. 



21. a*. 6 ~* _2 _i 3 « 

i , 29. xU2a J ' 

22. bx't £*. 33. 4_r* 

-1 '2 30. 7a~*x3a- 1 . of* 

23. 2a *. x 

24. JL. 27. -L. 81 . ^ ^ ^ 

2a l x ~5 a * #x 2 

35. ^/a 2 x^/a 8 . 38. yx+tyx*.' 

36. ^/a-'-^-^/a- 2 *. 39. 3 </a 8 -*-^/a 2 . 

37. ^/x x $te 2 . 40. J/a n x .J/a^^a 5 ". 



196 ALGEBRA. 

Find the value of 

41. 16*. 43. 125*. 45. 36~i. 47. 243$. 49. (fj)l. 

42. 4-1. 44. 8"*. 46. ^ 48. (ft)4. 60. tf&yl 

22L To prove that (a m ) n = a"" 1 is true for all values of m 
and n. 

Case I. Let n be a positive integer. 
Now, whatever be the value of m 

(a m ) n = a m >a m • a m • • • to n factors 

«m+m+«M— • to n terms __ •»»»» 

Case II. Let m be unrestricted as before, and let n be 

a positive fraction. Replacing n by — , where p and <? are 

positive integers, we have (a m ) n =(a w )«. 

Now the o/th power of (a m )*= \(a m ) 9 \ 9 =(a m ) q , [Case L] 

= (a m y=a m *. [Case L] 

Hence by taking the o/th root of these equals, 

(a w ) « = ya mp = a « . [Art. 215.] 

Case III. Let m be unrestricted as before, and let n be 
any negative quantity. Replacing n by — r, where r is posi- 
tive, we have 

(a m ) n = (a-)-" = t^-9 [Art. 217.1 

( a "0 r 

= -^, [Case II.] 

a" 1 ** 



— WW" _ /fflWt 



= a-™ r = a 



Hence Prop, m., Art. 213, (a m ) n = a mn has been shown to 
be universally true. 

Examples. (1) (6*)* = 6* x * = 6*. 

(2) {(ar 2 ) 8 }" 4 = (ar 6 )- 4 = *«♦. 



(8) (x^) a2 - c, = x^ x(aW) = a^. 



THE THEORY OF INDICES. 197 

222. To prove that (ab) n = a n b n , whatever be the value of 
i? ; a and b being any quantities whatever. 

Case I. Let n be a positive integer. 

Now (ab) n = ab • ab-ab • • • to n factors 

= (a • a • a • • • to n factors)(& • b • b • • • to n factors) = a n b n . 

•' 

Case II. Let n be a positive fraction. Replacing n by ^, 

where p and q are positive integers, we have (a6) n = (o6) *. 

Now the gth power of (a&)« = j(a&)« }*=(<*&)*, [Art. 221.] 

= a'6 * = (a* b 9 ) . [Case I.] 
p p p 
Taking the gth root, (ab) q = a"b q . 

Case III. Let n have any negative value. Replacing n 
by — r, where r is positive, 

(abY == (ab)~ r = -^- = — = ar+b-* = a n 6 w . 

Hence the proposition is proved universally. 
This result may be expressed in a verbal form by saying 
that the index of a product may be distributed over its factors. 

Note. An index is not distributive over the terms of an expres- 
sion. Thus (a * + b*) 2 is not equal to a + b. Again (a 2 + 6 2 )* is 
equal to Va 2 4- b 2 , and cannot be further simplified. 

Examples. (1) (yz)*- e (zx) e (xy)- e = y a -cz a ~ e z e x e x- c y- e = y"- 2 *^ . 

(2) {(a - &)*}-« x {(a + b)-*Y = (a - &)-» x (a + &)-« 

= {(a - 6) (a + &)}-* = (a 2 - 6 2 )~". 



I Since in the proof of Art. 222 the quantities a 
and b are wholly unrestricted, they may themselves involve 
indices. 

Examples. (1) (x' l y~*y ■*- (x'fy" 1 )"* = x : &~ J -*- ac"*y*=a5*y- 1 . 

(2) f«L^ + ji^y = f5^ + .^v 



198 ALGEBRA. 

EXAMPLES XXII. b. 
Simplify and express with positive indices : 

2. (^FV)" 8 . " ^ 8a "V ' 9. (4a-2-9x 2 )~*. 






12. (^/a*-s- ^/a;) 1 -- 

3i 



27. ( 5 i2^ x J^y*. 



13. Vo 1 ^ x Va6^. , y~ 

14. toft-V* x (a-^-^c-*)"*. 

15. \/a*W x (a V 1 )" 6 . ^(a 8 68 + q 6 ) 

16. v^f - v^x. " v^fc 6 - a«6«)-i 

17. (a~*^a)-« x V5FV5F- 29. ( a «'-ivA + ^. 

, i _i a 

18. ^a»+*& 2 "* + (<*»& »)*. n te 

19. ^W 6 x (a + &)-». 8a (a^ri)^ + ^f-- 

20. {(a:-2/)-8}«^{(x + y)»} 8 . 81 {JtL Xfl i(„,| ,, 1 



fyafi 2 -™ 



1 



U 8 6-v ' U- 8 &v . ,_*_„, 

, a 32. C^- 1 ) 6 -!^-^) 



23.' (a U*Vax ^x^)i 



24 
25. 



. v^r+w x c« 2 - & 2 r* • 34. fj^y^fcjEiy 

/ «" 8 \~ 3 /\ «~*- ^/6 8 \- J 

trf "V a2c -i ' 35. txfl=2=xJL 

x & 3 c' 2" +1 x 2 B_1 4~* 

M . (°-*Jy + &*. 86 . *•« 4» + « 

\ x a j x 



t. Since the index-laws are universally true, all the 
ordinary operations of multiplication, division, involution, 
and evolution are applicable to expressions which contain 
fractional and negative indices. 



THE THEORY OF INDICES. 



199 



In Art. 200, we pointed out that the descending 
powers of x are 

• •• x*, x*, x, 1, -, -, —, .... 

x or or 

A reason is' seen if we write these terms in the form 

• . • 3/ « X • X . 2/ . »C f*v i*'' • " * * • 
_1 



Ex. 1. Multiply 3 x~* + * + 2 x* by X* - 2. 
Arrange in descending powers of x. 

x + 2x* + 3x~^ 
s*-2 

x* + 2x +3 

— 2a; — 4 x* — 6 x * 

x^-4x* + 3 -6x"i 

Ex. 2. Divide 16 «-» - 6 a" 2 + 5 a" 1 4- 6 by 1+2 a" 1 . 

2ar 1 + l)16a-8- 6«- 2 -f 5a" 1 + 6(8 a" 2 -7a" 1 + 6 
16a- 8 + 8<r* 



-14a-2+ 6<r l 
-14a" 2 - 7a- 1 



12 ar l + 6 
12 a- 1 + 6 



Ex. 3. Find the square root of 



— + ^ 8 -2s + ^+a*-4V(afy- 1 ). 



y 



y 



rl 



Use fractional indices, and arrange in descending powers of x. 



-J. r 



x8_ 4xV*+ 4xV 1 + «M- 2x + |(x*- 2xy 2 + 

X* 



2x*-2xy~* 



i-wi^ 



— 4x^T ? +4x 2 y- 1 
~4x^~ 2 + 4x 2 y" 1 



2x*-4xy * + 



2 



xM-2x + | 
A*- 



»-+f 



Note. In this example it should be observed that the introduction 
of negative indices enables us to avoid the use of algebraic fractions. 



200 ALGEBRA. 

EXAMPLES XXTT. o. 

1. Multiply 3x* - 6 + 8x~* by 4x* + 3x~* 

2. Multiply Sa s - 4a^ - a~* by 3a* + a~* -6a~l. 

3. Find the product of c* + 2 c~* - 7 and 5 - 3 c~* + 2 c*. 

4. Find the product of 5 -f 2 x 2- -f 3 x"* 20 and 4 x* — 3 ar» 

5. Divide 21x + x* + x* + 1 by 3x* + 1. 

6. Divide 16a - 3a* - 2a~* + 8a" 1 by 6a* + 4. 

7. Divide 16 a~ 8 + 6 a" 2 + 6 a" 1 - 6 by 2 a' 1 - 1. 

8. Divide 66* - 66 ' - 46"^ -46"5 -6 by 6*-26~* 

9. Divide 21 a 8 " + 20 - 27a* - 26a 2 * by 3a* - 6. 

10. Divide 8c~ n - 8c n + 6c 8 * - 3c-*» by 6c* - 3c-». 

Find the square root of 

11. 9x - 12 x* + 10 - 4x~* + x" 1 . 

12. 25 a* + 16 - 30a -24 a* + 49a*. 

13. 4x»H-9x-" + 28-24x~" -16x* 

14. 12a* + 4 -6a 8 * + a 4 * + 6a 2 *. 

16. Multiply a'-8a~^ + 4a"*-2a* by 4 a~* + a* + 4 a~*. 

16. Multiply 1 - 2 ,J/x - 2x* by 1 - {/«• 

17. Multiply 2 v^- a* --by 2a -3^ -a"*. 

a \a 

18. Divide #*•* + 2x* - 16x~*- — by x* + 4x~* + — . 

X y/X 

19. Divide l-V«-A + 2 « 2 by 1-ai 

a" 1 

* 

20. Divide 4 v^-8x* - 5 + ™ +3x~* by 2 a* - 2/x - -JU 
Find the square root of 

21. 9x-*-18x-Vy+^-6^(g) + !^. 

22. 4V^-12^(x 8 y)+26Vy-24^f^+16x"*^ 

23. 8lfi^+0 + 36 ^-(*V 1 -l)-168 : ^- 

V y 2 I y/y v 

W ^r 2 2x vy 



THE THEORY OF INDICES. 201 

>. The following examples will illustrate the formulae 
of earlier chapters when applied to expressions involving 
fractional and negative indices. 

* p _h _p * * * J> * P P P 

Ex. 1. (a*-6*)(a * + ft «)=a*~*-a ~ k b*+ah •-&»"« 

_* t * _? 

=za*b~*-a *&*. 

Ex. 2. Multiply 2x** - a* + 3 by 2x** + a* - 8. 
The product = {2 x 2 * - (a* - 3)} {2 x 2 * + (a* - 3)} 

= (2a5 2 ?) 2 —(«*» — 3) 2 = 4x 4 * - x 2 p + 6a* - 9. 

Ex. 3. The square of 3 x* - 2 - x~^ 

- 9x + 4 + x" 1 - 2 • 3x* . 2 - 2 . 3x- • x~^ + 2 . 2 .x~* 

= 9x + 4 + x" 1 - 12x* - 6 + 4x~^ 
= 9x - 12 x* - 2 + 4x~* 4- ar 1 . 

8n _8n » n 

Ex. 4. Divide a* + a * by a 2 + a" 5 . 

3n _3n n n 

The quotient = (a T + a~ y ) -*- (a 2 + a *) 

n n n n 

= {(«*)» + (a M + («* + a" 1 ) 

n n n n 

= (a 2 ) 2 -a 2 .a * + (a *) 2 = a» - 1 + or*. 

EXAMPLES XXII. d. 
Write the value of 

1. (x*-7)(x* + 3). 3. (7x-9r 1 )(7a + 92Ti). 

2. (4x-6x- 1 )(4x + 3ar 1 ). 4. (x~ - y n )(x~ m + y n ). 

1 a 

5. (a* -2 or*) 2 . 6. (a*+fl£) 2 . 7. (x2-Jx-«) 2 . 

8. (5 x°2/* - 3 x-°2T 6 ) (4 x«y* + 5 x~ a y h ), 

9. (ia*-a"b 2 - 10. (3x a y- 6 + 5x- a y 6 )(3x a y l -5x-«y- 6 ). 



202 ALGEBRA. 



11. (o*_ l-<r*)* is. {(a + 5)* 4- C« — ft)*} 2 - 

12. (a* - x~5 + x)«. 14. {(a + &)* _ (a - &)~ty. 



Write the quotient of 

15. x - 9a by x* + 3a*. 20. 1 - 8a~« by 1 - 2a* 1 . 

16. x*-27byx*_3. »■ ^-^y^ + s 8 . 

17. a**-16bya*-4. »' ^ " J ^ ^ + l 

18. x*» + 8 by x* + 2. 28. x$ - 1 by x* - 1. 

19. &-C-* by C-<H1. »*• * 6 " + 32 by x* + 2. 



n + ( i+ ^y 



Find the value of 

25. (x + x*-4)(x + x* + 4). ^ x-7x* , /i , 2 \-i 

26. (2x*+4+3x~J)(2x*+4-3x~*). X ~ ^ % ' 

27. (2-x* + x)(2 + x* + x). 8L g f - 4 for 2 

28. (a* + 7 + 3a"«)(a«-7-3a-*). ' fo; 2 + 4 + 4x~* 

a* + 2\/a& + 4&« " a& - ft 8 x/a-b 



CHAPTER XXIII. 

Surds (Radicals). 

T. A surd is an indicated root which cannot be exactly 
obtained. 

Thus V 2 > V$> J/a s , Va 2 + b 2 are surds. 

By reference to the preceding chapter it will be seen that 
these are only cases of fractional indices; for the above 
quantities might be written 

2*, 5^, J, (a 2 + b*)k 

Since surds may always be expressed as quantities with 
fractional indices they are subject to the same laws of 
combination as other algebraic symbols. 



A surd is sometimes called an irrational quantity; 
and quantities which are not surds are, for the sake of 
distinction, termed rational quantities. 



Surds are sometimes spoken of as radicals. This 
term is also applied to quantities such as Vo 2 , V9, V27, 
etc., which are, however, rational quantities in surd form. 

230. The order of a surd is indicated by the root symbol, 
or surd index. Thus -y/x, -%/a are respectively surds of the 
third and nth orders. 

The surds of the most frequent occurrence are those of 
the second order ; they are sometimes called quadratic surds. 
Thus -y/^ V a > Va-j-y are quadratic surds. 



A mixed surd is one containing a factor whose root 
can be extracted. 

203 



204 ALGEBRA. 

This factor can evidently be removed and its root placed 
before the radical as a coefficient. It is called the rational 
factor, and the factor whose root cannot be extracted is 
called the irrational factor. 



I. When the coefficient of the snrd is nnity, it is said 
to be entire. 



When the irrational factor is integral, and all rational 
factors have been removed, the surd is in its simplest form. 



When surds of the same order contain the same 
irrational factor, they are said to be similar or like. 

Thus 5V3, 2V3, |V 3 are like surds - 

But 3^/2 and 2^/3 are unlike surds. 



i. In the case of numerical surds such as ^/2, ^/5, •••, 
although the exact value can never be found, it can be deter- 
mined to any degree of accuracy by carrying the process of 
evolution far enough. 

Thus v 5 = 2 - 236068 ---; 

that is V 5 lies between 2.23606 and 2.23607 ; and therefore 
the error in using either of these quantities instead of -y/5 is 
less than .00001. By taking the root to a greater number 
of decimal places we can approximate still nearer to the true 
value. 

It thus appears that it will never be absolutely necessaiy to 
introduce surds into numerical work, which can always be 
carried on to a certain degree of accuracy ; but we shall in 
the present chapter prove laws for combination of surd 
quantities which will enable us to work with symbols such 
as y2, -y/5, -y/a, ••• with absolute accuracy so long as the 
symbols are kept in their surd form. . Moreover it will be 
found that even where approximate numerical results are 
required, the work is considerably simplified and shortened 
by operating with surd symbols, and afterwards substituting 
numerical values, if necessary. 



SURDS. 205 



REDUCTION OF SURDS. 

236. Transformation of Surds of Any Order into Surds of e 
Different Order having the Same Value. A surd of any order 
may be transformed into a surd of a different order having 
tne same value. Such surds are said to be equivalent. 

Examples. (1) ^J/2 = 2* = 2 T *= ^2*. 

(2) tya = ap = <m= ^a*. 

237. Surds of different orders may therefore be trans- 
formed into surds of the same order. This order may be 
any common multiple of each of the given orders, but it is 
usually most convenient to choose the least common mul- 
tiple. 

Ex. Express ^J/a 8 , -fyb 2 , {/a 6 as surds of the same lowest order. 
The least common multiple of 4, 3, 6 is 12 ; and expressing the 
given surds as surds of the twelfth order they become l^a 9 , y/b* t 



Surds of different orders may be arranged according 
to magnitude by transforming them into surds of the same 
order. 

Ex. Arrange ^3, (/6, ^10 according to magnitude. 
The least common multiple of 2, 3, 4 is 12 ; and, expressing the 
given surds as surds of the twelfth order, we have 

^3 =»/3« = ^729, 
*/6 = »/0* = 1^1296, 
^10 = ^J/10 8 = ?J/1000. 

Hence arranged in ascending order of magnitude the surds are 

V3, </10, «/6. 

EXAMPLES XXIII. a. 
Express as surds of the twelfth order with positive indices : 
1. a& 2. a- 1 -*- <T*. 3. Vox* x v^vT 






•7/1-14 \ a -i 



206 ALGEBRA. 

Express as surds of the nth order with positive indices : 
7. V*. 9. «*. U. 1Q; IZ. £* 

\a a -l x -n 

Express as surds of the same lowest order : 

15. V«> v^ 6 - 18 - $**. \^°- **• >A ^11, ^13. 

16. ^/a 8 , V«- 19 - *VaW, Vab* 22. <{#, V 3 » \/ 6 - 

17. #« 3 , #a«, ^e 6 . 20. Vas*, v^. 23. ^2, •#, ^4. 

239. Reduction of a Surd to its Simplest Form. The root 

of any expression is equal to the product of the roots of 
the separate factors of the expression. 

For Vab = (ab)n = a «&S [Art. 222.] 

Similarly, Vabc = -y/a • -y/b • -y/c ; 

and so for any number of factors. 

Examples. (1) ty\b = f/S .^5. 

(2) */<fib = tya* . f/b = a*yb- 

(3) V 5 ^ =V26.V2 = 5V2. 

Hence it appears that a surd may sometimes be expressed as the 
product of a rational quantity and a surd ; when the surd factor is 
integral and as small as possible, the surd is in its simplest form [Art. 
233] . 

Thus the simplest form of yl28 is 8^/2. 

Conversely, the coefficient of a surd may be brought under the 
radical sign by first raising it to the power whose root the surd ex- 
presses, and then placing the product of this power and the surd factor 
under the radical sign. 

Examples. (1) 7 y/b = V*9 • V 6 = V 246 - 
(2) afyb = ya* '-*/b = </a*b. 

In this form a surd is said to be an entire surd [Art. 232]. 
By the same method any rational quantity may be expressed in the 
form of a surd. Thus 2 may be written as V5, and 3 as v^27. 



SURDS. 207 

240. When the surd has the form of a fraction, we mul- 
tiply both numerator and denominator by such a quantity 
as will make the denominator a perfect power of the same 
degree as the surd, and then take out the rational factor as 
a coefficient. 

Examples. (1) VJ = Vf = V2x~i = £V2. 

/on ^Imx ^ labmx 1 . 

(2) \^-A(-^i--^v^^. 

EXAMPLES XXIII. b. 
Express in the simplest form : 

1. V 288 - 6. 2^720. 10. v'- 2187. 14. \f. 

2. V 147 - 7. 5V245. 11. VMa*. /o^ 

3. .J/256. . 15. -*/-£. 

4. #432. 8 * ^ 1029 - 12 ' ^™* 6 ' * b 

5. 3V150. »• ^ 126 « 13 - ^f !«• V&- 



17. V- 108xV- 18. ^x^ 2 "* 6 . 19- ^x«+^. 

20. Va 8 + 2 a?b + ab 2 . 21. v^8 tfy - 24 xfy 2 + 24 x 2 y 8 - 8 jc^. 

Express (1) as entire surds, (2) in simplest form : 



23. 14 V&- x 2 * a b *«* 

** 6 ^ 4 ' sin 2 « a £7x^ <u JL*]^ 

25. 6^. 

26. AV?. 



Sab 120c 2 81 - — 

^ 2c *9a 2 & 

28 






^0^ 31. f^JL-. so. <x + y;^— -* 

a/^-2-. b *8a 8 '* + y 



*2 « 



2 -X 2 

<&% 2 



y * x 



ADDITION AND SUBTRACTION OF SURDS. 

241. To add and subtract like surds : Reduce them to their 
simplest form, and prefix to their common irrational part the 
sum of the coefficients. 



208 ALGEBRA. 



Ex. 1. The sum of 3 V 20 *. 4 V 6 » — 

5 

o o 

Ex. 2. The sum of x V%x*a + y v'- y*a - z Vz*a 

= x*2xy/a + y(—y)-f/a — Z'Zy/a 
= (2s 2 -y 2 ~s 2 )^a. 

242. Unlike surds cannot be collected. 

Thus the sum of 5V2, -2y3, and V 6 is 5^/2-2^/3 
+ V6, and cannot be further simplified. 

EXAMPLES XXIII. c. 
Find the value of 

1. 3 V*5 - V20 + 7 y/b. 7. 3 */\Q2 - 7 */32 + ,J/1250. 

2. 4V63 + 6 V7-8V28. 8. 6 #^54 - 2 \^Tl6 + 4 ,J/686. 

8. v 44 - 6 \/ 176 + 2 V 99 - •• 4 V 128 + 4 V 75 ~ 6 V 162 - 

4. 2 ^363 - 6 V 24 3 + V 192 - 10. 6 V 24 - 2 V 54 - V 6 - 

5. 2^/189 + 3^875-7^/56. 11. ^252 - ^294 - 48 Vi- 

6. 5^81-7^/192 + 4^/648. 12. 3 V147 - £ VI ~ V*V- 



MULTIPLICATION OF SUKDS. 

243. To multiply two surds of the same order: Multiply 
separately the rational factors and the irrational factors. 

i i ii 

For a -y/x x b -y/y = ax n x by" = abx n y n 

i 

= ab (xy) n = ab Vxy. 
Examples. (1) 5 ^3 x 3 y7 = 15 V 2 1- 

(2) 2y/XX Sy/X = 6X. 

(3) </a + b x Va^b = ^(a + 6)(a-6)= v'a 2 - 6 2 . 

If the surds are not in their simplest form, it will save 
/abor to reduce them to this form before multiplication. 

Ex. The product of 5 V^ 2 , V 48 > 2 V 64 
= 6 • 4 V 2 X 4 ^3 X 2 • 3 V6 = 480 • V 2 ' V 3 ' V 6 = 480 X 6 = 2880 



SURDS. 209 

244. To multiply surds of different orders: Reduce them 
to equivalent surds of the same order, and proceed as before. 

Ex. Multiply 6 ^2 by 2 ^/b. 

The product = 5 ^/2 2 x 2^/5 s = 10 ^2 2 x \/p = 10 £/500. 



EXAMPLES XXIII. d. 



Find the value of 



1. 2V14XV21. 6. Vx + 2xVx-2. 1Q 2^/g ^ 

2. 3V8XV«. 7. </l68x#147. J ,- '- 

4. 2 V 15x3 V 5. 8.5^128x2^432. «^V. 3^2 a* 

5. 8^/12x3^24. 9. aVP x 6 2 Va. 12. i^x|^- 



DIVISION OF SURDS. 

Suppose it is required to find the numerical value of 
the quotient when ^/5 is divided by ^/7. 

At first sight it would seem that we must find the square 
root of 5, which is 2.236- • -, and then the square root of 7, 
which is 2.645- •-, and finally divide 2.236- •• by 2.645- ••; 
three troublesome operations. 

But we may avoid much of this labor by multiplying both 
numerator and denominator by y7, so as to make the 
denominator a rational quantity. Thus 



V*>_V? V7_ V5x7 _V35 
V 7 ~V 7>< V7~~ 7 7 

Now V 35 = 5 - 916 - 

m . m V5 = 5,916-- 

246. The great utility of this artifice in calculating the 
numerical value of surd fractions suggests its convenience 
in the case of all surd fractions, even where numerical 



210 ALGEBRA. 

n /h 

values are not required. Thus it is usual to simplify — y- 
as follows : * 

a^/b __ a-y/b x V c __ « V&c # 

The process by which surds are removed from the denomi- 
nator of any fraction is known as rationalizing the denomina- 
tor. It is effected by multiplying both numerator and 
denominator by any factor which renders the denominator 
rational. We shall return to this point in Art. 250. 

247. To divide surds : Express the result as a fraction and 
rationalize the denominator. 

Ex. 1. Divide 4V75 by 26^56. 

The quotient =ht™= 4 * 5 V 3 =h£ 
4 26V56 26.x 2^/14 5^/U 

__ 2 y3 x yu _ 2 y42 _ y42 

~by/UXy/li"bx 14" 35* 

^_ ^x> _ Vbc_ Vbc 
ty& ~ tyc* x {/c - f/<fi " c 

EXAMPLES XXIII. e. 
Find the value of 

1. V^-^V 2 - 4 - 21^84-4-8^/98. 7. 6^/14^2^21. 

2. 3V7-2V8. 5. 5V27h-3V24. 3^11 5 

8. 2V120-S-V 3 - 6 - -l 3 V 125 -*- 5 V 65 - * 2V98 "*" 7V22* 

3y48 . 6^84 1Q 3 PHT , J~18g~ . 

"' 6VH2 ' V 392 ' ' a-b^a-b ' *(a-6) 6 

Given ^2 = 1.41421, V 3 = 1-73205, V5 = 2 - 23 607, V 6 = 2 -*4949 
y/1 = 2.64575 : find to four decimal places the numerical value of 

18. 1 21 . _?5_. 
2^/3 y262 

19 . _U 83. ^ 



11. 


14 

V 2 ' 


14. 


48 


12. 


25 
V6 


15. 


60 


is. 


1C 


16. 
17. 


144 -s- V 6 - 

V2-5- V 3 - 



V500 \1576 

_1_ 28. -5-^ 

V243' 2 VS6 



SURDS. 211 



COMPOUND SURDS. 

248. Hitherto we have confined our attention to simple 
surds, such as -y/5, tya, -y/x + y. An expression involving 
two or more simple surds is called a compound surd ; thus 
2-y/a — 3y& ; -fya -f- -y/b are compound surds. A binomial, 
which has a surd in one or both of the terms, is called a 
binomial surd. 

249. Multiplication of Compound Surds. We proceed as 
in the multiplication of compound algebraic expressions. 

Ex. 1. Multiply 2y/x-b by S^/x. 

The product = 3^/x(2 ^/x - 5)= 6x - l&^/x. 

Ex. 2. Multiply 2 y/b + 3 ^/x by y/b- y/x. 
The product = (2^/5 + 3^/a) ( V 5 - V x ) 

= 2^5 • V 6 + 3 V 5 ' V x ~ 2 V 6 • V x ~ 3 V aj ' V* 
= 10~3x + \/5x. 

Ex. 8. Find the square of 2y/x + V7 — 4 x. 

(2 V* + V7 - 4*) 2 =(2V*) 2 + ( V7 - 4x)2 + 4 V* . V7-4x 

= 4x4-7 -4x + 4V7x-4x 2 
= 7 + 4V7x-4x 2 . 

EXAMPLES XXIII. f. 
Find the value of 

1. (3v/x-5)x2Vx. 8. (Zy/a-2y/x)(2y/a + Sy/x). 

2. (y/x — v 'a) x 2y/x. 9. ( V x + Vx — 1) x Vx — 1. 

8. ( V« + V 6 ) x ^ aly - 10. (Vx + a- Vx-a) x Vx-f a. 

4. (Vx + y- l)x Vx + y- 11- (Va + x-2Va) 2 - 

5. (2V3 + 3V2) 2 . 12. (2Va-Vl + 4a) 2 . 

6. ( v /7 + 5V3)(2V7-4 v /3). 18. ( VaT+x - V5"^x) 2 . 

7. (3 v /5-4v2)(2 v /5 + 3 v '2). 14. (Va + x - 2)(Va~+~x' - 1). 

15. (V 2 + V 3 -V 5 )(V2 + V 3 + V 5 )- 

16. (V& + 3 V2+V 7 )(V 5 + 3 V 2 -V 7 )- 



212 ALGEBRA. 

Write the square of 

17. V2x + a-V2s-a. 20. 3Va 2 + 6 2 - 2Va 2 - ft 2 . 

18. Vs 2 -2y 2 + Va; 2 + 2y 2 . 21. 3x^/2 - 3 ^7 -2« 2 . 

19. Vw + n + Vj»- ». 22. V4x 2 -f 1 - V4x 2 -1. 

250. If we multiply together the sum and the difference 
of any two quadratic surds, we obtain a rational product 
This result should be carefully noted. 

Examples. (1) (V« + V 6 )(\A* - v /& ) = (V ) 2 -(V&) 2 = « - &• 
(2) (3V5 + 4 v3)(3V5 - 4 V3) = (3 V6) 2 - (4 V3) 2 = 45 - 48 = - 3. 
Similarly, (4- VcTiT)(4+ Vo+6) = (4) 2 - (Vo+6) 2 =16-a-6. 

251. Definition. When two binomial quadratic surds 
differ only in the sign which connects their terms, they are 
said to be conjugate. 

Thus 3 y7 + 5 yll is c onjugate to 3 y7 - 5 yll. 
Similarly, a — Va 2 — ic 2 is conjugate to a -f Va 2 — a 2 . 
2%e product of two conjugate surds is rational. [Art. 250.] 

Ex. (3 y/a + Va - 9 a) (3 V<* - Vx-9a) 

= (3Va) 2 -(Vx-9a) 2 = 9a-(x-9a)=18a-x. 

252. Division of Compound Surds. If the divisor is a 
binomial quadratic surd, express the division by means of 

*a fraction, and rationalize the denominator by multiplying 
numerator and denominator by the surd which is conjugate 
to the divisor. 

Ex. 1. Divide 4 + 3^2 by 5 - 8^/2. 

The quotient = 4 + 3 V 2 = 4 + 3 V 2 x 6 + 8 V 2 

5-3V2 6-3V2 6 + 3^2 

__ 20 ■+ 18 + 12 V2 + 15 y/2 = 38 4^27 y/2 

26-18 7 

Ex. 2. Rationalize *he denominator of - • 



Va 2 + b 2 + a 



The expression b * x V « 2 + & 2 -a 

Va 2 + & 2 + a Va 2 + 6 2 - a 

(a 2 + ft 2 ) - o 2 



SURDS. 213 

Ex. 3. Divide ^— ^- by -^-^. 
The quotient =^^x^^ 

_ (V 3 ) 2 -(V*) 2 

""14-12 + 8V3-7V3 

= = 2 — -y/3, on rationalizing. 

2 + y/S 

Ex. 4. Given ^6 = 2.236068, find the value of 87 



7-2^5 
Bationalizing the denominator, 

87 _ 87(7-f2y5) 
7-2V5" 49-20 
= 3(7 + 2 V5) 
= 34.416408. 

It will be seen that by rationalizing the denominator we have 
avoided the use of a divisor consisting of 7 figures. 



In a similar manner, where the denominator involves 
three quadratic surds, we may by two operations render that 
denominator rational 



Ex. 



V2 



V2 + V 3 - V 6 

The expression = V^C V 2 + V 3 -f V 5) 

P (V2 + V3 - V&)( V 2 + V 3 + V5) 

_ 2 + y6 + yio _ (2-f v6 + yio)y6 

~~ 2^6 ~~ (2y6;v6 
_ 3 + y6 + yi5 

~ 6 

EXAMPLES XXIII. g. 
Find the value of 

1. (9V2-7)(9V2+7). 4. (2^11+5^)^11-5^2) 

2. (3 + 5V7)(3-6V7). 5. (V« + 2y/b)( y/a - 2^/b). 
8. (5V8-2V7)(6V8 + 2V7). 6. (Sc-2y/x)(3c + 2y/z). 

7. (Va+x— V«)(Va+a!+V a )' 

8. (V2j> + 3g-2V7)(V2/)+3g + 2 v /g). 

9. (Va + « + Va — sc) ( Va -f x — Va — a:). 
10. (6 Vx 2 - 3 y 2 + 7 a)(5Va;* -3^-7 a). 



214 ALGEBRA. 

11. 29^(11 + 3^/7). 16. (3 + V 5 )(V6-2)-K5-V5) 

12. 17 + (3V7 + 2 V3). yg , V« + ys 

13. (3^/2-1) -?- (3^2 + 1). yo-yx"^ ^ 

14. (2V3+7V2)-f-(5v«-4V2). lg 2yi5 + 8 , 8y3-6y5 

15. (2«-Vxy)^(2Vxy-y). " 6+ yi6 : 5y3-3y5* 

Rationalize the denominator of 

19 25y3-4y2 21 V7+ y2 ^ y 2 

7V3-5V2' ' 9 + 2yl4 # ' x + y / x 2_^2 

1 0y6 - 2 y7 gg 2y/3 + 3y2 ^ a* 



20. 



3V6 + 2V7 6 + 2 y6 Vx* + a*+a 

VI + x 2 - VI - x 2 „a 3+y6 



25. vx ^" — — — — . 28. 



VTT^ + VT^x 2 6V3-2V12-V32+V60 



2 Vq + 6 + 3Va - 6 ylO + y5 + y3 

2Vo+~&-V^& " V 3 + V 10 -V 6 



27 V9+^-3 (V3+y5)(V5 + y2) 

V9Tx2 + 3' ' V2+V3+V^ 

Given y2 = 1.41421, y3 = 1.73205, V 6 =2.23607 : find to four 
places of decimals the value of 

M. — L_. 32. ? + & 88. V 6 + V, 3 . 84. V5 - » . 



2+V3 V^-2 4 + V 16 9-4^/6 

86. M+"x3^4. 86. (2-V3)(7-4V3) + (3v8-6). 

-y/O — 1 3 + yO 

254. To find a factor which will rationalize any binomial 
surd. 

Case I. Suppose the given surd is -tya — -tyb. 
Let -{/a = x, -f/b = y, and let n be the L. C. M. of p and 
q ; then x n and y n are both rational. 
Now x n — y* is divisible by x — y for all values of n, and 

x n — y n =(x — y)(x n l -f #*- 2 y -f of 1-3 ^ H r-y*" 1 ). 

Thus the rationalizing factor is 

and the rational product is a** — y". 



SURDS. 215 

Case II. Suppose the given surd is tya + -f/b. 
Let x, y, n have the same meanings as before ; then 

(1) If n is even, x" — y n is divisible by x + y, and 
x n — t/ n =(x + y) (x*- 1 — x*~ 2 y + ••• + xy n ~ 2 — ^*" 1 ). 

Thus the rationalizing factor is 

af-i _ # n_2 y -f- --. -f xy n ~ 2 — y n ~ 1 ) 
and the rational product is x* — y n . 

(2) If n is odd, x n + y n is divisible by a? + y, and 

<c n + y n = (a; + y) (af 1-1 — # w ~ 2 y + ... — o^/** - 2 + y 1 ). 
Thus the rationalizing factor is 

a*- 1 — o^^y + . . . — a^*" 2 + y 1 " 1 ; 
and the rational product is a? n -f- y n . 

Ex. 1. Find the factor which will rationalize y/S + <{/5. 
Let x = 3% y = 5* ; then x 6 and t/ 6 are both rational, and 
x 6 - tf> = (x -f y) (x 5 — x*y + xhp — x 2 y s + xy* — y 6 ); 
fchus s substituting for x and y, the required factor is 

3$ — 3^ . 5! + 3$ . 5I _ 3$ . 5! + si . 5! __ 5! f 

or 3^ - 9 . 5* + 3^ . 6* - 16 + 8* • 5* - 6$; 

and the rational product is 3* - 5$ = 3 8 - 5 2 = 2. 

Ex. 2. Express (6* + 9*) -=- (5* - 9*) 

as an equivalent fraction with a rational denominator. 

To rationalize the denominator, which is equal to 5* — 3*, put 

6* = x, 3* = y ; then since 

ac* - y* =(a - y)(x* + xhj + xy* + y«), 

o 2 1 T 2 3 

the required factor is 6* + 5* • 3* + 5* . 3* + 3* ; 

and the rational denominator is 6^ — 3* = 5 2 — 3 = 22. 

+ , . (5* + 3*) (5$ + 6* • 8* + 6* • 3^ + 3*) 
. \ the expression = A^_l — /\ v ^ _l _l — L 

22 

5* + 2 . 5^ . 3* + 2 . 5^ . 3* + 2 . 5* ■ 3^ + 3^ 

22 

14 + 5^ . 3* + 5 • 3* + 5* - 3* 
11 



216 ALGEBRA. 

PROPERTIES OF QUADRATIC SURDS. 

255. The square root of a rational quantity cannot be partly 
rational and partly a quadratic surd. 

If possible let -y/n = a + V m > 

then by squaring, n = a 2 -fm + 2a y/m\ 

/*» n — a*—m 
"^ m= 2« ' 
that is, a surd is equal to a rational quantity, .which is 
impossible. 



>. If jr + -\/y = a 4- V*' t^ 11 ^^ x = a a °d/ = 6. 

For if a; is not equal to a, let x = a + m ; then 

that is, -^6 = m + ^y; 

which is impossible. [Art. 255.] 

Therefore x=za, 

and consequently, y = b. 

If therefore a? -f- -y/y = a+ -y/b 9 

we must also have x — -yty = a — ^/6. 

257. It appears from the preceding article that in any 
equation of the form 

»+Vy = a +V 6 (*)> 

we may equate the rational parts on each side, and also the 
irrational parts ; so that the equation (1) is really equivalent 
to two independent equations, x = a and y = b. 



25a If Va+V*=V jr +V>' tten will Va-V* 

For by squaring, we obtain 

a + V& = * + 2 V^ -f- y; 

.-. a = x + y, -y/b = 2Vxy. [Art. 257.] 
Hence a — -y/b = x — 2-y/xy + y, 

and -yJa—^/b = -y/x — yty. 



SURDS. 217 

259. To find the square root of a binomial surd. 



Suppose Va + yft = V® + V^ 5 
then as in Art. 258, 

x + y = a (1); 

2Vxy = ^/*> ( 2 > 

Since (x — y) 2 = (x -f- y) 2 — 4 xy 

= a 2 — 6 . . from (1) and (2); 

.\ x — y= Va 2 — b. 

Combining this with (1), we find 

2 2 



... vg+^-^±^a + ^-v(^-» . 



260. The values just found for x and y are compound 
surds unless a 2 — 6 is a perfect square. Hence the method 
of Art. 259 for finding the square root of a + -y/b is of no 
practical utility except when a 2 — b is a perfect square. 

Ex. Find the square root of 16 + 2 y/bb. 
Assume V16 + 2 ^65 = y/x + y/y. 

Then 16 + 2 ^65 = x + 2 Vsy + y ; 

.-. «+y = 16 (1), 

2 y/xy = 2 V&5 (2). 

Since (x - y) 2 = (x + y) 2 - 4xy 

= 16 2 -4x66. . . by (1) and (2), 
= 4x9. 
t*. JB — Jf = i 6 « • • • • • • • • • (3)« 

From (1) and (3) we obtain 

x = 11, or 5, and y = 5, or 11. 
That is, the required square root is y/\\ + y/S. 
In the same way we may show that 

V16 - 2 y/66 - y/\\ - v& 



218 ALGEBRA. 

JTote. Since every quantity has two square roots equal in magnl 
tude but opposite in sign, strictly speaking we should have 

the square root of 16 + 2 y/bb = ± ( ^11 + y/b), 

the square root of 16—2 y/bb = ± ( y/\\ — ^/5). 

However, it is usually sufficient to take the positive value of 'the 
square root, so that in assuming Va — y/b = y/x — y/y it is understood 
that x is greater than y. With this proviso it will be unnecessary in 
any numerical example to use the double sign at the stage of work 
corresponding to equation (3) of the last example. 

261. When the binomial whose square root we are seek- 
ing consists of two quadratic surds, we proceed as explained 
in the following example. 

Ex. Find the square root of y/Ylb — ^/147. 
Since V 175 ->/ 147 =V 7 (V 26 - V*V) = V 7 ^- V 21 ). 
.-. Vy/17b - V147 = Vi . V5 - V21. 

But V5- v 21 = VJ - V!; 

.-. V VH5 - V147 = </7( V* - >/!)• 

262. To find the square root of a binomial surd by inspec- 
tion. 

Ex. 1. Find the square root of 11 + 2 -y/30. 

We have only to find two quantities whose sum is 11, and whose 
product is 30, thus 

11 + 2 V30 = 6 + 5 + 2 V6~x~6 = ( y/6 + V&) 2 . 

.'. VlT+ 2 ^30 = yfl + y/b. 

Ex. 2. Find the square root of 53 — 12 yl0. 
First write the binomial so that the surd part has a coefficient 2 ; 
thus 53 - 12 vlO = 53-2 ^360. 

We have now to find two quantities whose sum is 53 and whose 
product is 360 ; these are 45 and 8 ; 

hence 53 - 12^10 = 45 + 8 - 2 V45x~8 

=(V r 45-V8) a / 

.\ V63 - 12 V10 = V45 - V 8 

= Zy/b - 2^/2. 



SURDS. 219 

Ex. 8. Find the square root of a + 6 + V2a6 + 6 a . 

Rewrite the binomial so that the surd part has a coefficient 2 ; thus 

a + b + V2a&+& 2 = a + b + 2 ^ ab + ** ■ 

We have now to find two quantities whose sum is (a + 6) and 
whose product is 2ab + b * ; these are 2q + 6 , and £; hence 






.-. Va + & + V2a& + & 2 =Ja + £+J|. 

Note. The student should observe that when the coefficient of the 
surd part of the binomial is unity, he can make this coefficient 2 if he 
will also multiply the quantity under the radical by J. 

263. Assuming va-\--y/b = x+^/y, the method of Art. 
258 gives us -\/a —^/b = x — ^/y. 

EXAMPLES XXIII. h. 
Find the square roots of the following binomial surds : 

1. 7-2^/10. 7. 41-24V2. 13. V 27 + 2 V^- 

2. 13 + 2v/80. 8. 83 + 12^/35. 14. V 32 -^ 24 - 

3. 8-2V7. 9. 47-4^33. 15. 3y5 + V40. 

4. 5 + 2^/6. 10. 2J+V 6 - le - 2q + 2V a^rp . 

5. 75 + 12V 2 !. 11. 4J-IV 3 * W- ax-2a Vax-a 2 . 

6. 18-8^5. 12. 16 + 6^/7. 18. a+x+V2ax+x* 

19. 2a-V3a 2 -2a&-6 a . 20. 1 + a 2 +(1 + a 2 + a*)* 

Find the fourth roots of the following binomial surds : 

21. 17 + 12 V 2 - 23. fV& + 3i. 25. 49-20^/6. 

22. 56 + 24V5. 24. 14+ 8 V 3 - 26. 248 + 32^60. 

Find, by inspection, the value of 



27. V3-2V2. 30. Vl9 + 8^3. 38. Vll + 4^/6. 

28. V4 + 2V3. 31. V8 + 2V15. 34. V16 - 4^/14. 

29. V6-2V6. 32. V9 - 2^/14. 35. V29 + 6^/22. 



220 ALGEBRA. 



Express with rational denominator 



36. 
37. 



-J: 38. - V 2 -^ . 40. — 1 

#3-V2 #3 + ^2 </6-^8 

1 . 39. V 8 + ^ 4 . 41. — &-. 

#3-1 V 8 -v^ 4 V^+^9 



42. — 1— . 43. Find value of J 6 + 2 V 3 . 

2+^7 ^33-19^3 

EQUATIONS INVOLVING SURDS. 

264. Sometimes equations are proposed in which the 
unknown quantity appears under the radical sign. Such 
equations are varied in character and often require special 
artifices for their solution. We shall consider a few of the 
simpler cases, which can generally be solved by the follow- 
ing method : 

Bring to one side of the equation a single radical term by 
itself: on squaring both sides this radical will disappear. By 
repeating this process any remaining radicals can in turn be 
removed. 

Ex. 1. Solve 2 y/x - V4x-ll = 1. 



Transposing, 2^/z — l = V4 x — 11. 

Square both sides ; then 4 x — 4^/se + 1 = 4 x — 11, 

4 y/X = 12, 

y/X = S\ 

•*. X — w. 

Ex. 2. Solve 2 + y/x - 6 = 13. 

Transposing, y/x — 5 = 11. 

Here we must cube both sides ; thus x - 6 = 1331 ; 
whence x = 1336. 

EXAMPLES XXIII. k. 

1. y/x - 5 = 3 5. V5x-1 = 2Vx + 3. 

2. v'4a;-7 = 6. 6. 2V3 -7x-3V8x- 12 =0. 



3. 7_Vx-4 = 3. 7. 2V5x-35 = 5V2x-7. 



ft. 13-V6x-4 = 7. 8. V9x 2 -llx-5 = 3x-2. 



SURDS. 221 

9. ^235 + 11=^/6. W- V3T^i + S = Vx + 11. 

10. V4x 2 -7x + l=2x-lf. 16. V9x-8 = 3<\/x~T4-2 

11. Vx + 25 = 1 + >/*• 17 * v^ixTS - ^/b = Vx~+3. 

12. V8x + 33-3 = 2V2x. 18. V25x^29- V4x-ll = S^/x. 
18. Vx + 3 + Vx = 5. 19. Vx + 4 ab = 2 a + V x * 

14. 10-V25 + 9x = 3 v /x. 20. y/x + V4 a + x = 2 V& + x. 



i. When radicals appear in a fractional form in an 
equation, we must clear of fractions in the ordinary way, 
combining the irrational factors by the rules already ex- 
plained in this chapter. 

Ex. Solve V9 4-2x-V2x=- — 5 

V9 + 2x 
Clearing of fractions, 

9 + 2x - V2x(9 + 2 x) = 5, 

4 + 2 x = V2x(9 + 2x). 
Squaring, 16 + 16x + 4 x 2 = 18x + 4x a , 

16 = 2x, 
x = 8. 

EXAMPLES XXIH. 1. 

lm e vj*-2i = 8 V* - n 9. VSTH + v* = A. 

3V*-14 4V«-13 V* 

2 . 9VJL-23 6 V x-17 10 . V x-Vx^8 = 



3V*-8 2V«-6 v Vx^8 

a Vg + 3 _ 3y /x-5 10 

V«-2""3Vx-13' U - Vx + 5+V«=— • 

d 5 Vs+3 _ Vx-f 9 t 

4 ' 2 "^T2-^xT7' "• 2VX-V4x^3=-^^ 

5 2yx-l = yx--2 * 

' 2V* + t. V*-* 18 ' Mb = - tT~: + V9x-32. 

6 6yx-7 6 = 7yx-26 

y/x-1 7yx-21* 14. ^-7 = 

7 12 yx-ll = 6yx-f 5 



V9x-32 



y/X + 7 

4yx-4$ ~~2yx + $ 15. (yx+ll)(yx-ll) + 110=a 



8. vT+^+V^=-7^=- 16. 2Vx = 1 2 " 6 ^ 



CHAPTER XXIV. 
Imaginary Quantities, 

266. An imaginary quantity is an indicated even root of 
a negative quantity. In distinction from imaginary quanti- 
ties all other quantities are spoken of as real quantities. 
Although from the rule of signs it is evident that a nega- 
tive quantity cannot have a real square root, yet quantities 
represented by symbols of the form V— a, V— 1, are of fre- 
quent occurrence in mathematical investigations, and their 
use leads to valuable results. We therefore proceed to 
explain in what sense such roots are to be regarded. 

When the quantity under the radical sign is negative, we 
can no longer consider the symbol -y/ as indicating a possi- 
ble arithmetical operation ; but just as -yja may be defined as 
a symbol which obeys the relation -y/a x -yja = a, so we shall 
define V— a to be such that V— « X V— a = — a, and we shall 
accept the meaning to which this assumption leads us. 

It will be found that this definition will enable us to bring 
imaginary quantities under the dominion of ordinary alge- 
braic rules, and that through their use results may be 
obtained which can be relied on with as much certainty as 
others which depend solely on the use of real quantities. 

267. Any imaginary expression not involving the oper- 
ation of raising to a power indicated by an exponent that is 
an irrational or imaginary expression, can be reduced to the 
form a + b V— 1, which may be taken as the general type of 
all imaginary expressions. Here a and b are real quantities, 
but not necessarily rational. An imaginary expression in 

222 



IMAGINABY QUANTITIES. 228 

this form is called a complex number. If a = 0, the form 
becomes &V— 1, which is called a pure imaginary expression. 



268. By definition, V 11 ! X V^I = - 1. 

.*. y/a • V— 1 X V a * V— 1 = a(— 1) ; 
that is, (^/a • V— l) 2 = — a. 

Thus the product -y/a . V— 1 may be regarded as equiv- 
alent to the imaginary quantity V— a. 

269. It will generally be found convenient to indicate the 
imaginary character of an expression by the presence of the 
symbol V— 1 which is called the imaginary unit ; thus 

yCT4 = V4x(-l)= 2V^l. 

V^Ta 2 = V7a 2 x(-l)=aV7V^T. 

270. We shall always consider that, in the absence of 
any statement to the contrary, of the signs which may be 
prefixed before a radical the positive sign is to be taken. 
But in the use of imaginary quantities the following point 
deserves notice. 

Since (— a) x (— b) = db, 

by taking the square root, we have 

V— a x V— 6 = ± Vab. 

Thus in forming the product of V— a and V— b it would 
appear that, either of the signs -f- or — might be placed 
before Vo&. This is not the case, for 

V— a xV-6=Va.V-l X y&* V— 1 

= VaftCV^)* = - Va&. 

271. In dealing with imaginary quantities we apply the 
laws of combination which have been proved in the case of 
other surd quantities. 

Ex. 1. a + by/-l±(c + dv r ^l) = a±c+(b±d)V^l. 

Ex. 2. The product of a + bV— 1 and c -f dV— 1 

= (a + b V^l) (c + dV^l) 
= ac — bd + (6c + ad) V^TL 



224 ALGEBRA. 



The symbol V— 1 is often represented by the lettei 
i ; but until the student has had a little practice in the use 
of imaginary quantities he will find it easier to retain the 
symbol V— 1. The successive powers of V— 1, or i, are as 
follows: 

(V^l) l =V^l, i = <; 

(V=T)' = -1, * = -l, 

(V^T) 8 = -V-1, *« = -*'; 

(V=T)< = 1, i* = l; 

and since each power is obtained by multiplying the one 
before it by V— 1, or i, we see that the results must now 
recur. 

273. U a + byP^l = 0, then a = 0, and 6 = 0. 

For, if a + b V 1 ^! = 0, 

then 6V— 1 = — a; 

.\ -& 2 = a 2 ; 

.-. a 2 + 6 2 = 0. 

Now a* and b 2 are both positive, hence their sum cannot 
be zero unless each is separately zero ; that is, a = 0, and 
6 = 0. 

274. U a + 6 V^l = c + </V^T., then a = c, and 6 = </. 

For, by transposition, a — c + (b — d) V—T = ; 
therefore, by the last article, a — c = 0, and 6 — d = ; 
that is, a = c and 6 = d. 

Thus in order that two imaginary expressions may be 
equal it is necessary and sufficient that the real parts should 
be equal, and the imaginary parts should be equal. 

The student should carefully note this article and make 
use of it as opportunity may offer in the solution of equa- 
tions involving imaginary expressions. 



IMAGINARY QUANTITIES. 223 

275. When two imaginary expressions differ only in the 
sign of the imaginary part, they are said to be conjugate. 
Thus a — &V— 1 is conjugate to a + &V— 1. 



Similarly -y/2 + 3 V— 1 is conjugate to y2 — 3V— 1. 

276. The sum and the product of two conjugate imaginary 
expressions are both reaL 

For a + frV^T+a — b->f^T = 2a. 

Again (a + b V^l)(a - bV^l) = a 2 - (- b 2 ) = a 2 + 6 s . 

277. If the denominator of a fraction is of the form 
a + 6 V— 1, it may be ratiooalized by multiplying the nu- 
merator and the denominator by the conjugate expression 
a — &V— 1. For instance, 

c + dV^ri = (c + dV^T) (o - by/—l) 

a + bV^i (a + bV^l) (a - bV^l) 

_ ac-\-bd -\-(ad — bc)V—l 

~ a 2 + b 2 

__ ac + bd a d — be r^j 

Thus, by reference to Art. 271, we see that the sum, dif- 
ference, product, and quotient of ttvo imaginary expressions is 
in each case an imaginary expression of the same form. 

278. Fundamental Algebraic Operations upon Imaginary 
Quantities. 

Ex. 1. Find value of V- a 4 + 5V-9a* - 2V-4a*. 

V^IT* = Va* (-1) = a 2 V^Tl 

5 V-9a 4 = 5V9fl 4 (-l) = 15a 2 V^l 

-2V-4a 4 =-2V4aH-l) = - 4a 2 V^l 

= 12a 2 V^T 
Ex. 2. Multiply 2\Z^3 by 3V^2. 

2\/^~3 = 2\/3\/^T; 

(2V§\/^I)(3V2V^1) = 6V6(V-T) 2 =-6V5. 
Q 



226 ALGEBRA. 



Ex. 3. Divide 2 + SV^l by 2 + V^l. 

2 + 3V^ ^ (2 + 3\/^)(2-v^) =: 7 + 4V^ _ 7+4>/^ 
2+V3I (2+V3I)(2-V^i) 4-(-l) 6 

27d. The method of Art. 262 may be used in finding 

the square root of a + b V— 1. 

> 

Ex. Find the square root of — 7 — 24 V— 1. 

- 7 - 24 V^T = - 7 - 2 V^TH. 
We have now to find two quantities whose sum is — 7 and whose 
product is — 144 ; these are 9 and — 16 ; 

hence - 7 - 24 V^T = 9 + (- 16)- 2 V9x(-16) 

= (V9_v^T6)a; 

... V - 7 - 24 V^T = ± (3 - 4 v^l). 

EXAMPLES XXIV. 

Simplify : 

1. V^8+V^T8. 8. 6 V^IS - 2 VTTJ. 

2. 4v^"27 + 3V^T2. 4. 2>/^20 + 3V^46 ->/^80. 

5. 2V- a 2 * 2 + 7V- 4 a 2 *; 2 + 12V- 36 a 2 * 2 . 

6. V^-V^f+V^fJ+v^f 9.(2V^2+\/^3X>^ :: 3- , ^ :: fi) 

7. (v^T3)(V- 12). 10. (2 + V^)(3 - \/^i). 

8. (2 + V^2) (1 - V^l). 11. (4 + V^~2) (2 - 8 V^6). 
12. (2>/^3 + 3V^2)(4>/^3-6V^2). 

18. V27 + V^3. 14. - V=l +( V^2 + V^TJ). 

15. (_V^ + >/32) + (2\/^-v^~2). 

Express with rational denominator: 



16. l 



18. 8 -f2\/3T , 3-2>/iri 



3->/^2 2-5v^T 2 + 6V^I 

fW 3v^"2 + 2v^~6 1tt a + zy/^1 a-sv^T 
17. : ' *«*• — • 

3v^2-2v^6 a-zV-1 a + xV^l 

Find the square root of 

20. - 6+12V^T. 21. ^11-60V^T. 22. -47+8V^"5 

Express in the form a + ib: 

88. §+*£ 24. yS-V 2 . 28. 1+1 

2-S< 2v/3-<v^ l-< 



CHAPTER XXV. 



Peoblems. 



280. In previous chapters we have given collections of 

problems which lead to simple equations. We add here 
a few examples of somewhat greater difficulty. 

Ex. 1. A grocer buys 15 lbs. of figs and 28 lbs. of currants for 
$2.60; by selling the figs at a loss of 10 per cent, and the currants 
at a gain of 30 per cent, he clears 30 cents on his outlay : how much 
per pound did he pay for each ? 

Let x, y denote the number of cents in the price of a pound of figs 
and currants respectively ; then the outlay is 

16 a + 28 y cents. 
.-. 15a + 28y = 260 (1). 

The loss upon the figs is — x 16 x cents, and the gain upon the 

ill • 

currants is — x 28 y cents ; therefore the total gain is 

^-^ cents; 
6 2 ' 

.-.^-^ = 30 (2). 

6 2 v ' 

From (1) and (2) we find that x = 8, and y = 5 ; that is, the figs 
cost 8 cents a pound, and the currants cost 5 cents a pound. 

Ex. 2. At what time between 4 and 6 o'clock will the minute-hand 
of a watch be 13 minutes in advance of the hour-hand ? 

Let x denote the required number of minutes after 4 o'clock ; then, 
as the minute-hand travels twelve times as fast as the hour-hand, the 

hour-hand will move over — minute divisions in x minutes. At 4 

12 

o'clock the minute-hand is 20 divisions behind the hour-hand, and 

finally is 13 divisions in advance ; therefore the minute-hand moves 

over 20 + 13, or 33 divisions more than the hour-hand. 

227 



228 ALGEBRA. 



Hence * * = — + 33, 

12 

lis = 33; 

.-. x = 36. 

Thus the time is 36 minutes past 4. 

If the question be asked as follows : " At what times between 4 and 
6 o'clock will there be 13 minutes between the two hands ? " we must 
also take into consideration the case when the minute-hand is 13 
divisions behind the hour-hand. In this case the minute-hand gains 
20 — 13, or 7 divisions. 

Hence x = — + 7, 

12 * 

which gives * = 7-^. 

V 
Therefore the timts are 7 — past 4, and 36' past 4. 

Ex. 3. Two persons A and B start simultaneously from two places, 
c miles apart, and walk in the same direction. A travels at the rate of 
p miles an hour, and B at the rate of q miles ; how far will A have 
walked before he overtakes B ? 

Suppose A has walked x miles, then B has walked x — c miles. 

A, walking at the rate of p miles an hour, will travel x miles in - 

P 
hours ; and B will travel x — c miles in x ~ c hours : these two times 

being equal, we have ^ 

x__ x — c 

P Q 
qx=px—pc; 

whence x = — — • 

p-q 

t)C 

Therefore A has travelled — — miles. 

p-q 

Ex. 4. A train travelled a certain distance at a uniform rate. Had 
the speed been 6 miles an hour more, the journey would have occupied 
4 hours less ; and had the speed been 6 miles an hour less, the journey 
would have occupied 6 hours more. Find the distance. 

Let the speed of the train be x miles per hour, and let the time 
occupied be y hours ; then the distance traversed will be represented 
by xy miles. 



PROBLEMS. 229 

On the first supposition the speed per hour is x + 6 miles, and tho 
time taken is y — 4 hours. In this case the distance traversed will be 
represented by (x + 6)(y — 4) miles. 

On the second supposition the distance traversed will be repre- 
sented by (x — 6) (y + 6) miles. 

All these expressions for the distance must he equal ; 

.\ xy =(as + 6)(y - 4) = (x - 6)(y + 6). 

From these equations we have 

xy = xy + 6y — 4x — 24, 

or 6y-4s = 24 (1); 

and aey = xy — 6 y + 6 x — 86, 

or 6a;-6y = 36 . . (2). 

From (1) and (2) we obtain x = 30, y = 24. 
Hence the distance is 720 miles. 

Ex. 5. A person invests $ 3770, partly in 3 per cent Bonds at $ 102, 
and partly in Railway Stock at $ 84 which pays a dividend of 4} pel 
cent; if his income from these investments is $136.25 per annum, 
what sum does he invest in each ? 

Let x denote the number of dollars invested in Bonds, y the number 
of dollars invested in Railway Stock ; then 

s + y = 3770 (1). 

The income from Bonds is $ jg> or fi ; and that from Railway 

Stock is $li?, or $ 3 - V. 
v 84' 66 

Therefore £ + || = i36} ........ (2). 

o4 oo 

From (2) x + } } y = 4632}, 

and by subtracting (1) \ \ y = 862} ; 
whence y = 28 x 37} = 1050 ; 

and from (1) x = 2720. 

Therefore he invests $ 2720 in Bonds and 4 1050 in Railway Stock. 

EXAMPLES XXV. 

1. A sum of $ 100 is divided among a number of persons ; if the 
number had been increased by one-fourth each would have received 
a half-dollar less : find the number of persons. 



280 ALGEBRA. 

2. I bought a certain number of marbles at four for a cent; 1 
kept one-fifth of them, and sold the rest at three for a cent, and 
gained a cent : how many did I buy ? 

8. I bought a certain number of articles at five for six cents ; if 
they had been eleven for twelve cents, I should have spent six cents 
less : how many did I buy ? 

4. A man at whist wins twice as much as he had to begin with, 
and then loses $ 16 ; he then loses four- fifths of what remained, and 
afterwards wins as much as he had at first : how much had he origin- 
ally, if he leaves off with $ 80 ? 

5. A number of two digits exceeds five times the sum of its digits 
by 9, and its ten-digit exceeds its unit-digit by 1 : find it. 

6. The sum of the digits of a number less than 100 is 6 ; if the 
digits be reversed the resulting number will be less by 18 than the 
original number : find it. 

7- A man being asked his age replied, " If you take 2 years from 
my present age the result will be double my wife's age, and 3 years 
ago her age was one-third of what mine will be in 12 years. " What 
were their ages ? 

8. At what time between one and two o'clock are the hands of a 
watch first at right angles ? 

9. At what time between 3 and 4 o'clock is the minute-hand one 
minute ahead of the hour-hand ? 

10. When are the hands of a clock together between the hours of . 
6 and 7 ? 

11. It is between 2 and 3 o'clock, and in 10 minutes the minute- 
hand will be as much before the hour-hand as it is now behind it : 
what is the time ? 

12. At an election a majority of 162 was three-elevenths of the 
whole number of voters : find the number of votes on each side. 

18. A certain number of persons paid a bill ; if there had been 10 
. more each would have paid $ 2 less ; if there had been 6 less each 
would have paid $2.50 more: find the number of persons, and what 
each had to pay. 

14. A man spends $ 100 in buying two kinds of silk at $4.50 and 
$4 a yard ; by selling it at $4.25 per yard he gains 2 per cent: how 
much of each did he buy ? 

15. Ten years ago the sum of the ages of two sons was one-third of 
their father's age: one is two years older than the other, and the 
present sum of their ages is fourteen years less than their father's age : 
how old are they ? 



PROBLEMS. 281 

16. A basket of oranges is emptied by one person taking half of 
them and one more, a second person taking half of the remainder and 
one more, and a third person taking half of the remainder and six 
more. How many did the basket contain at first ? 

17. A person swimming in a stream which runs 1} miles per hour, 
finds that it takes him four times as long to swim a mile up the stream 
as it does to swim the same distance down: at what rate does he 
swim ? 

18. At what times between 7 and 8 o'clock will the hands of a 
watch be at right angles to each other ? When will they be in the 
same straight line ? 

19. The denominator of a fraction exceeds the numerator by 4; 
and if 5 is taken from each, the sum of the reciprocal of the new frac- 
tion and four times the original fraction is 6 : find the original fraction. 

20. Two persons start at noon from towns 60 miles apart. One 
walks at the rate of four miles an hour, but stops 2J hours on the 
way ; the other walks at the rate of 3 miles an hour without stopping : 
when and where will they meet ? 

21. A, B, and C travel from the same place at the rates of 4, 5, and 
6 miles an hour respectively ; and B starts 2 hours after A. How 
long after B must C start in order that they may overtake A at the 
same instant ? 

22. A dealer bought a horse, expecting to sell it again at a price 
that would have given him 10 per cent profit on his purchase ; but he 
had to sell it for $ 50 less than he expected, and he then found that he 
had lost 15 per cent on what it cost him : what did he pay for the 
horse ? 

23. A man walking from a town, A, to another, B, at the rate of 4 
miles an hour, starts one hour before a coach travelling 12 miles an 
hour, and is picked up by the coach. On arriving at B, he finds that 
his coach journey has lasted 2 hours : find the distance between A 
and B. 

24. What is the property of a person whose income is $ 1140, when 
one-twelfth of it is invested at 2 per cent, one-half at 3 per cent, 
one-third at 4} per cent, and the remainder pays him no dividend ? 

25. A person spends one-third of his income, saves one-fourth, and 
pays away 5 per cent on the whole as interest at 7J per cent on debts 
previously incurred, and then has $110 remaining: what was the 
amount of his debts ? 

26. Two vessels contain mixtures of wine and water ; in one there 
is three times as much wine as water, in the other five times as much 
water as wine. Find how much must be drawn oft" from each to fill a 
third vessel which holds seven gallons, in order that its contents may 
be half wine and half water. 



232 



ST. There are two mixtures of wine and water, one of which con- 
tains twice as much water as wine, and the other three times as much 
wine as water. How much most there be taken from each to fill a 
pint cop, in which the water and wine shall be equally mixed ? 

28. Two men set oat at the same time to walk, one from A to B, 
and the other from BtoA,a distance of a miles. The former walks 
at the rate of p miles, and the latter at the rate of q miles an hoar: at 
what distance from. A will they meet ? 

29. A train runs from A to B in 3 hoars ; a second train runs from 
A to C, a point 15 miles beyond B, in 3} hoars, travelling at a speed 
which is leas by 1 mile per hoar. Find distance from A to B. 

80. Coffee is bought at 36 cents and chicory at 9 cents per lb.: in 
what proportion most they be mixed that 10 per cent may be gained 
by selling the mixture at 33 cents per lb.? 

81. A man has one kind of coffee at a cents per pound, and another 
at b cents per pound. How much of each must he take to form a 
mixture of a — b lbs., which he can sell at c cents a pound without 
loss? 

82. A man spends c half-dollars in buying two kinds of silk at a 
dimes and b dimes a yard respectively ; he could have bought 3 times 
as much of the first and half as much of the second for the same 
money. How many yards of each did he buy ? 

83. A man rides one-third of the distance from A to B at the rate 
of a miles an hour, and the remainder at the rate of 2 6 miles an hour. 
If he had travelled at a uniform rate of 3 c miles an hour, he could 
have ridden from A to B and back again in the same time. Prove 

that - = - + r- 

cab 

84. A, B, C are three towns forming a triangle. A man has to 
walk from one to the next, ride thence to the next, and drive thence 
to his starting-point. He can walk, ride, and drive a mile in a, 5, c 
minutes respectively. If he starts from B he takes a + c — b hours, 
if he starts from C he takes b + a — c hours, and if he starts from A 
he takes c + b — a hours. Find the length of the circuit. 

MISCELLANEOUS EXAMPLES IV. 

1. Distinguish between like and unlike terms. Pick out the like . 
terms in the expression a 8 — 3 ab + b 2 — 2 a 8 + 3 b 2 + 5 ab + 7 a 8 . 

2. Subtract - 2a 8 + Sa 2 b + 56* - 4ab 2 from -l-2a& 2 + 36 8 
and multiply the result by — 1 -f 2 a — b. 

8. Divide 8 a; 8 - 8 «2y + 4xy 2 -y*by 2x-y. 



MISCELLANEOUS EXAMPLES IV. 233 

4. If the number of dollars I possess is represented by + <*$ what 
will — a denote ? 

5. Factor the following expressions : 

(i.) a 2 -64, 
(ii.) a 8 - 27. 

6. Find the value of — - - h l 



Qx-2 2x-f 3se-l 

7. Solve Hn»«-LLiJ = l* + 7« + 5--6s, 

5 3 3 

8. There is a number of two digits which when divided by the unit 
digit gives a quotient 6 ; but if the digits be inverted the number is 
increased by 36 : find the number. 

9. Find the H. C. F. of 4s 8 - 16a 2 + 13z - 3 

and 3a 8 - 13s 2 + 13 x - 3. 

10. Simplify (v^o«)x(v^)x a~% -*- a** 

11. Find the value of 20 Vj + 14 Vf + 2 V2l - 7 VS + V± + v^. 

lis 



12. Simplify 



— » »+!'«. *+ 4 



X + J X & + 1 

13. Find the value of f~a 2 - j (36 -c)+ 6 2 - ^| - c 2 l when 
a = 2, 6 = 3, c = 4. 



14. Solve — £-+ 6 - c 



sc — a sc + 6 « — a 

15. Find the L. CM. of 1 - x, 1 - s 2 , 1 - x*, and (1 - z)*. 

16. Solve ^ + ^ = 3f, 

3 6 * 

17. Simplify ^_2,-{2*-2-7-g-4)+2--|}. 

18. Find the square root of x* + 8x* - 2s 8 + 16« 2 - 8x + 1. 

19. Solve the equations 

w 2 7 \ 3 42/ 



(U0 f (^. 1 ) + u + ^. 



**±I-7. 



80. Expand the following binomials : 

(i.) (x + 3a)*, 

(H.) (2s-|) 6 . 



234 ALGEBRA. 

21. The sum of the two digits of a number is 8 times their differ- 
ence ; if the digits be inverted, the number is diminished by 18 : find 
the number. 

22. Find the factors of (i.) x 2 - 9x - 36, (ii.) 2a£ - 3a; - 14, and 
(iii.) o*6* - 7 <MW + a* 

23. Rationalize the denominator of — — — — , and simplify 
VI3T4V3. 5 ~ 2V ^ 

24. Simplify *'+8*- ll*-8«+ 10. 

a* + 3a; 2 -6a;-8 
For what values of x will both numerator and denominator vanish ? 

25. Solve the equations 

'(i.) 2a + 3y + 4* = 31, 

a; + 4y + z = 18, 

3a; + y + 2s = 16. 



26. Simplify V^8 + V^£ - V^18 + V - 2 + 2 V^3. 

28. Find the middle term of the expansion of [ - -f - ] • 

\x a) 



29. Simplify 



& 8 a 8 6 a 



(«_*V« + $_i> I + 1 + .I 

\6 a)\b a ) a* b 2 ab 



ab 

80. Solve the equations 

(i.) a(x — a)— b(x — 6) = (a+ 6)(x — a — 6). 
(ii.) (a + 6)x — ay = a 2 , (a 2 + & 2 )« — a&y = a 8 . 

31. A sum of $ 10.10 is divided among 7 women and 10 men ; the 
same sum could have been divided among 23 women and 4 men. Find 
how much each woman and man receives. 

32. Find the cube root of 8a*+ 12x 5 + 18«*+ 13x 8 +9a a + 3a; +1, 

12 
and the square root of y 2 + 4 y + 10 H f- — . 

y y 2 



84. Simplify V69 - 24^6 + [( V^6 + V^~3) ( V^S + 2 V^T)]. 



CHAPTER XXVI. 
Quadbatic Equations. 

281. Suppose the following problem were proposed for 
solution : 

A dealer bought a number of horses for $ 280. If he had 
bought four less, each would have cost $ 8 more ; how many 
did he buy ? 

We should proceed thus : 

280 
Let x = the number of horses ; then = the number of 

x 
dollars each cost. 

If he had bought 4 less, he would have had x — 4 horses, 

280 
and each would have cost dollars. 

aj — 4 

. Q , 280 280 

• • o i = -; 

x x — 4 

whence x(x — 4) + 35(x — 4) = 35 x ; 

.-. a 2 - 4s + 35 a; - 140 = 35®; 

.-. a? 2 -4 a = 140. 

This equation involves the square of the unknown quan- 
tity ; and in order to complete the solution of the problem 
we must discover a method of solving such equations. 

282. Definition. An equation which contains the square 
of the unknown quantity, but no higher poiver, is called a 
quadratic equation, or an equation of the second degree. 

If the equation contains both the square and the first 
power of the unknown, it is called an affected quadratic ; if it 

236 



236 ALGEBRA. 

contains only the square of the unknown it is said to be a 
pure quadratic. 

Thus 2 x 2 — 5x =3 is an affected quadratic, 
and 5 x* = 20 is a pure quadratic. 

PUKE QUADRATIC EQUATIONS. 

283. A pure quadratic may be considered as a simple 
equation in which the square of the unknown quantity is 
to be found. 

Ex. Solve 9 - = 25 . 

x 2 - 27 z 2 - 11 

Multiplying across, 9 x 2 - 99 = 25 x 2 - 675 ; 
transposing, 16 x 2 = 676 ; 

.-. x 2 = 36 ; 
and taking the square root of these equals, we have 

x = ±6. 
Note. We prefix the double sign to the number on the right- 
hand side for the reason given in Art. 196. 

284. In extracting the square root of the two sides of the 
equation x 2 = 36, it might seem that we ought to prefix the 
double sign to the quantities on both sides, and write 
± x = ±6. But an examination of the various cases shows 
this to be unnecessary. For ± x = ± 6 gives the four cases : 

-f a; = + 6, -f a: = — 6, —aj = 4-6, — a? = — 6, 

and these are all included in the two already given, namely, 
x = + 6, x = — 6. Hence when we extract the square root 
of the two sides of an equation, it is sufficient to put the 
double sign before the square root of one side. 

EXAMPLES XXVI. a. 
Solve the following equations : 

1. 4x*+5 = x 2 + 17. 4 2a 2 -6 s 2 -4 5s*-10 _. 

23 .2 2 4 7 

8. 3x 2 +3=^ + 24. 

3 • 6. x 2 +2=(*- i y- x + 2 \ 

3. (x + 1)(* - 1) = 2x 2 - 4. x + 2 



QUADRATIC EQUATIONS. 237 

e. *^« + * + «=6. 

x + a x — a 

7 3 (a* - 1) 4(s* - 4) 3(0s 2 - 1) _ - 

x 2 -l x a + 3 (a;2-l)(x2 + 3) 

8. (2x-c)(x-fd) + (2ae-fc)(x-c*)=2cd(2c(J--l> 

AFFECTED QUADRATIC EQUATIONS. 

The equation x 2 = 36 is an instance of the simplest 
form of quadratic equations. The equation (x — 3) a = 25 
may be solved in a similar way ; for taking the square root 
of both sides, we have two simple equations, 

a?-3 = ±5. 
Taking the upper sign, x — 3 = + 5, whence x = 8 ; 
taking the lower sign, x — 3 = — 5, whence x = — 2. 
.\ the solution is a? = 8, or — 2. 

Now the given equation (x — 3) 2 = 25 
may be written sc 2 — 6 x -f (3) 2 = 25, 

or sc 2 — 6 x = 16. 

Hence, by retracing our steps, we learn that the equation 

ar*-6a; = 16 

can be solved by first adding (3) 2 to each side, and then 
extracting the square root; and we add 9 to each side 
because this quantity added to the left side makes it a 
perfect square. 

Now whatever the quantity a may be, 

x 2 -f 2 ax + a 2 = (x -f- a) 2 , 

and x 2 — 2 ax -f a 2 =(x — a) 2 ; 

so that, if a trinomial is a perfect square, and its highest 
power, x 2 , has unity for a coefficient, the term without x must 
be equal to the square of half the coefficient of x. 

Ex. 1. Solve 7 x = x 2 - 8. 

Transpose so as to have the terms involving x on one side, and the 
square term positive. 

Thus ^-705 = 8. 



288 ALGEBRA. 

Completing the square, x* - 7 x -f (}) 2 = 8 + ^ ; 
that is, (x - i? = ^ ; 

.-. x-|=±f; 
.-. ac = }± 1 = 8, or -1. 

Note. We do not work out (J ) 2 on the left-hand side. 

Ex. 8. Solve 32 - 3 x 2 = 10s. 

Transposing, 3 x 2 + 10 x = 32. 

Divide throughout by 3, so as to make the coefficient of x 2 unity. 

Thus x * + i£-x = *£; 

completing the square, x 2 + ±£ x + (i) a = V + ¥ ' 
that is (x -f f ) 2 = iji ; 

... a; = -|±Y = 2 » or-6J. 

Ex.3. Solve 7(x + 2a) 2 + 3a 2 = 6a(7x+23a). 

Simplifying, 7 x 2 + 28 ax + 28 a 2 + 3 a 2 = 35 ax -f 115 a 2 , 
that is, 7 x 2 - 7 ax = 84 a 2 . 

Whence x 2 — ax = 12 a 2 ; 

completing the square, x 2 — ax -f ( - ] = 12 a 2 4- y ; 

*u * • / a\ 2 49a 2 

that is, ( X 2/ = 4"' 

"* 2" ± T' 

.\ x = 4a, or — 3a. 

286. We see then that the following are the steps required 
for solving an affected quadratic equation. 

(1) If necessary, simplify the equation so that the terms in 
x 2 and x are on one side of the equation, and the term without 
x on the other. 

(2) Make the coefficient of x* unity and positive by dividing 
throughout by the coefficient ofx*. 

(3) Add to each side of the equation the square of half the 
coefficient of x. 

(4) Take the square root of each side. 

(5) Solve the resulting simple equations. 

287. The quadratic equations considered hitherto have 
had two roots. Sometimes, however, there is only one solu- 



QUADRATIC EQUATIONS. 289 

turn. Thus if a 2 — 2a?-hl = 0, then (x — 1)* = 0, whence 
x = 1 is the only solution. Nevertheless, in this and similar 
cases we find it convenient to say that the quadratic has two 
equal roots. 

EXAMPLES XXVI. b. 

1. 6a* + 14a; = 66. 7. 16 = 173 + 438. 18. 2132+223+6=0. 

3. 3x2+121=443. 8. 21+3 = 232. 14. 603 2 -163 = 27, 
8. 263 = 63 a + 21. 9. 932-143-63=0. 16. 1832-273-26=0. 

4. 832 + 3 = 30. 10. 123 a = 293- 14. 16. 632 = 83 + 21. 

5. 33 2 + 36=22x. 11. 2032 = 12-3. 17. 163 2 -2a3 = a». 

6. 3 + 22-632=0. 12. 193 = 16-832. 18. 2l3 2 = 2a3 + 8a*. 

19. 632 = 11*3 + 7*2. 23. (3+l)(23 + 3)=43 2 -22. 

20. 1232 + 23*3+10*2 = 0. 24. (33-6)(23-6)=3 a +2»-3. 

21. 123 3 -ex- 20c 2 = 0. 25. a*x*-2ax + a 2 = b. 

22. 2(3-3) = 3(3 + 2)(3-3). 26. cdx* = (fix + d?x - cd. 

27. 5£ZLi = 3£. 89. 63 II 7 = _3-6_ gl x+4 3^2 =6 ^ 

3 + 1 2 73-6 23-13 3-4 3-3 * 

«« 83 — 8 63 — 2 At% 3+3 23—1 /v AA 

28. -- — -=— — f. 30. J^LH — _ =0. 32. 



3-2 3 + 6 23-7 3-3 3-3 6 9-23 

38. -* +-2_ = 1?. 88. ^ + 5+5=4 

3 + 33 + 6 20 3 + C3 + 62 

34 6 8 _ 3 89 23 . 33-l _ 63-ll 

'3+1 3 + 2 23 + 4* * 3-1 3+2 3-2' 

36. ^!-^ + 5! == 0. 40. *^-*±| + 6* = 0. 

6 2 C C* 3+S3-3 7 

86. -J- + _I- = °±*. 41. 8»+l + «-8 = 1Z. 

C + 3 (J + 3 cd 3+8 33-1 12 

87 23+i_23 + 7 3 = Q ^ 21^-16^ 7flC=s6> 

33-2 33-4 4 - 332-4 



Solution by Formula. After suitable reduction and 
transposition every quadratic equation can be written in 
the form 

ax 2 +- bx +- c = 0, 

where a, b, c, may have any numerical values whatever. If 
therefore we can solve this quadratic, we can solve any. 



