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ELEMENTARY ALGEBRA
THE MACMILLAN COMPANY
NBW YORK • BOSTON • CHICAGO • DALLAS
ATLANTA • SAN FRANCISCO
MACMILLAN & CO., Limitbd
LONDON • BOMBAY • CALCITITA
MBLBOURNB
THE MACMILLAN CO. OF CANADA, Ltd.
TORONTO
^ ■J
JOHN WALUS (16161703)
Professor at the UniverBlQr of Oxfoi^ In 1686 he published it
Treatise of Algebra which was the means of making the science of
algebm more widely known in England.
ELEMENTARY ALGEBRA
^econD i^ear Course
BY
FLORIAN CAJORI
OOLORADO COLLJBOE
AND
LETITIA R. ODELL
NOBTH 8IDB HIGH 80HOOL, DBNYBB
THE MACMILLAN COMPANY
1916
• •• •, • ••••• •
• ' • • ••••••••
GOPTBIOHT, 1916,
bt the magmillan company.
Set up and electrotjrped. Published July, 19x6.
CAJORI
J. 8. Cashing Co. — Berwick A Smith Co.
Norwood, Mass., U.S.A.
PREFACE
This book contains a brief review of the fundamental opera
tions of algebra followed by a thorough presentation of the
topics usually included in the work of the third half year.
The material is so arranged that the choice of topics for review
or advance study may be easily made.
The book contains not only a large number of practical
problems but also practical applications of graphs. Material
for drill in the manipulation of exponents and radicals will be
found in Chapter VII. Great pains have been taken to make
the subject of logarithms accessible to beginners. The most
difficult part in learning to compute with logarithms is the
process of " interpolation." Of the various arrangements of
logarithmic tables which have been suggested, we have selected
the one which renders interpolation easiest. The chapter on
logarithms is introduced earlier than usual, on account of the
great practical importance of the subject. If desired, it can
be taken up later, after Chapter VI. The concept of disfunction
does not receive isolated and abstract treatment; it is pre
sented as a fundamental idea in proportion, variation, and
graphics. Its connection with problems of everyday life is
firmly established.
The aims of the First Year Book have been kept in view in
the preparation of this text. Emphasis is thrown upon clear
ness of exposition and the use of expressions which recall the
axiomatic processes involved. Continued stress has been laid
upon oral exercises.
911236
vi PREFACE
Several of the practical problems given in the text were
suggested by the perusal of an English book, T. Percy Nunn's
Exercises in Algebra (including Trigonometry), Part 1, 1913.
The authors have received help from several teachers. Spe
cial mention should be made of Mr. E. A. Cummings of the
North Side High School in Denver, who has worked most of
the exercises and offered valuable suggestions.
FLORIAN CAJORI.
LETITIA R. ODELL.
TABLE OF CONTENTS
OHAPraB PAGB
L Elembntabt Definitions and Operations .... 1
Fundamental Definitions 1
Addition and Subtraction 2
Order of Fundamental Operations 3
Multiplication and Division 6
Parentheses 7
Review Exercises 9
Equations 11
Simultaneous Linear Equations 13
Positive Integral Exponents .20
Fractional and Negative Exponents 21
Powers and Boots 25
Factoring 32
Fractions, their Multiplication and Division ... 37
Highest Common Factor and Least Common Multiple 43
Addition and Subtraction of Fractions .... 44
Quadratic Equations 48
Problems Involving Quadratics 52
n. More Advanced Theory and Operations .... 56
Fundamental Laws of Algebra 56
Remainder Theorem and Factor Theorem ... 59
Binomial Theorem (preliminary treatment) ... 65
Systems of Linear Equations. Determinants ... 67
Problems Involving Linear Equations .... 74
Division by Zero Impossible ...... 76
Equations with Fractions 77
Miscellaneous Practical Problems 80
vu
Viii TABLE OF CONTENTS
CEAVm PAGK
III. Proportion, Variation, Function 86
Proportion 86
Functions 87
Different Modes of Variation 88
Variation Shown in Graphs 90
Problems in Proportion and Variation . ... 02
Graphs Exhibiting Empirical Data .... 101
IV. Logarithms 102
Logarithmic Curve 103
Fundamental Theorem 106
Finding Logarithms 108
Finding Antilogarithms Ill
I*roblems 115
Exponential Equations 116
V. Quadratic Equations and their Propbrtils . . 119
Equations Quadratic in Foi^m 119
Relations between Roots and Coefficients . . . 120
Nature of the Roots 122
Graph of the Quadratic Equation ax^ \bz\c = y . 128
VI. Systems of Equations Solvable bt Quadratics . . 127
A System of Two Equations, One Linear . . 127
A System of Two Equations, Both Quadratic . . 134
Possibility of Solution by Quadratics .... 135
Problems 137
VII. Exponents, Radicals, Imaoinaries 139
Meanings of Different Kinds of Exponents . . . 139
Different Kinds of Numbrrs 140
Simplifying Radicals 141
Operations with Radicals 144
Square Root of a ±2v7) 148
Irrational Equations 149
Irrational Equations Quadratic in Form . . .152
Graphic Representation of Complex Numbers . . 154
VIII. Series and Limits 168
Arithmetical Series 168
Geometrical Series 163
Infinite Geometrical Series 170
Theory of Limits 173
TABLE OF CONTENTS ix
PA6B
SUPPLBMENT 177
The Highest Common Factor by the Method of Division . . 177
Solutions of Quadratic Equations 180
Mathematical Induction and Proof of the Binomial Theorem . 181
Applications of the Binomial Theorem 185
Table of Squares and Cubes, Square Roots and Cube Roots . 188
Review Exercises Selected from College Entrance Examinations 190
*
^ , J »* >
ELEMENTARY ALGEBRA
SECOND YEAR COURSE
CHAPTER I
ELSMSNTARY DEFINITIONS AND OPERATIONS
FUNDAMENTAL DEFINITIONS
1. In arithmetic, numbers are commonly represented by
HinduArabic numerals. In algebra, numbers are represented
also by letters.
Any combination of numerals, letters and symbols of
operation, which stands for a number, is called an algebraic
expression.
An algebraic expression may consist of parts which are
separated by the f or — signs ; these parts, with the signs
immediately preceding them, are called terms.
Thus, in + a^  2 a& + 3 &3, there are three terms, + a^ — 2 a&, +3 b^.
Each of the numbers which multiplied together form a
product is called a factor of the product.
A factor consisting of one or more letters is called a literal
factor.
Terms which have the same literal factors are called similar;
terms which do not have the same literal factors are called dis
similar.
In 4 a3&  3 ab^  10 a% the terms + 4 a^b and > 10 a^b are siinilar ;
the terms —Sdb^ and  10 a^b are dissimilar.
B 1
2 ' ELE^BmARY ALGEBRA
. • • .• ••: .••••:: : •• :
• All algebraic expression of one term is called a monomial, of
two terms a binomial, of three terms a trinomial, and of several
terms a polynomial.
When a number is the product of two factors, either factor
may be called the coefficient of the other factor.
Often the word coefficient is applied only to the factor which
is expressed in numerals.
Thus, in 6 xyz^, the numeral 6 is called the coefBcient of xyz^.
An exponent is a number placed at the right and a little
above another number, called the base.
When the exponent is a positive integer, it indicates how
many times the base is taken as a factor.
The exponent expresses the power to which the number is
raised.
Thus, the 4 in a* expresses the fourth power of a.
Find a* when o = 2. Has o* the same value as 4 a, when a = 2 ?
The absolute value of a number is its value regardless of its
sign.
Thus + 5 and — 5 have the same absolute value, 6.
ADDITION AND SUBTRACTION
2. Addition : If similar terms have like signs, find the sum
of the absolute values and prefix the common sign.
If similar terms have unlike signs, find the difference of the
absolute values and prefix the sign of the one which has the
greater absolute value.
If the terms are dissimilar, the addition is indicated in the
usual way.
Thus, the sum of 3 a and 5 & is 3 a + 6 &.
To check an example in addition :
Substitute numerals for the letters and find the sum. This
sum must equal the result of substituting the numerals in the
answer.
ELEMENTARY DEFINITIONS AND OPERATIONS 3
3. Subtraction: Conceive the sign of the subtrahend to be
changed, and then proceed as in addition. Check by substituting
numerals for the letters.
In what other way may subtraction be checked ?
WBITTBN BXEB0ISB8
4. Find the sum of :
1. 7a36f 6c; 4a + 5612c; 6a + 2b + Sc.
2. 4aj* — 3aj*y — 4a^*; 3a^ — 8a?y« — Saj*; 3aj* — 6a5«y.
3. 5a» — 6a"6" — 7c»; 2a" 43a»6» + 8c»; — 5a"6» — 2c".
4. 3(a  b)+4:{a + b) ; 7(a  6) 6(a + 6) ;  6(a  b)
hll(a + 6).
6. a43a«49 a*4a46; 7 a^ }• 3a^ ^2 a^ + 10a + 7 ,
4a*8a»5.
6. From 2 m — 3 m« + 4 n take m — 3 mw — 9 n.
7. From 2a^'2ab + Sb^ take a^ — ab — b^.
8. From 4 4 3c 8d9e take 7c 4 5c 10 2d.
9. From a* 4 2 a*^*' — 6b^ take 2 a' — 5 a'ft*' + Sfe*'.
10. From 3(aj + y)—7(x — y) take 4(a: — y) — 6(a? 4 3^).
11. From (a? — y)a — (« 4 y)b take (a? 4 y)a + (a — y)6.
12. What number added to 7 aj^ — 5 a?y* — 6 a?y will give
2a^2 — 3aj*y44ajy?
ORDER OF FUNDAMENTAL OPERATIONS
5. The values of many algebraic expressions depend upon
the order in which the operations of addition, subtraction, mul
tiplication, and division are performed. To avoid confusion it
has been found necessary to adopt certain rules, so that an ex
pression shall always be interpreted in the same way and
4 ELEMENTARY ALGEBRA
^ represent the same value. The following rules are agreed
upon:
I. A succession of multiplications and divisions shall be
performed in the order in which they occur.
Thus 6. 8 + 4. 8 = 48 +4. 3 = 12. 3 = 36.
II. A succession of additions and subtractions shall be
performed in the order in which they occur.
III. A succession of multiplications, divisions, additions,
and subtractions shall be performed in accordance with I and
II ; the multiplications and divisions being performed before
any additions and subtractions.
Thus, 20 + 6. 743. 6 = 20 + 42 4 16 = 43;
17 + 15 + 88. 2*.4 = 17 46 16 + 4 = 17 + 64 = 18.
6. If there are parentheses, apply the preceding rules to the
expressions within the parentheses first ; then to the resulting
expression as a whole.
The forms ( ), [ ], { } go by the general name of " paren
theses" but they are designated by special names when it is
necessary to distinguish one from the other.
Thus, [ ] is called a " bracket," { } a " brace," but ( ) is
always called a " parenthesis."
The " vinculum " is also used to denote aggregation, thus
a + &*■ fit — ^«
ORAL BXEB0I8B8
7. Simplify:
1. 162 414. 7. 2857.2 45.
2. 3a?42^48aj43y. 8. ^xiy^^x + ^y.
3. 2a7a3a48a. 9. 15 • 3 45 4 10 4.
4. 10 5m 15 4 20m. 10. 24 4 5.510  7(15^2).
6. 5(7_2)4[8i3]. 11. 48 43(18^9.3)^9.
6. 25 ^ 5 4 15 4 40. 12. (12a7a)(13a20a):
ELEMENTARY DEFINITIONS AND OPERATIONS 5
MULTIPLICATION
8. To muUiply one term by another : Multiply the numerical
coefficients ; annex the letters, giving to each letter in the product
an exponent equal to the sum of its exponents in the two factors.
Law of signs: Like signs give plus, unlike signs give minus.
The degree of a term is the sum of the exponents of the
literal factors.
Thus 4 Qcy^afi is of the sixth degree.
An expression is homogeneottSy if all its terms are of the
same degree.
In a^ + 2 a^ — 6 x^ every term is of the fourth degree ; the expression
is homogeneous.
To multiply a polynomial by a polynomial: Arrange the
terms in both polynomials according to the Ascending or
descending powers of some letter; multiply each term of
the multiplicand by each term of the multiplier and add the
partial products.
Multiplication may be performed by detached coefficients.
For example, (2ic*7aj»5aj*46aj 3)(6 a^  4 a? 4 6).
275+63
64+5
124230 + 3618
_ 8+28 + 2024 + 12
10  35  25 +3015
1250+ 8 + 2167 + 4215
The product is 12 aj« 50 a* + 8 aj* + 21 aj»  67 a^ + 42 a?  15.
If any powers of x are lacking in either polynomial, the
terms in question must be represented by zeros.
Thus, in multiplying ac" + 2 05 — 4 by 05* + 05* + 1, we must write
1+0 + 2 4and 1+0+1 + + 1.
Check an example in multiplication by substituting numerals
for the letters.
6 ELEMENTARY ALGEBRA
JDIVISIOH .
* ' •
9. To divide oiie term by another : Divide the numerical
coefficients ; annex the letters, giving to each letter in the
quotient an exponent equal to its exponent in the dividend
minus its exponent in the divisor. Observe the law of signs :
Like signs give plus, unlike signs give minus.
To divide one polynomial by anothsr :
I. Arrange the terms.
II. Divide the first term of the dividend by the first term of
the divisor and obtain the first term in the quotient.
III. Multiply the divisor by this term in the quotient.
IV. Subtract the product from the dividend.
V. Treat the remainder as a new dividend and proceed
as before.
YI. Keep each new dividend arranged in the same order as
the first dividend.
Divide 14 a:*  27 ax* + 21 aW  32 a* + 12a8a; by 2x8  8 a« + 4 a«.
2a;23ax + 4a«
14jr*  27 oa* + 21 a'^a + 12 a»x 32(1*
14 re*  21 ax» + 28 a%c2
7a:23ax8a«
 60*8 7a2a;2 + i2a»a532(i*
 6aofi+ 9 qagg . 12 flgg
 16 a2a;2 I 24 a'te  32 a*
16a2aja + 24a8'x32a*
By detached coefficients this division may be performed as follows :
14 _ 27 + 21 + 12  32
14  21 + 28
23+4
738
 6
 6 +
7 + 12
912
32
^_
16 + 24
16 + 24 
32
32
The quotient is 7 a;^  3 ox — 8 a^.
Division may be checked by substituting numerals for the
letters, care being taken to avoid a zero divisor.
ELEMENTARY DEFINITIONS AND OPERATIONS 7
ORAL BXBBOI8B8
iricaJ 10. 1. In what other way may examples in multiplication
the and division be checked ?
dend 2. If there is a remainder in a division, what must be done
gns:' with it in checking?
3. If the multiplicand and the multiplier are arranged
according to the descending powers of some letter, how will
the product be arranged ?
11 of 4. If the multiplicand and the multiplier are homogeneous,
what will be true of the product ? Of what degree will it be ?
I
eed
as
2 I
• I
I
I
WRITTBN BXBRCISB8
11. Expand and check :
1. (7ah2a2442a»)(3a3 + 4a«).
2. (aj*« — 2 aj«3/* f y»)(af — ^).
Divide and check :
3. a^Sa^b + Sab^b^hyab.
4. ixfi — ]^ hy X — y, and by a? 4 y.
6. 12 aj+< + 20 af+»  107 ic»+2 ^ 37 0^+1 f 33 af by 6 aj»  11 a^.
6. m* f w' 4 i>' — 3 mnp hy m^n+p.
7. a»&' + c» + 3a6cby a^+V + c^f aftac + ^w.
8. 2,*2^h2/»2/«H.5y_^byy.
PARENTHESES
12. I. A parenthesis preceded by a j sign may be removed
without changing the signs of the terms within the parenthesis.
II. A parenthesis preceded by a — sign may be removed,
le provided that the sign of each term within the parenthesis be
changed.
L
8 ELEMENTARY ALGEBRA
WRITTEN BXEBOI8BS
13. Simplify:
1. 18 (72) 3.
2. 5a4(6a4a)(9a2a).
3. 8 a:  [2 aj 4 (3 aj  y) f 7] + [5 «  2 y + 3].
4. 7m4[2n j4 m (3 m5n) 8 ii{ f 9m].
6. 10c {4d(9c5d)  {(2c d) (4cf 7d)}.
6. (ajf 2)(aj3)5(a^3« + 2)10(aj + 3).
7. {5a«3[5+(af 2)(a5)7] a(al)\
9. (a + 6)(a  6)  (a  6)2 4 (a 4 2^)'.
10. 2{ 4(7 3) f (8 2) 5(93 + 5)1 411.
11. m{[— (1— m)— l]m— {m— (54m)(44m)}.
14. Terms may be inclosed in a parenthesis by reversing
the rules for removing parentheses.
When terms are inclosed in a parenthesis preceded by a 4
sign, the signs of the terms are not changed.
When terms are inclosed in a parenthesis preceded by a —
sigriy tjiersigns of all the terms are changed,
WRITTEN EXERCISES
15. Collect the coefficients of a? in a parenthesis preceded
by a 4 sign, and the coefficients of ^ in a parenthesis pre
ceded by a — sign :
1. a>x — ay '\ bx — by, 4. 2x ^3y — dx {■ dy — fx.
2. 7nx + na; 4 f^/y — ^y. 5 py — 9^ —px{ qy.
3. abx + cdy — cdx— aby, 6.7 aaj+3 6y 44 bx— 5 ay. ■
ELEMENTARY DEFINITIONS AND OPERATIONS 9
REVIEW EXERCISES
16. liA = 2a^^ab+7b', 5 =  5a*+ 2a646«,
C == S a" + 10 ab 6 b% D = 4a«9 6« + 3a5,
find the values of :
1. ^ + 5 + C4A 4. B A + GD.