240 ALGEBRA. 

Transposing, . aa? + bx = — c; (1) 

b c 

dividing by a, a? + -x = — • 

a a 

Completing the square by adding to each side [ — ] , 

a \& a J 4cr a 

that is, (x + AY = £=i 

\ 2aJ 4er 

extracting the square root, 

b ±y(&«-4a C ) 

_ -6±y(6 8 -4ac) 

" _ 2^ * 

Note. The student will observe that b, the first term of the 
numerator of the fraction, is the coefficient of x in equation (1) with 
Us sign changed, and that 4 ac, under the radical, is plus or minus 
according as the signs of a and c in equation (1) are like or unlike. 



— 4ac 



Instead of going through the process of completing 
the square in each particular example, we may now make 
use of this general formula, adapting it to the case in 
question by substituting the values of a, b, c. 

Ex. Solve 6 z 2 + 11 a - 12 = 0. 

Here a = 6» b = 11, c = - 12. 

. _ -11 ±V(U)g-4.5(-12) 
" X ~ 10 

_ -ll±ya61 _ -ll±10 _4 or _ 8# 
10 10 ""6' 

290. In the result «, -**V( y -*»>) f 

2a 

it must be remembered that the expression -y/(b 9 — 4 ac) is 
the square root of the compound quantity 6 2 — 4 ac, taken as 
a wJiole. We cannot simplify the solution unless we know 
the numerical values of a, b, a It may sometimes happen 
that these values do not make 6 2 — 4 ac a perfect square. In 
such a case the exact numerical solution of the equation 
cannot be determined. 



QUADRATIC EQUATIONS. 241 

Ex. 1. Solve 6a* 2 - 16 & + ll=0. 

We have 15± V(- 15)« -4-5. 11 

2-5 
__ 15^^/5 
10 
Now v^ = 2.236 approximately. 

. x = 15 ± 2.236 = 1 7236 ^ Qr 12764> 

These solutions are correct only to four places of decimals, and 
neither of them will be found to exactly satisfy the equation. 

Unless the numerical values of the unknown quantity are required 
it is usual to leave the roots in the form 

15 + y/5 15-y/e 

10 ' 10 

Ex.8. Solve a^-Sx + SrrO. 

We have 



-_3±V(-3)*-4.1 .5 

*- 2 



= 3±V9-20 = 3± v cr li 
2 2 

But — 11 has no square root exact or approximate [Art. 196]; so 
that no real value of x can be found to satisfy the equation. In such 
a case the roots are said to be imaginary or impossible [Art. 266]. 

291. Solution by Factoring. The following method will 
sometimes be found shorter than either of those already 
given. 

Consider the equation x* -f -J x = 2. 

Clearing of fractions, 3 a 2 -f 7 a? — 6 = . . . . (1); 
by resolving the left-hand side into factors, we have 

(3a>-2)(a> + 3) = 0. 

Now if either of the factors 3 x — 2, x -f 3, be zero, their 
product is zero. Hence the quadratic equation is satisfied 
by either of the suppositions 

3a>-2 = 0, or a? + 3 = 0. 

Thus the roots are -J, — 3. 

From this we see that when a quadratic equation has been 
simplified and brought to the form of equation (1), its solution 
can be readily obtained if the expression on the left-hand 



242 ALGEBRA. 

side can be resolved into factors. Each of these factors 
equated to zero gives a simple equation, and a correspond- 
ing root of the quadratic. 

Ex. 1. Solve 2 x 2 - ax + 2 bx = ab. 

Transposing, so as to have all the terms on one side of the equation, 

we have 

2 x 2 - ax + 2 bx - ab = 0. 

Now 2x a -ax + 2 bx - ab = x(2x- a)+ 6(2x-a) 

= (2x-a)(x + 6). 

Therefore (2x - a)(x + 5) = ; 

whence 2x — a = 0, or x + b = 0. 

.». x =-, or —6. 
2 

Ex. 8. Solve 2(x 2 - 6) = 3(x - 4). 
We have 2x a - 12 = 3x - 12 ; 
that is, 2x 2 =3x (1) 

Transposing, 2 x 2 — 3 x = 0. 

x(2x-3) = 0. 

,\ x = 0, or 2 x — 3 = 0. 

Thus the roots are 0, § . 

Note. In equation (1) above we might have divided both sides 
by x and obtained the simple equation 2 x = 3, whence x = f , which 
is one of the solutions of the given equation. But the student must 
be particularly careful to notice that whenever an x is removed by 
division from every term of an equation it must not be neglected, 
since the equation is satisfied by x = 0, which is therefore one of the 
roots, 

292. Formation of Equations with Given Roots. 

It is now easy to form an equation whose roots are known. 

Ex. Form the equation whose roots are 3 and J. 

Here x = 3, or x = \ ; 

.•. x — 3 = 0, or x — \ = ; 
both of these statements are included in 

(x-3)(x-i) = 0, 
or 2x 2 -7x + 3 = 0. 

From this it also appears that the factors of a trinomial, in the form 
ax 2 + bx + c, can be obtained by placing the expression equal to zero, 
solving the resulting quadratic equation (Art. 288), and subtracting 
each root separately from x. We shall return to the subject of thifi 
article in Chapter xxx. 



QUADRATIC EQUATIONS. 248 

293. Values found for the Unknown Quantity which do not 
satisfy the Original Equation. 

From the following example it will be seen that in solv- 
ing certain equations values may be obtained which will 
not satisfy the original equation. 

Ex. Solve Va;4-6 + V3a; + 4 = vl2a;-f 1. 

Squaring both sides, 

x+6+3x+4+ 2V(aj + 6)(3ac + 4)= 12« + 1, 
Transposing and dividing by 2 f 

V(s + 6)(3a; + 4)=4a;-4 (1). 

Squaring, (x + 6)(3x -f 4)= 16s 2 - 32a 4- 16, 

or 13x 2 -61x-4 = 0, 

(a;-4)(13a; + l) = 0; 

.-. x = 4, or - ^ 

If we proceed to verify the solution by substituting these values in 
the original equation, it will be found that it is satisfied by x = 4, but 
not by x = — fa. But this latter value will be found on trial to satisfy 
the given equation if we alter the sign of the second radical ; thus, 

Vx~+l - V3x+4 = Vl2 x + 1. 
On squaring this and reducing, we obtain 



-V(a + 6)(3s + 4) = 4a;-4 # ( 2 ); 

and a comparison of (1) and (2) shows that in the next stage of the 
work the same quadratic equation is obtained in each case, the roots 
of which are 4 and — fa, as already found. 

From this it appears that when the solution of an equation requires 
that both sides should be squared, we cannot be certain without trial 
which of the values found for the unknown quantity will satisfy the 
original equation. 

In order that all the values found by the solution of the equation 
may be applicable, it will be necessary to take into account both signs 
of the radicals in the. given equation. 

EXAMPLES XXVI. O. 

Solve by the aid of the formula in Art. 288 : 

1. 3s 2 = 15 -4s. 5. 5s 2 +4 + 21x = 0. 9. 35 + 9s-2sc 2 = 0. 

2. 2s 2 + 7x=16. 6. z 2 + ll = 7:r. 10. 3s 2 =a;-f-l. 

3. 2z 2 + 7 -9z = 0. 7. 8s 2 = a; + 7. 11. 3x 2 + 5a = 2. 

4. « 2 =3x + 6. 8. 5z 2 =17a;-10. 12. 2x 2 + 5aj-33 = 



244 ALGEBRA. 

Solve by resolution into factors : 

13. 6x* = 7 + x. 17. 4x a = 1 kx + 3. 81. 26x* = 5x + 6. 

14. 21+8x 2 = 26x. 18. x 2 -2 = ffx. 88. 35-4x = 4x*. 

15. 26x-21 + llx 2 =0. 19. 7x 2 = 28-96x. 83. 12x 2 -llax=36a 2 

16. 5x 2 +26x+24=0. 80. 96 »? = 4 a; + 15. 84. 12x 2 +36a 2 =43ox. 

85. 356 2 = 9x 2 +66x. 88. x 2 - 2ax + 8x = 16a. 

86. 36 x 2 - 35 6 2 = 12 bx. 89. 3x 2 - 2ax- 5x = 0. 

87. x 2 -2ax + 4a& = 26x. 30. ax 2 + 2x = te. 

Solve : 

38. V3 x + 10 + VxT~2 = V10 x + 16. 

88 3x-4 x-1 1 =0 36. V2x + 6-Vx~+4 = Vx^"4. 
x + 1 3x + 4 2 

fl /P w * a. 1 _ « + & 

•*' to 2 -^|L==<fe-cx 2 . 87 * aT^ + 6T^"'"^~- 

35. x 2 + 2cx-2dx = 2ca , -d 2 . 38. 2cx 2 +2d 2 (x-fc)= ax(x+5c) 

39 a? + m _j_ g — m = a? 2 -f m 2 , ag 2 — m 2 
* x — m x + w x 2 — ro a x 2 + i» a 

x + a + 6 a x 



41. Vx -f f + \/Sx -f J = V6x + }. 

42. Vx + 3 + V2x+ 1 = 2V3x-l. 

AO x-2 , 3x-ll 4x+13 

43. ■ H = ' • 

x-3 x-4 x + 1 

44 ^ ~~ n 4. m — n = & -f n* — 2 n 
2m + x 2k + x k + m + x 

45. (a - 6)x 2 + (b - c)x + c - a = 0. 

46. a(o-c)x 2 +o(c-a)x + c(a-o) = 0. 

47. Va — x + Vo — x = Va + b — 2 as. 

48. _L- + _L- = _i- + 1 



a — x 6 — x a — c 6 — c 

49. Vx-p + Vx-q = P + q • 

vx — q vx — p 

50. V(x - 2)(x - 3)+ 5Ap^| = Vx 2 + 6x + 8. 

'x — 8 



CHAPTER XXVII. 

Equations in Quadratic Form. 

294. An equation in the form ax* n -f- bx* = c, n being a 
positive or negative integer or fraction, is in quadratic form. 

Thus x 4 + 4x* = 117, x* + 7a£ = 44, and a* + af * = a are 
equations in quadratic form. 

We give a few examples showing that the ordinary rules 
for quadratic equations are applicable to those in quadratio 
form. 

Ex. 1. Solve x 4 - 13 x 2 = - 36. 

By formula [Art. 288] rf = » WP»)' - «(«») 

_ 13 ± V169 '-144 
2 

= 13j^ = 9or4; 
2 ' 

. •. x = ± 3 or ± 2. 

Ex.8. Solve 2x^--3x* = 2. 

„ x \ 3 ± V9 + 16 
By formula x 75 = j 

= 3±6 =2or _l. 
4 2 

Raising to the third power, x = 8, or — J. 

Ex. 3. Solve 2 af* - 9 x~^ = -4. 

■d * i -4 9 ± V~81 - 32 ± 7 , - 
By formula, x * = — — — = . = 4 or *. 

4 4 

Raising to the fourth power, 

x~ 1 = 256or I ^; 

that is, - = 256 or ^ ; 

x 

•'• x=*J*orl6. 
245 



246 



EXAMPLES XXVIL 



1. x*-13x* + 36 = 0. la 2v± + 2x -* = 5 . 

2. x* + 7x* = 8. a , , 

- . . . , .„ U. 6x*=7z*-2x~* 

3. x* — i«i*c- = 2lo. 

4. 8x* + 65x*+8=0. 12. x» + 6=5x\ 

1 * 

IS. 8x*-x«-2 = 0. 



5. 3V«-3x"i-8. 



s 



,f 



6. 27x*-l =26*T. u. «.* = S ,4. 

7.. x 4 - 74 x* = -1226. 



14. 6Vx = 6x^-13. 

8. x--2x-» = 8. "• 1 + 8^+^ = 0. 

9. 9 + ar* = lOaT 2 . 16. 8x*_ 8x"S = 68. 

i. Any equation which can be thrown into the form 
ax 2 -f bx -f c + p Vox 2 4- 6a? + c = g 



may be solved as follows. Putting y = Voa? 2 + bx -f- c, we 
obtain 

y 2 +py — q = o. 

Let r 2 and r 2 be the roots of this equation, so that 



•yjax 2 -\-bx-\-c = rj, Vase 2 + bx + c = r 2 ; 

from these equations we shall obtain four values of x. 

When no sign is prefixed to a radical, it is usually under- 
stood that it is to be taken as positive ; hence, if r x and r 2 
are both positive, all the four values of x satisfy the original 
equation. If, however, r x or r 2 is negative, the roots found 
from the resulting quadratic will satisfy the equation 



ax 2 + bx + c — p-y/ax 2 + bx -f- c = q 9 
but not the original equation. 

Kx. 1. Solve x 2 - 5s + 2Vx 2 -5x + 8 = 12. 
Add 3 to each side ; then 

x a - 6x + 3 + 2\^-""5T+3 = 15. 

Putting Vx 3 — 5 c + 3 = y, we obtain y 2 -f 2 y — 15 = : whence 
y = 3 or - 5. 

Thus Vx* -6x + S =+ 3, or Vx 2 - 6x + 3 = -6. 



EQUATIONS IN QUADRATIC FORM. 247 

Squaring and solving the resulting quadratics, we obtain from the 
first x = 6 or — 1 ; and from the second x = ^- The first 

pair of values satisfies the given equation, but the second pair satisfies 
the equation 

X 2 _ 5a- _ 2Vx 2 -5x + 3 = 12. 

Ex. 2. Solve 3 x 2 - 7 + 3V3x 2 -16x+21 = 16 x. 

Transposing, 3 x 2 - 16 x - 7 + 3V3x 2 - 16x-f 21 = 0. 
Add 28 to each side ; then 

3x 2 - 16x + 21 + 3V3x 2 - 16a + 21 = 28. 
Proceeding as in Ex. 1, we have 

j/ 2 -f 3 y = 28 ; whence y = 4 or — 7. 
Thus V3x 2 - l~6lcT2l = 4 or V3X 2 - 16x + 21 = - 7. 
Squaring and solving, we obtain 

x = 5, J, or . 

The values 6 and J satisfy the original equation. The other values 
satisfy the equation 

3 x a _ 7 _ 3V3x 2 -16x-f 21 = 16x. 

296. Occasionally equations of the fourth degree may be 
arranged in expressions that will be in quadratic form. 

Ex. Solve jc*-8x» + 10x 2 -f24x + 5 = 0. 
This may be written x 4 - 8x 3 + 16x 2 - 6x 2 + 24x = - 5, 
or (x 2 - 4 x) 2 - 6 (x 2 - 4 x) = - 6 ; 



u * i o a 6 ± V36 — 20 6 + 4 c - 
by formula, x 2 — 4x = ^ = ^ = 6 or 1 ; • 

J ' 2 2' 

whence x = 6, — 1, or 2 £ \o. 

The student will notice that in such examples he should 
divide the term containing ar* by twice the square root of 
the first term and then square the result for the third term. 
In this case a third term of 16 a? 2 is required, therefore we 
write the term 10 ar* of the original equation in the form 
16^-6^. 

297. Equations like the following are of frequent occur 
rence. 



248 ALGEBRA. 



Ex. Solve ^-6 _5x_ = 6 

x x*-G 

Write y for x ; thus 
* x 



y-f 5 = 6, or y2-6y + 6 = 0; 

if 

whence y = 6, or 1. 

... ^« = 5 ,or^l«=l; 
x x 

that is, x 2 - 5s - 6 = 0, or x 2 - x - 6 = 0. 

Thus x = 6, - 1 ; or x = 8, - 2. 

EXAMPLES XXVII. b. 
Solve the following equations : 

1. x 2 + x+l= — —• 5. x 2 + 2 Vx 2 + 6x = 24-6x 

a' 2 + x 

2. ^_ + ^ll = 2 j. «. ( z _«V + 4x _24 = 5 . 
X 1 — 1 X \ xj X 

*' ( a; "^) 2 " 4 ( a; ' f x) = 5 ' 7 * 2 7« f -4 = 26^. 

4. h -s — ^ = -r- # •• #* + o as — - = o. 

x x 2 -3 2 x 2 + 3x 



9. 3x 2 -4x + V3x 2 -4x-6 = 18. 



10. 2x 2 -2x + 2V2x 2 -7x + 6 = 6x-6. 

11. x 2 + 6 Vx 2 - 2x + 5 = 11 + 2x. 

12. 2 Vx 2 -Gx + 2 + 4x + l = x 2 -2x. 

13. V4x 2 -f 2x -f 7 = VI X- + 6x - 119. 

14. 3x(3-x)=ll -4 Vx 2 -3x + 5. 

15. x 2 - x + 3 \/2 a* - 3x + 2 = - + 7. 

2 



16. 2x 2 -2x~17 + 2 V2x 2 -3x+7=x. 

IT. 2x 2 + 3x+l = — 

2x 2 + 3x 

18. x 4 -8x3-12x 2 +112x = 128. 

19. x 4 + 2x 8 -3x 2 -4x-96 = 0. 

20. x*~ 10x8+ 30x 2 -25x + 4 = 0. 
91. x*-14x 8 + 61x 2 -84x + 20 = 0. 



CHAPTER XXVIII. 



Simultaneous Equations, Involving Quadratics. 



We shall now consider some of the most useful 
methods of solving simultaneous equations, one or more of 
which may be of a degree higher than the first ; but no fixed 
rules can be laid down which are applicable to all cases. 

299. Equations solved by finding the Values of (x +/) and 

Ex. 1. Solve x + y = 15 (1), 

ay = 36 (2). 

From (1) by squaring, x 2 -f 2 xy + y 2 = 225 ; 
from (2), 4xy = 144; 

by subtraction, x 2 — 2 xy -f y 2 = 81 ; 

by taking the square root, x — y = ± 9. 

Combining this with (1) we have to consider the two cases, 



» + y=15, 1 a + y = 15, \ 
x - y = 9. i x — y = — 9. / 






from which we find x = 12, \ x = 3, 1 

y = 8. J y = 12. ) 






Ex. 2. Solve x - y = 12 


• • (1). 




• • (2). 


From (1), x 2 - 2sy -f y 2 = 144 ; 




from (2), 4sy = 340; 




by addition, a 2 -f 2 sey + y 2 = 484 ; 




by taking the square root, x -f y = ± 22. 




Combining this with (1) we have the two cases, 




a + y = 22,j z + y = -22,l 
x-y = 12.) x-y= 12.1 






Whence x = 17, ) a; = — 5, \ 

y= 5. i y=-17.i 






249 





250 ALGEBRA. 

300. These are the simplest cases that arise, but they 
are specially important since the solution in a large number 
of other cases is dependent upon them. 

As a rule our object is to solve the proposed equations 
symmetrically, by finding the values of x + y and x — y. 
Prom the foregoing examples it will be seen that we can 
always do this as soon as we have obtained the product of 
the unknowns, and either their sum or their difference. 

Ex. 1. Solve x 2 + y 2 =74 (1), 

xy = S5 (2). 

Multiply (2) by 2, then by addition and subtraction we have 

x* + 2 xy + y 2 = 144. 
x*-2xy + y 2 = 4. 
Whence x -f y = ± 12, 

x-y=± 2. 
We have now four cases to consider ; namely, 

x + y = 12, 
x- y 

From which the values of x are 7, 6, — 6, — 7 ; 
and the corresponding values of y are 5, 7, — 7, — 5. 

Ex. 2. Solve x 2 + y* = 185 (1), 

x + y =11 (2). 

By subtracting (1) from the square of (2) we have 

2xy = 104; 
.\ a^ = 62 (3). 

Equations (2) and (3) can now be solved by the method of Art. 299, 
Ex. 1 ; and the solution is 

X = 13, or 4, \ 
y = 4, or 13. J 

examples xxvni. a. 

Solve the following equations : 

1. x -f y = 28, 3. x + y = 74, 5. x — y = 8, 

xy = 187. xy = 1113. xy = 513. 

2. x + y = 61, 4. x — y = 5, 6. xy — 1075, 

xy — 618. sey = 126. x — y = 18. 



= 12,| z + y= 12,1 « + y=-12 f l a+y = -12,l 
= 2. J 3-^ = - 2. J x-y= 2./ aj-y=- 2. J 



SIMULTANEOUS EQUATIONS. 251 

7. ay = 923, 15. x* + y* = 66, 28. x-y = 3, 

x + y = 84. xy = 28. x 2 ~3xy + y 2 =- 19. 

*• -.::;£ "*.■£:!?• «•*-+*-'«. 

9. x-y = -22, 17. x + y = 15, x + y-U. 

xy = 3848. a 2 + j/ 2 = 126. 35- AC* -10=1. 

10. ay = -2193, 18. x-y = 4, x 2 ~4xy+y 2 =62. 
x + y = -8. x2 + y 2 = 106. ! x 

11. x-y = -18, 19. x 2 + 2^ = 180, *' x + y = 2 ' 

xy = 1368. x-y = 6. x + y = 2. 

12. xy = -1914, 20. x 2 + y 2 =185, 

x + y = -66. x-y = 3. 27. - + - = ^» 

18. x 2 + y* = 89, 21. x + y = 13, ' 

xy = 40. x 2 + y 2 = 97. xy ~" 

14. x 2 + y 2 = 170, 22. x + y = 9, 28. ax+6y = 2, 
xy = 13. x 2 + xy + y 2 = 61. a&xy = 1. 

99. z*+pxy + y*=p + 2, 
gx 2 + xy + gy 2 = 2g+l. 

301. Equations which can be reduced to One of the Cases 
already considered. Any pair of equations of the form 

' *±P*y + f = « (1), 

x±y = b (2), 

where p is any numerical quantity, can be reduced to one of 
the cases already considered ; for, by squaring (2) and com- 
bining with (1), an equation to find xy is obtained ; the solu- 
tion can then be completed by the aid of equation* (2). 

Ex.1. Solve x»-y» = 999 (1), 

s-y = 3 (2). 

By division, x 2 + xy + y 2 = 333 (3); 

from (2), x 2 - 2xy + y 2 = 9 ; 

by subtraction, 3 xy = 324, 

xy = 108 (4) 

From C2) and {4), * = 12 » or ~ 9 » \ 

' ' y = 9, or - 12. J 



252 ALGEBRA. 

Ex. a. Solve x* + xy + y* = 2613 (IX 

z* + zy + j/* = 67 (2). 

Dividing (1) by (2), x 2 - xy + y 2 = 39 (8). 

From (2) and (3), by addition, x* + y 2 = 53 ; 
by subtraction, xy = 14 ; 

whence * = ± J. ± M ^ Art g^ B L] 

y = ±2, ±7. * L 

Ex.3. Solve --- = J (1), 

x y 3 x ' 

1 + 1-5 m 

• x 2 + y*~9 W% 

From (1), by squaring, 1 - 1. + 1 - 1 . 

^ ^ y 2 9 

by subtraction, — = - ; 

xy 9' 

addingto(2), l + A+l = i ; 

* 2 xy y 2 

.-. i+i = ±l. 



a; y 



Combining with (1), i = -, or - 1, 

x o 3 

! =4or-? ; 
y 8' 8' 



.% a; = |, or - 3, ) 
y = 3, or — f . 1 



EXAMPLES XXVIII. b. 



1 a£+y« = 407 l 3. x + y = 23, 5. x-y = 4, 

x + y = 11. a* + ys = 3473. x 8 - y 8 = 988. 

2. x* + y8 = 637, 4. x 8 - y 8 = 218, 6. x 8 - 3^ = 2197, 

x + y = 13. a; — y = 2. x — y = 13. 

7. a* + afy 2 + y* = 2128, 9. X* + x¥ + y 4 = 9211, 

x 2 + xy + y 2 = 76. a; 2 - xy + y 2 = 61. 

8. a* + x*y 2 + y* = 2923, 10. aj* + x*y 2 + y* = 7371, 

x 2 - xy + y 2 = 37. x 2 - xy + y* = 63. 



SIMULTANEOUS EQUATIONS. 258 

a 2 y 2 676' y x ro se 2 -a;y+y 2 = 21. 



» 



1 + 1 = ??. se-y = 4. j x 

* y 24 15 34 ^15 19 ' ^ + ^ =lTiir 

^ 1 , 1_61 * * + * *y * 4.1-11 

a* V~900' *+y = 8. a. + y- x *- 

ay = 80. 16. « 8 -y 8 = 66, j 1 

13. * + a=2i ^ r 

y x " 17. 4(s 2 + y 2 )=17zy, !_I = l. 

z + y = Q. x—y = 6. z y 

302. Homogeneous Equations of the Same Degree. The 

following method of solution may always be used when the 
equations are of the same degree and homogeneous. 

Ex. Solve z 2 + xy + 2f/* = 74 (1), 

2x 2 + 2sy + y 2 = 73 (2). 

Tut y = mx y and substitute in both equations. Thus 

s 2 (l + m + 2m 2 )=74 . ..*... (8), 

and & 2 (2 + 2m-t-m 2 )=78 (4).. 

By division, l + m + 2m 2 = 74 

J * 2 + 2m + m 2 73 

.•. 73 + 73m + 146 w 2 = 148 + 148m + 74ttf; 

.-. 72m 2 - 75m -76 = 0, 

or 24m 2 - 25m -25 = 0; 

.•. (8m + 6)(3m-6) = 0; 

.«. m=-{, or f. 

(1.) Take m = — f , and substitute in either (8) or (4). 

From (3) * 2 (1 - f. + J J) = 74 ; 

. ~2 64 x 74 ^. 
.-. » 2 = — ^— = 64; 

.«. x = ±8; 

••. y = mz = — { x ± 8 = =F 6. 
(li) Take m = J ; then from (3), s 2 (l + J + ^) = 74* 

* " 74 -"' 
.*. a: = ±3; . y = mx = J x ±3=±6» 



254 ALGEBRA. 

The student will notice that, having found the values of 
x, we obtained those of y from the equation y = mx, using, 
in each case, the value of m employed in finding those par- 
ticular values of x. 

303. Equations of which One is of the First Degree and the 
Other of a Higher Degree. We may from the simple equa- 
tion find the value of one of the unknowns in terms of the 
other, and substitute in the second equation- 
Ex. Solve 3s-4y = 5 (1). 

Sx*-xy- 3^ = 21 (2). 

5 _l 4 v 
From (1) we have x = ^ " ; 

o 

and substituting in (2), ?^+i^ _ «? + *£ _ 3 «■ = 21 ; 

.-. 76 + 120y + 48y 2 -15y-12y 2 --27jf a = 189; 
9y 2 + 1052/- 114=0; 
3y 2 + 35y-38 = 0; 
.-. (y-l)(3y + 38)=0; 

.-. y = l, or -*fi; 
and by substituting in (1), x = 3, or — iji. 

304. Symmetrical Equations. The following method of 
solution may always be used when the given equations are 
symmetrical, that is, when the unknown quantities in each 
equation may be interchanged without destroying the 
equality. The same method may generally be employed 
with advantage where the given equations are symmetrical 
except with respect to the signs of the terms. 

Ex. Solve s 4 + y 4 = 82 (1), 

x — ?/ = J . . • . . • • . ( /* 
Put x = u -f 0, and y = u — v ; 

then from (2) we obtain t? = 1. 

Substituting in (1), (ti + l) 4 + (u - 1)* = 82 ; 

.-. 2(tt 4 + 6i«2+ 1)=82; 
u* + 6 m 2 - 40 = ; 



SIMULTANEOUS EQUATIONS. 265 



whence u 2 = 4, or — 10 ; 

and u = ± 2, or ± V— 10. 



Thug, je = w + t> = 3, -1, 1±V^10; 



y = « - t> = 1, -3, - 1 ± V-10. 

Note. We may assume x + y = 2 m and as — y = 2 1>, m and t? be- 
ing any unknown quantities, whence we obtain x = u -f «, and 
y = w — v, the values used in the above. 



305. Miscellaneous Cases. The examples we have given 
will be sufficient as a general explanation of the methods 
to be employed; but in some cases special artifices are 
necessary. 

Ex. 1. Solve aty 2 - 6a; = 34 - 3y (1), 

3xy + y = 2(9 + x) (2). 

From (1), xhf* -6x + Sy = 34; 

from (2), 9xy-6x+3y = 54; 

by subtraction, x*y 2 — 9xy = - 20, 

„, 9 ± V81 - 80 9 ±1 - - 
xy = -^ — =_^_=6or4. 

(i.) Substituting xy = 6 in (2) gives y — 2 x = 3. 

From these equations we obtain x = 1, or — f, 1 

y = 6, or — 2. / 

(ii.) Substituting xy = 4 in (2) gives y — 2 a; = 6. 
From these equations we obtain x = — — ^ v , 1 
and y = 3 ± ^17. J 

Ex.2. Solve ^ + 2/s + s 2 = 49 (1), 

« 2 -Mx + x 2 = 19 ....... (2), 

x 2 + xy + y 2 = 39 (3). 

Subtracting (2) from (1), 

y*-x* + z(y-x)=S0; 

that is, ( y _ x )( iC + 2/ + 2 :) = 30 (4). 

Similarly from (1) and (3), 

(z-x)(x + y + «)= 10 (6) 



256 ALGEBRA. 

Hence from (4) and (6), by division, 

z — x 
whence y = 3s — 2& 

Substituting in equation (3), we obtain 

x*-3zz + 3z* = lS. 
From (2), x 1 + xz + z 2 = 19. 

Solving these homogeneous equations, we obtain 

x = ± 2, « = ± 3 ; and therefore y = ± 6 ', 

or x = ± -ii-, « = ± — ; and therefore y = =F — • 

examples xxvni. a 

1. 6x-y = 17, 6. 3x-y = ll, 9. 6x+y = 3, 

xy = 12. 3x*-y 2 = 47. 2x 2 -3xy-y 2 = 1. 

2. x 2 + xy = 15, 6. x-3y=l, 10. 3x 2 -6y 2 = 28, 
y* + xy = 10. r» 2 — 2a?2/4-9y 2 = 17. 3xy-4y 2 = 8. 

3. a; -y = 10, 7. x + 2y = 9, 11. 3x 2 -y 2 = 23, 
x 2 -2xy-3y 2 = 84. 3y 2 -5x 2 = 43. 2x 2 -xy = 12. 

4. 3 x + 2 = 16, 8. x 2 + y 2 = 5, 12. x 2 +*y+y 2 = 3J, 

xy = 10. 2xy-y 2 = 3. 2x 2 -3xy+2y 2 =2}. 

13. x 2 — 3xy + y 2 + l = 0, 14. 7xy-8x 2 =10, 

3x 2 - xy + 3 y 2 = 13. 8y 2 - 9xy = 18. 

15. x 2 -2xy = 21, 17. x 8 + y 8 = 152, 19. x 8 - y 8 = 208, 

zy + t = 18. xfy + xy 2 = 120. xy(x - y) = 48. 

16. x 2 + 3xy = 54, 18. x 8 - y 8 = 127, 20. x 2 y 2 + 6xy = 84, 
xy + 4 y 2 = 115. x*y - xy 2 = 42. x + y = 8. 

21. x 2 + 4y 2 + 80 = 16x + 30y, 22. 9x 2 +y 2 -63x-21y+ 128=0, 

xy = 6. xy=4. 

23.1 + 1 = 45, 25. x*+x 2 y 2 +y*=931, 38. x + y = 7 + Viy, 
x * y 2 4 ' a .2_ icy+y 2 =: i9. x 2 + y 2 = 133-xy. 

1_1 = 3 y * 

x y 2* 26- x 2 +xy+y 2 =84, 29. 3x 2 -6y 2 = 7, 

OA 1 , 1__243 x-Vxy+y=6. 3xy-4y 2 = 2. 

x 8 y 8 "^ - ' 

1 1_9 27. x+Vxy+y=66, 30. 5y*-7x 2 = 17 

i + y = 2' x 2 +xy+y 2 =2276. axy-Oa^O. 



SIMULTANEOUS EQUATIONS. 267 

31. 8a! 2 +165=iejc.y, 85. x* + y*=272, 89. x + y = 1072. 
7 xy + 3 y* = 1 :J2. x - y = 2. x * + y* = 16. 

32. 3x 2 +xy+y 2 =15, 36. x 5 - y 6 = 992, , . 
31xy-3x a -5y-=45. x-y = 2. 40. xy J +yx* = 20, 

33. x 2 +y 2 --3=3xy, 37. xy-6xy 2 =-9, x-+y* = 85. 
2x 2 -6+j/ 2 =0. xy-y=2. ^ , 

34. x*+y* = 706, 88. 2x 8 + 2y8 = 9xy, 4X ' * * + y = 6 ' 

x + y = 8. x + y = 3. 6(x~*+y~*) =6. 

43. ^+<^ = y» 44. 4x 2 +5y=6+20xy-85y»+2x, 

'« + !!= 10. 7x-lly=17. 

^ V^-Vy 4 ^±V*" 45 . ^2 + 400 = 41*,,, 

x 2 + y 2 = 706. y 2 +4x* = 6xy. 

46. 9x 2 +33x-12 = 12xy-4y 2 + 22y, 

x 2 - xy = 18. 

47. xy + a& = 2ax, 60. x 2 + y 2 - z 2 = 21, 
x*y 2 + a 2 & 2 = 2 6 2 y 2 . Sxz + 8y* - 2xy = 18, 

x + y — z = 5. 

48. (x 2 - y 2 ) (x - y) = 16 xy, 51. x-y-« = 2, 

(x*-y*)(x 2 -y 2 ) = 640x 2 y 2 . x 2 + y 2 -s 2 = 22, 

xy — o. 

19. 2*» - *j, + ? = 2j,, "■ * |+ ** + » = 18 « 

y 2 + y« + yx = — 12, 



CHAPTER XXIX. 
Problems leading to Quadratic Equations. 

306. We shall now discuss some problems which give 
rise to quadratic equations. 

Ex. 1. A train travels 300 miles at a uniform rate ; if the rate had 
been 5 miles an hour more, the journey would have taken two hours 
less : find the rate of the train. 

Suppose the train travels at the rate of x miles per hour, then thr 

... 300 , 
time occupied is — hours. 

x 300 

On the other supposition the time is , g hours ; 

"^+6 "IT w$ 

whence x 2 -f 5 x — 760 = 0, 

or (x + 30) (as - 25) = 0, 

.-. x=25, or -30. 

Hence the train travels 25 miles per hour, the negative value being 
inadmissible. 

It will frequently happen that the algebraic statement of the ques- 
tion leads to a result which does not apply to the actual problem 
we are discussing. But such results can sometimes be explained by 
a suitable modification of the conditions of the question. In the 
present case we may explain the negative solution as follows : 

Since the values * = 25 and — 30 satisfy the equation (1), if we 
write — x for x, the resulting equation, 

300 300 



— x+ 5 —or 



-2 (2), 



will be satisfied by the values x = — 25 and 30. Now, by changing 

300 300 

signs throughout, equation (2) becomes — = — — h 2 ; 

x — 6 x 
and this is the algebraic statement of the following question : 

A train travels 300 miles at a uniform rate ; if the rate had been 

5 miles an hour less, the journey would have taken two hours more : 

find the rate of the train. The rate is 30 miles an hour. 

268 



PROBLEMS. 25& 

Ex. 2. A person, selling a horse for $72, finds that his loss per 
cent is one-eighth of the number of dollars that he paid for the horse 
what was the cost price ? 

Suppose that the cost price of the horse is x dollars ; then the loss 
on ftlOO is ft* 

Hence the loss on ft x is x x -^-, or — dollars; 

800 800 

.*. the selling price is x dollars. 

800 

Hence x — £- = 72. 

800 

or x 2 - 800s + 67600 = ; 

that is, (x - 80)(x - 720) = ; 

' .•. s = 80, or 720; 

and each of these values will be found to satisfy the conditions of the 
problem. Thus the cost is either $80, or $ 720. 

Ex. 3. A cistern can be filled by two pipes in 33J minutes ; if the 
larger pipe takes 15 minutes less than the smaller to fill the cistern, 
find in what time it will be filled by each pipe singly. 

• 

Suppose that the two pipes running singly would fill the cistern 
in x and x — 16 minutes. When running together they will fill 

(I _| \ of the cistern in one minute. But they fill — , or -— 
x s-15/ J 33J' 100 

of the cistern in one minute ; hence 

1 1 _ 3 

x a- 16 100' 

100(2 x - 15) = Sx(x - 16), 

3« a -246ajf 1600 = 0, 

(s-76)(3a-20)=0; 

.•. x = 76, or 6$. 

Thus the smaller pipe takes 75 minutes, the larger 60 minutes. 
The other solution, 6|, is inadmissible. 

Ex. 4. The small wheel of a bicycle makes 135 revolutions more 
than the large wheel in a distance of 260 yards ; if the circumference 
of each were one foot more, the small wheel would make 27 revolu- 
tions more than the large wheel in a distance of 70 yards : find the 
circumference of each wheel. 



260 ALGEBRA. 

Suppose the small wheel to be x feet, and the large wheel y feet in 
circumference. 

780 780 

In a distance of 260 yards, the two wheels make — and — revo- 

x y 

lutions respectively. 

Hence 780 _ 780 = 136 

* V 

or --- = £• (1). 

x y 62 v ' 

Similarly, from the second condition, we obtain 

210 210 



x+1 y+i 



= 27, 



or —i L_ = _?_ (2) 

« + 1 y + 1 70 w 

From(l), x = 62 ? ; 

Wf 52 + 9y' 

whence * + 1 = fll *+M. 

9y + 52 

Substituting in (2), 9y + 62 — = — , 

e w ' 6ly + 52 y + 1 70' 

70 x 9y2 - 9(61 y + 62) (y + 1), 

9y 2 -113y-62 = 0, 

0r-18)(9|f + 4)=0; 

.•. y = 13, or — $. 

Putting y = 13, we find that x = 4. The other value of y is inad- 
missible ; hence the small wheel is 4 feet, the large wheel 13 feet in 
circumference. 

Ex. 6. On a river, there are two towns 24 miles apart. By rowing 
one half of the distance, and walking the other half, a man performs 
the journey down stream in 5 hours, and up stream in 7 hours. Had 
there been no current, each journey would have taken 5f hours. Find 
the rate of his walking, and rowing, and the rate of the stream. 

Suppose that the man walks x miles per hour, rows y miles per 
hour, and that the stream flows at the rate of z miles per hour. 

With the current the man rows y -f z miles, and against the current 
y — z miles per hour. 



PROBLEM8. 26\ 

Hence we have the following equations - 

12 j_ 12 -ft IN 

— + — — = 5 v l), 

x y + z 

^ + -^- = 7 (2), 

x y-z 

— I ~6| (3). 

x V 

From (1) and (3), by subtraction, — — i .... (4). 

y y + z IS 

Similarly, from (2) and (8) , — 1 = - . . , - ;6). 

y-z y 9 w 

From (4) lBz = y(y + z) (6); 

and from (5) 9 z = y(y — z) (7). 

From (6) and (7), by division, 2 = ¥ + g ; 

y — z 

whence y = 3 z ; 

,\ from (4) z = l\\ and hence y = 4J, « = 4. 

Thus the rates of walking and rowing are 4 miles and 4} miles per 
hour respectively ; and the stream flows at the rate of 1 J miles per 
hour. 



EXAMPLES XXIX. 

1. Find a number whose square diminished by 119 is equal to ten 
times the excess of the number over 8. 

2. A man is five times as old as his son, and the sum of the squares 
of their ages is equal to 2106 : find their ages. 

3. The sum of the reciprocals of two consecutive numbers is \% : 
find them. 

4. Find a number which when increased by 17 is equal to 60 times 
the reciprocal of the number. 

5. Find two numbers whose sum is 9 times their difference, and 
the difference of whose squares is 81. 

6. The sum of a number and its square is nine times the next 
higher number : find it. 

7. If a train travelled 5 miles an hour faster, it would take one 
hour less to travel 210 miles : what time does it take ? 

8. Find two numbers the sum of whose squares is 74, and whose 
sum is 12. 

9. The perimeter of a rectangular field is 500 yards, and its area 
is 14400 square yards : find the length of the sides. 



262 ALGEBRA. 

10. The perimeter of one square exceeds that of another by 100 
feet ; and the area of the larger square exceeds three times the area 
of the smaller by 325 square feet : find the length of their sides. 

11. A cistern can be filled by two pipes running together in 22 1 
minutes ; «the larger pipe would fill the cistern in 24 minutes less than 
the smaller one : find the time taken by each. 

12. A man travels 108 miles, and finds that he could have made 
the journey in 4 \ hours less had he travelled 2 miles an hour faster : 
at what rate did he travel ? 

13. 1 buy a number of foot-balls for $ 100 ; had they cost a dollar 
apiece less, I should have had five more for the money : find the cost 
of each. 

14. A boy was sent for 40 cents 1 worth of eggs. He broke 4 on his 
way home, and the cost therefore was at the rate of 3 cents more 
than the market price for 6. How many did he buy ? 

15. What are the two parts of 20 whose product is equal to 24 
times their difference ? 

16. A lawn 50 feet long and 34 feet broad has a path of uniform 
width round it ; if the area of the path is 540 square feet, find its 
width. 

17. A hall can be paved with 200 square tiles of a certain size ; if 
each tile were one inch longer each way it would take 128 tiles : find 
the length of each tile. 

18. In the centre of a square garden is a square lawn ; outside this 
is a gravel walk 4 feet wide, and then a flower border 6 feet wide. If 
the flower border and lawn together contain 721 square feet, find the 
area of the lawn. 

19. By lowering the price of apples and selling them one cent a 
dozen cheaper, an applewoman finds that she can sell 60 more than 
she used to do for 60 cents. At what price per dozen did she sell 
them at first ? 

20. Two rectangles contain the same area, 480 square yards. The 
difference of their lengths is 10 yards, and of their breadths 4 yards : 
find their sides. 

21. There is a number between 10 and 100 ; when multiplied by 
the digit on the left the product is 280 ; if the sum of the digits be 
multiplied by the same digit, the product is 55 : find it. 

22. A farmer having sold, at $75 each, horses which cost him x 
dollars apiece, finds that he has realized x per cent profit on his 
outlay : find x. 

28. If a carriage wheel 14f ft. in circumference takes one second 
more to revolve, the rate of the carriage per hour will be 2} miles less ; 
how fast is the carriage travelling ? 



PROBLEMS. 263 

24. A merchant bought a number of yards of cloth for $ 100 ; he 
kept 6 yards and sold the rest at $ 2 per yard more than he gave, and 
received $ 20 more than he originally spent : how many yards did he 
buy? 

25. A broker bought as many shares of stocK as cost him $ 1875 ; 
he reserved 15, and sold the remainder for $ 1740, gaining $4 a share 
on their cost price. How many shares did he buy ? 

26. A and B are two stations 300 miles apart. Two trains start 
simultaneously from A and B, each to the opposite station. The 
train from A reaches B nine hours, the train from B reaches A four 
hours after they meet : find the rate at which each train travels. 

27. A train A starts to go from P to Q, two stations 240 miles 
apart, and travels uniformly. An hour later another train B starts 
from P, and after travelling for 2 hours, comes to a point that A had 
passed 45 minutes previously. The pace of B is now increased by 5 
miles an hour, and it overtakes A just on entering Q. Find the rates 
at which they started. 

28. A cask P is filled with 50 gallons of water, and a cask Q with 
40 gallons of brandy ; x gallons are drawn from each cask, mixed and 
replaced ; and the same operation is repeated. Find x when there are 
8 J gallons of brandy in P after the second replacement. 

29. Two farmers A and B have 30 cows between them ; they sell 
at different prices, but each receives the same sum. If A had sold 
his at B's price, he would have received $ 320 ; and if B had sold his 
at A's price, he would have received $ 245. How many had each ? 

30. A man arrives at the railroad station nearest to his house 1J 
hours before the time at which he had ordered his carriage to meet 
him. He sets out at once to walk at the rate of 4 miles an hour, and, 
meeting his carriage when it had travelled 8 miles, reaches home 
exactly 1 hour earlier than he had originally expected. How far 
is his house from the station, and at what rate was his carriage 
driven ? 



CHAPTER XXX. 

Theory of Quadratic Equations, 
miscellaneous theorems. 

307. In Chapter xxvi. it was shown that after suitable 
reduction every quadratic equation may be written in the 
form 

ax? + bx + c = (1), 

and that the solution of the equation is 

x= -b±Vb*-4 ac (2) 

We shall now prove some important propositions con- 
nected with the roots and coefficients of all equations of 
which (1) is the type. 

NUMBER OF THE ROOTS. 

308. A quadratic equation cannot have more than two roots. 

For, if possible, let the equation ax? -f bx + c = have 
three different roots r lf r 2 , r 3 . Then since each of these 
values must satisfy the equation, we have 

ar 1 2 + br 1 + c = (1), 

ar 2 2 + br 2 + c = (2), 

ar/+&r 3 + c = (3). 

From (1) and (2), by subtraction, 

a(Ti 2 — r 9 *)+b(r! — r,)= ; 

divide out by r x — r 2 , which, by hypothesis, is not zero ; 
then 

. a(n-{ r 2 )+b = 0. 

264 



THEORY OF QUADRATIC &QUATIONS. 265 

Similarly from (2) and (3), 

.\ by subtraction, a(r x — r 3 ) = ; 

which is impossible, since, by hypothesis, a is not zero, and 
r x is not equal to r 8 . Hence there cannot be three different 
roots. 

309. The terms ' unreal,' ' imaginary,' and ( impossible' 
are all used in the same sense; namely, to denote expres- 
sions which involve the square root of a negative quantity, 
such as 

V— 1, V— 3, V— a. 

It is important that the student should clearly. distinguish 
between the terms real and rational, imaginary and irra- 
tional. Thus ^25 or 5, 3£, — £ are rational and real; -y/7 
is irrational but real; while V— 7 is irrational and also 
imaginary. 

CHARACTER OF THE ROOTS. 

310. In Art. 307 denote the two roots in (2) by r x and r* 



r l — n 9 r 2 — n J 

2a 2a 

then we have the following results : 

(1) If b 2 — 4 ac, the quantity under the radical, is positive, 
the roots are real and unequal. 

(2) If b 2 — 4 ac is zero, the roots are real and equal, each 

reducing in this case to — -—• 

2 a 

(3) If b 2 — 4 ac is negative, the roots are imaginary and 
unequal. 

(4) If b 2 — 4 ac is a perfect square, the roots are rational 
and unequal. 

By applying these tests the nature of the roots of any 
quadratic may be determined without solving the equation* 



266 ALGEBRA. 

Ex. 1. Show that the equation 2 x 2 ~ 6 x+ 7= cannot be satisfied 
by any real values of x. 

Here a = 2, b = — 6, c = 7 ; so that 

fr 2 - 4 ac =(- 6) 2 - 4 . 2 . 7 = - 20. 

Therefore the roots are imaginary. 

Ex. 2. For what value of k will the equation Sx 2 — 6 as + ifc = 
have equal roots ? 

The condition for equal roots gives 

(_6) 2 -4.3.* = 0, 
whence k = 3. 

Ex. 8. Show that the roots of the equation 

x 2 - 2 ax + a 2 - 6 2 + 2 be - c 2 = are rational. 

The roots will be rational provided (— 2 a) 2 — 4 (a 2 — & 2 +26c— c 2 ) 
is a perfect square. But this expression reduces to 4 (6 s — 2 be + c 2 ) 
or 4(6 — c) 2 . Hence the roots are rational. 

RELATIONS OF ROOTS AND COEFFICIENTS. 

-,, q. — b + V& 2 — 4 ac „ — b — V& 2 — 4ac 
311. Since r 1= = -J-— , r 2 = , 

2a 2a 

we have by addition 

, _ -.&4-V& 2 -4ac-5-V& 2 -4ac _ 6 m 

Za a 

and by multiplication we have 

(_ b + V& 2 -4oc) (_ 6 -V& 2 -4ac) 

4 a 2 



_ ( _ &) 2 _ (& 2 -4ac) _4ac_c (2) 

4a 2 4a 2 a w 

By writing the equation in the form o^H — x-\ — =0, these 

a a 

results may also be expressed as follows : 

In a quadratic equation where the coefficient of the first term 
is unity, 

(i.) the sum of the roots is equal to the coefficient of x 
with its sign changed ; 

(ii.) the product of the roots is equal to the third term. 

Note. In any equation the term which does not contain the 
unknown quantity is frequently called the absolute term. 



THEORY OF QUADBATIC EQUATIONS. 267 

FORMATION OF EQUATIONS WITH GIVEN BOOTS. 

b c 

312. Since = r x + r% and - = r x r^ 

% OL Of 

b c 
the equation a? + -x4-— =s may be written 

a a 

aj 2 -(r 1 -hr 2 )a?-hr 1 r 2 = (1). 

Hence any quadratic may also be expressed in the form 

x 2 — (sum of roots) x + product of roots = . (2). 
Again, from (1) we have 

(x — r^)(x — r 2 ) = (3). 

We may now form an equation with given roots. 

Ex. 1. Form the equation whose roots are 3 and — 2. 
The equation is (x - 3) (x + 2) = 0, 

or x* - x - 6 = 0. 

Ex. 2. Form the equation whose roots are \ and — $ . 

The equation is (x — J) (x + $) = ; 

that is, (7x - 3)(6<e + 4)= 0, 

or 36je 2 + 13s -12=0. 

When the roots are irrational it is easier to use the following 
method. 

Ex. 8. Form the equation whose roots are 2 + y/S and 2 — y/S. 

We have sum of roots = 4, 

product of roots = 1 ; 
.•. the equation is x 2 — 4as + 1 = 0, 

by using formula (2) of the present article. 

313. The results of Art. 311 are most important, and they 
are generally sufficient to solve problems connected with 
the roots of quadratics. In such questions the roots should 
never be considered singly, but use should be made of the 
relations obtained by writing down the sum of the roots, 
and their product, in terms of the coefficients of the equation. 



268 ALGEBRA. 

Kx. 1. If a and b are the roots of x 2 — px + q = 0, find the value 

of 0) «' + &*» ( 2 ) a 8 + & 8 . 

Wo have a 4- b = p, 

aft = q. 

.-. aP+b 2 =(a + b) 2 -2ab=iP-2q. 

Again, a 8 + 6 8 = (a + b) (a 2 + 6* - ab) 

= i>{(a + &)' - 3a6} = p(p 2 - 3g). 

Kx. 9. If a, 6 are the roots of the equation lz 2 + mz+ n = 0, find 
tho equation whose roots are -, — 

We have sum of roots = - + - = "*" , 

& a a& 

product of roots = ? x - = 1 ; 

& a 

••• by Art. 312 the required equation is 

or abx 2 - (a 2 + b 2 )x + a& = 0. 

As in the last example a 2 + 6 2 = ffl2 ~ 2 nZ , and a& = 5, 

t « 

.-. the equation is 5 ^ _ m»-2nl + n = Q 

or nZx 2 - (m 2 - 2 nl)x + n* = 0. 

Ex. 8. Find the condition that the roots of the equation 

ax 2 + bx + c = 

should be (1) equal in magnitude and opposite in sign, (2) reciprocals. 
The roots will be equal in magnitude and opposite in sign if their 

sum is zero ; therefore = 0, or 6 = 0. 

a 

Again, the roots will be reciprocals when their product is unity ; 

therefore - = 1, or c = a. 
a 

Ex. 4. Find the relation which must subsist between the coeffi- 
cients of the equation px 2 + qx -f r = when one root is three times 
the other. 

We have a + & = — &, <*& = -; 

p p 

but since a = 3 6, we obtain by substitution 

4& = -2, 3& 2 = £. 
P P 



THEOKY OF QUADRATIC EQUATIONS. 269 

From the first of these equations b? = ^ ^ , and from the second 

lOp 2 3p 
or 3g 2 = 16pr, 

which is the required condition. 

314. The following example illustrates a useful applica- 
tion of the results proved in Art. 310. 

Ex. If g is a real quantity, prove that the expression ^— ^ — — - — - 

can have all numerical values except such as lie between 2 and 6. 
Let the given expression be represented by y, so that 

s» + 2s-ll 
2( X _3) *' 

then multiplying across and transposing, we have 

a 2 -f 2z(l - y)+ 6y - 11 =0. 

This Is a quadratic equation, and if x is to have real values, 
4(1 — y) 2 — 4(6 y — 11) must be positive; or simplifying and dividing 
hy 4, y 2 — Sy + 12 must be positive ; that is, (y — 6)(y — 2) must be 
positive. Hence the factors of this product must be both positive 
or both negative. In the former case y is greater than 6; in the 
latter y is less than 2. Therefore y cannot lie between 2 and 6, but 
may have any other value. 

In this example it will be noticed that the expression 
#* — 8^ + 12 is positive so long as y does not lie between 
the roots of the corresponding quadratic equation 

y*-$y + 12 = 0. 

This is a particular case of the general proposition investi- 
gated in the next article. 

315. For all real values of x the expression ax 2 + bx + c 
has the same sign as a, except when the roots of the equation 
ax 2 + bx -f c = are real and unequal, and x lies between them. 

Case I. Suppose that the roots of the equation 

aa? + bx + c = 
are real ; denote them by r x and r 2 , and let r t be the greater. 



270 ALGEBRA. 



Then aa? + bx + c 



\ a a) 



= a \v?-(r x + r 2 )x + r x r 2 \ [Art 311.] 

= <z(a? — r x ) (x — r 2 ). 

Now if a? is greater than r x or less than r^ the factors 
x — Vu x — r 2 are either both positive or both negative ; 
therefore the expression (x — r^) (x — r 2 ) is positive, and 
aa?-\-bx+c has the same sign as a. But if x lies between 
r x and r^ the expression (x — rj) (x — r 2 ) is negative, and 
the sign of aa? + bx + c is opposite to that of a. 

Case II. If r x and r 2 are equal, then 

da? -4- bx + c = a (x — rj)*, 

and (a — i^)* is positive for all real values of x\ hence aa? 
+ bx + c has the same sign as a. 

Case III. Suppose that the equation aa?+ bx + c = has 
imaginary roots ; then 

aa? + bx -fc = a-<aj 2 + -# + -£• 

( a a) 



=«{(-£K-^}> 



but since 6 s — 4 ac is negative [Art. 310], the expression 

\ 2 ay 4 a 2 

is positive for all real values of x\ therefore aot? + bx + c 
has the same sign as a. 

EXAMPLES XXX. S. 

Find (without actual solution) the nature of the roots of the follow- 
ing equations : 

1. a* 3 + s -870 = 0. 3. Jo? = 14 -3a 2 . 5. 2x = ib 2 + 6. 

2. 8 + 6x = 6 x 2 . 4. x 2 + 7 = 4aj. 6. (<e+2) a =4x-fl5. 

Form the equations whose roots are 

7. 5, —3. 9. a + b, a — b. U. fa, —Jo. 

8. - 9, - 11. la J, J. 1* 0, j. 



THEORY OF QUADRATIC EQUATIONS. 271 

13. If the equation & + 2(1 + h)z + IP a has equal roots, what 
is the value of jfc ? 

14. Prove that the equation 3 mx* — (2 m + 8 n)g + 2n s has 
rational roots. 

15. Without solving the equation 3 as 2 — 4x — 1=0, find the sum, 
the difference, and the sum of the squares of the roots. 

16. Show that the roots of a(x 2 — 1) = (b — c)x are always reaL 
Form the equations whose roots are 

17. S + V 5 > 8 -V 6 - W. -2-rV 3 , -2-^/8. 18. -£, \. 

5 6 

. v a-6 a + 6 26 2a 

If a, & are the roots of the equation px 2 -f qx + r = 0, find the 
values of 

23. a 2 + & 2 . 25. a 2 & + a& 2 . 27. a 6 6 2 + o 2 6*. 

24. (a-&) 2 . 26. a 4 + 6*. 28. ^ + £ 

& a 

29. If a, 6 are the roots of x* — pa + q = 0, and a 8 , &* the roots of 
x 2 — Px -f § = 0, find P and Q in terms of p and q. 

30. If c, d are the roots of z 2 - ox + 6 = 0, find the equation 

whose roots are — , -• 

d 2 c 2 

81. Find the condition that one root of the equation ax 2 +bx+c=0 
may be double the other. 

82. Form an equation whose roots shall be the cubes of the roots 
of the equation 2x(x — a)= a 2 . 

88. Prove that the roots of the equation 

(a + b)x 2 -(a + 6 + e)x + £ = 
are always real. 2 

84. Show that (a + b + c)x 2 - 2 (a + b) x + (a + & - c) = has 
rational roots. 

/j.2 15 

85. Show that if x is real the expression ^ cannot lie between 

3 and 5. 2 *~" 8 

86. If x is real, prove that — ~ t " can have all values except 

7 r qA 2 2! — 1 

such as lie between 2 and — {. 



MISCELLANEOUS THEOREMS. 
31& The Remainder Theorem. If any algebraical expre. 
jT+p^ +p_J"-*+lh*'- l +— + p m -&+p* 
X* liiebied by x — a, the remainder trifl be 

U'+JV»" '+Pfl' *+p&"*+—+p m t a+p n . 
I'ivide the given expression by i-a till a remainder is 
sdmimed which does not involve x. Let Q be the quotient, 
,tud H the remainder; then 

x* + pvtT-' +2>rC"- 2 + —+P. -,Z +p m = Q(x - «)+ R. 
Siiu-i' if does not contain z, it will remain unaltered what- 
ever value we give to £. 
I'ut x — a, then 
a-+i»!«* '+P20" ■*+— +P- 1 a+.P. = QxO + .R, 

,\ B = a B +p I a" _l +JV*""* H r-JVi« +P.J 

which proves the proposition. 

Kroui tins it appears that when an algebraic expression 
with integral exponents is divided by x — a, the remainder 
tutu be obtained at once by writing a in the place of x in 
(lie given expression. 

Ex. The remainder when j'-Sr'lr-i [a divided by 1+2 is 
(_2)*-2(-2)* + (-2)-7; 
ilmi U, ltt + 18 - a - 7, or 23. 

317. In the preceding article the remainder is zero when 
tliii given expression is exactly divisible by x—a; hence we 

The Factor Theorem, If an;/ rational and integral expres- 
sion containing j' becomes eqnal to ichen a is written for x, 
lii tliciMI? by x-a, [See Art. 105.) 

"n ,/fnd thu condition that tf + px + q may be a 

I. Im I'vidi'tit that any such general expression can- 
[wrftH't Miuaro unless some particular relation 



MISCELLANEOUS THEOREMS. 273 

subsists between the coefficients p and q. To find the 
necessary connection between p and q is the object of the 
present question. 

Using the ordinary rule for square root, we have 



x* ^ 



2* + f 



px + q 



If therefore x*+px+q be a perfect square, the remainder, 

P 2 
o — ^j-, must be zero. Hence the condition is determined 
4 

by placing this remainder equal to zero and solving the 
resulting equation. 

319. Symmetry . v An expression is said to be symmetrical 
with respect to the letters it contains when its value is 
unaltered by the interchange of any pair of them; thus 
x + y+z> be + ca + ab, X s + y* + z 3 — xyz are symmetrical 
functions of the first, second, and third degrees respectively. 

It is worthy of notice that the only symmetrical expres- 
sion of the first degree in x, y, z is of the form M(x+y+z), 
where M is independent of x, y, z. 

320. It easily follows from the definition that the sum, 
difference, product, and quotient of any two symmetrical 
expressions must also be symmetrical expressions. The 
recognition of this principle is of greut use in checking 
the accuracy of algebraic work, and in some cases enables 
us to dispense with much of the labor of calculation. In 
the following examples we shall assume as true a principle 
which will be demonstrated in Chap. xlii. 

Ex. 1. Find the expansion of (x + y + z)*. We know that the 
expansion must be a homogeneous expression of three dimensions, 

T 



274 ALGEBRA. 

and therefore of the form a* + y 8 + z* + A(x*y + xt/* + tfz + yz 2 + z^z 
-f zx 2 ) + Bxyz, where A and B are quantities independent of x, y, z. 

Pat z = 0, then A, the coefficient of xhj, is equal to 3, the coefficient 
of x*y in the expansion of (as + J/) 8 - 

Put x = y = z= 1, and we get 27=3+(3x6) + #; whence S = 6. 

Thus (x + y + s)* 

= x* + tp + z* + 3x*y + Sxi/ 2 +Sy*z + Syz* + Sz 2 x + 3zx 2 + 6xyz. 

Ex. 2. Find the factors of 

(6 8 + c8)(6 - c; + (c« + a*)(c - a) + (a* + 6 8 )(a - 6). 

Denote the expression by E ; then E is an expression involving a, 
which vanishes when a = 6, and therefore contains a — 6 as a factor 
[Art. 317]. Similarly it contains the factors b — c and c — a\ thus 
j£ contains (6 — c)(c — a) (a — 6) as a factor. 

Also since 2? is of the fourth degree, the remaining factor must be 
of the first degree ; and since it is a symmetrical expression involving 
a, 6, c, it must be of the form m(a + b + c). [Art. 319.] 

.•. E.= m(b — c)(c — a)(a — b)(a + b + c). 

To obtain w we may give to a, 6, c any values that we find most 
convenient ; thus by putting a = 0, 6 = 1, c = 2, we find m = 1, and 
we have the required result. 

Note. For further information on the subject of Symmetry, the 
reader may consult Hall and Knight's Higher Algebra, Chap, xxxiv. 

EXAMPLES XXX. b. 

Without actual division find the remainder when 
l.aP — 6a; 2 + 6 is divided by x — 6. 

2. 3x 5 +llsc* + 90x 2 -19a+ 53 is divided by s + 5. 

3. X s -7x 2 a + Sxa? + 15a 8 is divided by x + 2a.. 

Without actual division show that 

4. 32 a; 10 - 33 x 5 + 1 is divisible by x - 1. 

5. 3a* + 5s 8 - 13a; 2 - 20a; + 4 is divisible by x* - 4. 

6. X* + 4a; 8 - 6x 2 - 36x - 36 is divisible by a; 2 - a; -6. 

Resolve into factors : 

7. a*-6x 2 + 11a; -6. 11. a*-39x+70. 

8. a; 8 -6a; 2 -2a; + 24. 12. x* - 8a; 2 - 31a; -22. 

9. a* + 9a; 2 4-26x + 24. 13. 6a; 8 + 7a; 2 - x - 2. 
10. a* - a; 2 - 41 x + 106. 14. 6a* + a; 2 - 19x + 6. 



MISCELLANEOUS THEOREMS. 275 

Find the values of x which will make each of the following expres- 
sions a perfect square : 

15. a* + 60? + 13a 2 +13s-l. 16. ac* + 6a* + lla£ + 3a; + 31. 

17. a*-2os 8 + (a 2 + 2&)x 2 -3a&a; + 2& 2 . 

18. 4p 2 x* - 4pqx* + (Q* 2 + 2p 2 )a; 2 - 6 pqx + ^-. 

19 a 2 a* q&ag* , 2qcs 8 . 9ft 2 a; 2 5&ca? . gc a 
9 2316 2 

20. z* + 2a3 8 + 3a 2 a; 2 + ca; + d. 

Find the values of x which will make each of the following expres- 
sions a perfect cube : 

21. 8s 8 - 36a; 2 + 66s -89. 22. ^-^ + 4a*aj 2 -28a 6 . 

27 3 

28. m 8 ** - 9 w 2 nx* + 39 mn 2 x 2 - 61 n 8 . 

24. If n be any positive integer, prove that 6 2 * 1 — 1 is always divisible 
by 24. 

Find the factors of 

25. a(6 -c) 8 + 6(c-a) 8 + c(a-&) 8 . 

26. a(b - c) 2 + b(c - a) 2 + c(a - ft) 2 + Babe 



CHAPTER XXXI. 



Indeterminate Equations of the First Degree. 



.. In Art. 167 we saw that if the number of unknown 
quantities is greater than the number of independent equa- 
tions, there will be an unlimited number of solutions, and 
the equations will be indeterminate. By introducing con- 
ditions, however, we can limit the number of solutions. 
When positive integral values of the unknown quantities 
are required, the equations are called simple indeterminate 
equations. 

The introduction of this restriction enables us to express 
the solutions in a very simple form. 

Ex. 1. Solve 7 x + 12 y = 220 in positive integers. 
Transpose and divide by the smaller coefficient ; thus, 

s = 31-y + ^l; 

Since x and y are to be integers, we must have 

3 — by . . 

— — £ = integer. 

Now multiplying the numerator by such a number that the division 
of the coefficient of y may give a remainder of unity, in this case 3, we 
have 

9 — 16 v . «. 
--£ = integer ; 

that is, 1 — 2 y H — ^ = integer ; 

and therefore ™ = integer. 

276 



INDETERMINATE EQUATIONS. 277 

Let ~y = m, an integer ; 

then y = 2 - 7 m (1). 

Substituting this value in the original equation, we obtain 

7s + 24 -84m = 220; 

.-. s = 28-f Urn (2). 

Equation (1) shows that m may be or have any negative integral 
value, but cannot have a positive integral value. 

Equation (2) shows in addition that m may be 0, but cannot have a 
negative integral value greater than 2. Thus the only positive integral 
values of x and y are obtained by placing m = 0, — 1, — 2. 

The complete solution may be exhibited as follows : 

m = 0, - 1, - 2, 
x = 28, 16, 4, 
y= 2, 9, 16. 

Ex.2. Solve bx — 14 y = 11 in positive integers . • • • (1). 
Proceeding as in Example 1, we obtain 

s = 2 + 2y+i£-±i; 

5 

,-. x - 2y - 2 = ilLti = integer. 

5 

Now multiplying the numerator by 4, we obtain 

WjL±i= integer; 



that is, 8 y + * "*" = integer. 

5 



y + 4 = 

5 

Let y = m, an integer ; 

5 

.•. y = 6m —4, > 
and from (1), <e = 14 m — 9. I 

This is called the general solution of the equation, and by giving 
to m any positive integral value, we obtain an unlimited number of 
values for x and y : thus we have 

m = 1, 2, 3, 4 ••• 

y = l, 6, 11, 13... 

a; = 6, 19, 33, 47-. 



278 ALGEBRA. 

From Examples 1 and 2 the student will see that there 
is a further limitation to the number of solutions according 
as the terms of the original equations are connected by -f- 
or — . If we have two equations involving three unknown 
quantities, we can easily combine them so as to eliminate 
one of the unknown quantities, and can then proceed as 
above. 

Ex. 8. In how many ways can $ 6 be paid in quarters and dimes ? 
Let x = the number of quarters, y the number of dimes ; then 

4 10 
or 6<e-f2y = 100; 

.-. 2&+f + y = 50; 

••. x — 2p, 

and y = 50 — 5 p. 

Solutions are obtained by giving to p the values 1, 2, 3, ..., 9 ; and 
therefore the number of ways is 9. If, however, the sum be paid 
either in quarters or dimes, p may also have the values and 10. If 
p = 0, then x = 0, and the sum is paid entirely in dimes ; if p = 10, 
then y = 0, and the sum is paid entirely in quarters. Thus if zero 
values of x and y are admissible, the number of ways is 11. 

EXAMPLES XXXI. 
Solve in positive integers: 

1. 3s + 8y = 103. 3. 7x + 12^ = 152. 5. 23 x + 25 y = 915. 

2. 5z + 2y = 53. 4. 13x+lly = 414. 6. 41 x + 47 y = 2191. 

Find the general solution in positive integers, and the least values 
of x and y which satisfy the equations : 

7. 6«-7y = 3. 9. 8z-21y = 33. 11. 19y-23x = 7. 

8. 6aj-13y = l. 10. 17 y - 13 a = 0. 12. 77y-30x = 295. 

13. A farmer spends $ 752 in buying horses and cows ; if each horse 
costs $37, and each cow $23, how many of each does he buy ? 

14. In how many ways can $ 100 be paid in dollars and half-dollars, 
including zero solutions ? 

15. Find a number which, being divided by 39, gives a remainder 
16, and, by 56, a remainder 27. How many such numbers are there ? 



CHAPTER XXXII. 

Inequalities, 

Any quantity a is said to be greater than another 
quantity b when a — b is positive ; thus 2 is greater than 
—3, because 2 — (—3), or 5, is positive. Also b is said to 
be less than a when b — a is negative ; thus —5 is less than 
—2, because —5 — (—2), or —3, is negative. 

In accordance with this definition, zero must be regarded 
as greater than any negative quantity. 

323. The statement in algebraic language that one ex- 
pression is greater or less than another is called an in- 
equality. 

324. The sign of inequality is >, the opening being placed 
towards the greater quantity. Thus, a > b is read " a is 
greater than b." 

325. The first and second members are the expressions on 
the left and right, respectively, of the sign of inequality. 

.326. Inequalities subsist in the same sense when corre- 
sponding members in each are the greater or the less. Thus, 
the inequalities a > b and 7 > 5 are said to subsist in the 
same sense. 

In the present chapter, we shall suppose (unless the con- 
trary is directly stated) that the letters always denote real 
anc. positive quantities. 

327. Inequality Unchanged. An inequality will still hold 
after each side has been increased, diminished, multiplied, or 
divided by the same positive quantity. 

279 



280 ' ALGEBRA. 

For, if a > b, then it is evident that 

a-hc>6-hc; 

a — c>b — c; 

ac>bc\ 

» 

a. b 

C G 

328. Term Transposed. In an inequality any term may be 
transposed from one side to the other if its sign be changed. 

If a-c>b, 

by adding c to each side, 

a > b + c. 

329. Members Transposed. If the sides of an inequality be 
transposed, the sign of inequality must be reversed* 

For if a > b, then evidently b < a. 

330. Signs Changed, ijf £Ae signs of aU the terms of an 
inequality be changed, the sign of inequality must be reversed. 

When a>b, then a — b is positive, and b — a is nega- 
tive ; that is, — a — (— b) is negative, and therefore 

— a < — 5. 

* 

331. Negative Multiplier. If the sides of an inequality be 
multiplied by the same negative quantity, the sign of inequality 
must be reversed. 

For, if a>b, then — a < — 6, and therefore 

— ac < — be 

332. Inequalities Combined. If inequalities, subsisting m 
the same sense, be either added, or multiplied together, the 
results will be unequal in the same sense. 

For if a x > &!, a^ > b^ a 3 > b 8 ••• a m > b^ it is clear tha +> 

«i + «2 + a 8 + ••• + «» > h + &a + &s + — + &»> 
and c^c^cis •••a w > &1&2&8 ••• &m» 



INEQUALITIES. 281 

I. It follows from the preceding article that if o > 6 
then a" > b n , 

and a~ n < &•*, 

where n is any positive quantity. 

334. The subtraction of two inequalities subsisting in 
the same sense does not necessarily give an inequality sub- 
sisting in the same sense. 



k. The division of an inequality by another subsisting 
in the same sense does not necessarily give an inequality 
subsisting in the same sense. 

The truth of these last statements is readily seen by 
considering the inequalities 

5>4, 
3>2. 

Subtracting member for member would give 2 > 2. 
Dividing* member by member would give J > 2. 

Ex. 1. Find limit of se in the inequality 

8^6^16 

Clearing of fractions, we have 

15a;-26>8aj + ll. 

Transposing and combining, 

12a;>36; 

.-. x>B. 

Note. The word " limit " is here used as meaning the range of 
values that x can have under the given conditions. 

Ex. 2. If a, b, c denote positive quantities, prove that 

a 2 + b 2 + c 2 > be + ca + ab. 
For 6 2 + c 2 >26c, 

c 2 + a 2 >2co 

a 2 + b 2 > 2 ab. 
Whence by addition a 2 + b 2 + c 2 >bc + ca + ab. 



282 ALGEBRA. 

Ex. 8. If as may have any real value, find which is the greater, 
a£ + 1 or x* + x. 

x* + 1 -(a; 2 + *) = x* - x 9 -(* - l) = (* a - 1)(* - 1) 

= 0»-l)»(* + l). 

Now (as — l) a is positive, hence 

a* + l> or <x* + z 

according as x + 1 is positive or negative ; that is, according as 
x > or < — 1. 

If x = — 1, the inequality becomes an equality. 

EXAMPLES XX7TO. 

Find limit of x in the following three inequalities : 



46 5a; 
3-3 



1. nx-™<?£+Z\. 



2 x + 2)(x + 3)>(a; - 4)(a: - 5). 

3. >x -+ 5 ax — 5 ab > b 2 when a>b. 

4 Prove that (ab + xy) (&c + by)>4 abxy, 

5. Prove that (b + c) (c + a) (a + &)> 8 a&c. 

6. Show that the sum of any real positive quantity and its 
reciprocal is never less than 2. 

7. If a 2 + 6 2 = 1, and x 2 + y 2 = 1, show that ox + by < 1. 

8. If a 2 + 6 2 + c 2 = 1, and a; 2 + y 3 + s 3 = 1, show that 

ox + 6y + as < 1. 

9. Which is the greater 2_ + -^ or ^- ? 

6 2 a + b 

10. Show that (afy + y 2 z + z 2 x)(xf/ 2 + yz 2 + zx*)>$x*\P&> 

11. Find which is the greater, 3 ab 2 or a 8 + 2 6*. 

12. Prove that a 8 & + ab 9 < o 4 -f- 6*. 

18. Prove that 6 abc < 6c(6 -f c) + ca(c + o) + <*&((* + b). 

14. Show that W + c 2 a 2 + a 2 b 2 > abc(a + 5 + c). 

15. Show that 2(a 8 -f ft 8 + c 3 )> bc(b + c)+ ca(c + a)+ab,(a + b). 

Note. For further information on the subject of Inequalities the 
reader may consult Hall and Knight's Higher Algebra, Chapter xix. 



MISCELLANEOUS EXAMPLES. 283 

MISCELLANEOUS EXAMPLES V. 

1. If a = — 1, 6 = 2, c = 0, d = 1, e = — 3, find the value of 

q 8 (d — c) — V3 a e + ab 

d(c-a)-2ad 2 +y/*ab 

2. Simplify [3(a-& + c)-(a-&)(6 -<;) + {(« + 6-cx8- 6)}]. 

3. Solve 3s-4- 4 ( 7a; - 9 >=Jf6+3^V 

16 6\ 3 / 

4. A man's age is four times the combined ages of his two sons, one 
of whom is three times as old as the other ; in 24 years their combined 
ages will oe 12 years less than their father's age : find their respective 
ages. ^_ 

6. Solve (i.) 8V*-1 = 5 + 6. 

3Vjc + 7 

(ii.) V8a; + 17 - V2* = V2s+"6. 

7. Expand (2a -36 2 ) 6 . 

8. Simplify —1 —± z-^- 

2x+l 3(aj + i) 6x + 3 

9. If 3 is added to the numerator a certain fraction is increased by 
J ; if 3 is taken from the denominator the fraction reduces to J : find 
the fraction. 

10. Find the value of (i.) 3 V243 + 2 V^ + 4 V76 - Vj» 

V5+3V2 



(ii.) 



2 + V5 



11. Solve (i.) — g— + 8 =16. 

v ' 1-23 1+23 

(ii.) K«- 2 a?) + .1(1 -2 as) =5(| -a). 

12. Find the limit of x in the inequality 

(a;_4)(a;-5)>(a;-2)(a;-l). . 

18. The breadth of a rectangular space is 4 yards less than its 
length; the area of the space is 252 square yards: find the length of 
each side. 

14. Solve (i.) x 2 + y 2 = -V°, x + y = £ . 

(ii.) 2x 2 + xy = 4, 3xy + 4j/ 2 = 22. 

15. Find the factors of (i.) x 8 + 12 x 2 y - 45 xy\ 

(ii.) 3x 2 -31»y + 66y2. 



284 ALGEBRA. 

16. Simplify x + 1 + a? " 1 L^ 

17. Solve *+i + *+*=6, 2y + 6^2xj L 4 = 5 

3 6 y a; y 

18. Find two numbers whose sum is 22, and the sum of theii 
squares is 260. 



19. Simplify 






90. Solve (i.) — = 3 + — ^— . 

vx-1 z 

(ii.) Vx a + 4 * - 4 + y/x 2 + 4 a; - 1 =z 6. 

21. For what value of k wiL. the equation z 2 + 2(k + 2)x -f 9 jfc = 
have equal roots ? 

22. Simplify the fractions : 



,*• -„-(«=*£ 6 + 



l — - a + 



x a — b 

23. B pays $28 more rent for a field than A ; he has three-fourths 
of an acre more and pays $ 1.76 per acre more. C pays $ 72.60 more 
than A ; he has six and one-fourth acres more, but pays 26 cents pei 
acre less : find the size of the fields. 

24. Solve (i.) ^-19 = 11. 

v ' x -6 x 6 

nn 2x -j- 2a? - 6 = 25 

^ ,; s_4 s-3 T 
26. Find the value of (jzIiL^=?V+ /^l^V^BN" when n _ 3# 

V2 

26. Rationalize the denominator of — — — 

V2 + V3-V6 
and find a factor which will rationalize Vs — V2. 

27. Find the square root of (i.) 11 + 4 VS. 

(ii.) - 6 + 12V3T. 

I Find the factors of (i.) a 2 - 13 - 6 ax + 9 a; 2 . 

(ii.) 343 a* -2708. 



MISCELLANEOUS EXAMPLES V. 285 

29. Solve (i.) x» + y 8 = 18^/2, x + y = 3^2. 

x* y* & x y 2 

30. A rectangular fie}d is 100 yards wide. - If it were reduced to a 
square field by cutting an oblong piece off one end, the ratio of the 
piece cut off to the remainder would be less by ^ than the ratio of the 
remainder to the original field. Find the length of the field. 

31. Extract the square root of 

(i.) 4x* + 12afy + 133 2 y 2 + 6a^a + |f*. 

a* a x & 

82. Solve (i.) 2x + y + 3s = 13, 

x + 2y + 4s = 17, 
4« + 3y + 2s = 16. 

(ii.) -1-+— 1— + _I_ = :a 
1 " x Vx+1 V5-l 

33. Simplify the fractions 

lis 



(i.) 



1-— 1 i— 2 + 



2x 1 — a; « — 1 

f u n 2x*-7x 2 y + 5sy 2 -y» 
2x 8 + 5aj 2 y-6»y 2 + y 8 " 

84. Two places, A and B, are 168 miles apart, and trains leave A 
for B and B for A simultaneously ; they pass each other at the end of 
one hour and fifty-two minutes, and the first reaches B half an hour 
before the second reaches A. Find the speed of each train. 

35. Solve (i.) 3s 2 + 4a; + 2Vs 2 -f x + 3 = 30 -a?. 

(ii.) x* + y* = 706, x + y = 8. 

36. Form the equation whose roots are the squares of the sum and 
of the difference of the roots of 

2x 2 + 2(w + n)x + m* + n 2 = 0. 

87. Employ the method of Arts. 310, 320 in showing that 

(a + 6) 6 - a 5 - & 6 = 6ab(a + 6)(a 2 + ab + IP). 

88. Solve xy(Sx + y)=10, 27« 8 + y 8 = 36. 



CHAPTER XXXIII. 
Ratio, Proportion, and Variation. 

Definition. Ratio is the relation which one quan- 
tity bears to another of the same kind, the comparison being 
made by considering what multiple, part, or parts, one quan- 
tity is of the other. 

The ratio of A to B is usually written A : 2?. The quan- 
tities A and B are called the terms of the ratio. The first 
term is called the antecedent, the second term the consequent. 

337. Ratios are measured by Fractions. To find what 
multiple or part A is of B, we divide A by B ; hence the 

ratio A : B may be measured by the fraction — , and we 

shall usually find it convenient to adopt this notation. 