2. AB^OD. 6, GB + AD.
3. ^A+BC + D, 6. D^AGB.
If a = 1, 6 = 3, c = — 2, d = 4, n = 2, find the numerical
values ' of :
2a3c + 4d^ 3a^2a26.
7c 2d
8. ?^ + ^. 13. (a+6)«(cd)».
Q a^ + &^ 14. (c + d)(ccr>+5^
• c2hcP' ^"^
, ^ a" — 6* X6. 2 d ^ 3 c • dn — &c ^ d.
10. — •
11. 5a^(6c)Hdn. 16. Vfe^+VS^.
Find the sum of :
17. (a 4 b)x  (a  6)2/ and (a  6)aj + (a + ^)y.
18. a6 + ic; ia + i6^c;andfa + i6Hc
19. ajfy + i2;; a? + yi2;; andajy + 2.
20. [(a + c)aj 4 (^ + c)y]  [(a  c)a;  (&  c)y].
21. [5(.'r42/)7(ajy)][2(aj + y)3(«y)].
Simplify :
22. aj2(?/22^)[y^(»''«')] + [25' (y^aj«)].
23. aj22;f3yM?[3a?(5«y7w?)]5« + 4a?{
, 24. (iB« + 2iC^ 3af2l)(ajl).
10 ELEMENTARY ALGEBRA
26. (a"+* — 4 a« + 5 a"* + a'^^a + 1).
26. (a"4&")(a"— 6»).
27. (a'a6hi6»)(ai6).
28. (a«6'»+^ — 6 a'6"* + 4 a'»6"»i) « 2 aft*"*"*.
29. (aj»2r — af +y+^ 4 aj*y**) 5 af y*.
30. (a** 4 a** + 1) 5 (a**  a* f 1).
31. (9 a*»  26 6^) s (3 a« + 6 fe'j
32. (iaj»xV + M^ty')^(iaJiy).
TYPE FORMS IN MULTIPLICATION
17. 1. (a ± 6)« = a* ± 2 a6 + &».
Take the upper signs together, and the lower signs together.
2. (af 6c)«=a*4&*4c2 + 2a62ac26c.
3. (a + 6)(a — 6) = a2  6».
4. (aj + a)(aj 4 6) = a.** 4 (a + ^)aj 4 a5.
6. (ax 4 6)(ca5 4 cf) = ocaj* 4 {ad \ bc)x + hd.
6. (a± 6)« = a«±3a*643a6»±6«.
ORAL BXBROISB8
18. Write, by inspection, the va4ue8 of :
1. (a42)«. 6. (a?^h)\ 9. (a»6»)«.
2. {bSy. 6. (c'd')*. 10. (aj»46')*.
3. (2 a: 1)2. 7. (c2 47(P)«. 11. (m»2n2)«.
4 (3 a 4)*. 8. (5«3y2)«. 12. (8a6a?«)2.
13. (aj — y4«)'. 18. (aj44)(aj49).
14. {2x y z)\ 19. (a? 4 20)(aj  15).
16. (5aj — 2 2/4 25)*. 20. (aj  8)(aj  6).
16. (a? 33/ 4)*. 21. (a44c)(a — 3c).
17. (a?7)(aj + 3). 22. (2a4 c)(3a  d).
^
ELEMENTARY DEFINITIONS AND OPERATIONS 11
23. (3aj44)(2a:7). 31. (a  2c)(a 7 c).
24. (566c)(46+3c). 32. (a«4 26)(a«436).
26. (x^y)(a — b). 33. (a» — 9 c)(a» + 8 c).
26. (3y4)(3y48). 34. (6a4 6)«.
27. (a« + c)(a« — c). 36. (a? + yy.
28. (2a + 6)(2a6). 36. (3aj4)«.
29. (5aj« — y»)(3/»f 5aj«). 37. (xyf.
30. («• 4 3/*)(»'  3/*). 38. (2 a 4 1)'.
Complete the squares :
39. 4aj*±12aj + ? 42. 4m«±?f9w*.
40. 9a«±24a4? 43. a*'±?+366»».
41. ?±18 6c + 81c». 44. 25aj*20 4?
EQUATIONS
19. An equation expresses an equality.
In other words^ an equation is a statement that two expres
sions stand for the same number.
2»6 = 7a? — 4, and (a + 6)* = a*  2 a6 f 6* are equations.
The first is an equation of condition^ because it is true only
under the condition that a? = 2. The second is an equation
of identity^ because it is true whatever values a and b may
have.
In this chapter we deal with equations of condition. The
root of an equation, containing one unknown, is a number
which, when substituted for the unknown, "satisfies" the
equation by reducing both sides to identical numbers.
An equation of the form
ooj* + 6aj»^ 4 ... 4^ = 0,
where n is a positive integer, is said to be of the nth degree.
12 ELEMENTARY ALGEBRA
The degree of any equation is indicated by the exponent of
the highest power of the unknown.
Thus, a; > 7 = 2, is an equation of the 1st degree, called linear.
a;3 — 3 X = — 2, is an equation of the 2d degree, called quadratic.
x8.62(^ + lla; = 6, is an equation of the 3d degree, called cubic.
x« — 1 = 0, is an equation of the 4th degree, called quartic.
etc.
20. The equilibrium of a balance is not disturbed so long as
like changes in the weights are made simultaneously on bpth
sides. So in equations, we may add the same number to both
sides, or subtract the same number from both sides, or we may
multiply or divide both sides by the same number (except divi
sion by zero).
The equality is maintained during all these changes.
Solve »(« + 4)=a^ — 3a? + 5.
Remove the parenthesis, x^ \Ax = x^ — Sx\ 6,
Subtract x^ from both sides, 4x = — Sx + b.
Add 3 x to both sides, 7 a; = 5.
Divide both sides by 7, x = f .
Check : In the given equation, substitute ^ for x,
WBITTBN BXEBOI8E8
21. Solve and check :
1. (aj2)(a?3)=a:(« + 4).
3. (m — 4)*4(m44:)2 = m(2mfl).
4. (r 4 1)'  r(r — 1)= r2(r + 2)f 5 r  1.
6. y(t/ + l)+(yhl)(2^ + 2)=(t/ + 2)(t/+3)+y(t/f4)9.
6. «.[4 + {4(4 + 01]=0.
7. 6«(4«8){53«~(7aj4) = ll.
ELEMENTARY DEFINITIONS AND OPERATIONS 13
8. s3[s93{94(6«)sn = 3.
9. 20(1  X) 3(x  6) S[x + 8 458  3(1  «)}]== 6.
10. (x 4 8)(aj  5)(x 4 6) = (» + 2)(x 4 S)(x + 4).
Solve each of the following equations for each letter in terms
of the others :
11. A = iab. j3 ;S = (a + Z). ^** ^i^i=^^2
12. S = ia(b^by ' ^ 16. = 1(27' 32).
I
SIMULTANEOUS LINEAR EQUATIONS
22. If a liQear equation contains but one unknown, one
value may be found for that unknown.
If a linear equation contains two unknowns, an unlimited
number of simultaneous values may be found for them. For
that reason the equation is called indeterminate.
In « + y = 3, we have x = l, y = 2; a; = 2, y = l;aj=— 2, y = 5j
a: = 0, y = 3 ; a; = 3, y = ; etc.
If these pairs of values are
plotted, the points thus obtained
will be seen to lie on a straight
line, as in Fig. 1 ; hence the name
linear equation.
Two points determine the posi
tion of a straight line. The graph
of a linear equation in two un
knowns is most easily made by
locating the points at which the
line crosses the coordinate axes. ^^' ^•
It was seen above that in the equation ajy = 8, 05 = 8 when y = 0,
and x = when y = 3. In Fig. 1,
x = S, y = locates the point A.
35 = 0, y = 3 locates the point B,
The two points A and B determine the line.
I 1 t I I III l^*T^T— T— '
IIIIIIIIIII
zz_sszzzzzz
Q ^ ZX
14 ELEMENTARY ALGEBRA
BXEBOI8B8
23. Make graphs of the following equations :
1. aj4y=7. 6. 4a;~9y = 36. 11. y = 0.
2. aj — y = 2. 7. 5a? = 10 — 2y. 12. a: = 0.
3. re — 15 = 3 y. 8. aj = y. 13. x = 3.
4. 3aJ— 4y=*12. 9. aj — 3y = 0. 14. y = — 4.
6. 62y — a;=0. 10. 3aj — y = 6. 16.a; = — 2.
24. If the graphs of two linear equations in x and y inter
sect, the values of x and y at the point of intersection satisfy
both equations ; this value of x and of y, called the coordinates
of the point, constitute one solution.
Since two straight lines cannot intersect in more than one
point, two linear equations in x and y cannot have more than
one solution*
Two linear equations in x and t/, representing lines which
intersect in one point, are called independent.
WHi'ri'BN BXEB0I8BS
25. Make graphs of the following equations and determine,
from the graph, the coordinates of the point of intersection.
Check, by substituting the coordinates in the equations.
We are not able to draw figures that are absolutely accurate. Hence,
solutions obtained from graphs are usually only approximations to the
true values.
1. a; + y = 6, 4. a: = 3, 7. 3« = 2y,
re — y = l. 2a:— 3y = 9. a?f4 = 7y.
2. 3aj — y = 7, 6. y = — 2, 8. a? — 3^ = 0,
2a;y = 5. 6a: — 3y=16. 3a:f23^ = 6.
3. a? 4 3 2/ = 8, 6. a? = y — 6, 9. a? f 1 = 0,
2ajfy = l. 6y = a: + 10. y — a?=6.
ELEMENTARY DEFINITIONS AND OPERATIONS 15
26. The two linear equations,
2xh2y = 10,
can be satisfied by the same values of x and y ; for example,
jc = 1 and y = 4, or a? = 3 and y = 2, or x = 5 and y = 0, etc.
Moreover, by dividing both sides of the second equation by 2,
the second equation can be reduced to the first. The two equar
tions are therefore not independent ; they are really different
forms of one and the same equation and represent one and the
same straight line. They are called equivalent equations.
27. The two linear equations,
x — y=z5,
xy=l,
are called inconsistent^ for the reason that no finite value of x
and of y can be found, such that x— y is equal to 5 and also
equal to 7 ; no pair of finite values of x and y satisfy both V
equations. This is geometrically evident from the fact that
the graphs of the two equations are parallel lines.
Two linear equations in two variables x and y belong there
fore to one of three groups :
1. The two equations are independent and represent two
lines which intersect in one point, or
2. The two equations are inconsistent and represent two
lines which do not intersect (are parallel), or
3. The two equations are equivalent and represent one and the
same straight line.
WBITTBN BXBBGISBS
28. Make graphs of the following equations :
1. 3aj6y=8, 3. 2yH3aj = 0,
oj— 2y = 5. 9ajh6y = 15.
2. 5y2x = 4:, 4. 2a:4y = 10,
4a:10y=:8. x2y = 5.
16 ELEMENTARY ALGEBRA
6. a— 3y = 2aj7, 7. aj = 3, 9. a;h4 = 0,
3ajH2y=4aj + 5y7. y = 2. yl = 0.
6. 5aj — 7y = ll, 8. ajhy = 0, 10. aj = 0,
10aj14y = 20. ajy=sO. 3^ = 0.
11. Can you tell, before making the graphs, whether two
equations are equivalent, or not? Inconsistent, or not?
12. Can you tell whether the graphs will be parallel, or not?
13. Can you tell whether the graphs will go through the
origin, or not ?
29. Systems of independent linear equations may be solved
by a process called elimination.
The elimination may be accomplished in one of three ways :
I. Addition or subtraction. — Make the coeflB.cients of one
variable alike in both equations and then either add or sub
tract the sides of the equations.
II. Substitution. — Find the value of one variable in terms
of the other in one equation, and substitute that value in the
other equation.
III. Comparison. — Find the value of one variable in terms
of the other in both equations, then equate these values and
solve the resulting equation.
EXBBCISBS
30. 1. Solve by addition or subtraction :
5ajh3y = 69, (1)
4aj7y=20. (2)
Solution. Multiply (1) by 4, (2) by 5, 20 x + 12 y = 276, (8)
20a;36y=100. (4)
Subtract (4) from (3) , 47 y = 376,
y = 8.
Substitute in (1), a; = 9.
CTieck : Substitute in (1) and (2), 46 + 24 = 69,
36 66 =20,
ELEMENTARY DEFINITIONS AND OPERATIONS 17
2. Solve by substitution,
5 + 6 = *'
(1)
7 8 14'
(2)
Solution. From (1),
Substitute in (2),
6 y 1
7 8 14'
160 201, 7y= 4.
111^=164.
y = 12.
Substitute in (1),
f+2 = 4.
a; = 10.
3. Solve by comparison,
3 a  13 y = 1,
(1)
(2)
Solution, From (1),
^ 13
(3)
From (2),
y = 16a;9H.
(4)
Compare (3) and (4),
8^=15x9H,
3xl = 195x129,
192 a; = 128,
x = f
1
Substitute in (3),
y = A
Solve a,nd check :
4. 12 « + 11^ = 12,
•• !+"¥'
42 a h 22 y = 4(H^.
X
•
%
18 ELEMENTARY ALGEBRA
3 9 3
^2 4 20
PROBLEMS
31. 1. The sum of two numbers is 180 and their difference
is 12. Find the numbers.
2. One number is 27 more than another. Their sum is
143. What are the numbers ?
3. Of two numbers, twice the greater exceeds three times
the smaller by 4 ; but the sum of the greater and twice the
smaller is 44. Find the numbers.
4. Two pounds of sugar and three pounds of flour cost 19^,
and three pounds of sugar and five pounds of flour cost 30 ^.
What is the price of each per pound ?
5. Three yards of velvet and five yards of silk cost me
$ 15.75. If I had bought two yards more of velvet and two
yards less of silk, my bill would have been $ 18.25. What
is the cost of the velvet and silk per yard ?
6. In five hours A can walk 1 mi. more than B can walk
in 6 hours ; in seven hours A can walk 3 mi. more than B can
walk in 8 hours. How many miles an hour can each walk ?
7. A, B, C, D, together have $ 300. A has $ 10 more than
C ; B has $ 5 less than half as much as D ; and A and B have
together $ 5 less than twice as much as C. How much has each?
8. A number consists of two digits whose sum is 12. If
three times the sum of the digits be subtracted from the
number, the digits will be reversed. What is the number?
9. The digits of a threedigit number will be reversed, if
396 be added to the. number. The units' digit is 3 times the
hundreds', and if double the tens' digit be increased by 6, the
result will equal the units' digit. Find the number.
^ELEMENTARY DEFINITIONS AND OPERATIONS 19
10. In a purse there are 3 times as many quarters as nickels,
and twice as many half dollars as nickels. The total value of
the coins is $ 3.60. How many of each kind are there ?
11. Some boys bought a boat and found upon paying for it
that if there had been 2 more of them, each would have paid
a dollar less ; but if there had been 2 fewer, each would have
had to pay $ 1^ more. How much did the boat cost ?
12. A rectangle is four times as long as it is wide. If it
were 3 in. shorter and 2 in. wider, its area would be increased
15 sq. in. Find its dimensions.
13. It costs as much to sod a square piece of ground at 25 ff
a square yard as it does to fence it at $ 1 a yard. How long is
the side of the square ?
14. A walk is laid 3 ft. wide around a rectangular court,
which is 15 ft. longer than it is wide. The area of the walk is
960 sq. ft. Find the dimensions of the court.
15. A man invests part of $ 5480 at 5 % and the remainder
at 4 %. The total annual income is $244. How many dollars
has he in each investment ?
16. The yearly income from a 5 % investment is $ 97.70
more than that from a 6 % investment. The sum of the two
investments is $ 6420. How much is invested at 6 % ?
17. A boy weighing 100 lb. is 6 ft. from the fulcrum of a
seesaw. He balances a boy who is 8 ft. from the fulcrum.
What is the weight of the second boy ?
By careful measurement it has been ascer i—r
tained that, to maintain a balance, the lengths
of the arms of a lever and the corresponding
weights must conform to the law, hwi = hv32 ;
that is, the product of one weight and its distance from the fulcrum is
equal to the product of the other weight and its distance from the fulcrum.
18. Two boys together weigh 170 lb. They balance when
one is 8 ft. and the other 9 ft. from the fulcrum. How much
does each weigh ?
20 ELEMENTARY ALGEBRA
19. If the lever is 8 ft. long, how far from the fulcrum
will two weights, 30 lb. and 50 lb., have to be in order to
balance?
20. A weight of 180 lb. is carried between two men by
means of a pole. One man is 5 ft. from the weight, the other
is 4 ft. How many lb, does each man lift ?
POSITIVE INTEGRAL EXPONENTS
82. In the expression a* we say a is raised to the nth power.
When n is a positive integer :
I. The nth power of a is the product obtained by using a
n times as a factor.
a • a • a • • • to n factors = a*.
II. The nth power of a* is equal to a* • a* • a* • • • to n
factors = a*".
Hence (a*")" = a"".
III. The nth power of a^ft** is equal to (a* • a* • a* • • • to n
factors) (6' • 6" . 6' • • • to n factors) = a"*" • h^ = a'"*6'*.
Hence {oFbpy = af^'^hf*'*.
IV. The nth power of — is equal to
a* • a* • a* • • • to n factors a***
ftp . 5j» . 5p . . . to n factors 6**
Hence, ( — ) = — •
' \bpj bP'*
33. From the law of the signs in multiplication :
1. An even power of any real number is positive.
2. An odd power of any real number has the same sign as
the number itself.
ELEMENTARY DEFINITIONS AND OPERATIONS 21
BXBBCISBS
34. Write the values of :
1. (2a'6)'. jj /_^V
2. ilahf)*. ^ ^^
4. /'3<«f\*. 13. (2a»6c»)'.