In order to compare two quantities, they must be ex- 
pressed in terms of the same unit. Thus, the ratio of $ 2 

to 15 cents is measured by the fraction — — or — • 

J 15 3 

Note. Since a ratio expresses the number of times that one quan- 
tity contains another, every ratio is an abstract quantity, 

33a By Art. 136, ? = ^5 

mo 

and thus the ratio a: bis equal to the ratio ma : mb ; that 
is, the value of a ratio remains unaltered if the antecedent and 
the consequent are multiplied or divided by the same quantity. 



Comparison of Ratios. Two or more ratios may be 
compared by reducing their equivalent fractions to a com- 
mon denominator. Thus, suppose a : b and x : y are two 

ratios. Now, - = ^, and - = — ; hence the ratio a : b is 

b by y by 

286 



RATIO, PROPORTION, AND VARIATION. 287 

greater than, equal to, or less than the ratio x : y according 
as ay is greater than, equal to, or less than bx. 

340. The ratio of two fractions can be expressed as a 

ratio of two integers. Thus, the ratio - : - is measured by 

a c ad 

- -*- - or — ; and is therefore equivalent to the ratio ad : be 

b d be 

341. If either, or both, of the terms of a ratio be a surd 
quantity, then no two integers can be found which will 
exactly measure their ratio. Thus, the ratio -y/2 : 1 cannot 
be exactly expressed by any two integers. 

342. If the ratio of any two quantities can be expressed 
exactly by the ratio of two integers, the quantities are said 
to be commensurable ; otherwise, they are said to be incom- 
mensurable. 

Although we cannot find two integers which will exactly 
measure the ratio of two incommensurable quantities, we 
can always find two integers whose ratio differs from that 
required by as small a quantity as we please. 

Thus, *& = 2.236067-.. = .559016 ... ; 

4 4 

and, therefore, 

V5 . 559016 , 559017 . 

4 1000000 1000000 ' 

and it is evident that by carrying the decimals, further, any 
degree of approximation may be arrived at. 

343. Ratios are compounded by multiplying together the 
fractions which denote them. 

Ex. Find the ratio compounded of the three ratios 

2 a : 3 6, 6 ab : 6 c 2 , c : a. 

The required ratio = ^ x ^5 x - = ^l. 
* 36 6c 2 a 6c 

344. When two identical ratios, a : b and a : b, are com- 
pounded, the resulting ratio is a 2 : b 2 , and is called the 



288 ALGEBRA. 

duplicate ratio of a : b. Similarly, a 3 : b* is called the triptt 

cate ratio of a : b. Also, a? : b* is called the subduplicate 
ratio of a : b. 

Examples (1) The.duplicate ratio of 2 a : 3 b is 4 a 2 : 9 ft 2 . 

(2) The subduplicate ratio of 49 : 25 is 7 : 5. 

(3) The triplicate ratio of 2 x : 1 is 8 xP : 1. 

345. A ratio is said to be a ratio of greater inequality, or 
of less inequality, according as the antecedent is greater or 
less than the consequent. 

346. If to each term of the ratio 8 : 3 we add 4, a new 
ratio 12 : 7 is obtained, and we see that it is less than the 
former because ±p- is clearly less than -f. 

This is a particular case of a more general proposition 
which we shall now prove, 

A ratio of greater inequality is diminished, and a ratio of less 
inequality is increased, by adding the same quantity to both its 
terms. 

Let - be the ratio, and let + be the new ratio formed 
b b+x 

by adding x to both its terms. 

No - — a ~^~ x — ax ~ bx _ x(a — b) . 

b b -f- x ~~ b(b 4 -x) ~~ b(b + x) ' 

and a — b is positive or negative according as a is greater or 
less than b. 

Hence, if a is > b, - is > a ; 

b b+x 

and if ais<6, -is<— -*—, 

b b+x 

which proves the proposition. 

Similarly, it can be proved that a ratio of greater inequal- 
ity is increased, and a ratio of less inequality is diminished, 
by taking the same quantity from both its terms. 

347. When two or more ratios are equal, many useful 
propositions may be proved by introducing a single symbol 
to denote each of the equal ratios. 



RATIO, PROPORTION, AND VARIATION. 289 

The proof of the following important theorem will illus- 
trate the method of procedure. 

ace 



• • •« 



each of these ratios V/m" + y g " + /»"+ .--V 

V/>6 n 4- fflf 1 + rf n -f ..7 ' 
where /?, q, r x /?, are any quantities whatever. 

Let ? = ^ = * = ...=fc ; 

b d f 

then a = bk, c = d&, e =/&, ••• ; 

whence pa n =pb n k n , qc" = qd n k n , re n = rf n k n , ••• ; 

pa" + <?c n + re* H _ pb n k n + gd n fc n + r/ w fe n H — frn. 

" pb n + gd n + r/ n H "" i)&" + 5d* + r/"+— "" ' 

i 
'pa* + gc n + re" -f "A * _ z. _ ? _ J? _ 
jpb n + qd n + rf " + •-•) ~~ ~~6~~d ' 

By giving different values to p, q, r, n many particular 
cases of this general proposition may be deduced ; or they 
may be proved independently by using the same method. 
For instance, if 

-=- = -, each of these ratios = *±£±* . 
b d f b + d+f' 

a result which may be thus enunciated : When a series of 

fractions are equal, each of them is equal to the sum of all the 

numerators divided by the sum of all the denominators, 

Ex. 1. If 5 = ? find the value of 5JL=_?1. 
V 4 Tx + 2y 

— — 3 — — 3 
5x-Sy _ y _j4 _3^ 

7a + 2if-7* + 2 -21 + 2 -80' 

y 4 

Ex. 2. Two numbers are in the ratio of 5 : 8. If 9 be added to 
each they are iu the ratio of 8:11. Find the numbers. 
Let the numbers be denoted by 6 x and 8 x. 

Then bx±9 = S_ x = s 

8x + 9 11 
Hence the numbers are 15 and 24. 
u 



290 ALGEBRA. 

Ex. 3. If A: B be in the duplicate ratio of A + zi B + z, prove 
Jhat x 2 = AB. 



By the given condition, ( A±J* V = A ; 

\B + xj B 



.-. B(A + x) 2 = A(B + x) a , 

<4 2 B + 2 ^4Bx + £x 2 = AB 2 + 2 ^LRb + A&, 
x\A-B)=AB(A-By t 

.-. a; 2 = AB, 

since ^4 — 2* is, by supposition, not zero. 

EXAMPLES XXXin. a. 

Find the ratio compounded of 

1. The duplicate ratio of 4:3, and the ratio 27 : 8. 

2. The ratio 32 : 27, and the triplicate ratio of 3 : 4. 

3. The subduplicate ratio of 25 : 30, and the ratio 6 : 25. 

4. The triplicate ratio of x : y, and the ratio 2 y 2 : 3 x 2 . 

5. The ratio 3 a : 4 6, and the subduplicate ratio of 6* : a 4 . 

6. If x : y = 5 : 7, find the value of x + y :y — x. 

7. If - = 31, find the value of X ~ S V . 

y 2x — by 

8. If 6 : a = 2 : 5, find the value of 2a — 36:36 — a. 

9. If 5 = 3 , and * = 5, find the value of 3 «*-&y . 

64 2/7 4: by — 7 ax 

10. If 7 * — 4 y : 3 x + y = 6 : 13, find the ratio x : y. 

11. if 2 q2 ~ 3 6* = A, find the ratio a : 6. 

a 2 + & 2 41 

12. If 2 re : 3 y be in the duplicate ratio of 2 x — m : 3 y — m, prove 
that m 2 — Qxy. 

13. If P : Q be the subduplicate ratio of P — x : Q — a, prove that 
P« . 



3 = 



P+G 

14. If - = - = -j prove that each of these ratios is equal to 
6 d f 

Sl 2a 2 c + Sc s e + 4e 2 c 



« 



2 & 2 cZ + 3 d^e + 4/ 2 <Z 

15. Two numbers are in the ratio of 3 : 4, and if 7 be subtracted 
from each the remainders are in the ratio of 2 : 3 : find them. 

16. What number must be taken from each term of the ratio 27 : 36 
that it may become 2:3? 



RATIO, PROPORTION, AND VARIATION. 291 

17. What number must be added to each term of the ratio 87 : 29 
that it may become 8:7? 

18. If -£— = — 2_ = _ !!_, showthati> + g + r = 0. 

b — c c — a a — b 

19. if _£_ = _JL_ = _?_-, show that x - y + z = a 

b + c c + a. a — b 

20. If ^ = - = — , show that the square root of 

a«b-2c*e + Sa*c*e* . g 2 ace . 

21. Prove that the ratio la+mc+ne : Z&+md+n/ will be equal to 
each of the ratios a:b, cid, e:f, if these be all equal ; and that it will 
be intermediate in value between the greatest and least of these ratios 
if they be not all equal. 

22. If P x ~ a y = cx ~<m = z±JL, t hen will each of these fractions 

cy — az by — ax x + z 

be equal to -, unless b + c = 0. 

y 

23. If 2 * ~ 3 y = ±=-3L = x + 8 * , prove that each of these ratios 

oz + y z — x 2y — 3x 

is equal to -; hence show that either x = y,"or z = x + y. 
V 

PROPORTION. 

348. Definition. Four quantities are said to be in 
proportion when the ratio of the first to the second is equal 
to the ratio of the third to the fourth. The four quantities 
are called proportionals, or the terms of the proportion. 

Thus, if - = -, then a, b, c, d are proportionals. This is 
b d 

expressed by saying that a is to b as c is to d, and the pro- 
portion is written 

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

The terms a and d are called the extremes, b and c the means. 

349. If four quantities are in proportion, the product of the 
extremes is equal to the product of the means. 

Let a, b, c, d be the proportionals. 



292 ALGEBRA. 

Then by definition - = - ; 

o a 

whence ad = be 

Hence if any three terms of a proportion are given, the . 

fourth may be found. Thus if a, c, d are given, then b= — • 

c 

Conversely, if there are any four quantities, a, b, c, d, 
such that ad = be, then a, b, c, d are proportionals ; a and d 
being the extremes, b and c the means ; or vice versd. 

350. Continued Proportion. Quantities are said to be in 
continued proportion when the first is to the second, as the 
second is to the third, as the third to the fourth ; and so on. 
Thus a, b } c, d, ••• are in continued proportion when 

? — ^ — £ — ... 

bed 

If three quantities a, b, c are in continued proportion, then 

a\b — bic\ 
.-. ac = b*. (Art. 349., 

In this case b is said to be a mean proportional between a 
and c; and c is said to be a third proportional to a and b. 

351. If three quantities are proportionals, the first is to the 
third in the duplicate ratio of the first to the second. 

Let the three quantities be a, b, c; then 7 = — 

b c 

XT a a b a a a* 

c b c b b 0* 
that is, a:c = a % :V. 

352. The products of the corresponding terms of two or more 
proportions form a proportion. 

If a\h*=cid and e :f= g : h, then will 

ae: bf= eg: dh. 



RATIO, PROPORTION, AND VARIATION. 



293 



Gob. 


If 


a:b = c : d, 


and 




b:x = d:y f 


then 




a : x = c : y. 



353. Transformations that may be made in a Proportion. 

If four quantities, a, b, c, d form a proportion, many other 
proportions may be deduced by the properties of fractions. 
The results of these operations are very useful, and some 
of them are often quoted by the annexed names borrowed 
from Geometry. 



(1) If a : b = c : d, then b : a = d : c. 

For? = £; therefore l-*-- = l-*4; 
b d b d 



[Inversion.] 



that is 



or 



b d 

— ~~ * 

a c ' 
b:a = d:& 



(2) If a : b = c : d, then a : c = b : d. 

For ad = 6c : therefore — = — : 

cd cd 



[Alternation.] 



that is, 



or 



c"~d ; 
a:c = b:d. 



(3) If a:6 = c:d, then a + b:b = c + d:d [Composition.] 

For? = -: therefore ® + l = £ + l; 
6 d 6 d 



d 



that is, 

or a + btb = c + d:d. 

(4) If a : 5 = c : d, then a — 6:6 = c — d:A [Division.] 

For a = -; therefore --1=--1; 
b d' b d ' 



that is, 
or 



a — 6 _ c — d . 
6 ~~ d ; 
a — 6 : 6 = c — d:d 



294 ALGEBRA. 

(5) If a : b = c : d, then a + & : a — 6 = c + d : c — d. 

[Composition and Division.] 

For by (3) «±» = l±i; and by (4) ^ = C -^i 
.\ by division, <L+£ _, £_+« . 

or <* + &•* « — & = c + d : c — d. 

Several other proportions may be proved in a similar way. 

Ex. 1. If a : b = c : d = e :/, 

show that 2a 2 + 3cP - 6e 2 : 2o 2 + 3d 2 - 6/* s a* : 6/. 

Let 2 = £ — £ = fc • then a = ofc, c = d%, e ss/fc ; 

. 2a 2 -f 3c 2 - 5e 2 ^ 2& 2 ifc 2 + Sffjfc 2 - 5/2fc 2 __ -, _a «__ge 
"" 26 2 + 3(F-5/ 2 2o 2 + 3d 2 -6/ 2 6 / bf 

or 2a 2 + 3d 2 - 6e 2 : 2ft 2 + 3d 2 - 6/ 2 = ae : bf. 

Ex.2. If 

(3a + 6o + c + 2<r)(Sa-6&-c + 2d) 

= (Sa-6& + c-2d)(3a + 6o-c-2d), 
prove that a, &, c, d are in proportion. 

We have 3q + 6o + c + 2ri = 8q + 6ft-«-2tf [ArL 349 , 

^ ... ... ., 2(3a + c) 2(3a-c) 

Composition and division, ^6 + 24) = 2(6ft-2«V 

Alternation, |5jt5 = ^±14 

^ 3a-c 6&-2d 

Again, composition and division, ~ = — -^ ; 

2c 4a 

whence a:b = c:d. 

s 2 + s-2 _ 4a; 2 + 5ac-6 
x-2 

Division, -^— = -^- ; 

g-2 5x-6' 

whence, dividing by a 2 , which gives a solution a? = 0, [Art. 291, note.] 

1 4 



Ex. 3. Solve the equation 

x-2 6z-6 



x-2 bx-Q 
and therefore the roots are 0,-2. 



; whence, * = — 2 : 



RATIO, PROPORTION, AND VARIATION. 296 

EXAMPLES XXXTH. b. 
Find a fourth proportional to 
1. a,ab,c. 2. a 2 , 2a&, 8 ft 2 . 8. as 8 , gy, 5aty. 

Find a third proportional to 
4. a 2 b, ab. 5. a*, 2 s 2 . a 3 s, 6 a*. 7. 1, x. 

Find a mean proportional between 
8. a 2 , b 2 . 9. 2 s 3 , 8a^ 10. 12 ox 2 , So*. it 27 a 2 **, 8 ft. 

If a, 6, e be three proportionals, show that 
12. o:a + 6 = a — 6 : a — c. 
18. (6 2 + 6c + c 2 )(ac-6c + c 2 )=6* + «c 8 + C». 
If a : b = o: d, prove that 

14. ab + cd:ab - cd = a 2 + c 2 :a 2 - c 2 . 

15. a 2 + ac + c 2 : a 2 - ac + c 2 = & 2 + bd + d 2 : b 2 - &d + d 2 

16. a : & = V3a 2 + 6c 2 : VS^ + Sd 2 . 

1W a, 6 c , d.„ 

17. - + - : a = - -f - : c. 
p q >p q 

18. & + g. «& = * + c «* 



a 6 a 2 + b 2 c d c 2 + d 2 

Solve the equations : 

19. 3«-l:6«-7 = 7x-10:9a; + 10. 

20. aj-12:2/ + 3 = 2aj-19:6y-13 = 6:14. 

2i 3 2 -2a; + 3 _ a; 2 -3a; + 5 | 22 2s-l = a?-f4 

2&-3 3x-5 * a? + 2x-l a; 2 + a; + 4" 

23. If (a + 6 -3c- 3d)(2a- 26 - c + d) 

- = (2a + 2&-c-d)(a-6-8c + 3d) 
prove that a, &, c, d are proportionals. 

24. If a, b, c, d are in continued proportion, prove that 

aid = a* + b* + c*ib* + cfi + cP. 

25. If & is a mean proportional between a and c, show that 

4 a 2 — 9 b 2 is to 4 6 2 — 9 c 2 in the duplicate ratio of a to 5. 

26. If a, 6, c, d are in continued proportion, prove that b + c is a 
mean proportional between a -f b and c + d. 

27. If a + ft:&-fc=:c + d:d + a, 
prove that a = c, ora + & + c + d = 0. 



296 ALGEBRA, 



VARIATION. 

L Definition. One quantity A is said to vary directly 
as another B, when the two quantities so depend upon each 
other that if B is changed, A is changed in the same ratio. 

Note. The word directly is often omitted, and A is said to vary 
as B. 

355. For instance : if a train moving at a uniform rate 
travels 40 miles in 60 minutes, it will travel 20 miles in 30 
minutes, 80 miles in 120 minutes, and so on ; the distance 
in each case being increased or diminished in the same ratio 
as the time. This is expressed by saying that when the 
velocity is uniform the distance is proportional to the time, or 
more briefly, the distance varies as the time. 

356. The Symbol of Variation. The symbol <x is used to 
denote variation ; so that Ace Bis read "A varies as B" 

357. If A varies as B y then A is equal to B multiplied by 
some constant quantity. 

For suppose that a u 0% a 8 ..., b l9 b^ b s ... are corresponding 
values of A and B. 

Then, by definition, — = — ; — = — ; — = — ; and so on ; 

a x bi a 2 b 2 c^ b s 

.'. ^ = ^ = ^ = ..., each being equal to — 
&i b 2 b 3 B 

Hence ^ ^ is always the same ; 

the corresponding value of B 

that is, -- = m, where m is constant. 

.•. A = mB. 

358. Definition. One quantity A is said to vary in- 
versely as another B when A varies directly as the reciprocal 
of B. [See Art. 176.] 



RATIO, PROPORTION, AND VARIATION. 297 



Thus if A varies inversely as B, A = — , where m is con- 
stant. 

The following is an illustration of inverse variation : If 6 
men do a certain work in 8 hours, 12 men would do the 
same work in 4 hours, 2 men in 24 hours ; and so on. Thus 
it appears that when the number of men is increased the 
time is proportionately decreased; and vice versd. 

359. Definition. One quantity is said to vary jointly as 
a number of others when it varies directly as their product. 

Thus A varies jointly as B and C when A = mBC. For 
instance, the interest on a sum of money varies jointly as 
the principal, the time, and the rate per cent. 

360. Definition. A is said to vary directly as B and 

inversely as C when A varies as — • 

G 

361. Grouping the principles of Arts. 357-360, we have 
A = mB, ' if A varies directly as B, 

A = — 9 if A varies inversely as B. 

A = mBC, if A varies jointly as B and C, 

A = ^-, if A varies directly as B and inversely as C. 

362. If A varies as B when C is constant, and A varies as C 
when B is constant, then will A vary as BC when both B and 
C vary. 

The variation of A depends partly on that of B and partly 
on that of C. Suppose these latter variations to take place 
separately, each in its turn producing its own effect on A; 
also let a, b, c be certain simultaneous values of A, B, O. 

1. Let C be constant while B changes to b ; then A must 
undergo a partial change and will assume some intermediate 
value a', where 

A -^ a) 



a 



800 ALGEBKA. 

13. If 6x — y « 10 a; — 11 y, and when x = 7, y = 5, find the equa 
tion between x and y. 

14. If the cube of x varies as the square of y, and if x= 3 when 
y = 6, find the equation between x and y. 

15. If the square root of a varies as the cube root of 6, and if 
a = 4 when 6 = 8, find the equation between a and b, 

16. If ?/ varies inversely as the square of x, and if y = 8 when 
a? = 3, find a; when y = 2. 

17. If x oc y + 0» where o is constant, and x = 15 when y = 1, and 
re = 35 when y = 5 ; find re when y = 2, 

18. If a + & a a — &, prove that a 2 + b 2 cc ah; and if a oc 6, prove 
that a 2 — b 2 cc ab. 

19. If y be the sum of three quantities which vary as «, x 2 , x* 
respectively, and when x = 1, y = 4, when « = 2, y = 8, and when 
x = 3, y = 18, express y in terms of x. 

20. Given that the area of a circle varies as the square of its radius, 
and that the area of a circle is 154 square feet when the radius is 
7 feet ; find the area of a circle whose radius is 10 feet 6 inches. 

21. The area of a circle varies as the square of its diameter : prove 
that the area of a circle whose diameter is 2} inches is equal to the 
sum of the areas of two circles whose diameters are 1} and 2 inches 

respectively. 

22. The pressure of wind on a plane surface varies jointly as the 
area of the surface, and the square of the wind's velocity. The press- 
ure on a square foot is 1 pound when the wind is moving at the rate 
of 15 miles per hour : find the velocity of the wind when the pressure 
on a square yard is 16 pounds. 

23. The value of a silver coin varies directly as the square of its 
diameter, while its thickness remains the same ; it also varies directly 
as its thickness while its diameter remains the same. Two silver 
coins have their diameters in the ratio of 4 : 3. Find the ratio of 
their thicknesses if the value of the first be four times that of the 
second. 

24. The volume of a circular cylinder varies as the square of the 
radius of the base when the height is the same, and as the height 
when the base is the same. The volume is 88 cubic feet when the 
height is 7 feet, and the radius of the base is 2 feet : what will be the 
height of a cylinder on a base of radius 9 feet, when the volume is 
396 cubio feet ? 



CHAPTER XXXIV. 

Arithmetical, Geometrical, and Harmonical 

Progressions. 

364. A succession of quantities formed according to some 
fixed law is called a series. The separate quantities are 
called terms of the series. 

ARITHMETICAL PROGRESSION. 

365. Definition. Quantities are said to be in Arithmeti- 
cal Progression when they increase or decrease by a common 
difference. 

Thus each of the following series forms an Arithmetical 
Progression: 

3, 7, 11, 15, ... 

8, 2, -4, -10, ... 

a, a + d 9 a + 2d, a + Sd\ •-. 

The common difference is found by subtracting any term 
of the series from that which follows it. In the first of the 
above examples the common difference is 4 ; in the second 
it is — 6 ; in the third it is d. 

366. The Last, or nth Term, of an A. P. If we examine 
the series 

a, a + d, a + 2d, a + 3 d, ••• 

we notice that in any term the coefficient of d is always less b$ 
one than the number of the term in the series. 

801 



302 ALGEBRA. 

Thus the 3d term is a + 2 d\ 

6th term is a + 5d; 
20th term is a + 19d; 
and, generally, the jpth term is a + (p — l)d. 

If n be the number of terms, and if I denote the last, ox 
nth term, we have 

J = a + (n-l)d 

367. The Sum of n Terms in A. P. Let a denote the first 
term, d the common difference, and n the number of terms. 
Also let I denote the last term, and S the required sum ; then 

8 = a + (a + d)+(a + 2 d) + — + (l - 2 d)+ (I - d)+J. 

and, by writing the series in the reverse order, 

S = l+(l-d)+(l-2d) + — + ( a + 2d) + (a + d)+a. 

Adding together these two series, 

2£ = (a + J)+(a + J) + ( a + D+"* to n terms = n (a -f J), 

... <S = |(a + J) *. (1). 

Sinoe J = a+(n-l)d (2); 

.-. 5 = ?|2o+(n-l)d| . . . (3). 

368. In the last article we have three useful formulas (1), 
(2), (3) ; in each of these any one of the letters may denote 
the unknown quantity when the three others are known. 

Ex. 1. Find the 20th and 35th terms of the series 

38, 36, 34, •••• 

Here the common difference is 36 — 38, or — 2. 

.-. the 20th term = 38 + 19 (- 2) = 0; 
and the 35th term = 38 + 34( - 2) = - 80. 

Ex. 2. Find the sum of the series 5|, 6}, 8, — to 17 terms. 
Here the common difference is 1 \ ; hence from (3) 

The sum = ^{2 x ^ + 16 x 1J} 

= Y(H + 20)s 17 * 81 = 263}. 

9 



ARITHMETICAL PROGRESSION. 303 

Ex. 3. The first term of a series is 5, the last 45, and the sum 400 

Ind the number of terms, and the common difference. 

If n be the number of terms, then from (1), 

400 = 5(6 + 46); 

whence n = 16. 

If d be the common difference, 

46 a the 16th term a 5 + 16 d , 
whence d s 2 §. 



EXAMPLES XXXIV. a 

1. Find the 27th and 41st terms in the series 6, 11, 17, •••• 

2. Find the 13th and 109th terms in the series 71, 70, 69, .... 
8. Find the 17th and 54th terms in the series 10, 11}, 13, •••. 

4. Find the 20th and 13th terms in the series —3,-2,-1, •••. 

5. Find the 90th and 16th terms in the series —4, 2.6, 9, •••• 

6. Find the 37th and 89th terms in the series —2.8, 0, 2.8, •••. 

Find the last term in the following series : 

7. 6, 7, 9, ... to 20 terms. 10. .6, 1.2, 1.8, ••• to 12 terms. 

8. 7, 3,-1, •••to 15 terms. 11. 2.7, 3.4, 4.1, ••• to 11 terms. 

9. 13}, 9, 4}, — to 13 terms. 12. x, 2aj, 3as, — to 25 terms. 
18. a — d, a + d, a + 3 d, ••• to 30 terms. 

14. 2 a — 6, 4 a — 3 6, 6 a — 5 b, ••• to 40 terms. 

Find the last term and sum of the following series : 

15. 14, 64, 114, ... to 20 terms. 18. J, —J, — J, — to 21 terms. 

16. 1, 1.2, 1.4, •.. to 12 terms. 19. 3}, 1, —1}, ••• to 19 terms. 

17. 9, 5, 1, ... to 100 terms. 20. 64, 96, 128, ••• to 16 terms. 

Find the sum of the following series : 

21. 5, 9, 13, -. to 19 terms. 26. 10, 9f, 9}, — to 21 terms. 

22. 12, 9, 6, — to 23 terms. 27. p, 3p, 6p, — to p terms. 
28. 4, 6J, 6}, •>. to 37 terms. 28. 3 a, a, —a, ••• to a terms. 

24. 10}, 9, 7}, ... to 94 terms. 29. a, 0, —a, ••• to a terms. 

25. — 3, 1, 6, ... to 17 terms. SO. — 8g, — g, q, ••• to p terms 



804 ALGEBRA. 

Find the number of terms and the common difference when 

31. The first term is 3, the last term 90, and the sum 1395. 

32. The first term is 79, the last term 7, and the sum 1075, 

33. The sum is 24, the first term 9, the last term —6. 

34. The sum is 7 14, the first term 1, the last term 58|. 

85. The last term is —16, the sum —133, the first term —8. 

36. The first term is —75, the sum —740, the last term 1. 

37. The first term is a, the last 13 a, and the sum 49 a. 

38. The sum is —320 a;, the first term 3g, the last term —35 a;. 

369. If any two terms of an Arithmetical Progression be 
given, the series can be completely determined ; for the data 
furnish two simultaneous equations, the solution of which 
will give the first term and the common difference. 

Ex. Find the series whose 7th and 51st terms are —3 and —355 

respectively. 

If a be the first term, and d the common difference, 

—3 = the 7th term = a 4- 6 d ; 
and — 355 = the 51st term = a + 50 d\ 

whence, by subtraction, —352 = 44 d; 
.*. d — — 8 ; and, consequently, a = 45. 

Hence the series is 45, 37, 29 .... 

370. Arithmetic Mean. When three quantities are in 
Arithmetical Progression, the middle one is said to be 
the arithmetic mean of the other two. 

Thus a is the arithmetic mean between a — d and a -+■ d. 

371. To find the arithmetic mean between two given quan- 
tities. 

Let a and b be the two quantities ; A the arithmetic mean 
Then, since a, A, b, are in A.P., we must have 

b — A = A — a, 

each being equal to the common difference ; 

whence ^ = «±i. 



ARITHMETICAL PROGRESSION. 305 

372. Between two given quantities it is always possible 
to insert any number of terms such that the whole series 
thus formed shall be in A. P. ; and by an extension of the 
definition in Art. 370, the terms thus inserted are called 
the arithmetic means, 

Ex. Insert 20 arithmetic means between 4 and 67. 

Including the extremes the number of terms will be 22 ; so that we 
have to find a series of 22 terms in A. P., of which 4 is the first and 
67 the last. 

Let d be the common difference ; 

then 67 = the 22d term, = 4 + 21 d ; 

whence d = 3, and the series is 4, 7, 10, — 61, 64, 67 ; 
and the required means are 7, 10, 13, ••• 58, 61, 64. 

373. To insert a given number of arithmetic means between 
two given quantities. 

Let a and b be the given quantities, m the number of 
means. 

Including the extremes the number of terms will be 
m + 2 ; so that we have to find a series of m + 2 terms in 
A. P., of which a is the first, and b is the last 

Let d be the common difference ; 

then b = the (m 4- 2)th term 

= a-f-(m + l)d', 

whence d = — — . ; 

m -f 1 

and the required means are 

. b — a „ . 2(b— a) „ , m(b — a) 

a-{ — , a+ K / ; "'0+ N i / 

m + 1 m + 1 m-f-1 

Ex. 1. Find the 30th term of an A. P. of which the first term to 
17, and the 100th term - 16. 
Let d be the common difference ; 

then - 16 = the 100th term 

= 17 + 99d; 

• • a — — ^. 

The 30th term » 17 + 29(- J)= 7*. 

x 



306 



ALGEBRA. 



Ex. 2. The sum of three numbers in A. ?. is 33, and their product 
is 792 : find them. 

Let a be the middle number, d the commc-i difference ; then the 
three numbers are a — d, a, a + d. 

Hence a — d + a + a + d = 3&; 

whence a = 11 ; and the three numbers are 11 — d, 11, 11 + d. 

.% 11(11 + d)(ll-d) = 792, 

121 - <P = 72, 

d = ±7; 

and the numbers are 4, 11, 18. 

Ex. 3. How many terms of the series 24, 20, 16, ••• must be taken 
that the sum may be 72 ? 

Let the number of terms be n ; then, since the common difference 
is 20 - 24, or — 4, we have from (3), Art. 367, 

72 = £{2 x 24 + (n - 1)(- 4)} 

= 24n-2n(n- 1) ; 
whence n 2 - 13 n + 36 = 0, 

or («-4)(n-9)=0; 

.*. n = 4 or 9. 

Both of these values satisfy the conditions of the question ; for if 
we write down the first 9 terms, we get 24, 20, 16, 12, 8, 4, 0, — 4, 
— 8 ; and, as the last five terms destroy each other, the sum of 9 
terms is the same as that of 4 terms. 



Ex. 4. An A. P. consists of 21 terms ; the sum of the three terms 
in the middle is 129, and of the last three is 237 ; find the series. 
Let a be the first term, and d the common difference. Then 

237 = the sum of the last three terms 

=:a + 20d + a + 19d + a + 18d = 3a + 67d; 

whence a-fl9d = 79 (1). 

Again, the three middle terms are the 10th, 11th, 12th ; 

hence 129 = the sura of the three middle terms 

= a + 9d + a + 10d + a + ll(J = 3a + 30d; 

whence a + 10d = 43 (2) 

From (1) and (2), we obtain d = 4, a = 3. 

Hence the series is 3, 7, 11, ••• 83. 



ARITHMETICAL PROGRESSION. 307 

EXAMPLES XXXIV. b. 

Find the series in which 

1. The 27th term is 186, and the 45th term 312. 

2. The 6th term is 1, and the 31st term — 77. 

3. The 15th term is — 26, and the 23rd term — 41. 

4. The 9th term is - 11, and the 102nd term - 160J. 

5. The 15th term is 25, and the 29th term 46. 

6. The 16th term is 214, and the 61st term 739. 

7. The 3rd and 7th terms of an A. P. are 7 and 19 ; find the 15th 
term. 

8. The 54th and 4th terms are — 125 and ; find the 42nd term. 

9. The 31st and 2nd terms are \ and 7f ; find the 69th term. 

10. Insert 15 arithmetic means between 71 and 23. 

11. Insert 17 arithmetic means between 93 and 69. 

12. Insert 14 arithmetic means between — 71 and — 2 \, 

13. Insert 16 arithmetic means between 7.2 and — 6.4. 

14. Insert 36 arithmetic means between 8J and 2J. 

How many terms must be taken of 

15. The series 42, 39, 36, ... to make 315 ? 

16. The series - 16, - 15, - 14, ... to make - 100? 

17. The series 16|, 151, 16, ... to make 129 ? 

18. The series 20, 18f , 17J, ... to make 162J ? 

19. The series - 10J, - 9, - 7J, ... to make - 42 ? 

20. The series - 6$, - 6|, - 6, ... to make - 52i ? 

21. The sum of three numbers in A. P. is 39, and their product is 
2184: find them. 

22. The sum of three numbers in A. P. is 12, and the sum of their 
squares is 66 : find them. 

23. The sum of five numbers in A. P. is 75, and the product of the 
greatest and least is 161 : find them. 

24. The sum of five numbers in A. P. is 40, and the sum of their 
squares is 410 : find them. 

25. The 12th, 85th, and last terms of an A. P. are 38, 267, 395 re- 
spectively : find the number of terms. 



308 



ALGEBRA* 



GEOMETRICAL PROGRESSION. 

374. Define tion. Quantities are said to be in Geometri- 
cal Progression when they increase or decrease by a constant 
factor. 

Thus each of the following series forms a' Geometrical 
Progression : 

3, 6, 12, 24, ... 

a, ar, ar 2 , ar 9 , ... 

The constant factor is also called the common ratio, and it 
is found by dividing any term by that which immediately 
precedes it. In the first of the above examples the common 
ratio is 2 ; in the second it is — £,; in the third it is r. 

375. The Last, or /ith Term, of a G. P. If we examine the 

series 

a, ar, ar 2 , ar 9 , ar 4 , ... 

we notice that in any term the index ofris always less by one 
than the number of the term in the series. 

Thus the 3rd term is ar 2 ; 

the 6th term is ar 5 ; 
the 20th term is ar 32 ; 
and, generally, the pth term is ar p ~\ 

If n be the number of terms, and if I denote the last, or 
nth term, we have I = ar*~\ 

Ex. Find the 8th term of the series — £, J, — }, ... 
The common ratio is \ -s- (— £), or — f ; 
,\ the 8th term = - \ x (- f ) 7 



376. Geometric Mean. When three quantities are in Geo- 
metrical Progression the middle one is called the geometric 
mean between the other two. 



GEOMETRICAL PROGRESSION. 309 

377. To find the geometric mean between two given quan- 
tities. 

Let a and b be the two quantities; G the geometric mean. 
Then since a, G, b are in G. P., 

G a 

each being equal to the common ratio; 

.-. G* = db\ 
whence G = Va6. 

378. To insert a given number of geometric means between 
two given quantities. 

Let a and b be the given quantities, m the number of 
means. 

There will be m + 2 terms ; so that we have to find a 
series of m + 2 terms in G. P., of which a is the first and 
b the last. 

Let r be the common ratio ; 
then 6= the (m+2)th term = ar™* 1 ; 



• 
• • 


r"* +1 = 


b 
a 5 


i 




.*. r = 


/6N 


m+l 



(!)• 



Hence the required means are ar, ar 2 , •••ar jn , where r has 
the value found in (1). 

Ex. Insert 4 geometric means between 160 and 5. 
We have to find 6 terms in G. P. of which 160 is the first, and 5 the 
sixth. 

Let r be the common ratio ; 

then 6 = the sixth term = 160 r 6 ; 

• i* — J±* 
• • ~ — u^ > 

whence, by trial, r = J ; 

and the means are 80, 40, 20, 10. 



310 ALGEBRA. 

379. The Sum of n Terms in G. P. Let a be the first term, 
r the common ratio, n the number of terms, and S the sum 
required. Then 

S = a + ar + at* + ••• + ar" -2 -f- ar"" 1 ; 
multiplying every term by r, we have 

rS = ar -f- ar 2 + ••• -f ar*~* + ar" -1 + at* . 
Hence by subtraction, 

rS — S = ar n — a; 
.\ (r-l)# = a(r"-l); 

. ^ = ^--l) (1) 

r — 1 
Changing the signs in numerator and denominator 

s== q^-j^ (2) 

1 — r 

Note. It will be found convenient to remember both forms given 
above for S, using (2) in all cases except when r is positive and greater 
than 1. 

Since ar 91 " 1 = J, it follows that ar* = rl, and formula (1) may be 
written 

r — 1 

Ex. 1. Sum the series 81, 54, 36, ••• to 9 terms. 
The common ratio = f \ = f , which is less than 1 ; 

hence the sum = 81 ** ~ ( P^ = 243{1 - (})»} 

= 243 - Vr* = 236ff . 

Ex. 2. Sum the series $, — 1, }, ••• to 7 terms. 
The common ratio = — f ; hence by formula (2) 

'2 2 



EXAMPLES XXXIV. C. 

1. Find the 5th and 8th terms of the series 3, 6, 12, ••-. 

2. Find the 10th and 16th terms of the series 256, 128, 64, .... 

3. Find the 7th and 11th terms of the series 64, — 82, 16, •••. 



GEOMETRICAL PROGRESSION. 311 

4. Find the 8th and 12th terms of the series 81, - 27, 9, .... 

5. Find the 14th and 7th terms of the series ^ T , ^j, ^, .... 

6. Find the 4th and 8th terms of the series .008, .04, .2, •••. 

Find the last term in the following series : 

7. 2, 4, 8, ... to 9 terms. 10. 3, - 3 2 , 3 8 , ... to 2 n terms. 

8. 2, - 6, 18, ... to 8 terms. 1L *» *» * - to * terms ' 

9. 2, 3, 4J, ... to 6 terms. 12 ' *' *' £» '" to ^ tenns - 

13. Insert 3 geometric means between 486 and 6. 

14. Insert 4 geometric means between J and 128. 

15. Insert 6 geometric means between 56 and — ^. 

16. Insert 6 geometric means between fj and 4|. 

Find the last term and the sum of the following series : 

17. 3, 6, 12, •-. to 8 terms. 20. 8.1, 2.7, .9, ... to 7 terms. 

18. 6, — 18, 54, ... to 6 terms. 21. 7 V, 3^, J, ... to 8 terms. 

19. 64, 32, 16, .- to 10 terms. 22. 4}, 1£, J, ... to 9 terms. 

Find the sum of the series : 

23. 3, — 1, |, ... to 6 terms. 30. 2, — 4, 8, ••• to 2p terms. 

24. i, i, f, ••• to 7 terms. _, 1,3 . Q . 

23 9 ' 31. -=-t, 1, —-,..« to 8 terms. 

25. — |, J, — f, ... to 6 terms. \/o ^/3 

26. 1, — J, J, .«• to 12 terms. 32. y/a, ^/a s t y/aP t — to terms. 

27. 9, - 6, 4, ... to 7 terms. 1 8 . 7 . WMa 

33. — , — 2, — , ... to 7 terms. 

28. §, — £, ^ ... to 8 terms. \/2 ^/2 

29. 1, 3, 3 2 , ... to p terms. 34. y/2, ^/Q, 3^,... to 12 terms. 

380. Infinite Geometrical Series. Consider the series 

..111 
19 2' 2* 2? " # 

1-4 

2» / i\ -i 

The sum to n terms = 7- = 2f 1 — ~ ]=2 — -— • 

1 _1 \ 2"/ 2 

2 

• 

From this result it appears that however many terms be 
taken the sum of the above series is always less than 2. 
Also we see that, by making n sufficiently large, we can 



812 ALGEBRA. 

make the fraction — - as small as we please. Thus by taking 

a sufficient number of terms the sum can be made to differ 

by as little as we please from 2. 

In the next article a more general case is discussed- 
SSI. Sum to Infinity. From Art. 379 we have 

e _ a(l — r") __ _a w* 

1 — r 1 — r 1 — r 

Suppose r is a proper fraction ; then the greater the value 
of n the smaller is the value of r", and consequently of 

; and therefore by making n sufficiently large, we can 

1_ r a 

make the sum of n terms of the series differ from * __ by 

as small a quantity as we please. 

This result is usually stated thus : the sum of an infinite 
number of terms of a decreasing Geometrical Progression is 

; or more briefly, the sum to infinity is 



1—r 1—r 

382. Recurring decimals furnish a good illustration of 
Infinite Geometrical Progressions. 

Ex. Find the value of .423. 

.423 = .4232323 ... = A + -?*- + - 2 - 3 — + ... = i. + M + .??-+. ... 

10 1000 100000 10 10* 10* 

io io 3 v io 2 10 4 ) io io» i_JL 

___£, 23 100 = i. , 23^ = 419 10» 

10 10 s " 99 10 990 990' 

which agrees with the value found by the usual arithmetical rule. 

EXAMPLES XXXIV. d. 

Sum to infinity the following series : 

1. 9, 6, 4, ... 3. i, }, h - *• J, f, fa - 7. .9, .03, .001, .~ 

2. 12,6,3,... 4. i, -J, i,... 6. J, -1, f,.- 8. .8, -.4, .2, ... 

Find by the method of Art. 382, tho value of 
9. .3. 10. .16. 11. .24. 12. .378. 13. .037. 



HARMONICAL PROGRESSION. 818 

Find the series in which 

14. The 10th term is 320 and the 6th term 20. 

15. The 5th term is f £ and the 9th term is J. 

16. The 7th term is 625 and the 4th term — 5. 

17. The 3d term is T % and the 6th term - 4|. 

18. Divide 183 into three parts in G. P. such that the sum of the first 
and third is 2^ times the second. 

19. Show that the product of any odd number of consecutive terms 
of a G. P. will be equal to the nth power of the middle term, n being 
the number of terms. 

20. The first two terms of an infinite G. P. are together equal to 1, 
and every term is twice the sum of all the terms which follow. Find 
the series. 

Sum the following series : 

21. y 2 + 26, y 4 + 46, y 6 + 6&, ... to n terms. 

22. l + 2 ^ 2 , 1, ?Jllv^, ... to infinity. 
3-2v>2 3 + 2^2 J 

28- V!i iV 2 . Wh to infinity. 

24. 2 n — $, 4 n + £, 6 n — ^ ••• to 2 n terms. 

25. The sum of four numbers in G. P. is equal to the common ratio 
plus 1, and the first term is ^ 7 . Find the numbers. 

26. The difference between the first and second of four numbers 
in G. P. is 96, and the difference between the third and fourth is 6. 
Find the numbers. 

27. The sum of $ 225 was divided among four persons in such a 
manner that the shares were in G. P., and the difference between the 
greatest and least was to the difference between the means as 21 to 6. 
Find the share of each. 

28. The sura of three numbers in G. P. is 13, and the sum of their 
reciprocals is ^. Find the numbers. 

HARMONICAL PROGRESSION. 

383. Definition. Three quantities, a, 6, c, are said to 

be in Harmonical Progression when - = , ~~ * 

c b — c 

Any number of quantities are said to be in Harmonical 
Progression when every three consecutive terms are in 
Harmonical Progression. 



814 



ALGEBRA. 



384. The Reciprocals of Quantities in Harmonical Progression 
are in Arithmetical Progression. 

By definition, if a, b, c are in Harmonical Progression, 

a_ a — b m 
c. b — c 9 

.'. a(b — c) = c(a — b) f 

dividing every term by abc, 

c b b ct 

which proves the proposition. We may therefore define an 
Harmonical Progression as a series of quantities the recipro- 
cals of which are in Arithmetical Progression. 

385. Solution of Questions in H. P. Harmonical proper- 
ties are chiefly interesting because of their importance in 
Geometry and in the Theory of Sound: in Algebra the 
proposition just proved is the only one of any importance. 
There is no general formula for the sum of any number of 
quantities in Harmonical Progression. Questions in H. P. 
are generally solved by inverting the terms, and making 
use of the properties of the corresponding A. P. 

Ex. The 12th term of an H. P is |, and the 19th term is ^ : find 
the series. 

Let a be the first term, d the common difference of the correspond- 
ing A. P. ; then 

6 = the 12th term = a + lid; 

and \ 2 - = the 19th term = a + 18<f ; 



whence 



d = 



— i 

"8* 



a 



4 
S' 



Hence the Arithmetical Progression is J, }, 2, J, 
and the Harmonical Progression is f , f, £, $,'••• 



386. Harmonic Mean. When three quantities are in 
Harmonic Progression the middle one is said to be the 
Harmonic Mean of the other two. 



THE PROGRESSIONS. 815 

387. To find the harmonic mean between two given quan- 
tities. 

Let a, b be the two quantities, H their haxmonio mean; 

then -, — , - are in A. P.; 
a H o 

" H a b H' 

H a b' 

p.__ 2ab t 

~~a + b 

388. Relation between the Arithmetic, Geometric, and Har- 
monic Means. If A, G, H be the arithmetic, geometric, and 
harmonic means between a and b, we have proved 

A = «±> ...... (1). 

G = Vo5 (2). 

2T=-^. (3). 

a + b v J 

Therefore AH=«±*.^~ 

2 a + b 

= ab=GP\ 
that is, O is the geometric mean between A and H. 

389. Miscellaneous Questions in the Progressions. Miscel- 
laneous questions in the Progressions afford scope for, 
much skill and ingenuity, the solution being often very 
neatly effected by some special artifice. The student will 
find the following hints useful. 

1. If the same quantity be added to, or subtracted from, 
all the terms of an A. P., the resulting terms will form an 
A. P., with the same common difference as before. [Art 
365.] 



816 



ALGEBRA. 



2. If all the terms of an A. P. be multiplied or divided 
by the same quantity, the resulting terms form an A. P., 
but with a new common difference. [Art. 365.] 

3. If all the terms of a G. P. be multiplied or divided by 
the same quantity, the resulting terms form a G. P. with 
the same common ratio as before. [Art. 374.] 

4. If a, b, c, d ••• be in G. P., they are also in continued 
proportion, since by definition 

? — - — -— ... — 1 

bed r 9 

Conversely, a series of quantities in continued proportion 
may be represented by x, xr, xr 2 , 

Ex. 1. Find three quantities in G. P. such that their product is 
843, and their sum 30}. 
a 



Let - , a, ar be the three quantities ; 
r 



then we have 

and 

From (1) 



- x a x ar = 343 



(1), 

(2). 



a 8 = 343 ; 
.-. a = 7; 
hence from (2) 7(1 + r + r 2 ) = V" **• 

Whence we obtain r = 3, or J; 

and the numbers are J, 7, 21. 



Ex. 2. If a, 6, c be in H. P., prove that 
also in H. P. 

Since -, , - are in A. P., 
a b c 



a 



b + c 1 c + a' a + & 



are 



a±b±e a + b + c a + b + c ^ ^ A p 
'6c 



a 



... H- 6 -^, l + «±°, l + «±* are in A.P., 

b c 



a 



... L±2, *L±*, «+* are in A. P.; 
' b ' 



a 
a 



c 

c 



6 + c c + a a + 6 



are in H. P, 



THE PROGRESSIONS. 817 

Ex. 8. The nth tens af an A. P. Is j|+ 2: find the sum of 49 
terms. 

Let a be the first term, and I the last ; then by putting n = 1, 
and n = 49 respectively, we obtain 

a = i + 2, 1 = ^ + 2; 

= 4£ x 14 = 843. 

Ex. 4. If a, b, c, d, e be in 6. P., prove that b + d is the geomet- 
ric mean between a + c and c + e. 

Since a, 5, c, d, « are in continued proportion, 

- = - = - = -• 
6 c d e' 

•% each ratio = ?-±4== 5 i* . [Art 347.] 

b + d c+e u 

Whence (6 + d) 2 = (a + c) (« + e). 



EXAMPLES XXXIV. e. 

1. Find the 6th term of the series 4, 2, 1J, ... . 

2. Find the 21st term of the series 2J-, 1||, 1^, •••. 

3. Find the 8th term of the series 1 J, 1-}^, 2^, ••• • 

4. Find the nth term of the series 3, 1}, 1, .... 

Find the series in which 

5. The 15th term is fa and the 23d term is fa. 

6. The 2d term is 2, and the 31st term is fa. 

7. The 39th term is fa and the 64th term is fa 

Find the harmonic mean between 

8. 2 and 4. 9. 1 and 13. 10. } and fa 

11. i and 1. 12. and — 1— 13. x + y and x — y 

a b x + y x — y 

14. Insert two harmonic means between 4 and 12. 

15. Insert three harmonic means between 2f and 12. 

16. Insert four harmonic means between 1 and 6. 



318 ALGEBRA. 

17. If G be the geometric mean between two quantities A and B, 
show that the ratio of the arithmetic and harmonic means of A and G 
is equal to the ratio of the arithmetic and harmonic means of G and B. 

18. To each of three consecutive terms of a G. P., the second of 
the three is added. Show that the three resulting quantities are in 
H.P. 

Sum the following series : 

19. 1 + If + 3-^ -f — to 6 terms. 

20. 1 + 1} + 2} + ... to 6 terms. 

21. (2a + x) + Sa+(4a—x)-\ — top terms. 

22. If — 1J + $ • to 8 terms. 

23. 1$ + 1J + $ H to 12 terms. 

24. If x — a, y — a, and z — a be in G.P., prove that 2(y — a) is 
the harmonic mean between y — x and y — z. 

25. If o, 6, c, d be in A. P., a, e, /, d in G. P., a, g, A, d in H. P. 

respectively ; prove that ad = ef = bh = cgr. 

26. If a 2 , 6 2 , C 2 be in A. P., prove that 6+c, c+a, a+ 6 are in H. P. 



CHAPTER XXXV. 
Pebmutations and Combinations. 

390. Each of the arrangements which can be made by 
taking some or all of a number of things is called a permu- 
tation. 

Each of the groups or selections which can be made by 
taking some or all of a number of things is called a com- 
bination. 

Thus the permutations which can be made by taking the 
letters a, b, c, d two at a time are twelve in number ; namely, 

ab 9 ac, ad, be, bd, cd, 
ba, ca, da, cb, db, dc; 

each of these presenting a different arrangement of two 
letters. 

The combinations which can be made by taking the letters 
a, b, c, d two at a time are six in number ; namely, 

ab, ac, ad, be, bd, cd; 

each of these presenting a different selection of two letters. 

Erom this it appears that in forming combinations we are 
only concerned with the number of things each selection 
contains ; whereas in forming permutations we have also to 
consider the order of the. things which make up each arrange- 
ment ; for instance, if from four letters a, b, c, d we make 
a selection of three, such as abc, this single combination 
admits of being arranged in the following ways : 

abc, acb, bca, bac, cab, cba, 

and so gives rise to six different permutations. 

819 



f 



320 ALGEBRA. 

391. Fundamental Principle. Before discussing the gen. 
eral propositions of this chapter the following important 
principle should be carefully noticed. 

If one operation can be performed in m ways, and (when it 
has been performed in any one of these ways) a second opera- 
tion can then be performed in n ways; the number of ways oj 
performing the two operations will be m x n. 

If the first operation be performed in any one way, we 
can associate with this any of the n ways of performing the 
second operation ; and thus we shall have n ways of per- 
forming the two operations without considering more than 
one way of performing the first ; and so, corresponding to 
each of the m ways of performing the first operation, we 
shall have n ways of performing the two ; hence the product 
m x n represents the total number of ways in which the two 
operations can be performed. 

Ex. Suppose there are 10 steamers plying between New York and 
Liverpool: in how many ways can a man go from New York to 
Liverpool and return by a different steamer ? 

There are ten ways of making the first passage ; and with each of 
these there is a choice of nine ways of returning (since the man is 
not to come back by the same steamer); hence the number of ways 
of making the two journeys is 10 x 9, or 90. 

This principle may easily be extended to the case in 
which there are more than two operations each of which 
can be performed in a given number of ways. 

Ex. Three travellers arrive at a town where there are four hotels ; 
in how many ways can they take up their quarters, each at a differ- 
ent hotel ? 

The first traveller has choice of four hotels, and when he has made 
his selection in any one way, the second traveller has a choice of 
three ; therefore the first two can make their choice in 4 x 3 ways ; 
and with any one such choice the third traveller can select his hotel 
in 2 ways ; hence the required number of ways is 4 x 3 x 2, or 24. 

392. To find the number of permutations of n dissimilar 
things taken rata time. 

This is the same thing as rinding the number of ways ii 



PERMUTATIONS AND COMBINATIONS. 821 

which we can fill r places when we have n different things 
at our disposal. 

The first place may be filled in n ways, for any one of 
the n things may be taken ; when it has been filled in any 
one of these ways, the second place can then be filled in 
n — 1 ways ; and since each way of filling the first place 
can be associated with each way of filling the second, the 
number of ways in which the first two places can be filled 
is given by the product n(n — 1). And when the first two 
places have been filled in any way, the third place can be 
filled in n — 2 ways. And reasoning as before, the number 
of ways in which three places can be filled is n(n— l)(n— 2). 

Proceeding thus, and noticing that a new factor is intro- 
duced with each new place filled, and that at any stage the 
number of factors is the same as the number of places filled, 
we shall have the number of ways in which r places can be 
filled equal to 

n(n — l)(rt — 2) ••• to r factors. 

We here see that each factor is formed by taking from n a 
number one less than that which applies to the place filled by 
that factor; hence the rth factor is n— (r— 1), or n-r-r+1. 

Therefore the number of permutations of n things taken 
r at a time is 

n(n — l)(n — 2)^*(n — r + 1). 

Cor. The number of permutations of n things taken all 
at a time is 

n(n — l)(n — 2)--« to n factors, 

or n(n - l)(w — 2)— 3.2-1. 



It is usual to denote this product by the symbol 



is read " factorial n" Also n ! is sometimes used for [n. 



n, which 



L We shall in future denote the number of permuta- 
tions of n things taken r at a time by the symbol "P^ so 
that 

n P r = n(n - l)(n - 2) ...(t* - r 4- 1); 

also n P n = \n. 



622 



ALGEBRA. 



In working numerical examples it is useful to notice that 
the suffix in the symbol n P r always denotes the number of 
factors in the formula we are using. 

Ex. 1. Four persons enter a carriage in which there are six seats : 
in how many ways can they take their places ? 

The first person may seat himself in 6 ways ; and then the second 
person in 5 ; the third in 4 ; and the fourth in 3 ; and since each of 
these ways may be associated with each of the others, the required 
answer is 6 x 5 x 4 x 3, or 360. 

Ex. 2. How many different numbers can be formed by using six 
out of the nine digits 1, 2, 3, ••• 9 ? 

Here we have 9 different things, and we have to find the number of 
permutations of them taken 6 at a time ; 

.*. the required result = *Pe ^ ' ^ - * 

=9x8x^x6x5x4= 60480. 

I i 1 
394. To find the number of combinations of n dissimilar 

things taken rata time. 

Let n C r denote the required number of combinations. 

Then each of these combinations consists of a group of 
r dissimilar things which can be arranged among themselves 
in [r ways. [Art. 392, Cor.] 

Hence n C r x \r is equal to the number of arrangements of 
n things taken r at a time ; that is, 

n G r x [r = n P r = n(rt - l)(n - 2)... (n-r + 1); 

... nC ^ n(n-l)(n-2)...(n-r + l) # # ^ 

Cor. This formula for n O r may also be written in a 
different form; for if we multiply the numerator and the 
denominator by \n — r we obtain 



n(n — l)(n — 2)...(n — r -f 1) X \n — r 



or 



[n 



(2); 



\r \n — r ' \r \n — r 

since n(n — l)(n — 2)-«-(w — r + 1) x |n — r = |n. 

It will be convenient to remember both these expressions 
for n C r , using (1) in all cases where a numerical result is 
required, and (2) when it is sufficient to leave it in an 
algebraic shape. 



PERMUTATIONS AND COMBINATIONS, 823 

Note. If In formula (2) we put r = w, we have 



1210 12 

but n C n = 1, so that if the formula is to be true for r=n, the symbol 
[0 must be considered as equivalent to 1. 

Ex. From 12 books in how many ways can a selection of 5 be 
made, (1) when one specified book is always included, (2) when one 
specified book is always excluded ? 

(1) Since the specified book is to be included in every selection, 
we have only to choose 4 out of the remaining 11. 

Hence the number of ways = U C 4 = ll * 10 * 9 x 8 = 330. 

J 1x2x3x4 

(2) Since the specified book is always to be excluded, we have to 
select the 5 books out of the remaining 11. 

Hence the number of ways = U C 5 = llxl0x9x8x7 = 462. 

J 1x2x3x4x6 



395* The number of combinations of n things rata time is 
equal to the number of combinations of n things /?— r at a time. 

In making all the possible combinations of n things, to 
each group of r things we select, there is left a correspond- 
ing group of n — r things ; that is, the number of combina- 
tions of n things, r at a time is the same as the number of 
combinations of n things n — r at a time; 

. n/~1 n/~1 

This result is frequently useful in enabling us to abridge 
arithmetical work. 

Ex. Out of 14 men in how many ways can an eleven be chosen ? 

The required number = i4(7ii = 14 C 8 = 14 x 13 x 12 = 364. 

1x2x3 

If we had made use of the formula 14 Cn, we should have had to 
reduce an expression whose numerator and denominator each con- 
tained 11 factors. 

396. In the examples which follow it is important to 
notice that the formula for permutations should not be used 
until the suitable selections required by the question have 
been made. 



324 ALGEBRA. 

Ex. 1. From 7 Englishmen and 4 Americans a committee of 6 Is 
to be formed: in how many ways can this be done, (1) when the 
committee contains exactly 2 Americans, (2) at least 2 Americans ? 

(1) The number of ways in which the Americans can be chosen is 
4 Ca ; and the number of ways in which the Englishmen can be chosen 
is 7 C4. Each of the first groups can be associated with each of the 
second ; hence - 

the required number of ways = 4 C?2 X 7 C* 

= JL X JL = JI 210 ' 

1212 lij? 121213 

(2) We exhaust all the suitable combinations by forming all the 
groups containing 2 Americans and 4 Englishmen ; then 3 Americans 
and 3 Englishmen ; and lastly 4 Americans and 2 Englishmen. 

The sum of the three results gives the answer. Hence the required 
number of ways = 4 C 2 x 7 C 4 + 4 C 8 x 7 C 8 + 4 C 4 x 7 3s 



<K" I? I? L*|3 ' [3 [3|4 [216. 

/' v = 210 + 140 + 21 = 371. 

In this example we have only to make use of the suitable formula 
for combinations, for we are not concerned with the possible arrange- 
ments of the members of the committee among themselves. 

Ex. 2. Out of 7 consonants and 4 vowels, how many words can be 
made each containing 3 consonants and 2 vowels ? 

The number of ways of choosing the three consonants is 7 C 8 , and 
the number of ways of choosing the two vowels is *C%; and since 
each of the first groups can be associated with each of the second, the 
number of combined groups, each containing 3 consonants and 2 
vowels, is 7 C 3 x 4 C 2 . 

Further, each of these groups contains 5 letters, which may be 
arranged among themselves in [5 ways. Hence 

17 14 
the required number of words = -=- x -=- x |5 

= 5 x [7 = 25200. 

EXAMPLES XXXV. a. 

1. Find the value of *p 4 , 7 P 6 , *C 5 , ™C 2 s. 

2. How many different arrangements can be made by taking (1) five, 
(2) all of the letters of the word soldier t 

3. If »<7 8 : n " 1 C r 4 = 8 : 6, find n. 



PERMUTATIONS AND COMBINATIONS. 325 

4. How many different selections of four coins can be made from a 
bag containing a dollar, a half-dollar, a quarter, a florin, a shilling, a 
franc, a dime, a sixpence, and a penny ? 

5. How many numbers between 3000 and 4000 can be made with 
the digits 9, 3, 4, 6 ? 

6. In how many ways can the letters of the word volume be 
arranged if the vowels can only occupy the even places ? 

7. If the number of permutations of n things four at a time is 
fourteen times the number of permutations of n — 2 things three at a 
time, find n. 

8. From 5 teachers and 10 boys how many committees can be 
selected containing 3 teachers and 6 boys ? 

9. If 20 (7 r = 20 a-io, find rc m i8C r . 

10. Out of the twenty-six letters of the alphabet in how many ways 
can a word be made consisting of five different letters two of which 
must be a and e ? 

11. How many words can be formed by taking 3 consonants and 2 
vowels from an alphabet containing 21 consonants and 5 vowels ? 

12. A stage will accommodate 5 passengers on each side : in how 
many ways can 10 persons take their seats when two of them remain 
always upon one side and a third upon the other ? 

397. Hitherto, in the f ormulae we have proved, the things 
have been regarded as unlike. Before considering cases in 
which some one or more sets of things may be like, it is 
necessary to point out exactly in what sense the words like 
and unlike are used. When we speak of things being 
dissimilar, different, unlike, we imply that the things are 
visibly unlike, so as to be easily distinguishable from each 
other. On the other hand, we shall always use the term 
like things to denote such as are alike to the eye and cannot 
be distinguished from each other. For instance, in Ex. 2, 
Art. 396, the consonants and the vowels may be said each to 
consist of a group of things united by a common character- 
istic, and thus in a certain sense to be of the same kind ; 
but they cannot be regarded as like things, because there is 
an individuality existing among the things of each group 
which makes them easily distinguishable from each other. 
Hence, in the final stage of the example we considered each 



826 



ALGEBRA. 



group to consist of five dissimilar things and therefore 
capable of [5 arrangements among themselves. [Art. 392, 
Cor.] 

398. To find the permutations of n things, taking them all 
at a time, when p things are of one kind, q of another kind, r 
of a third kind, and the rest all different. 

Let there be n letters ; suppose p of them to be a, q of 
them to be b, r of them to be c, and the rest to be unlike. 

Let x be the required number of permutations ; then if in 
any one of these permutations the p letters a were replaced 
by p unlike letters different from any of the rest, from this 
single permutation, without altering the position of any of 
the remaining letters, we could form \p new permutations. 
Hence if this change were made in each of the x permuta- 
tions, we should obtain x x\p permutations. 

Similarly, if the q letters b were replaced by q unlike 
letters, the number of permutations would be x x \p X (g. 
. In like manner, by replacing the r letters c by r unlike 
letters, we should finally obtain x x \p_ X [q x \r permutations. 

But the things are now all different, and therefore admit 
of \n permutations among themselves. Hence 

xx [g x\q x[r = [w; 

12 - 



that is, 



a? = 



\2\q\r' 

which is the required number of permutations. 

Any case in which the things are not all different may be 
treated similarly. 

Ex. 1. How many different permutations can be made out of the 
letters of the word assassination taken all together ? 

We have here 13 letters of which 4 are s, 3 are a, 2 are i, and 2 are 
n. Hence the number of permutations 

= tig 

lil§L2l2 
= 13.11.10.9.8.7.3.5 
= 1001 x 10800 = 10810800. 



PERMUTATIONS AND COMBINATIONS. 327 

Ex. 2. How many numbers can be formed with the digits 1, 2, 3, 
4, 3, 2, 1, so that the odd digits always occupy the odd places ? 
The odd digits 1, 3, 3, 1 can be arranged in their four places in 

,-=^-ways (1). 

The even digits 2, 4, 2 can be arranged in their three places in 

13 

^ways (2). 

I* 

Each of the ways in (1) can be associated with each of the ways 
in (2). 

14 13 
Hence the required number = -=- x == = 6 x 3 = 18. 

399. To find the number of permutations of n things rata 
time, when each thing may be repeated once, twice, ... up to r 
times in any arrangement. 

Here we have to consider the number of ways in which r 
places can be filled when we have n different things at our 
disposal, each of the n things being used as often" as we 
please in any arrangement. 

The first place may be filled in n ways, and, when it has 
been filled in any one way, the second place may also be 
filled in n ways, since we are not precluded from using the 
same thing again. Therefore the number of ways in which 
the first two places can be filled is n x n or n 2 . 

The third place can also be filled in n ways, and therefore 
the first three places in n s ways. 

Proceeding thus, and noticing that at any stage the index 
of n is always the same as the number of places filled, we 
shall have the number of ways in which the r places can be 
filled equal to n r . 

Ex. In how many ways can 5 prizes be given away to 4 boys, when 
each boy is eligible for all the prizes ? 

Any one of the prizes can be given in 4 ways ; and then any one 
of the remaining prizes can also be given in 4 ways, since it may be 
obtained by the boy who has already received a prize. Thus two 
prizes can be given away in 4 2 ways, three prizes in 4 8 ways, and so 
on. Hence the 6 prizes can be given away in 4 6 , or 1024 ways. 



828 



ALGEBRA. 



400. To find the total number of ways in which it is possible 
to make a selection by taking some or all of n things. 

Each thing may be dealt with in two ways, for it may 

either be taken or left ; and since either way of dealing with 

any one thing may be associated with either way of dealing 

with each one of the others, the number of ways of dealing 

with the n things is 

2 x 2 x 2 x 2 ... to n factors. 

But this includes the case in which all the things are left, 
therefore, rejecting this case, the total number of ways is 
2--1. 

This is often spoken of as " the total number of combina- 
tions " of n things. 

Ex. A, man has 6 friends ; in how many ways may he invite one or 
more of them to dinner ? 

He has to select some or all of his 6 friends ; and therefore the 
number of ways is 2 6 — 1, or 63. 

This result can be verified in the following manner. 

The guests may be invited singly, in twos, threes, ... ; therefore the 
number of selections 

= 6 + 15 4 20 + 15 + 6 + 1 = 63. 



401. To find for what value of r the number of combinations 
of n things rata time is greatest. 

_. n(n-l)(n-2)-..(n-r + 2)(n-r + l) 
bmce L *- 1.2.3...(r-l)r ' 

_ n(n-l)(n-2)..-(n-r + 2) . 
311(1 °- 1 - 1.2.3...(r-l) ' 



.-. "C r = "(7 r _ 1 x n ~ r + 1 - 



The multiplying factor 



r 

n— r-f-1 



may be written — ^ ± } 



r r 

which shows that it decreases as r increases. Hence as r 
receives the values 1, 2, 3 ; ««« in succession, n C r is continu- 



PERMUTATIONS AND COMBINATIONS. 329 

n -4- 1 

ally increased, until — — 1 becomes equal to 1 or less 

than 1. 

Now^ti-l>l,solongas?^ti>2; that is, 5±1 > r . 
r r 2 

We have to choose the greatest value of r consistent with 
this inequality. 

(1) Let n be even, and equal to 2 m ; then 

Ti + 1 2m -fl _ . , 

and for all values of r up to m inclusive this is greater than 

tit 
r. Hence by putting r = m = -, we find that the greatest 

number of combinations is n CL 



(2) Let n be odd, and equal to 2 m -f- 1 ; then 

n + 1 2m + 2 . i 

and for all values of r up to m inclusive this is greater than 
r; but when r=m-fl, the multiplying factor becomes equal 
to 1, and 

n C m+ i = n G m ; that is, n C n +i = n C n -i 5 

2 2 

and therefore the number of combinations is greatest when 
the things are taken ^T , or — - — at a time ; the result 

being the same in the two cases. 

EXAMPLES XXXV. b. 

1. Find the number of permutations which can be made from all 
the letters of the words, 

(1) irresistible, (2) phenomenon, (3) tittle-tattle. 

2. How many different numbers can be formed by using the seven 
digits 2, 3, 4, 3, 3, 1, 2 ? How many with the digits 2, 3, 4, 3, 3, 0, 2 ? 



880 ALGEBRA. 

3. How many words can be formed from the letters of the word 
Simoom, so that vowels and consonants occur alternately in each word ? 

4. A telegraph has 5 arms, and each arm has 4 distinct positions, 
including the position of rest: find the total number of signals that 
can be made. 

5. In how many ways can n things be given to m persons, when 
there is no restriction as to the number of things each may receive ? 

6. How many different arrangements can be made out of the let- 
ters of the expression a 6 6 8 c 6 when written at full length ? 

7. There are 4 copies each of 3 different volumes; find the 
number of ways in which they can be arranged on one shelf. 

8. In how many ways can 6 persons form a ring ? Find the num- 
ber of ways in which 4 gentlemen and 4 ladies can sit at a round table 
so that no two gentlemen sit together. 

9. In how many ways can a word of 4 letters be made out of the 
letters «, 6, e, c, d, o, when there is no restriction as to the number of 
times a letter is repeated in each word ? 

10. How many arrangements can be made out of the letters of the 
word Toulouse, so that the consonants occupy the first, fourth, and 
seventh places ? 

11. A boat's crew consists of eight men of whom one can only row 
on bow side and one only on stroke side : in how many ways can the 
crew be arranged ? 

12. Show that "+ 1 G r = n C r + "C r _i. 

13. If 2 »C 8 : "(72=44:3, find n. 

14. Out of the letters A, B, (7, p, q, r, how many arrangements can 
be made beginning with a capital ? 

15. Find the number of combinations of 50 things 46 at a time. 

16. If 18 C r = 18 a+ 2 , find 'C 6 . 

17. In how many ways is it possible to draw a sum of money from 
a bag containing a dollar, a half-dollar, a quarter, a dime, a five-cent 
piece, a two-cent piece, and a penny ? 



CHAPTER XXXVI. 

Probability (Chance). 

402. Definition. If an event can happen in a ways and 
fail in b ways, and each of these ways is equally likely, the 

probability, or the chance, of its happening is , and of 

£ a + b 

its failing is • Hence to find the probability of an 

event happening, divide the number of favorable ways by the 
whole number of ways favorable and unfavorable. 

For instance, if in a lottery there are 7 prizes and 25 
blanks, the chance that a person holding 1 ticket will win 
a prize is -^, and his chance of not winning is -ff . 

Instead of saying that the chance of the happening of an 

event is , it is sometimes stated that the odds are atob 

a + b 

in favor of the event, or b to a against the event. 

Thus in the above the odds are seven to twenty-five in 
favor of the drawing of a prize, and twenty-five to seven 
against success. 

403. The reason for the mathematical definition of proba- 
bility may be made clear by the following considerations : 

If an event can happen in a ways and fail to happen in b 
ways, and all these ways are equally likely, we can assert 
that the chance of its happening is to the chance of its fail- 
ing as a to b. Thus if the chance of its happening is repre- 
sented by ka, where k is an undetermined constant, then 
the chance of its failing will be represented by kb. 

.\ chance of happening + chance of failing = k(a + b). 

831 



832 



ALGEBRA. 



Now the event is certain to happen or to fail ; therefore the 
sum of the chances of happening and failing must represent 
certainty. If therefore we agree to take certainty as our unit, 
we have 

1 = A; (a + b\ 6r 7c = — — , 

a + 6 

.\ the chance that the event will happen is — — , 

rr a + b 



and the chance that the event will not happen is 



a + 6 



Cor. If p is the probability of the happening of an event, 
the probability of its not happening is 1 —p. 



404. The definition of probability in Art. 402 may be 
given in a slightly different form which is sometimes useful. 
If c is the total number of cases, each being equally likely 
to occur, and of these a are favorable to the event, then the 

probability that the event will happen is -, and the proba- 

c 

bility that it will not happen is 1 • 



Ex. 1. (a) From a bag containing 4 white and 5 black balls a man 
draws a single ball at random. What is the chance that it is black ? 

A black ball can be drawn in 5 ways, since any one of the 5 black 
balls may be drawn. In the same way any one of the 4 white balls 
may be drawn. 

Hence the chance of drawing a black ball is . or — 

8 4 + 5' 9 

(b) Suppose the man draws 3 balls at random. What are the 
odds against these being all black? 

The total number of' ways in which 3 balls can be drawn is 9 (7s, 
and the total number of ways of drawing 3 black balls is 5 Cs ; there- 
fore the chance of drawing 3 black balls 



_ 6 tfs. 


_6 


.4. 


3 


_ 5 


»C 3 


9 


.8- 


• 7 


42 



Thus the odds against the event are 37 to 5. 



PROBABILITY. 333 

Ex. 2. From a bag containing 5 red balls, 4 white balls, and 5 
black balls, 6 balls are drawn at random. What is the chance that 
8 are white, 2 black, and 1 red ? 

The number of combinations of 4 white balls, taken 3 at a 

4*3*2 
time, is — - — '— or 4. In the same manner the number of combina- 

1 • A • O K A 

tions of 6 black balls, taken 2 at a time, is ^-^ or 10. Since each of 

1.2 

the 4 combinations of white balls may be taken with any one of the 
10 combinations of black, and with each of the combinations so 
formed we may take any one of the 5 red balls, the total number of 
combinations will be 4 . 10 . 6 or 200. But the number of combinations 

of the entire number of balls, taken 6 at a time is 14 • 13 • 12 • 11 • 10 -9 

1*2.3*4*6.6 
or 3003, hence the chance that 3 white, 2 black, and 1 red ball will 

be drawn at one time is ^ftfj. 

Ex. 3. A has 3 shares in a lottery in which there are 3 prizes and 
6 blanks ; B has 1 share in a lottery in which there is 1 prize and 2 
blanks. Show that A's chance of success is to B's as 16 to 7. 

A may draw 3 prizes in 1 way ; he may draw 2 prizes and 1 

3*2 

blank in — — x 6 ways ; he may draw 1 prize and 2 blanks in 8 x 

o r 1*2 

— — ways ; the sum of these numbers is 64, which is the number of 
1*2 

ways in which A can win a prize. Also he can draw 3 tickets in 

9*8 7 

— , or 84 ways ; therefore A's chance of success = f J = Jf . 

B's chance of success is clearly J ; therefore A's chance : B's 
chance = if : * = 16 : 7. 

6 « 6 • 4 
Or we might have reasoned thus : A will get all blanks in — - — — , 

1 • 2 • o 
or 20 ways ; the chance of which is f f , or ^ ; therefore A's chance 
of success = 1 — ^ T = £f . 

405. From the examples given it will be seen that the 
solution of the easier kinds of questions in Probability 
requires nothing more than a knowledge of the definition 
of Probability, and the application of the laws of Permu- 
tations and Combinations. 

EXAMPLES XXXVI. 

1. A bag contains 5 white, 7 black, and 4 red balls ; find the 
chance of drawing : (a) One white ball ; (6) Two white balls ; 
(c) Three white balls ; (d) One ball of each color ; (e) One white, 
two black, and three red balls. 



S34 



ALGEBRA. 



2. If four coins are tossed, find the chance that there should be 
2 heads and 2 tails. 

3. One of two events must happen : given that the chance of the 
one is two-thirds that of the other, find the odds in favor of the other. 

4. Thirteen persons take their places at a round table. Show 
that it is 5 to 1 against 2 particular persons sitting together. 

5. There are three events A, B, C, one of which must, and only 
one can, happen ; the odds are 8 to 3 against A, 5 to 2 against B. 
Find the odds against C. 

6. A has 3 shares in a lottery containing 3 prizes and 9 blanks ; 
B has 2 shares in a lottery containing 2 prizes and 6 blanks. Com- 
pare their chances of success. 

7. There are three works, one consisting of 3 volumes, one of 4, 
and the other of 1 volume. They are placed on a shelf at random. 
Prove that the chance that volumes of the same works are all together 

8. The letters forming the word Clifton are placed at random 
in a row. What is the chance that the two vowels come together ? 

9. In a hand at whist what is the chance that the four kings are 
held by a specified player. 

10. There are 4 dollars and 3 half-dollars placed at random in a 
line. Show that the chance of the extreme coins being both half- 
dollars is \. 



1. Simplify 



MISCELLANEOUS EXAMPLES VI. 
6 — c , c — a . a — b 



+ 



a 2_( 6 _ c )2 &2_( c _ a) 2 tf-fr-fy* 

2. Extract the square root of 

(i.) 4x* + 6x* + ^a^ + i6ac + 25. 

(ii.) x*-^ + 2<xW + ^-2ax'< + a*. 
a 8 a 6 

3. A number of 3 digits exceeds 25 times the sum of the digits 
by 9 ; the middle digit increased by 3 is equal to the sum of the other 
digits, and the unit digit increased by 6 is equal to twice the sum of 
the other 2 digits : find the number. 

4. Find the value of 

5. Solve (i.) 2 = V?+Z±V2^s. 

V2 + x -V2-x 
(ii.) V3s-ll+V3o; = V12a;-23. 



MISCELLANEOUS EXAMPLES VI. 886 

«. Solve 2 (* + g > + *&±*1 = 6, 
x t + b x + a 

7. The sum of a certain number of terms of an A. P. is 45, and 
the first and last of these terms are 1 and 17 respectively. Find the 
number of terms and the common difference of the series. 

8. Solve (i.) 2x ~~ S -l + 2x - l = 0. 

K ' x-2 6 1-x 

(ii.) Vl2a^~6 + VSx-1 = V27a-2. 

9. Find the value of the seventh term in the expansion of 
(a + x) n when a = i, x = J, n = 9. 

10. A man starting from A at 11 o'clock passed the fourth mile- 
stone at 11.30 and met another man (who started from B at 12) at 
12.48; the second man passed the fourth milestone from A at 1.40: 
find the distance between A and B, and the second man's rate. 

11. Showthat x* + lSa 2 x>bax 2 + 9a 9 i if x>a. 

12. Extract the cube root of 

44 a 8 + 63 s 2 + afi + 27 + 6x* + 21 & + 54 se. 

13. Solve (L) x-y = S, (ii.) 2« 2 - 9xy + 9^ = 6, 

a 2 + xy 4-y 2 = 93. 4a 2 - lOxy + 11 y 2 = 86. 

14. Find a mean proportional between ~ — — — — > "*" * 

4 _-x/^~3 a ^ 

and the reciprocal of — • 

F 16 a 8 6 

15. Two vessels, one of which sails 2 miles an hour faster than the 
other, start together upon voyages of 1680 and 1152 miles respec- 
tively ; the slower vessel reaches its destination one day before the 
other : how many miles per hour did the faster vessel sail ? 

16. Solve (i.) x* = 8 + 7 a 8 . 

(ii.) x*» + b 2 =:c?-2bx*. 

17. Two numbers are in the ratio 2:7; the numbers obtained by 
adding 6 to each of the given numbers are in the duplicate ratio of 2 : 3. 
Find the numbers. 

18. Solve (i.) 2 bx 2 + 2 b = 4s + V*x. 

m\ s + 4 ,3«+10 23 + 3 
v J 2a + 3 2a: x-l 



(iii.) \/aT+3 + v^T8=V4« + 2L 
(iv.) x* + xy + y = 137 , 
y 2 + xy + x = 206. 



336 



ALGEBRA. 



19. Simplify 



[a^P-aFPT 



2-V 6 

20. Find the sides of a rectangle the area of which is unaltered if its 
length be increased by 2 feet while its breadth is diminished by 1 foot, 
and which loses $ of its area if its length be diminished by 2 feet and 
its breadth by 4 feet. 

21. The first term of a 6. F. exceeds the second term by 1, and the 
sum to infinity is 81 : find the series. 

22. Find the number of permutations which can be made from all 
the letters of the word Mississippi. 

23. Solve (i.) Vx + 2 + VTaT+I - V9x + 7 = 0. 

2z-3 



(ii.) 
(iii.) 



Vx - 2 + 1 
2 



= 2VaT-2-l. 
3 4 



+ 



a-6+Va; Vx-2 Vx + S 
(iv.) Vse — a — Vx — b — Vb — a. 