^'^'^'^ 14. (_a*6'c«)».
7. (0^2'')". ^^ f^^_f\\
8. (a^+'ft*"")'. A ^^^*^ /
19. (a*6")»».
20. (6a^yh)^.
10.
(23^)'
MEANING OF FRACTIONAL EXPONENTS
35. When an exponent n is a positive integer, we know that
a* = a • a • a • • • to w factors. That is, the exponent n indicates
that a is taken n times as a factor.
What is the meaning of a* ? It would be absurd to say that
a* signifies a taken ^ times Vs a factor. A number can be
taken ^s a factor only a whole number of times.
For the purpose of attaching a meaning to a* we stipulate
that we shall be able to multiply a' by a* according to the
same rule by which a* is multiplied by a^ The product of a*
and a* is found by adding the exponents; that is, a^>a^ = a*.
If we add the exponents in the multiplication of a' by a%
we obtain, 1 1 1 1 1
/
22 ELEMENTARY ALGEBRA
It is seen that a' is one of the two equal factors of a, or the
square root of a.
Hence a^ is another way of writing the square root of a.
Likewise, the meaning of a* is found by taking a* four times
as a factor, thus, i s s «
a* • a* • a* • a* = a*.
Hence a* means one of the four equal factors of a' or the
fourth root of aK
In general, to find a meaning for a«, where p and g are pos
itive integers, we take a« as a factor q times, and add the
exponents. We obtain,
   . j» J + — r+ ••• to c terms — _
a« . a« • a* • • • to g factors = a« « « = a« = a'.
It is seen that a« is one of the q equal factors of a'. Hence
t
a« means the gth root of a^ Thus,
The numerator of a positive fra^ional exponent indicates the
power of the base and the denominator indicates the root of that
power.
In finding the value of 27*, we may square 27, which gives
729, and then take the cube root of 729, which is 9. Or we
may take the cube root of 27, which is 3, and square 3, getting
9 as the result, that is,
27* = a/272 = {V27y.
In general, we have.
All irrational numbers which are expressed by the use of
radical signs can be expressed by the use of fractional ex
ponents. In fact, the simplification of expressions usually can
be effected more easily by the latter. Thus,
■v/16 a^h^d^ = 16*a*6^c* = 2 a^l^c.
X
X
ELEMENTARY DEFINITIONS AND OPERATIONS 23
MEANING OF A ZERO EXPONENT
36. We proceed to assign a meaning to aP, where a is not
itself zero. For simplicity we stipulate, as before, that in the
multiplication of two powers having equal bases, we find the
product by adding the exponents.
Accordingly, a"* >a^=z a""*"® = a"*.
Divide both sides by a*, 2!!j_?L — ?5!!.
a" a"*
Simplify, a* = 1.
Therefore, any number y except 0, vnth a zero exponent is equal
tO'l,
MEANING OF NEGATIVE EXPONENTS
37. Let m represent a positive integer or a positive fraction.
What is the meaning of a"^, where a is not zero ? Working on
the assumption (§ 35 and § 36) that in multiplication we are
permitted to add the exponents^ we obtain,
a"* • a~*" = a'*""' = a\
Since a° = 1, a* • a"~ = 1.
Divide by a**, a~** = — •
a"*
Therefore, any number with a negative exponent is equal to 1
divided by that number with a positive exponent.
Thus,
a»6*
1
.6*
6*
c
6*.
• c
a6«
a^b^c
" 1
a*
1 ~
— • c
6«
c
Therefore, a factor may be moved from the numerator
to the denominator of a fraction, or from the denomi
nator to the numerator, provided the sign of its exponent be
changed.
24 ELEMENTARY ALGEBRA
2
+ 6 _5
Care must be taken to transfer only /actors. In ^ — ^^, ar^ is not a
c
factor of the numerator. "*" is not equal to , • Since a~^ = ^ «
i + 6
it follow that ?:^±^ = ?l_ = li^.
c c a^
PRINCIPAL ROOTS
88. Since 6 • 6 = h 36, and (  6)( 6) = + 36, it follows that
both h 6 and — 6 are square roots of 36. The two square
roots of 36 are usually written in the form ± 6. Similarly the
two square roots of a^ are written ± a.
The positive square root of a number is called the principal
square root Unless otherwise stated, we shall consider only
the principal square root, and write V36 = 6. When V has
no sign before it, or has the + sign before it, the principal
square root is always understood.
When we write — V we mean the negative square root.
Thus, \/36 or (86)^ stands for h 6,  V86 or (36)* stands for — 6,
±V86 = ±(36)1=±6.
It can be shown that there are 3 different cube roots of a
number, four different fourth roots of a number, and, in general,
n different nth roots 'of a number.
For example, there are three different cube roots of &' ; namely, 6,
^^ "*" &, "" ^ ""^^^ 6. The fractional coeflacients in the last two
2 2
roots involve the imaginary number V— 3. Only one of the three roots is
real, namely, the root b. We call h ihe principal root.
In elementary algebra the principal root is the only root
usually considered.
I. .If a is a positive number, then the principal nth root of a
is its positive value ; we designate it by a" or ^a. When n is
an odd number, then this principal root is the only real root.
ELEMENTARY DEFINITIONS AND OPERATIONS 25
1
When n is even there is still another real root, namely, — a* or
Thus, the principal cube root of 64 is (64) t or v^ = 4 ; this is the only
cube root of 64 that is real.
The principal fourth root of 81 is (81) i or \^ = 8. There is another
real fourth root of 81 ; namely, —3, for we see that ( — 3) ( — 3) ( — 3) (— 3)
= 81.
II. If a is a negative number, and n is an odd integer, there
is no positive root ; there is a negative root and that is taken
as the principal nth root.
Thus, the principal cube root of — 27 is C— 27)^ or ^—27 =  3.
III. If a is a negative number, and n is an even integer, then
all the roots are imaginary. Imaginary numbers will be dis
cussed more fully later.
Thus, v^— 64 or ( — 64) • represents imaginary roots.
POWERS AND ROOTS
39. It is readily seen that
(a^y^^c^.a^^a^'C^^a."^
This result may be obtained at once by multiplying the
exponents 3 and 4. In this process algebraic expressions are
raised to powers ; it is called involution.
In general, (a*)* = a"*"
((aryy = a~»»
11
It is agreed that each of the symbols a% Va, 6*, Vb, (a6)n,
^^/ab shall represent only one nth root, namely, the principal
26 ELEMENTARY ALGEBRA
nth root. For the principal real roots the following formulas
can be shown to be true :
(a"")* = a"* or Va"*" = a"*.
Ill
a* • 6" = {ahy or Va • Vh = Va6.
? = r?Yor^
y/a _ */a ^
5^ W V6 
Express each of these formulas in words.
•
The square root of 2 cannot be exactly expressed by the
HinduArabic numerals. One can approximate its value by
extracting the square root to two, three, or more decimal
places, thus: V2 = 1.41+, V2 = 1.414+, V2 = 1.4147+, and so
on. The radical V2 or 2*, and other radicals of the same
kind, V3 or 3i, ^5 or 5^^, etc., whose values can be found ap
proximately, but not exactly, represent numbers called irra
tional numbers.
ORAL BXBBCISBS
40. Express with fractional exponents and simplify :
1. V3 win*. 5. 2V2a^. 9. \^5 . Vo^.
2. A/2a^h\ ^ 10. v/5.^35.
7. 3a/^.
3. V21a'c\ ^^'_ 11. ^.■^.
, 8. 7\/. 7,
4. ay/^M. ^y 12. 6V2^^*.
Express with radical signs :
13. 2(7 y)l 15. Sh^yl 17. 5(x + 2/)*.
14. Sajij/* ^®* "1* 18 4(aj*y2)*.
4.27*.
ELEMENTARY DEFINITIONS AND OPERATIONS 27
Simplify :
19. 49i. 26 2^ *'■ (3)"'(^)'
20. 81*. ■ 8* ' 82 (M)*.
21. 50 . 16i. ''' ^f!„r ^'' <^>"'_,
, , 27. ^^ 84. a».()*.
22. 16i.26i. 4'a>
23 iis) • 29. a»(a^ + 6). ^_,^_,
24. (^)*. 80. 36^343*. **• '^i;=r*
37. £l^. 38. «1±^.
m~* + b~* Multiply numerator and denomi
' mr* — 6~*' nator by tfibf.
Find the principal roots of :
40. Vxy. .g 1 256 a^b* »/ >^a"
41. V4^. >'289m" >/««««•
8.
42. V9^. 60. </256^. 56. Vl25a!>Y.
43. V26^. ^sr^^ "• A^
44. VIOO ai6. ^^' XUoO'riOwie'
h^^
si 512 a^
400 a^y «
45. ^8a36i2. ^ 58. \/— ^
46. ■\/64.afifz'^.
x^
47. ^ 343a36i« . 53. V/^?^. ^^' ^^^^'
48. V^/1^. e,.^ 60. #^'.
\ 729 2/3 54. ^J/729^y%24; \ 76r
Find the ^loo reaZ fourth roots of :
61. 10,000 a2&3c*(a + ^)* «3. j\ x*y'^^(x  yy.
62. 625 7/i8n«p3(a2  52)2. e4. .OOSl(x  yyzhi}^.
28 ELEMENTARY ALGEBRA
Find the real fifth root of :
65. 32iB6(m+7ij9)«y»24. 66. 243 a»(6  c)i«ar«2*.
67. What kind of roots of positive numbers have two real
values ? Which of those two real roots is the principal root ?
68. What kind of roots of positive or negative numbers
have only one real value ?
69. What kind of roots of negative numbers have no real
value?
SQUARE ROOT
41. Since (a h 6)2 = a^ + 2 a6 + d«,
Va2h2adH62 = a + 6.
Thus, by inspection, we can extract the square root of a tri
nomial which is a perfect square.
Or we may find the square root in this way :
a2h2a6h6* ah^
a*
2a6h&*
Trial divisor, 2 a
Complete divisor, 2 a + 6
Thus the square root is a + 6.
This simple case will enable us to devise a rule which is
applicable to more complicated cases. The procedure is as
follows :
I. Extract the square root of the first term and subtract its
square from the polynomial, leaving 2 a6 H bK
II. In 2aby we see the factor b which we know is the
second term of the root. This factor b may be obtained by
dividing 2ab by 2 a. Thus we call 2 a the trial divisor.
Since a is the part of the root already found, we see that
the trial divisor is double the root already found. After b h^s
been found, add it to the trial divisor and we have 2 aft + 6,
the complete divisor.
ELEMENTARY DEFINITIONS AND OPERATIONS 29
III. Multiply 2ab + bhj by and subtract.
This process may be extended to finding the square root of
any polynomial.
Find the square root of x* + 21xa86a:f 86 — Oa*. Arrange the
terms according to the ascending or descending powers of some letter,
x*6a« + 21ic236a; + 36 \x^Sx + e
— 6ar« + 21x3 — 36x + 86, 1st remainder
6x8+ 9x2
12 x^  86 X + 86, 2d remainder
12x286x + 86
Ist trial divisor, 2 (x^) = 2 x^
Ist complete divisor, 2 x^ 3 x
2d trial divisor, 2(xa  3x)= 2x2 6x
2d complete divisor, 2 x^ — 6 x + 6
* 8d remainder
The procedure is as follows :
I. The square root of x* is x'. Here a = x^ ; 2 a = 2x2, the first trial
divisor.
II. 2 a6 = — 6x8 ; divide — 6x8 by 2x2 and we obtain — 8x, the value
of &. Thus the first complete divisor is 2 x2 — 3 x.
III. Multiply 2x2 8x by —8x and we obtain —6x8 + 9x2.
Subtract this product from the 1st remainder and we obtain
12x286x+86.
IV. Treat this remainder as we did the first remainder and proceed
as before.
Since the third remainder isO, x2 — 3x + 6is the required square root.
BXBBCI8BS
42. Find the square root of :
1. 4 a«  28 a»6« h 49 6".
2. 4a«12a6 + 4ac + 9626&chc*.
3. ai2 — 2icyh6a:Hy2 — 6yh9.
4. 4ic*20ic»30aj + 9H37a^.
6. 9aj«2ic*3aj* + lh 10 aj' 12 a^2 a.
6. 9 a« h 9 d« h 24 a'^ft h 24 ab^  8 a^ft*  8 a«6* 50 a»6».
7. ^5aj + 25. 8. ^'2 + i.
4 9 a*
30 ELEMENTARY ALGEBRA
^ 4 4 , , 187 , 15 ^25
9. m* — rnr \ ——nr :wi4r'
9 48 4 4
tix A 4a , a* , 2a' 4a' , a*
11. a' H a5, to three terms.
12. 1 — a, to four terms.
SQUARE ROOT OF ARITHMETICAL NUMBERS
43. 1« = 1, 2* = 4, ..., 9* =81, 10« = 100, ..., 90« = 8100,
100* = 10000, 1000» = 1,000,000.
This is sufficient to show that a perfect square consisting of
one or two digits has one digit in the root ; one consisting of
three or four digits has two digits in the root ; one consisting
oijive or six digits has three digits in the root ; and so on. In
other words, the square of a number has twice as many inte
gral digits, or one less than twice as many, as the number
itself.
Hence, to find the square root we must separate the number
into periods of two digits each, towards the left and right from
the decimal point The period farthest to the left may have
one or two digits, whereas the one farthest to the right must
have two, a cipher being annexed, if necessary, to complete it.
The square root has therefore one digit corre9ponding to every
period in its square. Separating the square number into
periods enables one to find one digit in the root at a time.
This is seen more clearly if we consider that the square of
any number of tens, say 4 tens (40), ends in two ciphers (1600) ;
hence the two digits on the right are not needed to find the
tens' digit (4), and are set aside until the unit's digit is to be
found.
Likewise, the square of any number of hundreds (400), ends
in four ciphers (160,000), and all of these may be set aside until
the hundreds' digit is found and they are needed for the tens'
and units' digits.
ELEMENTARY DEFINITIONS AND OPERATIONS 31
After pointing off the number into periods, the method is
similar to the one used for polynomials. The procedure is
based on the formula d^{2ab ^b^ = (a\ by.
When dividing by the .trial divisor, exclude, for brevity, the
righthand digit in the remainder.
The reason for this is seen in the example which follows :
the trial divisor 4 is equal to 40 of the next lower units. Now
14 divided by 4 gives the same digit as 148 s 40.
Point off one decimal place in the root for each period in
the decimal.
Find the square root of 648.643241.
648.643241 1 23.421
4
148 1st remainder.
129
1964 2d remainder.
1866
9832 3d remainder.
9364
46841 4th remainder.
46841
1st trial divisor, 2 (2) = 4
1st complete divisor, 43
2d trial divisor, 2(23) = 46
2d complete divisor, 464
3d trial divisor, 2(234) = 468
3d complete divisor, 4682
4th trial divisor, 2(2342)= 4684
4th complete divisor, 46841
6th remainder.
Point off three decimal places in the root.
Thos, the square root is 23.421.
To find the square root of a fraction, which is not a perfect
square : Reduce the fraction to a decimal and find the square
root of the decimal.
EXERCISES
44. Find the square root of :
1. 6889. 6. 567,009.
2. 2025. 7. 2777.29.
3. 161,604. 8. 88.454025.
4. 337,561. 9. .00698896.
6. 7.3984. 10. .059049.
32 ELEMENTARY ALGEBRA
Find the square root, to 3 decimal places, of :
11. 2. 13. 5. 15. 3.
12. 3. 14. 1.005. 16. f
FACTORING
45. Factoring is the process of finding two or more expres
sions whose product is a given expression.
An integral expression is one that contains no fractions.
A rational integral algebraic expression is one containing no
fractions, no negative exponents, and no indicated roots.
2 + 3 aaj + aj'* is a rational integral expression ; J x* + J x + 1, a;* + 6,
y/a + x + y, X* + 10, are not rational integral expressions.
A prime factor of an integer is an integral factor which is
exactly divisible only by itself or one.
Thus, and 13 are prime factors of 65.
A prime factor of a rational integral expression is a rational
integral factor which cannot itself be resolved into rational
integral factors other than itself ojid one.
Thus, a — 6 and a^ + aft + 6* are prime factors of a^ — 6«.
In the following exercises we shall confine ourselves to find
ing 2>rime factors of rational integral expressions. In § 83
irrational factors will be required.
I. Type form : aft + ac — ad.
ab + a4i —ad=: a(b 4 c — d).
46. Factor.:
1. y^2y' + Sy*y. 3. a(x + y)b(x + y).
2. 80x»y*160 ajy 240 o^, 4. (a?  y)c (« + y)c.
6. m{c — d)— n{c — 6i)\p{c — (Z).
6. a{m — n) + b{m — n).
7. m(a — &) — w(6 — a).
ELEMENTARY DEFINITIONS AND OPERATIONS 33
8. a(xSy)+b(Sy''X),
® (a? — y)— 2a(y — aj)H36(y — a).
10. c(a — & — c)+d(6+ c — a).
11. Type form: ax \ bx \ ay + by.
ax { bx +ay + by == a{x ^ y)'\' b(x '\' y)=^ix+y)(a + b).
47. Factor:
1. Sx + Sy{ax + ay. 6. 2ac — 36c — 2ad436d.
2. 005 — ay — &aj + 6y. 6. a"c" — &"c" f a"d" — 6*d*.
3. 2aaj — 2 6a5 + & — a. 7. a'— a6Hac— ac+6c— c*.
4. 4a— 46 + ac — &c. 8. aVh6V — aV — ^V*
III. Type form :
tf ± 2a6H 6* and fl^hft* + c* + 2a6  2ac 26c
a*±2a6h6«=(a±6)*.
a* h 6* + c* 4 2 a6  2 oo  2 6c = (a + 6  c )*.