24. Find the condition that one root of ax 2 + bx + c = shall be n 
times the other. 

25. Find the value of x 8 — 3x 2 — 8x + 16 when x = 3 + i. 

26. Given log 648 = 2.81157, log 864 = 2.93661, find the logarithm 
of 3 and of 5. 

27. Two trains run, without stopping, over the same 36 miles of 
rail. One of them travels 15 miles an hour faster than the other and 
accomplishes the distance in 12 minutes less. Find the speeds of the 
two trains. 

28. Extract the square root of 

9x*- 2x s y + ^x 2 t^-~2xy i + 9^. 

29. Find, by logarithms, the value of 



fl5(.318)^\A 
I 16 f ' 



am n« l'i. 1 + ax -1 a -1 — x" 1 flwc* 1 
30. Simplify , , ■ x —, — - ■*- — — • 

a" 1 ^" 1 a~ l x — ax" 1 x — a 



31. The men in a regiment can be arranged in a column twice as 
deep as its breadth ; if the number be diminished by 206, the men can 
be arranged in a hollow square three deep having the same number of 



MISCELLANEOUS EXAMPLES VI. 837 

men in each outer side of the square as there were in the depth of the 
column ; how many men were there at first in the regiment ? 

32. Solve (i.) 2 x 2 + xy + y 2 = 37, 

$x 2 + ±xy + y 2 = 7S. 

(ii.) 27 X s + y 8 = 152, 

Zx 2 y + xy 2 = 40. 

33. Simplify 8$ + #(2 x 4-6) - 1/2 + 4"* - (32)"*. 

34. A man bought a field the length of which was to its breadth as 
8 to 5. The number of dollars that he paid for 1 acre was equal to the 
number of rods in the length of the field ; and 13 times the number of 
rods round the field equalled the number of dollars that it cost. Find 
the length and breadth of the field. 

35. Solve (i.) x 2 + xy + Sy 2 = 14 + 2 y/2, 

2x 2 + xy + by 2 - 24 + 2 y/2. 

(ii.) 2s + 3y = 10, 

6x 2 + a + y = 4£. 

36. Find two numbers whose sum added to their product is 34 t 
and the sum of whose squares diminished by their sum is 42. 

37. Find the sixth term in the expansion of each of the following 
expressions : 

1 9 

(i.) (.a + 9b-*)\ (ii.) (*«-£i) • <?H0 (f-^) '• 

38. A varies directly as B and inversely as C ; A = f when B = ^ 
and C = A : find B when A = V 48 and O = y/7&. 

39. Solve (i.) Vx + 12 + y/x + 12 = 6. 

(ii.) x 2 + yVxy= 9, 
tf> + x Vxy = 18. 

40. Form an equation whose roots shall be the arithmetic and 
harmonic means between the roots of x 2 — px + q = 0. 

z 



CHAPTER XXXVII. 

Binomial Thboebm. 

406. It may be shown by actual multiplication that 

(a + b)(a + c)(a + d)(a + e) 

= a 4 + (b + c + d + e)a s + (be + bd + be + cd + ce + de)a* 

+(bcd + bce + bde + cde)a + bcde (1) 

We may, however, write this result by inspection ; for the 
complete product consists of the sum of a number of par- 
tial products each of which is formed by multiplying 
together four letters, one being taken from each of the four 
factors. If we examine the way in which the various par- 
tial products are formed, we see that 

(1) The term a 4 is formed by taking the letter a out of 
each of the factors. 

(2) The terms involving a 8 are formed by taking the 
letter a out of any three factors, in every way possible, and 
one of the letters b, c, d, e, out of the remaining factor. 

(3) The terms involving a 2 are formed by taking the 
letter a out of any two factors, in every way possible, and 
two of the letters b, c, d, e, out of the remaining factors. 

(4) The terms involving a are formed by taking the letter 
a out of any one factor, and three of the letters b 9 c, d, e, 
out of the remaining factors. 

(5) The term independent of a is the product of all the 
letters b, c, d, e. 

Ex. Find the value of (a - 2) (a + 3) (a - 6)(a + 9). 

The product 

r= a 4 +(-2 + 3-6 + 9)a 8 + (- 6 + 10 - 18 - 15 + 27 - 45)a* 

+ (30 - 64 + 90 - 136)a +270 
= a* + 6 a 8 - 47 a 2 - 69 a + 270. 

338 



BINOMIAL THEOREM. 839 

407. If in equation (1) of the preceding article we sup- 
pose c = d = e = b, we obtain 

(a + b) 4 = a 4 +±a s b + 6a 2 b 2 + 4a& 8 + & 4 . 

We shall now employ the same method to prove a formula 
known as the Binomial Theorem, by which any binomial of 
the form a -f- b can be raised to any assigned positive inte- 
gral power. 

408. To find the expansion of (a + b) n when n is a positive 
integer. 

Consider the expression 

(a + b)(a + c)(a -f d) ••• (a + k) 9 

the number of factors being n. 

The expansion of this expression is the continued product 
of the n factors, a + b, a + c 9 a + d 9 ••• a + 7c, and every 
term in the expansion is of n dimensions, being a product 
formed by multiplying together n letters, one taken from 
each of these n factors. 

The highest power of a is a n , and is formed by taking the 
letter a from each of the n factors. 

The terms involving a"" 1 are formed by taking the letter 
a from any n — 1 of the factors, and one of the letters 
by c, d, ••• k from the remaining factor; thus the coefficient 
of a n_1 in the final product is the sum of the letters 
b, c, d y ••• k ; denote it by S^ 

The terms involving a n ~ 2 are formed by taking the letter 
a from any n — 2 of the factors, and fruo of the letters 
b, c, d, ••• k from the two remaining factors ; thus the coeffi- 
cient of a n ~ 2 in the final product is the sum of the products 
of the letters 6, c, d, ••• k taken two at a time ; denote it by #2. 

And, generally, the terms involving a n ~ r are formed by 
taking the letter a from any n — r of the factors, and r of 
the letters b, c, d, ••• A; from the r remaining factors; thus 
the coefficient of a n ~ r in the final product is the sum of the 
products of the letters b, c, d, ••• k taken r at a time ; denote 
it by flU 



Now suppose c, d, ■■• ft, each equal to 6; then Si becomes 
*Cfi; -S, becomes "Cj6 5 ; S 3 becomes "Ogft*; and so on; thus 

(a+6)"=a-+-(7 1 a- , 6+"C J a- ! ft ! +"O s a- 8 6 8 + ... +"C„6")j 
substituting for "G u "C? fc •■• we obtain 

(a + &)- = O" + na—'ft + \~ 2 a*-*b* 



s±^*>^, 



+ 7?.. •' ■rt»+-T» 



the series containing n + 1 terms. 

This is the Binomial Theorem, and the expression on the 
right side is said to be the expansion of (a + ft)". 

409. The coefficients in the expansion of (« + ft)" are very 
conveniently expressed by the symbols "Ci,"Cj,'"(7 3 ... "(7„. We 
shall, however, sometimes further abbreviate them by omit 
ting n, and writing O u O a C a ... (7„. With this notation we 
have 

(a + by = a" -\ Ctt-'ft + Caa""^' + Crf-'ft* H h C n b: 

If we write ~ b in the place of ft, we obtain 
(a - 6)" = a" + a 1 tr- 1 (-^)+ 0&-\- by 

+ C#-«(- ft) s + ... + C„<- &)" 

= tf> _ C^-'ft + C^"-% s - <V"W + — +(- 1)"C»6". 

Thus the terms in the expansion of (a + 6)" and (« — ft)" 

are numerically the same, but in (a — ft)" they are alternately 

positive and negative, and the last term is positive or nega. 

tive according as n is even or odd. 



BINOMIAL THEOREM. 841 

Ex. 1. Find the expansion of (a + y) 8 . 
By the formula, the expansion 

= cfi + 6 0ia 6 y + *C 2 aty 2 + «C 8 a*y* + *C±cJY + tCsay* f *C<f!p 

= a 6 + 6a 5 y + 15aty 2 + 20afy 8 + 15a 2 y^ + 6ay* + i/*, 

on calculating the values of 6 Ci, 6 C2, 6 C f 8 t •••• 

Ex. 2. Find the expansion of (a — 2 x) 7 . 

(a - 2 s) 7 = a 7 - 7 da 6 (2 &) + 7 C 2 a 5 (2 a) 2 - 7 C 8 a 4 (2 a;) 8 + ... to 8 terms. 

Now remembering that n C r = "(/„_, after calculating the coefficients 
up to 7 Cs, the rest may be written down at once; for 1 C^ = 1 Cs; 
7 Cs = 7 C2 ; and so on. Hence 

(a - 2 s) 7 = a 7 - 7 a 6 (2 a) + ^ a 5 (2 s) 2 - ^4 a*(2 a) 8 + - 
v J 1.2 1-2.3 

= a 7 - 7 a 6 (2 a) + 21 a\2 x) 2 - 35 a*(2 a:) 8 + 35 a 8 (2 a) 4 

- 21 a 2 (2 x) & + 7 a(2 a) 8 - (2 a;) 7 
= a 7 - 14 trig + 84a 6 a 2 - 280 a 4 x 8 + 560 a 8 ** 

- 672 «2 X 5 + 448 a#6 _ i 2 8 x 7 . 



410. The (r + i)th or General Term. In the expansion of 
(a + b) n , the coefficient of the second term is n C x ; of the third 
term is n C 2 ; of the fourth term is B <7 3 ; and so on ; the suffix 
in each term being one less than the number of the term to 
which it applies ; hence n C r is the coefficient of the (r -f l)th 
term. This is called the general term, because by giving to 
r different numerical values any of the coefficients may be 
found from n C r ; and by giving to a and b their appropriate 
indices any assigned term may be obtained. Thus the 
(r -f l)th term may be written 

"(7 r a-'&', or <»-l)(»-g)-(»-r + l) „-^. 

[r* 

In applying this formula to any particular case, it should 
be observed that the index of b is the same as the suffix of C, 
and that the sum of the indices of a and b is n, 

♦See Art. 392, Cor. 



342 



ALGEBRA. 



Ex. 1. Find the fifth term of (a + 2x*)W. 
Here (r + 1)= 6, therefore 
the required term = 17 C 4 a 18 (2 s*) 4 

1.2.3.4 

= 38080 a 18 * 12 . 
Ex. 2. Find the fourteenth term of (3 — a) 1 *. 
Here r + 1 = 14, therefore 
the required term = 15 Ci3(3) 2 (— a) M 

= "C 2 x(-9o u ) 
= - 945 a 18 . 



[Art. 895.] 



411. Simplest Form of the Binomial Theorem. The most 
convenient form of the binomial theorem is the expansion of 
(1 + x) n . This is obtained from the general formula of Art. 
408, by writing 1 in the place of a, and x in the place of 6. 
Thus 

(1 + x) n = 1 + n dx + n C#? + ... + "Off + ... + n C n x* 

the general term being — '— . ' "^ — '-&. 

412. The expansion of a binomial may always be made to 
depend upon the case in which the first term is unity ; thus 

(a + &)• = J afl + ?\ I "= a n (l + c) n , where c = -• 

* 

Ex. Find the coefficient of x 16 in the expansion of (x 2 — 2 a) 10 . 

We have (x 2 - 2 a;) 10 = x 20 ( 1 - - Y°; 

C9\io 
1 — ] , we 

have in this expansion to seek the coefficient of the term which con- 
tains — • 
a* 

Hence the required coefficient = 10 C4(— 2) 4 

_ lOjJM^ 16 _ 886a 
1*2.3.4 



BINOMIAL THEOREM. 848 



PROOF BY MATHEMATICAL INDUCTION. 

413. By actual multiplication we obtain the following 
identities : 

(a + by = a* + 2db + b*, 

(a + bf = a? + 3a 2 b + 3ab* + W, 

(a + by = a + 4a% + 6a 2 b 2 + 4a5 3 + b\ 

Selecting any one of these, and rewriting so as to exhibit 
the laws of formation of exponents and coefficients, we have 

1 • 1 • J 1 • A • 6 

+ rM^ a ^ Alt - 216 >- 

If these laws of formation hold for (a + b) n , n being any 
positive integer, then 

(a + b) n = a n + na n ' l b + "^"I^ a-V 

, m(n — 1) (n — 2) „ .,. ... 

+ 1.2.3 a b +- • • ' W- 

Multiplying each side of the assumed identity by (a + b) 
and combining terms, we obtain 

(a + b) n+1 = a n+1 + (n + 1) a*6 + *^? "t^ an ~ 1&2 

+ «(n+J^-l) an _ 268+> _ _ (2) 

It will be seen that n in (1) is, in every instance, replaced 
by (n + 1) in (2). Hence if the theorem be true for any 
value of n, it will be true for the next higher value. We 
have shown by multiplication that the theorem is true when 
n successively equals 2, 3, and 4; hence it is true when 
n = 5, and so on indefinitely. The theorem is therefore 
true for all positive integral values of n. 



844 



ALGEBRA. 



414. In the expansion 

(a + b) n = a" + na*- l b + ^ ~ 1 ^ a»-y+ — 

we observe that in any term 

(1) The exponent of b, the second term of the binomial, 
is one less than the number of the term from the first. 

(2) The sum of the exponents is n. 

(3) The last factor of the denominator of the coefficient 
is the same as the exponent of the second term of the 
binomial. 

(4) The last factor of the numerator of the coefficient 
is the exponent of the first term of the binomial increased 

byl. 
Hence the (r + i)th or general term of (a + b) n is 

n(n-l)».(n-r+l) an ^ r 

1 • L • O ••• T 

» 

Ex. Find the 6th term in the expansion of (2 a + 6) 10 . 
Here n = 10, and r + 1 = 6. 

We have 10 ' 9 '_ 8 ' 7 ' 6 (2 a) W = 3 ' 2 ' 7 ' 6 (2 a) W 



1-2.3.4.5 



= 252(2)6a 5 & 6 = 8064 a 6 6 6 . 



Note. The student should observe that the coefficient contains 
the same number of factors in both numerator and denominator. 



EXAMPLES XXXVIL a. 
Expand the following binomials ; 

1. (a + 2)*. 6 

2. . (* + 3) 6 . 

3. (a + xV. 

4. (a-x) 6 . 

5. (l-2y) 6 . 



1 (-!)'• 



T^rite in simplest form : * 

10. The 4th term of (1 + x) u . 

11. The 6th term of (2 - y) 8 . 

12. The 6th term of (a — 6 6) 7 . 

13. The 15tb term of (2 as - l) 17 . 



.. (** + f) 4 - 
1 («+?)•. 



14. The 7th term 

15. The 6th term 



- (• -IT 



9 



BINOMIAL THEOREM. 345 

16. Find the value of (x - V3) 4 + (x + V 3 ) 4 - 



17. Expand ( vT^x 2 + 1)* - ( VT=& - 1)*. 

18. Find the coefficient of x 12 in (a 2 + 2 &)*>. 

19. Find the coefficient of x in [a; 2 — — ) • 

20. FiDd the term independent of x in ( 2 a; 2 — - ) • 

21. Find the coefficient of a;" 20 in (— - — V 6 . 

415. Equal Coefficients. Jn the expansion of (1 + «) B ^e 
coefficients of terms equidistant from the beginning and end 
are equal. 

The coefficient of the (r + l)th term from the beginning 
is tt O- 

The (r -f l)th term from the end has n + 1 — (r + 1), or 
n — r terms before it ; therefore counting from the begin- 
ning it is the (n—r-\- l)th term, and its coefficient is n C n .. r , 
which has been shown to be equal to n C r [Art. 395]. Hence 
the proposition follows. 

416. Greatest Coefficient. To find the greatest coefficient in 
the expansion of (1 + x) n . 

The coefficient of the general term of (1 -f x) n is n C r ; and 
we have only to find for what value of r this is greatest. 
By Art. 401, when n is even, the greatest coefficient is 

"C n ; when n is odd, it is n O«_i, or n C n +i ; these coefficients 

2 "T" "T" 

being equal. 

417. Greatest Term. To find the greatest term in the expan- 
sion of (a -f- b) n . 

We have (a + b) n = a n fl + -Y; 

therefore, since a n multiplies every term in [ 1 -f - j , it will 
be sufficient to find the greatest term in this latter expansion. 



846 



ALGEBRA. 



Let the rth and (r + l)th be any two consecutive terms. 
The (r -fr- l)th term is obtained by multiplying the rth term 



by 



n-r-f 1 b 

1 . _ • 



£; that is, by f*±A - l) £. [Art. 410.] 



The factor 



n + 1 



1 decreases as r increases ; hence the 



(r -+- l)th term is not always greater than the rth term, but 
only until ( ^-± 1 j_ becomes equal to 1, or less than 1. 



Now 



that is, 



| r »±l_i > \6 > i |80 i on gag»±l_l>5. 
\ r Ja r b 

5 + l>5 + l,or^±^>r . . . (1). 
r b a + b 



If LILT — 2_ be an integer, denote it by p\ then if r =p 
a + b 

the multiplying factor becomes 1, and the (p -+- l)th term 
is equal to the pth. ; and these are greater than any other 
term. 

jf \ n -r ) be not an integer, denote its integral part by 
a + 6 
q\ then the greatest value of r consistent with (1) is q\ 
hence the (q + l)th term is the greatest. 

Since we are only concerned with the numerically greatest 
term, the investigation will be the same for (a — b) n ; there- 
fore in any numerical example it is unnecessary to consider 
the sign of the second term of the binomial. Also it will 
be found best to work each example independently of the 
general formula. 

m 

Ex. Find the greatest term in the expansion of (1 + 4 as) 8 , when x 
has the value %. 

Denote the rth and (r + l)th terms by T r and T r +i respectively ; then 



_ 8-r+l 

ir+l = 



• 4sx T r = ^lx^x T r ; 
r 3 



hence 
that is, 



r r+ i> T rt so long as ? — ? x |>1 ; 

r o 

86-4r>3r, pr 36>7r. 



BINOMIAL THEOREM. 347 

The greatest value of r consistent with this is 5 ; hence the greatest 
term is the sixth, and its value 

= *C B x (J)« = *C 8 x ($)* = Aft t±. 

418. Sum of the Coefficients. To find the sum of the coeffi- 
cients in the expansion o/(l -f x) n . 

In the identity 

(1 + x ) n = 1 + Cis + C#? + Ctf? + ... + Off; 
put a: = 1 ; thus 

= sum of the coefficients. 
Cob. Ci+ C 2 + C 3 + ..- + (7 n = 2"-l; 

that is, the total number of combinations of n things taking 
some or all of them at a time is 2 n — 1. [See Art. 400.] 

419. Sums of Coefficients equal. To prove that in the 
expansion of (1 + #) n > ^ e sum °f ^ e coefficients of the odd 
terms is equal to the sum of the coefficients of the even terms. 

In the identity 

(1 + x) n = 1 + O x x + C#? + C#* + ... + C n x«, 
put x = — 1 ; thus 

= 1-C 1 4-C 2 -C 8 +-C 4 -C' fi +...; 
.*. 1 + Ci + Gi + ••• = Ci + Ci + Cj 4- ■••• 

420. Expansion of Multinomials. * The Binomial Theorem 
may also be applied to expand expressions which contain 
more than two terms. 

Ex. Find the expansion of (5c 2 + 2x — l) 8 . 
Regarding 2 x — 1 as a single term, the expansion 

= (a; 2 ) 8 + 3(s 2 ) 2 (2s - 1)+ 3x 2 (2s - 1)« + (2x- 1)« 

= xP + 6« 6 + 9iB 4 --4x 8 — 9* 2 + Qx — 1, on reduction. 

421. Binomial Theorem for Negative or Fractional Index. 

For a full discussion of the Binomial Theorem when the 
index is not restricted to positive integral values the student 



348 



ALGEBRA. 



is referred to Chapter xlv. It is there shown that when x 
is less than unity, the formula 

,. N + . n(n — 1) 9 , n(n — V)(n — 2) , 

(l + x)» = l + nx+ 1>2 + 1-2-3 - rf +"" 

is true for any value of n. 

When n is negative or fractional the number of terms in 
the expansion is unlimited, but in any particular case we 
may write down as many terms as we please, or we may find 
the coefficient of any assigned term. 



Ex. 1. Expand (1 + x)~ 3 to four terms. 
a+aO- 8 =l + (-3)aH 



( _3)(-3-l)^ | (-3)(-3-lX-3-,2) xa 



-- l - S X + 6x 2 - 10z 3 + ... 



1*2.3 



Ex. 2. Expand (4 -f 3 x) * to four terms. 

(4 + 3*)* = 4!(l + 3 f) f =8(l + ^)* 

= 8 L 1+ 2-X + _ lT2-VTJ + TT2T3 \T) + "'J 

Q r, , 3 Sx , 3 9a? 1 27 x» , T 

= o 1H — • 1 — • • • - -+-••• 1 

L 2 4 8 16 16 64 J 

= 8 + 9 x + f i x* - AV * 8 + ' * - 



422. In finding the general term we must now use the 
formida 

n(n - l)( 7i-2)...(7i-r + l) ^ 

i XT 

[r 

written in full; for the symbol n C r cannot be employed when 
n is fractional or negative. 

Ex. 1. Find the general term in the expansion of (1 + xy, 

i(i-l)(i-2)...G-r+l) 



The (r + l)th term =. 



Lr 



-XT 



' _ l(--l)(-3) (-5) ...(-2r + 3) 

2-ll 



af. 



BINOMIAL THEOREM. 349 

The number of factors in the numerator is r, and r — 1 of these 
are negative ; therefore, by taking — 1 out of each of these negative 
factors, we may write the above expression 

Ex. 2. Find the general term in the expansion of (1 — g)~*. 
TheCr-f^thtermz^C-^C-^C-^-C-^^ + ^^x^ 

= (- 1)r 3.4-5...(r + 2) ( _ 1 y af 

, n8r 3.4.5-..(r + 2) 
v J l-2.3...r 
^ ( r + 1)^ + 2) ^ 
1-2 
by removing like factors from the numerator and denominator. 



I. The following expansions should be remembered : 

(l-a^-^l + a + a^ + ar'H \-af-\ . 

(l-x)- 2 = l+2x + 3x? + 4:a?-{ 1- (y -f- 3_) asr H 

1 • 2 

424. The following example illustrates a useful appli- 
cation of the Binomial Theorem. 

Ex. Find the cube root of 126 to 5 places of decimals. 
(126)* =(5« + l)* = 6(1+31) 

= 5(1-1 — • • H — • ••• 1 

\ 3 6» 9 5« 81 5» / 

_ fi , 1 1 11.1 1 

s o +- • — — — • — -+- — • — — ... 

3 52 9 5* 81 5* 

K , 1 22 1 2* 1 27 

3 10 3 9 10 5 81 107 

K , .04 .00032 , .0000128 
= o H ••• 

3 9 81 

= 5 + .013333 .000035 — + — 

= 5.01329, to five places of decimals. 



EXAMPLES XXX Vn. T>. 

In the following expansions find which is the greatest term : 

1. {x + gy when x = 4, y = 3. 4. (a-46) 1 * when a = 12, 6 = 2, 

2. (x _ j,)« when x = 8, y = i. ». (7i+2j/)» when *=8, y=14. 
8. (1 + x)* when * = }. 6. (2& + S)" when* = j, n = 15. 

7. In the expansion of (1 + »)* the coefficients of the (2 r + 1) Hi 
and (r + 6)tli terms are equal : find r. 

8. Find n when the coefficients of the 10th and 26th terms of 
(1 + x)" are equal. 

9. find the relation between r and n in order that the coefficients 
of (r + 3)th and (2 r - 3)th terms of (1 + a) 8 " may be equal. 



10. Find the coefficient of x" in the expansion of I 



1.1V 



11. Find the middle term of (1 + »)** in its simplest form. 
IS. Find the sum of the coefficients of (x + j/) lt . 
18. Find the sum of the coefficients of (3 x + c)*- 

14. Find the rth term from the beginning and the rtli term from 
the end of (a -t- 2i)". 

15. Expand {a* + 2 a + 1)' and (a£ - 4s + 2)». 

Expand to four terms the following expressions : 

16. (1 + x)*. IB. (l+3a:)-". 8S. (2 + x)~*. 

17. (1+*)*. 20. (l-x*)-: 23. (1+2*)"*. 
IS. (l+x)i 91. (1 + 8*)-*. «■ (a-2*)"*. 

Write in simplest form : 

2fi. The 5th term and the 10th term of (1 + a;)~*. 

26. The 3d term and the 11th term of (1 + 2x)^. 

27. The 4th term and the (r + l)th term of (1 + x)-*. 
38. The 7th term and the (r + l)th term of (1 - *)*. 
SB. The (r + l)th term of (a - 6x)"', and of (1 — nx)*. 
Find to four places of decimals the value of 

30. vT22. 81. \/620. SB. v'Sl. 83. 1 +VSfi 



BINOMIAL THEOBEM. 851 

Find the value of 
84. (x + V2) 4 + (* - V 2 ) 4 - 36. (v^ + l^-CV* 2 - 1 ) 6 - 

35. (Vx 2 -a 2 +s)G-(Vx^a*--x)<\ 37. (2-vT^x)«+(2+ VT^e) 6 . 

38. Find the middle term of [± + ^V°. 

\x a/ 

39. Find the middle term of (l - £\ M . 

40. Find the coefficient of x w in (x 2 + — V*. 

41. Find the coefficient of x 1 * in (ax* - 6x) 9 . 

42. Find the coefficients of x 82 and x" 17 in (x* --\^ 16 - 

43. Find the two middle terms of ^3a -^V. 

Write in simplest form 

44. The 8th term of (l+2&)~*. 49. The (r+l)th term of (1-x)-*. 

45. The 11th term of (l-2x 8 )^". 50. The (r+l)th term of (1+x)*. 

46. The 10th term of (l+3a 2 ) J A 51. The (r+l)thterinof (l+x) J s\ 

47. The 6th term of (3 a - 2 6)" 1 . 52. The 14th term of (2i°-2 7 x) ^ 

48. The (r+l)thtermof (1-x)" 2 . 53. The 7th term of (38+6*x)~*\ 



425. Definition. The logarithm of any number to a 
given base is the index of the power to which the base must 
be raised in order to equal the given number. Thus if 
a' = N, x is called the logarithm of Nto the base a. 

Examples. (1) Since 3< = 81, the logarithm of 81 to base 8 is 4. 
(2) Since 10' = 10, 10* = 100, 10" = 1000, — 
Hie natural numbers 1, 2, 3, •■■ are respectively the logarithms of 10, 
100, 1000,- to base 10. 

426. The logarithm of 2f to base a is usually written 
log.-??", so that the same meaning is expressed by the two 
equations 

a' = N; x = log. N. 
Ex. Find the logarithm of 32^/4 to base 2y2. 
Let x be the required logarithm ; then, by definition, 
(2^2)" = 32^/4; 
... (2. 2^)" = 2* -2*; 
... 2$" = 2***; 
hence, by equating the indices, \ x = ^ ; 

... a = ^ = 3.a. 

427. When it is understood that a particular system of 
logarithms is in use, the suffix denoting the base is omitted. 
Thus in arithmetical calculations in which 10 is the base, 
we usually write log 2, log 3, ■■• instead of logii>2, log w 3, -■-. 

Logarithms to the base 10 are known as Common Loga- 
rithms ; this system was first introduced in 161f> by Briggs, 
a contemporary of Napier the inventor of Logarithms. 



LOGARITHMS. 853 

PROPERTIES OF LOGARITHMS. 

428. Logarithm of Unity. The logarithm of lis 0. 

For a°=l for all values of a ; therefore log 1 = 0, what- 
ever the base may be. 

429. Logarithm of the Base. The logarithm of the base 
itself is 1. 

For a 1 = a; therefore log a a = l. 

430. Logarithm of Zero. The logarithm ofO, in any system 
whose base is greater than unity, is minus infinity. 

For a- 00 = — = 0. 

a 00 

Also, since a 4 * = oo, the logarithm of + oo is + oo. 

431. Logarithm of a Product. The logarithm of a product 
is the sum of the logarithms of its factors. 

Let MN be the product ; let a be the base of the system, 
and suppose 

x = log a M, y = log a N; 

so that a x = M, a 9 = 2V. 

Thus the product MN= a* x a* = a x+9 ; 

whence, by definition, log a M N = x + y = log a M+ log a N. 

Similarly, log tt MNP = log a M+ log a JV+ log a P; 

and so on for any number of factors. 

Ex. log42 = log (2 x 3 x 7)= log2 + log3 + log7. 

432. Logarithm of a Quotient. The logarithm of a quotient 
is the logarithm of the dividend minus the logarithm of the 
divisor. 

M 

Let — be the fraction, and suppose' 

N 

x=\og a M, y = log a N; 

so that a* = M, a' = N. 

2a 



854 ALGEBRA. 

Thus the fraction ^ = - = <*•-»; 

N a' ' 

M 

whence, by definition, log.-— = x — y = log. M — log. 1ST. 

Ex. log (2\ ) = log Jf = log 16 - log 7 

= log (3 x 6) - log 7 = log 8 + log 6 — log 7. 

433. Logarithm of a Power. The logarithm of a numbei 
raised to any power, integral or fractional, is the logarithm of 
the number multiplied by the index of the power. 

Let log. (M p ) be required, and suppose 

x = log. M, so that a* = M\ 
then M p = (a') p = oJ"; 

whence, by definition, log a (M p ) =px; 

that is, log a (M p ) = p log. Ml 

m 

Similarly, log. (M~ r ) = i log° M. 

r 

Ex. Express the logarithm of -*^- in terms of log a, log 6, and log a 

(fib 2 

= § log a — (log c 6 + log b 2 ) = } log a — 6 log c — 2 log 6. 

434. From the equation 10* = -#", it is evident that 
common logarithms will not in general be integral, and that 
they will not always be positive. 

For instance, 3154 > 10 3 and < 10 4 ; 

.•. log 3154 = 3 -f- a fraction. 
Again, .06 > 10~ 2 and < 10" 1 ; 

.\ log .06 = — 2 -f- a fraction. 
Negative numbers have no common logarithms. 

435. Definition. The integral part of a logarithm is 
called the characteristic, and the decimal part, when it is so 
written that it is positive, is called the mantissa. 

The characteristic of the logarithm of any number to the 
base io can be written by inspection, as we shall now show. 



LOGARITHMS. 855 

436. The Characteristic of the Logarithm of Any Number 
Greater than Unity. It is clear that a number with two 
digits in its integral part lies between 10 1 and 10 2 ; a num- 
ber with three digits in its integral part lies between 10* 
and 10 3 ; and so on. Hence a number with n digits in its 
integral part lies between 10"" 1 and 10 n . 

Let N be a number whose integral part contains n digits ; 
then 

Jjf = lQ(n-l)+ a fraction . 

.\ log JV= (» — 1) -f* a fraction. 

Hence the characteristic is n — 1 ; that is, the characteris- 
tic of the logarithm of a number greater than unity is less by 
one than the number of digits in its integral part, and is 
positive. 

437. The Characteristic of the Logarithm of a Decimal Frac- 
tion. A decimal with one cipher immediately after the 
decimal point, such as .0324, being greater than .01 and less 
than .1, lies between 10~ 2 and 10~ ! ; a number with two 
ciphers after the decimal point lies between 10~ 8 and 10~ 8 ; 
and so on. Hence a decimal fraction with n ciphers immedi- 
ately after the decimal point lies between 10~ (n+1) and 10"*. 

Let D be a decimal beginning with n ciphers ; then 

J) = 10-<n+l)+a fraction . 

.\ log D = — (» + 1) 4- a fraction. 

Hence the characteristic is — (n 4- 1) ; that is, the charac- 
teristic of the logarithm of a decimal fraction is greater by 
unity than the number of ciphers immediately after the deci- 
mal point and is negative. 

438. Advantages of Common Logarithms. Common loga- 
rithms, because of the two great advantages of the base 10, 
are in common use. These two advantages are as follows : 

(1) From the results already proved it is evident that the 
characteristics can be written by inspection, so that only the 
mantissse have to be registered in the Tables. 



356 ALGEBRA. 

(2) The mantissae are the same for the logarithms of alV 
numbers which have the same significant digits; so that it is 
sufficient to tabulate the mantissae of the logarithms of 
integers. 

This proposition we proceed to prove. 

439. Let N be any number, then since multiplying or 
dividing by a power of 10 merely alters the position of the 
decimal point without changing the sequence of figures, it 
follows that Nx 10 p , and JV-*-10*, where p and q are any 
integers, are numbers whose significant digits are the same 
as those of JV. 

Now log(JY"xlO")=logiV r -|-i>loglO = logJY"-hi> . (1). 
Again, log(iV~-*- 10*) = log N-q log 10 = logN-q . (2). 

In (1) an integer is added to log N, and in (2) an integer 
is subtracted from log N\ that is, the mantissa or decimal 
portion of the logarithm remains unaltered. 

In this and the three preceding articles the mantissae 
have been supposed positive. In order to secure the advan- 
tages of Briggs' system, we arrange our work so as always to 
keep the mantissa positive, so that when the mantissa of any 
logarithm has been taken from the Tables the characteristic 
is prefixed with its appropriate sign, according to the rules 
already given. 

440. In the case of a negative logarithm the minus sign 
is written over the characteristic, and not before it, to indi- 
cate that the characteristic alone is negative, and not the 
whole expression. Thus 4.30103, the logarithm of .0002, is 
equivalent to — 4 -f .30103, and must be distinguished from 
— 4.30103, an expression in which both the integer and the 
decimal are negative. In working with negative logarithms 
an arithmetical artifice will sometimes be necessary in order 
to make the mantissa positive. For instance, a result such 
as —3.69897, in which the whole expression is negative, 
may be transformed by subtracting 1 from the character- 
istic and adding 1 to the mantissa. Thus, 

- 3.69897 = - 4 + (1 - .69897) = 4.30103. 



LOGARITHMS. 357 

Ex. 1. Required the logarithm of .0002432. 

Ill Seven-Place Tables we find that 3859636 is the mantissa of log2432 
(the decimal point as well as the characteristic being omitted); and, 
by Art. 437, the characteristic of the logarithm of the given number is 

.-. log.0002432 = 4.3869636. 
This may be written 6.3859636 - 10. 



Ex. 2. Find the value ofV.00000165, given 

log 165 = 2.2174839, log 697424 = 5.8434968. 
Let x denote the value required ; then 

log z = log (.00000165)* = i log (.00000165)= £(6.2174839); 

the mantissa of log. 00000165 being the same as that of log 166, and 
the characteristic being prefixed by the rule. 

Now K^2174839)=£(l(j + 4.2174839) = 2.8434968 

and .8434968 is the mantissa of log 697424 ; hence x is a number con- 
sisting of these same digits, but with one cipher after the decimal 
point. [Art. 437.] 

Thus x = .0697424. 

441. Logarithms transformed from Base a to Base 6. Sup- 
pose that the logarithms of all numbers to base a are known 
and tabulated. 

Let iV be any number whose logarithm to base b is 
required. 

Let y = \og b N f so that b> = JV; 

.-. log a (^)=log tt JV; 
that is, y log tt b = log a JV; 

•^ = l^ Xl0gaJV; 
or v log h N=- -xlog a JV" (1) 

Now since N and b are given, log N and log a b are known 
from the Tables, and thus logj^may be found. 

Hence to transform logarithms from base a to base b we 

multiply them all by ; this is a constant quantity, and 

is given by the Tables j it is known as the modulus. 



858 ALGEBRA. 

Cob. If in equation (1) we put a for JV, we obtain 
log»a=- rXlog a a = 



log. b - log. b' 

.'. log b a x log. 6 = 1. 

442. Logarithms in Arithmetical Calculation. The fol- 
lowing examples illustrate the utility of logarithms in 
facilitating arithmetical calculation. 

Ex. 1. Given log 3 = .4771213, find log{(2.7)« x (.81)* + (90)*}. 
The required value = 3 logf J -f f log^V - i h>g90 

= 3(log3» - 1) + | (log 3* - 2)- f (l°g 3* + 1) 

= (» + ¥ - f) log 3 -(3 + | + J) 

=f J log3 - m = 4.6280766 - 5.85 = 2.7780766. 

The student should notice that the logarithm of 5 and its powers 
can always be obtained from log 2 ; thus 

log 5 = log J 2 ° = log 10 - log2 = 1 - log2. 

Ex. 2. Find the number of digits in 875 16 , given 

log 2 = .3010300, log 7 = .8450980. 
log (875 w) = 16 log (7 x 125) = 16 (log 7 + 3 log 5) 
= 16(log7+3-31og2) 
= 16 x 2.9420080 = 47.072128 ; 
hence the number of digits is 48. [Art. 436.] 

EXAMPLES XXXVIII. a. 

1. Find the logarithms of ^/32 and .03126 to base -f/2, and 100 
and .00001 to base .01. 

2. Find the value of log 4 512, logs .0016, loggi^ log 49 343. 

3. Write the numbers whose logarithms to bases 25, 3, .02, 1, — 4, 
1.7, 1000, are J, - 2, - 3, 5, -1,2, - f respectively. 

Simplify the expressions • 

4 . i^igggat 5. io g { («-^(^\n 

6. Find by inspection the characteristics of the logarithms of 3174, 
625.7, 8.502, .4, .374, .000135, 23.22065. 



LOGARITHMS. 859 

7. The mantissa of log 37203 is .5705780: write the logarithms of 
37.203, .000037203, 372030000. 

8. The logarithm of 7623 is 3.8821259: write the numbers whose 
logarithms are .8821259, £8821269, 7.8821269. 

Given log 2 = .3010300, log 3 = .4771213, log 7 = .8450980, find the 
value of 

9. log 729. 10. log 8400. 11. log. 256. 
12. log 5. 832. 13. log #392. 14. log. 3046. 

15. Show that log {$ -f log f {$ - 2 log J = log 2. 

16. Find to six decimal places the value of 

logtt}-21ogTtt + logW> 

17. Simplify log {(10.8)* x (.24)* -«- (90)-*}, and find its numerical 
value. 

18. Find the value of 

log ( Vl26 . Vl08 -*- vT008 • #162). 



19. Find the value of log i / 588 * 768 t 

^686 x 972 

20. Find the number of digits in 42**. 

(Q1 \ 1QQ0 
— J is greater than 100000. 

22. How many ciphers are there between the decimal point and 

(9M000 
-) ? 

23. Find the value of #.01008, having given 

log 398742 = 5.6006921. 

24. Find the seventh root of .00792, having given 

log 11 = 1.0413927 and log 600.977 = 2.6998179. 

25. Find the value of 2 log J J -f log^- 3 log J{. 



860 



ALOEBBA. 



N 





1 


2 


3 


4 


5 


6 


7 


8 


9 


10 


0000 


0043 


0086 


0128 


0170 


0212 


0253 


0294 


0334 


0374 


11 
12 
13 
14 


0414 
0792 
1139 
1461 


0463 
0828 
1173 
1492 


0492 
0864 
1206 
1523 


0531 
0899 
1239 
1553 


0569 
0934 
1271 
1684 


0607 
0969 
1303 
1614 


0645 
1004 
1335 
1644 


0682 
1038 
1367 
1673 


0719 
1072 
1399 
1703 


0765 
1106 
1430 
1732 


15 


1761 


1790 


1818 


1847 


1875 


1903 


1931 


1959 


1987 


2014 


16 
17 
18 
19 


2041 
2304 
2553 

2788 


2068 
2330 
2577 
2810 


2095 
2365 
2601 
2833 


2122 
2380 
2625 
2856 


2148 
2405 
2648 

2878 


2176 
2430 
2672 
2900 


2201 
2455 
2695 
2923 


2227 
2480 
2718 
2945 


2253 
2504 
2742 
2967 


2279 
2529 
2765 
2989 


20 


3010 


3032 


3064 


3076 


3096 


3118 


3139 


3160 


3181 


3201 


21 
22 
23 
24 


3222 
3424 
3617 
3802 


3243 
3444 
3636 
3820 


3263 
3464 
3655 
3838 


3284 
3483 
3674 
3856 


3304 
3502 
3692 
3874 


3324 
3622 
3711 
3892 


3345 
3541 
3729 
3909 


3365 
3560 
3747 
3927 


3386 
3579 
3766 
3945 


3404 
3598 
3784 
3962 


25 


3979 


3997 


4014 


4031 


4048 


4065 


4082 


4099 


4116 


4133 


26 
27 
28 
29 


4150 
4314 
4472 
4624 


4166 
4330 

4487 
4639 


4183 
4346 
4502 
4654 


4200 
4362 
4518 
4669 


4216 

4378 
4533 
4683 


4232 
4393 
4548 
4698 


4249 
4409 
4564 
4713 


4265 
4425 
4579 

4728 


4281 
4440 
4594 
4742 


4298 
4456 
4609 

4757 


30 


4771 


4786 


4800 


4814 


4829 


4843 


4857 


4871 


4886 


4900 


31 
32 
33 
34 


4914 
6051 
6185 
5315 


4928 
6065 
5198 
5328 


4942 
6079 
6211 
5340 


4955 
5092 
5224 
5353 


4969 
5105 
5237 
5366 


4983 
6119 
5250 

5378 


4997 
5132 
5263 
5391 


5011 
6145 
5276 
5403 


6024 
5159 
5289 
6416 


5038 
5172 
5302 
5428 


35 


5441 


5453 


6465 


5478 


5490 


5502 


5514 


5527 


5539 


5551 


36 
37 

38 
39 


5563 
5682 
5798 
6911 


5575 
6694 
5809 
5922 


5587 
6705 
6821 
5933 


5599 
5717 
5832 
5944 


6611 
5729 
5843 
5955 


5623 
5740 
6865 
5966 


5635 
5752 
5866 
5977 


6647 
5763 

5877 
5988 


5658 
6775 
5888 
5999 


5670 
5786 
5899 
6010 


40 


6021 


6031 


6042 


6053 


6064 


6076 


6085 


6096 


6107 


6117 


41 
42 
43 
44 


6128 
6232 
6335 
6435 


6138 
6243 
6345 
6444 


614l> 
6253 
6355 
6454 


6160 
6263 
6365 
6464 


6170 
6274 
6375 
6474 


6180 
6284 
6385 
6484 


6191 
6294 
6396 
6493 


6201 
6304 
6406 
6503 


6212 
6314 
6415 
6513 


6222 
6325 
6425 
6522 


45 


6532 


6542 


6551 


6661 


6571 


6580 


6590 


6599 


6609 


6618 


46 
47 
48 
49 


6628 
6721 
6812 
6902 


6637 
6730 
6821 
6911 


6646 
6739 
6830 
6920 


6656 
6749 
6839 
6928 


6665 
6758 
6848 
6937 


6675 
6767 
6857 
6946 


6684 
6776 
6866 
6955 


6693 
6785 
6876 
6964 


6702 
6794 

6884 
6972 


6712 
6803 
6893 
6981 


50 


6990 


6998 


7007 


7016 


7024 


7033 


7042 


7050 


7059 


7067 


51 
52 
53 
54 


7076 
7160 
7243 
7324 


7084 

.7168 

7251 

7332 


7093 
7177 
7269 
7340 


7101 
7185 
7267 
7348 


7110 
7193 

7275 
7356 


7118 
7202 
7284 
7364 


7126 
7210 
7292 
7372 


7135 

7218 
7300 
7380 


7143 
7226 
7308 

7388 


7152 
7235 
7316 
7396 



LOGARITHMS. 



361 



N 





1 


2 


3 


4 


5 


6 


7 


8 


9 


55 


7404 


7412 


7419 


7427 


7435 


7443 


7461 


7469 


7466 


7474 


56 
57 
58 
59 


7482 
7559 
7634 
7709 


7490 
7566 
7642 
7716 


7497 
7574 
7649 
7723 


7605 

7582 
7657 
7731 


7513 
7689 
7664 
7738 


7520 
7597 
7672 
7745 


7628 
7604 
7679 
7762 


7636 
7612 
7686 
7760 


7543 

7619 
7694 
7767 


^551 
7627 
7701 

7774 


60 


7782 


7789 


7796 


7803 


7810 


7818 


7825 


7832 


7839 


7846 


61 
62 
63 
64 


7863 
7924 
7993 
8062 


7860 
7931 
8000 
8069 


7868 
7938 
8007 
8075 


7875 
7945 
8014 

8082 


7882 
7952 
8021 
8089 


7889 
7969 
8028 
8096 


7896 
7966 
8035 
8102 


7903 
7973 
8041 
8109 


7910 
7980 
8048 
8116 


7917 

7987 
8055 
8122 


65 


8129 


8136 


8142 


8149 


8166 


8162 


8169 


8176 


8182 


8189 


66 
67 
68 
69 


8195 
8261 
8326 
8388 


8202 
8267 
8331 
8395 


8209 
8274 
8338 
8401 


8216 
8280 
8344 
8407 


8222 
8287 
8351 
8414 


8228 
8293 
8367 
8420 


8235 
8299 
8363 
8426 


8241 
8306 
8370 
8432 


8248 
8312 
8376 
8439 


8254 
8319 
8382 
8445 


70 


8451 


8457 


8463 


8470 


8476 


8482 


8488 


8494 


8500 


8506 


71 
72 
73 

74 


8513 
8673 
8633 
8692 


8519 
8579 
8639 
8698 


8525 
8585 
8645 
8704 


8531 
8591 
8651 
8710 


8537 
8597 
8657 
8716 


8643 
8603 

8663 
8722 


8549 
8609 
8669 
8727 


8555 
8615 
8676 
8733 


8561 
8621 
8681 
8739 


8567 
8627 
8686 
8745 


75 


8751 


8766 


8762 


8768 


8774 


8779 


8785 


8791 


8797 


8802 


76 

77 
78 
79 


8808 
8865 
8921 
8976 


8814 
8871 
8927 
8982 


8820 
8876 
8932 
8987 


8825 
8882 
8938 
8993 


8831 
8887 
8943 
8998 


8837 
8893 
8949 
9004 


8842 
8899 
8954 
9009 


8848 
8904 
8960 
9015 


8854 
8910 
8965 
9020 


8859 
8915 
8971 
9025 


80 


9031 


9036 


9042 


9047 


9053 


9058 


9063 


9069 


9074 


9079 


81 
82 
83 
84 


9085 
9138 
9191 
9243 


9090 
9143 
9196 
9248 


9096 
9149 
9201 
9253 


9101 
9154 
9206 
9258 


9106 
9159 
9212 
9263 


9112 
9165 
9217 
9269 


9117 
9170 
9222 
9274 


9122 
9175 
9227 
9279 


9128 
9180 
9232 
9284 


9133 
9186 
9238 
9289 


85 


9294 


9299 


9304 


9309 


9315 


9320 


9325 


9330 


9335 


9340 


86 
87 
88 
89 


9345 
9395 
9445 
9494 


9350 
9400 
9450 
9499 


9355 
9405 
9455 
9604 


9360 
9410 
9460 
9509 


9365 
9415 
9465 
9513 


9370 
9420 
9469 
9518 


9375 
9425 
9474 
9523 


9380 
9430 
9479 
9528 


9385 
9435 
9484 
9533 


9390 
9440 
9489 
9538 


90 


9542 


9547 


9552 


9557 


9562 


9566 


9571 


9676 


9681 


9686 


91 
92 
93 
94 


9590 
9638 
9685 
9731 


9595 
9643 
9689 
9736 


9600 
9647 
9694 
9741 


9606 
9652 
9699 
9745 


9609 
9657 
9703 
9750 


9614 
9661 
9708 
9764 


9619 
9660 
9713 
9759 


9624 
9671 
9717 
9763 


9628 
9676 
9722 

9768 


9633 
9680 
9727 
9773 


95 


9777 


9782 


9786 


9791 


9795 


9800 


9805 


9809 


9814 


9818 


96 
97 
98 
99 


9823 
9868 
9912 
9956 


9827 
9872 
9917 
9961 


9832 
9877 
9921 
9965 


9836 
9881 
9926 
9969 


9841 
9886 
9930 
9974 


9845 

9890 
9934 
9978 


9850 
9894 
9939 
9983 


9854 
9899 
9943 
9987 


9859 
9903 
9948 
9991 


9863 
9908 
9952 
9996 



USB OF THE TABLE. 

_«■« 360-361 we S>* ve a four-pla 06 table con 
,>n "^^(jggjg of the common logarithms of all 
*„" 00 to 1000. 

«nd the logarithm of a number. 

the number consists of three figures, as 56.7. 

ltuiuo headed N find the first two significant 

i -i line with these and in the column baring at 

third figure will be found the mantissa. Thus 
ith r>G and in the column headed 7 we find 7536. 
j,..]) is the decimal part of the logarithm, prefix 
'ristic [Art. 436], and we have 
log 56.7 = 1.7536. 

in common logarithms the mantissa remains 
when the number is multiplied by an integral 
0, we change one or two-figure numbers into 

numbers by addition of ciphers before looking 
itissEe. The mantissa of log 56 will be that of 
y change in the logarithm being in the charac- 

log 560 = 2.7482, 
log 56 = 1.7482. 
le manner log 7 has for mantissa that of log 700. 
log 700 = 2.8451, 
log 7=0.8451. 
se the logarithm of a number of more than 
i, as 62543, is required. Since the number lies 
■00 and 62600, its logarithm lies between their 
In the column headed 2f we find the first two 
on a line with these and in the columns headed 
find the mantisste .7959 and .7966. Prefixing 
ristic [Art. 436], we have 
log 62600 = 4.7966, 
log 62500 = 4.7959. 



LOGARITHMS. 868 

Therefore while the number increases from 62500 to 
62600, the logarithm increases .0007. Now our number is 
ffo of the way from 62500 to 62600 ; hence if to the log- 
arithm of 62500 we add -^ of .0007, a nearly correct 
logarithm of 62543 is obtained. 

Thus log 62543 = 4.7959 

.0003 correction 
= 4.7962 

(d) Suppose the logarithm of a decimal, as .0005243, is 
required. The number lies between .0005240 and .0005250. 
In the column headed N we find the first two significant 
figures, 52 ; on a line with these and in the columns headed 
4, and 5, we find the mantissse .7193 and .7202. Prefixing 
the characteristic [Art. 437], we have 

log .0005250 = 4.7202 

log .0005240 = 4.7193 

differences .0000010 .0009 

Now .0005243 is .0000003 greater than .0005240 ; hence 

log .0005243 equals log .0005240 plus *^?^ or ^- of .0009 

(the difference of logarithms) ; 

that is, log .0005243 = 4.7193 

.0003 (nearly) 

= 4.7196 

In practice negative characteristics are usually avoided by 
adding them to 10 and writing —10 after the logarithm. 
Thus in the above example 4.7196 = 6.7196 - 10. 

445. The increase in the logarithms on the same line, as 
we pass from column to column, is called the tabular differ- 
ence. In finding the logarithm of 62543, we assumed that 
the differences of logarithms are proportional to the differ- 
ences of their corresponding numbers, which gives us results 
that are approximately correct. For greater accuracy we 
must use tables of more places. 



364 ALGEBRA. 

446. To find the number corresponding to a logarithm. 

(a) Suppose a logarithm, as 1.7466, is given to find the 
corresponding number. 

Look in the table for the mantissa .7466. It is found in 
the column headed 8 and on the line with 55 in the column 
headed JV. Therefore we take the figures 558, and; as the 
characteristic is 1, point off two places, obtaining the num- 
ber 55.8. 

(b) Suppose a logarithm, as 3.7531, is given to find the 
corresponding number. 

The exact mantissa, .7531, is not found in the table, there- 
fore take out the next larger, .7536, and the next smaller, 
.7528, and retain the characteristic in arranging the work. 

Thus, the number corresponding to 3.7536 is 5670 
and the number corresponding to 3.7528 is 5660 

differences .0008 10 

Now the logarithm 3.7531 is .0003 greater than the loga- 
rithm 3.7528, and a difference in logarithms of .0008 corre- 
sponds to a difference in numbers of 10 ; therefore we should 
increase the number corresponding to the logarithm 3.7528 by 

.0003 3 « 1A 
— — — or - of 10. 

.0008 8 

Thus the number corresponding to the logarithm 

3.7531 = 5660 

3.7 correction 
= 5663.7 

(c) Suppose a logarithm, as 8.8225 — 10 or 2.8225, is given 
to find the corresponding number. 

Take out the mantissse as in the previous example. 

The number corresponding to 2.8228 is .0665 [Art. 437.] 
The number corresponding to 2.8222 is . 0664 

differences .0006 .0001 

Now the logarithm 2.8225 is .0003 greater than the loga- 
rithm 2.8222, and a difference in logarithms of .0006 corre- 
sponds to a difference in numbers of .0001; therefore we 



LOGARITHMS. 365 

should increase the number corresponding to the logarithm 

2.8222 by ^f or ? of .0001. 
J .0006 6 

Thus 

the number corresponding to the logarithm 2.8222 = .0664 
the number corresponding to the logarithm 2.8225 = .0664 

Correction, .00005 
= .06645 

EXAMPLES XXXVIII. b. 

Find the common logarithms of the following : 

1. 50. 4. .341. 7. 12345. 

2. 203. 5. 0.045. 8. 0.010203. 
8. 6.73. 6. 5206. 9. 354.076. 

Find the numbers corresponding to the following common loga- 
rithms: 

10. 1.8156. 12. 4.0022. 14. 3.8441. 

11. 2.1439. 18. T.9131. 15. 7.4879-10. 



447. Cologarithms. The logarithm of the reciprocal of a 
number is called the cologarithm of that number. 

Thus colog 210 = logyfr = logl - log 210. 

Since log 1 = 0, we write it in the form 10 — 10 and then 
subtract log 210, which gives 

colog 210 = (10 - 2.3222) - 10 = 7.6778 - 10. 

Hence 

Rule. To find the cologarithm of a number, subtract the 
logarithm of the number from 10 and write — 10 after the 
result. 

448. The advantage gained by the use of cologarithms is 
the substitution of addition for subtraction 



366 



ALGEBRA. 



Ex. Find by use of logarithms the value of 



4.20 



7.42 x .068 



log; 



4.26 



= log 4.26 + log — -f log 1 



7.42 x .068 " ' "7.42 " ".068 

= log 4.26 + colog 7.42 + colog .068 
= .6204 +(9.1296 - 10)+ 1.2366 
= 10.9956 - 10. 

The number corresponding to this logarithm Is 9.9. 
In finding colog .068 we proceed as follows : 

colog .058 = log -L- = 10 - [log .068] - 10 (Art. 447), 

.058 

= 10 - [8.7634 - 10] - 10 (Art. 437), 

= 10 - 8.7634 + 10 - 10 = 1.2366. 

449. Exponential Equations. Equations in which the un- 
known quantity occurs as an exponent are called exponential 
equations, and are readily solved by the aid of logarithms. 

Ex. Find the value of x in 15* = 28. 

Taking the logarithms of both sides of the equation, we have 

log 15* = log 28; 
/. x log 15 = log 28. 
jB = j2£|| = 1^72 =1>2806 
log 15 1.1761 

examples xxxvm. a 

Find by use of logarithms : 
. 24.051 X .02456 

' .006705 x .0203* 
-« 145.206 x (- 7.664) 

' 448.1 x(-.2406)(- 47.85)' 

3. (742. 8024) ^ 

4. (-.0012045)* 

5. y/TS x 1/M2 x v^a^O. 
(18)2 x .78 x (3.4562)* 

a V£8149 x 80.80008 

#8283 x (.0006412)*' 
7. 845692.1 x .845856. 



8. .00010101 x (7117.1)«. 

9. (285.42)M x (5.672)» 
V20 x VM x y/- 124.89* 

10. «/ l2.876x VMSx (.005157y 
\ 29.029 x (52.81)* X (.4)» 

11. 3*+* = 405. 

12. 10 6 " 8 * = 2*-a». 

13. 128*-* x 187 a. _ 1458. 

14. 2* x 6*- 2 = S 2 * x 7 1 -*. 

15. 2*+y = 6v, 3* = 3 x 2»+h 

16. &-*-' = 4-*, 22*-i = 8*-*. 

* Treat negative quantities occurring in logarithmic work as posi- 
tive. When the numerical result is obtained, determine its sign by the 
ordinary rules of multiplication and division. 



CHAPTER XXXIX. 

Interest and Annuities.. 

450. Questions involving Simple Interest are easily solved 
by the rules of Arithmetic ; but in Compound Interest the 
calculations are often very laborious. We shall now show 
how these arithmetical calculations may be simplified by 
the aid of logarithms. Instead of taking as the rate of 
interest the interest on $ 100 for one year, it will be found 
more convenient to take the interest on $ 1 for one year. 
If this be denoted by $ r, and the amount of $ 1 for 1 year 
by $ R, we have R = 1 + r. 

451. * To find the interest and amount of a given sum in a 
given time at compound interest. 

Let P denote the principal, R the amount of $ 1 in one 
year, n the number of years, I the interest, and M the 
amount. 

The amount of P at the end of the first year is PR ; and, 
since this is the principal for the second year, the amount 
at the end of the second year is PR x R or PR?. Similarly 
the amount at the end of the third year is PR?, and so on ; 
hence the amount in n years is PR"; that is, 

M=PR*; 
and therefore I = P(R" - 1). 

Ex. Find the amount of $ 100 in a hundred years, allowing com- 
\ pound interest at the rate of 6 per cent, payable quarterly ; having 
given 

log 2 = .3010300, log 3 = .4771213, log 14.3906 = 1.15808. 

The amount of $ 1 in a quarter of a year is $ (1 + \ • yfo) or $ fj. 

867 



368 ALGEBRA. 

The number of payments is 400. If M be the amount, we have 

Jf=100(tt)«oo; 
.-. log M = log 100 + 400 (log 81 - log 80) 
= 2 + 400(41og3- 1 -31og2) 
= 2 + 400 (.0063962) = 4.16808 ; 
whence M= 14390.6. 

Thus the amount is $ 14390.60. 
Note. At simple interest the amount is $ 600. * 

452. To find the present value and discount of a given sum 
due in a given time, allowing compound interest. 

Let P be the given sum, V the present value, D the dis- 
count, R the amount of $ 1 for one year, n the number of 
years. 

Since Fis the sum which, put out to interest at the present 
time, will in n years amount to P, we have 

P = VR*\ 
... V=PM H , 
and D = P-V=P(1- 22-). 

ANNUITIES. 

463. An annuity is a fixed sum paid periodically under 
certain stated conditions ; the payment may be made either 
once a year or at more frequent intervals. Unless it is 
otherwise stated, we shall suppose the payments annual. 

454. To find the amount of an annuity left unpaid for a 
given number of years, allowing compound interest. 

Let A be the annuity, R the amount of $ 1 for one year, 
n the number of years, M the amount. 

At the end of the first year A is due, and the amount of 
this sum in the remaining n — 1 years is AR n ~ lm , at the 
end of the second year another A is due, and the amount of 
this sum in the remaining n — 2 years is AR*~* ; and so on. 

.-. M = AR n - 1 + AR n ~ 2 + ... +AR 2 + AR + A 

■s A(l + R + R 2 + — to n terms) = A^- — ^ 

R — 1 



INTEREST AND ANNUITIES. 369 

455. To find the present value of an annuity to continue foe 
a given number of years, allowing compound interest. 

Let A be the annuity, R the amount of $ 1 in one year, 

n the number of years, V the required present value. 

The present value of A due in 1 year is AR' 1, 
the present value of A due in 2 years is AR' 2 
the present value of A due in 3 years is AR~*; and so on. 

[Art. 452.] 

Now V is the sum of the present values of the different 
payments ; 

.-. F= AR' 1 + AR' 2 + AR" Z + ... to n terms 

^AR'^-Z-^A 1 - 1 *-". 

Note. This result may also be obtained by dividing the value of 
M, given in Art. 464, by R n . [Art. 451.] 

Cor. If we make n infinite we obtain for the present 
value a perpetual annuity 

A A 



V= 



R-l r 



EXAMPLES XXXIX. 

1. If in the year 1600 a sum of $ 1000 had been left to accumulate 
for 300 years, find its amount in the year 1900, reckoning compound 
interest at 4 per cent per annum. Given 

log 104 = 2.0170333 and log 12885.5 = 4.10099. 

2. Find in how many years a sum of money will amount to one 
hundred times its value at 5} per cent per annum compound interest. 
Given log 1065 = 3.023. 

o. Find the present value of $ 6000 due in 20 years, allowing com- 
pound interest at 8 per cent per annum. Given 

log 2 = .30103, log 3 = .47712, and log 12875 = 4.10976. 

4. Find the amount of an annuity of $ 100 in 15 years, allowing 
compound interest at 4 per cent per annum. Given 

log 1.04 = .01703, and log 180076 = 6.26646. 

5. What is the present value of an annuity of $1000 due in 30 
years, allowing compound interest at 5 per cent per annum ? 

2b 



860 



ALGEBRA. 



N 





1 


2 


3 


4 


5 


6 


7 


8 


9 


10 


0000 


0043 


0086 


0128 


0170 


0212 


0253 


0294 


0334 


0374 


11 
12 
13 
14 


0414 
0792 
1139 
1461 


0453 
0828 
1173 
1492 


0492 
0864 
1206 
1523 


0531 
0899 
1239 
1553 


0569 
0934 
1271 
1584 


0607 
0969 
1303 
1614 


0645 
1004 
1335 
1644 


0682 
1038 
1367 
1673 


0719 
1072 
1399 
1703 


0755 
1106 
1430 
1732 


15 


1761 


1790 


1818 


1847 


1875 


1903 


1931 


1959 


1987 


2014 


16 
17 
18 
19 


2041 
2304 
2553 

2788 


2068 
2330 
2577 
2810 


2095 
2355 
2601 
2833 


2122 
2380 
2625 
2856 


2148 
2405 
2648 

2878 


2175 
2430 
2672 
2900 


2201 
2455 
2695 
2923 


2227 
2480 
2718 
2945 


2253 
2504 
2742 
2967 


2279 
2629 
2765 
2989 


20 


3010 


3032 


3054 


3075 


3096 


3118 


3139 


3160 


3181 


3201 


21 
22 
23 
24 


3222 
3424 
3617 
3802 


3243 
3444 
3636 
3820 


3263 
3464 
3665 

3838 


3284 
3483 
3674 
3856 


3304 
3502 
3692 

3874 


3324 
3522 
3711 
3892 


3345 
3541 
3729 
3909 


3365 
3560 
3747 
3927 


3385 
3579 
3766 
3945 


3404 
3598 
3784 
3962 


25 


3979 


3997 


4014 


4031 


4048 


4066 


4082 


4099 


4116 


4133 


26 
27 
28 
29 


4160 
4314 
4472 
4624 


4166 
4330 

4487 
4639 


4183 
4346 
4502 
4654 


4200 
4362 
4518 
4669 


4216 
4378 
4533 

4683 


4232 
4393 

4548 
4698 


4249 
4409 
4564 
4713 


4265 
4425 
4579 

4728 


4281 
4440 
4594 
4742 


4298 
4456 
4609 
4757 


30 


4771 


4786 


4800 


4814 


4829 


4843 


4857 


4871 


4886 


4900 


31 
32 
33 
34 


4914 
6051 
6185 
5315 


4928 
5065 
5198 
5328 


4942 
5079 
5211 
5340 


4955 
5092 
5224 
5353 


4969 
5105 
5237 
5366 


4983 
5119 
6250 
6378 


4997 
5132 
5263 
5391 


5011 

5145 

'5276 

5403 


5024 
6159 
5289 
6416 


5038 
6172 
6302 

5428 


35 


6441 


5453 


5465 


5478 


5490 


5502 


6514 


5527 


5539 


6551 


36 
37 

38 
39 


5563 
5682 
5798 
6911 


5575 
5694 
5809 
5922 


5587 
5705 
5821 
6933 


5599 
5717 
5832 
5944 


5611 
5729 

5843 
5955 


5623 
5740 
5855 
6966 


6635 
6752 
5866 
5977 


5647 
5763 

5877 
5988 


5668 
6776 
5888 
5999 


5670 
5786 
6899 
6010 


40 


6021 


6031 


6042 


6053 


6064 


6075 


6085 


6096 


6107 


6117 


41 
42 
43 
44 


6128 
6232 
6335 
6435 


6138 
6243 
6345 
6444 


614^ 
6253 
6355 
6454 


6160 
6263 
6365 
6464 


6170 
6274 
6375 
6474 


6180 
6284 
6385 
6484 


6191 
6294 
6396 
6493 


6201 
6304 
6405 
6603 


6212 
6314 
6415 
6613 


6222 
6325 
6425 
6522 


45 


6532 


6542 


6551 


6561 


6571 


6580 


6590 


6599 


6609 


6618 


46 
47 
48 
49 


6628 
6721 
6812 
6902 


6637 
6730 
6821 
6911 


6646 
6739 
6830 
6920 


6656 
6749 
6839 
6928 


6665 
6758 
6848 
6937 


6675 
6767 
6857 
6946 


6684 
6776 
6866 
6955 


6693 
6786 
6875 
6964 


6702 
6794 
6884 
6972 


6712 
6803 
6893 
6981 


50 


6990 


6998 


7007 


7016 


7024 


7033 


7042 


7060 


7059 


7067 


51 
52 
53 
54 


7076 
7160 
7243 
7324 


7084 

.7168 

7251 

7332 


7093 
7177 
7259 
7340 


7101 
7185 
7267 
7348 


7110 
7193 
7275 
7366 


7118 
7202 
7284 
7364 


7126 
7210 
7292 
7372 


7135 
7218 
7300 
7380 


7143 
7226 
7308 
7388 


7152 
7235 
7316 
7396 



LOGARITHMS. 



361 



N 





1 


2 


3 


4 


5 


6 


7 


8 


9 


55 


7404 


7412 


7419 


7427 


7435 


7443 


7451 


7459 


7466 


7474 


56 
57 
58 
59 


7482 
7559 
7634 
7709 


7490 
7666 
7642 
7716 


7497 
7574 
7649 
7723 


7505 
7682 
7657 
7731 


7513 

7689 
7664 
7738 


7620 
7597 
7672 
7745 


7528 
7604 
7679 
7752 


7536 
7612 
7686 
7760 


7543 
7619 
7694 
7767 


^651 
7627 
7701 

7774 


60 


7782 


7789 


7796 


7803 


7810 


7818 


7825 


7832 


7839 


7846 


61 
62 
63 
64 


7853 
7924 
7993 
8062 


7860 
7931 
8000 
8069 


7868 
7938 
8007 
8076 


7876 
7946 
8014 
8082 


7882 
7962 
8021 
8089 


7889 
7959 
8028 
8096 


7896 
7966 
8035 
8102 


7903 
7973 
8041 
8109 


7910 
7980 
8048 
8116 


7917 
7987 
8065 
8122 


65 


8129 


8136 


8142 


8149 


8156 


8162 


8169 


8176 


8182 


8189 


66 
67 
68 
69 


8195 
8261 
8326 

8388 


8202 
8267 
8331 
8395 


8209 
8274 
8338 
8401 


8216 
8280 
8344 
8407 


8222 
8287 
8361 
8414 


8228 
8293 
8357 
8420 


8235 
8299 
8363 
8426 


8241 
8306 
8370 
8432 


8248 
8312 
8376 
8439 


8254 
8319 
8382 
8445 


70 


8461 


8467 


8463 


8470 


8476 


8482 


8488 


8494 


8500 


8506 


71 
72 
73 

74 


8513 
8573 
8633 
8692 


8619 
8679 
8639 
8698 


8525 
8585 
8645 
8704 


8631 
8591 
8651 
8710 


8537 
8597 
8657 
8716 


8543 
8603 
8663 

8722 


8549 
8609 
8669 
8727 


8565 
8615 
8675 
8733 


8561 
8621 
8681 
8739 


8567 
8627 
8686 
8746 


75 


8751 


8756 


8762 


8768 


8774 


8779 


8785 


8791 


8797 


8802 


76 

77 
78 
79 


8808 
8865 
8921 
8976 


8814 
8871 
8927 
8982 


8820 
8876 
8932 
8987 


8825 
8882 
8938 
8993 


8831 
8887 
8943 
8998 


8837 
8893 
8949 
9004 


8842 
8899 
8954 
9009 


8848 
8904 
8960 
9015 


8854 
8910 
8965 
9020 


8859 
8915 
8971 
9026 


80 


9031 


9036 


9042 


9047 


9053 


9058 


9063 


9069 


9074 


9079 


81 
82 
83 
84 


9085 
9138 
9191 
9243 


9090 
9143 
9196 
9248 


9096 
9149 
9201 
9253 


9101 
9154 
9206 
9258 


9106 
9159 
9212 
9263 


9112 
9165 
9217 
9269 


9117 
9170 
9222 
9274 


9122 
9175 
9227 
9279 


9128 
9180 
9232 
9284 


9133 
9186 
9238 
9289 


85 


9294 


9299 


9304 


9309 


9316 


9320 


9325 


9330 


9335 


9340 


86 
87 
88 
89 


9345 
9395 
9445 
9494 


9350 
9400 
9450 
9499 


9355 
9405 
9455 
9504 


9360 
9410 
9460 
9509 


9365 
9415 
9465 
9513 


9370 
9420 
9469 
9518 


9376 
9425 
9474 
9523 


9380 
9430 
9479 

9628 


9385 
9435 
9484 
9533 


9390 
9440 
9489 
9538 


90 


9542 


9547 


9552 


9557 


9562 


9566 


9571 


9576 


9581 


9586 


91 
92 
93 
94 


9590 
9638 
9685 
9731 


9596 
9643 
9689 
9736 


9600 
9647 
9694 
9741 


9606 
9652 
9699 
9745 


9609 
9657 
9703 
9760 


9614 
9661 
9708 
9754 


9619 
9660 
9713 
9759 


9624 
9671 
9717 
9763 


9628 
9675 
9722 
9768 


9633 
9680 
9727 
9773 


95 


9777 


9782 


9786 


9791 


9795 


9800 


9805 


9809 


9814 


9818 


96 
97 
98 
99 


9823 
9868 
9912 
9956 


9827 
9872 
9917 
9961 


9832 
9877 
9921 
9965 


9836 
9881 
9926 
9969 


9841 
9886 
9930 
9974 


9845 

9890 
9934 
9978 


9850 
9894 
9939 
9983 


9854 
9899 
9943 
9987 


9859 
9903 
9948 
9991 


9863 
9908 
9962 
9996 



372 ALGEBRA. 

461. Value of Any Term. In the series 

]>y taking x small enough we may make any term as large as 
we please compared with the sum of all that follow it; and by 
taking x large enough we may make any term as large as we 
please compared with the sum of all that precede it. 

(i.) The ratio of any term, as a n x*, to the sum of all that 
follow it is 

» or 



i 



«»+i« n+1 + a H+ <pT +2 + — a n+l x + a n+ ^ H 

When x is indefinitely small, the denominator can be made 
as small as we please ; that is, the fraction can be made as 
large as we please. 

(ii.) Again, the ratio of the term a^ to the sum of all 
that precede it is 

«n^ Qr <K . 

where y = - • 
x 

When x is indefinitely large, y is indefinitely small; 
hence, as in the previous case, the fraction can be made as 
large as we please. 

462. The following particular form of the foregoing 
proposition is very useful. 
In the expression 

a n x n + a^x"- 1 -\ hai& + Oq, 

consisting of a finite number of terms in descending powers 
of x, by taking x small enough the last term Oq can be made 
as large as we please compared with the sum of all the 
terms that precede it, and by taking x large enough the first 
term a n x n can be made as large as we please compared with 
the sum of all that follow it. 

Ex. 1. By taking n large enough we can make the first term of 
n 4 — 5n 8 — 7n + 9as large as we please compared with the sum of all 



LIMITING VALUES AND VANISHING FRACTIONS. 373 

the other terms ; that is, we may take the first term w* as the equiva- 
lent of the whole expression, with an error as small as we please pro- 
vided n be taken large enough. 

Ex. 2. Find the limit of 3g8 ~ 2g2 ~ 4 when (1) x is infinite ; (2) x 
w is zero. T 

(1) In the numerator and denominator we may disregard all terms 
but the first ; hence 

limit 3g8-2s 2 -4 _3g8_3 
s = oo 6a£-4a; + 8 5a^ .6* 

(2) When a; is indefinitely small toe way disregard all terms but the 

last; hence the limit is ^--, or — • 

' 8 2 



VANISHING FRACTIONS. 

463. Suppose it is required to find the limit of 

a? -f- ax — 2 a 2 

a 2 — a 2 
when x= a. 

If we put a = a + ft, then ft will approach the value zero 
as x approaches the value a. 
Substituting a + ft for <c, 

a? + a3?-2a 2 = 3aft + ft 2 = 3a + ft . 
ar'-a 2 2ah + K 2 2a + h' 

and when ft is indefinitely small the limit of this expression 
isf. 

There is, however, another way of regarding the question; 
for 

x 2 + ax — 2a 2 _ (a? — a) (x -f 2 a) _ a? + 2q 

x 2 — a 2 (x — a)(x + a) x + a' 

and if we now put a? = a the value^of the expression is f , 

as before. 

x 2 -4- (xx — 2 (x 
If in the given expression — -*—z we put x = a 

x 2 — a 2 
before simplification, it will be found that it assumes the 
form %, the value of which is indeterminate [Art. 183] ; also 
we see that it has this form in consequence of the factor 



874 ALGEBRA. 

x — a appearing in both numerator and denominator. Now 
we cannot divide by a zero factor, but as long as x is not 
absolutely equal to a, the factor x — a may be removed, and 
we then find that the nearer x approaches to the value a, 
the nearer does the value of the fraction approximate to f, 
or in accordance with the definition of Art. 458, 

when x = a, the limit of *? + ax -J a% is |. 

' x 2 - a 2 2 

464. Vanishing Fractions. If f(x) and f(x) are two func- 
tions of x, each of which becomes equal to zero for some 

particular value a of x, the fraction ^Jr4 takes the form $, 
and is called a vanishing fraction. 

Ex. 1. If x = 3, find the limit of 

x*-5x 2 + 7x-H 
x»-x 2 -6x-3 ' 

When x = 3, the expression reduces to the indeterminate form # ; 
but by removing the factor x — 3 from numerator and denominator, 

the fraction becomes - ""** — When x = 3, this reduces to £, 

x 2 + 2x+l ** 

which is therefore the required limit. 

Ex. 2. The fraction VSx ~~ a ~ Vx + a becomes 2 when x = a. 

x — a 

To find its limit, multiply numerator and denominator by the surd 
conjugate to V3x — a — Vx + a ; the fraction then becomes 
(3x-a)-(x + a) Qr 2 



(x — a) ( V3 a — a + Vx + a) V3 x — a + Vx + a 

whence by putting x = a we find that the limit is • 

V2a 

1 — 8 /x 

Ex. 3. The fraction ^— becomes - when x = 1. 

1-^/x 

To find its limit, put x = 1 + h, and expand by the Binomial 
Theorem. Thus the fraction 

" 1 _ ( i + h) h " 1 -(1 + lh - ^ 2 + •••) " - * + A* - ... 
Now ft = when x = 1 ; hence the required limit is {. 



LIMITING VALUES AND VANISHING FRACTIONS. 375 

465. We shall now discuss some peculiarities which may 
arise in the solution of a quadratic equation. 
Let the equation be 

ax 2 + bx + c = 0. 
If c = 0, then ax 2 + bx = 0; 

whence x = 0, or ; 

a 

that is, one of the roots is zero and the other is finite. 

If 6 = 0, the roots are equal in magnitude and opposite 
in sign. 

If a = 0, the equation reduces to bx + c = ; and it 
appears that in this case the quadratic furnishes only one 

root, namely, • But every, quadratic equation has two 

b 

roots, and in order to discuss the value of the other root we 
proceed as follows : 

Write - for x in the original equation and clear of frac- 

y 

tions; thus, <xf + by + a = 0. 

Now put a = 0, and we have 

ctf + by = 0; 

the solution of which is y = 0, or — ; that is, x = oo, or 
c c 

b 

Hence, in any quadratic equation orir root ivitt become in- 
finite if the coefficient ofx 2 becomes zero. 

This is the form in which the result will be most fre- 
quently met with in other branches of higher Mathematics, 
but the student should notice that it is merely a convenient 
abbreviation of the following fuller statement : 

In the equation ax 2 + bx + c = 0, if a is very small, one 
root is very large, and as a is indefinitely diminished this 
root becomes indefinitely great. In this case the finite root 

approximates to as its limit. 

b 



376 



ALGEBRA. 



EXAMPLES XL. 

Find the limits of the following expressions : 

(1) when 05=oo. (2) when 05 = 0. 

4 (x-3)(2-5x)(3x + l) 
(2s -1)8 

2. (3* a -D 2 5 . l ~* l ~ x 

x* + 9 



x (2x-3)(3-5x) 

7x 2 -6x + 4 



2x 8 -l 2x 2 

- (3-x)(x + 5)(2-7x) 
(7x-l)(x+l)* 



8 (3 + 2x«)(x-5) 
(4x 8 -9)(l + x)' 

Find the limits of 

7. *±X W hen x = -1. 8. V»-> ^ + V« ^j« wh en * 
a?-l Va; 8 - i o 2 

9. (tf-^+fr- jO* when , = a . 
(a 8 -* 8 )* +(o-*)i 

10. V« , + « + ^-V« , -« + ^. when * = a 

Va+x — Va — x 



= 2a 



CHAPTER XLI. 

CONVERGENCY AND DIVERGENCY OF SERIES. 

466. We have, in Chapter xxxiv., defined a series as an 
expression in which the successive terms are formed by 
some regular law ; if the series terminates at some assigned 
term, it is called a finite series ; if the number of terms is 
unlimited, it is called an infinite series. 

In the present chapter, we shall usually denote a series by 
an expression of the form 

Wi + w 2 + ^8 + ••• + tt * + "•• 

467. Definitions. Suppose that we have a series con- 
sisting of n terms. The sum of the series will be a function 
of w; if n increases indefinitely, the sum either tends to 
become equal to a certain finite limit, or else it becomes 
infinitely great. 

An infinite series is said to be convergent when the sum of 
the first n terms cannot numerically exceed some finite 
quantity, however great n may be. 

An infinite series is said to be divergent when the sum of 
the first n terms can be made numerically greater than any 
finite quantity by taking n sufficiently great. 

TESTS FOR CONVERGENCY. 

468. When the Sum of the First n Terms of a Given Series 
is Known. If we can find the sum of the first n terms of a 
given series, we may ascertain whether it is convergent or 
divergent by examining whether the series remains finite, or 
becomes infinite, when n is made indefinitely great 

377 



878 ALGEBRA. 

For example, the sum of the first n terms of the series 

1— of 



l+x + a? + a?+ ••• is 



l-» 



If a; is numerically less than 1, the sum approaches to the 

finite limit , and the series is therefore convergent. 

1 — x 

If x is numerically greater than 1, the sum of the first n 
terms is — ^— , and by taking n sufficiently great, this can 

X —~~ JL 

be made greater than any finite quantity ; thus the series is 
divergent. 

If x = 1, the sum of the first n terms is n, and therefore 
the series is divergent. 

If x = — 1, the series becomes 

1-1+1-1+1-1 + .... 

The sum of an even number of terms is zero, while the 
sum of an odd number of terms is 1 ; and thus the sum 
oscillates between the values and 1. This series belongs 
to a class which may be called oscillating or periodic conver- 
gent series. 