48. Factor:
1. 9r*24r« + 16a». 3. a* + a + i.
2. a*»14a»h49. 4. (ajy)*2(aj3^)+l.
6. 4(c  3)2  12(c2  9) + 9(c + Sy.
6. sfi\'f/^'{z* — 2xy^23DZ + 2yz.
7. a* + 6*4c'2a6 + 2ac — 26a
8. l + a* + 6* + 2a262a6.
9. 4a* + 6* + c^4a6 + 4ac26c.
10. m*4n»h9 — 2mw 6m + 6w.
IV. Type form: tf6*.
49. a«6* = (a + 6)(a6).
a*2a6 + 6«c*=(a6)«c* = (a6hc)(a6c).
a* — c* — 2cd— cP = a*— (c + d)* = (ahc + d)(a — c — d).
a«2a6 + 6*c» + 2cd(f» = (a6)«(cd)«
=(a — 6 + c — d)(a — 6 — c h d).
34 ELEMENTARY ALGEBRA
50. Factor:
1, c8~64. 6. 100aWa*hl2a6366».
2, 121c«144d*. 6. 36y» + 84y + 4992!«.
3. aj* — m* — 2mn — w*. 7. m« + 10m» + 25 — 81n*.
4. 49a«64 + 166&«. 8. (a  3)«  (6  2)*.
9. 46» — a?* + 4a^ha*44a& — 4y«.
10. 26(f*l9c*d«10ad + a2 + 6cd.
V. Type form : jfi + bx + c,
51. Factor:
1. aj*h5a; + 6. 6. r» — 4 r« — 77 ««.
2. a;25a? + 6. 7. b^c^ ^ 7 bey { 10 yK
3. yj 7^46. 8. c^d* H 7 cc««  120 (f*.
4. 2/2 + 72/98. 9. <«23«*60.
6. 2!2 + 102 + 9. 10. a*2/® ■ 16 ic'y^ + 39 2^
VI. Type form : or^ + 6x + c.
52. ao?* + &« h c = (pa? + g)(ra? 4 «), wherein pr =: a, ps { qr
= 6, g« = c.
By this method p, g, r, « must be found by trial, so that the
three conditions just named will be satisfied.
Factor 6a^ — 5x—6.
Here a = 6f 6 = — 6, c = ^6.
If possible, we must find numbers p, q, r, 8, such that pr = 6^ gs = — 6,
p« + gr = — 5.
pr = 6 suggests the values jp = 3, r = 2, orjp = 1, r = 6, etc.
g« = — 6 suggests the values g = 3, « = — 2, or g = 2, s = — 3, etc.
Try the different sets of values of p and r, whose product is 6, with
each pair of values of g and s, whose product is — 6, until a combination
is found which satisfies ps { qr= —6.
ELEMENTARY DEFINITIONS AND OPERATIONS 35
Try the sets of values p = 3, r = 2, g = 2, a = — 8.
Examine (3x 42)(2x8)
I I
9a;
We see that the sum of the two cross products + 4 a; and ~ 9 x is — 6 a;,
which is the middle term of 6 a^ — 5 x — 6. Hence
6a^6a?6=(3x + 2)J2x — 3).
53. Factor:
1. 6aj* + 5a? — 6. 6. 15 a? + 14 oa?  8 a*.
2. 5aj«414a?3. 7. 17 a«aj*  36 oa; + 4.
3. 5aj214a?3. 8. 20r» + 19r« + 35*.
4. 8y*14y39. 9. 9 aj»  36 a^  13 y*.
6. 18m* + 37m + 19. 10. 28m248m + 17.
VII. Type form: a* + o^ft* + &*•
a* + a»6« + 6* = a* h 2a*b^ + 6'  a^b^ = (a« + 6')*  a«&»
= (a« + a6 f 6*)(a«  aft + b^)
54. Factor:
1. a;*H«* + l. 6. a?* — 14 aj*y* + y*.
2. 9m* + 8mV44w*. 7. m* — 38mVHn*.
3. 16a* + 20a«6« + 96*. 8. 4 a*  13 a'6« + 9 6*.
4. r8 + r*5* + «*. 9. 25c* + 26c«cP + 9d*.
6. a*  5 a«6* 4 4 6*. 10. 49a* + 110a*62 + 81 6*.
55. In factoring, first take out monomial factors. Then
inspect the resulting polynomial, ascertaining to which type
form it belongs, and factor it accordingly. In the final form,
all factors should be prime.
36 ELEMENTARY ALGEBRA
ORAL BXBROI8BS
56. Factor:
1. rn^a}. 9. 209j5i«*. 17. a^ + yhi.
2. ^a^— y^. 10. m* — n*. 18. aj* — ^a?.
3. a« + 6a!> + 96*. 11. p^j:^. 19. a5»4ajy».
4. 9aj» 6ajy + y* 12. aj» + 2aj35. 20. 4a»a6*.
6. 4aj*12a?5+9«*. 13. aj«2a?35. 21. a^69&».
6. oaj*— 16ay«. 14. m*2m — 3. 22. 5a' + a*6.
7. ca^ + cy\ _ 15. p^Qp^ 5. 23. a*  2 a*6» + &*.
8. y«5y + 6. 16. 4c*±4cd + cP. 24. 9a*±12a6+4&*.
WBITTBN BXBBOI8BS
67. Factor:
1. 4a* — 20a6. 6. a*hc — ah*c. 9. 16 + 8 c + c«.
2. 6»d». 6. 25r*36«*. 10. l~a«.
3. m* — 9n*. 1. f ^^y — z. 11. a^^ — 6*.
4. 4a*.8a«6. 8. mV— 2mn*4l. 12. a*  5 a6 + 6 6».
13. cflj f cy — cte — dy. . 24. m* — 2 mn + n* + m — n.
14. 6*.66« + 5. 26. v^'{7*'\r^ + r.
16. 64&». 26. aj* + 2aj« + l.
16. 7 a? — 12 — aj*. 27. c*cP — c — d.
17. a*®— a*. 28. 6* — 26c + c» + 5& — 5c.
18. 3aj» + 3aj«27aj27. 29. 60* + 13a6 + 66*.
19. a?*" + 2af15." 30. a?"  ajyi*.
20. aj»6aj*412aj— 8. 31. r* 4r» + 6r» — 4r + 1.
21. a^^a^\, 32. (a? — y)*  9(aj + y)*.
22. r»«*2««<». 33. a* + 12 a6 + 36 6*.
23. 264A:* + 4^*^^«. 34. (a + 6)*  7(a + 6) + 12.
36. (a?  13 aj h 42)(aj»+ 3 a?  10).
36. (6c)*+3612(6c).
ELEMENTARY DEFINITIONS AND OPERATIONS 37
37. (a + («)*+ (6 + c)*  2(o + d)(6 + c).
38. 25 a*  41 a*6* + 16 6*.
39. a*2a6 + 6«2ac + 26c+<^.
40. c" — (?•.
41. a?49a? + 360.
42. i^47r»+l.
43. ab—abc^.
44. m* + 64n*.
46. (m — n)* — 1 + a(m — n f 1).
46. 3a*10a686«h3adH2W9ac6&c.
47. a?*3iB* + 4.
48. 4a* + 816*.
49. 4a*~37a26«H96*.
60. 64y*+128.v»2!2 + 81»*.
«
FRACTIONS
58. KfrcLCtion is the indicated quotient obtained by divid
ing one number by another.
The fundamental principle of operations with fractions is —
Both numercUor and denomincUor of a fraction may be multiplied
or divided by the saws number, vnthout changing the value of the
fraction.
Thus, I = li, 2^ = ?, 8(^ + y ) =2_.
8 16 2a a 4(x+y)(xy) xy
38 ELEMENTARY ALGEBRA
BXEROISES
59. Reduce to lowest terms :
1. — — nn' O.
6a»&c* 3aj* + 3a;18
2 ?i^. 7 (am«ift)v
2a26 g 4(mny
* a«6« ' * n*(m2)«'
 5 ggg — 10 gy g a* — 2 a — 1
aj* — 5ajy46y«' " a» — 20*41
g 8(a^6*) j^ aa^c«92a56c+&V
24a? + 246' ' 62.a«926c6o+c*'
60. From the laws of signs in multiplication and division it is
evident that
a — a a . — a a a
6 6' 6 b ' b b
From these relations we see that
1. The signs in both numerator and denominator may be
changed without changing the value of the fraction. .
2. The sign of the numerator and of the fraction may be
changed without changing the value of the fraction.
3. The sign of the denominator and of the fraction may be
changed without changing the value of the fraction.
61. The following principles may be used effectively in
operations with fractions.
1. The signs of an even number of factors may be changed
* without changing the sign of the product. Explain.
Thus, a • 6 • c = ( — a) • (— 6) • c = abc.
2. The signs of an odd number of factors may be changed,
provided the sign of the product is changed. Explain.
Thus, G'b c= — (— a) 'b ' c = abc ; a'b 'C=^ (—«)(— b)(—c).
ELEMENTARY DEFINITIONS AND OPERATIONS 39
MULTIPLICATION AND DIVISION OF FRACTIONS
62. In multiplication and division of fractions in algebra
we have the same rules of operation as in arithmetic;
namely,
1. To find the product of two fractionsy mtdtiply the numerators
together for a new numerator , and the denaHrainaJtora together for a
new denominator.
2. To find the quotient y invert the divisor and then proceed as
in the multiplication of a fraction by a fraction.
It is easy to establish the truth of these rules by means of
the equation.
For example, the second rale may be proved as follows :
Let
a c
then
a = &£, c = dy,
and
ad = hdx^ he = hdy.
Dividing equals by equals,
ad bdx X
be hdy y
But
x_a . c
y h' d'
Hence
a , c __ad
h d he
This proves the rule for the division of one fraction by another.
Proceeding in a similar manner, prove the rule for the
multiplication of fractions.
Multiply m±^ by ^^'^y\
hx {hy ax\ ay
40 ELEMENTARY ALGEBRA
BXEROISBS
63. Perform the indicated operations :
^ a b d ^ r'S r* — «*
1, ■ • — I — • 6. • •
c c T^ + ra r* — rs
^ 2a Sbx 2h ^ x^3x{2 x'^Sx
X a 3aa a^ — 6x ijfi\x'42
• 262'^6— l' ' a + b'^(a + by
. 16m*n 8 m' ^ a^ + ah a\b
45 fl^ ^ ^ <^* — y* {'^ — yy
«
jj a^~4a; + 3 gg12a?435 5ar10
aj27aj + 10' aj2H2a?3 '«2__3a.*
. a^ + oft a'c + 2 a^c + aft^c
a2 4 62 a^c — b^c
13.
m
*— n* j6 (m' — m^n + mn^ — n*)
14.
3 m2 43 n2 m^ — 2 mn + n«
18ag9a , 2a2f5a3
r2l "^r^llr + io'
gg — c2 t 2 a& 4 ft^ ?f^lAzi£.
a2— c2 — 2 6c — 62'a + 5_c*
2a2 4a10 3a2 44a4
17.
a2_4 23a
ELEMENTARY DEFINITIONS AND OPERATIONS 41
05*— J/* x^ — xy a?^ — 2 a?y + y^
18.
19.
20.
(xhyy x^y a^\2xy + y^
9 gg  16 feg , 6 gg  8 a5
3g2llg620&2 • g25g6 *
&2136 + 42 a2 + 6g27
a53 67.g + 21 * g281
COMPLEX FRACTIONS
64. A complex fraction is one which contains one or more
fractions in its numerator, or denominator, or both.
Complex fractions may be simplified in two ways :
1. Simplify the numsrator and denominator, then divide the
numerator by the denominator,
2. Multiply both numeraJbor and denominator by the I. c. d. of
their fractional terms.
The pupil is probably familiar with the first method from
his study of arithmetic. In simplifying complex fractions
arising in algebra, the second method is usually the easier.
It should be thoroughly m^^stered.
Simplify, iZljk.
i ~i r
The l.c.m. of 2, 8, 4, 9 is 86.
Multiplying the nomerator of the complex fraction by 86 gives 18 — 12.
Multiplying the denominator of the complex fraction by 36 gives 9 — 4.
Hence, Izil = l^Jl? = ? .
JJ 94 6
BIXBBCISES
65. Perform the indicated operations and simplify :
1.
L o + ft *J a»6»
42 ELEMENTARY ALGEBRA
3 ^ 'f ^ M
■ 1  a* \1  a 1 4 ay
4. (2aj*6a? + 3)!/'iiy
5. fl + i + lY?+l?+^y?iy:£.
7.
\n sj \m r)
\» y ^J \y ^
10. (m^
11.
i+i
12.
a
b
13.
I a
" + 6
11
a &
m + 1
V(m + 1)« m + i;
14.
2 + 2
a
r
15.
1+^
s
a + 2> q~ &
16.
a — b a+ 6
a — 6 , a4 ^
a 6 a+b 'a—b
ELEMENTARY DEFINITIONS AND OPERATIONS 43
a?3 —
17. ^•
a? + 4
a«
+ ab
19
a«
6*
X V.
a
a + b
a —
b
r
3
r
r
20.
S + r
r
+'
r
18. 3.
x + y{ —
y r + 3 r
HIGHEST COMMON FACTOR AND LEAST COMMON MULTIPLE
66. Only expressions which contain no fractions or radicals
will be considered here.
The highest common factor (h. c. f .) of two or more expres
sions is the product of the prime factors that occur in each
expression, every factor being taken the least number of times
it occurs in any one expression.
This definition indicates the process of finding the h. c. f .
The lowest common multiple (1. c. m.) of two or more expres
sions is the product of all the prime factors that occur in the
expressions, every factor being taken the greatest number of
times it occurs in any one expression.
This definition indicates the process of finding the 1. c. m. Observe
that the 1. c. m. is the least expression which is exactly divisible by each
of the given expressions, while the h.c.f. is the expression of highest
degree by which each of the given expressions is exactly divisible.
EXERCISES
67. 1. Find the h. c . f . and 1. c. m. of
4aa: 4a3^, Sa*** Say, 12a;» 122^.
Factoring, 4 ax — 4 ay = ^a(x — y) .
8 a%e2  8 a^j^ = 2^a^(x \y){xy).
12 ac»  12 y8 = 28 . ^x  y){V^ + «y + V^)
The h. c. f. = 22(a;  y) = 4(x  y).
The 1. cm. = 2« . 3 aHx  y)(x + y)(x2 + xy + y2)
= 24a2(x + y)(«»y»).
44 ELEMENTARY ALGEBRA
Find the h. c. f. and 1. c. m. of :
2. 9 a6c, 18 a«6c», 27 a»6*c.
3. 16 aj»y V,  30 a^i^z, 45 «V»'
4. (a  6)*, a*  6«.
5. 3a«+6a6 + 36«,9a«96«.
6. 2(a  6)»(a 4 6), 3(a f 6)»(a  6).
7. a^5»+6,aj* + 2aj8.
8. 3a»3a6«, 3a(a+&)«.
9. 3aaj»15aaj418a, 6aV + 24a*a;126a«.
10. sfi — y\ X — y, ax ^ ay ^ ex { cy.
11. a(a  1)«, a«(a*  1), 2 a(a« + 2 a  3).
12. 5(aj»  y»), 10(aj + 3^)«, 15(x  y)«(a; + y).
13. A«6A14, ^'10^ + 21, A«49.
14. 16(ab 4 b), 8 a(a«  6«), 24 ab(a} + 2ab + 6*).
16. m\m — 1)*, m(m' — 1), m\fn}n — w).
16. (y + 4)(y216),y«y20,ajy + 4ajay4a.
ADDITION AND SUBTRACTION OF FRACTIONS
68. The process is the same as in arithmetic. If the frac
tions do not have the same denominator, they must be reduced
to fractions which do have the same denominator. The lowest
common denominator (1. c. d.) is obtained by finding the lowest
common muLtiplk of the given denominators.
1. Perform the indicated addition and subtraction :
1 — X . a{x 5a — X
' "^ — •
2 ax 3 aa^ 6 a*x
ELEMENTARY DEFINITIONS AND OPERATIONS 45
The 1. c. d. is 6 a^^. The redaction of the fractiomi to the 1. c. d. is
effected as follows :
1 — « Sax 3 ax — 3 ax^
6a^^i2ax =3 ox,
2 ax Sax 6 a^^
6a%c«^3ax»=p2a, a+ . 2a^ 2a^ + 2ax ,
Sax^ 2 a QaH^
6 a^x^ i 6 a^x = «,
5a — X X _ 6 ox — x^
6a*e * x"~ 6a2x2
We obtain ^~^ 4. <^^ + ^ _ 5a — x _ 3 ax— 3 ax^H2 a'^H2 ax— 5 ax\oi^
2ax 3ax2 Ca^x 60^x2
^2a«3e^^ + x^ ^ns.
In practice, much of the work can be done mentally and need not be
written down.
2. Perform the indicated subtractions :
^ 1 J^
ab 2a 46*
The 1. c. d. is 4 ab(a — b) . Reduce the fractions to the 1. c. d. :
1 4ab 4ab
4 a6(a — 6) + (a — 6) = 4 ab,
a — b 4a6 4a6(a— fr)
4a6(a6)^2a =26(a6), 1 . 2^(<» ^) = 2a6  2 6^
^ ^ '^ ^' 2a 26((r6) 4a6(a6)
4a6(a6)f.46 =a(ab\ i • <»(«^) = <»" " «^ .
^ ^ "V« >'» 4^ a(a6) 4a6(a&)
Subtracting and simplifying, ^^^t.^^^7,^\ 4n».
4 ao(a — 0)
3. Perform the indicated additions :
a* — 1 aj(aj + 1) 1 — a?
The lowest common denominator is x(x^ — 1). Before reducing the
last fraction to one with the denominator x(x2  1), it is a convenience to
be able to write x  1 in place of 1 — x. As 1 — x differs from x — 1 only
46 ELEMENTARY ALGEBRA
in algebraic sign, we can do this, provided we change also the sign of the
fraction or else change the sign of the numerator. The former change is
simpler. Thus we have,
3^2 1
a;2  1 x(x + l) x1
The reduction to the lowest common denominator, x(x* — 1), is effected
as follows :
x(x2l)+a;(a;+l)=a; 1,
««  1 z »(aJ«  1)
2 x—l 2a; — 2
x{z+l) « — 1 a(a:« — 1)
x(a«l)i.(xl) =x(« + l), — ^ • x(x_±lX__3i^±x_^
^ ^ ^ ^ ^ ^' x^l xix+1) «(a;2_i)
Weobtaini + — ^ L = 3xH 2x 2 x« x
x2l^a;(a+l) x1 x(a^l)
4 X  2  x« .^.
= x(x«l)  ^'^^
4. Perform the indicated addition and subtraction :
1 1 + 1
(a  6)(c  a) (a  6)(c  6) (c  6)(a  c)
Here the factor (a — b) occurs twice, and both times with the same
order of the letters. The same is true of (c^b). But in the first de
nominator occurs the factor (c — a), in the last denominator occurs {a—c).
It is a convenience to have (c — a) in both denominators. Since (a — c)
differs from (c — a) only in sign, we write in the last denominator (c — a)
and at the same time change the sign of the fraction. We obtain
(aft)(c_a) (o&)(c6) (c6)(ca)
The 1. c. d. = (a — 6)(c — a)(c — 6). Reducing to a common denom
inator and adding,
e — b i
= 0, Ana,
ELEMENTARY DEFINITIONS AND OPERATIONS 47
BXEBCISBS
69. Perform the indicated operations :
, 2x.^x . 6,6
1 TT^—r* 4.  +
3 4 a?l aj + l
2. • 5.
y X X — y X — 2/
3. JL 8_. 6. ^^^ + 5i^.
aj + 1 x + 1 a + b a—^b
„ a — b , b — c , c — a
7. 1 f •
ab be ac
(y »)(»») («»)(» y) (»  2/)(2/  «)
g __3 7 4 20a
' l2a l + 2a l4a«*
10. ^+ 1 1
11.
a + 6 (a + 6)» a« — 6«
11 .V
2(a) y) 2(3! +y) as'y'
12. J + ^ll ?+ 1
m+w m + 3n m — n m — 3n
j„ g 6 , a' + ft' — oft
a*  62 a* + 62 ■ ft4 _ ^4 (a _^ ft)(a2 _. 52)
14. ?t^ + 1±1 + ^+^
{xyXxz) (y^z){yx) (z'^x)(zy)
io« — — f
(abXac) (6.c)(6.a) (acXcb)
48 ELEMENTARY ALGEBRA
,^ a*42af4 a*2a44,4.a»
iv. — — f — *
a + 2 2a a2
17. l^+.,l^ 1^+ 1
4a + 4 44a 1212a«33a
18. ^i—.^J— ^+ 1
a* — a; — 6 aj^+6a5 + 8 12 — as — a^
19. a;.yg ^ y^g,a; ^ ggy
(ajy)(aj2) (y  z){y  x) (z  x)(z  y)
20. —
PiPgXpr) q(q p) (q  r) pqr
21. :^f ^ ^^
2a a + 3 a» + a6
22. ^ +^ nlf.
a? + l 1 — a^ 05—1
23. ^ ^ ,+ ^
24.
2a:aj» aj*+3a;+2 1  a?*
2 _3 4
(a2)(a3) (3a)(al) (la)(2'a)
1.1.1
25. — H: — H
(a — 6)(6 — c) (c — a){b — a) (a— c)(6 — c)
QUADRATIC EQUATIONS
70. The equation ao* + 6a5 + c = is a complete gt^odrottc
equation. It is a quadratic equation because the highest power
of the unknown x is the second ; it is complete because it con
tains a term involving the unknown x to the first power and a
term c (called the ahsolvie term) which is free from x.
ELEMENTARY DEFINITIONS AND OPERATIONS 49
Quadratic equations of tlie forms
are called incomplete quadratic equationSj because either the term
involving x to the first power or the absolute term c is absent.
71. Incomplete quadratic equations are easily solved.
Take the form »* = c.
Extracting the square root of both sides of the equation, x = ± Vc.
The form ic* + 6a; = is solved by factoring, but may be solved
also by the method of " completing the square," to be explained
later.
Factoring, we obtain, x(x + 6) = 0,
Make the first factor equal to zero, x = 0.
Make the second factor equal to zero, x + & = 0.
x=— h,
72. The solution of quadratic equations leads to two values
of the unknown quantity. As both of these values are usually
of interest and importance in the solution of problems, it is
customary, in the extraction of square roots, to write down
both results, the principal value and also the second value. This
is indicated by the use of the symbol ± . Thus, in " «= ± Vc,"
+ Vc is the principal root, — Vc is the second root.
Since, in finding a? = ± Vc, the square root of both sides of
the equation has been extracted, it might be claimed that the
sign ± should be written on both sides, giving
± aj=:^ Vc.
But this result is the same as when we write a? = ± Vc.
For, the equation ± « = ± Vc means here
(1) +a;= + \/c. (8) ■\x= — y/c,
(2) x = >/c. (4) — a;=H>/c.
Of these four sets, the first two are the same, and the last two are the
same. Hence, %=i±y/c gives all the values of x.
E
50 ELEMENTARY ALGEBRA
73. A complete quadratic equation may be solved in three ways :
(1) Bj factoring,
(2) By completing the square,
(3) By substitution in a formula obtained by the method of
completing the square.
1. With our present knowledge of factoring, the first method is
applicable only when the roots of the quadratic equation are ra
tional. The method depends upon the principle that, if the prod
uct of two or more factors equals zero, one factor must equal zero.
Solve 3iB« = 2 + 5a;.
Transpose all the terms to the first side, Sar^ — 5a; — 2 = 0.
Factor, (3 a; + 1) (x  2) = 0.
Place the first factor equal to zero, 8 x + 1 = 0.
x = J.
Place the second factor equal to zero, a; ^ 2 = 0.
a; = 2.
The required roots are 2, — .
II. The second method depends on the type form of a tri
nomial which is a perfect square : a^ ±2ah\ ¥.
If the b^ is lacking, we may take the square root of the first
term, double it, divide the middle term by this, and square the
quotient.
Take, for illustration, x^ ± 6x, To complete the square, we take the
principal square root of x^, which is x ; double it, 2 x ; divide 5 x by 2 x,
{ ; square j, ^.
Hence x^±6x + ^\aa, perfect square.
This results in the following rule for the solution of a quad
ratic equation :
> 1. Transpose all terms containing a^ and x to the left side of the
equation; all others to the right side.
2. Dimde both sides by the coefficient of a?.
3. Add to both sides the square of half the coefficient of x.
4. Extras the square root of both sides and solve the resulting
equations.
ELEMENTARY DEFINITIONS AND OPERATIONS 51
Solve 3aj«5aj2 = 0.
Transpose, Sx^ — bx = 2.
Divide by 3, a;2 _ j j.  .
Add(J.})2, x24a; + jj =
Take the square root, x — f =; d: J.
Whence, « = 4 ± {.
a = 2,  }.
^n<., 2,  J.
III. The third method depends upon the formula derived
from the solution of the type form of the complete quadratic.
Solve aa^ + bx^ c=:0.
Transpose c, ox^ H 6a; = — c. ^' . , (^^
Divide by a, ^
'^^ *aiL^__c
(2 'a)
Addfl.^V
»
a  a
o 4a^ 4a2
/',
Take the square root, » + ;^ = ± ^^^  4 gc
2a 2a
 »:fc yy  4 oc •
2a
In using the formula, transpose all terms of the given equa
tion to the left side. Why ?
Solve 3 aj*  5aj = 2.
Transpose the 2, 3 a;8 . 5 » — 2 = 0.
Here a = 8, 6 = — 5, c = — 2.
.^^^6±v^6TM^i±7^2ori/
6 6*
BXBBCISES
74. Solve ; if a numerical equation has irrational roots, aiy
proximate their values, to three decimal places. Use the table
of square roots in § 197.
1. x^ + 4a; = 6. 3. «« — 8a: = — 11.
2. 3iB«4.aj14 = 0. 4. a?»2a;15 = 0.
52 ELEMENTARY ALGEBRA
5. aj2+6aj = 9. 10. «* — 2 ooj = 6»  a*.
6. aj* + aj4l = 0. n. m»aj«  2 ma? = n  m^.
7. x^ — ax=c. „u
12. aa? f ^^ = (6 + c)a;«.
8. 6a?» — caj=:d. 6 + c
9. 3aj»h»47 = 0. 13. (3 a:  5)» = 4 ic
14. 12«»413aj35 = 0.
16. i^+ 1 1
05 — 1 05 — 2 a?— 3
16. (a: + 2)(aj2)=7(ajf 2)6.
17. (icf 3)22(a;43)Hl = 0.
18. 5aj(aj3)2(aj26) = (a;H3)(a; + 4).
19. l«^^±2 ^?^5 ^, (<^'f)(y;^^) = 2y.
2a; + l a;3 c* 4 cP ^
(a? 4 m)* (a; — 7i)* 2y — a+2c
23. (ajl)(a:+2)(aj* + ajl)=0.
24. (a?*7a:+12)(ar^+3a;+2)=0.
PROBLEMS
Ascertain in each problem whether both roots of the quadratic equation
are applicable to the problem.
75. 1. Find two consecutive numbers whose product is 992.
2. Find the length and breadth of a rectangle whose area is
375 sq. in., and whose length exceeds its breadth by 10 in.
3. Express 71 as the sum of two numbers whose product is
448.
4. If the length of a square be increased by 2, and the width
be increased by 3, the area of the resulting rectangle is 40.
Determine the length of a side of the square.
ELEMENTARY DEFINITIONS AND OPERATIONS 53
5. The height of a triangle exceeds its base by 8 ; if the
area of the triangle is 1209, what is its base ?
6. The diagonal of a square is 2 ft. longer than the side.
Find the side.
7. A cylinder 12 ft. in height has a capacity 125 cu. ft.
Determine the diameter of its base.
8. When the edges of a cube are each increased by 6 in.,
the volume is increased by 936 cu. in. Find the dimensions of
the original cube.
9. The sum of the numerator and denominator of a fraction
is 77. If the numerator is increased by 111 and the denomi
nator is increased by 40, the fraction is doubled. Find the
fraction.
10. Find two numbers which differ by 2, the cubes of which
differ by 296.
11. The radius of one circle is twice the radius of another.
Find the radii of both, if the difference of their areas is 75.
12. The difference of the volumes of two spheres is 100 cu.
in. ; the difference of their radii is 5 in. Find their radii,
correct to two decimal places.
4irr8
The volume of a sphere is
3
13. A woman paid $ 64 for silk. If she had bought 4 yards
less for the same money, she would have paid $ 1 J more per
yard. How many yards did she buy ?
14. The longer leg of a right triangle, exceeds the shorter
leg by 3 ft. The area of the triangle is 135 sq. ft. Find the
length of each leg.
15. A bookdealer sells a number of algebras for $ 87. Had
he reduced the price of each book by 12 ^, he would have sold
16 more books for the same sum of money. How many books
would he have sold at the reduced rate ?
54 ELEMENTARY ALGEBRA
16. If a man's daily wage had been $ 1 less, it would have
taken him 15 days longer to earn $ 180. How many days did
he work to earn $ 180 ?
17. A picture 10" X 14" is placed in a frame of uniform
width. If the area of the frame is equal to half the area of
the picture, how wide is the frame ?
Draw a figure.
18. Find two consecutive even numbers whose product is
528.
19. Find two consecutive odd numbers whose product is
8099.
20. If the difference between the parallel sides of a trape
zoid is 5 ft., and, the altitude of the trapezoid is equal to the
longer of the parallel sides, find the lengths of the parallel sides
when the area is 2375 sq. ft.
21. A flower bed is 15' x 20'. How wide a walk, must sur
round the bed, to increase the total area by 770 sq. ft. ?
22. A tinner makes a square box 3 in. deep, with a capacity
of 1587 cu. in. From each comer of a square sheet of tin a
3inch square is cut and the four rectangular parts of the tin
are turned up. What are the dimensions of the square sheet
of tin?
Draw a figure.
23. If a square has its length reduced by 7 in. and its width
by 10 in., what are the linear dimensions of the resulting rec
tangle, if its area is 8370 sq. in. ?
24. An oil tank can be filled by one pipe in 2 hours less
time than by another pipe. If both pipes are open 1^ hours,
the tank will be filled. In what time can the tank be filled by
each pipe ?
25. A number of postage stamps can be arranged in a rec
tangle, each side containing 60 stamps. If the same number
ELEMENTARY DEFINITIONS AND OPERATIONS 55
of stamps be arranged in two rectangles so that each side of
one rectangle will contain 12 more stamps than each side of
the other, how many stamps does a side of each of the latter
rectangles contain ?
26. A boat's crew can row at the rate of 9 miles an hour.
What is the speed of the current in the river if it takes them
2 hours and 15 minutes to row 9 miles up stream and back ?
27. Divide $ 1248 among three persons, so that the second
shall have $ 3 more than the first, and the third shall have as
many times the share of the second as there are dimes in the
first person's share.
28. The population of a city increases from 20,000 to 20,808
in two years. What is the annual rate of increase per hundred ?
29. A sum of $ 2000 drawing interest that is compounded
annually, amounts to $ 2142.45 in two years. Find the rate of
interest.
CHAPTER II
MORS ADVANCED THEORY AND OPERATIONS
FUNDAMENTAL LAWS OF ALGEBRA
76. The operations of algebra obey certain fundamental
laws which we have not formulated thus far. Nevertheless
we have so accustomed ourselves to follow them, that we find
it difficult to see how a new algebra might be made, in which
a different set of laws would prevail. We shall now explain
the laws which underlie our algebra.
If several positive and negative numbers are added or sub
tracted, it matters not in what order the operations are per
formed ; the numbers may be commuted at pleasure.
For example, 4 + 7 — 6 = 6,
or 7 + 46 = 6,
or 6 + 4 + 7 = 6.
This is called the commutative law for addition.* Using let
ters, the law may be stated thus,
a + b =:b + a.
* In our algebra, addition and subtraction may be represented geometrically
by the addition and subtraction of distances along a straight line. Let a and
b represent distances measured off toward
the right, then a + 6 and 6 + a both repre — ^ ,. ^ ^ ^
sent the same distance OC in Fig. 3; the Q I i C
commutative law is obeyed. \y a
Suppose, now, that a and b are assigned Fia. 3.
meanings entirely different from those given
above; suppose a means a rotation about the line OA as an axis, through
90 degrees, and b means a rotation about OB as an axis, through 90 degrees.
56
MORE ADVANCED THEORY AND OPERATIONS 57
Again, 5 + — 2 = 12,
or (5 + 9) 2 = 12,
or 6f(92)=12.
That is, the final result is the same, whether the 9 be asso
ciated with the 5 or with the — 2.
This is called the associative law for addition. Using letters,
it may be expressed thus,
a + b + c =(a \ b) + c
= a+(6 4c).
If an expression contains two or more factors, it matters
not in what order the multiplications are performed.
For example, 2.7 = 72.
This is called the commutative law for miUtiplication,* Using
letters, the law may be expressed thus :
a ' b = b • a.
Let a rectangle be the figure rotated (Fig. 4). Then a\b (i.e. the rotation
about OAy followed by a rotation about OB) brings the rectangle in a position
where " Alg/' is horizontal, as in Fig. 5. On the other hand, b\a (i.e. a
rotation about OB, foUowed by a rotation about OA) brings the rectangle in
Fig. 6.
a position in which ** Alg." is vertical (Fig. 6) . Since the final positions of the
rectangle are different, it follows that in this case, a+b^b+a. That is, the
commutative law is not obeyed. (The symbol ^ means " is not equal to.")
Whether the commutative law for addition is obeyed or not depends there
fore upon the definitions given to a and 6, and to the processes of addition
and subtraction.
* There is an advanced algebra, called quaternions, in which (;' =^ji; that
is, the commutative law for multiplication does not generally hold true.
Quaternions are used in the study of mathematical physics.
68 ELEMENTARY ALGEBRA
Again, it matters not how the factors are associated or grouped, for
6 > 6 . 3 = 90.
6. (6. 8)= 90,
(6 . 6) . 8 = 90.
This is called the associative law for mtUtiplication. Using
letters, the law may be expressed thus :
a 'b • c=r:a» {b • c)={a • 6) • c.
Again, a factor placed before or after a parenthesis contain
ing two or more terms may be distributed among the various
terms without any change in the final result.
That is, 6(9  4 + 6) = 6 X 9  6 X 4 + 5 X 6 = 66.
This is called the distributive law for multiplication. Using
letters, the law may be expressed thus :
a(6 + c)==a6 + ac.
raSTORICAL NOTE
77. It is a carious fact that, in the development of algebra, the funda
mental laws were the last things to be explained. The beginnings of
algebra can be traced back to ab6ut 2000 years before Christ, but not
until the nineteenth century were the fundamental laws of algebra formu
lated. In the earlier treatment the laws were tacitly assumed to be
true. For instance, it was assumed that ab = hat a{b \ c) = ab + ac,
(ab)c=:a(bc)f without special attention being called to this matter nor
special names being given to the relations assumed. The need of an
explicit statement of the fundamental laws came to be recognized when
it was perceived that, besides the algebra which we are studying, in which,
for example, ab is always equal to ba, there could be established other
algebras in which ab is not always equal to ba. Among those who helped
to perfect the science of algebra along these lines were the Englishmen,
George Peacock, D. F. Gregory, Augustus £>e Morgan, and Sir William
Rowan Hamilton ; the Frenchmen, F. J. Servois, and A. L. Cauchy ; the
Germans, Martin Ohm, Hermann Grassmann, and Hermann Hankel;
and the American, Benjamin Peirce. The names "commutative law,"
" distributive law," were first used by Servois in 1814. Among the first
to use the name ** associative law" was Sir William Rowan Hamilton.