469. When the Sum of the First n Terms of a Given Series 
is Unknown. There are many cases in which we have no 
method of finding the sum of the first n. terms of a series. 
We proceed therefore to investigate rules by which we can 
test the convergency or divergency of a given series without 
effecting its summation. • 

470. First Test. An infinite series in which the terms are 
alternately positive and negative is convergent if each term is 
numerically less than the preceding term. 

Let the series be denoted by 

Ui — u** + % 8 — u 4 + u 5 — w 6 -f- •-. 

where w, > u 9 > u^ > u A > u s • • . . 



CONVERGENCY AND DIVERGENCY OF SERIES. 379 

The given series may be written in each of the following 
forms : 

(u x — u 2 )+(u s — u 4 ) + (u 5 — We)+ .... (1), 

Hi— ( u 2 — u s) — ( M 4 — *%) — ( w e — «f)— • • • (2). 

From (1) we see that the sum of any number of terms is 
a positive quantity ; and from (2) that the sum of any num- 
ber of terms is less than Ux ; hence the series is convergent. 

For example, in the series 

the terms are alternately positive and negative, and each 
term is numerically less than the preceding one ; hence the 
series is convergent. 

471. Second Test. An infinite series in which all the terms 
are of the same sign is divergent if each term is greater than 
some finite quantity, however small. 

For if each term is greater than some finite quantity a, 
the sum of the first n terms is greater than na ; and this, 
by taking n sufficiently great, can be made to exceed any 
finite quantity. 

472. Before proceeding to investigate further tests of 
convergency and divergency, we shall lay down two impor- 
tant principles, which may almost be regarded as axioms. 

I. If a series is convergent it will remain convergent, and 
if divergent it will remain divergent, when we add or remove 
anyfijiite number of its terms; for the sum of these terms 
is a finite quantity. 

II. If a series in which all the terms are positive is con- 
vergent, then the series is convergent when some or all of 
the terms are negative ; for the sum is clearly greatest when 
all the terms have the same sign. 

We shall suppose that all the terms are positive unless 
the contrary is stated. 



880 ALGEBRA. 

473. Third Test. An infinite series is convergent if from 
and after some fixed term the ratio of each term to the precedr 
ing term is numerically less than some quantity which is itself 
numerically less than unity. 

Let the series beginning from the fixed term be denoted by 

Mi + t*2 + 1*8 + ^4+ •••; 

and let — <**,—< r, — < r. •••. 

tti v* «, ' 

where r < 1. 

Then u* +u^ + Uz + u A + ••• 

= tM 1 H 1 — • — — • — • — !-••• 1 

\ •«! tt, «! U| ttj U) / 

<i«i(l+r + t J + f J + ...); 

that is, < 5 — — . since r < 1. 
1 — r 

Hence the given series is convergent. 

474. In the enunciation of the preceding article the 
student should notice the significance of the words "from 
and after a fixed term." 

Consider the series 

1 + 2x + 3a?+ 4a? + ... + nx*-* +.... 

Here i = Jl=fl + l), } 

u nmml w-i \, w-iy 

and by taking n sufficiently large we can make this ratio 
approximate to x as nearly as we please, and the ratio of 
each term to the preceding term will ultimately be a?. 
Hence if x < 1, the series is convergent. 

But the ratio -^- will not be less than 1, until nx < 1 : 

w n _i n — 1 

that is, until n > 



1-x 

Here we have a case of a convergent series in which the 
terms may increase up to a certain point, and then begin to 



CONVERGENCY AND DIVERGENCY OF SERIES. 381 

decrease. For example, if x = J&, then = 100, and 

1 —»x 

the terms do not begin to decrease until after the 100th 
term. 

475. Fourth Test. An infinite series in which all the terms 
are of the same sign is divergent if from and after some fixed 
term the ratio of each term to the preceding term is greater 
than unity, or equal to unity. 

Let the fixed term be denoted by u v If the ratio is equal 
to unity, each of the succeeding terms is equal to u^ and 
the sum of n terms is equal to nu±\ hence the series is 
divergent. 

If the ratio is greater than unity, each of the terms after 
the fixed term is greater than u 1} and the sum of n terms is 
greater than nu x ; hence the series is divergent. 

476. In the practical application of these tests, to avoid 
having to ascertain the particular term after which each 

term is greater or less than the preceding term, it is con- 

u n 
venient to find the limit of when n is indefinitely 

increased ; let this limit be denoted by I. 

If I < 1, the series is convergent. [Art. 473.] 

If I > 1, the series is divergent. [Art. 475.] 

If I = 1, the series may be either convergent or divergent, 

and a further test will be required ; for it may happen that 



Un 



< 1, but continually approaching to 1 as its limit when n 

is indefinitely increased. In this case we cannot name any 
finite quantity r which is itself less than 1 and yet greater 
than I. Hence the test of Art. 473 fails. If, however, 

— — > 1, but continually approaching to 1 as its limit, the 

series is divergent by Art. 475. 

u n 
We shall use "Lim " as an abbreviation of the worda 

u 
"the limit of — — when n is infinite. " 



882 ALGEBRA. 

Ex. 1. Find whether the series whose nth term fe 0* + )»* ^ 
convergent or divergent. n 

Here *l_ = (* + *)*% «^ (n + l)(n-l) 2 

tin-i n 2 (n-1) 2 n 8 » 

.•. Lim = 05; 

Wn-l 

hence if x < 1 the series is convergent ; 

if x > 1 the series is divergent. 

u 
If x = 1, then Lim — — = 1, and a further test is required, 

Ex. 2. Is the series 

l 2 -f 22« + &x 2 + 42s 8 + ••• 
convergent or divergent ? 

Here Lim —5- = Lim 7 - = x. 

Wn-i (w — l) 2 x"- 2 

Hence if x < 1 the series is convergent ; 

if x > 1 the series is divergent. 

If x = 1 the series becomes l 2 + 2 2 + 3 2 + 4 s + ••., and is obviously 
divergent. 

Ex. 3. Tn the series 



a + (a + d)r+(a + 2d)r 2 + h(a + n-l .d)r"- 1 + ..., 

Lim = Lim — -7 -~ • r = r; 

thus if r < 1 the series is convergent, and the sum is finite. 

477. Fifth Test. If there are two infinite series in each of 
which all the terms are positive, and if the ratio of the corre- 
sponding terms in the two series is always finite, the two series 
are both convergent, or both divergent. 

Let the two infinite series be denoted by 

and v x + v t + v s + v A + •••. 

The value of the fraction 

Mi 4- M g + u s~\ hu n 

^4-^2 + ^3+ -• +V n 



4 

CONVERGENCY AND DIVERGENCY OF SERIES. 383 

lies between the greatest and least of the fractions 

and is therefore & finite quantity, L say; 

.\ Wi + w 2 + tt 8 H h «*. = Z>(i>i + v a + v 8 H f-v n ). 

Hence if one series is finite in value, so is the other ; if 
one series is infinite in value, so is the other ; which proves 
the proposition. 

478. Auxiliary Series. The application of the principle 
of the preceding article is very important, for by means of 
it we can compare a given series with an auxiliary series 
whose convergency or divergency has been already estab- 
lished. The series discussed in the next article will fre- 
quently be found useful as an auxiliary series. 

479. The infinite series -r- + ^-+7r + ir-\ is always 

\p 2 P 3* 4 P 

divergent except when p is positive and greater than 1. 

Case I. Let p > 1. 

The first term is 1 ; the next two terms together are less 
than — ; the following four terms together are less than ■— . 

the following eight terms together are less than — ; and so 

on. Hence the series is less than 

1,2.4.8. 
2* 4» 8* ' 

that is, less than a geometrical progression whose common 

ratio — is less than 1. since »>1; hence the series is 
convergent. 

Case II. Letp = l. 

The series now becomes l + i + -J- + i + -J-+»-. 
The third and fourth terms together are greater than £ or 
\ ; the following four terms together are greater than £ or 



384 ALGEBRA. 

\ ; the following eight terms together are greater than ^ 01 
\ ; and so on. Hence the series is greater than 

and is therefore divergent. [Art. 475.] 

Case III. Let p < 1, or negative. 

Each term is now greater than the corresponding term in 

Case II., therefore the series is divergent. 

Hence the series is always divergent except in the case 
when p is positive and greater than unity. 

Ex. Prove that the series - H 1 h ••• 4- n "t + ••• is divergent. 

14 9 n 2 

Compare the given series with 1 -\ \-~ + ••• H h •••• 

2 3 n 

Thus if u n and v n denote the nth tenns of the given series and the 
auxiliary series respectively, we have 

u n __n+\ . l_ w-f-l . 
v n n 2 n n 

hence Lim— = 1, and therefore the two series are both convergent or 

both divergent. But the auxiliary series is divergent, therefore also 
the given series is divergent. 

This completes the solution of Ex. 1. [Art. 476.] 

480. Convergency of the Binomial Series. To shoio that 
the expansion of (1 + x) n by the Binomial Theorem is con- 
vergent when x < 1. 

Let u r , u r+l represent the rth and (r + l)th terms of the 
expansion; then 

u r+l _ n — r 4- 1 



x. 



u r 



When r > n + 1, this ratio is negative ; that is, from this 
point the terms are alternately positive and negative when x 
is positive, and always of the same sign when x is negative. 

Now when r is infinite, Lim -^ = x numerically ; therefore 

u r 

since x < 1 the series is convergent if all the terms are of 
the same sign ; and therefore still more is it convergent when 
some of the terms are positive and some negative. [Art. 472.] 



CONVERGENCY AND DIVERGENCY OF SERIES. &86 

EXAMPLES XLI. 
Find whether the following series are convergent or divergent s 

x z + a z + 2a x + Sa 
x and a being positive quantities. 

2. -J— + -L- + -J— + -1- + .... 

8 1 11 1 . 

' W (« + l)(y + 1) (* + 2)(y + 2) (x + 3)(y + 3) ^ * ' 
a; and y being positive quantities. 

1.2 2.3 T 8-4 4.6 

1.2 3.4 T 6.6 7-8 T 
6 i +? + ? + £+ 8 - l+3aj + 6x 2 +7a^ + 9aj*+.... 

" \* \* \* '" 2 3 4 5 

7. Vi + V* + Vi + \/* + — 9 ' Tp + 2? + & + 4r*'"' 
10. i+? + ^ + . aj3 +... + _Jil_ + .... 

2610 n a + 1 

Note. For further information on the subject of Convergency and 
Divergency of Series the reader may consult Hall & Knight's Higher 
Algebra, Chapter xxi. 

2c 



CHAPTER XLII. 

Undetermined Coefficients, 
functions of finite dimensions. 

481. In Art. 105 it was proved that if any rational inte- 
gral function of x vanishes when x = a, it is divisible by 
x— a. 

Let Pfpf + p 1 x n ' l +p<pr~ 2 H \-p n 

be a rational integral function of x of n dimensions, which 
vanishes when x is equal to each of the unequal quantities 

0], a 2 , Og, ••• a n . 

Denote the function by f(x) ; then since f(x) is divisible 
by x — a^ we have 

f(x) = (x - flhXivf- 1 H ), 

the quotient being of n — 1 dimensions. 

Similarly, since /(a?) is divisible by x — o^wq have 

W*" 1 H = (x — a^iptfx?- 2 H ), 

the quotient being of n — 2 dimensions ; and 

jPo« n ~ 2 + ••• = (a — (h)(p<^~ s + — )• 
Proceeding in this way, we shall finally obtain after n 
divisions 

f(x)=p (x — al)(x — O2)(oj — a 8 ) ... (<c — a w ). 

482. If a rational integral function of n dimensions vanishes 
for more than n values of the variable, the coefficient of each 
power of the variable must be zero. 

Let the function be denoted by /(#), where 

886 



UNDETERMINED COEFFICIENTS. 887 

and suppose that f(x) vanishes when x is equal to each of 
the unequal values a^ Og, 03, ••• a n ; then 

f(x)=p (x - ax) (x - aj)(x - 03) ... (a? - a,). 

Let c be another value of x which makes f(x) vanish; then 
since /(c) = 0, we have 

Po(c- <h)(c -a^ic- Os)"'(c - a n ) = 0; 
and therefore p Q = 0, since, by hypothesis, none of the other 
factors is equal to zero. Hence f(x) reduces to 

Pi« w_1 +P2® n ~ 2 H-Ps^" 8 H hjPn. 

By hypothesis this expression vanishes for more than n 
values of x } and therefore p x = 0. 

In a similar manner we may show that each of the coeffi- 
cients jp 2 > P$t •••i>» must be equal to zero. 

This result may also be enunciated as follows : 

If a rational integral function of n dimensions vanishes for 
more than n values of the variable, it must vanish for every 
value of the variable. 

Cor. If the function f(x) vanishes for more than n values 
of x, the equation f(x) = has more than n roots. 

Hence also, if an equation of n dimensions has more than 
n roots it is an identity, 

Ex. Prove that 

(x — 6) (x — C) (x — c) (x — a) (x — a) (x — b) _ - 
(a - 6)(a - c) + (& - c)(b -a) + (c- a)(c - &) 

This equation is of two dimensions, and it is evidently satisfied by 
each of the three values a, b, c \ hence it is an identity. 

483. If two rational integral functions of n dimensions are 
equal for more than n values of the variable, they are equal 
for every value of the variable. 

Suppose that the two functions 

PqX" +Pi# n - 1 +jp a » ft ~ 2 -r- -~+Pn, 



qo** + QiX"' 1 + Q r 2» n " a H r-Qi 



n> 



are equal for more than n values of x ; then the expression 

(i>6 ~ flo)*" +(l>l ~ ft)**" 1 +(l>2 - ft)**" 2 + — +(Pn~ Qn) 



888 ALGEBRA. 

vanishes for mor$ than n values of x ; and therefore, by the 
preceding article, 

that is, p = go> Pi = ?b 1>2 = 2» •••*>» = ?n- 

Hence the two expressions are identical, and therefore are 
equal for every value of the variable. Thus 

If two rational integral functions are identically equal, we 
may equate the coefficients of the like powers of the variable. 

Cor. This proposition still holds if one of the functions 
is of lower dimensions than the other. For instance, if 

#>#" +PiX n ~ 1 +P2* n ~* +Ps® n ~ 8 + — +P» 
= q*x"- 2 + qzx"-* + — + q n , 

we have only to suppose that -in the above investigation 
q = 0, q x = 0, and then we obtain 

i>o = 0, Pi = 0, 1)2 = 22, l> 8 = ft>— P* = q»- 

484. The theorem established in the preceding article 
for functions of finite dimensions is usually referred to as 
the Principle of Undetermined Coefficients. The application 
of this principle is illustrated in the following examples. 

Ex. 1. Find the sum of the series 

1-2 + 2.3 + 3.4 +...+ n(n + 1). 
Assume that 

1-2+2. 3+3.4+... ln(n+l)=u4+JBw+Cw 2 +2>n 8 +.Efo 4 +-", (1) 

where A, B, C,D,E,-» are quantities independent of n, whose values 
have to be determined. Change n into n + 1 ; then 

1 .2 + 2.3+.-. + n(n + l) + (n+ l)(w + 2) 

= A + B(n + 1)+ C(n + l) 2 + Z>(w + l) 8 + E(n + l) 4 +.». (2) 
By subtracting (1) from (2), 
(n + l)(n + 2)= B \- C(2n + 1)+ D(Sn* + 3w + 1) 

+ ^(4 n 8 + 6 n? + 4 n + 1) + - .. 

This equation being true for all integral values of n, the coefficients 
of the respective powers of n on each side must be equal ; thus E and 
all succeeding coefficients must be equal to zero, and 

32>=1; 3Z> + 2C = 3; D+ C+B = 2; 
whence D = J, (7=1, B = 



UNDETERMINED COEFFICIENTS. 389 

Hence the sum = A + — + n 2 + in 8 . 

To find A, put n = 1 ; the series then reduces to its first term, and 

2 = A + 2, or A = 0. 

Hence 1 . 2 -f 2 . 3 + 3 . 4 + ... + n(n + 1)= J n(n + I)(n + 2). 

Note. It will be seen from this example that when the nth term is 
a rational integral function of n, it is sufficient to assume for the sum 
a function of n which is of one dimension higher than the nth term of 
the series. 

Ex! 2. Find the conditions that x 8 + px 2 + qx + r may be divisible 

by 

x 2 + ax + 6. 

The quotient will contain two terms ; namely, x and a term inde- 
pendent of x. Hence, we assume 

X s +px 2 + qx + r = (x -f k) (x 2 + ax + b). 

Equating the coefficients of the like powers of x, we have 

& + a=i>, ak + b = q, kb = r. 

From the last equation k = -; hence by substitution we obtain 

b 

£-M=p, and 9£+b = q; 



that is, r = 6(p — a), and ar = J^g — ft); 

which are the conditions required. 

EXAMPLES XLII. a. 

Find by the method of Undetermined Coefficients the sum of 

1. 12 + 32 + 52 _|_ 72 _|_ ... to n terms. 

2. 1.2-3 + 2 .3.4 + 3.4.6 + — ton terms. 

3. 1.2 2 + 2.3 2 + 3.4 2 + 4.5 2 + ... tonterms. 

4. I 8 + 3 8 + 6 8 + 7 8 + — to n terms. 

5. 1* + 2* + 3* + 4 4 + ... to n terms. 

6. Find the condition that x 8 — 3px + 2 q may be divisible by a 
factor of the form x 2 + 2 ax + a 2 . 

7. Find the conditions that ax 8 + bx 2 + ex + d may be a perfect 
cube. 



890 ALGEBRA. 

8. Find the conditions that cPx* + foe 8 + ex 2 + dx +f* may be a 
perfect square. 

9i Prove the identity 

a 2 (x-b)(x-c) &(x-c)(x-a) c 8 (s- q)(s- b) _ 
(a-ft)Ca-c) + (&-c)(6-a) + (c-a)(c-6) ' 



FUNCTIONS OF INFINITE DIMENSIONS. 

485. If the infinite series a -\-aiX + a 2 x 2 + a 9 x* + ••• is 
equal to zero for every finite value of x for which the series 
is convergent, then each coefficient must be equal to zero identi- 
cally. 

Let the series be denoted by S, and let S x stand for the 
expression a, + a& -f a&? + ••• ; then S = a + xS 1 , and 
therefore, by hypothesis, % + xSi = for all finite values of 
x. But since S is convergent, Si cannot exceed some finite 
limit ; therefore by taking x small enough, xS x may be made 
as small as we please. In this case the limit of S is a ; but 
S is always zero, therefore a must be equal to zero identi- 
cally. 

Removing the term cr , we have xSi = for all finite values 
of x\ that is, Oi + a^c + atf? + ••• vanishes for all finite 
values of x. 

Similarly, we may prove in succession that each of the 
coefficients o^ a% Og, ••• is equal to zero identically. 

486. If two infinite series are convergent and equal to one 
another for every finite value of the variable, the coefficients of 
like powers of the variable in the two series are equal. 

Suppose that the two series are denoted by 

Oq + OqX + ci&? + ^^ + ••• 
and A« + A x x + A#? + A&P+ •••; 

then the expression 

Oq — A* -f («i — A x )x + (02 — AJ)x* +(03 — A^x* + 



••• 



UNDETERMINED COEFFICIENTS. 



391 



vanishes for all values of x within the assigned limits \ 
therefore by the last article 

do — -4, = 0, a 1 — A 1 = 0, a 2 --4 2 = 0, 03— A 8 = 0, •••, 

that is, Oq = A& a x = A^ a 2 = A% <% = A& ••• ; 

hence the coefficients of like powers of the variable are equal, 
which proves the proposition. 



EXPANSION OF FRACTIONS INTO SERIES. 



487. Expand i + — — in a series of ascending powers 



of x. 
Let 



1 -j-a? — x 2 



2 + x* A + Bx+ Cx 2 + Dx*+..., 
1 + x — ar 



where A, B, C, D, •••, are constants whose values are to be 
determined; then 

2 + x* = A(l+x -x*)+Bx(l + x - x*)+ Cx*(l + x - x*) 

+ #3 8 (l+a-<C 2 )+... 
x*+ ••• 



= A + B 


x+ C 


x*+ D 


A 


B 


C 




-A 


-B 



Equating the coefficients of like powers of x, we have 
A = 2, B + A = 0, C+B-A = l, D+C-B = 0, 



.-. B = -2; 



.\ 0=5; 



.-. Z> = -7j 



thus 



2 + a?2 =2-2x + 5x*-7x*+-... 
l + aj-aj 2 



488. Both numerator and denominator should be arranged 
with reference to the ascending powers of the same quantity ; 
then dividing the first term of the numerator by the first 
term of the denominator determines the form of the expansion. 



S92 



ALGEBRA. 



Ex. Expand 



2 



in a series of ascending powers of x. 



2x 2 -3x» 

Dividing x°, of the first term of the numerator, by x 2 , of the first 
term of the denominator, we obtain x~ 2 for the first term of the ex- 
pansion ; therefore we assume 



2x a -3x 8 



= Ax~* + Bx' 1 + C+ Dx + ... ; 



then 



2=2A+2B 
-SA 



x+ 20 
-SB 



x*+ 2D 
-3C 



*» + 



thus 



Equating the coefficients of like powers of x, we have 

A = l; 2B-SA = 0, 2C-3B = 0, 2Z)-3C = 

.\ B = i; .-. C=i; .-. 2> = 

2 



0, 



2x 2 -3x» 



= z- 2 + faJ- 1 + f + V*+- 



1 . 3 _i9.27x , 

= — + — -4 + — — -f. ••• 

x 2 2x 4 8 



EXAMPLES XLH. b. 

Expand to four terms in ascending powers of x : 

1 -8x « 1 + x 



L 1 + 2ag . 2. 

1-x-x 2 l-x-6x 2 



3. 



4. 



3 + x 



5. 
6. 



1 + ax — ax 2 — x 8 
2-2x + 3x 2 



4 + x + x 2 



7. 
8. 



2 + x + x 2 2-x-x 2 

2 + x 



2x-3x 2 
c 



9. 
10. 



ax 



3x 2 + x* 

1 +X + X 2 
X + X 8 + X* 



EXPANSION OF SURDS INTO SERIES. 



489. Expand VI +x in ascending powers of x. 



Let VI +x = A + Bx+Cx 2 + Dx* + Ex A + 
By squaring both sides of the equation, we have 



l + x = A 2 + 2AB 



x + B 2 
+ 2 AC 



x* + 2AD 
+ 2BC 



x*+C* 
+ 2BD 

+ 2AE 



«* + 



••t 



UNDETERMINED COEFFICIENTS. 393 

Equating the coefficients of like powers of x 9 we have 
4 = 1; 2AB = 1, B? + 2AC = 0, 2AD + 2BC = 0, 
.-.* = *; .-. = -i; .'. D = tV: 

C , + 2JBD + 2-4jE = 05 

••• ^ = -Tfy5 

thus VT+i = i+?-?+^-^+ .... 

2 8 16 128 

Note. The expansion can be readily effected by the use of the 
Binomial Theorem [Art. 421]. 

EXAMPLES XLn. o. 
Expand the following expressions to four terms : 

1. VT^x. 3. Va 2 - x 2 5. (1 + x)*. 

2. Va-x. 4. v^2 + x. 6. (1 -J- x + x 2 )*. 

REVERSION OF SERIES. 

490. To revert a series y = ax + bx* + ex 3 + ... is to ex- 
press a; in a series of ascending powers of y. 
Eevert the series 

y = x-2x* + 3x*-4:X 4 +:. . . . . (1). 
Assume x = Ay + By* + Cy 8 + Dtf + ••• . 

Substituting in this equation the value of y as given in 
(1), we have 

x=A(x-2x 2 +3x s -4x 4 + ...) =A(x -2x 2 +3x*-4:X A +.:) 
+£(a?-2;c 2 +3a 3 -4a 4 + ...) 2 =B(x 2 +±x i --4:X $ +6x 4 +..-) 

+ C(a?-2s 2 +3a 8 -4# 4 + ...^C^-e* 4 ) 

+ Z>(a-2 x 2 +3x s -4x A + — y=D(x 4 + ... ) 

Equating the coefficients of like powers of a?, 

-4 = 1; J3-2J. = 0, C-4£ + 3^1 = 0, 
.-. 5 = 2; .-. = 5; 

D-6C + 10B-±A = 0, 

... Z> = 14. 

Hence a? = y + 2 s/ 2 + 5 y 3 + 14 #* -f- ••• > 



894 ALGEBRA. 

491. If the series bey = l + 2a + 33 2 + 4^ + .* 
put 3/ -1 = 3; 

then 3 = 2a; + 3a! , + 4a^+.". 

Assume x = Az -f ife 2 + Cfe 8 ■+- ••• and complete the wori 
as in Art 490 after which replace z by its value y — 1. 

EXAMPLES XLII. d. 
Revert each of the following series to four terms : 

1. y = a; + a;* + je 8 + a5* + .... 4. y=l +»+£.+??+£+ .... 

2 o 24 

2. y=x+3a; 2 +5a 8 +7iB*+.... ^ ^ 3.7 

5. y = as — — + — — —-+ •••• 
a .,-* ^-l* 8 ^j. 3 6 7 

O. tf^X — — + — — — +•••• 

2 4 8 6. y=oa;+6a5 2 +ciB 8 +daj*+.... 

PARTIAL FRACTIONS. 

492. A group of fractions connected by the signs of 
addition and subtraction is reduced to a more simple form 
by being collected into one single -fraction whose denomina- 
tor is the lowest common denominator of the given frac- 
tions. But the converse process of separating a fraction into 
a group of simpler, or partial, fractions is often required. 

For example, if we wish to expand — — in a series 

r r l-lx + Sx* 

of ascending powers of x, we might use the method of Art. 

487, and so obtain as many terms as we please. But if we 

wish to find the general term of the series, this method is 

inapplicable, and it is simpler to express the given fraction 

1 2 

in the equivalent form f- — • Each of the ex- 

JL — — X JL ~~ ij X 

pressions (1—a?)" 1 and (1 — 3 a?)" 1 can now be expanded 
by the Binomial Theorem, and the general term obtained. 

493. We shall now give some examples illustrating the 
decomposition of a rational fraction into partial fractions. 
For a fuller discussion of the subject the reader is referred 
to treatises on Higher Algebra, or the Integral Calculus. 



UNDETERMINED COEFFICIENTS. 3$5 

In these works it is proved that any rational fraction may 
be resolved into a series of partial fractions ; and that 

(1) To any factor of the first degree, as x — a, in the 
denominator there corresponds a partial fraction of the 

form • 

x — a 

(2) To any factor of the first degree, as x — b, occurring 
n times ' in the denominator there corresponds a series* of 
n partial fractions of the form 

-A- + -C-+...+ * . 
x - b (x- by T (a>- &)• . 

(3) To any quadratic factors, as x*+px + q, in the de- 
nominator there corresponds a partial fraction of the form 

Ax + B 

x*+px+q 

(4) To any quadratic factor, as x*+px + q, occurring n 
times in the denominator there corresponds a series of 
n partial fractions of the form 

Ax + B , Cx + D , . Bx + S 



(x*+px + q) (aP+px + q)* (x* + px + q) n 

Here the quantities A, B, (7, D, ••• B, S 9 are all inde- 
pendent of x. 

We shall make use of these results in the examples that 
follow. 

Ex. 1. Separate ^-^ into partial fractions. 

Since the denominator 2 x 2 + x — 6 = (z + 2) (2 x — 8), we assume 

5»-ll A , B 



2x 2 +s-6 s + 2 2s-3' 

where A and B are quantities independent of x whose values have to 
be determined. 

Clearing of fractions, 

5 a: - 11 = 4 (2 a; - 3) + B(x + 2) . 
Since this equation is identically true, we may equate coefficients of 
like powers of x ; thus 

2A + B=b, -3^ + 25 = -ll; 

whence -4 = 3, B = — 1. 

. 5s -11 = 3 1 

" 2z* + x-6 x + 2 2rc-8 



#96 ALGEBRA. 

Ex. 2. Resolve «*£+_* mt0 partial fraction*. 

(x-a)(x + 6) 

Assume mx + n = A + B m 

(x — a)(x + 6) x — a x + b 
.\ mx + n = A(x + b) -f B(x — o) .... (1) 

We might now equate coefficients and find the values of A and B x 
but it is simpler to proceed in the following manner : 

Since A and B are independent of x, we may give to as any value 
we please. 

In (1) put x — a = 0, or x = a ; then 

a + b ' 
putting 3 + 6 = 0, or x = - 6, B= mb-n 



a + b 



. mx + n _ 1 / m«-fn . TW& — n \ 
"(x — a)(x + 6) o+6\x-a x + b )' 

Ex. 3. Resolve — into partial fractions. 

(2x-l)(9-x 2 ) 

Assume 23X-11X 2 1 = ^l_ + ^_ + _C_ . (1) . 

(2x-l)(3 + a)(3-x) 2x-l 3+x 3-x w * 
.\ 23x - 11 a; 2 = A (3 + as) (8 - x) + 5(2 x - 1)(3 - x) 

+ C(2x-l)(3+x). 

By equating the coefficients of like powers of x, or putting in 
succession 2 x — 1 = 0, 3 -f x = 0, 8 — x = 0, we find that 

4=1, 5 = 4, C = -l. 
23a -11a: 2 1 ,4 1 



■" (2x-l)(9-x 2 ) 2x-l 3 + x 3-x 

Ex. 4. Resolve "3" into partial fractions. 

(x-2) 2 (l-2x) 

Assume 3x 2 + x-2 = ^_ + ^_ + C 

{x - 2) 2 (1 -2 a) l-2xx-2^(x-2) 2 ' 
.-. 3x 2 +x-2 = 4(x-2) 2 + 5(l-2x)(x-2) + C(l-2x). 
Let 1 -2x = 0, then 4 = -J; 

let x - 2 = 0, then C = —4. 

To find B, equate the coefficients of x 2 ; thus 

3 = 4-25; whence B = — J. 

, 3x 2 + x-2 _ 1 5 4 

° ' (x - 2) 2 (1 - 2 x) 3(1 - 2 x) 3(x - 2) (x - 2)*' 



UNDETERMINED COEFFICIENTS. 897 



42 — 19 x 
Ex. 5. Resolve into partial fractions. 

(x 2 + l)(s-4) 



(s 2 +l)(x-4) » 2 +l x-4' 
.-. 42 - 19s =(ite + £)(x - 4)+ (7(0? + 1). 

Let x = 4, then C = — 2 ; 

equating coefficients of as 2 , = A + O, and A = 2 ; 

equating the absolute terms, 42 = — 4 B -f C, and i? = — 11, 

. 42 -19s = 2as~ll 2 
" (s 2 + l)(a - 4) x 2 + 1 a - 4 

494. In all the preceding examples the numerator has 
been of lower dimensions than the denominator ; if this is 
not the case, we divide the numerator by the denominator 
until a remainder is obtained which is of lower dimensions 
than the denominator. 

Ex. Resolve "*~ — x ~~ into partial fractions. 
3x 2 -2s-l 

By division, 

<* + 6*-7 _. .,. 8x-4 

Sss»-2*-l S<#-2z-l 



3b 2 -2x-1 Sx + l a-1* 
6s3+5s 2 -7 = 2 3 _5_ _1_ 
3x 2 -2s-l 8s + l *-l 

495. The General Term. We shall now explain how reso- 
lution into partial fractions may be used to facilitate the 
expansion of a rational fraction in ascending powers of x. 

1 _1_ (K w 

Ex. 1. Find the general term of -^ _ when expanded in 

& l-3x + 2x 2 

a series of ascending powers of z. 

By Ex. 1, Art. 493, we have 

— L±!2 — = — I ?- = 7(1-2 a)"* - 6(1 - *w 

1-33 + 2& 2 l-2x 1 - * v ' v ' 

= 7[l+(2a;) + (2a0 2 +... + (2a0*+...] 

-6(1 +x + x 2 H h« r +— )• 

Hence the (r + l)th, or general term of the expansion, is 

7(2 «) r - 6 xr or [7(2)' - 6]af. 



898 ALGEBRA. 

Ex. 2. Find the general term of ^ *t "" n ■- when expanded 

(x-2) 2 (l-2x) *^ 

tn a series of ascending powers of x. 
By Ex. 4, Art. 493, we have 

3x 2 + x-2 _ 1 5 4 



{x -2)2(1 -2 x) 3(1 - 2 x) 8(x-2) (x-2)* 



+ 



3(1 -2x) 3(2 -a) (2 -a;)* 
1,5 4 



•0-*«>e(i-5) 4(1 -sy 

= 4(l-2,). 1+ |(l-.)^(l.|)- 
b-J[1+(2x) + (2x)«+». + (2«)'+...] 

Hence the (r + l)th or general term of the expansion is 

. 1 3%^ 2- ; 

496. The following example sufficiently illustrates the 
method to be pursued when the denominator contains a 
quadratic factor. 

Ex. Expand i* — in ascending powers of x and find 

(l + x)(l+x 2 ) 
the general term. 

Assume 7 -±* = ^- + Bx±C 

(l + x)(l + x 2 ) 1 + * l + s a 

.«. 7 + x = .4(l+x 2 ) + CBx + C)(l+x). 

Let 1 + x = 0, then ^4 = 3 ; 

equating the absolute terms, 1 = A+ C, whence (7 = 4; 

equating the coefficients of x 2 , = A + I?, whence 5 = — 3. 

7 + x _ 3 4-3x 



(1 + xXl+x 2 ) 1 + 05 1 + x 2 

= 3(1 + x)-i + (4 - 3x)(l + x 2 )-i 

= 8{1 - x + x 2 h(- iyxp +».} 

+(4 - Sx){l - x 2 + x* +(- l)** 2 * +...$, 



UNDETERMINED COEFFICIENTS. 399 

To find the coefficient of of ; 

(1) If r is even, the coefficient of or in the second series is 

r r 

4(— 1)2; therefore in the expansion the coefficient of ar is 3+4(— 1)2. 

(2) If r is odd, the coefficient of xr in the second series is 

f— 1 H-l 

— 8(— 1) a , and the required coefficient is 3(— 1) * — 3. 

EXAMPLES XLII. e. 

Resolve into partial fractions : 

1 7s-l 2 46 + 13a; 8 l + 3s + 2s* 

l-5x + 6x 2 ' * 12x 2 -llx-16* ' (l-2x)(l-x 2 ) 

^ x 2 ~10a; + 13 g 26x 2 + 208x 



(x - l)(x 2 - 5x + 6) (x 2 + l)(x + 6) 

. 2x* + x 2 -x-3 2x 2 -llx + 6 

0. — ; ; — -• V» 



6. 
7. 



x(x-l)(2x + 3) (x-3)(x 2 + 2x-6) 

9 - 3x 8 -8x 2 + 10 t 

(x-l)(x + 2) 2 " ' (x-l) 4 

x*-3x 8 -3x 2 +10 n 5x 8 +6x 2 + 5x 

(x + l) 2 (x-3) " * (x 2 -l)(x+l)*' 



Find the general term of the following expressions when expanded 
in ascending powers of x. 

12 l + 3x 13 5x + 6 14 x 2 + 7x + 3 

l + llx+28x 2 ' * (2+x)(l-x)' * x 2 + 7x+10* 

15. 2x^4 19> 2x+l 



16. 



(1 - x 2 )(l - 2x) (x - l)(x 2 + 1) 

4 + 3x + 2x 2 " -^ 1-X + 2X 2 



(l_x)(l+x-2x a ) (1-*) 



— *^a 



17 3 + 2x-x 2 21 1 

(1 + x) (1 - 4 x)*' ' (l-ax)(l-6x)(l-cx) 

18. , t + lx ^ 8* 3 " 2x2 



(2 + 3x)(l + «)« (2 - 3x + x 2 )* 



CHAPTER XLII1. 
Continued Fractions. 
497- An expression of the form a -\ 



a continued fraction ; here the letters a, b, c, ■•• may denote 

any quantities whatever, but for the present we shall only 

consider the simpler form o, -\ ^ , where a u Oj, 

a,, ■-• are positive integers. This will be usually written in 
the more compact form 

<h+ <h + 

498- When the number of quotients a„ a* % ••• is finite 
the continued fraction is said to he terminating; if the 
number of quotients is unlimited the fraction is called an 
infinite continued fraction. 

It is possible to reduce every terminating continued frac- 
tion to an ordinary fraction by simplifying the fractions in 
succession beginning from the lowest. 

499. To convert a given fraction into a continued fraction. 

Let — be the given fraction ; divide mbyn; let % be the 

p the remainder; thus 

- = a i - r -Z = a l + -i 



CONTINUED FRACTIONS. 401 

divide n by p, let a^ be the quotient and q the remainder ; 
thus 

*■ . Q .1 

P P P 

q 

divide p by q, let a 3 be the quotient and r the remainder ; 
and so on. Thus 

m 1 , 1 1 



a 8 + ««. 

If m is less than n, the first quotient is zero, and we put 

m_l 
n ~~ n 

m 

and proceed as before. 

It will be observed that the above process is the same as 
that of finding the greatest common measure of m and n ; 
hence if m and n are commensurable, we shall at length 
arrive at a stage where the division is exact and the 
process terminates. Thus every fraction whose numerator 
and denominator are positive integers can be converted into 
a terminating continued fraction. 

Ex. Reduce { f J to a continued fraction. 

Find the greatest common measure of 832 and 159 by the usual 

process, thus: 

159)832(5 
795 

37)159(4 
148 

11)37(8 
33 

4)11(2 
8 

We have the successive quotients g 

5, 4, 3, 2, 1, 3 ; hence J)3(g 

832 = 6 , J__L J_ J_ 1. 3 

159 4+ 3+ 2+ 1+ 3* u 

2d 



402 ALGEBRA. 

500. Convergents. The fractions obtained by stopping 
at the first, second, third, ••• quotients of a continued frac- 
tion are called the first, second, third, ••• convergents, because, 
as will be shown in Art. 506, each successive convergent is 
a nearer approximation to the true value of the continued 
fraction than any of the preceding convergents. 



501. To show that the convergents are alternately less and 
greater than the continued fraction. 

Let the continued fraction be a^ -j- 



The first convergent is a 1} and is too small because the part 

1 1 1 

••• is omitted. The second convergent is aH — , 

<h + <h + <h 

and is too great because the denominator a^ is too small. 

The third convergent is a x -\ , and is too small because 

ay] — is too great; and so on. 

When the given fraction is a proper fraction, a } = ; if 
in this case we agree to consider zero as the first conver- 
gent, we may enunciate the above results as follows : 

The convergents of an odd order are all less, and the con- 
vergents of an even order are all greater, than the continued 
fraction. 

502. To establish the law of formation of the successive 

convergents. 

Let the continued fraction be denoted by 

«2+ «8+ «4+ 

then the first three convergents are 

% a&t + 1 q 8 (q 1 q 2 -H)+ *i 
V a* ' Os-Oj + l * 



CONTINUED FRACTIONS. 408 

and we see that the numerator of the third convergent may 
be formed by multiplying the numerator of the second con- 
vergent by the third quotient, and adding the numerator 
of the first convergent ; also that the denominator may be 
formed in a similar manner. 

Suppose that the successive convergents are formed in a 
similar way ; let the numerators be denoted by p^ p^ p& •••, 
and the denominators by q Y , q* q 8 , • ••. 

Assume that the law of formation holds for the nth con- 
vergent ; that is, suppose 

The (n + l)th convergent differs from the nth only in 
having the quotient a n -\ in the place of a*; hence the 

(n + l)th convergent 



_(■ 
1 



«»H }Pu-l+Pn- 



<**+!/ q n +l(«»P»-l +Pn-2) + P*-l 



cWv 



= *"&•+***, by supposition. 

<*n+lQn + 0«-l 

If therefore we put 

Pn+l = <*>n+lPn +Pn-19 &.+1 — <K+lQn + Qn-» 

we see that the numerator and denominator of the (n + l)th 
convergent follow the law which was supposed to hold in the 
case of the nth. But the law does hold in the case of the 
third convergent, hence it holds for the fourth, and so on; 
therefore it holds universally. 

Ex. Reduce f f f to a continued fraction and calculate the successive 
convergents. 

ByArt.499,^ = 2 + -!-_LJL_i_!. 
J '313 6+1 + 1 + 11 + 2 

The successive quotients are 2, 6, 1, 1, 11, 2. 
The successive convergents are f , if, if-, f| , f f$, fjf . 



404 ALGEBRA. 

Explanation. With the first and second quotients take the first 
and second convergents, which are readily determined. Thus, in this 
example, 2 is the first convergent, and 2 -f | or ^ the second con- 
vergent. The numerator of the third convergent, 15, equals the numer- 
ator of the preceding convergent, 13, multiplied by 1, the third quotient, 
plus 2, the numerator of the convergent next preceding but one. The 
denominator is formed in a similar manner : thus 7 = 1x6 + 1. 

The fifth convergent = n ( 28 )+ 15 = §23 

ft 11(13)+ 7 150 

503. If the fraction is a proper fraction, we may consider 
zero as the first convergent, and proceed as follows : 

Reduce -fifr to a continued fraction, and calculate the suc- 
cessive convergents. 
Proceeding as in Art. 499, 

227)84(0 
00 

84)227(2 
168 

59)84(1 
59 
25)59(... 

We obtain Q+A-*^ljL*\. 

2+1 + 2+2+1 + 3+-2 

The successive quotients are 0, 2, 1, 2, 2, 1, 3, 2. 
Writing £ for the first convergent and arranging the work 
as show in the example of the preceding article, we have 

Quotients 02122 1 32. 

Convergents £ £, \, f, fa, tf> ffa, -fa. 

504. It will be convenient to call a n the nth partial quo- 
tient ; the complete quotient at this stage being 



«» + 



We shall usually denote the complete quotient at any stage 
by k. 



CONTINUED FRACTIONS. 405 

We have seen that 

Pn _ 0>nPn-l + Pn-% 

Let the continued fraction be denoted by x ; then x differs 
from. — only in taking the complete quotient k instead of the 
partial quotient a* ; thus 

505. To show that if ^ be the nth convergent to a continued 
fraction, then 

Pn<fn-\ — Pn-l<Jn = (— !)"• 

Let the continued fraction be denoted by 

<h + <h + a 4 + 
then i> w <fc-i— p w -i?»= (a* A-i+P«-*)ft.-i— Pu-i(M+-i+9+4) 

«(-l) , p n . 2 g w _ 8 -l) ll _8g n - 2 ), similarly, 



-(-^""■(Aft-^fti). 

But ftgx - jhfc =(Oi«2 + 1)— a, . (^ = 1 =(- l) f ; 
hence p.gr^ - j^g. = ( - 1)" 

When the continued fraction is less than unity, this result 
will still hold if we suppose that c^ = 0, and that the first 
convergent is zero. 

Note. When we are calculating the numerical value of the suc- 
cessive convergents, the above theorem furnishes an easy test of the 
accuracy of the work. 

Cor. 1. Each convergent is in its lowest terms; for if p n 
and q n had a common divisor it would divide p n g n .\ — p n -iQm 
or unity ; which is impossible. 



406 ALGEBRA. 

Cor. 2. The difference between two successive convergent* 
is a fraction whose numerator is unity, and whose denominator 
is the product of the denominators of these convergents; for 

*Pn Pn-l PnQn l~Pn-lQn * 



506. Each convergent is nearer to the continued fraction 
than any of the preceding convergents. 

Let x denote the continued fraction, and -A * -^-> -^~ 

Q* 9*+l <7n+2 

three consecutive convergents ; then x differs from only 

Qn+2 

in taking the complete (n -f 2)th quotient in the place of 
a w+2 > denote this by k ; thus 

kPn+i+P~ 



x = 



&&.+1 + Qn 



Xr ^Pn = KPn+ l Qn ~ Pn Qn+l) = % 

~ <2n Qn(Jcq n+l + q n ) QnQcq n+1 + q n )' 

Pn+1 Pn + iqn~Pnq n +l_ * 



and 42+1 ~ x = /7 _ - 

Now k is greater than unity, and q n is less than q n +i; hence 

on both accounts the difference between -^ and x is less 

Pn Qn+1 

than the difference between — and x; that is, every con- 

IB 

vergent is nearer to the continued fraction than the next 
preceding convergent, and therefore nearer than any pre- 
ceding convergent. 

Combining the result of this article with that of Art. 501, 
it follows that 

The convergents of an odd order continually increase, but 
are always less than the continued fraction ; 

The convergents of an even order continually decrease, but 
are always greater than the continued fraction. 

* The sign ^ means " difference between." 



CONTINUED FRACTIONS, ' 407 

507. To find limits to the error made in taking any con- 
vergent for the continued fraction. . 

■k©t 7T 9 7^~> 7T^~ be three consecutive convergents, and 

9n Qn+l Qn+2 

let k denote the complete (n + 2)th quotient; 
then % fc^i|l>. . 



X~~ — 



Qn q n Qcq n+1 + q n y 



u (*« + t) 



Now k is greater than 1, therefore the difference between 
the continued fraction x, and any convergent, — , is less than 



1 _„_._.., 1 



Q 



-, and greater than 



Again, since q n+t > g n , the error in taking -^ instead of x 



1 . , .... M 1 



a 



is less than — and greater than 



?■" ° 2fl» 



»+l 



508. From the last article it appears that the error in 

taking — ^ instead of the continued fraction is less than 

11 1 
, or — - -; that is, less than -; hence 

?»?«+i Qn(a n+ iq n + q n -i) a«+i?* 2 

the larger a n+1 is, the nearer does -^ approximate to the 

continued fraction ; therefore, any convergent which imme- 
diately precedes a large quotient is a near approximation to 
the continued fraction. 

Again, since the error is less than — -, it follows that in 

order to find a convergent which will differ from the con- 
tinued fraction by less than a given quantity -, we have 

a 

only to calculate the % successive convergents up to &-, where 
q n 2 is greater than a. q » 



408 ALGEBRA. 

509. The properties of continued fractions enable us to 
find two small integers whose ratio closely approximates to 
that of two incommensurable quantities, or to that of two 
quantities whose exact ratio can only be expressed by large 
integers. 

Ex. Find a series of fractions approximating to 3,14159. 
In the process of finding the greatest common measure of 14159 
and 100000, the successive quotients are 7, 15, 1, 25, 1, 7, 4. Thus 

3.14169 = 3 + — — - — — -. 

7+15+1+25+1+7+4 

The successive convergent^ are 

h V. HI. «l. — 

This last convergent which precedes the large quotient 25 is a very 

near approximation, the error being less than -, and there- 

** \ 25 x (113)2' 

fore less than -, or .000004. 

25 x (100) 2 

510. Any convergent is nearer to the continued fraction than 
any other fraction whose denominator is less than that of the 
convergent. 

Let x be the continued fraction, ^, ^l 1 two consecu- 

Qn Qn-1 

tive convergents, - a fraction whose denominator a is less 
than q n . 

If possible, let - be nearer to x than % then - must be 

s q n * 

nearer to x than ^5=1 [Art. 506] ; and since x lies between 
Pj± and -?2=1, it follows that - must lie between *2 and ?^=1* 

Qn Qn-l * Qn 2—1 

Hence 

*_*•=!<&_*»=!> that is <-^-; 

* Qn-1 Qn Qn-l QnQn-l 

Qn 

that is, an integer less than a fraction ; which is impossible. 
Therefore ^ must be nearer to the continued fraction than -• 

Qn 9 



CONTINUED FRACTIONS. 409 

EXAMPLES XLIII. a. 
Calculate the successive convergents to 

1 2+i-l- J- 111 

1+ 3+ 5+ 1+ 1+ 2* 

2 1 1 1 1 1 1 1 T 

' 2-f 2+ 3+ 1+ 4+ 2+ 6* 

3. 3+ -LJ-mi. 

^3 + i + 2+ 2+ 1+ 9 

Express the following quantities as continued fractions and find the 
fourth convergent to each : also determine the limits to the error made 
by taking the third convergent for the fraction. 

*• 1H- & ihtf- 8 - - 37 - 10 - - 3029 - 

*• Hh 7. fjfg. 9. 1.139. 11. 4.316. 

12. Find limits to the error in taking £ #} yards as equivalent to a 
metre, given that a metre is equal to 1.0936 yards. 

13. Find an approximation to 

1 + J- llll 



3+ 5+ 7+ 9+ 11 + 
which differs from the true value by less than .0001. 

14. Show by the theory of continued fractions that ffi differs from 
1.41421 by a quantity less than tttot* 

RECURRING CONTINUED FRACTIONS. 

511. We have seen that a terminating continued fraction 
with rational quotients can be reduced to an ordinary frac- 
tion with integral numerator and denominator, and there- 
fore cannot be equal to a surd ; but we shall prove that a 
quadratic surd can be expressed as an infinite continued 
fraction whose quotients recur. We shall first consider a 
numerical example. 

Ex. Express -^19 as a continued fraction, and find a series of frac- 
tions approximating to its value. 



v™ + * -g | V19-2 ^g 



6 



8 8 V 19 + 2 ' 



410 



ALGEBRA. 



V19 + 2 ^ 1 V19-3 _ 1 2 

6 5 V 19 + 3 

a/19 + 3 -_ 3 Vl9-3 _ 3 5 

2 2 V 19 + 3 

V^19 + 3 =1 V19-2 =1 3 

6 5 V 19 + 2 

yi9 + 2 = 2 yl9-4 = 2 1 

3 3 V!9 + 4 

Vl9 + 4 = 8+(Vl9-4) = 8+... 
after this the quotients 2, 1, 3, 1, 2, 8 recur ; hence 

^^4-f-L-i-J-JL-LJ-.... 

v ^2+ 1+ 3+ 1+ 2+ 8+ 

It will be noticed that the quotients recur as soon as we come to a 
quotient which is double the first. 

Explanation. We first find the greatest integer in -y/19 ; this is 4, 
and we write -y/19 = 4 + ( ^/19 — 4). We then express ^/19 — 4 as an 
equivalent fraction with a rational numerator. Thus 

/19 1 _ (Vl9-4)(yl9 + 4) _ 3 
Viy_4 ~ V19 + 4 - V19 + 4 



The work now stands 

V19'= 4 + 



V19 + 4 



= 4 + 



V19 + 4 
3 



We begin the second line with ^ , the denominator of this 

o 

complex fraction, which is itself a fraction with a rational denomi- 
nator. The greatest integer in this fraction is 2, and we write 

Vl9 + 4 ^ 2 V19-2 
3 3 

We then multiply numerator and denominator by the surd conjugate 

to -v/19 — 2, so that after inverting the result , we again begin 

v V 19 + 2 

a line with a rational denominator. The same series of operations is 

performed in each of the following lines. 

The first seven convergents formed as explained in Art. 502 are 

f h V. tt. tt. w. w- 



CONTINUED FRACTIONS. 411 

The error in taking the last of these is less than , and is 

1 i (326)3' 

therefore less than or , and still less than .00001. Thus 

(320) 2 102400' 

the seventh convergent gives the value to at least four places of deci- 
mals. 



512. Every periodic continued fraction is equal to one of 
the roots of a quadratic equation of which the coefficients are 
rational. 

Let x denote the continued fraction, and y the periodic 
part, and suppose that 

„ . 1 1 111 
x = a H • • • , 

6 + c+ h+ k+ g 

and y=m-\ ••• , 

n-f- u+ v+ y 

where a, b, c, ••• h, k, m, n, ••• u, v are positive integers. 

P P f 
Let ■£, — f be the convergents to x corresponding to the 

quotients h, k respectively; then since y is the complete 

quotient, we have x = ^~r- — — ; whence y = ~ — —.* 
^ ' q'y + q q'x—p' 

T T f 

Let -, -7 be the convergents to y corresponding to the 



8 



r f y + r 



quotients u, v respectively ; then y = -j 

if I 

Substituting for y in terms of x and simplifying, we obtain 
a quadratic of which the coefficients are rational. 

The equation s f y 2 -\-(s—r r )y—r = 0, which gives the value 
of y, has its roots real and of opposite signs ; if the positive 

p'v ~f~ p 

value of y be substituted in x = , , on rationalizing the 

qy ~r q a\ /g 

denominator the value of x is of the form — -£3t — f where 

C 

A, B, C are integers, B being positive since the value of y 

is real. 



412 ALGEBRA. 

Ex. Express 1+— -i- — ~-— as a surd. 

2+ 3+ 2+ 3+ 

Let x be the value of the continued fraction ; then 

3-1= A- 1 

2+ 3+(x-iy 
whence 2x 2 + 2se - 7 = 0. 

The continued fraction is equal to the positive root of this equation 

and is therefore equal to ^ — ^— 



EXAMPLES XLin. b. 

Express the following surds as continued fractions, and find the 
sixth convergent to each : 



1. v 3 - 


6. V 13 - 


11. 3^/5. 


14. 1 • 


2. V 5 - 


7. V 14 - 


12. 4^10. 


V33 


3. V 6 - 


8. V 22 - 


13. l • 


15. Vf- 


4. V 8 - 


9. 2^3. 


V21 


16. VA- 


5. vfll. 


10. V2. 







17. Find limits of the error when % 8 - is taken for ^/17. 

18. Find limits of the error when f £f is taken for y/28. 

19. Find the first convergent to ^/lOl that is correct to five places 
of decimals. 

20. Find the first convergent to ^/15 that is correct to five places 
of decimals. 

Express as a continued fraction the positive root of each of the 
following equations : 

21. z 2 +2x-l=0. 22. a 2 -4a;-3=0. 23. 7 » 2 -8 s-3^=0. 
24. Express each root of x 2 — 5 x + 3 = as a continued fraction. 

Ill 



25. Find the value of 3 + 



6+ 6 + 6+ 



26. Find the value of — — — — .... 

1+ 3+ 1+ 3+ 

27. Find the value of 3 + — — =^- 

1+ 2+ 3+ 1+ 2+ 3+ 

28. Find the value of 5 + — X l l 



\+ i+ 14. 104- 



CHAPTER XLIV. 
Summation of Series. 

513. Examples of the summation of certain series (Arith- 
metic and Geometric) have occurred in previous chapters. 
We will now consider methods for summing other series. 

514. Recurring Series. A series ^ + t^-h w 2 + ^8+ •••* 
in which from and after a certain term each term is equal to 
the sum of a fixed number of the preceding terms multiplied 
respectively by certain constants, is called a recurring series. 
A recurring series is of the 1st, 2d, or rth order, according 
as 1, 2, or r constants are required as multipliers. 

515. Scale of Relation. In the series 

1 + 2<c + 3s 2 + 4S 8 + 5a* 4 + ..., 

each term after the second is equal to the sum of the two 
preceding terms multiplied respectively by the constants 2 x 
and — x 2 ; these quantities being called constants because 
they are the same for all values of n. Thus 

5x* = 2x • 4s 3 +(- a 2 ) . 3^ 

that is, u 4 = 2 xu s — ofu^ 

and generally, when n is greater than 1, each term is con- 
nected with the two that immediately precede it by the 

equation 

u n = 2 xu n ^ - aft^.* 

or, u n — 2 xu n _i -f A n _2 = 0. 

In this equation the coefficients of u n , %„__!, and u n _^ taken 
with their proper signs, form what is called the scale oj 
relation. 

413 



414 ALGEBRA. 

Thus the series 

l + 2x + 3x* + 4x* + 5x 4 +»* 
is a recurring series in which the scale of relation is 

1 — 2a? + «* 

516. To find any term when the scale of relation is given. 

if the scale of relation of a recurring series is given, any 
term can be found when a sufficient number of the preceding 
terms are known. As the method of procedure is the same, 
however many terms the scale of relation may consist of, the 
following illustration will be sufficient : 

If 1 —px— qx 2 — rx* 

is the scale of relation of the series 

«o + <h x + <W? + <W? + ••• 

we have 

a n x* =px. • a^iX"' 1 + qx* • a n _&?-* + rx 3 • a n _&T-* 9 

or a H = pa n _i + qa n _ 2 + ra w _ 8 ; 

thus any coefficient can be found when the coefficients of the 
three preceding terms are known. 

517. To find the scale of relation. If a sufficient number 
of the terms of a series be given, the scale of relation may 
be found. 

Ex. Find the scale of relation of the recurring series 

2 -f bx + 13a 2 + 35 s 8 + 97 x* + 275a 5 + 793 x* + .... 

This is plainly not a series of the first order. If it be of the second 
order, to obtain p and q we have the equations 

13 = bp + 2q, and35 = 13i) + 5gr; 

whence p = 6, and q = — 6. By using these values of p and q, we can 
obtain the fifth and sixth coefficients ; hence they are correct, and the 
scale of relation is 

l-5& + 6a; 2 . 

If we could not have obtained the remaining coefficients with these 
values of p and q, we would have assumed the series to be of the third 
order, and formed the equations 

86 = 13p-f 5g + 2r, 
97 = 36j> + 130 + 6r, 
275 = 97p + 35g + 13r; 



SUMMATION OF SERIES. 415 

whence values for p, q, and r would have been obtained, and trial with 
the seventh and following coefficients Would have shown whether they 
were correct. 

518. If the scale of relation consists of 3 terms it involves 
2 constants, p and q ; and we must have 2 equations to deter- 
mine p and q. To obtain the first of these we must know at 
least 3 terms of the series, and to obtain the second we must 
have one more term given. Thus to obtain a scale of 
relation involving two constants we must have at least 4 
terms given. 

If the scale of relation be 1 — px — qx* — ra 8 , to find the 3 
constants we must have 3 equations. To obtain the first of 
these we must know at least 4 terms of the series, and to 
obtain the other two we must have two more terms given; 
hence to find a scale of relation involving 3 constants, at 
least 6 terms of the series must be given. 

Generally, to find a scale of relation involving m constants, 
we must know at least 2 m consecutive terms. 

Conversely, if 2 m consecutive terms are given, we may 
assume for the scale of relation 

1 —piX—pspt? —pi?? — ••• -*p m of*. 

519. The Sum of n Terms of a Recurring Series. The 

method of finding the sum is the same whatever be the 
scale of relation ; for simplicity we shall suppose it to con- 
tain only two constants. 
Let the series be 

and let the sum be S ; let the scale of relation be 1 — px— qx*\ 
so that for every value of n greater than 1, we have 

NOW S = «o + <*>l x + GfrB 8 + •'• + C^n-l^" 1 ) 

— pxS = — patfc — pcbiX* • — pa^^f 1 " 1 — pcin^x*, 

— qx*S = — qa&? — • • • — qa n _ipp- 1 — qa n __#? — qa^x"* 1 , 



416 ALGEBRA. 

Hence 

(l-px-qa^S=a^+(a 1 -pa )x^(pa n _ 1 +qa n _ 2 )af l --qa n _ l af^\ 

for the coefficient of every other power of x is zero in con 
sequence of the relation 

. O _ ao+(qi-p«o )a? (pa n -i + qa n -^ + gan-i*** 1 

Thus the sum of a recurring series is a fraction whose de- 
nominator is the scale of relation. 

520. If the second fraction in the result of the last article 
decreases indefinitely as n increases indefinitely, the formula 
for the sum of an infinite number of terms of a recurring 
series of the second order reduces to 

s __ cip + fa - pao)x 
1 — px — qx 2 

If we develop this fraction in ascending powers of x as 
explained in Art. 487, we shall obtain as many terms of the 
original series as we please ; for this reason the expression 

Oo + (^ — pa^x 
1 —px — qx 2 

is called the generating function* of the series. The summa- 
tion of the series is the finding of this generating function. 
If the series is of the third order, 

s _ «o + (<h — pa )x+ (a 2 —pai — gct^x* 

1 —px — qx 2 — TCB 8 

. From the result of Art. 519, we obtain 

l—vx — nj = «o + «i» + <W? + — 4- « n -i^ + 



px-gx 2 

' . tP«n-i + go^)** + qa—&+\ 
1 —px — qx 2 ' 



Sometimes called the generating fraction. 



SUMMATION OF SERIES. 417 

from which we see that although the generating function 

ao+(<h— P<h)v 
1 — px—qx 2 

may be used to obtain as many terms of the series as we 
please, it can be regarded as the true equivalent to the 
infinite series 

Oo -f- (hx + a&? + •••, 

only if the remainder 

(ff«n-i + gfl„-2K + g^n-ia**' 
1-px-qx* 

vanishes when n is indefinitely increased ; in other words 
only when the series is convergent. 

522. The General Term, When the generating function 
can be expressed as a group of partial fractions the general 
term of a recurring series may be easily found. 

Ex. Find the generating function, and the general term, of the 
recurring series 

1- 7a; -a 2 -43a* 

Let the scale of relation be 1 - px — qx 2 ; then 

-l + 7;>-g = 0, -43+p + 7g = 0; 

whence p = 1, q = 6 ; and the scale of relation is 

1-x-Qx 2 . 

Let 8 denote the sum of the series ; then 

fl=l-7&- x 2 -43aj*-... 

-xS= - x + 7x*+ «*+••• 

-6x*&= - 6a£ + 42a* +••• 

.-. (l-x-Qtf^S^l-Sx, 

s = l-8as . 

which is the generating function. 

1 ft T 

If we separate — — Into partial fractions, we obtain 

1 — x — Qx 2 

2 1 



l+2x 1 — 3a: 
2 k 



418 ALGEBRA. 

By actual division, or by the Binomial Theorem. 
2 



l + 2x 



= 2 [1 - 2x + (2x) 2 + (- l) r (2*) r + ...] 



- t-L- = -[1 + 3x + (3x) 2 + - + (3s) r + "0 

1 — oX 

Whence the (r + l)th, or general term, is 

[2(2') ( - l) r - 8 r ] x" = {( - l)^ 1 - 3Taf. 

EXAMPLES XLIV. a. 

Find the generating functions of the following series. 

1. 1 + 6x + 24x 2 +84x 8 + —. 

8. 2 + 2x-2x 2 + 6x 8 -14x 4 + —. 

3. 3-16x + 42x 2 -94x 8 +—. 

4. 2-6x + 4x 2 + 7x 8 -26x 4 + —. 

5. 4 + 6x + 7x 2 + 113*+ .... 

6. 1 + x + 2x2 + 20? + 3a* + 3s 5 + 4a! 8 + 4»* + .... 

7. 1 + 3x+7x 2 + 13x 8 + 21x* + 31x 6 +.... 

8. 1 -3x + 6x 2 -7x 8 + 9x 4 ~ llx& + .... 

Find the generating function and the general term in each of the 
following series: 

9. 1 + 5x + 9x 2 +18x 8 + —. 11. 2 + 3x + 5x 2 + 9x 8 + —. 
10. 2-x + 5x 2 -7x8+ .... 12. 7-6x + 9x 2 + 27x* + «... 
13. 3 + 6x + 14x 2 + 36x 8 + 98x 4 + 276x 5 + .... 

THE METHOD OF DIFFERENCES. 

523. Let u n denote some rational integral function of n, 
and let u v u^ u s , u 4 , ••• denote the values of u n when for n 
the values 1, 2, 3, 4, ••• are written successively. 

From the series u l} u% u s , u^ u 5 , • • • obtain a second series 
by subtracting each term from the term which immediately 
follows it. 

The series w 2 — u l9 % — u 2 , u 4 — w 3 , u 5 — u^ «•• thus found 
is called the series of the first order of differences, and may be 
conveniently denoted by Duj, Du 2 , Du Sf Du^ «... 

By subtracting each term of this series from the term that 
immediately follows it, we have Du 2 — Du h Du$ — Du^ 



SUMMATION OF SERIES. 419 

Du A — Du& ... which may be called the series of the second 
order of differences, and denoted by D 2 u l9 D 2 u 2 , A%> .... 

From this series we may proceed to form the series of the 
third, fourth, fifth, • • • orders of differences, the general terms 
of these series being A w r> D^i^ D^ • •• respectively. 

524. Any Required Term of the Series. From the law of 
formation of the series 



Mfc ih> ^ U 4> U 5> U 19 

Du lf Duz, Dtta, Du^ Du s , • 

A^u A^» A^sj A w * 
A^u A w » A^» ••• 



••• 



••• 



it appears that any term in any series is equal to the term 
immediately preceding it added to the term below it on the 
left. 

Thus u 2 = u 1 + Dut, and Du 2 = Du x + D 2 u Y . 

By addition, since u 2 + Du 2 = u 3 , we have 

u 8 = u x + 2 Dv^ + D 2 u v 

In an exactly similar manner by using the second, third, 
and fourth series in place of the first, second, and third, we 
obtain Du 3 = Du Y + 2 D 2 u x + A w i« 

By addition, since u 3 + Du^ = u^ we have 

u A = u Y -f 3 JDui + 3 A^i + A w i* 

So far as we have proceeded, the numerical coefficients 
follow the same law as those of the Binomial Theorem. 
We shall now prove by induction that this will always be 
the case. For suppose that 

u n+ i = ui + fdht l + n &~£) D 2 u x + ... + •CrD.fr + ... + A«i; 

then by using the second to the (n + 2)th series in the place 
of the first to the (n + l)th series we have 

#*W = -Dmi + »A*i + ^"^ A^i + - + "CUiAwi 

+ ••• + A+iUi. 



u M 



420 ALGEBRA. 

By addition, since u %+l + Bu n+1 = u n+29 we obtain 
•w» + 2=%+(n+l)Z)w 1 4- — +( n C r + w C r _,)2) r tt 1 + - +D n+l u 1 . 

Bat»ft + -CU = ( n "^ + 1 + l)x"g^«^x«CLi 

_ (n+l)n(n-l) - (^+T-r+l) _^ 
~ 1.2.3...(r-l)r ~" 0r * 

Hence if the law of formation holds for u H+1 it also holds 
for w B+2 , but it is true in the case of u^ therefore it holds for 
u 5 , and therefore universally. Hence 

= t*i +(n - l)Du x + ^.2 D * u * + - + D »-i u i- 

If we take a as the first term of a given series, d h d^ d s ... 
as the first terms of the successive orders of differences, any 
term of the given series is obtained from the formula 

a H = a + (n-l)d 1 + ( n - 1 )( n - 2 ) d t 

+ («-l)(n-2)(n-3 ) 4t+ . t<> 

525. The Sum of n Terms of the Series. Suppose the 
series u^ u^ U& ••• is the first order of differences of the 
series 

Vi, Va, v& V+, -.., 

then ^ = (v n+1 -v n ) + (v n - v^) H h (v 2 - v x )+ v„ 

identically ; 

.'• v»+i = w n + w n _i H h «, + w, + 1^. 

Hence in the series 

0, V* Vg, V4, v 5 - 
W D W fc W » M 4" # 

the law of formation is the same as in the preceding article ; 



SUMMATION OF SERIES. 421 

that is, t*i 4- «a + w 3 + ••• -+• u n 

If l£ 

If, as in the preceding article, a is the first term of a given 
series, d^ e^, d 3 , ••• £/ie % /?rs$ terms of the successive orders of 
differences, the sum of n terms of the given series is obtained 
from the formula 

c ~ i ^ (n — 1) , , n(n — l)(n — 2) , 

l£ l£ 

+ J>(*-l)(*-2)(>-8) <t+ ..; > 

I* 

Ex. 1. Find the 7th term and the sum of the first seven terms of 
the series 4, 14, 30, 52, 80, •••♦ . 

The successive orders of differences are 

10, 16, 22, 28, 

6, 6, 6, 

0, 0. 
Here n = 7, and a = 4. 

Hence, using formula, Art. 524, the 7th term 

- 4 + 6 • 10 + ?-!$ . 6 = 154. 
1.2 

Using formula, Art. 525, the sum of the first seven terms 

-7.4+^5. 10 + ^^.6 = 448. 
1.2 1.2-3 

Ex. 8. Find the general term and the sum of n terms of the series 

12, 40, 90, 168, 280, 432, .... 

The successive orders of difference are 

28, 50, 78, 112, 152, ... 

22, 28, 34, 40,... 

6, 6, 6, ... 

0, 0, ... 

Hence the nth term [Art. 524] 

-1? \»Hn i )1 22(n-l)(n-2) Q(n - l)(n - 2)(n - 3) 

v J |2 [8 

= n 8 + 5 n 2 + 6 n. 



422 ALGEBRA. 

Using the formula for the sum of n terms we obtain 
S -13n I 28n(ii-l) , 22n(n-l)(n-2) 6 n(n-l)(n-2)(n-3l 

el (3n» + 26 n a + 69 » + 46) 
12 

= A n (« + l)(3n 9 + 23n + 46). 



». It will be seen that this method of summation 
will only succeed when the series is such that in forming 
the orders of differences we eventually come to a series in 
which all the terms are equal. This will always be the 
case if the nth term of the series is a rational integral 
function of n. 

PILES OF SHOT AND SHELLS. 

527. Square Pile. To find the number of shot arranged in 
a complete pyramid on a square base. 

The top layer consists of a single shot ; the next contains 
4 ; the next 9, and so on to n\ n being the number of layers : 
hence the form of the series is 

l 8 , 2 2 , 3 2 , 4 2 ,.-,n 2 . 
Series 1, 4, 9, 16, •••,«*. 

1st order of differences 3, 5, 7, 
2d order of differences 2, 2, 

3d order of differences 0. 

Substituting in Art. 525, we obtain 

xy_„ , »(*-!) n(n-l)(n-2) Q _ n(n+l)(2n+l) 
*~ n + 1.2 t6+ H2T3 J " 6 ' 



I. Triangular Pile. To find the number of shot arranged 
in a complete pyramid the base of which is an equilateral trv 
angle. 



SUMMATION OF SERIES. 423 

The top layer consists of a single shot ; the next contains 
3 ; the next 6 ; the next 10, and so on, giving a series of the 
form 

1, 1+2, 1+2+3, 1+2+3+ V" 

Series 1, 3, 6, 10, 

1st order of differences 2, 3, 4, 

2d order of differences 1, 1, 

3d order of differences 0. 

Hence 

, n(*-l) n(n-l)(n-2) n(n + l)(n+2) 

529. Rectangular Pile. To find the number of shot arranged 
in a complete pile the base of which is a rectangle. 

The top layer consists of a single row of shot. Suppose 

this row to contain m shot ; then the next layer contains 

2(m + 1) ; the next 3(ra + 2), and so on, giving a series of 

the form 

m, 2m + 2, 3m+6, 4m+12, ••• 

1st order of differences m+2, m+4, m+6, 

2d order of differences 2 2, 

3d order of differences 0. 

Now let I and w be the number of shot in the length and 
width, respectively, of the base ; then m = I — w + 1. 
Making these substitutions, we have 

n(n + l)(3*--tfl + l) 
6 



EXAMPLES XLIV. b. 

1. Find the eighth term and the sum of the first eight terms of the 
series 1, 8, 27, 64, 125, .... 

2. Find the tenth term and the sum of the first ten terms of the 
series 4, 11, 28, 56, 92, ... 



424 ALGEBRA. 

Find the number of shot in : 

8. A square pile, having 15 shot in each side of the base. 

4. A triangular pile, having 18 shot in each side of the base. 

6. A rectangular pile, the length and the breadth of the base con- 
taining 50 and 28 shot respectively. 

6. An incomplete triangular pile, a side of the base having 25 shot, 
and a side of the top 14. 

7. An incomplete square pile of 27 courses, having 40 shot in each 
side of the base. 

8. Find the ninth term and the sum of the first nine terms of the 
series 1, 3 + 6, 7 + 9 + 11, —. 

The numbers 1, 2, 3, ••• are often referred to as the natural 
numbers. 

9. Find the sum of the squares of the first n natural numbers. 

10. Find the sum of the cubes of the first n natural numbers. 

11. The number of shot in a complete rectangular pile is 24395 ; If 
there are 34 shot in the breadth of the base, how many are there in its 
length ? 

12. The number of shot in the top layer of a square pile is 169, and 
in the lowest layer is 1089 ; how many shot does the pile contain ? 

18. Find the number of shot in a complete rectangular pile of 15 
courses, having 20 shot in the longer side of its base. 

14. Find the number of shot in an incomplete rectangular pile, the 
number of shot in the sides of its upper course being 11 and 18, and 
the number in the shorter side of its lowest course being 80. 

Find the nth term and the sum of n terms of the series : 

15. 4, 14, 30, 52, 80, 114, .... 

16. 8, 26, 54, 92, 140, 198, .... 

17. 2, 12, 36, 80, 150, 252, .... 

18. 8, 16, 0, - 64, - 200, - 432, .... 

19. 30, 144, 420, 960, 1890, 3360, .... 

20. What is the number of shot required to complete a rectangular 
pile having 15 and 6 shot in the longer and shorter side, respectively, 
of its upper course ? 

21. The number of shot in a triangular pile is greater by 150 than 
half the number of shot in a square pile, the number of layers in each 
being the same : find the number of shot in the lowest layer of the 
triangular pile. 



SUMMATION OF SERIES. 425 

82. Find the number of shot in an incomplete square pile of 16 
courses when the number of shot in the upper course is 1005 less than 
in the lowest course. 

88. Show that the number of shot in a square pile is one-fourth the 
number of shot in a triangular pile of double the number of courses. 

INTERPOLATION. 

530. The process of introducing between the terms of a 
series intermediate values conforming to the law of the 
series is called interpolation. An important application is in 
finding numbers intermediate between those given in loga- 
rithmic and other mathematical tables. For this purpose 
we may employ the formula used in finding the nth term 
by the Differential Method, giving fractional values to n. 

Ex. Given log 40 = 1.6021, log 41 = 1.6128, log 42 = 1.6232, log 43 
= 1.6336, ...find log 40. 7. 

Series 1.6021, 1.6128, 1.6232, 1.6336, 

1st order of differences, .0107, .0104, .0103, 

2d order of differences, - .0008, - .0001, 

3d order of differences, + .0002. 

Substituting in formula of Art. 524, we have 

lo g 40.7 = 1.6021 + £(.0107)+ 1( - » ) (^L 3 ) 

7 / 3\/ 18 W . 0002 X 

+ M ioA ioA 15 i 

= 1.6021 + .00749 + .000031 + .000009 = 1.6096 +. 

Here log 40 is the first term (n = 1) ; log 41 is the second term 
(n = 2); hence in introducing the intermediate term log 40.7 we give 
to n a value 1.7. 

EXAMPLES XLIV. C. 

1. Given log 8 = 0.4771, log 4 = 0.6021, log 5 = 0.6990, log 6 
= 0.7782, ... ; find log 4.4. 

2. Given log 51 = 1.7076, log 52 = 1.7160, log 53 = 1.7243, log 54 
= 1.7324, ... ; find log 51.9. 

8. Given ^/h = 2.236, #6 = 2 - 449 > V 7 = 2645 > V 8 = 2 - 828 J find 
x/O, V7A, and V7?74. 

4. Given #51 = 3.7084, #52 = 8.7325, #53 = 3.7563, ... ; find 
#61718. 



CHAPTER XLV. 



Binomial Thbobbm. Ant Index. 

531. In Chapter xxxvii. we investigated the Binomial 
Theorem when the index was any positive integer ; we shall 
now consider whether the formulae there obtained hold in 
the case of negative and fractional values of the index. 

Since, by Art. 411, every binomial may be reduced to one 
common type, it will be sufficient to confine our attention to 
binomials of the form ( 1 + x)\ 

By actual evolution we have 

(l+x) i =VT : Fx=l+$x-$x> + &<x? 5 

and by actual division, 

(1 -*)-» = _ 1— = l + 2a>+3s 2 + 4a*+...; 

and in each of these series the number of terms is unlimited. 
In these cases we have by independent processes obtained 

an expansion for each of the expressions (1 + x)* and 
(1 + x)~*. We shall presently prove that they are only 
particular cases of the general formula for the expansion of 
(1 + x) n , where n is any rational quantity. 
This formula was discovered by Newton. 

532. Suppose we have two expressions arranged in 
ascending powers of x, such as 

l+^ + g^L^ a< i + «(«-lX^- 2 ) rf + ... (1), 
and i + ^ + '^^^+ ^-lX"- 2 ) ^^.-. <2> 

426 



AT 



BINOMIAL THEOREM. ANY INDEX. 42r 

The product of these two expressions will be a series in 
ascending powers of x ; denote it by 

1 + Ax + Bx* + Cx* + Dx 4 + —; 

then it is clear that A, B, (7, ••• are functions of m and n, 
and therefore the actual values of A, B y (?,••• in any par- 
ticular case will depend upon the values of m and n in that 
case. But the way in which the coefficients of the powers 
of x in (1) and (2) combine to give A, B, C, ••• is quite inde- 
pendent of m and n ; in other words, whatever values m and 
n may have A, B, C, "-preserve the same invariable form. 
If therefore we can determine the form of A, B, (?,••• for 
any value of m and n, we conclude that A, B, C, ••• will 
have the same form for all values of m and n. 

The principle here explained is often referred to as an 
example of " the permanence of equivalent forms " ; in the 
present case we have only to recognize the fact that in any 
algebraic product the form of the result will be the same 
whether the quantities involved are whole numbers, or frac- 
tions ; positive, or negative. 

We shall make use of this principle in the general proof 
of the Binomial Theorem for any index. The proof which 
we give is due to Euler. 

533. To prove the Binomial Theorem when the index is a 
positive fraction. 

Whatever be the value of m, positive or negative, integral or 
fractional, let the symbol f(m) stand for the series 

then/(n) will stand for the series 

^ ^ 1-2 ^ 1.2-3 ^ 

If we multiply these two series together the product will 
be another series in ascending powers of x, whose coeffU 
dents will be unaltered in form whatever m and n may be. 



/ 



428 ALGEBRA. 

To determine this invariable form of the product we may 
give to m and n any values that are most convenient ; for 
this purpose suppose that m and n are positive integers. 
In this caseffrn) is the expanded form of (1 + «)", and/( n ) 
is the expanded form of (1 + x) n ; and therefore 

Am) xf(n)=(l + x)~ x (1 + a?)" =(1 + »)«-+•, 

but when m and n are positive integers, the expansion of 
(1 + x) 9 ** is 

1 +(m + n)a? + ^ — ^ '\ — ^- *ar + •••. 

1 • 2 

This then is the form of the product of f(m) xf(n) in aU 
cases, whatever the values of m and n may be; and in 
agreement with our previous notation, it may be denoted by 
f(m -h ri) ; therefore for all values ofm and n 

f(m)xf(n)=f(m + n). 

Also A™)xA n )xf(P)=A™> + w)x/(p) 

—f(m + n +p), similarly. 

Proceeding in this way we may show that 

f(m) xfln) xf(p) '" *° & factors=/(w+tt+p+"' to 7c terms). 

Let each of these quantities, m, n, p, •••, be equal to -, 
where h and k are positive integers ; 



but since h is a positive integer, /(A) = (l + a?)*; 

.•.(i+«0* =/(|); 

but /(- ) stands for the series 

1 + * x+ k VL-l a > + ... i 



BINOMIAL THEOREM. ANY INDEX. 429 



.•.(1+^=1+^+^4^ 



Of/ -4- .*•- 

k ' 1.2 . ^ ' 

which proves the Binomial Theorem for any t positive frac- 
tional index. 

534. To prove the Binomial Theorem when the index is 
any negative quantity. 

It has been proved that 

/(m)x/(ri)=/(m + n) 

for all values of m and n. Replacing m by — n (where n 
is positive), we have 

/(- «) X /(*)=/(- » + n)=/(0)= 1, 
since all terms of the series except the first vanish ; 

buty(n)=(l +x) n for. any positive value of *; 

•'• (TT^ =/( - w) ' 

or (l+s)- n =/(-n). 

But/(— n ) stands for the series 

1 +(_ n)x + (- n)(- n - 1) ^ + ... . 

1 • z 

.-. (1 + »)- = 1 +(- n)a> + (~ »)(- » ~ 1) ^ + ... . 
which proves the Binomial Theorem for any negative index. 



It should be noticed that when x < 1, each of the 
series /(m), /(n), /(m -f n) is convergent, and /(m + n) is 
the true arithmetical equivalent of /(m) x f(n). But when 
a? > 1, all these series are divergent, and we can only assert 
that if we multiply the series denoted by /(m), by the 
series denoted by f(ri), the first r terms of the product will 
agree with the first r terms of f(m + n), whatever finite 
value r may have. 



i 



CHAPTER XLVL 

Exponential and Logarithmic Series. 

>. The advantages of common logarithms have been 
explained in Art. 438, and in practice no other system is 
nsed. But in the first place these logarithms are calculated 
to another base and then transformed to base 10. 

In the present chapter we shall prove certain formulae, 
known as the Exponential and Logarithmic Series, and give a 
brief explanation of the way in which they are used in con- 
structing a table of logarithms. 

537. To expand a* in ascending powers of jr. 
By the Binomial Theorem, if n > 1. 



H)" 



=l+nx l+vxj™- 1 ) . J, ns(«s-l) (n»-2) . 1 . 



x 



u -i) j.w. j 



=l + aH — Ntj — + — j3 H-— . . . (1). 

By putting x = 1, we obtain 



( 1+ ;)"= 1+1+ TT+MM+™ 

- ( i+ r=i( i+ 0'}'- 



(2) 



480 



1 



EXPONENTIAL AND LOGARITHMIC SERIES. 431 

hence the series (1) is the xth power of the series (2) ; that is, 

1+x+ I V + \ »A »/ 



12 ' [3 + 



and this is true however great n may be. If, therefore, n be 
indefinitely increased, we have 



l + x + 



ti + - = ( 1+1+ tt'")'' 



•••. 



The series 1 + 1+I + 1 + I+... 
is usually denoted by e; hence 

Write ex for $c, then 

If 12 

Now let e c = a, so that c = log,a; by substituting for c 

we obtain 

. h . i , ^(log««) s , a? (log, a) 8 f 

a'=l + a?log.a + <% + 13 +'"• 

This is the Exponential Theorem. 

538. The series 

+ IT |8 + |4 + ' 

which we have denoted by e, is very important, as it is the 
base to which logarithms are first calculated. Logarithms 
to this base are known as the Napierian system, so named 
after Napier, the inventor of logarithms. They are alsr 



432 ALGEBRA. 

called natural logarithms from the fact that they are the 
first logarithms which naturally come into consideration in 
algebraic investigations. 

When logarithms are used in theoretical work it is to 
be remembered that the base e is always understood, just as 
in arithmetical work the base 10 is invariably employed. 

From the series the approximate value of e can be deter- 
mined to any required degree of accuracy ; to 10 places of 
decimals it is found to be 2.7182818284. 

Ex. 1. Find the sum of the infinite series 

1 +1 + 1 + _!+.... 

I? 14 IS 

Wehave « = 1 + 1 + I + i + 1+ ...; 

II 15- li 

and by putting x = — 1 in the series for e*, we obtain 

-i i ill 1.1 

hence the sum of the series is \{e -he' 1 ). 

a — bx 



Ex. 2. Find the coefficient of & in the expansion of 
= (a — bx)e-* 



e* 



The coefficient required = ^~~ ' • a — \ '„ • b 

\r \r J= l 

= Lzl£(a + rb). 

lr 

539. To expand log, (1 + x) in ascending powers of x 
From Art. 537, 

a^l+.log.a+^I^.V^I^-f-.... 



EXPONENTIAL AND LOGARITHMIC SERIES. 433 

In this series write 1 + x for a; thus (1 + x) 9 
=l+i/log.(l+ a! ) + |{log.(l+a ! )} 2 +|{log.(l + »)} , + -.(l> 
Also by the Binomial Theorem, when x < 1 we have 
(l+*y=l+yx+y^a?+ ^-^- 2 ^ + - . (2). 

Now in (2) the coefficient of y is 

•+^+ fc i%i a *+ ( -ft:K-^ +-., 

/ij2 /jj3 /g4 

that is, a? — 77 + ^- — T"i • 

I 6 4 

Equate this to the coefficient of y in (1) ; thus we have 

log. (1 +»)=*- 2 + 3""i + '"' 
This is known as the Logarithmic Series, 

540. Except when x is very small the series for log, (1 + x) 
is of little use for numerical calculations. We can, however, 
deduce from it other series by the aid of which Tables of 
Logarithms may be constructed. 

541. In Art. 539 we have proved that 

log, (1 + «)=«-- + ! ; 

changing x into — a?, we have 

log,(l - x) = - x - - - . 

By subtraction, 

log -r^f= 2 ( a!+ f + f + -> 

Put 1 + x = n + 1 so that x = ; we thus obtain 

2f 



[ 



434 ALGEBRA. 

From this formula by putting n = 1 we can obtain log, 2. 
Again by putting n = 2 we obtain log, 3 — log, 2 ; whence 
log, 3 is found, and therefore also log, 9 is known. 

Now by putting n = 9 we obtain log, 10 — log, 9 ; thus the 
value of log, 10 is found to be 2.30258509 .... 

To convert Napierian logarithms into logarithms to base 

10 we multiply by -~, which is the modulus [Art. 

log, 10 1 

441] of the common system, and its value is fr^^z^z^: — > 

2.30258509 ••• 

or .43429448 ••• ; we shall denote this modulus by M. 

By multiplying the last series throughout by M we obtain 
a formula adapted to the calculation of common logarithms. 

Thus 

M log,(n + 1)— M log,n = 

(2n + 1^3(2n + l) 8 5(2w + l)*^ j* 
that is, log M (n + 1) — log 10 n = 

of M U M I 

(2n + l 3(2n + l) 3 5(2n + l) 5 > 

Hence if the logarithm of one of two consecutive num- 
bers be known, the logarithm of the other may be found, 
and thus a table of logarithms can be constructed, 

EXAMPLES XLVI. 



L Show that 



(l)r*=l -£ + *_*+-. 

|» I* I* 

(2) <i *~ 1 = l + l + — + 1 + 
(} 2e + [3 + [6 + (7 + 



•••- 



2. Expand log VI + x in ascending powers of as. 
8. Prove that log,2 = J + & + A + A + •••• 

4. Showthatlog 10 (^) = ^(* + f + f + ...). 

5. Provethat log J-±^- = 4a + 4x 2 + 2 rt 8 a^ + 20 «* + .... 

1 — 3x 3 

6. Show that if ae>l, logVx 2 - 1 = logs --L--JL- ^-i-»- 

2x 2 4s* Qsfi 



CHAPTER XLVIL 
Determinants. 

Consider two homogeneous linear equations 

a^x + by = 0, 

a>& + b$ = ; 

multiplying the first equation by b 2 , the second by b h sul> 
tracting and dividing by x, we obtain 

aj) 2 — a $i = C^- 
This result is sometimes written 



a 2 



hi 

b 2 



= 0, 



and the expression on the left is called a determinant. It 
consists of two rows and two columns, and in its expanded 
form or development, as seen in the first member of (1), each 
term is the product of two quantities; it is therefore said 
to be of the second order. The line afi 2 is called the prin* 
cipal diagonal, and the line bia 2 , the secondary diagonal 

The letters a lf b x , a 2 , b 2 are called the constituents of the 
determinant, and the terms ajb^ ajb x are called the elements. 



THE VALUE OF THE DETERMINANT AFTER CERTAIN 

CHANGES. 



543. Since 


a 2 b 2 


= a x b 2 — 


■ C&2&1 = 


bi b 2 



it follows that the value of the determinant is not altered by 
changing the rows into columns, and the columns into rows. 

Again, it is easily seen that 



<h t>i 


— 


b x a x 


, and 


<h &i 


= — 


C&2 b 2 


<h h 




b 2 a 2 




a 2 b 2 




<h h 



435 



436 



ALGEBRA. 



that is, if we interchange two rows or two columns of the deter- 
minant, we obtain a determinant which differs from it only in 
sign. 



Let us now consider the homogeneous linear equa- 
tions 

<*& + b$ + c& = 0, 

«a» + b& + c& = 0, 

a& + b& + c# = 0. 

By eliminating x, y, z, we obtain 

<h (b& — b z cj) + b x (c&s — CgOa) + <h (aj> s — ajb^) = 0, 



b % % 


+ &i 


% a* 


+ Cl 


Oa b 2 


h <* 




Cs Og 




<h h 



or a x 



This is usually written 



= 0. 



= 0, 



Oj ?> 2 °2 

03 6 8 Ca 

and the expression on the left being a determinant which 
consists of three rows and three columns is called a deter- 
minant of the third order. 



By a rearrangement of terms, the expanded form 
of the above determinant may be written 



or 



hence 



«i|&a 


h 


+ a 2 


fi 


8 &1 


+ <h 


&i 


&s 


<<a Cs 




<% <a 




c, c, 




Oi bi c x 


=— 


Oi a$ a s 


• 




a% ©2 ^2 




b x b 2 & 8 






03 


63 ( 


<3 




C 


1 C 2 


<« 





that is, ^e value of the determinant is not altered by changing 
the rows into columns, and the columns into rows. 



DETERMINANTS. 



437 



Minors. From the preceding article, 



<h 


h 


<h 


h 


<h 


h 






= a. 



b 2 
b 8 






+ <h 



"3 






+ <h 



hi 

b 2 



c 2 



= «! 



& 2 <h 


— a» 


b x c x 


+ <h 


b x c x 


b* <* 




b a <* 




b 2 C2 



(!)• 



Also from Art. 544, 
= <h b 2 



a>i 


61 


Ci 


a 2 


&2 


% 


<h 


b* 


c* 






-h 



a 2 



c 2 



+ <h 



a* 



'& 



(2). 



We shall now explain a simple method of writing down 
the expansion of a determinant of the third order, and it 
should be noticed that it is immaterial whether we develop 
it from the first row or the first column. 

From equation (1) we see that the coefficient of any one 
of the constituents a l9 Oj, Ogis that determinant of the second 
order which is obtained by omitting the row and column in 
which it occurs. These determinants are called the minors 
of the original determinant, and the left-hand side of equa 
tion (1) may be written 

where A^ A 2 , A s are the minors of a„ a 2 , a 3 respectively. 
Again, from equation (2), the determinant is equal to 

where A l} B l9 C x are the minors of Oj, 6 b c x respectively. 
547. The determinant Oj b x c x 

&2 ®2 ^2 
0-8 *8 *V 

= «! (& 2 Cjj — b^) + b x (c^Os — Cga 2 ) + Ci (a^s — a s b 2 ) 
= — b x (a 2 c s — a 8 c 2 ) — a x (c 2 b s — C3&2) — <*i (Ms — M2) 5 



hence 



a x 


hi 


C\ 


= — 


bi 


a,! 


Ci 


a 2 


b 2 


<k 




b 2 


a 2 


<ht 


«8 


h 


c* 




b* 


<h 


Cs 



438 



ALGEBRA. 



Thus it appears that if two adjacent columns, or rows, oj 
the determinant are interchanged, the sign of the determinant 
is changed, but its value remains unaltered. 

If for the sake of brevity we denote the determinant 

Oi bi Ci 

C&2 b 2 Cg 

Oj b 8 Cb 

by (aficfij), then the result we have just obtained may be 
written 

(pia>&) = — (oi&a^). 

Similarly we may show that 

(ciaA) = — (a&bs) = + (afiiCs). 

548. Vanishing of a Determinant. If two rows or two col- 
umns of the determinant are identical the determinant vanishes. 

For let D be the value of the determinant, then by inter- 
changing two rows or two columns we obtain a determinant 
whose value is — D; but the determinant is unaltered; 
hence Z> = — D, that is D = 0. Thus we have the follow- 
ing equations, 

OxAx — a 2 A 2 + a%A z = 2>, 

b x A x — b*A 2 + &a4* = 0, 
C\A\ — c 2 A 2 -f- c%A.% = U. 

549. Multiplication of a Determinant. If each constituent 
in any row, or in any column, is multiplied by the same factor, 
then the determinant is multiplied by that factor. 



For 



ma x Ox Cj 
ma 2 b 2 C2 
mo^ b s <% 

= max • A x — ma 2 • A 2 + ma^ • A 3 = m(a 1 A 1 — a 2 A 2 + a^A^) ; 
which proves the proposition. 

Cor. If each constituent of one row, or column, is the 
same multiple of the correspondiug constituent of another 
row, or column, the determinant vanishes. 



DBTBEMINANTS. 



439 



560. A Determinant expressed as the Sum of Two Other 

Determinants. If each constituent in any row, or column, 
consists of two terms, then the determinant can be expressed 
as the sum of two other determinants. 

Thus we have 



a x + d x b\ Cj 


^ 


<h b\ Ci 


+ 


a\ b x C\ 


a 2 + d\ b 2 <% 




Oj & 2 ^ 




d 2 b 2 <% 


a 8 + d\ b 8 Cs 




a 8 b s Cg 




ds h <% 



for the expression on the left, 

= K + d^Ai - (a 2 + d\)A 2 + (a 8 + ds)A s 

= (a 1 A 1 — a 2 A 2 + a 3 A s ) + (d 1 A 1 — d^ + d\A^\ 

which proves the proposition. 

In like manner if each constituent in any one row, or 
column, consists of m terms, the determinant can be 
expressed as the sum of m other determinants. 

Similarly, we may show that 
<h + dx bx + ex C+ 



a 2 -\- ^2 

08+^8 

a 2 b 2 

«8 &3 



h + € s 



C2 



+ 



ax 
a 2 

«3 



e 2 



C2 



<*. 


&1 Cl 


+ 


* «i 


<h 


(2s 


&a c 2 




c? 2 e 2 


<* 


dg 


&8 <* 




a\ e$ 


% 



In general if the constituents of the three columns con- 
sist of m, n, p terms, respectively, the determinant can be 
expressed as the sum of mnp determinants. 



Ex. 1. Show that 



b + c a — 6 
c + a b — c 
a + b c — a 



a 
b 
c 



= 3a6c-a 8 -6 8 - A 



The given determinant 



b 


a 


a 


— 


b 


b 


a 


+ 


c 


a 


a 


— 


c 


b 


a 


c 


b 


b 




c 


c 


b 




a 


b 


b 




a 


c 


b 


a 


c 


c 




a 


a 


c 




b 


c 


c 




b 


a 


c 



440 



ALGEBRA. 



Of these four determinants the first three vanish, Art. 548; thus 
the expression reduces to the last of the four determinants ; hence 
its value 

= - {c(c* -ab)-b(ac- ft 2 ) + a(a* - be)} 

= 3a&c-a«-&»-c». 



Ex. 2. Find the value of 


67 


U 


> ! 


21 


. 










39 13 14 






81 24 26 




We have 


67 19 21 


— 


10 + 57 19 21 


= 


10 19 21 


+ 


57 19 21 


39 13 14 




+ 39 13 14 




13 14 




39 13 14 


,81 24 26 




9 + 72 24 26 




9 24 26 




72 24 26 


= 


10 19 21 
13 14 


— 


10 19 19 + 2 
13 13+1 




10 19 2 
13 1 








9 24 26 




i 


) ! 


24 


24 + 2 




9 24 2 





= 10 



13 1 
24 2 



+ 9 



19 2 
13 1 



= 20 -63 =-43. 



551. Simplification of Determinants. Consider the deter- 
minant 

<h+pt>i + Q<h &i <>i 
a 2 +pb 2 + q<h b 2 ^ 

<h+pbs + q<hi h <* 
as in the last article we can show that it is equal to 



<h 


h 


Ci 


Og 


*>2 


c 2 


a 3 


b 8 


Ps 



+ 



+ 



qcy 


h 


G\ 


q% 


h 


C 2 


q% 


h 


0s 



pb x b x Cx 
pb 2 b 2 Ca 
pb s b s Cg 

and the last two of these determinants vanish [Art. 549, 
Cor.]. Thus we see that the given determinant is equal to 
a new one whose first column is obtained by subtracting 
from the constituents of the first column of the original 
determinant equimultiples of the corresponding constituents 
of the other columns, while the second and third columns 
remain unaltered. 
Conversely, 



ill bi Ci 


— 


»i +Ph + qci 


h Cx 


Oi 2 o 2 c 2 




a>2 +pb 2 + go,, 


b 2 Ca 


G&3 t>3 C3 




a t +p& 8 + qcs 


h <% 



DETERMINANTS. 



441 



and what has been here proved with reference to the first 
column is equally true for any of the columns or rows; 
hence it appears that in reducing a determinant we may 
replace any one of the rows or columns by a new row or 
column formed in the following way : 

Take the constituents of the roto or column to be replaced, 
and increase or diminish them by any equimultiples of the 
corresponding constituents of one or more of the other rows or 
columns. 

After a little practice it will be found that determinants 
may often be quickly simplified by replacing two or more 
rows or columns simultaneously : for example, it is easy to 
see that 



ax+pbx 


b x - qc x 


<h 


= 


a x 


b x Cj 


a*+p&i 


&a— qCs 


c* 




<k 


b 2 c. 


<h+Ph 


h-QCa 


<* 




«3 


b 8 Cs 



but in any modification of the rule as above enunciated, care 
must be taken to leave one row or column unaltered. 

Thus, if on the left-hand side of the last identity the 
constituents of the third column were replaced by <% + ra l9 
Co + ra*, C3 + ra& respectively, we should have the former 
value increased by 

<h+pbi bi — qcx ra x 
Og +pb 2 b 2 — qc2 rat 

and of the four determinants into which this may be re- 
solved there is one which does not vanish, namely 

pb\ — q<h. Wi 
pb 2 — qc 2 ra^ 

ph — Q<* fa* 



Ex. 1. Find the value of 



29 26 22 
25 31 27 
63 64 46 



442 



ALGEBRA* 



The given determinant 



3 


26 


-4 


— - 


-6 31 


-4 




54 


-8 




-12 


1 


1 26 







-3 6 







1 


2 



=-3x4x 



1 26 1 

-2 31 1 

3 64 2 



= -12 I - 



= -12x 



1 26 1 

-3 6 

12 



3 5 
1 2 



= 132. 



Explanation. In the first step of the reduction keep the second 
column unaltered ; for the first new column diminish each constituent 
of the first column by the corresponding constituent of the second ; 
for the third new column diminish each constituent of the third col- 
umn by the corresponding constituent of the second. In the second 
step take out the factors 3 and — 4. In the third step keep the first 
row unaltered ; for the second new row diminish the constituents of 
the second by the corresponding ones of the first ; for the third new 
row diminish the constituents of the third by twice the corresponding 
constituents of the first. The remaining steps will be easily seen. 



Ex. 2. Show that 



a — b — c 2a 2a 

2b b-c-a 26 
2c 2c c — a — b 



= (a+6 + c)». 



The given determinant 

a+b+c a+6-fc a+b+c 
26 b-c-a 26 
2c 2c c—a—b 



= (a+6+c)x 



11 1 

26 6-c-a 26 
2c 2c c—a—b 



= (a + 6 + c) x 



10 

26 -6-c-a 

2c -c-a-6 



= (a + 6 + c) x 



— 6 — c — a 

— c — a — 6 



= (a+6 + c) 8 . 



Explanation. In the first new determinant the first row is the 
sum of the constituents of the three rows of the original determinant, 
the second and third rows being unaltered. In the third of the new 
determinants the first column remains unaltered, while the second and 
third columns are obtained by subtracting the constituents of the first 
column from those of the second and third respectively. The remain- 
ing transformations are sufficiently obvious. 



DETERMINANTS. 



443 



EXAMPLES XLVn. a. 



Calculate the values of the determinants : 



1. 



2. 



3. 



1 1 1 

35 37 34 
23 26 25 


13 16 19 

14 17 20 

15 18 21 


13 3 23 . 
30 7 53 
39 9 70 


a h g 
h b f 
g f c 


• 



5. 



& 



7. 



8. 



1 * — y 
— z 1 x 
y — x 1 

11 1 

1 1 + s 1 
1 1 l + y 
a — b & — c c — a 
b — c c — a a — b 
c — a a — b b— c 

b + c a a 

b c + a b 
c c a + b 



9. Without expanding the determinants, prove that 



a 


b 


c 


— 


X 


y 


z 




P 


Q 


r 





y 


b 


Q 


— 


X 


a 


P 




z 


c 


r 





Solve the equations : 



10. 



a 


a 


X 


m 


m 


m 


b 


X 


b 



= 0. 



11. 



x y z 
p q r 
a b c 



16 -2s. 11 10 

11 -3a 17 16 

7-x 14 13 



Prove the following identities : 



12. 



13. 



14. 



b+c c + a a + b 
q + r r+p P + q 
y + z z + x x + y 



= 2 



a b o 
p q r 
x y z 



1 
1 
1 

1 
a 



a a 2 

b b 2 

c c 2 

1 1 

b c 



= (6 -*c)(c — d)(a— 6). 



= (6 -c)(c- «)(a - b)(a + b + c). 



a 8 6 8 c 8 



Calculate the value of the determinants : 



15. 



3 6 9 

4 7 10 
6 8 11 



16. 



1 

x 



1 

y 



l 



X 2 jfi S* 



= 0. 



444 



ALGEBRA. 



APPLICATION TO THE SOLUTION OF SIMULTANEOUS 
EQUATIONS OF THE FIRST DEGREE. 



The properties of determinants may be usefully 
employed in solving simultaneous linear equations. 
Let the equations be 

<h* + b$ 4- <a* + d x = 0, 

a& + b$ + <& + <*s = 0; 

multiply them by A x , — A& A a respectively and add the 
results, A h A*, A s being minors of a^ a* ty in the deter- 
minant 

D = a x b x Ci | . 

<h b 2 <* j 

«8 &8 <* I 

The coefficients of y and z vanish in virtue of the relations 
proved in Art. 548, and we obtain 

(a x A x — a 2 A 2 + a 8 A 3 )x + (d x A x — d a A 2 + oV4 8 ) = 0. 

Similarly we may show that 

(b x B x - b t/ B 2 + b 3 B s )y + (d x B x - d 2 B 2 + d s B 8 ) = 0, 
and (cxCi — C2C2 + CsC s )z + (d x C x — d 2 Ca + d 8 C 8 ) = 0. 
Now a x A x — 02^1 2 + a*^ = — (b x B x — b u B 2 + b 8 B 8 ) 

= CiCi — CgQ + CgQ = #> 
hence the solution may be written 





X 




d x 


h 


c x 


d 2 


b* 


Ca 


d 8 


h 


<* 





-y 




d x 


a x 


d| 


d 2 


Cly 


*2 


d s 


<h 


<i> 





X 




| d x 


a, 


&I 


d 2 


a* 


6, 


d s 


03 


«Si 





-1 


O! 


b x (^ 


02 


b 2 (^ 


Og 


h Ps 



or more symmetrically 
x -y 



-1 



b x c x d x 




Ox Ci d*! 




a x &j d\ 




a x b x Cj 


b 2 <% dj 




G&2 C2 (l 2 




0-2 O2 C*2 




«2 &2 c* 


ft* <% <^g 




Oj Cj "8 




«3 63 d$ 




a« &a C3 



{ 



DETERMINANTS. 



445 



Ex. Solve 



g + 2t/ + 3*-13a0, 
** + *+ *- 7 = 0, 
3* + 4* + 3#-21=:0. 



We have 
x 



2 3 


-13 


1 1 


- 7 


4 3 


-21 





-9 


1 
2 
3 


3-13 
1 - 7 
3-21 



* 




-1 


1 2 -13 

2 1-7 

3 4 -21 




12 8 

2 1 1 

3 4 3 



or 








x _ 
-8~ 


24 


-16 


— J 
8 


* 


whence x = 


It 


V 


= 3, 


and z 


= 2. 


• 







Explanation. The denominator of g is a determinant formed by 
taking the coefficients m each column except that of x. In the same 
manner for the denominators of v and z we omit the columns of 
coefficients of y and z. The last determinant is formed by taking the 
three columns of coefficients of x, y, and z. 



Suppose we have the system of four homogeneous 
linear equations 

OiX + bjy + c x z + d x u = 0, 
a& 4- &# + c& + d 2 u = 0, 

OgB + &«y + <& + d s u = o, 
cyv + &# + <v + d«tt = 0. 

From the last three of these, we have, as in the preceding 
article, 





X 




b t 


c* 


d 2 


6 8 


C8 


d 8 


h 


c 4 


d 4 



- -» _ 



— tt 



02 


<* 


<** 


08 


& 


<*8 


a 4 


<?4 


d< 



<h 


&2 <h 




<h 


b 8 ds 




a 4 


& 4 d 4 





a* b % c 2 
a* 6 4 c 4 



Substituting in the first equation, the eliminant is 



<*i 



©2 ^2 ^*2 
*8 ^8 ^ 

64 c 4 d 4 



-61 



&2 C$ (*2 

a A c A d. 



+ <>i 



Ot b 2 d? 
a* h ds 

a A ^4 &4 



-* 



C&2 &2 ^ 
«8 &8 <% 

a 4 64 C| 



=»a 



446 



AMEBRA. 



This may be more concisely written in the form 



Oi &i Cj dx 

a, 6, Cg d, 

<h h <* ^ I 

a 4 &4 c 4 d 4 ] 



= 0; 



the expression on the left being a determinant of the fourth 
order. 

Also we see that the coefficients of a^ b^ c^ d x taken 
with their proper signs are the minora obtained by omitting 
the row and column which respectively contain these con- 
stituents. 



t. More generally, if we have n homogeneous linear 
equations 

«i#i + b x x 2 + c^ -f- • •• -f £*£« = 0, 

a&i + b&i + c&z H h AW» = 0, 



«n»i + b n x 2 + c n ajg + — + &„a?. = 0, 

• 

involving n unknown quantities x lf a? 2 . z& ••• a? n , these quan- 
tities can be eliminated and the result expressed in the form 

a x b x c, • • • k x I = 0« 
a 2 &2 c 2 ••• ^2 



«n &n <>n—K 



The left-hand member of this equation is a determinant 
which consists of n rows and n columns, and is called a 
determinant of the nth order. 

The discussion of this more general form of determinant 
is beyond the scope of the present work ; it will be sufficient 
here to remark that the properties which have been estab- 
lished in the case of determinants of the second and third 
orders are quite general, and are capable of being extended 
to determinants of any order* 



DETERMINANTS. 447 

Signs of the Terms. Although we may always de- 
velop a determinant by means of the process described above, 
it is not always the simplest method, especially when our 
object is not so much to find the value of the whole deter- 
minant, as to find the signs of its several elements. 



The expanded form of the determinant 

a x bi Ci 

0*2 ®2 Cg 
% *3 C 3 

= Onbfo — a>J>&2 + a 2 ^i — aA^ + ^A^ — <^>2<h 5 

and it appears that each element is the product of three 
factors, one taken from each row, and one from each column ; 
also the signs of half the terms are + and of the other half 
— . When written as above the signs of the several elements 
may be obtained as follows. The first element a^tfe, in which 
the suffixes follow the arithmetical order, is positive; we 
shall call this the leading element ; every other element may 
be obtained from it by suitably interchanging the suffixes. 
The sign + or — is to be prefixed to any element according 
as the number of inversions of order in the line of suffixes 
is even or odd ; for instance in the element a 3 6 2 Ci, 2 and 1 
are out of their natural order, or inverted with respect to 3 ; 
1 is inverted with respect to 2 ; hence there are three inver- 
sions and the sign of the element is negative ; in the element 
ajbfo there are two inversions, hence the sign is positive. 

557. The determinant whose leading element is aj)^^ ••• 
may thus be expressed by the notation 

the S ± placed before the leading element indicating the 
aggregate of all the elements which can be obtained from it 
by suitable interchanges of suffixes and. adjustment of signs. 
Sometimes the determinant is still more simply expressed 
by enclosing the leading element within brackets; thus 
(o 1 5 2 C3^4 •••) is used as an abbreviation of 2 ± Oqb^A 

* The Greek letter Sigma. 



448 



ALGEBRA. 



Ex. In the determinant (ai^cs^s) what sign is to be prefixed to 
the element ai&sCiffces ? 

Here 3, 1, and 2 are inverted with respect to 4 ; 1 and 2 are inverted 
with respect to 3, and 2 is inverted with respect to 5 ; hence there are 
six inversions and the sign of the element is positive. 



558. Determinant of Lower Order. If in Art. 554, each 
of the constituents b 19 c D ••• k x is equal to zero, the deter- 
minant reduces to a Y A x ; in other words it is equal to the 
product of a x and a determinant of the (n — l)th order, and 
we easily infer the following general theorem : 

If each of the constituents of the first row or column of a 
determinant is zero except the first, and if this constituent is 
equal to m, the determinant is equal to m times that deter- 
minant of lower order which is obtained by omitting the first 
column and first row. 

Also since by suitable interchange of rows and columns 
any constituent can be brought into the first place, it follows 
that if any row or column has all its constituents except 
one equal to zero, the determinant can be immediately ex- 
pressed as a determinant of lower order. 

This is sometimes useful in the reduction and simplifica- 
tion of determinants. 



Ex. 



Find the value of 












30 


11 


20 


38 




6 


3 










11 


-2 


36 


8 




19 


6 


17 


22 



Diminish each constituent of the first column by twice the corre- 
sponding constituent in the second column, and each constituent of 
the fourth column by three times the corresponding constituent in the 
second column, and we obtain 



8 


11 


20 


5 





3 








16 


-2 


36 


9 


7 


6 


17 


4 



and since the second row has three zero constituents, this determinant 



= 3 







DETERMINANTS. 






449 


8 20 5 


= 3 


8 20 5 


= 3 


10 


= -3 


8 5 


= 9. 


15 36 9 




8 19 5 




8 19 5 




7 4 




7 17 4 




7 17 4 




7 17 4 









EXAMPLES XLVII. b. 



Calculate the values of the determinants : 



1. 


1 


1 1 


1 


• 




5. 


3 2 1 


4 


• 






1 


2 3 4 








15 29 2 14 






1 


3 6 10 








16 19 3 17 






1 


4 10 20 








33 39 8 38 




3. 


7 


13 10 6 


• 




6. 


1 + a 1 1 1 




5 


9 7 4 


m 






1 1+6 1 1 




8 


12 11 7 








1 1 1+c 1 




4 


10 6 3 








111 1+d 


3. 


a 
1 
1 
1 


111 
all 
1 a 1 
11a 


• 




7. 


x y z 
x z y 
y z x 
z y x 


• 


4. 





1 1 


1 


8. 


x y z 


• 




1 


6 + c a 


a 




—x c b 






1 


b c + a 


b 




—y—c a 






1 


c c a + b 




— z —6 —a 




Sc 


lve the equations : 








9. 


ix 


-5y + 2s = ll, 




12. 


x + y+ s = l, 




2x 


+ 3y- s = 20, 






ax + by+ cz = k, 




7x 


-4y + 3s = 33. 






a 2 x + b*y + c*z = &. 


10. 




* + £ + * = 7, 
2 3 4 




13. 


ax 4- by+ cz = k, 




x + 2y + 3s = 48, 






a 2 x + b 2 y + c 2 s = ifc 2 , 




X 

3 


-^ + * = 4. 
3 3 






a*x + 6 8 y + c*z = #*. 


11. 


2x 


- y+3z-2u 


= 14, 


14. 


x+ y+ z+ w = l, 




X 


+7y+ z— u 


= 13, 




ax+ 6y+ C2?-|- du = A:, 




Sx 


+ 6y — 5s + 3w 


= 11, 




a 2 x + ft 2 y + c 2 z + a"% = A; 2 , 




4x 


-3y + 2< 


z- 


-tt 


= 21. 




a 8 x + & 8 y + c 


5 *H 


-d 3 t 


i = lfl. 



2o 



CHAPTER XLVIII. 
Theory of Equations. 

559. General Form of an Equation of the nth Degree. Let 

p n a?'-\-p l &'~ 1 4-p 1 X"~ 1 -\- ••■ -t-P» i^+P. be a rational integral 
function of x of n dimensions, and let us denote it by f{x) ; 
then f(x) = is the general type of a rational integral equa- 
tion of the nth degree. Dividing throughout byj>o,we see 
that without any loss of generality we may take 

3f t + p 1 af- > +p 2 3r- 2 +-+p^ 1 x+p K = 

as the general form of a rational integral equation of any 
degree. 

Unless otherwise stated the coefficients p Ps, --P* will 
always be supposed rational. 

If any of the coefficients p„ p^p^, ---p, are zero, the equa- 
tion is said to be incomplete, otherwise it is called complete. 

560. Any value of x which makes f{x) vanish is called a 
root of the equation f(x)= 0. 

561. We shall assume that every equation of the form 
fiV)=0 has a root, real or imaginary. The proof of this 
proposition will be found in treatises on the Theory of 
Equations; it is beyond the range of the present work. 

562. Divisibility of Equations. If a is a root of the equa- 
" )= 0, then is f(x) exactly divisible by x — a. 

!e the first member by x — a until the remainder no 
contains x. Denote the quotient by Q, and the 
ler, if there be one, by R. Then we have 

f(x)=Q(x-a)+B = Q. 



THEORY OP EQUATIONS. 451 

Now since a is a root of the equation x = a, therefore 

Q(a-a)+.K = 0, 

hence E = ; 

that is, the first member of the given equation is exactly 
divisible by x — a. 



\. Conversely, if the first member of f(x) = is exactly 
divisible by x — a, then a is a root of the equation. 

For, the division being exact, 

Q(s-a)=0, 

' and the substitution of a for x satisfies the equation; hence 
a is a root. 

DIVISION BY DETACHED COEFFICIENTS. 

564. The work of dividing one multinomial by another 
may be abridged by writing only the coefficients of the 
terms. The following is an illustration. 

Ex. Divide 3x*-$x*-5x*+26x 2 -S3x+2G by 3»-2x 2 -4a;+8. 

1 + 2 + 4-8)3-8- 5 + 26-38 + 26(3-2 + 8 
3 + 6 + 12-24 



-2+ 7 + 
-2- 4- 


2-38 

8 + 16 


3- 

3 + 


6-17 + 26 
6 + 12-24 



-5+2 

Thus the quotient is 3sc 2 — 2 x + 3 and the remainder is — 5 x + 2. 

It should be noticed that in writing the divisor, the sign of every 
term except the first has been changed; this enables us to replace 
the process of subtraction by that of addition at each successive stage 
of the work. 



HORNER'S METHOD OF SYNTHETIC DIVISION. 



>. For convenience we again give an explanation of 
Horner's Method of c 7*ithetic Division, which has already 
been considered in Art. 63. 



A 



3- 


-8- 5 + 26-33 + 26 




6 + 12- 


-24 






- 4- 


- 8 + 16 








6 + 12- 


-24 



452 ALGEBRA. 

Let us take the example of the preceding article. The 
arrangement of the work is as follows : 

1 
2 
4 

-8 

'3-2+ 3+ 0- 5+ 2 

Explanation. The column of figures to the left of the vertical 
line consists of the coefficients of the divisor, the sign of each after the 
first being changed ; the second horizontal line is obtained by multi- 
plying 2, 4, — 8 by 3, the first term of the quotient. We then add 
the terms in the second column to the right of the vertical line ; this 
gives — 2, which is the coefficient of the second term of the quotient. 
With the coefficient thus obtained we form the next horizontal line, 
and add the terms in the third column ; this gives 3, which is the co- 
efficient of the third term of the quotient. 

By adding up the other columns we get the coefficients of the terms 
in the remainder. 



In employing this method in the following articles 
our divisor will be of the form x ±a, which enables us to 
still further simplify the work, as the following example 
shows : 

Ex. Find the quotient and remainder when 3 x 1 — sfi+Sl aj*+21 a&+£ 
is divided by x + 2. 

3 -1 31 21 6 | -2 

-6 14 -28 -6 12 -24 6 
3 -7 14 3 -6 12 - 3 11 

Thus the quotient is 8x* - 7 x 6 + 14«* + 3a* - 6x 2 + 12s - 3, and 
the remainder is 11. 

Explanation. The first horizontal line contains the coefficients of 
the dividend, zero coefficients being used to represent terms correspond- 
ing to powers of x which are absent. The divisor is written at the 
right of this line with its sign changed (Art. 564) and 1, the coefficient 
of a;, omitted. The first term of the third horizontal line, which con- 
tains the quotient, is the result of dividing 3, the coefficient of x 7 in 
the dividend, by 1, the coefficient of x in the divisor. This is then 
multiplied by the divisor — 2, and the resull 1,-6, the first term of 
the second horizontal line ; the sum of — 1 and — 6 gives — 7, the 



{ 



THEORY OF EQUATIONS, 458 

second term of the quotient, which multiplied by — 2 gives 14 for the 
second term of the second horizontal line ; the addition of 14 and 
gives 14 for the third term of the quotient, which multiplied by — 2 
gives — 28 for the third term of the second line, and so on. 

m 

567. Number of Roots. Every equation of the nth degree 
has n roots, and no more. 

Denote the given equation by/(#) = 0, where 

f(x) = x" +p l x n ~ 1 + j^*- 2 H hp n . 

The equation /(a?) = has a root, real or imaginary ; let this 
be denoted by a x ; then f(x) is divisible by x — a l9 so that 

f(x)=(x-a l )f 1 (x), 

where f x (x) is a rational integral function of n — 1 dimen- 
sions. Again, the equation f x (x) = has a root, real or im- 
aginary ; let this be denoted by a 2 ; then/^a;) is divisible by 

x — ay, so that 

fi(x)=(x-a 2 )f 2 (x), 

where f 2 (x) is a rational integral function of n — 2 dimensions. 

Thus f(x) = (x — a x ) (x — a 2 )f 2 (x). 

Proceeding in this way, we obtain 

/(») = (» — a^)(x — Oa)— (x — a n ). 

Hence the equation f(x) = has n roots, since f(x) vanishes 
when x has any of the values a lf a^ a 8 , • • • a n . 

Also the equation cannot have more than n roots ; for if x 
has any value different from any of the quantities a^ a^ 
<*&"• a n> a U the factors on the right are different from zero, 
and therefore f(x) cannot vanish for that value of x. 

In the above investigation some of the quantities a^ a 2 , 
Og, • • • a n may be equal ; in this case, however, we shall sup- 
pose that the equation has still n roots, although these are 
not all different. 

568. Depression of Equations. If one root of an equation 
is known it is evident from the preceding paragraph that we 
may by division reduce or depress the equation to one of the 
next lower degree containing the remaining roots. So if k 



454 ALGEBRA. 



roots are known we may depress the equation to one of the 
(n — A;)th degree. All the roots but two being known, the 
depressed equation is a quadratic from which the remaining 
roots are readily obtained. 



Formation of Equations. Since /(aj)=(x— a$ (x— a 2 ) 
• ••(» — a n ) [Art. 567], we see that an equation may be formed 
by subtracting each root from the unknown quantity and placing 
the continued product of the binomial factors thus formed equal 
toO. 

Ex. Form the equation whose roots are 1, — 2, and J. 

(x-l)(s + 2)(x-i) = 0; 

.-. 2x 8 + x 2 -5x + 2 = 0. 

EXAMPLES XLVHI. a. 

1. Show that 4 is a root of x 8 - 5x 2 - 2x + 24 = 0. 

2. Show that + 8 is a root of x 8 - 7x 2 + 7x + 15 = 0i 

3. Show that - \ is a root of 6x 8 + 17x 2 - 4x - 3 = 0. 

4. Show that | is a root of 10x 8 - 3 x 2 - 9x + 4 = 0. 

5. One root of X s + 6x 2 - 6x - 63 = is 3 ; find the others. 

6. One root of x 8 - 23 x 2 + 166 x - 378 = is 7 ; find the others. 

7. One root of x 8 — 2 x 2 + 6 x — 9^ 7 = is f ; what are the others ? 

8. Two roots of x 4 - 15x 2 -f 10s + 24 = are 2 and 3 ; find the 
others. 

9. Tworootsof x 4 - 3X 8 -21x 2 + 43x + 60 = are3 and 6; find 
the others. 

10. One root of x 3 + 2 ax 2 -f 5 a 2 x -f 4 a 8 = is — a ; what are the 
others ? 

11. Form the equation whose roots are — 1, — 2, and — 6. 

12. Form the equation whose roots are — 2, — 3, + £, and — \. 

570. Relations between the Roots and the Coefficients. Let 

us denote the equation by 

and the roots by a, b, c, • • • k ; then we have identically 

a" +.p 1 a n - 1 +p 2 af 1 - 2 H \-Pn-&+p n 

= (a? — a)(a? — b)(x — <;)••• (a? — ft); 



THEORY OF EQUATIONS. 455 

hence, by multiplication, we have 

x n +p l x n ~ l + p& n - 2 H hl?„-i» +p n 

= a>» - S^- 1 + S^*- 2 + (- 1)"- 1 ^.!* + (- l) w # n . 

Equating the coefficients of like powers of x in this 
identity, we have 

i> 2 = 8* 
-Ps = S* 

In which S t stands for the sum of the roots a, ft, c ••• k\ S $ 
stands for the sum of the products of the roots taken two 
at a time, and so on to S n , which equals the continued 
product of all the roots. That is: 

(1) The coefficient of the second term with its sign changed 
equals the sum of the roots. 

(2) The coefficient of the third term equals the sum of 
all the products of the roots taken two at a time. 

(3) The coefficient of the fourth term with its sign changed 
equals the sum of all the products of the roots taken three 
at a time, and so on. 

(4) The last term equals the continued product of all the 
roots, the sign being + or — according as n is even or odd. 

571. It follows that if the equation is in the general form : 

(1) The sum of the roots is zero if the second term is 
wanting. 

(2) One root, at least, is zero if the last term is wanting. 

572. The student might suppose that the relations estab- 
lished in the preceding article would enable him to solve 
any proposed equation ; for the number of the relations is 
equal to the number of the roots. A little reflection will 
show that this is not the case; for suppose we eliminate 
any n— 1 of the quantities a,b, c, ••• k, and so obtain an equa- 
tion to determine the remaining one; then since these quan- 
tities are involved symmetrically in each of the equations, 



456 ALGEBRA. 

it is clear that we shall always obtain an equation having 
the same coefficients ; this equation is therefore the original 
equation with some one of the roots a, b, c, ••• A; substituted 
for x. 
Let us take for example the equation 

a 8 +P1X 2 +p& + p s = 0; 

and let a, b, c be the roots ; then 

a + b + c = — pi, 

ab + ac + be = +p* 

dbc = — p s . 

Multiply these equations by a 2 , — a, 1 respectively and 
add; thus a 8 = — p x a 2 — p^a — p& 

that is, a 8 +p x a 2 + p& + jp s = 0, 

which is the original equation with a in the place of x. 

The above process of elimination is quite general, and 
is applicable to equations of any degree. 

573. If two or more of the roots of an equation are con- 
nected by an assigned relation, the properties proved in Art. 
570 will sometimes enable us to obtain the complete solution. 

Ex. 1. Solve the equation 4 se 8 — 24 x 2 -f 23 x + 18 = 0, having given 
that the roots are in arithmetical progression. 

Denote the roots by a — &, a, a + b ; then the sum of the roots is 
3 a ; the sum of the products of the roots two at a time is 3 a 2 — b 2 ; 
and the product of the roots is a (a 2 — ft 2 ); hence we have the equations 

3a = 6, 3a 2 -6 2 = ^, a(a 2 -6 2 ) = -f; 

from the first equation we find a = 2, and from the second b = ± f , 
and since these values satisfy the third, the three equations are con- 
sistent. Thus the roots are — £, 2, f . 

Ex. 2. Solve the equation 24 X s — 14 x 2 — 63 « + 45 = 0, on© root 
being double another. 

Denote the roots by a, 2 a, 5 ; then we have 

Sa + b = fa 2a 2 + 3a6=-Y> 2a 2 6=--^. 
From the first two equations, we obtain 

8a 2 -2a-3 = 0; 
.•. a = } or — J, and b = — £ or $ } . 



THEORY OF EQUATIONS. 457 

It will be found on trial that the values a = — J, b = \ } do not 
satisfy the third equation 2 a 2 b = — *fi ; hence we are restricted to the 
values a = f, b = — J. 

Thus the roots are J, J, — J. 

574. Although we may not be able to find the roots of 
an equation, we can make use of the relations proved in 
Art. 570 to determine the values of symmetrical * functions 
of the roots. 

Ex. Find the sum of the squares and of the cubes of the roots of the 
equation x 8 — px 2 4- qx — r = 0. 

Denote the roots by a, 6, c ; then a + b + c=p, bc + ca + ab = q» 
Now a 2 + & 2 + c 2 =(a 4- b + c) 2 - 2(bc + ca + ab) = p 2 - 2q. 
Again, substitute a, b, c for x in the given equation and add ; thus 

a s + 58 + c s - p (a 2 + ft 2 + c 2 )+ g(a + 6 + c)- 3r = ; 

/. a 8 + 6 8 + c* = p(p 2 -2q)-pq + 3r=pfi-Spq + Zr. 

EXAMPLES XL VIII. b. 
Form the equation whose roots are : 

1. h h ±V3. 2- 0, 0, 2, 2, -3, -8, 

3. 2, 2, - 2, - 2, 0, 5. 4. a + 6, a - 6, - a + 6, - a - 6. 

Solve the equations : 

5. a* - 16 x 9 + 86s 2 - 176* + 105 = 0, two roots being 1 and 7. 

6. 4x 8 + 16 x 2 — 9 as — 36 = 0, the sum of two of the roots being 
zero. 

7. 4 x 8 + 20 x 2 - 23 x + 6 = 0, two of the roots being equal. 

8. 3 x 8 — ' 26 x 2 + 52 x — 24 = 0, the roots being in geometrical pro- 
gression. 

9. 2 x 8 — x 2 — 22 x — 24 = 0, two of the roots being in the ratio of 
3:4. 

10. 24 x 8 + 46 x 2 + 9x — 9 = 0, one root being double another of 

the roots. 

11. 8x* - 2X 8 - 27 x 2 + 6x + 9 = 0, two of the roots being equal 
but opposite in sign. 

* A function is said to be symmetrical with respect to its variables 
when its value is unaltered by the interchange of any pair of them ; 
thus x + y + z, be + ca -f ab, x 8 + y s + z* — xyz are symmetrical func- 
tions of the first, second, and third degrees respectively. [See Art 
319.] 



458 ALGEBRA. 

12. 64 x»- 89 a£- 26 z+ 16 = 0, the roots being in geometrical pro- 
gression. 

13. 32x*- 48x 2 + 22s-3 = 0, the roots being in arithmetical pro- 
gression. 

14. 6a* - 29a* + 40a* -7x- 12 = 0, the product of two of the 
roots being 2. 

15. a* - 2a* - 21x 2 + 22 a; + 40 = 0, the roots being in arithmetical 
progression. 

16. 27a* - 196s 8 + 494x a - 620a; + 192 = 0, the roots being in 

geometrical progression. 

17. 18b 3 + 81 x 2 + 121a; + 60 = 0, one root being half the sum of 
the other two. 

18. Find the sum of the squares and of the cubes of the roots of 
a* + qx* + rx + * = 0. 

575. Fractional Roots. An equation whose coefficients are 
integers, . that of the first term being unity, cannot have a 
rational fraction as a root. 

If possible suppose the equation 

«" +Pv^' 1 +p<P?~ 2 + ••• +p n -&* +P*-ia? +p n = 
has for a root a rational fraction in its lowest terms, repre- 
sented by -. Substituting this value for x and multiplying 

through by b n ~ x , we have 

? +Pia n - 1 +p*r*b + — +Pn-iab n - 2 +Pn& n - 1 =0. 



Transposing, 

- 2l =p 1 a n ~ 1 +p<fT-*b + ... +p n -!ab»-* +p n b*-\ 
o 

This result is impossible, since it makes a fraction in its 
lowest terms equal to an integer. Hence a rational fraction 
cannot be a root of the given equation. 

576. Imaginary Roots. In an equation with real coeffi- 
cients imaginary roots occur in pairs. 

Suppose that f(x) = is an equation with real coefficients, 
and suppose that it has an imaginary root a + ib ; we shall 
show that a — ib is also a root. 



THEORY OF EQUATIONS. 459 

The factor of f(x) corresponding to these two roots is 

(x — a — ib)(x — a + ib), or (x — a) 2 + b*. 

Let f(x) be divided by (x — a) 2 -f 6 2 ; denote the quotient 
by Q, and the remainder, if any, by Ex + B* ; then 

f(x)= Q \(x - a) 2 + 6 2 | + Bx+ B'. 

In this identity put x = a + ib> then /($) = by hypothesis; 
also (a;-a) 2 + & 2 = 0; hence B (a + t&) + B' = 0. 
Equating to zero the real and imaginary parts, 

Ba + B' = 0, Bb = 0; 

and b by hypothesis is not zero, 

.-. JR = and .#' = 0. 

Hence /(a?) is exactly divisible by (x — a) 8 -f- W f that is, by 

(x — a— ib)(x — a -f- ifyi 

» 

hence x = a — ib is also a root. 

577. In the preceding article we have seen that if the 
equation f(x) = has a pair of imaginary roots a ± ib, then 
(x — a) 2 + 6 2 is a factor of the expression f(x). 

Suppose that a±ib, c±id, e ± ig, ••• are the imagi- 
nary roots of the equation f(x) = 0, and that * <£ (x) is the 
product of the quadratic factors corresponding to these 
imaginary roots; then 

+ ( x )={( x -a)* + b^{(x-c)' + (F}\(x-ey + g>\.... 

Now each of these factors is positive for every real value of 
x ; hence <j> (a?) is always positive for real values of x. 

578. As in Art. 576 we may show that in an equation with 
rational coefficients, surd roots enter in pairs; that is, if 
a -+• -y/b i s a T0 °t ^ nen a ~~ V^ * 9 a ^ so a r0 °k 

Ex. 1. Solve the equation 6 a* — IS a£ — 86a£ — £ + 8 = 0, having 
given that one root is 2 — y/S. 

* The Greek letter Phi. 



460 ALGEBRA. 

Since 2 — -y/3 is a root, we know that 2 + ^/3 is also a root, and 
corresponding to this pair of roots we have the quadratic factor 
* 2 -4* + l. 

Also 

6s* - 13 a* - 86aJ» - z + 3 = (a* - 4x + l)(6a? + 11 s + 3) ; 
hence the other roots are obtained from 

6z 2 + llx + 3 = 0, or (3s + l)(2a; + 3) = 0; 

thus the roots are — J, — J, 2 + ^/3» 2 — -y/3. 

Ex. 2. Form the equation of the fourth degree with rational coeffi- 
cients, one of whose roots is y/2 + V— 3. 

Here we must have y/2 -f V^3, v/2— v^3 as one pair of roots, 
and — v^2 + V— 3, — ^/2 — V— 3 as another pair. 

Corresponding to the first pair we have the quadratic factor 

a; 2 — 2y/2 x + 6, and corresponding to the second pair we have the 

quadratic factor 

aP + 2^2x4-6. 

Thus the required equation is 

(x 2 + 2y/2x + 5)(x 2 - 2v/2x + 5)= 0, 
or (x 2 + 5) 2 -8s 2 = 0, 

or 3 4 + 2& 2 + 26 = 0. 

EXAMPLES XLVIH. C. 
Solve the equations : 

1. 3aj* - 10x 8 + 4a; 2 - x - 6 = 0, one root being * -rV-8 

2. aj* - 36x 2 + 72x - 36 = 0, one root being 3-^3. 

3. x 4 + 4a 8 + 6x 2 + 2% - 2 = 0, one-root being - 1 + V^T. 

4. x* + 4b 8 + 6a; 2 +4x + 5 = 0, one root being V^T. 

TRANSFORMATION OF EQUATIONS! 

579. The discussion of an equation is sometimes simpli- 
fied by transforming it into another equation whose roots 
bear some assigned relation to those of the one proposed. 
Such, transformations are especially useful in the solution of 
cubic equations. 



THEORY OP EQUATIONS. 461 

580. To transform an equation into another whose roots are 
those of the original equation with their signs changed. 

Let f(x) = be the equation. 

Put — y for x\ then the equation /(—#)= is satisfied 
by every root of f(x) = with its sign changed; thus the 
required equation is /(— y)= 0. 

If the given equation is 

then it is evident that the required equation will be 

3f -Piy n ~ l +P2V n ~ 2 +(- ly^Pn-iV +(- l»n = 0; 

therefore the transformed equation is obtained from the 

original equation by changing the sign of every alternate term 

beginning with the second. 

Note. If any term of the given equation is missing it must be 
supplied with zero as a coefficient. 

Ex. Transform the equation sc 4 — 17 x 2 — 20 x — 6 = into another 
which shall have the same roots numerically with contrary signs. We 
may write the equation thus : 

s* + Ox 8 - 17 x a - 20 a; - 6 = 0. 

By the rule, we have 

aj* - Ox 8 - 17s 2 + 20s - 6 = 0, 

or jB*-17a; a + 20x-6 = 0. 

581. To transform an equation into another whose roots are 
equal to those of the original equation multiplied by a given 
factor. 

Let f(x) = be the equation, and let q denote the given 
quantity. Put y = qx, so that when x has any particular 

value, y is q times as large; then # = -> and the required 
equation is 

/V\ w /V\ n-1 /V\*~ 2 /V\ 

Multiplying by q n , we have 

sf +Piqir~ 1 +i? 2 g 2 2r~ a +— +Pn-i<T r ~ 1 y +Pn<f =& 



I 

{ 



462 ALGEBRA. 

Therefore the transformed equation is obtained from the 
original equation by multiplying the second term by the given 
factor^ the third term by the square of this factor^ and so on. 



The chief use of this transformation is to clear an 
equation of fractional coefficients. 

Ex. Remove fractional coefficients from the equation 

y 
Put x = - and multiply each term by g 8 ; thus 

By putting q = 4 all the terms become integral, and on dividing by 
2, we obtain 

583. To transform an equation into another whose roots exceed 
those of the original equation by a given quantity. 

Let f(x) = be the equation, and let h be the given quan- 
tity. Assume y = x+h, so that for any particular value 
of x, the value of y is greater by h ; thus x = y — h, and the 
required equation is f(y — h)=0. 

Similarly if the roots are to be less by h, we assume 
y = x — h, from which we obtain x = y + h, and the required 
equation is f(y -f h) = 0. 



If n is small, this method of transformation is 
effected with but little trouble. For equations of a higher 
degree the following method is to be preferred : 

Let f{x) = p Q X n + P^X"- 1 + p 2 X"-* -\ \-p n -lX+Pn, 

put x = y + h, and suppose that f(x) then becomes 

QoV n + ?i2/ n_1 + ?22T~ 2 + — + Qn-iV + q n > 
Now y = x — h; hence we have the identity 

= q (x - h) n + qjfx - h)"" 1 +•••+ g„-i(a - h)+ q n ; 

therefore q n is the remainder found by dividing f(x) by 
x — h ; also the quotient arising from the division is 

q^x - ft)"- 1 + qi (x - h)»- -* + ... + q^. 



J 



THEORY OP EQUATIONS. 463 

Similarly g w _! is the remainder found by dividing the last 
expression by x — h, and the quotient arising from the divi- 
sion is 

^»-»)^Hft(»-*)"" 8 +---+s--i; 

and so on. Thus q n , q n _ ly q n _ 2 > ••• may be readily found by 
Synthetic Division. The last quotient is Qq, and is obviously 
equal to jv 

Hence to obtain the transformed equation, 

Divide f(x) by x±h according as the roots are to be greater or 
less by h than those of the original equation, and the remainder 
will be the last term of the required equation. Divide the quo- 
tient thus found by x ± h, and the remainder will be the coeffi- 
cient of the last term but one of the required equation; and 
so on. 

Ex. Find the equation whose roots exceed by 2 the roots of the 
equation 

4 a* + 32 a; 8 + 83 a; 2 + 76a; +21 = 0. 

The required equation will be obtained by substituting x — 2 for x 
in the proposed equation ; hence in Horner's process we employ x + 2 
as divisor, and the calculation is performed as follows : 

4 32 83 76 21 [+2 

4 24 35 6 1 9 

4 16 3 |0 

4 8 |-13 

4 |0 

4 

Thus the transformed equation is 

4a* - 13a: 2 + 9 = 0, or (4a; 2 - 9)(a0 - 1)= 0. 

The roots of this equation are + f , — f , +1, — 1 ; hence the roots 
of the given equation are 

585. To transform a complete equation into another which 
wants an assigned term. 

The chief use of the substitution in the preceding article 
is to remove some assigned term from an equation. 



464 AUQBBRA. 

Let the given equation be 

Po«" + Pi* n ~ l +2>2*"~ 2 + — +P»-iV+Pu = 0; 
then if y = x — h, we obtain the new equation 

Po(y + *)• +i>i(y + A)- 1 +i>*(y + A)" -2 + — +jp» = 0, 

which, when arranged in descending powers of y, becomes 

+ { W(W [ ^ 1 W , +»-l)A*+ft}r- i + -=0l 

If the term to be removed is the second, we put npji+pi=0, 

so that h = ; if the term to be removed is the third, 

we put Pq 

n( V 1 W 2 +(n - l)ft* +1>2 = 0, 

and so obtain a quadratic to find h ; and, similarly, we may 
remove any other assigned term. 

Sometimes it will be more convenient to proceed as in the 
following example. 

Ex. Remove the second term from the equation 

pa? + qx 2 + rx + 8 = 0. 

q 
Let o, 6, c be the roots, so that a + 6 -r c = — -• Then if we 

increase each of the roots by h^-» in the transformed equation the 

Sp q q 
sum of the roots will be equal to — - + - ; that is, the coefficient of 

the second term will be zero. * p 

Hence the required transformation will be effected by substituting 

z—£- for a; in the given equation. 
op 

As the general type of a cubic equation can be reduced to a more 

simple form by removing the second term, the student should carefully 

notice that the transformation is effected by substituting x minus the 

coefficient of the second term divided by the degree of the equation, 

for x in the given eauation. 



THEORY OP EQUATIONS. 465 

586. To transform an equation into another whose roots are 
the reciprocals of the roots of the proposed equation. 

Let /(&) = be the proposed equation ; put y = -, so that 

1 /1\ x 

x = -; then the required equation is /[ - ) = 0. 

One of the chief uses of this transformation is to obtain 
the values of expressions which involve symmetrical func- 
tions of negative powers of the roots. 

Ex. If a, 6, c are the roots of the equation 

sc 8 — px* -f qx — r = 0, 

find the value of \ + \ + 4- 

a* b* & 

Write - for z, multiply by y 8 , and change all the signs ; then the 

y 



resulting equation 
has for its roots 



ry*-qy 2 +py-l = 0, 

1 1 I. 
a b c 



hence *s! = 2, sA = £. 

a r ab r 

... z-Utf-gJ* 

a 2 r 2 

587. Reciprocal Equations. If an equation is unaltered 

by changing x into -, it is called a reciprocal equation. 

x 

If the given equation is 

the equation obtained by writing - for x, and clearing of 
fractions, is 

PS? + Pn-ltf 1 "* 1 + Pn-*>?~* + ' ' * + P$? + P& + * = 0. 

If these two equations are the same, we must have 

Pn-l „ P*-2 ^ P* „ Pi ^ 1 , 

Pl= j-> Pi= p? " ,f p ~ 1= f: p ~ 1= f: p * = jJ 

* 2 - stands for the sum of all the terms of which - is the type, 
a a 

2h 



466 ALGEBRA. 

from the last result we have p n = ± 1, and thus we have 
two classes of reciprocal equations. 

(i.) If p n = 1, then 

Pl=Pn-h P*=Pn-2, Pt^Pn-to mmm 3 

that is, the coefficients of terms equidistant from the begin- 
ning and end are equal. 

(ii.) If p n = — 1, then 

Pi = — P«-l> Pi = — i>n-2> P$ = ~ i>«-8> — 5 

hence if the equation is of 2 m dimensions p m = — p m , or 
p TO = 0. In this case the coefficients of terms equidistant 
from the beginning and end are equal in magnitude and 
opposite in sign, and if the equation is of an even degree 
the middle term is wanting. 

588. Standard Form of Reciprocal Equations. Suppose 
that f(x) = is a reciprocal equation. 

If f(x) = is of the first class and of an odd degree it 
has a root — 1 ; so that fix) is divisible by x + 1. If <£(a?) 
is the quotient, then <f>(x) = is a reciprocal equation of the 
first class and of an even degree. 

If fix) = is of the second class and of an odd degree, 
it has a root -f- 1 ; in this case fix) is divisible by x — 1, and 
as before <f>(x) = is the reciprocal equation of the first 
class and of an even degree. 

If fix) = is of the second class and of an even degree, 
it has a root + 1 and a root — 1 ; in this case fix) is divisi- 
ble by x 2 — 1, and as before <£(#) = is a reciprocal equa- 
tion of the first class of an even degree. 

Hence any reciprocal equation is of an even degree with its 
last term positive, or can be reduced to this form; which may 
therefore be considered as the standard form of reciprocal 
equations. 

589. A reciprocal equation of the standard form can be reduced 
to an equation of half its dimensions. 

Let the equation be 

oar 2 ™ + bx**- 1 + co 2 "*-* H h leaf* H h ex 2 + bx + a = : 



THEORY OF EQUATIONS. 467 

dividing by of" and rearranging the terms, we have 

«(^+^)+^- 1 +^i)+{^- 2 +^i)+-+*=o- 

hence writing z for x + -, and giving to p in succession the 

x 

values 1, 2, 3 ••• we obtain 

rf + 3 = *-25 

or 

& + ^ = z(f-2)-z = *-$z\ 

a> 

ar 

and so on ; and generally x™ -\ — ■ is of m dimensions in z, 

af* 

and therefore the equation in z is of m dimensions. 

590. To find the equation whose roots are the squares of 
those of a proposed equation. 

Let/(a?)=0 be the given equation; by putting y = x 2 , 
we have x = ^/y, and therefore the required equation is 

/( Vy>= o. 

Ex. Find the equation whose roots are the squares of those of the 

equation 

x 8 + p\x 2 -f p& -f pz = 0. 

Putting x = y/y, and transposing, we have 

whence (y 2 + 2 p 2 y + j> 2 2 ) y = Pi 2 !/ 2 + 2 pip$y + jtf, 

or y 8 + (2j>a -i>i 2 )y 2 + (j* 2 - 2i>ip 8 )y -jtfsOt 



468 ALGEBRA. 



EXAMPLES XLVm. d. 

1. Transform the equation xP — 6x* -f Ox 8 — 9x 2 + 5oj — 1 = 
into another which shall have the same roots with contrary signs. 

2. Transform the equation 2b 8 — 4x 2 + 7a: — 3 = into another 
whose roots shall be those of the first multiplied by 3. 

8. Transform the equation x 8 — 7x — 6 = into another whose 
roots shall be those of the first multiplied by — }. 

4. Transform the equation se 8 — 4 x 2 -f J a; — £ = into another 
with integral coefficients, and unity for the coefficient of the first term. 

5. Transform the equation 3 as* — Sa^ + x 2 — x-f 1 = into another 
the coefficient of whose first term is unity. 

6. Transform the equation x* + 10 x 8 -f 39 a? + 76 x + 65 = into 
another whose roots shall be greater by 4. 

7. Transform the equation x* — 12a* + 17a? -9x + 7 = Into 
another whose roots shall be less by 3. 

8. Diminish by 1 the roots of the equation 

2x*-13x 2 + 10x-19 = 0. 

9. Find the equation whose roots are greater by 4 than the corre- 
sponding roots of x* + 16 a 8 + 72 a 2 + 64 x - 129 = 0. 

10. Solve the equation 3x 8 -22x 2 -f 48x- 32 = 0, the roots of 
which are in harmonical progression. 

11. The roots of x 8 — 11 x 2 + 36 a: — 86 = are in harmonical pro- 
gression ; find them. 

Remove the second term from the equations : 

12. afl-Qx^ + lOx-S^O. 

13. x* + 4x 8 + 2x 2 -4x-2 = 0. 

14. Transform the equation x 8 — - — =0 into one whose roots 

4 4 

exceed by $ the corresponding roots of the given equation. 

15. Diminish by 3 the roots of the equation 

a£-4a* + 3x 2 -4x + 6 = 0. 

16. Find the equation each of whose roots is greater by unity than 
a root of the equation X s — 5 x 2 + 6 x — 3 = 0. 

17. Find the equation whose roots are the squares of the roots of 

x* + x 8 + 2x 2 + x+l = 0. 

18. Form the equation whose roots are the cubes of the roots of 

x* + 3x 2 + 2 = 0. 



THEORY OP EQUATIONS. 469 

If a, 6, c are the roots of z* + qx + r = 0, form the equation whose 
roots are 

19. kcr\ Jdr\ Jfcc-i. 21. ±±±, c -±±, 2_±i. 

a 2 ' ft 2 c 2 

80. 62c2 , cV , a262 . 83. 6c + 1, ca + i, aft +1 

a o c 

Solve the equations : 

23. 2x*'+a; 8 -6a; 2 + ic + 2 = 0. 

24. «*-10x 8 + 26a?-10aj + l=0. 

25. a^-Sa^ + Oo^-O^ + Saj-lrrO. 

26. 4a*-24a* + 57x*-73s8 + 57a*-24a; + 4 = 0. 



DESCARTES' RULE OF SIGNS. 

591. When each term of a series has one of the signs + 
and — before it, a continuation or permanence occurs when 
the signs of two successive terms are the same : and a change 
or variation occurs when the signs of two successive terms 
are opposite. 



Descartes 9 Rule. In any equation, the number of pos- 
itive roots cannot exceed the number of variations of sign, and 
in any complete equation the number of negative roots cannot 
exceed the number of permanences of sign. 

Suppose that the signs of the terms in a multinomial are 

4-H 1 1 1 ;we shall show that if 

this multinomial is multiplied by a binomial whose signs 

are H , there will be at least one more change of sign in 

the product than in the original multinomial. 

Writing only the signs of the terms in the multiplica 
tion, we have 

+ + + + - + - 

+ - 

+ + + + - + - 

4- + - + + + - + - + 



470 ALGEBRA. 

a double sign, spoken of as an ambiguity, being placed wher- 
ever there is a doubt as to whether the sign of a term is 
positive or negative. 

Examining the product we see that 

(i.) An ambiguity replaces each continuation of sign in 
the original multinomial ; 

(ii.) The signs before and after an ambiguity or set of 
ambiguities are unlike ; 

(iii.) A change of sign is introduced at the end. 

Let us take the most unfavorable case and suppose that 
all the ambiguities are replaced by continuations ; from (ii.) 
we see that the number of changes of sign will be the same 
whether we take the upper or the lower signs ; let us take 
the upper ; thus the number of changes of sign cannot be 
less than in 

+ + + — -- + - + -+, 

and this series of signs is the same as in the original multi- 
nomial with an additional change of sign at the end. 

If then we suppose the factors corresponding to the nega- 
tive and imaginary roots to be already multiplied together, 
each factor x— a corresponding to a positive root introduces 
at least one change of sign ; therefore no equation can have 
more positive roots than it has changes of sign. 

To prove the second part of Descartes' Rule, let us sup- 
pose the equation complete and substitute — y for x ; then 
the permanences of sign in the original equation become 
variations of sign in the transformed equation. Now the 
transformed equation cannot have more positive roots than 
it has variations of sign, hence the original equation cannot 
have more negative roots than it has permanences of sign. 

Whether the equation f(x) = be complete or incomplete 
its roots are equal to those of /(— x) but opposite to them 
in sign; therefore the negative roots of /(sc)=0 are the 
positive roots of /(— x)= ; but the number of these posi- 
tive roots cannot exceed the number of variations of sign in 
/(—a); that is, the number of negative roots of /(a?)=0 
cannot exceed the number of variations of sign in /(— »). 



THEORY- OF EQUATIONS. 471 

We may therefore enunciate Descartes? Rule as follows: 

An equation f(x) = cannot have more positive roots than 
there are variations of sign in f(x), and cannot have more 
negative roots than there are variations of sign in /(—a?). 

Ex. Consider the equation x 9 + 5 £ 8 — X s + 7 x -f- 2 = 0. 

Here there are two changes of sign, therefore there are at most two 
positive roots. 

Again /( — x) = — x 9 + 6 X s + X s — 7 x + 2, and here there are three 
changes of sign, therefore the given equation has at most three nega- 
tive roots, and therefore it must have at least four imaginary roots. 

593. It is very evident that the following results are in- 
cluded in the preceding article. 

(i.) If the coefficients are all positive, the equation has 
no positive root ; thus the equation x s + x s + 2x-\-l = 
cannot have a positive root. 

(ii.) If the coefficients of the even powers of x are all of 
one sign, and the coefficients of the odd powers are all of 
the contrary sign, the equation has no negative root ; thus 
the equation 

x 7 + «* - 2x* + a 8 - 3a? + 7x - 5 = 

cannot have a negative root. 



examples XLvni. e. 

Find the nature of the roots of the following equations : 

1. s* + 2x 8 -13x 2 -14x + 24=0. 

2. x*-l0x* + 35x 2 - 60b + 24 = 0. 

3. 3x* + 12x 2 + 6a;-4 = 0. 

4. Show that the equation 2x 7 — x* + 4s 8 — 5 = has at least four 

imaginary roots. 

5. What may be inferred respecting the roots of the equation 
:c io_4 a j6 + x 4_2a;-3 = 0? 

6. Find the least possible number of imaginary roots of the equa- 
tion x 9 - x* + x* + x 2 + 1 = 0. 



THEORY OP EQUATIONS. 473 



• • 



If. 



={(« - ay + r(x - ay-*h + .-.}{*(*) + ft0'(aO+n7 *"(*) + ••}» 

lf- 

In this identity, by equating the coefficients of h, we have 
f(x) =r(x- a) r - l 4> (x) + (a? - a) r <£ '(»). 

Thus f (x) contains the factor x — a repeated r — 1 times ; 
that is, the equation f'(x) = has r — 1 roots equal to a. 

Similarly we may show that if the equation /(#)= has 
8 roots equal to b, the equation f f (x)= has $ — 1 roots equal 
to b ; and so on. 

From the foregoing proof we see that if f(x) contains a 
factor (x — a) r , then /'(#) contains a factor (x — a) r_1 ; and 
thus f(x) and /'(#) have a common factor (a — a)*"" 1 . There- 
fore if /(a?) and f(x) have no common factor, no factor in 
/(as) will be repeated ; hence the equation f(x) = has or 
has not equal roots, according as f(x) and f\x) Jiave or have 
not a common factor involving x. 

596. It follows that in order to obtain the equal roots of 
the equation f(x) = 0, we must first find the highest common 
factor of f(x) and /'(#), and then placing it equal to zero, 
solve the resulting equation. 

§ 

Ex. Solve the equation re*- 11 x*+ 44 a£- 76 x + 48 = 0, which has 
equal roots. 

Here /(«) = x* - 11a* + 44a; 2 - 76a; + 48, 

f(x) = 4a* - 33a; 2 + 88a; - 76 ; 

and by the ordinary rule we find that the highest common factor of 
f(x) and f(x) is * — 2 ; hence (x — 2) 2 is a factor of f(x); and 

/(*) = (x - 2) 2 (a; 2 - 7 x + 12) 

= (a;-2) 2 (a;-3)(a;-4); 
thus the roots are 2, 2, 3, 4. 



474 ALGEBRA. 



LOCATION OF THE ROOTS. 

597. If the variable x changes continuously from a to b the 
function f(x) will change continuously from f(a) to f(b). 

Let c and c + h be any two values of x lying between a 

\ and &. We have 

* 

' /(c + *) -/(«) = TO +£r<f>) + ». +p/*(c); 

[2 [n 



and by taking ft small enough, the difference between f(c+h) 
and /(c) can be made as small as we please ; hence to a small 
change in the variable x there corresponds a small change in 
the function f(x), and therefore as x changes gradually from 
a to 5, the function f(x) changes gradually from /(a) to /(by 



It is important to notice that we have not proved 
that f(x) always increases from /(a) to /(&), or decreases 
from /(a) to /(b), but that it passes from one value to the 
other without any sudden change ; sometimes it may be in- 
creasing and at other times it may be decreasing. 

599. If f(a) and f(b) are of contrary signs then one root of 
the equation f(x) = must lie between a and b. 

As x changes gradually from o to b, the function f(x) 
changes gradually from f(a) to /(&), and therefore must 
pass through all intermediate values ; but since /(a) and 
f(b) have contrary signs the value zero must lie between 
them ; that is, f(x) = for some value of x between a and 6. 

It does not follow that f(x) = has only one root between 
a and b ; neither does it follow that if /(a) and f(b) have 
the same sign f(x) = has no root between a and b. 

600. Every equation of an odd degree has at least one real 
root whose sign is opposite to that of its last term. 

In the function f(x) substitute for x the values + oo, 0, 
— oo, successively, then 

/(+«)= + 0O, /(0)=P W , /(- 00) =-0O. 



THEORY OF EQUATIONS. 476 

If p n is positive, then f(x) = has a root lying between 
and — oo, and if p n is negative f(x) = has a root lying 
between and + oo. 

601. Every equation which is of an even degree and has its 
last term negative has at least two real roots, one positive and 
one negative. 

For in this case 

/(+oo)= + co, f(0)=p n /(-oo)= + ooi 

but p n is negative ; hence f(x) = has a root lying between 
and + oo, and a root lying between and — oo. 

602. If the expressions f(a) and f(b) have contrary signs, 
an odd number of roots of f (jr) — will lie between a and b ; 
and if f(a) and f(b) have the same sign, either no root or an 
even number of roots will lie between a and b. 

Suppose that a is greater than 5, and that c, d, e, ••• k 
represent all the roots of f(x) = 0, which lie between a and 
b. Let <f> (x) be the quotient when f(x) is divided by the 
product (x — c)(x — d) (x — e) • • • (x — k) ; then 

f(x) = (x — c) (x — d) (x — e) ••• (x — k) <£ (x). 

Hence f(a) = (a — c) (a — d) (a — e) ••• (a — k) $ (a). 

/(&) = (b - c) (6 - d) (b - e) ... (b-k)<t> (b). 

Now <£ (a) and <£ (6) must be of the same sign, for other- 
wise a root of the equation <£ (x) = 0, and therefore of 
f(x) = 0, would lie between a and b [Art. 599], which is 
contrary to the hypothesis. Hence if /(a) and f(b) have 
contrary signs, the expressions 

(a — c)(a — d)(a — e) ••• (a — k), 
(b - c) (b - d) (6 - e) ... (6 - &) 

must have contrary signs. Also the factors in the first 
expressions are all positive, and the factors in the second 
are all negative ; hence the number of factors must be odd, 
that is, the number of roots c, d, e, • • • k must be odd. 

Similarly if /(a) and f(b) have* the same sign the num- 



476 ALGEBRA. 

ber of factors must be even. In this case the given con* 
dition is satisfied if c, d, e, ••• k are all greater than a, or 
less than 6; thus it does not necessarily follow that f(x) =0 
has a root between a and b. 



EXAMPLES XLVm. f. 

1. Find the successive derived functions of 2 a*— a£— 2a?+5x— 1. 

Solve the following equations which have equal roots : 

2. «*-9x 2 + 4x+12 = 0. 8. a^-ea^+^aP-lOa + SsO. 

4. a* -13a* + 67 a* -171 a 2 + 216s -108 = 0. 

5. a^-a^ + 4a; 2 -3x + 2 = 0. 

6. 8s* + 4a*-18a? + llx-2=:0. 

7. Show that the equation 10a£ — 17x 2 + x + 6 = has a root 
between and — 1. 

8. Show that the equation a* — 5a^ + 3a; 2 + 36a; — 70 = has a 
root between 2 and 3, and one between — 2 and — 3. 

9. Show that the equation jc 4 — 12a; 2 + 12x — 3 = has a root 
between — 3 and — 4, and another between 2 and 3. 

10. Show that x 5 + 6a* — 20s 2 - 19a; — 2 = has a root between 
2 and 3, and a root between — 4 and — 5. 

STURM'S THEOREM AND METHOD. 

603. In 1829, Sturm, a Swiss mathematician, discovered 
a method of determining completely the number and situa- 
tion of the real roots of an equation. 

604. Let f(x) be an equation from which the equal roots 
have been removed, and let f x (x) be the first derived func- 
tion. Now divide /(a?) by fi(x), and denote the remainder 
with its signs changed by f 2 (x). Divide f x (x) by f 2 (x) and con- 
tinue the operation, which is that of finding the H.C.F. of 
f(x) and f Y (x), except that the signs in every remainder are 
changed before it is used as a divisor, until a remainder is 
obtained independent of x ; the signs in this remainder must 
also be changed. No other changes of sign are allowed. 



THEORY OF EQUATIONS. 477 

The expressions f(x), fi(x), / 2 (#), •••/»(#) are called 
Sturm's Functions, 

Let Qu Q 2 > ••• Qn-i denote the successive quotients obtained; 
then the steps in the operation may be represented as 
follows • 

/(*) =Qi /i(») -/•(«), 
/»(*) =Q« /•(») -/4(*)» 

• ••••••••• 

/»- 2 («)= Q«-i/»-iO»)-/»0»). 

From these equalities we obtain the following: 

(1) Two consecutive functions cannot vanish for the same 
value of jr. 

For if they could, all the succeeding functions would 
vanish, including f n (x), which is impossible, as it is inde- 
pendent of x. 

(2) When any function except the first vanishes for a par- 
ticular value of jr, the two adjacent functions have opposite 
signs. 

Thus in f % (x) = Qf z (x)-f A (x) if f 8 (x) = 0, we have 

ft(x)=-f 4 (x). 

We may now state Sturm's Theorem. 

If in Sturm's Functions we substitute for x any particular 
value a and note the number of variations of sign; then assign 
to x a greater value 5, and again note the number of variations 
of sign; the number of variations lost is equal to the number 
of real roots of f(x) which lie between a and b. 

(1) Let c be a value of x which makes some function 
except the first vanish ; for example, f r (x), so that f r (c)= 0. 
Now when x = c, f r -i(x), and/^a) have contrary signs, and 
thus just before x—c and also just after x = c, the three 
functions f r -\(x) 9 f r (x), fr+i(n) have one permanence of sign 
and one variation of sign, hence no change occurs in the 



*.: i-' 



.1 .!l 

M 



478 ALGEBRA. 

number of variations when x passes through a value which 
makes a function except f(x) vanish. 

(2) Let c be a root of the equation J{x)=0 so thatf(c)=0. 
Let A be any positive quantity. 

Now /(c+*)=/(c)+V f (c) + ^/ , (c) + -.-, [Art. 594.] 

and as c is a root of the equation f(x) = 0, y(c) = 0, hence 

y(c+*) = v , (c)+^y fr (c)+.... 