AnODSTUS DB MOBOUI (1806lti71)
Was professor of matheinaticB In London and wrote papers on Che tounda
tioDS of algebra and on loglo. To as English Journal, the Aihenieum, he con
tributed amaxing articles on circle squarers. These articles were aCterwarde
ooUected iu a book, entitled A Budget of Paradozet.
« 41 W
V
• »
• w ». ^ C » .
V «.
k >. V. i.
MORE ADVANCED THEORY AND OPERATIONS 59
REMAINDER THEOREM AND FACTOR THEOREM
78. Divide aj* — 5a^ + 7aj — 2bya? — a.
X — a
x2 +(a  6)x + (a*  6 a + 7)
(a  6)a;2 + 7 x  2
(a~6)a;'^(a26a)ag
(a2 _ 5 a + 7)a;  2
(q^  6a + 7)a;  q8 + Sgg  7 q
a86a2 + 7a2
Observe that the remainder, a' — 5 a^ + 7 a — 2, is the same
as the dividend, if in the dividend a is substituted for x. This
illustrates the
Remainder Theorem : If a rational integral expression in x is
divided by x— a, the remainder is the same as the original ex
pression with a substituted for x.
Now aj' — 5aj* + 7a? — 2 would be exactly divisible by a? — a,
if a»  5 a2 + 7 a  2 = 0.
Hence such a polynomial will be exactly divisible by a
binomial of the form a: — a, if , when a is substituted for x,
the polynomial vanishes, i,e. equals zero. This illustrates the
Factor Theorem : If a rational integral expression in x becomes
zero when a number a is substituted for x, then x — a is a factor
of the expression.
Thus, the expression a:P \2sfi —05—2, fora;=l, becomes 1 + 2—1—2=0,
hence x — 1 is a factor of the expression.
Again, for a; = — 2, this expression becomes 64 — 64 + 2 — 2 = 0, hence
X — (— 2)ora; + 2isa factor of it.
A third factor can be found by dividing x' + 2a^ — x — 2by the prod
uct of a; — 1 and a; + 2 ; that is, by x^ + x — 2. The quotient is
a:* + x' + a;^ + x+l. Hence we have,
a:6 + 2a*x2 = (x l)(x + 2) (xl^ \ ofi + x^ + x ^ 1).
In the example above we took x = 1 and a; = — 2. Notice
that 1 and — 2 are both integral factors of the absolute term
60 ELEMENTARY ALGEBRA
— 2 in the expression aj'42a^ — aj — 2. There is a theorem
bearing on this point which we shall use along with the Factor
Theorem. It relates to rational integral expressions,
a?" + oo;"! + 6aj— 2 H \' kx + k.
The theorem, which we give without proof, is as follows :
When a rational integral eacpression has 1 as the coefficient of
the highest power ofx, and the otJier coefficients a, 6, "•, h, k are
aU integral numbers (either positive or negative) , then, in search
i'f^g for rational va^vss of x that will make the given escpression
zero, only integral factors of the absolute term k need he tried.
Factor sc* + 6 x« + 42 a;  49.
The coefficient of a:* is 1 ; the other coefficients, 6, 42, — 49 are integers.
Hence both conditions are satisfied, and we need to try only integral
factors of 49 ; namely, ± 1, ± 7, ± 49.
When « = !, 1+6 + 42 — 49 = 0. Hence, by the Factor Theorem,
a; — 1 is a factor of the expression.
When* a: =— 1, 1 — 6 — 42 — 49 ^ 0; a; + lis noJ a factor.
Whenx = 7, 2401 + 2068 + 294  49 z^fc 0, a: — 7 is noJ a factor.
When X = — 7, 2401 — 2068 — 294 — 49 = 0, a; + 7 is a factor.
In the same way, a; — 49 and a; + 49 are found not to be factors.
Dividing the expression by (x — l)(x + 7) yields another factor, aj2+7.
Hence, a^ + 6x» + 42x 49= (x l)(x + 7)(x2 + 7).
BXBBOISES
79. 1. Divide ^a^ — 1 a^ {• ba^ — x — l\yj x — 6, then com
pare the remainder with the dividend. How does this illus
trate the Kemainder Theorem ?
2. Divide 5a!^ — 4aj8 + 6a? + 2bya? — a and illustrate the
Remainder Theorem.
3. Using the Factor Theorem, factor
aj4 _ 5 aj3 + 9 aj2  15 a? + 18.
4. Factor oj^ — 17 oj^ + 3 a? + 54.
* The symbol ^ means * * is not equal to." See § 76, footnote.
MORE ADVANCED THEORY AND OPERATIONS 61
6. What is the remainder when 2aj3 — 3a;2 + 5aj— 1 is
divided byaj — c? By x — b? By a? — 1?
6. What is the remainder when 5a^{'2x^{Sx + 6 is
divided by x fw ? By a? + 1 ?
7. Is a? + 2 an exact divisor of aj* — » — 6 ?
8. If a rational integral expression, with 1 as the coefficient
of the highest power of a?, and with all other coefficients
integers, is exactly divisible by a? — a, where a is an integer,
what relation is a to the last term ?
9. Find a factor ofa?*^ — aj*faj3 — aj^faj — 1.
10. Find a factor of aj3 — 9 aj2 + 17 a? — 6.
11. Find a factor of 2 aj*  a^  2 aj2  3 a?  10.
12. Find a factor of a^ — 13 a; f 12.
13. Factor a^ — a**.
If a is substituted for x, then a* — a^ = 0. Hence, by the Factor
Theorem, x — a is a factor of ar^ — a*. Division of a^ — a^ by x — o
yields a second factor. Hence
05*^ — a* = (x — a) (x* + x'a 4 «^a^ 4 ««* + «*).
The quotient may be obtained by the following special method of short
division :
x^4x = x*. x*^x•a = x*a. x^a ^ x • a = x^a^.
x^a^ t X • a = xa**. xa^ h x • a = (i*.
14. Factor a;*^ 4 a^
If —a is substituted for x, then (— a)* + a^ = — a* + a^ = 0. Hence
by the Factor Theorem, x^(— a) orx + aisa factor of x^ + a*. By
division, we find a second factor. Hence
x^ 4 a^ = (x + a) (x* — x^a 4 sc^a^ — xa* 4 a*) •
The special rule of division given in Ex. 13 applies to this case, except
that the signs in the quotient are alternately 4 and — .
62 ELEMENTARY ALGEBRA
16. Factor a^ + a*.
If ± a is substituted for x, we obtain ( i a)® + o* = o* + o* t^ 0.
Hence, neither a; + a nor a; — a is a factor of x* + o^. But afi + <jfi may
be considered as the sum of two cubes (x^)^ + (a^)' and can therefore be
factored, as j^ + 0^ is factored.
16. Explain the following results pertaining to rational and
real factors : , . , . i. t. u i. j
a* f 52 cannot be factored.
a* 4 &* cannot be factored.
a« + 6« = (a2 H 62) (a*  a^b^ 4 ^O
a' 4 6' cannot be factored.
Note. The sum of the same two even powers of two numbers may be
factored, if the exponent is the product of an odd and an even factor.
One factor is the sum of the numbers with exponents equal to the even
factor ; the other factor may be found by inspection or by long division.
Thus, in a^^ + 6^^, the exponent 12 = 8 • 4, where 3 is odd, 4 even.
One factor of a^^ + h^^ is a* + 6*. The other factor is a* — a*6* + &».
Hence, a^ + h^^= (a* + 6*)(a8  a*6* + b^).
If numerical cubic equations of the form ofi + dx^ + 6x + c = 0, a, 6, c
being positive or negative integers, have at least one root rcUional, then
the equation can be solved vnth the aid of the Factor Theorem.
0^ + 4 x^ — 2x — 5 becomes zero when — 1 is substituted for x. Hence
X 4 1 is a factor, the quotient x' 4 3 x — 6 is another factor.
Write the cubic equation in the form (x + 1) (x^ + 8 x— 5) = 0.^
Make the first factor equal to zero, x 4 1 = 0,
and X = — 1.
Make the second factor equal to zero, os^ + Sx — 6 = 0,
Q_i_ "v/29
and solve the quadratic equation, x = — .
Hence the roots of the cubic are 1, 3 + \/29 3\/29
'22
17. Solve (a?  l)(a;2 4 6 a; 4 8) = 0.
Solve the following cubic equations :
18. aj»4 5aj247a;43 = 0. 20. aj» 6aj2  8a? 4 7 = 0.
19. a;'45aj23aj22 = 0. 21. a;» 4 18 02^32 a? 4 55=0.
MORE ADVANCED THEORY AND OPERATIONS 63
\
80. Factor:
1. aj2 — 16y2.
2. 25a26l
3. 4a^ — 9y2.
4. 05^ + 3^.
6. 7? — j^.
6. aj'  1.
7. 142/*.
8. aa;2 — ayK
9. aj2 _ 100.
10. iB» — 1000.
11. m» 4 8.
12. m'8.
ORAL REVIEW
13. 27 a'.
14. 27 + a».
16. m^ + n*.
16. m*^ — w^
17. a* — 6*.
18. a* + &*.
19. 2^ — 1.
20. 2^y».
21. 05*40*.
22. x*—a?.
23. 10 aj'  10.
24. 10 01? + 10.
25. iB»43ar2 4.3aj4. 1.
26. 2/»3y243yl.
27. a* 4 2 a^* 4 3^.
28. 0^4 82/«.
29. 7? — 21v?.
30. a?* — 162^.
31. 10a^420ajy4102^^
32. «« — 1.
33. a:«4l.
34. a?— (2y + a;)'.
36. aj2 — (2y2)2.
36. 2a* — 2a.
WRITTEN REVIEW
81. Factor:
1.. 8 a« 4 hK
2. ci«di2,
3. ci2 + c2i2.
4. 27 a^ 1252^.
5. 27a^2^.
6. 8m«427n».
7. 27a^64 2/».
8. m*^ — n^^
9. a*  h\
10. c«464d«.
11. aj^— (2a?— y)*.
12. aj»+(y + «)».
13. a*+4a^2a;24a;4l.
14. aj«4a:*7aj2 + 28.
16. aj*3aj» + 7a?13a?46.
16. a?— a^ — 5 a^x 4 5 a\
17. m' 6m* 4 12m— 8.
18. ajio— yo^
19. a.** 4 35*2^* + ^.
Hint. X* + ojV 4 y*
= x* + 2icV4y*a:*y*.
20. a*4a' + l.
21. 4a?*43a;y 492^.
22. a* — 7 a262 4 64.
23. 49p*H68p2g*4369*.
24. 64aj* + 55a^.y2 4252^.
64 ELEMENTARY ALGEBRA
IRRATIONAL AND IMAGINARY FACTORS
82. Thus far all factoring has been confined to the discovery
of rationed factors. But it is sometimes advantageous to re
solve expressions into factors which involve irrational or even
imaginary terin?»>^ Expressions like a' — 2 6*, a* — 3 b*, a? + 6^,
or a* + 6* can be factored, if the restriction is removed that
the factors must be real and rational.
BXBROISBS
83. 1. Factor a» — 3.
This may be done by the type form a^— 62=:(a f 6)(a — 6), where
62 = 3 and h = VS. We obtain
a23=(a + V3)(aV8).
2. Factor aj2 f y\
This, too, may be factored by the type fonn a* — 6^ =(a + 6)(a — 6).
Write «2 f y2 in the form x^{y^). Since V y2 := yy/ZTi = fy,
where % = V— 1, we have,
x^ + y^^(x + %y){x'iy).
Kesolve into irrational or imaginary factors :
3. m«5. 7. a* + 62. 11. 9a« + 166*.
4. a* 12. 8. 7^bsK 12. a*36^
5. a2 4.4. 9. 2a26». 13. a2— 6.
6. m2h9n2. 10. aj^ — 8y*. 14. a* — 46*.
16. By factoring, find all four roots of the equation
a^l = 0.
16. By factoring, find all four roots of the equation
a^16 = 0.
MORE ADVANCED THEORY AND OPERATIONS 65
THE BINOMIAL THEOREM
84. By multiplication, we find that
(a ± by= a±b.
(a ± 6)»= a» ± 3a26 + 3a5* ± ¥.
(a ± 6/= a^ ± 4a»6 4 ea^fe^ ± 4a6» 4 6*.
(a ± 6)«= a» ± 5 a*6 + 10 a.»62 ± lo a^b" 4 5 a5* ± 6».
(a ± 6)«=:a» ± 6a»6 4 15 a*^^ ± 20a'&» 4 l^a^b* ± 6ab^ +b^,
etc., wherein a and 6 represent the first and second terms,
respectively, of any binomial.
A careful study of these products will show that they follow
certain laws, by which they may be written down without
recourse to laborious multiplications.
In the products we observe the following laws :
I. The first term isTyraiaed to the same power as that of the
binomial. In each succeeding term the easponent of a decreases
byl.
II. The factor b does not appear in the first term. The ex
ponent of b in the second term is 1 and increases by 1 in each
succeeding term,
III. The coefficient of the first term is 1. The coefficient of
any term after the first is found by multiplying the coefficient of
the preceding term by the exponent of a in that term^ and divid
ing by 1 more than the exponent ofb, '
IV. If the binomial is a i b, the signs in the product are.
all plus; if the binomial is a — 6, the signs are alternately 4
and — .
V. The number of terms is one more than the exponent of the
binomial,
VI. Each term is of the same degree as the binomial,
p
66 ELEMENTARY ALGEBRA
Using these laws, find without actual multiplication the
product of (a 4 by.
The first term is by I, a',
1x7
The second term has by III the coefficient or 7, by I the factor j
cfi^ by II the factor &. Hence the term is +1 <fib.
The third term haa by III the coefficient l^i^ or 21, by I the factor
a^, by II the factor h\ Hence the term is + 21 a^h^, '
The fourth term has by III the coefficient ^^ ^ ^ or 35, by I the fac
o
tor a^, by II the factor &*. Hence the term is + 36 a*&*, and so on. By
y the final result has 8 terms. We obtain,
(a + by aT + 7 (i«6 + 21 <fib^ + 86a*6» + 86a«6* + 21 a^lfi + 7 a6« + b\
All terms are of the seventh degree, as required by VI.
85. If n is a positive integer, then the products found above
may all be represented by one general formula, as follows :
(a + by = a 4 m»6 + 5^?^^ a^fe^ + n{n^l){n^2) ^,.3^,
This formula is called the Binomial Theorem. By actual
multiplication it was shown above to be true for all positive
integral valueg of n, up to 6. We found the expansion of
(a + by on the tacit assumption that the formula holds for
n = 7. Really we had no right to take for granted, without
proof, that the laws given above do hold true for n>6.
Frequently certain relations hold true up to a certain point,
but no further.
For example, the first three integers 1, 2, 3 are all prime numbers. A
careless reasoner might be tempted to juntp to the conclusion that all ;
integers are prime numbers, which is, of course, not true.
That the Binomial Theorem is true for any positive integral
value of n will, be proved later in § 193.
MORE ADVANCED THEORY AND OPERATIONS 67
Expand (2 a; — 3 yy.
Here a = 2 x, b = Sy. For conyenience in expanding, place 2 x and
8^ in parentheses, thus (2 x) and (3y). By Law IV of § 84, the terms
in the expansion are alternately +, — . We obtain,
(2a;3y)*=(2a;)*4(2x)8(8y)+6(2a;)2(3y)24(2aj)(3y)»+(3y)*.
Simplify each term in the expansion :
(2x  3y)*= 16 X*  96x8y + 216 xV  216xy» + 81 y*.
EZEBOISES
86. Expand:
1. (xhyY 4. (aSby. 7. (ox — yY^.
2. (a 6)8. 6. (c2d2)4^ 8. (3a2 + 2 6»)».
3. (2x + yf. . 6. (3a 62)6^ 9. (m — 2ny.
10. (2r33s2y.
11. Write the first three terms of (a—b)^,
12. Write the first five terms of (x + y)^.
13. Compare the coefficients of the first and last terms in
the given expansions in § 84 ; of the terms next to the first
and last.
14. What is the third term of {mx f ny)^ ?
16. Write the 9th term of (2 a^  3 by.
SYSTEMS OF LINEAR EQUATIONS. PETERMINANTS
87. Systems of n independent linear equations ir. n xm
knowns may be solved by eliminating tho same unknown from
all the equations. The resulting system contains in general
one less equation and one less unknown.
From this resulting system eliminate a second unknown, and
continue this process until one equation containing only one
unknown is obtained.
68 ELEMENTARY ALGEBRA
BXBBOISES
88. 1. Solve the system of equations
« H y + « = 9, (1)
xh2y^z = S, (2)
2aj33^f 42 = 7. (3)
SoluHon. Subtract (2) from (1),  y + 2 « = 1. (4)
Subtract (8) from 2 . (1), 6 y  2 = 11. (6)
Add (4) and (6), * 4 y = 12,
y = 3.
Substitute 8 for y, in (4), «; = 2.
Substitute 2 for z, and 3 for y, in (1), x = 4.
Check by substituting the answers in all three equations (1), (2), and (3).