If A be taken very small, we may disregard the terms con- 
taining its higher powers and obtain 

f(c + h) = hf'(c), 

and as h is a positive quantity, f(c -f h) and /'(c) have the 
same sign. That is, the function just after x passes a root 
has the same sign as f(x) at a root. 

In a like manner we may show that f(c — h) = — ft/'(c), 
or that the function just before x passes a root has a sign 
opposite to f(x) at a root. Thus as x increases, Sturm 1 s 
Functions lose one variation of sign only when x passes 
through a root of the equation f(x) = 0. 

There is at no time a gain in the number of variations of 
sign, hence the theorem is established. 

605. In determining the whole number of real roots of 
an equation f(x) = we first substitute — oo and then + oo 
for x in Sturm's Functions : the difference in the number of 
variations of sign in the two cases gives the whole number 
of real roots. 

By substituting — oo and for x we may determine the 
number of negative real roots, and the substitution of -f oo 
and for x gives the number of positive real roots. 

606. When -f- oo or — oo is substituted for x f the sign of 
any function will be that of the highest power of x in that 
function. 



THEORY OF EQUATIONS. 479 

607. Let us determine the number and situation of the 
real roots ofa^-hSa? 2 — 9a? — 4 = 0. 

Here./i(a?) = 3 a? 2 + 6x - 9. 

itow any positive factor may be introduced or removed in 
fi!iding^(aj),/ 8 (aj), etc., for the sign of the result is not affected 
by so doing ; hence multiplying the original equation by 3, 
we have 

3a? 2 + 6a?-9)3ar' + 9a? 8 -27a?-12(a? + l 

3^ + 60^- 9a? 

3a? 2 - 18a? -12 

3a? 2 + 6x- 9 

3) -24a;- 3 

— 8 a? — 1 .% f 2 (x) = 8 x + 1. 

8 a? + 1 )24 x 2 + 48 a? - 72 (3 x + 5 
24a? 2 + 3a? 

9 )45a?-72 
5a?- 8 

8 

40 a? -64 
40a? + 5 

-69 .\/ 8 (a?)=69. 
We therefore have 

X»)=« 8 + 3a? 2 -9a?-4, 
/x(a?)= 3a? 2 + 6x -9, 
/ a (a?)=8a? + l, 
/ 8 (»)=69. 

M fi(*) /*(*) fsW 
When a? = — oo we have - — + — + 3 variations. 

When a? = + oo we have + + + + no variations. 

When a? = we have — — + + 1 variation. 

Hence the number of real roots is 3, of which one*is posi- 
tive and two are negative. 

To determine the situation of these roots we substitute 
different numbers, commencing at and working in each 
direction, thus : 



.1 



480 







ALGEBRA 


L* 






/C») 


/i(*) 


m 


/.(») 




« = --QO 


— 


+ 


— 


+ 


3 variations. 


x = — 5 


— 


+ 


— 


+ 


3 variations. 


a? = — 4 


+ 


+ 


— 


+ 


2 variations. 


a? = -3 


+ 


± 


— 


+ 


2 variations. 


a? = -2 


+ 


— 


— 


+ 


2 variations. 


* = -l 


+ 


— 


— 


+ 


2 variations. 


a? = 


— 


— 


+ 


+ 


1 variation. 


a? = l 


— 


± 


+ 


+ 


1 variation. 


a> = 2 


— 


+ 


+ 


+ 


1 variation. 


a> = 3 


+ 


+ 


+ 


+ 


no variation. 


a? = oo 


+ 


+ 


+ 


+ 


no variation. 



Thus one root lies between — 4 and — 5 ; a second lies 
between and — 1, and the third lies between 2 and 3. 

EXAMPLES XLVIIL fir. 
Determine the number and situation of the real roots of : 

1. se 8 -4a; 2 -6x + 8=0. 6. a* - 4a; 8 + x 2 + 6x + 2 = 0. 

2. 2a* - lis 2 + 8& -16 = 0. 6. a*-a* + a;-l=0. 



3. x 8 -7x + 7 = 0. 



7. a£-9x 2 + 23a; -16 = 0. 



4. s 4 -4x* + 6x 2 -12a; + 2 = 0. 8. x 6 + s 8 - 23* + 2a;- 1 = 0. 



GRAPHICAL REPRESENTATION OF FUNCTIONS. 

Coordinates. 

608. Two lines drawn at right angles to each other as in 
Fig. 1 form a simple system of lines of reference. Their 

intersection, O, is called the 
origin. Distances from O along 
XX' are called abscissas; dis- 
-P tances from XX' on a line paral- 

i lei to TT 1 are called ordinates. 



\IU 



X'r 



-a 



a 



I 
\ 



O 



Fio. 1. 



-6 



P' 



609. Abscissas measured to 
the right of the origin are con- 
sidered positive, and to the left, 
negative. Ordinates measured 
above XX 1 are considered posis 
tive, and when taken below XX' 
are negative. 



i 



THEORY OF EQUATIONS. 



481 



610. The abscissa and ordinate of a point are called the 
co-ordinates of that point, and the lines XX and YY are 
called Co-ordinate Axes, Axis of X, and Axis of Y, or Axis 
of Abscissas, and Axis of Ordinates. 

611. Any point in the plane can be given by means of 
these co-ordinates : thus the point P of Fig. 1 is located by 
measuring the distance a to the right of on the axis of 
abscissas, and then taking a distance b vertically upwards. 

Since a and b can be either positive or negative, a point 
P is found by taking a- positive and b negative ; P" is found 
by taking a negative and b negative ; and P" 1 is found by 
taking a negative and b positive. 

Abscissas and ordinates are generally represented by x 
and y respectively. Thus for the point P, x = a, and y = b ; 
for P\ x = a, and y = — b, etc. 

612. Instead of writing "the point whose co-ordinates 
are 5 and 3," a more concise form is used : thus the point 
(5, 3) means that the point will be found by taking an 
abscissa of 5 units and an ordinate of 3. 

Locate the points (3, - 2) ; (5, 8) ; (- 4, 4) ; (- 8, - 3). 

GRAPH OF A FUNCTION. 

613. Let f(x) be any 
rational integral func- 
tion of a?, and let us 
place it equal to y. If 
we give a series of nu- 
merical values to x we 
can obtain corresponding 
values for y. Now lay- 
ing off the values of x 
as abscissas, and the cor- 
responding values of y 
as ordinates, we have a 
series of points which lie 
upon a line called the 
graph of the given func- 
tion. 




Fig. 2. 



482 



ALGEBRA. 



Ex. 1. Construct the graph of 2 x — 1. 

Let 2 x — 1 = y. Giving to x successive values, we obtain the 
corresponding values of y as follows: 

*" x = -2, y = -6. 

z = — 1, y = — 3. 

05 = 0, y = — 1. 

x=l, y=l. 

x = 2, y = 3. 

a; = 3, y = 6. 

Locating these points as 
explained in Art. 611 and 
drawing a line through 
A r them, we have in this case 
the straight line AB y Fig. 
2, as the graph required. 

Ex. 8. Plot the equation 
x 2 -2x-4. 

Putting y=x 2 — 2x— 4, we 
obtain the following values : 

x = — 2, y = 4. 
re = — 1, y = — 1. 
x = 0, y = — 4. 




35 = 1, 


y =- 5. 


a = 2, 


y = — 4. 


x = 3, 


y = -l. 


x = 4, 


y = 4. 



The points lie on the 
line ABC, Fig. 3, which is 
the graph of the given 
equation. 

Ex. 3. Plot the graph of 
x 8 — 2x. 

Assuming y = as 8 — 2 x, 
we have 

x = — 2, y = — 4. 
x = — 1, y = 1. 
x = 0, y = 0. 
x=l, y = — 1. 
Y* Fig. 4. x = 2, y = 4. 

The line ABCD, Fig. 4, is the required graph. • 

By taking other values between those assumed, we may locate 
the curve with greater precision. 



THEORY OF EQUATIONS. 483 

614. In giving values to x it is evident that the substitu- 
tion of a root gives #=0, that is, the ordinate, or distance 
from the Axis of Abscissas, is 0; hence where the graph 
cuts the Axis of X we have the location of a real root, and 
the graph will cross the Axis of X as many times as the 
equation has real and unequal roots. If the roots of the 
equation be imaginary, the curve will not touch the Axis 
of X 

EXAMPLES XL VIII. h. 

Construct the graphs of the following functions : 

1. 2x-3. 3. ^-6. 5. a^-2. 7. s*-3s 2 + 3. 

2. a^+2aj+l. 4. x?+x-l. 6. &-bx+Z. 8. a^+3x 2 +6ac-12. 

SOLUTION OF HIGHER NUMERICAL EQUATIONS. 

Commensurable Roots. 

615. A real root which is either an integer or a fraction 
is said to be commensurable. 

By Art. 582 we can transform an equation with fractional 
coefficients into . another which has all of its coefficients 
integers, that of the first term being unity : hence we need 
consider only equations of this form. Such equations can- 
not have for a root a rational fraction in its lowest terms 
[Art. 575], therefore we have only to find the integral roots. 
By Art. 570 the last term of f(x) is divisible by every 
integral root, therefore to find the commensurable roots of 
f(x) it is only necessary to find the integral divisors of the 
last term and determine by trial which of them are roots. 

616. Newton's Method. If the divisors are small numbers 
we may readily ascertain by actual substitution whether 
they are roots. In other cases we may use the method of 
Arts. 562 and 566 or the Method of Divisors, sometimes 
called Newton's Method. 

Suppose a to be an integral root of the equation 



484 ALGEBRA. 

By substitution we have 

a n +pia n - 1 + j> a a— * +— +p m _ia +p n = 0. 
Transposing and dividing throughout by a, we obtain 



Pn 

— = -p^j fta — -Aar- - a—, 



■ — • ** #»*-> #»»— 1 



in which it is evident that — must be an integer. Denot* 

ing — by Q and transposing — jv-i» 
a 

Q + Pn-i = Pi<* n ~* —Pi<**~* - <r~ l * 

Dividing again by a gives 
a 



= — ••• — Pid*-* — Pia""* — (T" 1 . 



Again, as before, the first member of the equation must 
be an integer. Denoting it by Q a and proceeding as before, 
we must after n divisions obtain a result 

Qn-l+Pl 1 

a - 1- 

Heuce if a represents one of the integral divisors of the 
last term we have the following rule : 

Divide the last term by a and add the coefficient of x to the 
quotient 

Divide this sum by a, and if the quotient is an integer add 
to it the coefficient of x 2 . 

Proceed in this manner, and if a is a root of the equation 
each quotient will be an integer and the last quotient will be — 1. 

The advantage of Newton's method is that the obtaining 
of a fractional quotient at any point of the division shows 
at once that the divisor is not a root of the equation. 

Ex. Find the integral roots of x* + 4 a* - x 2 - 16 x - 12 = 0. By- 
Descartes' Rule the equation cannot have more than one positive root, 
nor more than three negative roots. 

The integral divisors of - 12 are ± 1, ± 2, ± 3, ±4, ±6. Sub- 
stitution shows that — 1 is a root, and that + 1 is not a root. 



THEORY OP EQUATIONS. 486 

To ascertain if 2 is a root, arrange the work as follows : 

1 + 4- 1-16- 12 12 

-1-6-11- 6 

- 2 - 12 - 22 
Hence 2 is a root. 

Explanation. The first line contains the coefficients of the original 
equation, and the divisor 2. Dividing the last term, — 12, by 2 gives 
a quotient — 6 ; adding — 16, the coefficient of x, gives — 22. Divid- 
ing — 22 by 2 gives — 11 ; adding — 1, the coefficient of x 2 , gives —12. 
Dividing — 12 by 2 gives — 6 ; adding -f 4, the coefficient of x 8 , gives 
— 2, which divided by 2 gives a final quotient of — 1, hence 2 is a 
root. 

Since the equation can have no more than one positive root, we 
will only make trial of the remaining negative divisors, thus : 

1 + 4 - 1 - 16 - 12Lz-_e 1 + 4-1-16-12 [j-4 

+ 2 + 3 

- 14 - 13 

Hence — 6 is not a root. Hence — 4 is not a root. 

• 

1 +4-1-16-121-J* 1 + 4-1- 16 -12 1-2 

-Ul + 4-f 4 -1-2 + 5+ 6 

+ 3 + 3-12 +2 + 4-10 

Hence — 3 is a root. Hence — 2 is a root. 

EXAMPLES XLVIII. 1. 

Solve the following equations, which have one or more integral 
roots: 

1. x 8 -9x 2 + 26x-24 = 0. 11. x* + 8x* + 9x 2 - 8x- 10 = 0. 

2. x 8 -x-2x 2 + 2 =0. 12. x 4 + 2X 8 - 7x 2 - 8x + 12 = 0. 
8. 2x 8 + 5x 2 -llx + 4 = 0. 13. x 4 -3x 2 - 6x - 2 = 0. 

4. 4x 8 -20x 2 + 31x-14 = 0. 14. x 4 - 2x 8 - 12x 2 + 10x+ 3 = 0. 

5. x 8 -2x 2 -29x + 30 = 0. 15. fix 4 - x 8 - 17x 2 + I6x -4 = 0. 

6. x8-8x 2 + 6x+14 = 0. 16. x l -2x*-13x 2 +14x+24=0. 

7. x 4 -2x 8 -7x 2 + 8x+12 = 0. 17. x*+4x 8 -22x 2 - 4x + 21 =0. 

8. x 4 -4x 8 -14x 2 +36x+45=0. 18. x* + 2X 8 - 7x 2 -8x+ 12 = 0. 

9. x*- 3x 2 -42x-40 = 0. 19. x 4 - 6x 2 - 16x + 21 = 0. 

10. x*-10x 2 -20x-16 = 0. 20. x*+ 4x 8 - x 2 - 16x - 12 = Q, 

21. 2x 4 -x 8 -29x 2 + 34x + 24 = 0. 

22. x 5 -3x*-6x 8 + 15x 2 + 4x-12 = 0. 



9 



486 ALGEBRA. 

617. The Cube Roots of Unity. 

Suppose x = -^1 ; then X s = 1, or a£ — 1 =s 0; 
that is (x - l)(x* + x + 1) = 0. 

.-. either x — 1 = 0, or a 2 + a? + 1 = 0; 



whence a? = 1, or x = — : — =j- • 

It may be shown by actual involution' that each of these 
values when cubed is equal to unity. Thus unity has three 
cube roots, 

-1+V33 -i,yi8 t 

lj 2 ' 2 ' 

two of which are imaginary expressions. 

Let us denote these by a and b ; then since they are the 
roots of the equation 

a? + x + 1 = 0, 

their product is equal to unity ; 

that is, ab = 1 ; 

.•. a*b = a 2 ; 
that is, b = a 2 , since a 8 = 1. 

Similarly we may show that a = 6*. 

618. Since eacA of the imaginary roots is the square of the 
other, it is usual to denote the three cube roots of unity 
by 1, co, co 2 .* 

Also co satisfies the equation a? + x + 1 = ; 

.-. l + co + co 2 = 0; 

that is, the sum of the three cube roots of unity is zero. 

Again co • co 2 = co 8 = 1 ; 

therefore (1) the product of the two imaginary roots is unity; 
(2) every integral power of co 8 is unity. 

* The Greek letter Omega. 



I 



THEORY OP EQUATIONS. 487 

CARDAN'S METHOD FOR THE SOLUTION OF CUBIC 

EQUATIONS. 

619. The general type of a cubic equation is 

x* + Px*+Qx + 22 = 0, 
but as explained in Art. 585 this equation can be reduced 
to the simpler form X s + qx + r = 0, 

which we shall take as the standard form of a cubic equation. 

620. We proceed to solve the equation x 3 + qx + r = 0. 
Let x = y + z ; then 

a? = tf i + z s + 3yz(y + z)==i/ i + z s + 3yzx, 
and the given equation becomes 

y* + *? + (3yz + q)x + r = o. 

At present y, z are any two quantities subject to the con- 
dition that their sum is equal to one of the roots of the given 
equation ; if we further suppose that they satisfy the equa- 
tion 3 yz + q = 0, they are completely determinate. We 

thus obtain 

y> + *8 = -r (1), 

s» = -_2!_ (2); 

27 V W 

hence 3/* — -2-r = — r, or f + rtf = ^« 

11 y° 11 

Solving this equation, 

Substituting in (1), * = -£-^£ + £ (4). 

We obtain the value of x from the relation x = y + z; thus 



x 



-H+^}M-§-^}'-<» 



488 



ALGEBRA. 



The above solution is generally known as Cardan's Solu- 
tion, as it was first published by him in the Ars Magna, in 
1545. Cardan obtained the solution from Tartaglia; but 
the solution of the cubic seems to have been due originally 
to Scipio Ferreo, about 1505. 

In this solution we assume x = y + z> and from (2) find 

z = — 2-, hence to solve a cubic equation of the form 
3y 

ar + qx+rs=Q 

we substitute y — ^r-for x. 

* Sy J 

Ex. Solve the equation X s - 16 a; = 126. 

1*1** V — I -^ — l or V + ■ f° r s» then 
\ 3y J y 

y y 8 v 



or 



whence 



y8 +1^=126, 



tf - 126 y» = - 126. 
.\ y 8 = 126, 
,\ y = 6. 

But x = y + - = 6. 

Dividing the given equation x* — 16 x — 126 = by « — 6, we 
obtain the depressed equation 

X 2 + 6a + 21=0, 

the roots of which are - 3 + 2\/^"3, and - 3 - 2 V^S. 

Thus the roots of X s - 15s = 126 are 6, - 3 + 2 V^8, and - 3 
-2>/^~3. 



BIQUADRATIC EQUATIONS. 

621. We shall now give a brief discussion of some of the 
methods which are employed to obtain the general solution 
of a biquadratic equation. It will be found that in each of 
the methods we have first to solve an auxiliary cubic equa- 
tion; and thus it will be seen that as in the case of the 
cubic, the general solution is not adapted for writing down 
the solution of a given numerical equation. 



THEORY OF EQUATIONS. 489 

The solution of a biquadratic equation was first 
obtained by Ferrari, a pupil of Cardan, as follows : 

Denote the equation by 

x 4 + 2px* + qx 2 + 2rx + s = 0;. 

add to each side (ax + b) 2 , the quantities a and b being 
determined so as to make the left side a perfect square; 
then 

x 4 + 2 pa* + (q + a*)x 2 + 2(r + ab)x + s + b 2 =(ax + b)*. 

Suppose that the left side of the equation is equal to 
(x 2 +px + A;) 2 ; then by comparing the coefficients, we have 

p 2 -h 2 ft = q -f a 2 , pk = r + ab, ft 2 = s + 6 2 ; 

by eliminating a and 6 from these equations, we obtain 

(pk - rf = (2 A; + p 2 - g) (ft 2 - «), 

or 2 ft 8 — gft 2 + 2 (pr — s) ft -f-p 2 s — gs — r 2 = 0. 

From this cubic equation one real value of k can always 
be found [Art. 600] ; thus a and b are known. Also 

(a 2 +px + ft) 2 = (ax + 6) 2 5 
.'. x 2 +px + A; = ± (ax -f- 6) ; 

and the values of x are to be obtained from the two 
quadratics 

a 2 + ( P - «) a + (ft - 6) = 0, 

and a 2 -f- (p + a) a? + (ft + b) = 0. 

Ex. Solve the equation 

s* -2a 8 - 5x 2 + 10x-3 = 0. 

Add a 2 x 2 + 2 abx + & 2 to each side of the equation, and assume 
x 4 - 2x» + (a 2 - 5) a 2 + 2 (a& + 5) x + 6 2 - 3 = (x 2 - as + Aj) 2 : 
then by equating coefficients, we have 

a 2 = 2& + 6, ab = -k-5, 6 2 = A^ + 8; 

.-. (2fc+6)(& 2 + 3) = (ft + 5) 2 ; 

.-. 2&8 + 5ifc 2 -4&-7 = 0. 

By trial, we find that k = — 1 ; hence a 2 = 4, 6 2 = 4, ab = — 4. 
But from the assumption, it follows that 

(x 2 - x + &) 2 = («s + b) 2 . 




490 ALGEBRA. 

Substituting the values of ft, a and ft, we have the two equations 

a?-a;-l=± (2x-2); 
that is & — 3x + 1 =0, and x 2 + z - 3 = ; 

whence the roots are ^ > ^ v . 

623. The following solution was given by Descartes in 
1637. 

Suppose that the biquadratic equation is reduced to the 
form 

a* 4 + qtf + rx + *= 0; 

assume at + q7?+ rx + *= (<x? + kx+r)(p? — kx + m)i 

then by equating coefficients, we have 

I + m — ft 2 = q, k(m — Z)= r, Zm = *. 

From the first two of these equation, we obtain 

2m = fe 2 + g + 7, 2Z = &* + g-£: 

A: A; 

hence substituting in the third equation, 

(ft 8 + qk + r)(k? + qk - r)= 4*A?, 

or fc6 + 2gfc4+(g* _4*)fc 2 -r a = 0. 

This is a cubic in fc 2 which always has one real positive 
solution [Art. 600] ; thus when k 2 is known the values of I 
and m are determined, and the solution of the biquadratic is 
obtained by solving the two quadratics 

a? + kx + 1 = 0, and a? — kx + m = 0. 
Ex. Solve the equation 

Assume a* - 2s 2 + 8a; - 3 = (x 2 + kx + l)(j& - kx + m) j 
then by equating coefficients, we have 

l + ffi-tfs-2, fc(ro-Z)=8, Zro = -3; 
whence we obtain (& 8 - 2 k + 8)(A* - 2 A - 8) = - 12*?, 
or # -4ifc* + 16* 3 -64=0. 



THEORY OP EQUATIONS. 491 

This equation is clearly satisfied when & 2 — 4 = 0, or k = ± 2. It 
will be sufficient to consider one of the values of k ; putting k = 2, we 
have 

m + 1 = 2, m — Z = 4 ; that is, I = — 1, m = 3. 
Thus x* - 2x* + 8x - 3 =(x 2 + 2x - l)(x* - 2x + 3); 
hence x 2 + 2x-l = 0, and a? -2x + 3 = 0; 

and therefore the roots are — 1 ± y/2, 1 ± V— 2. 

624. The general algebraic solution of equations of a 
degree higher than the fourth has not been obtained, and 
Abel's demonstration of the impossibility of such a solution 
is generally accepted by mathematicians. If, however, the 
coefficients of an equation are numerical, the value of any 
real root may be found to any required degree of accuracy 
by the method of Art. 626. 

EXAMPLES XLVIIL k. 

Solve the following equations : 
1. x8-18x = 36. 8. x 8 -6x 2 + 8x-18 = 0. 

3. x» + 72x- 1720 = 0. 9. 8x*-36x + 27 = 0. 
8. «8 + 68x- 816 = 0. 10. x 8 - 16* -4 = 0. 

4. x« + 21x + 342 = 0. 11. x* + 8x» + 9*2-8* - 10 = 0. 

5. 28x 8 -9x 2 + l=0. 12. x* + 2x«- 7x 2 - 8* + 12=0. 

6. x 8 - 15s 2 -33* + 847 = 0. 13. x*- 3X 2 - 6* -2 = 0. 

7. 2x» + 3x 2 + 3x + l = 0. 14. x*-2x»- 12x2 + 10x + 3=0. 

INCOMMENSURABLE BOOTS. 

625. The incommensurable roots of an equation cannot 
be found exactly. If, however, a sufficient number of the 
initial figures of the root have been found to distinguish it 
from the other roots we may carry the approximation to the 
exact value to any required degree of accuracy by a method 
first published in 1819 by W. G. Horner. 



492 ALGEBRA. 



HORNER'S METHOD OF APPROXIMATION. 

Let it be required to solve the equation 

a*-3x t -2x + 5 = (1). 

By Sturm's Theorem there are 3 real roots and one of 
them lies between 1 and 2 ; we will find its value to four 
places of decimals, which will sufficiently illustrate the 
method. 

Diminishing the roots of the equation by 1 [Arts. 583, 
584], we have 

1 -3 -2 +5 II 
1-2-4 

-2 -4 1 

1 -1 

-1 -5 
1 



The transformed equation is 

y»-5y + l = (2). 

Equation (1) has a root between 1 and 2. The roots of 
equation (2) are each less by 1 than those of equation (1) ; 
hence equation (2) has a root between and 1. This root 
being less than unity the higher powers of y are each less 
than y. Neglecting them, we obtain an approximate value 
of y from — 5 y -f 1 = 0, or y = .2. 

Diminishing the roots of (2), the first transformed equa- 
tion, by .2, we have 

1 ± o -5 +1 L2 

.2 .04 .992 

.2 - 4.96 .008 

.2 m 

.4 - 4.88 
.2 

.6 



THEORY OF EQUATIONS. 493 

The transformed equation is 

2 8 + .62 9 -4.88« + .008 • • • . (3). 

Equation (2) has a root between .2 and .3; the roots of 
equation (3) are less by .2 than those of equation (2); 
hence equation (3) hqs a root between and .1. Neglect- 
ing in equation (3) the terms involving the higher powers, 
as was done in the case of the first transformed equation, 
we have * 

- 4.88 a; + .008 = 0, or s = .001. 

Diminishing the roots of (3). the second transformed equa- 
tion, by .001, we have 

1.001 



+ .6 
.001 


-4.88 
.000601 


+ .008 

- .004879399 


.601 
.001 


- 4.879399 
.000602 


.003120601 


.602 
.001 


- 4.878797 





.603 

The transformed equation is 

v 8 + .603 v 2 - 4.878797 v + .003120601 .... (4). 

Equation (3) has a root between .001 and .002 ; the roots of 
equation (4) are less than those of equation (3) by .001 j 
therefore equation (4) has a root between and .001. 
Neglecting the terms involving v 8 and v 4 , and solving 

- 4.878797 v + .003120601 = 0, 

we have v = .0006. Diminishing the roots of (4), the third 
transformed equation, by .0006, we find this to be the correct 
figure for the fourth decimal place ; hence 1.2016 is the 
value, to the fourth decimal place, of the root which lies 
between 1 and 2. 

Denoting the coefficients of the successive transformed 
equations by (A), (B), (C), etc., the work is more compactly 
arranged thus : 



THEORY OF EQUATIONS. 495 

627. It sometimes happens that the division of the last 
term of the first transformed equation by the coefficient of x 
in that equation gives a quotient greater than unity. In that 
case, as where the signs of these terms are alike, we obtain 
another figure of the root by the method used to obtain the 
integral part of the root. 



If in any transformed equation after the first the 
signs of the last two terms are the same, the figure of the root 
used in making the transformation is too large and must 
be diminished until these terms have unlike signs. 

629. If. in any transformed equation the coefficient of the 
first power of the unknown quantity is zero, we may obtain 
the next figure of the root by using the coefficient of the second 
power of the unknown quantity as a divisor and taking the 
square root of the result. 

630. Negative incommensurable roots may be found by 
transforming the equation into one whose roots shall be posi- 
tive [Art. 580], and finding the corresponding root. This 
result with its sign changed will be the root required. 

631. Any Root of Any Number. By Horner's Method we 
can find approximately any root of any number ; for placing 
y/a equal to x we have for solution the equation #* = a, or 
x* — a = 0. 

EXAMPLES XL VIII. 1. 

Compute the root which is situated between the given limits in the 
following equations : 

1. x 8 + K t 2 + 6 x - 120 = ; root between 2 and 3. 

2. x 3 — 2x — 6=0; root between 2 and 3. 

3. x* - 2 X s + 21 x - 23 = ; root between 1 and 2. 

4. x* + x - 1000 = ; root between 9 and 10. 

5. X s + % 2 4- x — 100 = ; root between 4 and 6. 

6. 2 x s + 5 x'* - 4 x — 10 = ; root between 1 and 2. 

7. x*-46x 2 -2&x+l$ = Q; root between and 1. 



496 ALGEBRA. 

8. x 8 + x — 3 = 0; root between 1 and 2. 

9. x 8 + 2x-20 = 0; root between 2 and 3. 

10. x 8 + 10 x 2 + 8x - 120 = 0; root between 2 and 3. 

11. 3 x 8 + 6 x - 40 = ; root between 2 and 3. 

12. x 4 - 12 x 2 4- 12 x - 3 = ; root between - 3 and — 4. 

13. x 6 -4x* + 7x»-863=0; root between 4 and 5. 

Find the real roots of the following equations : 

14. ae*-3x-l=0. 16. x 4 -8x 8 + 12x 2 + 4x-8=0. 

15. x 8 -22x-24 = 0. 17. x* + x 8 + x 2 + 3x - 100 = 0. 

Find to four decimals, by Horner's Method, the value of the 
following z 

18. #11. 19. #13. 80. #6. 81. #7. 

MISCELLANEOUS EXAMPLES VII. 



1. Simplify b-{b -(a + &)- [b -(& - a - &)] + 2a}. 

8. Find the sum of 
a + b - 2 (c + d), b + c - 3 (d + a) and c + a" - 4 (a 4- 6). 

3. Multiply ix + £ybyx-Jy. 

4. If x = 6, y = 4, * = 3, find the value of y/2 z + 3 y + z. 

5. Find the square of 2 — 3 x + x 2 . 

6. Solve £±-? + ^4 = 2. 

x— 1 x— 6 

7. Find the H. C.F. of a 8 - 2 a - 4 and a 8 - a 2 —4. 

,. slmpllfy J^ + JL^_4±|j. 

a + b a — b a 2 — ft 2 

?x + |=13 

9. Solve ? 4 

1*-* = 3 
3 8 

10. Two digits, which form a number, change places when 18 is 
added to the number, and the sum of the two numbers thus formed 
is 44 : find the digits. 

11. If a = 1, b = - 2, c = 3, d = - 4, find the value of 

q 2 & 2 + fr 2 c -f d(a - b) 
10 a -(c+6) 2 



A 



MISCELLANEOUS EXAMPLES VII. 497 

"12. Subtract — x 2 + y 2 — z 2 from the sum of 

is 2 + iy 2 , i2/ 2 + J* 2 , and \z*-\x*. 

13. Write the cube of x -f 8 y. 

14 . Simplify *£% x *^ x t 

x 2 + y 2 xy + y l x 

15. Solve f (2s-7)-i(&-8) = ^4±i+4. 

16 

16. Find the H.C.F. and L.C.M. of 

s* + a; 3 -f2a;-4 and a; 8 + 3 x 2 - 4. 

17. Find the square root of 4 a 4 + 9(1 - 2 a) + 3 a 2 (7 - 4 a). 

18. Solve 2 3 

X ~ 2 + 3 






19. Simplify (— 2 — W 

\x+ a x-a/ 



x 2 + a? 



x 2 + ax 

20. When 1 is added to the numerator and denominator of a cer- 
tain fraction the result is equal to f ; and when 1 is subtracted from 
its numerator and denominator, the result is equal to 2: find the 
fraction. 

21. Show that the sum of 12a+6 6— c, — la — 6+c and a+6+6c, 
is six times the sum of 25 a + 13 b — 8 c, — 13 a — 13 b — c, and 
- 11 a + b + 10 c. 

22. Divide a 2 - ay + i 3 § y 2 by a; - Jy. 

23. Add together 18 { ^ - ^ (^ + *H , 

24. Find the factors of (i.) 10 x 2 + 79 x - 8. (ii.) 729 a* -y«. 

25. Solve ^ + ^±3 = 3-^?. 

6 17 11 

26. Find the value of 

(6a - 3 b)(a -b)-b{Sa- c(4 a - 6)- 6 2 (a + c)}, 
when a = 0, 6 = — 1, c = £. 

27. Find the H. C. F. of 7 a^-lOa^-^s+lO and 2 s 8 -x 2 -2 s+1. 

9Q Q,mr,iif v x ' 2 - 7 sy + 12 y 2 . a* - 5 xy 4 -4 y 2 

28. Simplify ^ ^ + fl ^ ^ ^ + xy _ 2y2 . 

2k 



ick when the hands 

value of 
1. 
i%+V by x>-8y». 

8 a -1)]. 

1 — x + 8 is divided 



, and xP-nfy+xs/'- 
d half-dollars, and 
times that of the 

#»i-6» is divided 

ilueof 

j. 

4** + 3x- 10 and 



MISCELLANEOUS EXAMPLES VII. 499 

48. Simplify — ? ^-^- 

F * s-4 «-6 (x-2)(a;-8) 

49. Find the L. C. M. of 

8 a 8 + 38 s 2 + 69 a + 30 and 6 x* - 13 a? - 13 x + 30. 

50. A boy spent half of his money in one shop, one-third of the 
remainder in a second, and one-fifth of what he had left in a third. 
He had 20 cents at last : how much had he at first ? 

51. Find the remainder when a 7 - 10a° + 8s 6 - 7 X s + Sx- 11 
is divided by x 2 — 6 x + 4. 

52/simplify4{a-|(6-^}{l(2a-6)+2(6-c)}. 
53. If a = $$> 6 = 1, c = f, prove that 



(a - y/b)(y/a + 6) Va-6 = 



3c* 



Va-<? 

54. Find the L.C.M. of s 2 -7a;+12, 8s 2 -6x-9, and 2 a 2 -6 aj- 8. 

55. Find the sum of the squares of ax + by, bx — ay, ay + bx, 
by — ax'; and express the result in factors. 

56. Solve ? + 2 = 3*zll£ = s + ll = i. 

6 4 4 8 16 

57. Simplify q ° + b * - A±A - * { ±rl L_l. 

F J a*-6* a 2 -6 2 2U 2 + 6* a-6i 

58. 8ol¥e*-(3«-^±i^ = |(2« + 67) + |(l+|y 

59. Add together the following fractions : 

2 -4x a 2 -a; 2 



x 2 + xy + y 2 x*-y* y\x - y) 2 x*y — y* 

60. A man agreed to work for 30 days, on condition that for every 
day's work he should receive $2.50, and that for every day's absence 
from work he should forfeit $ 1.50 ; at the end of the time he. *«ceived 
$ 51 : how many days did he work ? 

61. Divide5^+27-l^ 2 -4a^ + ^ , -§^by^+3^a, 

4 4 8 4 2 



62. Find the value of 

4j (2 ,_ x) _35[3£^_^{ 3 ,-|(7x-4 ! ,)}] 

when x = — J ana y = 2. 



500 



ALGEBRA. 



68. Simplify 



10s- 11 10s - 1 x 2 - 2 x + 5 

8 (x 2 - 1 ) 3 (x 2 + x -f 1) (x 8 - 1 ) (x + 1) ' 



64. Find the cube root of «** a* - 3 -^ ** + 3o6 ^ _ » ^ 

6 8 6 c c 8 

65. Solve 4x-17 lQg-18 as 8 aB -«) + 6«-4 



x-4 " 2x-3 2x-7 x-1 

66. Find the factors of (i.) aJ+SxHx+S. (ii) x 2 -2xy--323y 2 

K* + y)+2*=r2l 

67. Solve 3x-£(y + «V-66 

*+ l(* + y-*;=88j 

aa Qi^i;*, g + 2y -? x 2 + 63 xy -f 7 y* 

68. Simplify - - o-— 3~ # 

f x — y 2 x 2 + 3 xy - 3d y 2 

69. Find the square root of -(3 b - 2 c - 2 a) 8 {2 (a + c) - 3 6}. 

70. The united ages of a man and his wife are six times the united 
ages of their children. Two years ago their united ages were ten 
times the united ages of theii children, and six years hence their 
united ages will be three times the united ages of the children : how 
many children have they ? 

71. Find the sum of 

a?-8xy~f y 2 , 2y 2 -|y 8 + * 2 , xy- \tf + y*> and 2xy- Jy». 

**{*. From {( a +6)(a-x)-(a-6)(6-x)} subtract (a+5) 2 -26x. 

73. Ifo = 5, 6 = 4, c = 3, find the value of 

\/6 abc + (b + c)» + (c 4- a) 8 +(a + &) 8 -(a + 6 + s) s - 

74. Find the factors of 

(i.) 3X 8 4- 6x 2 - 189x. (ii.) a 2 + 2a& + 6 2 + a + 6. 



75. Solve 



l>x = gyl 
CP + g)*-(g-.p)y = r i 



76. Simplify 



x +i 



x 2 -^ 



2x 2 + xy+^ 4 ( x8 -f) 



77. Solve ^^4 + 



2x-15 



x + 7 2(x + 7) 2x-6 

<r4_/>.2 9 r 4- 2 . < 

78. Reduce - - ^ to its lowest terms. 

2x 8 -x-l 



MISCELLANEOUS EXAMPLES VH. 501 



79. Add together the fractions : 

1 . 1 



and 



2x 2 -4x + 2' 2x 2 + 4x + 2' 1 - x 2 

80. A number consists of three digits, the right-hand one being 
zero. If the left-hand and middle digits be interchanged, the number 
is diminished by 180 ; if the left-hand digit be halved, and the middle 
and right-hand digit be interchanged, the number is diminished by 
336 : find the number. 

81. Divide 1 - 5s + ¥fx* - iJt«* - ¥** by 1 - x - i$x*. 

82. If p = 1, q = J, find the value of 

(P 2 + q 2 )-(p ~ q) Vy 2 4- 2pq + tf 

83. Multiply —-5x 2 + | + 9 by --x + 3. 

84. Find the L. C. M. of 

(a 2 6 - 2 a& 2 ) 2 , 2 a 2 - 3 ab - 2 6 2 , and 2(2 a 2 + a&) 2 . 

85. Solve 2 -*±* = ±«±i + §*±8 

b + 1 4x + 4 3x+l 

86. Reduce — 5 ** "" * 4 g2 + 16 — to its lowest terms. 

3x 8 -2x 2 + 16x-48 

• 87. Find the square root of 

4 a* + 9^a 2 + \\ + 12 a(a 2 + l)+ 18. 

88. Solve JL + ^M-z^a + b 
2 a ob 



. • 



3«_22 = 6(6 _ 0) 
a 6 

89. Multiply 

90. A bag contained ten dollars in dimes and quarters; after 
17 dimes and 6 quarters were taken out, three times as many quarto li- 
as dimes were left : find the number of each coin. 

91. Find the value of 

6(a-6)-2{3a-(a + 6)} + 7{(a-26)-(6a-26)}, 
when a = — J 6. 

92. Divide 3x* - 6x» + 7x 2 - llx - 13 by 3x-2. 



502 



ALGEBRA. 



98. * ind the L. C. M. of 
15(.p* 4- g 8 ), 5(p3 -pq + g 2 ), 4(p* +pq + g*), and 6(p* - & 

94. Resolve into factors : 



(i.) o«-8&i* 

95. Solve *±-« = -^i2« r 

jk + 6 x + a + ft 

96. Simplify 
35 a 2 6 2 c 2 - 49 &*c» 



(U.) -aJ» + 2x-l + aj*. 



00 



65 a b bc- 91 a 8 6 2 c 2 
97. Solve 



(ii) y*-7y» + 8y«-12v 
v y 2y 2 -2y-60 



7x-9y + 4« = 16 

3 2 

2o5-3y + 4«-5 = 

98. Simplify ^-V- fc-ai + S X J+2> 

* 2/ + 1 

99. Find the square root of 

4 a 2 - 12 aft - 6 5c + 4 oc -f 9 6 2 + c* 

4a 2 + 9c 2 - 12 ac 

• * 

100. An express leaves New York at 3 p.m. and reaches Albany 
at 6 ; the, ordinary train leaves Albany at 1.30 p.m. and arrives at New 
York at 6. If both trains travel uniformly, find the time when they 
will meet. 

101. Soi7e (i.) .Gx + .75s - .16 = x - .585a + 6. 



(ii.) 



37 



+ -^rr = 



se 2 -5x + 6 x-2 S-x 
102. Simplify (i.) o a + x — ; +—J±- x 



(ii.) (1 + x)* -*- 



+ 



2a* 



1 + 



ax + x 2 a* + a?x 2 +«* 
x 



1-S + 



X 



1 + x + a> J 



103. Find the square root of 

also the cube root of the result. 

104. Divide 1 — 2 » by 1 + 3 a to 4 terms. 



* t 



MISCELLANEOUS EXAMPLES VH. 503 

106. I bought a horse and carriage for $ 450 ; I sold the horse at a 
gain of 6 per cent, and the carriage at a gain of 20 per cent, making 
on the whole a gain of 10 per cent: find the original cost of the 
horse. 

106. Find the divisor when (4 a 2 + 7 ab 4- 6 6 2 ) 2 is the dividend, 
8 (a 4- 2 b) 2 the quotient, and b 2 (9 a + 11 b) 2 the remainder. 

107. Solve (i.) 6x (x - 3) = 2 (x - 7). 

(ii.) 1 + 6= 8 " 2 



(x-l)(x-2) x-2 x-1 



ab 



108. If x = a + 6 + i^i, m dy = <l±± + 

4 (a + 6) 4 a + b 

prove that (x - a) 2 - (y - ft) 2 = ft 2 . 

109. Find the square root of 

25 5 6 

a — x 8a 



110. Solve o a + x + , o o o _ 

a 2 4- ax + x 2 a 2 - ax + x 2 x (a 4 + a*x 2 + x 4 ) 

111. Subtract _*±J? — from * + 4 



x 2 4- x - 12 x 2 - x - 12' 

and divide the difference by 1 H ^ x "~ — *_. 

J x 2 + 7 x + 12 

113. Find the H. C. F. and L. C. M. of 

2x 2 4- (6a-106)x-30a6 and 3x 2 - (9a + 156)x + 45a&. 
113. Solve (i.) 2cx 2 -a6x + 2a6d = 4cax. 

(ii.) - 2&= x2 8*- 1 



2(x + 3) '* x 2 -9 4(x-3) 
114. If a = 1, 6 = 2, c = 3, a* = 4, find the value of 
a h 4- b c + c d 



3(a« + &> + c°)(- + i+2:Y 

\a b c/ 



b* + c* + d° + (a + &)(& 4- c) 

115. I rode one-third of a journey at 10 miles an hour, one-thinl 
more at 9, and the rest at 8 miles an hour ; if I had ridden half the 
journey at 10, and the other half at 8 miles per hour, I should have 
been half a minute longer on the way : what distance did I ride ? 

116. The product of two factors is (3x + 2y) 8 - (2x + 3yy, and 
one of the factors is x — y : find the other factor. 

117. If a 4- 6 = 1, prove that (a 2 - ft 2 ) 2 = a 8 4- & 8 - ab. 

118. Resolve into factors : 

(i.) x 8 4- y 8 4- Sxy (x + y). (ii.) ro 8 - n 8 - m (m 2 -n 2 ) + n(m- ri) 2 . 



504 



ALGEBRA. 



119. Solve (i.) * 8 -y 8 = 28\ (ii.) x 2 -6xy + lly* = 9\ 
x* + xy + y* = 7 /• x-3y=lf* 

180. Find the square root of 

(a - 6)* - 2 (a 2 + 6 2 )(a - 6) 2 + 2 (a 4 + &*)• 

121. Simplify the fractions 



00 



a 



2__ 



a*-l 



a + 



(ii) 



km--;) 



2 



26 



a+ 1 

122. Find the H. C. F. of 

a?b + b 2 c - abc - a& 2 and ax* + ab -a*- bx*. 

123. A village had two-thirds of its voters Republicans : in an elec- 
tion 25 refused to vote, and 60 went over to the Democrats ; the voters 
were now equal. How many voters were there altogether ? 



124. Solve (i.) 



x a 



a + b 



+ (a - 6) = 



2 ax 
a + b 



(ii.) § + 2 = 6 /I__L\ = 2 . 
125. Simplify W (l + 4^ + (t .i^^ 

k ' ; (x + 1)* - (x - 1)* 



126. Divide 

x* + (a-l)ic 8 -(2a + l)x 2 + (a 2 + 4a - 6)x + 3a + 6 
by x 2 -3x + a + 2. 

127. Resolve into factors : 

4 

(i.) x 2 + bxy - 24 y 2 + x - Sy. (ii.) a 8 

x 

128. Find the square root of p 2 — 3 q to three terms. 

129. Solve (L)«zl5-«^ = *^1-^i|. 

w x - 6 a; - 7 a; - 2 x - 3 

(ii.) ax + 1 = by + 1 = aj/ + bx. 

130. Find the H. C. F. of 

3 X 2 + (4 a _ 2 &)x - 2 a& + a 2 and x 8 + (2 a - b)x* - (2 ab - a 2 )x - a 2 b. 

131. Simplify 



, n (x°) 8 (x h y (x«) 8 - 

' x*+ c x c +° x° + * 



(ii.) x^y V4Y + ^- 



x' 



MISCELLANEOUS EXAMPLES VII. 506 

132. At a cricket match the contractor provided dinner for 27 per- 
sons, and fixed the price so as to gain 12£ per cent upon his outlay. 
Six of the cricketers being absent, the remaining 21 paid the fixed 
price for their dinner, and the contractor lost $3: what was the 
charge for the dinner ? 

138. Prove that x(y + 2) -f - + - is equal to a, if 

y x 

aj = — ^— and y = q "~ » 
y+1 2 

184. Find the cube root of 

x* _ 12 a? + 64s - 112 + — ™+4- 

x re 2 a 8 

135. Find the H. C. F. and L. C. M. of 

x* + 2ax* + a*z+2a* and x* - 2 ax 2 + a 2 ^ - 2 a 8 . 

136. Simplify 

(i.) 42 / *s-8y 3s-4y ^ 66 { 3s-2y 2s-3|Q 

m . 46 + a q-45 q a -3& 2 
^ * ; 36 + a a-36 a 2 -96 2 " 

137. Resolve 4 a 2 (s 8 + 18 ab 2 ) - (32 a 6 + 9 W) into four factors. 



138. Solve (i.) bVSx - 1 = V76a;-29. 

(ii.) _?IL. = 70, -**- = 84, -J^_ = 140. 

139. Show that the difference between 
x + _*_ + _*- and -J2L-+ 6 ■ 



a — a 35 — 6 a — c a — a x — 6 s — c 

is the same whatever value x may have. 

140. Multiply x% + 2 y$ + 3 s$ by ** - 2 j/* - 3«* 

141. Walking 4 J miles an hour, I start 1£ hours after a friend 
whose pace is 3 miles an hour ; how long shall I be in overtaking him ? 

142. Express in the simplest form 

(9».32x-?-l-27» 
w v „ + 4*) x 16 *. (ii.) g^ 

143. Find the square root of 

y x My \» 



506 ALGEBRA. 

144. Simplify 



x* 



«.)(-* 1 \ s 8 -! ft - !)«(«+ IP + 

W V»-1 x + 1/ a* + l ** + {* + l 

(ii) [ ^-y 4 . <** + <*? ] x f a 5 -q 8 y2 i q*-2q»y+qy ) 

146. Find the value of 

(i.) V8 + V60-V18+V58. (ii.) V36 + 14 y/Q. 

146. Solve (i.) En5 - *n« = 2 ( q ~ 6 ) . 

a: — ax — 6x— (a + 6) 

rin 2s + 3y = in 

v ' 4x 2 + 9xy + 9y2 = n/ 

147. Show that 

( q + ft) 8 - c* (5 + c)» - a* (c + o) 8 - ft 1 
(a + b)—c b + c — a c + a — b 

is equal to 2(o + 6 + c) 2 + a 2 -f b 2 + c 2 . 

148. Divide a -s + 4a*x^ -4a*x* 
by q* + 2 q*x* - x*. 

149. Find the square root of 

(a - l) 4 + 2(a* + 1)- 2(q 2 + i)( a _ i)«. 

150. How much are pears a gross when 12 more for a dollar lowers 
the price five cents a dozen ? 

151. Show that if a number of two digits is six times the sum of its 
digits, the number formed by interchanging the digits is five times their 
sum. 

152. Find the value of 

1 1_ 1 

(a - 6)(6 - c) (b - c)(a -c)~ (c- a)(6 - a) 

153. Multiply 

qj K ^ 12 + 41x + 36x 2 26a; -8a 2 -14 





f T *>*> 




4 + 7x 


- ujr u 


8 


-4* 


154. 


Ifx- 


1_ 

X 


1, prove that 


* 2 + \ 

X 2 


= 3, and x 8 — 


*-* 


155. 


Solve 




(ii.) 2x 2 - 
2xy - 


23 

e + 4 

-3^ = 
-3y2 = 


|(x + 5), 

231 
3/' 





MISCELLANEOUS EXAMPLES VII. 507 

_i 

156. Simplify (i.) lfv^O -3V6 -Vi- 0*0 ^(-^r) + IL r 

157. Find the H. C. F. of f p 2 _ jj^ + (3 p _ 1)aj « p Q, _ i) and 

p(p + I)* 2 - (p 2 - 2p - l)x - Q> - 1). 

158. Reduce to its simplest form 

ax 4-- x* -4--i- 



y y 2 

150. Find the square root of 

(i.) 1 - 22»+i + 42»- (ii.) 9» - 2 • 6* + 4». 

160. A clock gains 4 minutes a day. What time should it indicate 
at 6 o'clock in the morning in order that it may he right at 7.15 p.m. 
on the same day ? 

161. If x = 2 + V2, find the value of x 2 + 4* 

x 2 

162. Solve (i.) a^+« = **^l£. Hi.) VTT^-hVn^ = a 
.163. Simplify ^ - + & * c * 



(6 « a )(c - a) (c - 6)(a - 6) (a - c)(6 - c) 

164. Find the product of iV5, J$/2, ^80, {/5, and divide 

8-4^5 b 3^/5 - 7 
V5 + 1 5+V7 

165, Resolve 9 xty* - 576 y 2 - 4 x 8 -f 256 x 2 into six factors. 

1 «i- 

1M . v . ,, x (x + a)- x(x + 2o) 

M. Simplify (i.) (g + ; )(x , q) ^ (g2 J a2)(g+ ; g)g » 

exV r 3(m-w)x . r4 (r-8) r^-a 2 ^ ] 
W m+» * L 7(r+s) ' I 21x2/ 2 " i "4(w 2 -n 2 ) J J 

167. Simplify (i.) (a *y»+f+^— 2i_. 

(ii.) V14 - V132. 

168. Find the H. C. F. and L. C. M. of 

20x* + x 2 -l, 25x* + 5x 8 -x-l, 25x*-lOa0 + l. 



508 



ALGEBRA. 



108. Solve 



(i.) a + x + V2ax + 30 = b. 

1 



(ii.) x + 9| + 



x 11 
7 8 



= & 



170. The price of photographs is raised $ 3 per dozen, and customers 
consequently receive ten less than before for $ 5 : what were the prices 
charged ? 



171. If 



K)'= 



3, prove that a 8 + — = 0. 



178. Find the value of 



x + 2 a , x — 2a . 4ab 



when « = 



ab 



2b-x 26-fx x 2 -4V* a + b 

173. Reduce to fractions in their lowest terms 

rn (1 1 1\ . / g-fy + g 1 \ , 

^° Kx^y* zj^yxt + yt + zt-xy-yz-zx x + y + z)* ' 

<w('-.-?j + .-fi)(*+A-.-?i)- 

174. Express as a whole number 



175. Simplify 



(*■) 



n 



(27)* +(16)* — t +^? 

(8)-* (4)"* 
(H.) ^97 - 56 V8. 



n 



1 - x» 1 - x n 



176. Solve (i.) «=i5 + «^ = ?Li^ + « + «« 

X — 3a x — 4a a; — 4a x — 3a 

(ii.) 3x 2 + xy + 3y 2 = 8J "» 
8x 2 -3xy + 8y 2 = 17f J 

«»~ t?- ^ *v * * aPx 2 + 2 aft 2 * 8 + bht 

177. Find the square root of ^ — » * 

a 2 * + 2 a m x n + x 2 " 

178. Simplify 

(i .) A x ^3x^x^2. ,.,, I cO^xV^F" !" 

Va v* a -i w 1 3VF ^ J " 

179. A boat's crew can row 8 miles an hour in still water : what 
is the speed of a river's current if it take them 2 hours and 40 min- 
utes to row 8 miles up and 8 miles down ? 



MISCELLANEOUS EXAMPLES VII. 509 

180. If a = x 2 — yz, 6 = y 2 — zx, c = z 2 — xy, prove that 

a 2 — be — x(ax + by + cz). 

181. Find a quantity such that when it is subtracted from each 
of the quantities a, 6, c, the remainders are in continued proportion. 



.Simplify (i0(, + y __L_) X g5£ 



182. 

V x + v- 

x + y 

AH 2(7s-4) s-10 2(4s-l) 

v ' J 6a 2 -7x + 2 6x 2 -a>-2 4a; 2 -1 
188. Find the sixth root of 
729 - 2916 x 2 + 4860 a* - 4320 a* + 2160 a* - 676 x™ + 64 x"». 

184. Simplify 

(i.) \= + 



a? + Va; 2 - 1 x - Va; 2 - 1 



(ii.) ^16+</81-V^~6l2+^I92-7J/9. 
185. Solve (i.) 1— - — = «. 



6- 



6- 5 



6-a; 



(ii.) aJ 2 y 2 + 192 = 28a;y> 
a; + y = 8 J 
186. Simplify 

6 — c . c — o , a — 6 

+ t^ — : ^ + 



a 2 _(5 _ C )2 6 2 _( c _ a )2 C 2_( a _ 6 )8 

187. Solve (i.) x - 16} + — ^— = 6. 

05 — 16} 

(ii.) 2(x + y- 1 ) = 3(3?-! - y) = 4. 

188. If a^ = «6(a + &) and x 2 - a;y + y 2 = a 8 + 6 8 , prove that 



(h)(hH 



189. Find the H. C. F. of 

(2a 2 - 3a - 2)a; 2 +(a 2 + 7 a + 2)a; - a 2 - 2a 
a nd (4 a 2 + 4 a + 1 ) x 2 - (4 a 2 + 2 a ) x + a 2 . 

190. Multiply V2a; + V2(2a;-1) — 

V2x 

by -J— + V2(2a;-1)-V2a;. 

V2a; 



510 ALGEBRA. 

191. Divide a«8* + M<* + eta* - a«&* -- Wc* - <*a* 
by a*& + & s c + <*a-aft s -&c*-ca t . 

. 192. Simplify 

7 1 10x-l 



00 



2(x + 1) 6(x - 1) 3(x»+x+ 1) 
Vx~+a Vx — a > w Vx 8 — a 1 



(iL ) jVx + fl,Vx-fl> 
« Vx — a Vx + a > 



V(x + a)* — ax 

198. K p be the difference between any quantity and its reciproca. 
q the difference between the square of the same quantity and the square 
of its reciprocal, show that 

P*(p* + 4) = g* 

191. A man started for a walk when the hands of his watch were 
coincident between three and four o'clock. When he finished, the 
hands were again coincident between five and six o'clock. What was 
the time when he started, and how long did he walk ? 

195. If n be an integer, show that 7 2 "+ 1 + 1 is always divisible by 8. 

196. Simplify v q/ v q/ 



{'+$(-$ 



197. Find the value of 

rn 7 + 3y5 7-3y5 
W 7-3v5 7+3^5 

v } Vl+^-Vl^ 62 + 1 

198. If a + 6 + c + a* = 2s, prove that 

4(a& + cd)* - (a 2 + 6 2 - c 2 - d 2 ) 2 = 16 (* - a) (s - 6) (s - c) (* - d)- 

199. A man buys a number of articles for $ 6, and sells for $ 5.40 
all but two at 5 cents apiece more than they cost ; how many did he 
buy? 

200. Find the square root of 

2(81 x* + y*)- 2(9x 2 + s^XSx- y) 2 + (3x-y) 4 * 
901. If x : a :: y : b :: z : c, prove that 
(6c + ca + a6) 2 (x 2 + y 2 + s 2 ) = (te + ex + oy) 2 (a 2 + 6* + c 3 ). 

202. If a man save $ 10 more than he did -the previous year, and if 
he saved $ 20 the first year, in how many years will his savings amount 
to 91700? 



MISCELLANEOUS EXAMPLES VH. 511 

208. -Given that 4 is a root of the quadratic x a — 6x + q = 0, find 
the value of q and the other root. 

204. A person having 7 miles to walk increases his speed one mile 
an hour after the first mile, and finds that he is half an hour less on 
the road than he would have been had he not altered his rate. How 
long did he take ? 

205. If (a+& + c)x = (-a + & + c)y=(a-ft + c)»=(a + &-c)t0, 

show that - H f- — = — 

y z w x 

206. Find a Geometrical Progression of which the sum of the first 
two terms is 2f , and the sum to infinity 4£. 



( x + iH 1 - if 

207. Simplify x y/ v ' 



208. A man has a stable containing 10 stalls ; in how many ways 
could he stable 5 horses ? 

209. In boring a well 400 feet deep the cost is 27 cents for the first 
foot and an additional cent for each subsequent foot ; what is the cost 
of boring the last foot, and also of boring the entire well ? 

210. If a, b are the roots of x 2 + px + q = 0, show that p, q are the 
roots of the equation 

x 2 + (a + b - ab)x - ab(a + 6) = 0. 

211. Extract the square root of 7 — 80 V— 2. 

212. If % + z — ? = x determine the ratios xxyiz 

y x z — y 

218. If a, 6, c are in H. P., show that 



\a b c)\c b a) ft 2 ac 



214. Find the number of permutations which can be made from all 
the letters of the words 

(i.) Consequences, (ii.) Acarnania. 

215. Expand by the Binomial Theorem (2 a — 3 x) 6 ; and find the 
numerically greatest term in the expansion of (1 -f x)% if x = f , and 
» = 7. 

216. When x = — , find the value of 

4 

1 + 2* , l-2x 



1 + V1 + 23 l-Vl-2» 



512 



ALGEBRA. 



817. Simplify 



x 2 -6c 



afi-ea 



x*-*6 



(a - 6)( _ C ) • (ft - C )(6 _ a) T (c - a)(e - 6) 

818. Solve the equations 

(i.) (x 2 - 5x + 2)« = a? - 6x + 22. 

(ii.) (, 2+ l) a +4 (^ + I)=12. 

819. Prove that 

(y - z) s + (x - y) 8 + 3(x - y)(x - z)(y -«) = («- *)». 

820. Out of 16 consonants and 5 vowels, how many words can be 
formed each containing 4 consonants and 2 vowels ? 

881. If b — a is a harmonic mean l>etween c — a and <f — a, show 
that' d — c is a harmonic mean between a — c and 6 — c. 

228. In how many ways may 2 red balls, 3 black, 1 white, 2 bine 
be selected from 4 red, 6 black, 2 white, and 6 blue ; and in how 
many ways may they be arranged ? 

823. The sum of a certain number of terms of an arithmetical 
series is 36, and the first and last of these terms are 1 and 11 respec- 
tively : find the number of terms, and the common difference of the 
series. 

824. Expand by the Binomial Theorem 

(i.) (2 - ^V. (ii.) (1 - f x)* to five terms. 



285. Solve 



x 2 — xy + x = 85. 
xy — y 2 + V = 15. 



886. Simplify ^^ x ii^I^i^H and find the value of 
3V27 4VT5 7V48 

: , given that V 6 = 2 - 230 - 



887. By the Binomial Theorem find the cube root of 128 to six 
places of decimals. 

228. There are 9 books, of which 4 are Greek, 3 are Latin, and 
2 are English ; in how many ways could a selection be made so as to 
include at least one of each language ? 

229. Simplify 

n\ V45X 8 - V80s 8 + VbcPx 

a — x 

(in [ x ^ + x ~^ x*-x~* ( . J x* + 2x~* x*-2x~* l 



MISCELLANEOUS EXAMPLES VII. 5lS 

230. (i.) Form the quadratic equation whose roots are 5 ± y/6. 
(ii.) If the roots of x 2 — px + q = are two consecutive integers, 
prove that p 2 — 4g — 1=0. 

281. Solve x 9 + 1 = 81(^ + y); x* + x = 9(y« + 1). 

232. Find log 16 128, log 4 VI28, log 2 £ ; and having given 

log 2 = .3010300 and log 3 = .4771213, 
find the logarithm of .00001728. 

233. A and B start from the same point, B five days after A ; 
A travels 1 mile the first day, 2 miles the second, 3 miles the third, 
and so on ; B travels 12 miles a day. When will they be together ? 
Explain the double answer. 

234. Solve the equations : 

(i.) 2* = 8H-i, 9* = 3*-». 
(ii.) «■ = y 2 ", 2» = 2 x 4», x + y + z = 16. 

235. The sum of the first 10 terms of an arithmetical series is to 
the sum of the first 5 terms as 13 is to 4 ; find the ratio of the first 
term to the common difference. 

236. Find the greatest term in the expansion of (1 — z)~* when 

*r — 12 
* — T3« 

237. Five gentlemen and one lady wish to enter an omnibus in 
which there are only three vacant places; in how many ways can 
these places be occupied (1) when there is no restriction, (2) when 
one of the places is to be occupied by the lady ? 

238. (i.) Given log2=.801030, log3=.477121, and log7=.845098, 

find the logarithms of .005, 6.3, and (AW^. 

(ii.) Find a; from the equation 18 8 -^' 8 =(64 v /2) a •- , . 

239. If P and Q vary respectively as y* and y* when z is constant, 

and as z* and z % when y is constant, and if x = P + Q, find the equa- 
tion between x, y, z; it being known that when y = z = 64, x = 12 ; 
and that when y = 4 z = 16, x = 2. 

240. Simplify 

log W + 2 lo &¥ - lo g-W + 1o &tVt- 

241. If the number of permutations of n things 4 at a time is to the 
number of combinations of 2 n things 3 at a time as 22 to 3, find n. 

242. If - + - = — - — + — - — , prove that 2 b is either the arith- 

a c 2b — a 2b — c 

metic mean between 2 a and 2 c, or the harmonic mean between a 

and a 

2l 



614 



ALGEBRA. 



MS. If n C r denote the number of combinations of n things taken 
r together, prove that 

*+*Cr+l = «C r+ , + "O-i + (2 x «C r ). 
844. Find (L) the characteristic of log 54 to base 3. 

(ii.) logio (.0125) K (iii.) the number of digits in 3« 
Given logi 2 = .30103, log w 3 = .47712. 

345. Write the (r + l)th term of (2 as 2 - a*)*, and express it in 

its simplest form. 

246. At a meeting of a debating society there were speakers; 
6 spoke for the affirmative, and 4 for the negative. In how many 
ways could the speeches have been made, if' a member of the affirma- 
tive always spoke first, and the speeches were alternately for the 
affirmative and the negative ? 

347. Form the quadratic equation whose roots are 



a + b + Va 2 + 6 2 and 



2ab 



a + b + Va 2 + 6 2 



348. A point moves with a speed which is different in different 
miles, but invariable in the same mile, and its speed in any mile varies 
inversely as the number of miles travelled before it commences this 
mile. If the second mile be described in 2 hours, find the time taken 
to describe the nth mile. 

349. Solve the equations : 

(i.) x\b - c)+ax(c - a)+a\a - 6)= 0. 
(ii.) (a 2 - px + p 2 ) (qx +pq+ p 2 ) = qx* + p 2 q* + p*. 

250. Prove by the Binomial Theorem that V8 is the value, to 
infinity, of 

, . 3 . 3«5 , 3«5'7 . 

4 4-8 4-8-12 

251. A and B run a mile race. In the first heat A gives B a start 
of 11 yards and beats him by 57 seconds ; in the second heat A gives 
B a start of 81 seconds and is beaten by 88 yards : in what time could 
each run a mile ? 

252. A train, an hour after starting, meets with an accident which 
detains it an hour, after which it proceeds at three-fifths of its former 
rate and arrives 3 hours after time ; but had the accident happened 
50 miles farther on the line, it would have arrived 1 J hours sooner z find 
the length of the journey. 



MISCELLANEOUS EXAMPLES VII. 515 

268. Expand for 4 terms by the method of Undetermined Co* 
2 



efficients 



8x 2 -2x* 



254. A body of men were formed into a hollow square, three deep, 
when it was observed that with the addition of 25 to their number a 
solid square might be formed, of which the number of men in each 
side would be greater by 22 than the square root of the number of men 
in each side of the hollow square : required the number of men. 



255. Expand into a series Va 2 + 6 2 . 

256. Solve the equation V2x-1 + V3x-2 s= V4x-3 + V6x-4 

257. Separate 7 ^ a + 22 ^ + 5 mt0 partia i fractions. 

(x + 3)(x 2 -l) 

258. Solve the equation x* — 6 a; 2 — 6x — 6=0. 

259. Find the generating function of 

l + 5x + 7x 2 + 17x 8 + 31x* + ... # 

q /m Ai2 __ A 

260. Separate — - — n r mt0 partial fractions. . 

(x 2 + l)(x 2 -x-2) 

261. Solve the equation x 4 + 3 a? 2 = 16 x + 60. 

262. Express |Jf as a continued fraction and find the fourth con- 
vergent. 

268. What is the sum of n terms of the series 1, 8, 27, 64, ••• ? 

264. The sum of 6 terms of the series 1 — xV— 1 — afi + — is equal 
to 65 times the sum to infinity ; find x. 

265. Convert 2^/5 into a continued fraction. 

266. Find limits of the error when f £f is taken for VS8. 

267. Sum to infinity the series 

3 - x _ 2o? 2 - 16x 8 - 28s* - 676s 6 + ..* 

268. Find value of -L -i- -^- -!- -~- -i—. 

3+2+1+3+2+1 + 

269. Solve, by Cardan's Method, the equation 

x 8 - 30 x +133=0. 

270. Solve the equation x 8 - 13 x 2 + 15 x + 189 = 0, having given 
that one root exceeds another root by 2. 

271. Solve the equation x 4 — 4x 2 + 8x + 36 = 0, having given that 
one root is 2 + V— 3. 



516 



ALGEBRA. 



872. Sum to infinity the series 

4 - 9x + 16 a? 2 - 26 a* + 86 a* - 40 a* + .... 

278. Solve the equation a* - 123c 8 + 47 a* — 72* + 96 = 0. 



274. Solve the equation 



= 0. 



4* 6a; + 2 8& + 1 
6* + 2 9x + 3 12x 
8oj+1 12s 16s + 2 

275. Solve the equation 2 a 5 + a* + x + 2 = 12 a* + 12**. 

276. Given log2 = .30108, and log3 = .47712, solve the equations : 

(1.) 6- = y - 6-. (ii.) V& + V6^ = ff. 

277. Find the ralue of 1 +-L _L JL * .. in the form of a 
quadratic surd. 3+2+3+2+ 



278. Separate 



a* + 7a*-s-8 



(x 2 + x+l)(a; a -3x-l) 



into partial fractions. 



279. Find the general term when a 8a? ~~ 8 is expanded in ascend* 
ing powers of x. x —*■*— 4 

>. Solve the equations : 

v x + y + Vx — y 
(iL) V2x 2 + 1+V2x 2 -1 = 2 



V3-2a* 
(ilL) ** - 4a* -10a* + 40a* + 9x- 86 =0. 



CHAPTER XLIX. 
Graphical Representation of Functions. 



Definition. Any expression which involves a 
variable quantity x, and whose value is dependent on that 
of x, is called a function of jr. 

Thus,3a?-f 8, 2x* + 6x — 7, a 4 — 3x* + x* — 9 are func- 
tions of x of the first, second, and fourth degree respectively. 



The symbol f(x) is often used to denote briefly a 
function of x. If y =f(x), by substituting a succession of 
numerical values for x, we can obtain a corresponding suc- 
cession of values for y which stands for the value of the 
function. Hence, in this connection it is sometimes con- 
venient to call x the independent variable, and y the depend- 
ent variable. 

634. Consider the function x(9 — a? 2 ), and let its value be 
represented by y. 

Then, when x = 0, y = 0x9= 0, 

when x = l, y = lx8= 8, 

when x = 2, y=2x5= 10, 

when x = 3, y = 3x0= 0, 

when x = 4, y = 4 x (— 7) = — 28, 

and so on. 

By proceeding in this way we can find as many values of 
the function as we please. But we are often not so much 
concerned with the actual values which a function assumes 
for different values of the variable as with the way in which 
the value of the function changes. These variations can be 
very conveniently represented by a graphical method which 
we shall now explain. 

517 



518 



ALGEItUA. 



l 4 ^ 



I 



Y 
Fig. 1. 



Two straight lines XOX 1 , TOY 1 are taken inter- 
secting at right angles in 0, thus dividing the plane of the 
paper into four spaces XOY, YOX', X'OY', Y'OX, which 

are known as the first, sec- 
ond, third, and fourth quad- 
rants respectively. 

The lines XOX 1 , YOY' are 
usually drawn horizontally 
and vertically ; they are 
I i ■ taken as lines of reference 

and are known as the axis of 
x and the axis of / respec- 
tively. The point is called 
the origin. Values of x are 
measured from along the 
axis of x, according to some 
convenient scale of measure- 
ment, and are called abscissas, positive values being drawn 
to the right of along OX, and negative values to the left 
of along OX'. 

Values of y are drawn (on the same scale) parallel to the 
axis of y, from the ends of the corresponding abscissas, and 
are called ordinates. These are positive when drawn above 
XX', negative when drawn below XX'. 

636. The abscissa and ordinate of a point taken together 
are known as its coordinates. A point whose coordinates 
are x and y is briefly spoken of as "the point (x, y)." 

The coordinates of a point completely determine its posi- 
tion in the plane. Thus, if we wish to mark the point (2, 3), 
we take x = 2 units measured to the right of 0, y = 3 units 
measured perpendicular to the #-axis and above it. The re- 
sulting point P is in the first quadrant. The point (—3, 2) 
is found by taking x = 3 units to the left of 0, and y = 2 
units above the aj-axis. The resulting point Q is in the 
second quadrant. Similarly the points (—3, — 4), (5, -— 5) 
are represented by R and S in Fig. 1, in the third and fourth 
quadrants respectively. 



GRAPHICAL REPRESENTATION OP FUNCTIONS. 519 

This process of marking the position of a point in reference 
to the coordinate axes is known as plotting the point. 

637. In practice ft is convenient to use squared paper; 
that is, paper ruled into small squares by two sets of equi- 
distant parallel straight lines, the one set being horizontal 
and the other vertical. After selecting two of the intersect- 
ing lines as axes (and slightly thickening them to aid the 
eye), one or more of the divisions may be chosen as our unit, 
and points may be readily plotted when their coordinates are 
known. Conversely, if the position of a point in any of the 
quadrants is marked, its coordinates can he measured by the 
divisions on the paper. 

In the following pages we have used paper ruled to tenths 
of an inch, but a larger scale will sometimes be more con- 
venient. See Art. 657. 



X 



Ex. Plot the points (5, 2), 
(-3,2), (-3,-4), (5, -4) : 
on squared paper. Find the 
area of the figure determined 
by these points, assuming the 
divisions on the paper to be ~ 
tenths of an inch. 

Taking the points in the 
order given, it is easily seen 
that they are represented by ~ 
P, §, i?, S in Fig. 2, and that I 
they form a rectangle which 
contains 48 squares. Each of 
these is one-hundredth part of 
a square inch. Thus, the area of the rectangle is .48 of a square inch. 



£- 



Fig. 2. 



EXAMPLES XLIX. a. 

[ The folloioing examples are intended to be done mainly by actual 
measurement on squared paper ; where possible^ they should also be 
verified by calculation.'] 

Plot the following pairs of points and draw the line which joins them : 

1. (3,0), (0,0). 2. (-2,0), (0, -8). 

3. (3, -8), (-2, 6). 4. (5,5), (-2, -2). 

5. (-2, 6), (1,-3). 6. (4,6), (-1,5). 



520 ALGEBRA. 

7. Plot the points (3, 3), (-3, 3), (- 3, - 3), (3, - 3), and find 
the number of squares contained by the figure determined by these 
points. t 

8. Plot the points (4, 0), (0, 4), (-4, 0), (0, - 4), and find the 
number of units of area in the resulting figure. 

9. Plot the points (0, 0), (0, 10), (5, 6), and find the number of 
units of area in the triangle. 

10. Show that the triangle whose vertices are (0, 0), (0, 6), (4, 3) 
contains 12 units of area. Show also that the points (0, 0), (0, 6), 
(4, 8) determine a triangle of the same area. 

11. Plot the points (5, 6), (- 6, 6), (6, - 6), (- 6,-6). If one 
millimeter is taken as unit, find the area of the figure in square 
centimeters. 

12. Plot the points (1, 8), (— 3, — 9), and show that they lie on a 
line passing through the origin. Name the coordinates of other 
points on this line. 

13. Plot the eight points (0, 6), (3, 4), (6, 0), (4, - 3), (- 5, 0), 
(0, — 6), (— 4, 8), (—4, — 8), and show that they are all equidistant 
from the origin. 

14. Plot the two following series of points : 

(i.) (5, 0), (5, 2), (6, 5), (5, -1), (6, -4); 
(ii.) (-4, 8), (- 1, 8), (0, 8), (3, 8), (6, 8). 

Show that they lie on two lines respectively parallel to the axis of y, 
and the axis of x. Find the coordinates of the point in which they 
intersect. 

15. Plot the points (13, 0), (0, - 13), (12, 6), (- 12, 6), (- 13, 0), 
(—6, —12), (5, —12). Find their locus, (i.) by measurement, 
(ii.) by calculation 

16. Plot the points (2, 2), (- 3, - 3), (4, 4), (- 5, - 6,) showing 
that they all lie on a certain line through the origin. Conversely, 
show that for every point on this line the abscissa and ordinate are 
equal. 

GRAPH OF A FUNCTION. 

638. Let f(x) represent a function of x, and let its value 
be denoted by y. If we give to a? a series of numerical 
values, we get a corresponding series of values for y. If 
these are set off as abscissas and ordinates respectively, we 
plot a succession of points. If all such points were plotted, 



GRAPHICAL REPRESENTATION OF FUNCTIONS. 521 



we should arrive at a line, straight or curved, which is known 
as the graph of the function f(x), or the graph of the equa- 
tion y = f(x). The variation of the function for different 
values of the variable x is exhibited by the variation of the 
ordinates as we pass from point to point. 

In practice, a few points carefully plotted will usually 
enable us to draw the graph with sufficient accuracy. 

639. The student who has worked intelligently through 
the preceding examples will have acquired for himself some 
useful preliminary notions which will be of service in the 
examples on simple graphs which we are about to give. In 
particular, before proceeding farther he should satisfy him- 
self with regard to the following statements : 

(i.) The coordinates of the origin are (0, 0). 
(ii.) The abscissa of every point on the axis of y is 0. 
(iii.) The ordinate of every point on the axis of x is 0. 
(iv.) The graph of all points which have the same abscissa 
is a line parallel to the axis of y. (e.g. x = 2.) 

(v.) The graph of all points which have the same ordinate 

is a line parallel to the axis of x. (e.g. y = 5.) 

(vi.) The distance of any point P(x, y) from the origin is 
given by OF 2 = x 2 + y 2 . 



Ex. 1. Plot the graph of y 
When x = 0, y = ; thus 

the origin is one point on the 

graph. 

Also, when 

05 = 1, 2, 3, ... -1, 

— 2, — 3, •••, 
y = 1, 2, 3, ... — 1, 

— 2, — o, ••«. 

Thus the graph passes 
through O, and represents a 
series of points each of which 
has its ordinate equal to its 
abscissa, and is clearly repre- 
sented by POP in Fig. 3. 



= x. 




Fig. 8. 



x 3 2 1 -1 -2 -S 
y 6 6 4 8 2 1 



By joining these points we obtain a line M N parallel to that in 
Example 1. 

The results printed in heavier type should be specially noted and 
compared with the graph. They show that the distances OK, OM 
(usually called the intercepts on the axes) are obtained by separately 
putting x = 0, if = in the equation of the graph. 

Note. By observing that in Example 2 each ordinate is 3 units 
greater than the corresponding ordinate in Example 1, the graph of 
y — x + 3 may be obtained from that of y = x by simply producing 
each ordinate 3 units in the positive direction. 

In like manner the equations 



k + 5, y = 



-6, 



represent two parallel lines on opposite sides of y = x and equi- 
distant from it, as the student may 
easily verify for himself. 

Ex. S. Plot the graphs represented 
by the following equations : 
00 !> = 23i 



(«■) y 

(hi.) y 



Here we only give the diagram, 
which the student should verify in 
detail for himself, following the 
method explained in the two preced- 
ing examples. Fra. 4. 
EXAMPLES XLIX. b. 

[In the following examples Nos. 1-18 are arranged in groups of 
three ; each group shonld be represented on the same diagram so as to 
exhibit clearly the ■pnnitifin of the three yro.)>u< ri-.i'ztirely to each other.'] 

d by the following equations : 



Plot tiie graphs 



= 5 x + 6. 



GRAPHICAL REPRESENTATION OF FUNCTIONS. 523 

7. y + x = 0. 8. y + x = 8. 9. y + 4 = x. 

10. 4s = 3y. 11. 3y = 4x + 6. 12. 4y+3x = 8. 

13. a - 5 = 0. 14. y — 6 = 0. . 15. 6 y = 6 se. 

16. 3 x + 4 y = 10. 17. 4 a; + y = 9. 18. 6 a; - 2 y = 8. 

19. Show by careful drawing that the last three graphs have a 
common point whose coordinates are 2, 1 . 

20. Show by careful drawing that the equations 

z + y = 10, y = se — 4 
represent two straight lines at right angles. 

21. Draw on the same axes the graphs of x = 5, x = 9, y = 8, y = 11. 
Find the number of units of area enclosed by these lines. 

22. Taking one tenth of an inch as the unit of length, find the 
area included between the graphs oi x = 7, x=— 3, y = — 2, y = S. 

23. Find the area included by the graphs of 

y = x + Q, y — x — 6, y=— x + 6 t y= — x — 6. 

24. With one millimeter as linear unit, find in square centimeters 
the area of the figure enclosed by the graphs of 

y=2x+$, y = 2x-8, y=-2x + 8, y=-2x — 8. 

640. The student should now be prepared for the fol- 
lowing statements : 

(i.) For all numerical values of a the equation y = ax rep- 
resents a straight line through the origin. 

(ii.) For all numerical values of a and b the equation 
y = ax + b represents a line parallel to y = ax, and cutting 
off an intercept b from the axis of y. 

641. Conversely, since every equation involving x and y 
only in the first degree can be reduced to one of the forms 
y = ax, y = ax + b, it follows that every simple equation con- 
necting two variables represents a straight line. For this reason 
an expression of the form ax-\-b is said to be a linear func- 
tion of x, and an equation such as y=ax+b, or ao;-f-6y-f-c=0, 
is said to be a linear equation. 

Ex. Show that the points (3, — 4), (0, 4), (12, 8) lie on a straight 
line, and find its equation. 



524 



ALGEBRA. 



Assume y = ax + b as the equation of the line. If it passes through 

the first two points given, their coordinates must satisfy the above 

equation. Hence 

- 4 = 3 a + 6, 4 = 9 a + 6. 

These equations give a = J , b = — 8. 

Hence y = J x — 8, or 4 x — 3 y = 24, 

is the equation of the line passing through the first two points. 
Since z = 12, y = 8 satisfies this equation, the line also passes through 
(12, 8). This example may be verified graphically by plotting the 
line which joins any two of the points and showing that it passes 
through the third. 



APPLICATION TO SIMULTANEOUS EQUATIONS. 

642. It was shown in Art. 167 that in the case of a simple 
equation between x and y, it is possible to find as many- 
pairs of values of x and y as we please which satisfy the 
given equation. We now see that this is equivalent to saying 
that we may find as many points as we please on any given 
straight line. If, however, we have two simultaneous equa- 
tions between x and y, there can be only one pair of values 
which will satisfy both equations. This is equivalent to 

saying that two straight lines 
can have only one common 
point. 

Ex. Solve graphically the equa- 
tions : 

3 x + 7 y = 27, 5 x + 2 y = 16. 

If carefully plotted these two equa- 
tions will be found to represent the 
lines in the diagram. On measur- 
ing the coordinates of the point at 
which they intersect it will be found 
that x = 2, y = 3, verifying a solution 
Fig. 5. by method of Art. 171. 

643. It will now be seen that the process of solving 
two simultaneous equations is equivalent to finding the 
coordinates of the point (or points) at which their graphs 
meet. 




GRAPHICAL REPRESENTATION OF FUNCTIONS. 525 

644. Since a straight line can always be drawn by joining 
any two of its points, in solving linear simultaneous equa- 
tions graphically, it is only necessary to plot two points of 
each line. The points where the lines meet the axes will 
usually be the most convenient to select. 

645. Two simultaneous equations lead to no finite solu- 
tion if they are inconsistent with each other. For example, 
the equations ^ 

are inconsistent, for the second equation can be written 

x 4- 3 y = 2|, which is clearly inconsistent with x -f- 3 y = 2. 

The graphs of these two equations will be found to be 

two parallel straight lines which have no finite point of 

intersection. 

Again, two simultaneous equations must be independent. 

The equations 

H 4* + 3y = l, 16a? + 12y = 4 

are not independent, for the second can be deduced from the 
first by multiplying throughout by 4. Thus any pair of 
values which will satisfy one equation will satisfy the other. 
Graphically these two equations represent two coincident 
straight lines which of course have an unlimited number of 
common points. 

EXAMPLES XLIX. c. 

Solve the following equations, in each case verifying the solution 
graphically : 

1. y = 2z + 3, 2. y = 3z + 4, 8. y = 4x, 

y + x = Q. y = x + $. 2x + y = l$. 

4. 2x — y = $, 5. 3z + 2y = 16, 6. 6y - 5a = 18, 

4x + 3y = 6. 5se-3y = 14. 4z = Sy. 

7. 2 x + y = 0, 8. 2 x - y = 3, 9. 2 y = 5 x + 15, 

y = f(x + 5). 3 x -5 y = 15. 3y-4x = 12. 

10. Prove by graphical representation that the three points (3, 0), 
(2, 7), (4, — 7) lie on a straight line. Where does this line cut the 
axis of y ? 



526 



ALGEBRA. 



11. Prove that the three points (1, 1), (— 3, 4), (5, — 2) lie on a 
straight line. Find its equation. Draw the graph of this equation, 
showing that it passes through the given points. 

12. Show that the three points (3, 2), (8, 8), ( — 2, — 4) lie on a 
straight line. Prove algebraically and graphically that it cuts the 
axis of x at a distance 1} from the origin. 

646, We shall now give some graphs of functions of 
higher degree than the first. 

Ex. 1. Plot the graph of 2 y = x*. 

Corresponding values of x and y may be tabulated as follows : 



X 

y 


• • • 

• • • 


3 


2.5 


2 
2 


1.5 


1 
.5 






-1 


-2 


-3 


• • • 


4.5 


3.125 


1.126 


.5 


2 


4.5 


• • • 



Here, in order to obtain a figure on a sufficiently large scale, it 
will be found convenient to take two divisions on the paper for our 

unit. 

If the above points are plotted 
and connected by a line drawn 
freehand, we shall obtain the 
curve shown in Fig. 6. This curve 
is called a parabola. 

There are two facts to be spe- 
cially noted in this example.' 

(i.) Since from the equation we 

have x = ± V2 y, it follows that for ' 
every value of the ordinate we have 
two values of the abscissa, equal 
in magnitude and opposite in sign. 
Hence the graph is symmetrical 
with respect to the axis of y\ so 
that after plotting with care enough 
points to determine the form of the graph in the first quadrant, its 
form in the second quadrant can be inferred without actually plotting 
any* points in this quadrant. At the same time, in this and similar 
cases beginners are recommended to plot a few points in each quadrant 
through which the graph passes. 

(ii. ) We observe that all the plotted points lie above the axis of x. 
This is evident from the equation ; for since x 2 must be positive for 




Fig. 6. 



GRAPHICAL REPRESENTATION OF FUNCTIONS. 527 



x* 
all values of x, every ordinate obtained from the equation y = — 

must be positive. 

In like manner the student may show that the graph of 2 y = — x 2 

is a curve similar in every respect to that in Figi 6, but lying entirely 

below the axis of x. 

Note. Some further remarks on the graph of this and the next 
example will be found in Art. 651. 

Ex. 2. Find the graph of y=2 x + — • 

4 

Here the following arrangement will be found convenient : 



X 

2x 

X 2 

4 

y 


3 
6 

2.25 

8.25 


2 
4 

1 

5 


1 
2 

.25 

2.25 










-1 


- 2 

-4 

1 
-3 


-3 


-4 
-8 

4 

-4 


-5 


-6 


-7 


-8 
-16 

16 




-2 


-6 


-10 


-12 


-14 


.25 


2.25 


6.25 


9 


12.25 


- 1.75 


-3.75 


-3.75 


-3 


-1.75 



From the form of the 
equation it is evident that 
every positive value of x will 
yield a positive value of y, 
and as x increases y also 
increases. Hence the por- 
tion of the curve in the first 
quadrant lies as in Fig. 7, 
and can be extended indefi- 
nitely in this quadrant. In 
the present case only two or 
three positive values of x 
and y need be plotted, but 
more attention must be paid 
to the results arising out of 
negative values of x. 

When y = 0, we have — + 2 x = ; thus the two values of x in the 

4 

graph which correspond to y = furnish the roots of the equation 

^- + 2x = 0. 
4 




Fig. 7. 



528 



ALGEBRA. 



647. If f(x) represents a function of x, an approximate 
solution of the equation f(x)=0 may be obtained by plotting 
the graph of y=f(x), and then measuring the intercepts made 
on the axis of x. These intercepts are values of x which 
make y equal to zero, and are therefore roots of f(x)= 0. 

648. If f(x) gradually increases till it reaches a value a, 
which is algebraically greater than neighboring values on 
either side, a is said to be a maximum value off(x). 

If f(x) gradually decreases till it reaches a value b, which 
is algebraically less than neighboring values on either side, 
b is said to be a minimum value of /(#). 

When y=f(x) is treated graphically, it is now evident 
that maximum and minimum values of f(x) occur at points 
where the ordinates are algebraically greatest and least in 
the immediate vicinity of such points. 

Ex. Solve the equation x 2 — 7se-fll = graphically, and find 
the minimum value of the function x 2 — 7 x + 11. 

Put y — x l — 1 x + 11, and find the graph of this equation. 



X 




li 


i 

5 


2 


3 


3.5 


4 


5 
1 


6 
5 


7 


y 


1 


- 1 


-1.25 


-1 


11 




The values of x which make the 
function x 2 — 7 x + 11 vanish are 
those which correspond to y = 0. 
By careful measurement it will be 
found that the intercepts OM and 
ON are approximately equal to 2.38 
and 4.62. 

The algebraical solution of 



gives 



x 2 -7z+ll =0 

x = i(±7V5). 



Fig. 8. 



If we take 2.236 as the approximate 
value of V5, the values of x will be 
found to agree with those obtained 
from the graph. 



GRAPHICAL REPRESENTATION OF FUNCTIONS. 529 

Again, x? - 7 x + 11 = (z - j)* - %. Now (x — J ) ! must be positive 
for all real values of 2 except 2 = }, In which case it vanishes, and 
the value of the function reduces to — f, which is the least value it 

The graph shows that when x = 3.6, y = — 1.25, and that this is 
the algebraically least ordinate in the plotted curve. 

649. The following example shows that points selected 
for graphical representation must sometimes be restricted 
within certain limits. 



(i.) y = ± V3fi - x* ; (ii.) x = ± VS6 - ff. 

In order that y ma; be a real 
quantity we see from (i.) that 
36— r? must be positive. Thus 
x can only have values between 
— 6 and 4- 6. Similarly from 
(ii.) it is evident that y must 
also lie between — 6 and + 6. 
Between these limits it will be 
found that all plotted points 
will lie at a distance 6 from 
the origin. Hence the graph 
is a circle whose centre is 
and whose radius is 6. 

This is otherwise evident, 
for the distance of any point 

P(x, ji) fro m the o rigin is given ^ B 

by OP ^Vx' + y'. [Art. 63!).] 
Hence the equation a 2 + y* = SO 
series of points all of which are a 

Note. To plot the curve from equation (ii.), we should select a 
succession of values for y and then find corresponding values of x. 
In other words we make y the independent and x the dependent 
variable. The student should be prepared to do this in some of the 
examples which follow. 



530 



ALGEBRA . 



EXAMPLES XLIX. d. 

1. Draw the graphs of y = z 2 , and x = y 2 , and show that they 
have only one common chord. Find its equation. 

x 2 

2. From the graphs, and also by calculation, show that y = — 

cuts x = — y 2 in only two points, and find their coordinates. - 

8. Draw the graphs of 



x 2 



x a 



(i.) 3, 2 =-4x; (ii.)y = 2x-^-; (iii.) y = ±. + x-2. 

4 4 

4. Draw the graph of y = x + x 2 . Show also that it may be 
deduced from that of y = x 2 , obtained in Example 1. 

5. Show (i. ) graphically, (ii.) algebraically, that the line y = 2 x— 3 

x 2 
meets the curve y = — h x — 2 in one point only. Find its coordinates. 

4 

6. Find graphically the roots of the following equations to 2 
places of decimals : 

(i.) £ 2 + a j_2 = 0; (ii.) x 2 -2x = 4; (iii.) 4x 2 -16x + 9 = 0; 
4 

and verify the solutions algebraically. 

7. Find the minimum value of x 2 — 2 x — 4, and the maximum 
value of 5 -f 4 x — 2 x 2 . 

8. Draw the graph of y =(x — l)(x — 2) and find the minimum 
value of (x— l)(x — 2). Measure, as accurately as you can, the 
values of x for which (x — l)(x — 2) is equal to 5 and 9 respectively. 
Verify algebraically. 

9. Solve the simultaneous equations. 

x 2 + y 2 = 100, x + y = 14 ; 

and verify the solution by plotting the graphs of the equations and 
measuring the coordinates of their common points. 

10. Plot the graphs of x 2 + y 2 = 25, 3 x + 4 y = 25, and examine 
their relation to each other where they intersect. Verify the result 
algebraically. 

650. It should be observed that when the symbols for zero 
and infinity are used in the sense explained in Arts. 181, 182, 
they are subject to the rules of signs which affect other 
algebraical symbols. Thus we shall find it convenient to 
use a concise statement such as " when #= + 0, y = -{- oc" 



GRAPHICAL REPRESENTATION OF FUNCTIONS. 531 

to indicate that when a very small and positive value is 
given to x, the corresponding value of y is very large and 
positive. 

651. If we now return to the examples worked out in Art. 
646, in Example 1, we see that when x= ± oo, y= + oo ; hence 
the curve extends upwards to infinity in both the first and 
second quadrants. In Example 2, when x = + oo, y = -f oo. 
Again y is negative between the values and — 8 of x. For 
all negative values of x numerically greater than 8, y is 
positive, and when a; = — oo, y = + oo. Hence the curve 
extends to infinity in both the first and second quadrants. 

The student should now examine the nature of the graphs 
in Examples XLIX. d. when x and y are infinite. 

Ex. Find the graph of xy = 4. 

The equation may be written in the form 

4 

x 

from which it appears that when x = 0, y = oo and when x = oo, y = 0. 
Also y is positive when x is positive, and negative when x is negative. 
Hence the graph must lie entirely in the first and third quadrants. 

It will be convenient in this case to take the positive and negative 
values of the variables separately. 

(1) Positive values : 



X 





1 

4 


2 


3 


4 

1 


5 

.8 


6 
1 


• • • 


oo 


y 


QO 


2 


1* 


• « • 






Graphically these values show that as we recede farther and 
farther from the origin on the x-axis in the positive direction, the 
values of y are positive and become smaller and smaller. That is, 
the graph is continually approaching the x-axis in such a way that 
by taking a sufficiently great positive value of x we obtain a point 
on the graph as near as we please to the x-axis, but never actually 
reaching it until x = oo- Similarly, as x becomes smaller and smaller 
the graph approaches more and more nearly to the positive end of 
the y-axis, never actually reaching it as long as x has any finite 
positive value, however small. 



532 



ALGEBRA. 



















1 






Y 




















• 




































































i 

-4- 




































































































1 












































1 




































































\\ 


















1 
























































































1 


















































(' 
















1 





















X 




1 


r\ 










*S 


r\ 




! 






















• 


^" 
















^ 


y 












































A 
























1 






















\ 






















1 






















\ 






















1 
4 






















\ 


































































' 


































































\ 


Y 


r 
i 












































1 


















' 



(2) Negative values : 



Fig. 10. 



X 

y 


-0 


-1 


-2 


-3 


-4 


-5 


• • • 


— 00 


— 00 


-4 


-2 


-It 


-1 


-.8 


• • • 


-0 



The portion of the graph obtained from these values is in the third 
quadrant as shown in Fig. 10, and exactly similar to the portion 
already traced in the first quadrant. It should be noticed that as 
x passes from -f- to — the value of y changes from + oo to — oo. 
Thus the graph, which in the first quadrant has run away to an 
infinite distance on the positive side of the y-axis, reappears in the 
third quadrant coming from an infinite distance on the negative side 
of that axis. Similar remarks apply to the graph in its relation to 
the x-axis. 



When a curve continually approaches more and more 
nearly to a line without actually meeting it until an infinite 
distance is reached, such a line is said to be an asymptote to 
the curve. In the above case each of the axes is an asymptote. 

653. Every equation of the form y = -, or xy = c, where 

x 

c is constant, will give a graph similar to that exhibited in 



GRAPHICAL REPRESENTATION OF FUNCTIONS. 533 

the example of Art. 651. The resulting curve is known as 
a rectangular hyperbola, and has many interesting properties. 
In particular we may mention that from the form of the 
equation it is evident that for every point (x, y) on the curve 
there is a corresponding point (— x, — y) which satisfies the 
equation. Graphically this amounts to saying that any line 
through the origin meeting the two branches of the curve in 
P and F is bisected at 0. 



In the simpler cases of graphs, sufficient accuracy 
can be obtained usually by plotting a few points, and there 
is little difficulty in selecting points with suitable coordi- 
nates. But in other cases, and especially when the graph 
has infinite branches, more care is needed. The most im- 
portant things to observe are (1) the values for which the 
function f(x) becomes zero or infinite ; and (2) the values 
which the function assumes for zero and infinite values of x. 
In other words, we determine the general character of the 
curve in the neighborhood of the origin, the axes, and 
infinity. Greater accuracy of detail can then be secured by 
plotting* points at discretion. The selection of such points 
will usually be suggested by the earlier stages of our work. 
The existence of symmetry about either of the axes should 
also be noted. When an equation contains no odd powers 
of x, the graph is symmetrical with regard to the axis of y. 
Similarly the absence of odd powers of y indicates symmetry 
about the axis of x. Compare Art. 646, Ex. 1. 

2 x -+- 7 
Ex. Draw the graph of y = - 1 — • [See fig. on next page.] 

x — 4 

2+1 
2 x + 7 x 

We have y — -r- = j, the latter form being convenient for 

infinite values of x. x 

(i.) When y = 0, x = - J, 1 

When y = oo, x = 4 ; J 

.". the curve cuts the axis of x at a distance —3.6 from the origin, 
and meets the line x = 4 at an infinite distance. 

If x is positive and very little greater than 4, y is very great and 
positive. If x is positive and very little less than 4, y is very great 
and negative. Thus the infinite points on the graph near to the line 



534 



ALGEBItA. 



1 = 4 have positive ordinate* to the right, and negative ordinates to 
the left of this line. 

(ii.) When x = 0, y=— 1.76,1 

When a; = oo, j = 2; J 

.'. the curve cuts the axis of y at a distance — 1.75 from the origin, 
and meets the line y = ~i at an infinite distance. 

By taking positive values of y very little greater and very little 
less than 2, it appears that the curve lies above the line y = 2 when 
x = + tB, and below this line when x =— to. 

The general character of the curve ia now determined : the lines 
PO*P(x = i) and Q0'Q'(y = 2) are asymptotes ; the two branches of 
the curve lie in the compartments PO'Q, P'O'Q', and the lower 
branch cuts the axes at distances — 3.5 and — 1.75 from the origin. 

To examine the lower branch in detail, values of x may be selected 
between — oo and — 3.5 and between — 3.5 and 4. 



• 


-„ 


- 


-i« 


-8 


-0 


-3.5 


-> 





* 


S 




4 


y 


s 


1.25 


.75 


.5 





-» 


-1.75 


-5.5 


-13 




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2 ■> 








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__ _" _." T 










1* P 



GRAPHICAL REPRESENTATION OF FUNCTIONS. 535 



The upper branch may now be dealt with in the same way, 
selecting values of x between 4 and oo. The graph will be found 
to be as represented in Fig. 11. 



When the equation of a curve contains the square or 
higher power of y, the calculation of the values of y corre- 
sponding to selected values of x will have to be obtained by 
evolution, or else by the aid of logarithms. We give one 
example to illustrate the way in which a table of four-figure 
logarithms may be employed in such cases. 

Ex. Draw the graph of 
y8 = jc(9-x 2 ). 

For the sake of brevity we 
shall confine our attention to 
that part of the curve which lies 
to the right of the axis of y, leav- 
ing the other half to be traced 
in like manner by the student. 

When x = 0, y = ; there- 
fore the curve passes through 
the origin. Again, y is posi- 
tive for all values of x between 
and 3, and vanishes when 
x = 3 ; for values of x greater 
than 3, y is negative and con- 
tinually increases numerically. Fig. 12. 



































- 
















Y 












































































































































































































































1 


x ; 








J 




















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y 


































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y' 


) 




















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X 


© O CO o o 


1 
1 

8 
8 

2 


2 


3 
9 






4 


5 


6 




X* 


4 


16 


25 


36 




9-x* 


5 


-7 


-16 


-27 




V s 


10 


-28 


-80 


-162 




log y3 


1 


1.4472* 


1.9031* 


2.2096* 




logy 


.3333 


.4824 


.6344 


.7365 




y 


• 2.15 


- 3.04 


-4.31 


-6.45 




* T~ t-~ 


■« • 




• , v 


~£ 4. 


i • 









sign is disregarded, but care must be taken to insert the proper 
signs in the last line which gives the successive values of y. 



ALGEBRA. 



These points will be sufficient to give a rough approximation to the 
curve. For greater accuracy a few intermediate values such as 
x = 1.5, 2.5, 3.5 ■■■ should be taken, and the resulting curve will be as 
in Fig. 12, in which we have taken two tenth* of an inch as our linear 
unit. 

MEASUREMENT ON DIFFERENT SCALES. 
656. For convenience on the printed page we have sup- 
posed the paper to be ruled to tenths of an inch, generally 
using one of the divisions as our linear unit. In practice, 
however, it will often be advisable to choose a unit much 
larger than this in order to get a satisfactory graph. For the 
sake of simplicity we have hitherto measured abscissas and 
ordinates on the same scale, but there is no necessity for 
so doing, and it will often be found convenient to measure the 
variables on differ- 

_ (- eut scales suggested 

" by the particular con- 

ditions of the ques- 
tion. 

As an illustration 
let us take the graph 

of k=2> givett in 

Art. 646. If with the 
same unit as before 
we plot the graph of 
y — a?, it will be 
found to be a curve 
similar to that there 
shown, but elongated 
in the direction of 
the axis of y. In 
fact, it will be the 
same as if the former 
graph were stretched 
to twice its length in 
the direction of the 
j-axis. 



GRAPHICAL REPRESENTATION OF FUNCTIONS. 537 



657. Any equation of the form y = ax*, where a is con- 
stant, will represent a parabola elongated more or less 
according to the value of a; and the larger the value of a 
the more rapidly will y increase in comparison with x. We 
might have very large ordinates corresponding to very small 
abscissas, and the graph might prove quite unsuitable for 
practical applications. In such a case the inconvenience is 
obviated by measuring the values of y on a considerably 
smaller scale than those of x. 

Speaking generally, whenever one variable increases 
much more rapidly than the other, a small unit should 
be chosen for the rapidly increasing variable and a large 
one for the other. Further modifications will be sug- 
gested in the examples which follow. 



On pages 
536 and 537 we 
give for compar- 
ison the graphs of 

y= a* (Fig. 13), 

and 
y = 8 x* (Fig. 14). 

In Fig. 13 the 
unit for x is twice 
as great as that 
for y. 

In Fig. 14 the 
as-unit is ten times 
the y-unit. 

It will be useful 
practice for the stu- 
dent to plot other 
similar graphs on 
the same or a larger 
scale. For exam- 
ple, in Fig. 14 the 
graphs of y = 16 x 2 



30 


Y 


































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i 

"" 1 


10 




















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i '/ 




1 


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1 


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1. 

Fig. 14. 



1-5 



- x>. Verify it from the graphs of 
S. Plot the graph of y = —, showing that it consists of two 



1. Discuss the general character of the graph of y — ^- where a 



8. Plot the graphs of 

(i.) S=l+h (ii.) S = 2 + ^- 
Verify by deducing them from the graphs of y = -, and y = — • 

6. Plot the graph of y = x* — 8 x. Examine the character of the 
curve at the points (1, — 2), (— 1, 2), and show graphically that the 
roots of the equation X s — 3 x = are approximately — 1. 782, 0, and 
1.732. 

7. Solve the equations : 

3 a; + 2 y = 16, xy = 10, 



. Plot the graphs of 

(i.) „ = !» = *; (II.) « 



9. Trace the c 



GRAPHICAL REPRESENTATION OF FUNCTIONS. 539 
Plot the graphs of 

"10. r = !±JS. u. y = I±if. 

l—X 1—x 

18. y = ^=i§. 18. y = fa-l)(»-2). 

a; — 4 x — 8 

16. y = x«-6x a + ll»-0. 17. lOy = x* - 6x a + x - 6. 

18. y = -*L. 19. y = 40 * ■ 80. y = *( 8 -*). 

y x a + 2 y x a + 10 y x + 6 

21. y = (*- 2 )(*- 3 ) . 22. y= («-l)(«-2)(« + l) 

x — 6 4 

28. y a = x a -6x + 4. 24. 4y a = x a (5 - x). 

25 „*- *(3-s)(s-8) gg g ^ (x + 7)(x-4)(x-10) 

x 2 + & ' x a + 6 

27. ^ = ^«JZfQ. 28. y* = ( 81 -* a )(* a ^. 

29. 5y 8 = x(x a -64). 30. 6y 8 = x a (36 - x 2 ). 

31. Plot the graphs of y = x 8 , and of y = 2 x 2 + x — 2. Hence find 
the roots of the equation x 8 — 2 x a — x + 2 = 0. 

32. Find graphically the roots of the equation 

x 8 - 4 x a - 5 x + 14 = 
to three significant figures. 

659. Besides the instances already given there are several 
of the ordinary processes of arithmetic and algebra which 
lend themselves readily to graphical illustration. 

For example, the graph of y = x°- may be used to furnish 
numerical square roots. For since x = ~Vy, each ordinate 
and corresponding abscissa give a number and its square root. 
Similarly cube roots may be found from the graph of y = x 3 . 

Ex. 1. Find graphically the cube root of 10 to 3 places of decimals. 

The required root is clearly a little greater than 2. Hence it will 
be enough to plot the graph of y = x 3 taking x = 2.1, 2.2, ••-. The 
corresponding ordinates are 9.26, 10.66, •••. 



540 



ALCEBEA. 



11 


y * i • • 


i ' ' • f i • ' ■ i ' i 4 JP 














• ^r % 


10 








^r 












c 






V 



























When 

x = 2, y = 8. 

Take the axes 
through this point 
and let the units 
for x and y be 10 
inches and .5 inch 
respectively. On 
this scale the por- 
tion of the graph 
differs but little 
from a straight line, 
and yields results to 
a high degree of 
accuracy. 
When y — 10, the measured value of x will be found to be 2.154. 

Ex. 2. Show graphically that the expression 4 x* + 4 x — 3 is 
negative for all real values of x between .5 and — 1.5, and positive 
for all real values of x outside these limits. [Fig. 16.] 

Put y = 4 x 2 + 4 x — 3, and proceed as in the example given in 
Art. 648, taking the unit for x four times as great as that for y. It 
will be found that the graph cuts the axis of x at points whose abscissas 
are .5 and — 1.5 ; and that it lies below the axis of x between these 
points. That is. the value of y is negative so long as x lies between 
.5 and - 1.5, and positive for all other values of x. 



2.1 
Tig. 15. 



2.15* 



2.2 



Or we may proceed as follows : 



Put y x -f 4 x 2 , and 
yi = — 4 x + 3, and plot 
the graphs of these 
two equations. At their 
points of intersection 
y l — y 2 , and the values 
of x at these points are 
found to be .5 and — 1.5. 
Hence for these values 
of x we have 

4x 2 = 4x + 3, 
or 4x 2 +4x — 3 = 0. 

Thus the roots of the 
equation 

4x 2 + 4x-3 = 





1 




!p 


1 




|y 








I : VI ' 










! ■ <\\ 




































SJ 5 








\ 


















TC 






\ ' \ 








J\ 








\r 










— i \ _ 








Vl x! 






X' 


r^ 










X 






•l-5\ 


-i 




o 7 


*\l 


*5 








\i 






















/ 


i 
















uX 1 






1 












TT Y'l 1 


1 1 





Fig. 16. 



GRAPHICAL REPRESENTATION OF FUNCTIONS. 541 

are furnished by the abscissas of the common points of the graphs of 
4a; 2 and — 4x + 3. 

Again, between the values .5 and — 1.5 for x it will be found 
graphically that y x is less than y 2 , hence y x — y 2 , or 4 x 2 + 4 x — 3 is 
negative. 

Both solutions are here exhibited. 

The upper curve is the graph of y = 4 x 2 ; PQ is the graph of 
y=-4x + 3; and the lower curve is the graph of y = 4 x 2 -f 4 x — 3. 

660. Of the two methods in the last example the first is 
the more direct and instructive; but the second has this 
advantage : 

If a number of equations of the form x 2 = px -f- q have to 
be solved graphically, y = x 2 can be plotted once for all on 
a convenient scale, and y=px-\-q can then be readily- 
drawn for different values of p and q. 

Equations of higher degree may be treated similarly. 

For example, the solution of such equations as 

as 3 = px + q, or x* = ax 2 + bx + c 

can be made to depend on the intersection of y = x 3 with 
other graphs. 

Ex. Find the real roots of the equations 

(i.) z8-2.5a;-3=:0; (ii.) x*-3x + 2 = 0. 

Here we have to find the points of intersection of 

(i.) y = x* y (ii.) y = x», 

y = 2.5x + 3; y = 3x-2. 

Plot the graphs of these equations, choosing the unit for z five 
times as great as that for y. 

It will be seen that y = 2.5 x + 3 meets y = x s only at the point for 
which x = 2. Thus 2 is the only real root of equation (i.). 

Again y = 3 x — 2 touches y = x 8 at the point for which x = 1, and 
cuts it where x = — 2. 

Corresponding to the former point the equation a; 8 — 3ie + 2 = 
has two equal roots. Thus the roots of (ii.) are 1, 1, — 2. 



. .. 


Y 




_ ..^,:f_. 








^ /^ 


_ 


Q^ 


__ — l.--.,^- 


_- = -=_. _! 


^ 


s*' 


Z--7- /■-'- 




Z ,5" 




-—.- *Z»=iJZ 




zd't-— - 




1 


~ 



66L In Art. 642 we hare given the graphical solution of 
two linear simultaneous equations. As the principle is the 
same for equations of any degree, the few examples of this 
kind on pages 531, 540 have been given without special 
explanation. It may, however, be instructive here to show 
the graphical solution of some of the equations discussed in 
Chap, xxviii. 

Ex. Solve the following equations graphically: 

(L)a-y = 21 (ii.) & + y* = n\ 

xy = 35 | xy = 36 J 

(Compare An. 300, Ex. 1.) 

Here xy — 36 is represented by a rectangular hyperbola [Art. 661] ; 
x — y = 2 is the line QS, and z? -f- y 1 = 74 is represented by the circle. 
The roots of (i.) are the coordinates of <J> and S ; that is, 



he roots of (ii.) are the coordinates of P, Q, S, and S; that is, 
= B,p = 7j 1 = 7,^=5; z=- 7,y =-6; a = -5,y=-7- 



GRAPHICAL REPRESENTATION OF FUNCTIONS. 543 



1 


Y 




r- A - - 




A- - 




V 




3 






^^ 


"^ z 


~? 




~£_ 


^o^r 


/ 


Zv v 


~£_ 


z ^^^ 


nj 


Z Ju ""^'"""'""~ 


T 








X J 


"2 i i 


\ 


^ i 


-^3 /. 




"^"^^ v_ ^_ 








^%s- Z 








f O V ^. 




"7 ^ A ^. 
















1 \ 


f -r 




Yi 



Fig. 18. 



EXAMPLES XLIX. f. 

1. Draw the graph of y = x 2 on a scale twice as large as that in 
Fig. 18, and employ it to find the squares of .72, 1.7, 3.4 ; and the 
square roots of 7.56, 5.29, 9.61. 

2. Draw the graph of y = Vx taking the unit for y five times as 
great as that for x. 

By means of this curve check the values of the square roots found 
in Example 1. 

3. From the graph of y = x* (on the scale of the diagram of 
Art. 659) find the value of \^9 and v^8 to 4 significant figures. 

4. A student who was ignorant of the rule for cube root required 

the value of V14.71. He plotted the graph of y = as 8 , using for x 
the values 2.2, 2.3, 2.4, 2.5, and found 2.45 as the value of the cube 
root. Verify this process in detail. From the same graph find the 

value of Viks. 

5. Find graphically the values of x for which the expression 
x 1 — 2 x — 8 vanishes. Show that for values of x between these limits 



544 



ALGEBRA. 



the expression is negative and for all other values positive. Find the 
least value of the expression. 

6. From the graph in the preceding example show that for any 
value of a greater than 1 the equation x 2 — 2 x + a = cannot have 
real roots. 

7. Show graphically that the expression x 2 — 4 x + 7 is positive 
for all real values of x. 

8. On the same axes draw the graphs of 

y = x 2 , y =x + 6, y = x — 6, y=— x + 6, y =— x— 6. 

Hence discuss the roots of the four equations 

x 2 - x — 6=0, x 2 — x + 6 = 0, x 2 + x - 6 = 0, x 2 + x + 6 = 0. 

9. If x is real, prove graphically that 6 — 4 x — x 2 is not greater 
than 9 ; and that 4 x 2 — 4 x + 3 is not less than 2. Between what 
values of x is the first expression positive ? 

10. Solve the equation x 8 = 3x 2 + 6x— 8 graphically, and show 
that the function x 8 — 3 x 2 — 6 x + 8 is positive for all values of x 
between — 2 and 1, and negative for all values of x between 1 and 4. 

11. Show graphically that the equation x 8 +px 4- q = has only 
one real root when p is positive. 

12. Trace the curve whose equation is y = 2 *. Find the approxi- 
mate values of 2 4,15 and 2 5,26 . Express 12 as a power of 2 approximately. 

Prove also that log 2 26.9 + log 2 38 = 10. 

13. By repeated evolution find the values of 10*, 10*, 10^, 10^* 

By multiplication find the values of 10 T ^, 10"&, 10 T \ 10^, lO 1 ^. 
Use these values to plot a portion of the curve y = 10* on a large 
scale. Find correct to three places of decimals the values of log 3, 
log 1.68, log 2.24, log 34.3. Also by choosing numerical values for 
a and 6, verify the laws 

log ab = log a + log b ; log - = log a — log 6. 

b 

[By using paper ruled to tenths of an inch, if 10 in. and 1 in. be 
taken as units for x and y respectively, a diagonal scale will give values 
ofx correct to three decimal places and values ofy correct to two.'] 

14. Calculate the values of x(9 — x) 2 for the values 0, 1, 2, 3, -..9 
of x. Draw the graph of x(9 — x) 2 from x = to x = 9. 

If a very thin elastic rod, 9 inches in length, fixed at one end, 
swings like a pendulum, the expression x(9 — x) 2 measures the 
tendency of the rod to break at a place x inches from the point of 
suspension. From the graph find where the rod is most likely to 
break. 



GRAPHICAL REPRESENTATION OF FUNCTIONS. 545 

15. The reciprocal of a number is multiplied by 2.26 and the 
product is added to the number. Find graphically what the number 
must be if the resulting expression has the least possible value. 

16. Show graphically that the expression 4 x 2 + 2 x — 8. 75 is positive 
for all real values of x except such as lie between 1.25 and — 1.75. 
For what value of x is the expression a minimum ? 

17. Find graphically the real roots of the equations : 

(i.) x 9 + x - 2 = 0. (ii.) a^-7x + 6 = 0. 

18. Draw the graphs of 

x + y=9±, xy = 12, x 2 -y 2 = 32, 

on the same axes. Hence find the solutions of the following pairs of 
simultaneous equations : 

(i.) x + y = 9J1 (ii.) x 2 - y 2 = 32 1 (iii.) x 2 -y 2 = 32 

xy = 12J x + y = 9jj* X y = 12 ' 

19. Draw the graphs of y = x 8 and y = 3 x 2 — 4 on the same axes, 
and find the roots of the equation x 8 — 3 2c 2 + 4 = 0. 

Show that the expression x 8 — 3 x 2 + 4 is negative for values of x 
less than — 1, and positive for all other values of x. 

20. From a graphical consideration of the following pairs of 
simultaneous equations : 

(i.) x 2 + y 2 = a, (ii.) x + y = a, 

xy = 6, xy = 6, 

explain why (i.) has either four solutions or none, while (ii.) has two 
solutions or none. 

21 . Draw the graphs of y = x 8 and y = x 2 4-3x— 3 on the same axes. 
Hence find the roots of the equation x 8 — x 2 — 3x + 3=0 to three 

places of decimals, and discuss the sign of the expression 
x 8 — x 2 — 3 x + 3 for different values of x. 

' • 

PRACTICAL APPLICATIONS. 

662. In all the cases hitherto considered the equation of 
the curve has been given, and its graph has been drawn by 
first selecting values of x and y which satisfy the equation, 
and then drawing a line so as to pass through the plotted 
points. We thus determine accurately the position of as 
2n 



546 ALGEBRA. 

many points as we please, and the process employed assures 
us that they all lie on the graph we are seeking. We could 
obtain the same result without knowing the equation of the 
curve provided that we were furnished with a sufficient 
number of corresponding values of the variables accurately 



Sometimes from the nature of the case the form of the 
equation which connects two variables is known. For ex- 
ample, if a quantity y is directly proportional to another 
quantity x it is evident that we may put y = ax, where a is 
some constant quantity. Hence in all cases of direct pro- 
portionality between two quantities the graph which exhibits 
their variations is a straight line through the origin. Also, 
since two points are sufficient to determine a straight line, it 
follows that in the cases under consideration, we only require 
to know the position of one point besides the origin, and 
this will be furnished by any pair of simultaneous values of 
the variables. 

Ex. 1. Given that 5.5 kilograms are roughly equal to 12.126 pounds, 
show graphically how to express any number of pounds in kilograms. 
Express 7 j pounds in kilograms, and i\ kilograms in pounds. 

Here measuring pounds horizontally and kilograms vertically, the 
required graph Is obtained at once by joining the origin to the point 
whose coordinates are 12.125 and 6.5. 



By measurement It will be found that 7 j pounds = 3.4 kilograms, 
and i\ kilograms = fl.37 pounds. 

Ex. B. The expenses of a school are partly constant and partly 
proportional to the number of boys. The expenses were $3250 for 
105 boys, and $3710 for 128. Draw a graph to represent the ex- 



GRAPHICAL REPRESENTATION OP FUNCTIONS. 547 

peases for any number of boys j find the expenses for 115 boys, and 
the number of boys that can be maintained at a cost of $3550. 

If the expenses for x boys are represented by $ y, it is evident that 
x and y satisfy a linear equation y = ax + b, where a and 6 are con- 
slants. Hence the graph is a straight line. 



Fib. ». 

As the numbers are large, it will be convenient If we begin 
measuring ordinates at 3000, and abscissas at 100. This'enables us 
to bring the requisite portion of the graph into a smaller compass. 
The points P and Q are determined by the data of the question, and 
the line PQ is the graph required. 

By measurement we find that when x = 115, y = 3450 ; and that 
when y = 3550, x = 120. Thus the required answers are $ 3450, and 
120 boys. 

663. Sometimes corresponding values of two variables 
are obtained by observation or experiment. In sucb cases 
the data cannot be regarded as free from error ; the position 
of the plotted points cannot be absolutely relied on ; and 
we cannot correct irregularities in the graph by plotting 
other points selected at discretion. All we can do is to 
draw a curve to lie as evenly as possible among the plotted 
points, passing through some perhaps, and with the rest 
fairly distributed on either side of the curve. As an aid to 
drawing an even continuous curve a thin piece of wood or 
other flexible material may be bent into the requisite curve, ' 
and held in position while the line is drawn. When the 
plotted points lie approximately on a straight Hue, the 
simplest plan is to use a piece of tracing paper or celluloid 
on which a straight line has been drawn. When this has 



been placed in the right position the extremities can be 
marked on the squared paper, and by joining these points 
the approximate graph is obtained. 

Ex. 1. The following table gives statistics of the population of a 
certain country, where P ia the number of millions at the beginning 
of each of the years specified. 



Tear 


1830 


1835 


1840 1860 


1800 I 1865 


1870 


1880 


P 


20 


22.1 


23.5 20.0 


34.2 ' 38.2 


41.0 


49.4 



Let t be the time in years from 1830. Plot the values of P 
vertically and those of I horizontally and exhibit the relation between 
Pand 1 by a simple curve passing fairly evenly among the plotted 
points. Find what the population was at the beginning of the 
years 1848 and 1875. 

The graph is given in Fig. 21 on tie opposite page. The popula- 
tions in 1848 and 1875, at the points A and B respectively, will be 
found to be 27.8 millions and 45.3 millions. 

n the following 



■ 


1 | 4 


6.8 8 


9.6 | 12 


14.4 


y 


4 , 8 

1 


12.2 ] 13 


15.3 20 


24.8 



Supposing these values to involve errors of observation, draw the 
graph approximately and determine the most probable equation 
between x and y. [See Fig. 22.] 

After carefully plotting the given points we see that a straight 
line can be drawn passing through three of them and lying evenly 
among the others. This is the required graph. 

Assuming y — ax + b for its equation, we find the values of a and 6 
by selecting two pairs of simultaneous values of x and y. 

Thus substituting / = 4, y = 8,and:r = 12. y = 20 in the equation, we 
obtain a = 1.6, 6 = 2. Thus the equation of the graph is y—. I. ox +2. 

664. In the last example as the graph is linear it can be 
produced to any extent within the limits of the paper, and 
so any value of one of the variables being determined, the 



GRAPHICAL REPRESENTATION OF FUNCTIONS. 549 



-1,4- 1 1 Ml 




















m\ 




































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<V 








































r 




3 




l\ 




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3 




L 




V 




1 



o 

CO 

co 



m 
oo 



o 
»*. 
oo 



m 
oo 



o 



lO 04 

m 
00 o 



o 
m 
oo 






m 
tn 
oo 



o 



o 
m 



o 
























































corresponding value of the other can be read off. When 
large values are in question this method is not only incon- 
venient but unsafe, owing to the fact that any divergence 
from accuracy in the portion of the graph drawn is increased 
when the curve is produced bevond the limits of the plotted 
points. The following example illustrates the method of 
procedure in such cases. 



Ex. 



In a certain machine P is the force in pounds required to 
weight of W pounds. The following corresponding values 
id W were obtained experimentally : 



P ■ 3.08* ! 3.9 i 6.8 


8.8 j 8.2 


11* 


13.3 


IT 21 1 36.25 ■ 6U2 


87.5 j 103.75 


120 


152.5 



By plotting these values on squared paper, draw the graph con- 
necting P and IF, and read off the value of P when W= 70. Also 
determine a linear law connecting P and IT ; find the force necessary 
to raise a weight, of 310 pounds, and also the weight which could be 
raised by a force of ISO.ti pounds. 

As the page is too small to exhibit the graphical work on a 
convenient scale, we shall merely indicate the steps of the solution, 
which is similar in detail to that of the last example. 

Plot the values of P vertically and the values of IF horizontally. 
It will be found that a straight line can be drawn through the points 
corresponding to the results marked with an asterisk, and lying 



GRAPHICAL REPRESENTATION OF FUNCTIONS. 551 

• 

evenly among the other points. From this graph we find that when 
W=70, P = 7. 

Assume P = aW + b, and substitute for P and W from the values 
corresponding to the two points through which the line passes. By 
solving the resulting equations we obtain a = .08, b = 1.4. Thus 
the linear equation connecting P and W is P= .08 TF + 1.4. 

This is called the Law of the Machine. 

From this equation, when W= 310, P= 26.2, and when P= 180.6, 
W = 2240. 

Thus a force of 26.2 pounds will raise a weight of 310 pounds; 
and when a force of 180.6 pounds is applied the weight raised is 
2240 pounds. 

Note. The equation of the graph is not only useful for determin- 
ing results difficult to obtain graphically, but it can always be used to 
check results found by measurement. 

665. The example in the last article is a simple illustra- 
tion of a method of procedure which is common in the 
laboratory or workshop, the object being to determine the 
law connecting two variables when a certain number of 
simultaneous values have been determined by experiment 
or observation. 

Though we can always draw a graph to lie fairly among 
the plotted points corresponding to the observed values, 
unless the graph is a straight line it may be difficult to find 
its equation except by some indirect method. 

For example, suppose x and y are quantities which satisfy 
an equation of the form xy = ax+by, and that this law has 
to be discovered. 

By writing the equation in the form 

2 + ^ = 1, oTau + bv = l: 
y x 

where u = -, v = -, it is clear that u, v satisfy the equation 
y x 

of a straight line. In other words, if we were to plot the 

points corresponding to the reciprocals of the given values, 

their linear connection would be at once apparent. Hence 

the values of a and b could be found as in previous examples, 

and the required law in the form xy = ax + by could be 

determined. 



552 



ALGEBRA. 



Again, suppose x and y satisfy an equation of the form 
x*y = c, where n and c are constants. 
By taking logarithms, we have 

n log x -f- logy = log c. 

The form of this equation shows that logo; and logy 
satisfy the equation to a straight line. If, therefore, the 
values of log x and log y are plotted, a linear graph can be 
drawn, and the constants n and c can be found as before. 

Ex. The weight, y grams, necessary to produce a given deflec- 
tion in the middle of a beam supported at two points, x centimeters 
apart, is determined experimentally for a number of values of x with 
results given in the following table : 



X 

y 


50 


60 


70 


80 


90 


100 


270 


150 


100 


60 


47 


32 



log a; 


logy 


1.699 


2.431 


1.778 


2.176 


1.845 


2.000 


1.903 


1.778 


1.954 


1.672 


2.000 


1.619 



Assuming that x and y are connected by the equation x n y = c, find 
n and c. 



From pages 360, 361 we obtain the values of log x 
and log y given here corresponding to the. observed 
values of x and y. By plotting these we obtain the 
graph given in Fig. 23, and its equation is of the form 

n log a; 4- log y = log c. 



To obtain n and c, choose two extreme points through which the line 
passes. It will be found that when 

log x = 1.642, log y = 2.6, 
and when log x = 2. 1, log y = 1.21. 

Substituting these values, we have 

2.6 + n x 1.642 = log c (i.), 

1.21 + nx 2.1 =logc (ii.); 

/. 1.39-0.468n = 0; 
whence n = 3.04. 

/. from (ii.) log c = 6.88 + 1.21 

= 7.59 ; 
/. c = 39 x 10 6 , from the tables. 
Thus the required equation is x*y = 39 x 10 6 . 



GRAPHICAL REPRESENTATION OP FUNCTIONS. 555 

The student should work through this example in detail on a 
larger scale. The figure below was drawn on paper ruled to tenths of 
an inob and then reduced to half the original scale. 



ALGEBRA. 



BXAMFLKS ZUZ. e- 

1. Given that 6.01 yards = 5.5 meters, draw the graph showing the 
equivalent of any number of yards when expressed in meters. 

Show that 22.2 yards = 20.3 meters approximately. 

S. Draw a graph showing the relation between equal weights in 
grains and grams, having given that 18.1 grains = 1.17 grams. 

Express (i.) 3.5 grams in grains. 

(ii. ) 3.00 grains as a decimal of gram. 

8. If 3.26 inches are equivalent to 8.28 centimeters, show how to 
determine graphically the number of inches corresponding to a given 
number of centimeters. Obtain the number of inches in a meter, and 
the number of centimeters in a yard. What is the equation of the 
graph? 

4. The following table gives approximately the circumferences of 
circles corresponding to different radii : 



a 


15.7 


20.1 


81.4 


44 


52.2 


• 


2.5 


3.3 


6 


7 


8.3 



Plot tbe values on squared paper, and from the graph determine the 
diameter of a circle whose circumference is 12.1 inches and the cir- 
cumference of a circle whose radius is 2.8 inches. 

6. For a given temperature, G degrees on a Centigrade thermom- 
eter are equal to F degrees on a Fahrenheit tbermometer. The 
following table gives a series of corresponding values of F and C : 



c 


-10 


-6 





5 


10 


15 


25 


40 


F 


14 


23 


32 


41 


60 


69 


77 


104 



Draw a graph to show the Fahrenheit reading corresponding to a 
given Centigrade temperature, and find the Fahrenheit readings cor- 
responding to 12.5° C. and 31" C. 

By observing the form of tbe graph find the algebraical relation be- 
tween F and C. 

6. At different ages the mean after-lifetime ("expectation of 
life") of males, calculated on the death rates of certain years, was 
given by the following table: 



GRAPHICAL REPRESENTATION OF FUNCTIONS. 555 



Age 


6 


10 


14 


18 


22 


26 


27 


Expectation 


60.38 


47.60 


44.26 


40.96 


37.89 


34.96 


34.24 



Draw a graph to show the expectation of life of any male be- 
tween the ages of 6 and 27, and from it determine that of persons 
aged 12 and 20. 

7. If W is the weight in ounces required to stretch an elastic 
string till its length is I inches, plot the following values of W and I : 



w 
I 


2.5 

8.5 


3.76 


6.25 


7.6 
9.3 


10 


11.25 


8.7 


9.1 


9.7 


9.9 



From the graph determine the unstretched length of the string, 
and the weight the string will support when its length is 1 foot. 

8. The highest and lowest marks gained in an examination are 
297 and 132 respectively. These have to be reduced in such a way 
that the maximum for the paper (200) shall be given to the first 
candidate, and that there shall be a range of 150 marks between the 
first and last. Find the equation between x, the actual marks gained, 
and y, the corresponding marks when reduced. 

Draw the graph of this equation, and read off the marks which 
should be given to candidates who gained 200, 262, 163 marks in the 
examination. 

9. A body starting with an initial velocity, and subject to an 
acceleration in the direction of motion, has a velocity of v feet per 
second after t seconds. If corresponding values of v and t are given 
by the annexed table, 



V 

t 


9 
1 


13 


17 


21 


25 


29 


33 


87 


41 


45 


2 


3 


4 


5 


6 


7 


8 


9 


10 



plot the graph exhibiting the velocity at any given time. Find from 
it (i.) the initial velocity, (ii.) the time which has elapsed when 
velocity is 28 feet per second. Also find the equation between v and t. 



556 



ALGEBRA. 



10. The connection between the areas of equilateral triangles and 
their bases (in corresponding units) is given by the following table : 



Area 


.43 


1.73 


3.90 


6.93 


10.82 


16.69 


Base 


1 


2 


3 


4 


5 


6 



Illustrate these results graphically, and determine the area of an 
equilateral triangle on a base of 2.4 feet. 

11. A body falling freely under gravity drops s feet in t seconds 
from the time of starting. If corresponding values of s and t at 
intervals of half a second are as follows : 



t 

s 


.5 
4 


1 
16 


1.5 


2 
64 


2.5 


3 


3.5 


4 


36 


100 


144 


196 


256 



draw the curve connecting s and t, and find from it 

(i.) the distance through which the body has fallen after 1.8"; 
(ii.) the depth of a well if a stone takes 3-16" to reach the bottom. 

12. A body is projected with a given velocity at a given angle 
to the horizon, and the height in feet reached after t seconds is 
given by the equation h = 64 1 — 16 1 2 . Find the values of h at 
intervals of Jth of a second and draw the path described by the 
body. Find the maximum value of h, and the time after projection 
before the body reaches the ground. 

13. The following table gives the sun's position at 7 a.m. on 
different dates: 



Mar. 23 


Apr. 8 


Apr. 20 


MayS 


May 27 


June 22 


July 18 


Aug. 5 


Aug. 25 


80° E. 


82° E. 


85° E. 


s:>° E. 


92° E. 


95 E. 


94° E. 


91° E. 


85° E. 



Show these results graphically, and estimate approximately the 
sun's position at the same hour on June 8. 

14. At a given temperature p pounds per square inch represents the 
pressure of a gas which occupies a volume of v cubic inches. Draw 
a curve connecting p and v from the following table of corresponding 
values ; 



GRAPHICAL REPRESENTATION OF FUNCTIONS. 557 



p 

V 


86 


30 


• 

25.7 


22.5 


20 


18 

• 


16.4 


15 


5 


6 


7 


8 


9 


10 


11 


12 



15. Plot on squared paper the following measured values of x and 
y 9 and determine the most probable equation between x and y : 



X 

y 


3 
2 


5 


8.8 


11 


13 
4 


15.5 


18.6 


23 


28 


2.2 


3.4 


3.8 


4.6 


5.4 


6.2 


7.25 



16. Corresponding values of x and y are given in the following 
table : 



X 

y 


i 

2 


3.1 


6 


9.5 


12.6 


16 


19 
9 


23 


2.8 


4.2 


5.3 


6.6 


8.3 


10.8 



Supposing these values to involve errors of observation, draw the 
graph approximately, and determine the most probable equation 
between x and y. Find the correct value of y when x = 19, and the 
correct value of x when y = 2.8. 

17. The following corresponding values of x and y were obtained 
experimentally : 



X 

y 


0.6 


1.7 


3.0 


4.7 


5.7 


7.1 


8.7 


9.9 


10.6 


11.8 


148 


186 


265 


326 


388 


436 


529 


662 


611 


652 



It is known that they are connected by an equation of the form 
y = ax + 6, but the values of x and y involve errors of measurement. 
Find the most probable values of a and 6, and estimate the error in the 
measured value of y when x — 9.9. 

18. In a certain machine, P is the force in pounds required to 
raise a weight of W pounds. The following corresponding values of 
P and W were obtained experimentally : 



568 



ALGEBBA. 



p 
w 


2.8 
20 


3.7 


4.8 


5.5 


6.5 


7.3 


• 

8 


9.5 


10.4 


11.75 


25 


31.7 


35.6 


45 


52.4 


57.6 


65 


71 


82.5 



Draw the graph connecting P and TP, and read off the value of JP 
when JP = 60. Also determine the law of the machine, and find from 
it the weight which could be raised by a force of 31.7 pounds. 

19. The following values of x and y, some of which are slightly 
inaccurate, are connected by an equation of the form y = ax* + b. 



X 

y 


l 


1.6 


3 
5 


3.7 


4 


5 


6.7 


6 


6.3 


7 


3.26 


4 


6.5 


7.4 


9.26 


10.6 


11.6 


14 


15-26 



By plotting these values, draw the graph, and find the most prob- 
able values of a and b. 

Find the true value of x when y = 4, and the true value of y when 
x = 6. 

90. The following table gives corresponding values of two variables, 
j and y: 



X 

y 


2.76 


3 

9.8 


3.2 
8 


3.5 


4.3 


4.6 
5.4 


5.3 
5 


6 
4.3 


7 


8 


10 


11 


6.5 


6.1 


4.1 


4 


3.9 



These values involve errors of observation, but the true values are 
known to satisfy an equation of the form xy = ax + by. Draw the 
graph by plotting the points determined by the above table, and 
find the most probable values of a and b. Find the correct values of 
y corresponding to x = 3.5, and x = 7. 

SI. Observed values of x and y are given as follows : 



X 


100 • 90 
SO ' 81.08 

t ■ 


70 


00 


50 

37.8 


40 


&w> 


40.7 



GRAPHICAL REPRESENTATION OF FUNCTIONS. 559 

Assuming that x and y are connected by an equation of the form 
xy n = c, find n and c. 

. The following values of x and y involve errors of observation: 



X 

y 


66.83 


63.10 


58.88 


61.52 


48.53 


44.16 


40.36 


144.5 


158.5 


177.8 


208.9 


236.0 


264.9 


309.0 



If x and y satisfy an equation of the form x n y = c, find n and c. 



MISCELLANEOUS APPLICATIONS OF LINEAR GRAPHS. 



When two quantities x and y are so related that a 
change in one produces a proportional change in the other, 
their variations can always be expressed by an equation of 
the form y = ax, where a is some constant quantity. Hence 
in all such cases, the graph which exhibits their variations 
is a straight line through the origin, so that in order to draw 
the graph it is only necessary to know the position of one 
other point on it. Such examples as deal with work and 
time, distance and time (when the speed is uniform), quan- 
tity and cost of material, principal and simple interest at a 
given rate per cent, may all be illustrated by linear graphs 
through the origin. 

Ex. 1. At 8 a.m. A starts from P to ride to Q, which is 48 miles 
distant. At the same time B sets out from Q to meet A. If A rides 
at 8 miles an hour, and rests half an hour at the end of every hour, 
while B walks uniformly at 4 miles an hour, find graphically 

(i.) the time and place of meeting ; 
(ii.) the distance between A and B at 11 a.m. ; 
(iii.) at what time they are 14 miles apart. 

In Fig. 24, on the following page, let the position of P be chosen 
as origin ; let time be measured horizontally from 8 a.m. (1 inch to 
1 hour), and let distance be measured vertically (1 inch to 20 miles). 

In 1 hour A rides 8 miles ; therefore the point D (1, 8) marks his 
position at 9 a.m. In the next half hour he makes no advance to- 
wards Q; therefore, the corresponding portion of the graph is DE. 
The details of J.'s motion may now be completed by the broken line 
PDEFGHKX. 



560 



ALGEBRA. 




GRAPHICAL REPRESENTATION OF FUNCTIONS. 561 

On the vertical axis, mark PQ to represent 48 miles and mark the 
hours on the horizontal line through Q. At 9 a.m. B has walked 
4 miles towards P. Measuring a distance to represent 4 miles down- 
wards we get the point R, and QR produced is the graph of 2?'s motion. 
It cuts ,4's graph at X. Hence the point of meeting is X, which is 
28 miles from P, and the time is 1 p.m. 

The distance between A and B at any time is shown by the dif- 
ference of the ordinates. Thus at 11 a.m. their distance apart is 
MG, which represents 20 miles. 

Lastly, NT represents 14 miles ; thus A and B are 14 miles apart 
at 11.30 a.m. 

Ex. 2. A, B, and C run a race of 300 yards. A and G start 
from scratch, and A covers the distance in 40 seconds, beating G by 
GO yards. B, with 12 yards start, beats A by 4 seconds. Supposing 
the rates of running in each case to be uniform, find graphically 
the relative positions of the runners when B passes the winning post. 
Find also by how many yards B is ahead of A when the latter has 
run three fourths of the course. 

In Fig. 26 let time be measured horizontally (0.6 inch to 10 seconds), 
and distance vertically (1 inch to 60 yards). O is the starting point 
for A and G\ take OP equal to 0.2 inch, representing 12 yards, on the 
vertical axis : then P is 5's starting point. 

,4's graph is drawn by joining O to the point which marks 40 sec- 
onds. From this point measure a vertical distance of 1 inch down- 
wards to Q. Then since 1 inch represents 60 yards, Q is C's position 
when A is at the winning post, and OQ is Cs graph. 

Along the time-axis take 1.8 inch to R, representing 36 seconds ; 
then PR is 2?'s graph. 

Through R draw a vertical line to meet the graphs of A and G in 
S and T respectively. Then S and T mark the positions of A and G 
when B passes the winning post. 

By inspection RS and ST represent 30 and 54 yards respectively. 

Thus B is 30 yards ahead of A, and A is 64 yards ahead of G. 

Again, since A runs three fourths of the course in 30 seconds, the 
difference of the corresponding ordinates of ^'s and 2Ts graphs after 
30 seconds will give the distance between A and B. By measurement 
we find VW =.0.46 inch, which represents 27 yards. 

It is recommended that the student draw a figure for himself on a 
scale twice as large as that given in Fig. 26. 

667. When a variable quantity y is partly constant and 

partly proportional to a variable quantity x f the algebraical 

relation between x and y is of the form ?/= ax-\- b, where a 

and b are constant. The corresponding graph will therefore 

2o 



502 



ALGEBRA. 



10 20 3 


) R 40 50 


500 — - t— — —-I-. ..--,.. . 

aBCOi id 8 


i i 


ZL. -*£to,w*- 


1 1 


t 


Jr r 




1 1 




L t 


2/0 ~" —.-.,-,.., . - 


I j s 7 




/A J 


V, 


LV £ 




/ 3 




-t < 




4 2ot 




t -/ 


/ 


fc- 2 




rt t 




JL 


A 1 U ~~ ft 


t 




1 




L 




1 




t 






If f 




A A 1 




1 / / 












/ / / 




/ y / 




tt 7 




A r t 




12Q — jl 'I ~ 




Jt-t-T 




3-Tl 








rH 








tiu 




1 ' / 




ill 








60 -f-jj 




ttt 




-Eu 




%TT 




50 til 




-flfc 




it 




P fct 




7 




nt _ 





Fig. 25. 



GRAPHICAL REPRESENTATION OF FUNCTIONS. 563 

be a straight line ; and since a straight line is completely 
determined when the positions of two points are known, it 
follows that, in all problems which can be illustrated by 
linear graphs, it is sufficient if the data furnish for each 
graph two independent pairs of simultaneous values of the 
variable quantities. 

Some easy examples of this kind have already been given 
on page 546 and in Examples XLIX. g. We shall now 
work out two more examples. 

Ex. 1. In a certain establishment the clerks are paid an initial 
salary for the first year, and this is annually increased by a fixed 
bonus, the initial salary and the bonus being different in different 
departments. A receives $ 1000 in his 10th year, and $ 1450 in his 19th. 
B, in another department, receives $ 1060 in his 6th year and $ 1260 in 
his 13th. Draw graphs to show their salaries in different years. In 
what year do they receive equal salaries? Also find in what year A 
earns the same salary as that received by Bfor his 21st year. 

In Fig. 26 let each horizontal division represent 1 year ; and let the 
salaries be measured vertically, beginning at 1000, with 1 division to 
represent $ 10. 

If the salary at the end of x years is denoted by $y, it is evident 
that in each case we have a relation of the form y = ax + b, where 
a and b are constant. Thus the variations of time and salary may 
be represented by linear graphs. 

Since no bonus is received for the first year, x = 9, when y = 1000, 
and x = 18, when y = 1450. Thus the points P and Q are determined, 
and by joining them we have the graph for !4's salary. Similarly, 
the graph for 2?'s salary is found by joining P' (4, 1050) and Q' 
(12, 1250). 

These lines have the same ordinate and abscissa at L, where 
c = 16, y = 1360. Thus A and B have the same salary when each 
has served 16 years, that is in their 17th year. Again 2?'s salary at 
the end of 20 years is given by the ordinate of M, which is the same as 
that of Q which represents A'a salary after 18 years. 

Thus A'a salary for his 19th year is equal to 5's salary for his 
21st year. 

Ex. 2. Two sums of money are put out at simple interest at 
different rates per cent. In the first case the amounts at the end of 
6 years and 15 years are $ 1300 and $1750 respectively. In the second 
case the amounts for 6 years and 20 years are $1660 and $2100. 
Draw graphs from which the amounts may be read off for any year, 
and find the year in which the principal with accrued interest will 



564 



ALGEBEA. 







1 


1 








^-t I 




45 1 E 


1450 


-J -^ 




4 A 




/ ' 






1400 .. ,., , - 








» 


















TfL 








J / 




- U- 




-J-L 


1300 " | ,,-.,— - 
on. . 


-l-i 


2 


-tl 




r i ■ ■ 

j 1 


o_.. ,. _. . . __,J 


1 t 


ny 








/: 








7 _ 












vA -, 




& 3 




$1 M 




11 fin . i.i, *°i- - SI 




1150 S j3_ 




3c ^Z- 




-J Jf- 




7 f_ 


, 






-A -J 




£ 3 - 




-4 t - 




ft t - 




3 / 1 








^ J^ _ 




I . 




t ir ._ 




1000 1 1 1 1 1 1 1 i 1 ,f 1 re ? 


SL 



5 p 10 15 

FiQ. 26. 



20 



i 



GRAPHICAL REPRESENTATION OF FUNCTIONS. 565 

amount to the same in the two cases. Also from the graphs read off 
the value of each principal. • 

When a sum of money is at simple interest for any number of 

years, we have 

Amount = Principal + Interest, 

where principal is constant, and interest varies with the number 
of years. Hence the variations of amount and time may be rep- 
resented by a linear graph in which x is taken to denote the number 
of years, and y the number of dollars in the corresponding amount. 

Here, as the diagram is inconveniently large, we shall merely indi- 
cate the steps of the solution which is similar in detail to that of the 
last example. The student should draw his own diagram. 

Measure time horizontally (1 inch to 10 years), and amount 
vertically (1 inch to $200) beginning at $ 1300. 

The first graph is the line joining L (6, 1300) and M *(15, 1750). 
The second graph is the line joining L' (5, 1650) and M' (20, 2100). 
In each of these lines the ordinate of any point gives the Amount 
for the number of years given by the corresponding abscissa. 

Again LM, and L'M' intersect at a point P where x = 25, y = 450. 
Thus each principal with its interest amounts to $ 2250 in 25 years. 

When x = there is no interest ; thus the principals will be 
obtained by reading off the values of the intercepts made by the two 
graphs on the y-axis. These are $ 1000 and $ 1500 respectively. 

Note. To obtain the result y = 200 it will be necessary to continue 
the y-axis downwards sufficiently far to show this ordinate. 

EXAMPLES XLIX. h. 

1. At noon A starts to walk at 6 miles an hour, and at 1.30 p.m. 
B follows on horseback at 8 miles an hour. When will B overtake 

(i.) when A is 5 miles ahead of B ; 
(ii.) when A is 3 miles behind B. 

[Take 1 inch horizontally to represent 1 hour, and 1 inch vertically 
to represent 10 miles. ] 

2. By measuring time along OX (1 inch for 1 hour) and distance 
along OY (1 inch for 10 miles) show how to draw lines 

(i.) from O to indicate distance travelled towards Y at 12 miles 
an hour ; 

(ii.) from Y to indicate distance travelled towards O at 9 miles 
an hour. 

If these are the rates of two men who ride towards each other 
from two places 60 miles apart, starting at noon, find from the graphs 



whim they are first 18 miles from each other. Also find (to the 
UNtmat minute) their time of meeting. 

3. Two bicyclists ride to meet each other from two places 95 
mlW apart. A starts at 8 a.m. at 10 miles an hour, and B starts 
nt U.30 a.m. at 15 miles an hour. Find graphically when and where 
llwj meet, and at what times they are 37] miles apart. 

4. A and B start at the same time to go from New York to Fairview, 
A walking 4 miles an hour, B riding 9 miles an hour. B reaches 
Knirview in 4 hours, and immediately rides back to N'ew York. After 
a hours' rest he starts again for Falrview at the same rate. How 
far from New York will he overtake A, who has in the meantime 
rusted (I J hoars ? 

B. At what distance from New York, and at what time, will a 
train which* leaves Sew York for Hyde Park at 2.33 p.m., and goes at 
the rate of 35 miles an hour, meet a train which leaves Hyde Park at 
1.45 p.m. and goes at the rate of 25 miles an hour, the distance be- 
tween New York and Hyde Park being 80 miles ? 
Also find at what times the trains are 24 miles apart. 

6. A, B, and C set out to walk from Chicago to Aurora at 5, 0, 
and 4 miles an hour respectively. C starts 3 minutes before and B 

7 minutes after A. Draw graphs to show (i.) when and where A 
overtakes C ; (ii.) when and where B overtakes A ; (iii.) C's position 
relative to the others after he has walked 45 minutes. 

[Take 1 inch horizontally to represent 10 minutes, and 1 inch to 
the mile vertically.] 

7. X and Y are two towns 35 miles apart. At 8.30 p.m. A starts 
to walk from X to Y at 4 miles an hour ; after walking 8 miles he 
rests for half an hour and then completes bis journey on horseback 
at 10 miles an hour. At 0.48 a.m. B starts to walk from Y to X at 
3 miles an hour j find when and where A and B meet. Also find 
at what times they are 6J miles apart 

8. A can beat B by 20 yards in 120, and B can beat C by 10 yards 
in 50. Supposing their rates of running to be uniform, find graphi- 
cally how much start A can give C in 120 yards so as to run a dead 
heat with him. It A, B, and C start together, where are A and C 
when B has run 80 yards P 

ft. A, B, and C run a race of 200 yards. A gives B a start of 

8 yards, and C starts some seconds after A. A runs the distance in 
25 seconds and beats C by 40 yards. B beats A by 1 second, and 
when he has been running 15 seconds he is 48 yards ahead of C. 
Find graphically how many seconds C starts after A. Show also 



GRAPHICAL REPRESENTATION OF FUNCTIONS. 567 

from the graphs that if the three runners started even they would 
run a dead heat. 

[Take 1 inch to 40 yards, and 1 inch to 10 seconds.] 

10. A cyclist lias to ride 75 miles. He rides for a time at 9 miles 
an hour and .then alters his speed to 15 miles an hour covering the 
distance in 7 hours. At what time did he change his speed ? 

11. A and B ride to meet each other from two towns X, and Y, 
which are 60 miles apart. A starts at 1 p.m., and B starts 36 minutes 
later. If they meet at 4 p.m., and A gets to Y at 6 p.m., find the time 
when B gets to X. Also find the times when they are 22 miles apart. 
When A is halfway between X and Y, where is B ? 

12. The distance between two towns X and Y is 119 miles ; if I 
were to set out at noon to cycle from X, riding 23 miles the first hour 
and decreasing my pace by 3 miles each successive hour, find graphi- 
cally how long it would take me to reach Y. Also find approximately 
the time at which I should reach a town 48 miles from Y. 

13. At 8 a.m. A begins a ride on a motor car at 20 miles an hour, 
and an hour and a half later B, starting from the same point, follows 
on his bicycle at 10 miles an hour. After riding 36 miles, A rests 
for 1 hour 24 minutes, then rides back at 9 miles an hour. Find 
graphically when and where he meets B. Also find (i.) at what 
time the riders were 21 miles apart, (ii. ) how far B will have ridden 
by the time A gets back to his starting point. 

14. I row against a stream flowing 1£ miles an hour to a certain 
point, and then turn back, stopping two miles short of the place 
whence I originally started. If the whole time occupied in rowing 
is 2 hours 10 minutes, and my uniform speed in still water is 4J miles 
an hour, find graphically how far upstream I went. 

[Take 1.2 of an inch horizontally to represent 1 hour, and 1 inch 
to 2 miles vertically.] 

15. One train leaves Albany at 3 p.m. and reaches New York at 
6 p.m. ; a second train leaves New York at 1.30 p.m. and arrives at 
Albany at 6 p.m. ; if both trains are supposed to travel uniformly, 
at what time will they meet ? Show from a graph that the time does 
not depend upon the distance between New York and Albany. 

16. At 7.40 a.m. the accommodation train starts from Hudson and 
reaches New York at 11.40 a.m. ; the express starting from New York 
at 9 a.m. arrives at Hudson at 11.40 a.m. : if both trains travel uni- 
formly, find when they meet. Show, as in Ex. 15, that the time is 
independent of the distance between New York and Hudson, and 
verify this conclusion by solving an algebraical equation. 



568 ALGEBRA. 

♦ 

17. A boy starts from home and walks to school at the rate of 
10 yards in 3 seconds, and is 20 seconds too soon. The next day he 
walks at the rate of 40 yards in 17 seconds, and is half a minute late. 
Find graphically the distance to the school, and show that he would 
have been just in time if he had walked at the rate of 20 yards in 7 
seconds. 

18. A body is moving in a straight line with varying velocity. 
The velocity at any instant is made up of the constant velocity with 
which it was projected (measured in feet per second) diminished by 
a retardation of a constant number of feet per second in every second. 
After 4 seconds the velocity was 320, and after 13 seconds it was 140. 
Draw a graph to show the velocity at any time while the body is in 
motion. 

A second body projected at the same time under similar conditions 
has a velocity of 460 after 5 seconds, and a velocity of 150 after 15 
seconds. Show graphically that they will both come to rest at the 
same time. Also find at what time the second body is moving 100 feet 
per second faster than the first, and determine from the graphs the 
velocity of projection in each case. 

19. In a certain examination the highest and lowest marks gained 
in a Latin paper were 153 and 51. These have to be reduced so that 
the maximum (120) is given to the first candidate, and the minimum 
(30) to the lowest. This is done by reducing all the marks in a 
certain ratio, and then increasing or diminishing them all by the 
same number. In a Greek paper the highest and lowest marks 
were 161 and 56 ; after a similar adjustment these become 100 and 
40 respectively. Draw graphs from which all the reduced marks 
may be read off, and find the marks which should be finally given 
to a candidate who scored 102 in Latin and 126 in Greek. 

Show also that it is possible in one case for a candidate to receive 
equal marks in the two subjects both before and after reduction. 
What are the original and reduced marks in this case ? 

MISCELLANEOUS GRAPHS. 

1. Plot the graphs of 

2y = 3(x-4), 3y = l-6z, 

obtaining at least five points on each graph. Find the coordinates of 
the point where they meet. 

2. Draw the graphs represented by 

y = 5-3z, y = l(z + 6); 
and find the coordinates of their point of intersection. 



GEAPHICAL BEPRESENTATION OF FUNCTIONS. 569 

3. By finding the intercepts on the axes draw the graphs of 

(i.) 15z + 20y = 6; (ii.) 12x + 21y = U. 

In (i.) take 1 inch for unit, and in (ii.) take six tenths of an inch as 
unit. In each case explain why the unit is convenient. 

4. Solve y = 10 X + 8, 7 & + y = 26 graphically. 
[Unit for x, one inch ; for y, one tenth of an inch.] 

5. From the graph of the expression 11 x + 6, find its value 
when x = 1.8. Also find the value of x which will make the 
expression equal to 20. 

6. With the same units as in Ex. 4 draw the graph of the 
function — — — -• From the graph find the value of the function 

when x = 1.8 ; also find for what value of x the function becomes 
equal to 8. 

7. Show that the straight lines given by the equations 

9y = 5x+ 66, 6ie + 2y + 10 = 0, x + Sy = ll, 
meet in a point. Find its coordinates. 

8. Draw the triangle whose sides are given by the equations : 

3 y - a; = 9, a; + 7 y = 11, 3 a; + y = 13 ; 
and find the coordinates of its vertices. 

9. Show graphically that the values of J x and y which satisfy 
theequations 5x = 2y-18, 6y = 6-7a;, 

also satisfy the equation x f y = 2. 

10. Draw the graphs of (i.) y = x 2 , (ii.) y = 8 x 2 . 
In (i.) take 0.4" as unit for x, 0.2" as unit for y. 
In (ii.) take 1" as unit for x, 0.1" as unit for y. 

11. On the same scale as in Ex. 10 (ii.) draw the graph of y = 16 x 2 . 
Show that it may also be simply deduced from the graph of 
Ex. 10 (ii.). 

12. Plot the graph of y = « 2 , taking 1 inch as unit on both axes, 
and using the following values of x : 

-0.4, -0.3, -0.2, -0.1, 0, 0.1, 0.2, 0.3, 0.4. 

13. Draw the graph of x = y 2 , from y = to y = 5, and thence find 
the square roots of 7 and 3.6. 

[Take 0.2" as unit for x, 1" as unit for y.] 



570 ALGEBRA. 

14. Draw the graph of y = 5 + x — a! 2 f or values of x from — 2 to 
+ 3, and from the figure obtain approximate values for the roots of 
the equation 5 + x — x 2 = 0. 

[Take 1" as unit for ar, 0.2" as unit for y.] 

15. Draw the graphs of 

(i.) 6z + 6y = 60, (ii.) 6y-z = 24, (iii.) 2z-y = 7; 
and show that they represent three lines which meet in a point. 

10. Solve the following equations graphically : 

(i.) x 2 + y 2 = 63, (ii.) x 2 + y 2 = 100, 

y — x = 5; a: + y = 14 ; 

(iii.) a? + t/ 2 = 34, (iv.) a? + y 2 = 36, 

2 a; + y = 11 ; 4 a; + 3 y = 12. 

[Approximate roots to be given to one place of decimals.] 

17. Solve the equation 3 -+■ 6 x = x 2 graphically, and find -the 
maximum value of the expression 3 + 6 x — x 2 . 

18. Draw the graphs of x 2 and 3 x -f 1. By means of them find 
approximate values for the roots of x 2 — 3 x — 1 = 0. 

19. If 24 men can reap a field of 29 acres in a given time, find 
roughly by means of a graph the number of acres which could be 
reaped in the same time by 15, 33, and 42 men respectively. 

20. The highest marks gained in an examination were 136, and 
these are to be raised so that the maximum is 200. Show how this 
may be done by means of a graph, and read off, to the nearest 
integer, the final marks of candidates who scored 61 and 49 
respectively. , 

21. Draw a graph which will give the square roots of all numbers 
between 25 and 36, to three places of decimals. 

[Plot the graph of y = x 2 , beginning at the point (5, 26), with 10" 
and 0.5" as units for x and y respectively.] 

22. I want a ready way of finding approximately 0.866 of any 
number up to 10. Justify the following construction. Join the 
origin to a point P whose coordinates are 10 and 8.66 (1 inch being 
taken as unit) ; then the ordinate of any point on OP is 0.866 of 
the corresponding abscissa. Head off from the diagram, 

0.866 of 3, 0.866 of 6.6, 0.866 of 4.8, and — ?— of 5. 
1 ' 0.866 

23. A starts from New York at noon at 8 miles an hour ; two hours 
later B starts, riding at 12 miles an hour. Find graphically at what 
time and at what distance from New York B overtakes A. At 
what times will A and B be 8 miles apart? If C rides after 2?, 



GRAPHICAL REPRESENTATION OF FUNCTIONS. 571 



starting at 3 p.m. at 15 miles an hour, find from the graphs 
(i.) the distances between A, B, and C at 5 p.m. ; 
(ii.) the time when G is 8 miles behind B. 

24. If and Y represent two towns 45 miles apart, and if A 
walks from Y to at 6 miles an hour while B walks from to Y at 
4 miles an hour, both starting at noon, find graphically their time 
aud place of meeting. 

Also read off from the graphs 

(i.) the times when they are 15 miles apart ; 
(ii.) 2?'s distance from Y at 6.16 p.m. 

25. At 8 a.m. i starts from P to ride to Q which is 48 miles 
distant. At the same time B sets out from Q to meet A. Ii A rides 
at 8 miles an hour, and rests half an hour at the end of every hour, 
while B walks uniformly at 4 miles an hour, find graphically 

(i.) the time and place of meeting ; 
(ii.) the distance between A and B at 11 a.m. ; 
(iii.) at what time they are 14 miles apart. 

26. The following table gives statistics of the population of a 
certain country, where P is the number of millions at the beginning 
of each of the years specified. 



Year 


1830 


1835 


1840 


1845 


1850 

• 


1855 


1860 


P 


20 


22 


24.6 


28 


31 


36 


41 



Let t be the time in years from 1830. Plot the values of P 
vertically and those of t horizontally and show the relation between 
P and t by a simple curve passing fairly evenly among the plotted 
points. Find what the population was at the beginning of the 
years 1847 and 1858. 

27. The salary of a clerk is increased each year by a fixed sum. 
After 6 years' service his salary is raised to $1280 and after 15 years 
to 8 2000. Draw a graph from which his salary may be read off for 
any year, and determine from it (i.) his initial salary, (ii.) the salary 
he should receive for his 21st year. 

28. Draw the graphs of y=x 2 and 2 y =x +3 on the same diagram. 
Deduce the roots of the equation 2 x 2 — x — 3 = 0. 

29. Taking 1 inch as unit, plot the graph of y = jc 8 — 3 aj, taking 
the following values of x : 

0, ±.2, ±.4, ±.6, ±.8, ±1, ±1.2, ±1.4, ±1.6, ±1.8, ±2. 
Find the turning points, and the value of the maximum or 
minimum ordinates between the limits given. 



572 



ALGEBRA. 



80. From the graph in Ex. 29 find to two places of decimals the 
roots of X s — 3 x = 0. 

81. Solve the following pairs of equations graphically : 

(i.) x + y = 15, (ii.) x = y = 8, (iii.) x* + 1/ 2 = 13, 
sy = 36 ; a#=18; sey = 6. 

82. An india-rubber cord was loaded with weights, and a measure- 
ment of its length was taken for each load as tabulated. Plot a graph 
to show the relation between the length of the cord and the loads. 



Load in pounds 


10 


12 


17 


21 


23 


25 


Length in centimeters 


36.4 


37.7 


40.6 


43.0 


44,3 


45.4 



What was the length of the cord unloaded ? 

33. The mean temperature on the first day of each month, on an 
average of 50 years, had the following values : 

Jan. 1, 37° ; May 1, 50° ; 

Feb. 1, 38°; June 1, 57°; 



Mar. 1, 40 c 



Sept. 1, 69°; 
Oct. 1, 54°; 
Nov. 1, 46° ; 
Dec. 1, 41°. 



Julyl, 62 3 ; 
April 1, 45°; Aug. 1, 62°; 

Represent these variations by means of a smooth curve. 

[The difference of length of different months may be neglected.] 

84. A manufacturer wishes to stock a certain article in many sizes ; 
at present he has five sizes made at the prices given below : 



Length in inches 


20 


27 


33 


45 


54 


Price in dollars 


11 


14.5 


20 


35 


48.5 



Draw a graph to show suitable prices for intermediate sizes, and 
find what the prices should be when the lengths are 30 inches and 
46 inches. 

35. Several tourists set out for a station 3 miles distant and go 
at the rate of 3 miles an hour. After going half a mile one of them 
has to return to the starting point ; at what rate must he now walk in 
order to reach the station at the same time as the others ? 

36. A motor car on its way to Bristol overtakes a cyclist at 9 a.m. ; 
the car reaches Bristol at 10.30 and after waiting 1 hour returns, meet- 
ing the cyclist at noon. Supposing the speeds of car and cyclist to be 
uniform, find when the cyclist will reach Bristol. Also compare the 
speeds of the car and cyclist. 



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