2. ajyh« = 20, S. 5x — 6y{'Tz = 105,
2aj3^« = 30, 3aj6y + 82; = 103,
3a. — 43^4.22; = 10. 3y — 4« = 44.
4. « I y f 2; 4 to = 10,
« — yH2« — 3«; = — 3,
3aj — 2y + 2— w = 5,
3aj3yf22 — 4m?=— 3.
89. In previous parts, three methods of elimination have
been explained : (1) by addition or subtraction, (2) by sub
stitution, (3) by comparison.
A fourth method of solution is by means of a formula, called
a determinant. In deriving this formula we use the old
methods of solution.
Solve : ax + by = c (1)
dx{ey=f. (2)
Multiply (1) by d, (2) by a, adx + bdy = cd (3)
adx + aey = c^f .. (4)
Subtract (3) from (4), (ae — bd)y = c^f— cd
ae — bd
Similarly, a;=£lzi_St.
ae — bd
MORE ADVANCED THEORY AND OPERATIONS 69
The numerator af—cd may be written
a c
d f
The term af is the product of a and / which lie on the diagonal from
the upper left comer to the lower right corner.
The term cd is the product of c and d which lie on the diagonal from
the upper right comer to the lower left comer.
The second product is subtracted from the first.
3 4
Similarly,
5 6
= 3x64x6.
The numerator ce — bf may be written
The denominator ae — bd may be written
Expressions written in this form are called determinants.
c
b
f e
kTI
a b
jXI.
d
e
Hence,
a? =
c b
f e
a b
d e
(5)
y =
a
c
d
f
a
b
d
e
(6)
The determinants in (5) and (6) can be used conveniently
as formulas for solving two linear equations in two unknowns.
The determinants can be written down at once, according to
the following rules :
I. In the two denominators the determinant is the same. It is
fortned by writing the coefficients of x and y as they stand in the
equations (1) and (2).
II. The determinant in the numerator of (p) is formed from
the denominator by replacing , the coefficients of x, by the abso
lute terms .
/
70
ELEMENTARY ALGEBRA
III. The determinant in the numerator of (6) ia formed by
b c
replacing , the coefficients ofy, by the absolute terms,
Example. Solve
3a?f 4y = 34,
5 a? — 3 y = — 11.
The denominator is
3 4
5 8
The numerator for the value of x is
34 4
11 3
3 34
611
The numerator for the value of ^ is
Hence,
34 4
■ ^ 84.(8)4.(ll) _ 102f44 ^
3(«3)4.6 920
flC =
11 3
8 4
6 8
68
29
= 2,
y =
8
34
6
11
3
4
6
8
3(^ 11)  84 . 6
29
 33  170  203
29
 29
= 7.
EXERCISES
90. Find the values of the following determinants :
1.
2.
3.
9 6
8 7
.9 .2
.5 .1
4. b
. 3 •
4
2
10
6.
9
a b
6.
la 86
46
3
a
7.
a»
62
61
8. a
be
ae
6c
9.
a2
6'
Solve, by determinants :
10. 2ajh32/ =4,
5 oj ♦ 6 y = 7.
11. aia; + 6iy=Ci,
a^ 4 62^ = C2.
MORE ADVANCED THEORY AND OPERATIONS 71
12. 2iB — 5y= — 13, 14. x + my = ^m*,
3 oj — 4 y = — 9. oj 4 ny = — w*.
13. aa?f 6y = a — 6, 16. aa; + 62/ = 2a6,
oa? — 6y = a f 6. 6a?  ay = a* + 6*.
Solve, by any method :
16. 3a;f 4y = 7, 20. 3a?f4y45 = 0,
4aj4y=s3. 6a? + 7y48 = 0.
17. aj2y = 8, 21. aaj + 6y=l,
3 a; + y = 9. ca?  dy = 1.
18. 8 a: — y = 34, 22. cx + dy=sa,
x + SyssoS. mx iny^b.
19. 3a? = 4y, 23. 2(a?2) = 3(y + 3),
5aj=6y4. 3(a? 2)2 = 6(y +3).
24. (rl)(a + 2)=(r3)(«l)48,
4(2rl)16(«2)=20.
25. (6 — a)aj4(a + % = a'4ftS
6a?2ay = 6«2a*.
26. m — w 33 6 — a,
am — ac = 6/1 — 6c.
27. (m 4 n)x f (m h />).y = m  n,
(?/i + p)a: +(m + n)y = m 4i>.
91. The method of solution by determinants may be ex
tended to systems containing three or more equations.
For example : aa? + 6y + caj = d, (1)
ex hfy + gz = hy (2)
kx + ly + rnz = n, (3)
72
ELEMENTARY ALGEBRA
Using the addition and subtraction method, we obtain after some
prolonged effort,
__ €(fm 4 hlc + ngb — cfn — gld — mhb
afm + 6/c + kgh —cfk — fjria — meb
(4)
__ g/im ^^ enc \ kgd — cftA; — gna — med
cj/w* + elc 4 A;(76 — c/A; ^gla — W€6
_ fl^n + el d + ifc^fe — djk — /iZg ~ ne6
afm 4 eZc 4 A;(76 — cfk — gla — meb
(5)
(6)
These numerators and
denominators may be
written down more con
veniently in the form of
1^ determinants. The six
terms of the denomi
nator may be written
a
b c
e
f 9
k
I m
provided that the f ollow
ing rule of expansion is
observed (Fig. 7) :
(1) Take the positive of the products of the terms marked
by the 4 arrows.
The three products are 4 <0», 4 e^c, 4 kgb,
(2) Take the negative of the products of the terms marked
by the — arrows.
The three products, with their signs changed, are — c/fc, — gla^ — meb*
It follows then that
a h c
e f 9
k I m
= afm \ elc 4 kgb — cfk — gla — meb.
MORE ADVANCED THEORY AND OPERATIONS 73
This process of expansion must be fixed thoroughly in the
mind. The numerators of (4), (5), and (6) can be expressed by
the determinants in the same way.
Three independent linear equations in three unknown
quantities can be solved by determinants according to the
following rules :
I. In the three denominators of the vahies of x, y, and z, the
determinant is the same. It is formed by writing the coefficients
ofx, y, z, as they stand in equations (1), (2), (3).
II. The determinant representing the numerator of (4) is
formed from the denominator by replacing the column containing
a, e, k, the coefficients ofx, by the absolute terms d, h, n.
III. The determinant representing the numerator of (5) is
formed from the denominator by replacing the column containing
b, f I, the coefficients of y, by the absolute terms d, h, n.
«
IV. TVie determinant representing the numerator of (6) is
formed from the denominator by replacing the column containing
c, g, m, the coefficients ofz, by the absolute terms d, h, n.
Example, Solve :
The denominator is
1
2
3
2xy\z = b,
3x4y422! = 3.
1 1
1
4
1
2
= lx(l)x2+2(4)xH3xlxllx(l)x3lx(4)xl
2x2xl=4
The numerator for x is
6 1
51
3 4
1
1
2
=6x(l)x2+6x(4)xl+3xlxll>^(l)x3lx(4)x6
2x6xl=12.
16 1
2 5 1
3 3 2
The numerator for y is
=lx6x2 + 2x3xl43xlx6lx 5x31x3x12x2x6= 8.
74
ELEMENTARY ALGEBRA
The numerator otz\&
1 1 6
216
84 8
= lx(l)x3+2x(4)x6+3x6xl6x(l)x86x(4)xl
8x2xl = 4.
Hence x =( 12) + ( 4)= 8.
y=(8)+(4)=2.
*=(4) + (4)=l.
BXBBOI8B8
92. Solve:
1 a + y + » = 9,
«  y + 2 = 3,
05 4 y — 2 = 1.
2. 2aj3y + 2;= — 1,
3a; + 2y — 2 = 4,
^x — y\ 5 2 = 17.
3. a + 26c = l,
8a44643c = 29,
3a564c = ll.
4. wi + n = 3,
j!> 4 m = 12,
n 4"i> = 6.
6. r + 2 « = 10,
3r4« = 12,
6 « + 6 « = 22.
6. aj + y + « = 2,
3aj — 4yf62=41,
9aj13y=57.
7. a.y» = ^,
2aj — 3y442 = l,
4aj46y + 2 = 4^.
B. aj + y42 = a,
aj  y + 25 = 6,
ic + y — « = c.
9. ax + by = m,
c^ I cf2 = n,
eaj 4/2 =1>.
10. r4» + «4w=sO,
2r + 3a44iJ45w= — 6,
3r4a+5^6u = 2,
47.4.55 — 2« — tt = 2.
PROBLEMS
93. 1. A man has 20 coins, all dollars and quarters, amount
ing to % 13.25. How many coins of each denomination has he ?
2. Weights of 13 and 17 lb. are fastened to the ends of a
9foot pole. Where must the pole be supported in order that
the weights will balance ?
MORE ADVANCED THEORY AND OPERATIONS 75
3. Two men carry a weight of 300 lb. by means of a pole
from which the weight is suspended. One man holds the pole
at a distance of 3.7 feet from the weight, the other at a dis
tance of 4.3 feet. How much of the weight does each man
carry ?
4. Two unknown weights balance when suspended 11 and
12 inches respectively from the fulcrum. If the weights are
interchanged, 23 ounces must be added to the lesser weight
to restore the balance. Find the two weights.
6. A farm laborer received $ 2.75 and board for every day
he worked, and paid $.75 for board every day he was idle.
After GO^days he received $154.50. How many days was
he idle?
6. A •man has $12,000 at interest in two investments.
From one he receives 5 % interest, from the other 6 % . His
income from these investments is $ 645 per year. How is the
money divided ?
7. Find the capital and the receipts of a railway company,
if its receipts are distributed as follows :
(1) Payment of a guaranteed dividend of 4 % on ^ of the
capital stock.
(2) Payment of a dividend of 3 % on the remaining capital
stock, this dividend amounting to $ 162,000.
(3) To working expenses and a reserve fund, which absorb
55 % of the entire receipts.
8. The area of a trapezoid is the product of half the sum
of the two bases and the altitude. Find the two bases, if the
altitude is 12 in., the difference between the bases is 3 in., and
the area of the trapezoid is 150 sq. in.
9. How many gallons of each of two liquids, one 25%
alcohol, and the other 75 % alcohol, must be taken for a 17
gallon mixture that is 45 % alcohol ?
76 ELEMENTARY ALGEBRA
10. How many ounces of silver, 75 % pure and 85 % pure,
must be mixed to give 20 ounces of silver, 83 % pure ?
11. Tin and lead occur in the following ratios : In plumber's
solder 1:2; in soft solder 2 : 1 ; in common pewter 4 : 1. How
many pounds of common pewter and plumber's solder must be
taken to make 10 lb. of soft solder ?
DIVISION BY ZERO IMPOSSIBLE
94. To avoid perplexing errors in algebraic transformations
it should be made very clear that it is not permissible to divide
by zero.
Let a and h be two fixed numbers. Both a and b^ie finite
numbers, because in elementary algebra there are no fixed num
bers that are infinite. ,
a CL * *
Letting  = a;, we define the quotient  as the value of x in
b b
the equation bx = a,
I. As long as a and b are fixed numbers different from zero,
it is always possible to find a value for x.
II. If a is not zero, but b is zero, we are asked to find a
number x which, multiplied by zero, gives a as a product.
But every fixed number, when multiplied by zero, gives the
product zero.
Hence no number x exists which satisfies the equation •x=a.
If, therefore, we perform operations which amount to divi
sion by zero, we must not be surprised if we obtain absurd
results.
Point out the error in the following :
Let x = c.
Multiply both sides oix = chy x^ x^ = ex.
Subtract c^ from both sides, x^ — c^ = cx— c^.
Factor, (x + c) (ic — c) = c(x — c). .
Divide by (x — c), x\ c = c.
Since x = c, 2 c= c.
Divide both sides by c, 2 = 1.
MORE ADVANCED THEORY AND OPERATIONS 77
EQUATIONS WITH FRACTIONS
95. When an equation involves fractions we usually begin
its solution by clearing it of fractions. Tbis operation rests
on the principle that when equals are multiplied by equals, the
results are equal.
1. Solve . ^ n = ^?^.
m + n m — n
Observe that this equation is meaningless and absurd when m ^n ,
for in that case it involves a division by zero. For the same reason we
cannot have m:=^ n^ for then m + n = 0. Let us assume that neither
denominator is zero.
Then the 1. c. d. = (m + n) {m — n). Multiply both sides by
(m + n)(riii — n).
We obtain, (m — n)x — n(m^ — n^) z= m (m + n).
Transpose, (m — n)x = n(m^ ^ n^)\ m(m \'n).
Divide both sides by (m  n) , x = n(m2~ n2)+ m (m f n) ^
m — n
2. Solve _^ + .^ZLi=:^. (1)
Observe that if jc = 1, or x^ = 1, this equation represents an absurdity,
since it involves division by zero.
On the supposition that x^ =^ 1, the 1. c. d. = 4{x^ — 1).
Multiply both sides by 4(x2l), 4x(x + l) + 4(x3) = 6(x2l). (2)
Simplify, 4x« + 4x + 4x— 12=5x25.
Transpose, x* — 8 x + 7 = 0.
Factor, (x — 7) (x  1) = 0.
Hence, x = 7, x = 1.
Substitute in the original equation (1) ; we find x = 7 satisfies it, but
X = 1 does not satisfy it, since it gives rise to division by zero. We
must reject x = 1 ; x = 7 is the only correct solution of (1).
It is of interest to notice that, while x=l does not satisfy (1),
it does satisfy (2). In other words, equation (2) is satisfied by
two values of a, while equation (1) is satisfied by only one
value of a?.
It is clear that the new root was introduced in the act of
multiplying both sides of (1) by 4:(x^ — 1). Such new roota
78 ELEMENTARY ALGEBRA
which do not satisfy the original equation, but do satisfy an
equation that is derived from it, are called extraiieoua roots.
When we multiply both sides of an equation by an expres
sion involving x, we obtain a new equation some of whose roots
may not satisfy the original equation and are not correct solu
tions of it ; they are extraneous roots. This may happen even
though all operations are performed free of error.
Hence, to make sure that our values of x are correct, we must
mihstitute the values found for x in the original equaJtioUy to see
whether it is satisfied or not.
BXBBOI8B8
96. Solve and check :
2 1 .43
1.
oj— 1 aj3 7« + 3 6«42
2. 5+4 = ^ + 6. 6. ^ = ^.
y y x\2 x + b
8 3 ^ 1 ^ 3ml _ 3m5
2r + l r1 * 3m + 4 3m2
1 1
7.
{x  l){x  4) (oj  2){x  6)
8. 4A__=7.?i^.
y4l y3
10. (m* — 4)y = m ~ 2.
m + n
x 1 16. ^4.5^=3.
11.  = 
a
a c
„ ex 1 16. —  — = —
^2 J="^' Ax^S 6ajl
18. ?l=a. 17. ^+ ^ '"^
0? — 2 a: — 3 a — 4
MORE ADVANCED THEORY AND OPERATIONS 79
5 4 1
18.
19.
05— 5 a; — 4 05 — 3
1111
20.
a?— 1 x—3 x — 5 x — 7
_3 1 ^ 5rH5
r43 3r r*9
Solve and tell, by substitution in the original equation,
which of the roots, if any, are extraneous. '
21 ^_+^_ = S.
x1 a!2 2
05 + 6 « — 7 (a! + 6)(a! — 7)
23. .i'^r^ =a;+3.
6a!« — 9a;46
a!«9
26. 3ir^^^y_
y 2yl
26. 2« 8 3
x + 1 2a!«a!3 2a!3
27.
3 _ a!» + 34a!31
2a;3 3a;44 6aj2.aj12
oo 1 7. 2 6
28. r — 6 =
y — 5 ^2 _ 52
80 ELEMENTARY ALGEBRA
m
MISCELLANEOUS PRACTICAL PROBLEMS
97. Applications of algebra, such as occar in the problems which
follow, are frequently made by builders and manufacturers. Certain
formulas which have been established by engineers and mathematicians
are used in computing the desired results. We shall assume the formulas
as true without giving the mode of deriving them. Care must be taken
to use in every problem the proper units of measure.
1. Find correctly, to two decimal places, the side of a
square whose area is 35.06 sq. in.
2. Compute the altitude of an equilateral triangle whose
sides are a inches.
3. If one side of a rectangle is a inches long, and one of
the adjacent sides is three times as long, find the length of the
diagonal.
4. A ladder of the length 9 a is placed against a wall, with
its foot at a distance 4 a from the wall. How high above the
ground is the top of the ladder ?
6. In placing blackboards in schoolrooms, architects deter
mine the height of the chalk rail above the floor by the
formula A = 26hf (^4), '
where h = height of the chalk rail in inches, g = the number
of the grade to which the pupils in the room belong. Find
the height of the chalk rail for pupils in the 5th grade ; also
for pupils in the 8th grade and in the 3d grade.
6. The Baldwin Locomotive Works in Philadelphia have
derived a formula, i2 = 3^, for finding the approximate
relation between the resistance, B, offered by a railway train,
and the velocity, the train traveling on a straight and level
track. In this formula B is the resistance in pounds per ton
weight of the train, v is the velocity of the train in miles per
hour. What is the resistance per ton of a train running at the
MORE ADVANCED THEORY AND OPERATIONS 81
rate of 40 mi. an hour ? What is the resistance of this train,
if it weighs 100,000 T. ?
7. The load that may be safely applied to an iron chain is
given by the formula, I = 7.11 d^, where I is the load in tons
(2000 lb.), and d is the diameter of the iron chain in inches.
Find the largest safe load that may be applied to a chain for
which (f is f in. Find d when Z = 13 T.
8. The distance which one can see from an elevation at the
sea has been found to be cZ = 1.23 V^, where d is the number
of miles one can see, and h is the elevation of the observer in
feet. How far is the horizon from a man standing at the sea
shore, whose eye is 6 ft. from the ground ?
• 9. In Ex. 8, how far can one see from a cliff 27 ft. high ?
67 ft. high? How high must a cliff be to afford a view of
20 mi. ?
10. From a cliff I can just see the lights of a seaport 12 mi.
across the sea. If these lights are 20 ft. above the sea, what
is the height of my eye ?
11. According to a rule sometimes used by architects, the
"rise" (r inches) in the steps of a stair
case (Fig. 8) is connected with the " tread " [ T
(t inches), by the formula r = ^ (24 — *), I
where r and t are also subject to the limita ' l \ u '*
tions, 7 > r > 5, 12 > « > 10. Explain the fw 8
formula in words. What is the " rise " when
the tread is 10^' ? Hf"?
12. Engineers have determined that the velocity (v feet per
minute) of a stream at the bottom of a river is connected with
the velocity ( V feet per minute) at the surface, by the formula
V = F+ 1 — 2VF. Find the velocity at the bottom when the
velocity at the surface is 6 ft. per minute, 8 ft. per minute,
95 ft. per minute.
G
82 ELEMENTARY ALGEBRA
13. A rectangular beam (Fig. 9) is supported at the ends
and loaded in the middle. The weight it will just bear, with
out breaking, is w; = , where w
is the weight in pounds, b the
breadth of the beam in inches, d
p Q is the depth of the beam in inches,
I the length of the beam in feet,
and A: a coefficient equal to 3470 for wrought iron, 2540 for cast
iron, 450 for red pine. Find the greatest weight which can be
hung in the middle of a wroughtiron bar 5 ft. long, 1 in, wide,
and 2 in. deep.
14. In £x. 13, find w when the beam is cast iron, also when
it is red pine.
16. In Ex. 13, how deep must a beam of red pine be, to sup
port 2 tons, if it is 6 ft. long and 3 in. wide ?
16. At a curve in a railway track the outer rail is raised
above the inner rail by an amount indicated by the formula
h = , where h is the number of inches that the outer rail
1.25 r'
is elevated above the inner, w is the width between the rails in
feet, V is the greatest speed of a train in miles per hour, r is
the radius of the curve in feet. If r = 2000 ft., w = 56^ in.,
V = 60 mi. per hour, what is ^ ?
17. The safe load which can be applied to a rope is i = ^iC*,
its breaking load is 6 = k^c^ where I and b are measured in
tons, and c (the circumference of the rope) in inches. For
common hemp ropes ki = .032, Ajj = .18 ; for best hemp rope,
ki = .100, k2 = .60 ; for steelwire rope, ki = .450, Aij = 2.8. Cal
culate the safe load (a) for a common hemp rope 3.5 in. in
circumference, (6) for a steelwire rope 2.5 in. in circumference,
(c) for a best hemp rope 4 in. in circumference. Calculate the
breaking load for each of these.
MORE ADVANCED THEORY AND OPERATIONS 83
18. Draw a graph showing the safe load of steelwire ropes
for different circumferences, up to 4 inches.
19. The breaking load of the best hemp rope is given by
the formula, 6 = .60 c*, where b is the breaking load in tons,
c is the circumference of the rope in inches. Draw a graph
showing the breaking load from c = .3 in. to c = 4 in. From
the graph determine the breaking load for three values of c
not used in constructing the graph.
20. In the flow of water through pipes, a certain head
is necessary merely to overcome the friction of the water
against the pipes. This head is given by the expression,
1/ 0043 \ s*
hssl .0036 4 )—  , where a is the speed of the water in
A y/s /32
feet per second, I is the length of the pipe in feet, d is the
diameter of the pipe in inches, h is the head of the water in
feet. Find the head required to overcome the friction in a
3inch pipe, 3000 feet long, in which water is running at the rate
of 3 feet per second.
21. A house is to rest on piles as a foundation. In the erection
of it one must know how much each pile can support. This is
computed from the performance of the " pile driver," a machine
that lifts a heavy " ram " and drops it on the pile. The max
imum load Z, in tons, that a pile will bear is obtained from the
formula, Z= ^ ^ — , where w is the weight of the ram in cwt.,
h is the height in feet from which the ram falls, d is the dis
tance in inches that the pile was driven in by the last blow,
p is the weight of the pile in cwt. If lo = 6, A = 4, p = 15,
d = 1^, what is the maximum load the pile will safely bear ?
22. A steam engine is operated by steam which enters a
cylinder C (Fig. 10) and presses against one side of the piston
P and moves it ; the motion is transmitted by the piston rod B
to the crank shaft i^i, which turns the wheel around. The rate
84
ELEMENTARY ALGEBRA
at which the engine does mechanical work depends upon the
size of the cylinder, the pressure of the steam, and the number
of revolutions per minute. The rate of doing work is measured
in horse powers, H. If d is the diameter of the cylinder in
/
V
Fig. 10.
inches, I the length of the piston stroke in feet, n the number
of revolutions of the wheel per minute, and p the mean steam
pressure per square inch, then ^^ ^JL^. ' If w= 22, d = 55",
I = 61", p = 20 lb., find H. If n = 25, p = 15 lb., I = ^ ft.,
what must d be, that the horse power may be 30 ?
23. The horse power required to move a ship is given approx
imately by the expression H = .0088 s^ (.05 A h .005 S), where
s is the speed in knots, A is the immersed cross section of the
ship in square feet, S is the wetted surface of the ship in
square feet. A and S remaining the same, state the nature
of the variation of ^ as a function of s. What horse power
is needed to move a steamer at 12 knots, when A = 230 and
fi' = 3100?
CHAPTER III
PROPORTION, VARIATION, FUNCTION
PROPORTION
98. A proportion expresses the equality of two fractions.
The fractions are sometimes called ratios.
Thus, — = £ is a proportion. It is written also a\h — cid.
b d
In a/b = c/dy the terms a and d are called the extremes, the terms b
and c the means. In the ratio a/b, a is sometimes called the antecedent,
b the conseqiient, but ordinarily there is no need of these terms.
A mean proportional between two numbers a and b is the number m
which satisfies the equation — = ^ .
m b
Since a proportion is really an equation, it is treated like an
equation.
Thus, in the proportion, ? = ^.
b d
1. Multiply both sides by bd, ad = be.
This relation is expressed by the theorem :
In a proportion, the product of the means is equal to the product of the
extremes,
2. When only three terms of a proportion are known numbers, the
fourth tenn may be found.
Thus, if a, b, and c are known, we obtain from ^ = ,
b d
ad= be
and d = .
a
3. Add 1 to both sides of the equation,  + 1 =  h 1,
b d
Whence, the new proportion,
85
a\b c + d
b d
86 ELEMENTARY ALGEBRA
BXBBOI8BS
99. 1. Prove that if the product of two numbers, e/, is equal
to the product of two other numbers, gh, one pair may be taken
as the means, the other pair as the extremes, of a proportion.
Prove that, if  = , then
b d
2.
a — 6 c — d
b d
7.
c d
3.
a — b c — d
a + b c4d
8.
6* d*
A
a c
9.
a» c»
'M»
a \b c4d
6» (f
6.
a c
10.
a* c"
a — b c — d
6" d"
6.
a 4 b _c + d
a^b c—d
11.
Simplify the following ratios by writing each in the form of
a fraction and reducing the fraction to its lowest terms :
12. 126:675. 16. (a  6)* : a*  6».
13. 69 a»c» : 46 a*6c. 16. (8m« + n») :16(2m4w).
14. a*6*:(a + 6)c. 17. 4(p^  8 g*) : (p  2 g).
18. (a* h a*6* + 6*) : (a«  a5 + &«).
19. (aj» — ay — 2i^=):3(a; — 2y).
22. (aJ«2ary4y*}:(«' + »yl3/*).
23. (a?* + 1) : (ic*  a? V2 f 1).
1
PROPORTION, VARIATION, FUNCTION 87
24. Find the mean proportional between 31.2 and 0.96.
26. Find the mean proportional between 3 a^c^ and 7 c^d^.
26. Determine which is the greater of the ratios 19 : 25, or
66 : 74.
27. Two numbers are in the ratio of 3 : 4, and their difference
is to their product as 1 is to 18. Find the numbers.
28. Find the last term of a proportion whose first three
terms are 27, 6, and 18.
29. Divide $ 644 among A, B, and C, so that A's share is to
B's in the ratio of 2 to 3, and B's share is to C's in the ratio of
6 to 7.
30. The income of an estate should be divided between A
and B in the ratio of 6 ; 4 ; $ 1100, however, is paid to A, and
$ 700 to B. Which has been unjierpaid, and by how much ?
FUNCTIONS
100. The formula 8 = 16.1 1^
is used in finding the distance (space) through which a body
falls from rest, in time t, where the time is measured in seconds
and the distance in feet.
During the fall, the time t and the distance 8 both increase ;
t and 8 are variable8y while the number 16.1, of course, is con
8tant.
For every value of t it is possible to compute », 8 being de
pendent for its value upon the value of ^ ; 8 is said to be a
function of t
A variable i8 a function of another variable when its value i8
determined for every value of the other variable.
We have used variables x and y in the drawing of graphs.
In the equation of the straight line, y = 3 a?  1, a; and y are
variables ; 3 and 1, of course, are constants ; y is a function of
oj, since for every value of x there exists a definite value of y.
88 ELEMENTARY ALGEBRA
ORAL EXERCISES
101. 1. The velocity of a body falling from rest is given by
the formula, ^ ^ ^^.2 t,
where v is the velocity in feet per second and t is the time of
fall in seconds.
Which are the variables ? Which is the constant ? Why is
V a function of ^ ?
2. The area of a circle is given by the formula
Why is u4 a function of r ? Is tt a constant or a variable ?
3. If a man's monthly salary is % 125, his earnings may be
expressed by the equation,
E = 125 w,
where n is the number of months during which he draws his
salary, and E the earnings expressed in dollars.
In this formula explain " constant," " variable," and " func
tion."
4. The area of a rectangle is given by the formula
A = hh,
where h and h are the " base " and " height " in linear units,
and A is the " area " in the corresponding square units. Here
A varies when h varies and also when h varies. That is, A is
a function of the two variablesy b and h. Find A, when 6=2,
^ = 3 ; also when 6 = 4, ^ = 2.
DIFFERENT MODES OF VARIATION
102. When one variable is a function of another variable,
the variation takes on different forms in different cases. In'
working problems it is necessary to know the exact mode of
variation, else absurd results will be obtained.
PROPORTION, VARIATION, FUNCTION 89
In geometry it is proved that the length of a circle varies as
its radius, that the area of a circle varies as the sgware of its
radius, that the volume of a sphere varies as the cvbe of its
radius.
In symbols, C : Ci = r : rj,
V: Fi = r^ : Vi^,
Hence, if the second radius is twice the first (ri = 2>), then
the length of the second circle is twice that of the first, the
area of the second circle is four times that of the first, the
volume of the second sphere is eight times that of the first.
Of special interest and importance are the problems in which
one variable varies directly as another, or inversely as another.
The words " directly " and " inversely " have the same signifi
cance in algebra as in arithmetic.
For example, at any moment, the length (L) of the shadow of a ver
tical post cast upon level ground depends upon the height {H) of the post ;
the taller the post, the longer the shadow. Here L varies directly as H ;
we wriiie. _ _ _ . _ . _ _ __ _ _ .
L:Li = n'.nuOT L = kJS, or Lccff,
where kis9, constant and where L x ^signifies ^' L varies asH,^^ These
three notations denote the same relation between L and H; all three in
dicate that L is a function of H,
The time {t)it takes a train to travel from New York to Chicago de
pends upon its speed (v) ; the greater the speed, the less the time. Here
t varies inversely as « ; we write
t.ti = Vi'.Vy OTt=k<,OTtcC,
V V
From t = k*we obtain vt = k; that is, if one variable varies inversely
V
as another, the product of the two variables is a constant.
Sometimes there are three variables. When one variable de
pends upon two other variables in such a way as to vary as the
product of the two variables, we say that the first variable varies
jointly as the two other variables.
90 ELEMENTARY ALGEBRA
Ex. 4 in § 101 was of this sort. If we doable the length of the base (6)
of a triangle and at the same time doable its height (A), the area (^) is
four times greater. In symbols, the variation is expressed thus,
AiA\=hhi h\hu or ji = ib • &A, or ji K hh.
There are many other kinds of variation. A variable may
vary directly as a second variable and inversely as a third, as in
dicated by the formula y = A;. Or, a variable may depend
z
upon three or more other variables. Thus, the price of lace
depends not only upon the number of yards, but also upon the
quality of the article, the cost of labor, the rate of profit ex
acted vby the seller. Practical problems involving variables as
functions of other variables and exhibiting many different
forms of variation were given in § 97.
VARIATION SHOWN IN GRAPHS
103. Consider a proportion in which two of the terms are
variables.
An automobile travels 29 miles in 2 hours ; at this rate, how
far does it travel in h hours ?
As the distances are proportional to the times of travel, we have the
proportion, d:29 = A:2,
where d indicates the number of miles and h the number of hours. Here
d and h are variables ; when h increases, d increases also, the value of d
being dependent upon that of h.
The relation between h and d can be expressed equally well
by changing the equation to the form
^ 29,
2
This shows that h and d vary in such manner that the number
d is always — as large as the number h\ d is a function of h.
*4
PROPORTION, VARIATION. FUNCTION 91
Thia variation is exhibited by tlie tollowing graph (Fig. 11).
Hours are measured along the horizontal line OX; miles aie
measured along the ver
tieal line Y. For cou
venieuce the divisions
representing bours are
taken ten times longer
than the divisions rep
resenting miles. The
line OA is constructed
ae follows :
When h = 0, then
ds=0. This determines
the point 0.
When ft = 4, then
d = 5S. This deter
mines the point A.
Through and ^
draw the line OA,
Using this graph we
can tell by inspection
the distance traveled
during a time not ex
ceeding 5 hours. For
instance, to tell the
distance traveled in
3.2 hr., we find on OX
the point C, then pro
ceed parallel to OT to
the point B; and from B to the left, parallel to OX. The
answer is 46*^ miles. Actual multiplication gives 46.4 miles.
How accurately can the distance be measured by this graph ?
How accurately can the time be measured by this graph ?
How can a graph be drawn that will give distances correctly
to ^ of a mile ?
92 ELEMENTARY ALGEBRA
EXBBOISB
104. 1. From the graph, Fig. 11, tell the distance traveled
in 2.5 hr., 3.4 hr., 4.1 hr., 4.9 hr.
2. From the same graph tell the time needed for traveling
24 mi., 50 mi., 35 mi., 65 mi., 48 mi.
3. Draw a graph that will show the railroad fare for trav
eling short distances at the rate of 2 cents a mile.
4. Draw a graph showing corresponding values of x and y
in the proportion x:y = 1.9 : 2.3.
6. A clerk in a New York store converts prices in " marks
permeter" into ^* dollar speryard.'' He draws a graph and
roughly " checks " his computations by it. If 1 m. = 1.1 yd.,
and 1 M. = $ .24, draw the graph.
Hint. 1 M. per 1 m. = l .24 per 1.1 yd. = $ ? per 1 yd.
6. One knot (nautical mile) is equal to 1.15 miles. Draw
a graph for changing from one unit to the other and then tell
from it the speeds, in miles per hour, of cruisers making 21,
22, 24, 27, 29 knots per hour.
7. In lifting weights with the aid of a small, welllubricated
screw jack, part of the power varies with the weights that are
raised, and the rest of the power is a fixed amount to overcome
friction. It was found by trial that 5 lb. was necessary to lift
a weight of 100 lb., and that 2.8 lb. was necessary to lift 50 lb.
Derive an equation and draw a graph showing the relation
between the power P and the weight W.
By the conditions of the problem, the relation between P and TTcan
be expressed by a linear equation of the form P= aW\ b, where P and
W are the variables, and a and b are constants to be found from our
data. Observe that aW is the part of the power which varies with the
weight, and that b is the constant part that is expended to overcome
friction. Notice that P is a function of W.
When P = 6 lb., W = 100 lb., hence 6 = 100 a + 6.
When P= 2.8 lb., Tr= 60 lb., hence 2.8 = 50 a + 6.
PROPORTION, VARIATION, FUNCTION
93
Solving for a and b we obtain a = .044, 5 = .6.
Substituting these values we obtain F = .044 Tf 4 .6,
which expresses ** the law of the machine." The graph of this equation
is shown in Fig. 12.
8. In Fig. 12, tell
by inspection the
power, P, necessary
to raise a weight, TT,
of 30 lb., of 40 lb., of
65 lb., of 75 lb. To
what fraction of a
pound can you esti
mate the power by
this graph?
9. In Fig. 12, tell
by inspection the
weight Wy which can
be raised when P is
2^ lb., 3 lb., 4 lb.
By this graph, can
W be determined as
closely as 1 lb. ? How
would you draw a graph yielding more accurate results ?
10. In a system of welllubricated pulleys a weight of
50 lb. is raised by a power of 10.7 lb., and a weight of 80 lb.
is raised by a power of 16.7 lb. If part of the power varies
as the weight, and part of the power is fixed, find the equation
or " law of the machine " which shows the relation between
P and W, Draw a graph exhibiting simultaneous values of
PandTT.
Y
100
/
90
J
f
/
80
/
J
f
70
/
/
&60
A
r
/
•;j6o
J
J
f
"M'a^
/
% 40
>
"^30
J
f
/
20
J
J
f
10
/
y
r
1
1
r^
J
/
5
~xl
T
'0'
.;
rr
in
P
3U
n(
1
lo~
Fig. 12.
94 ELEMENTARY ALGEBRA
PROBLEMS
105. 1. If XQcy, and x^^ when y = 6, find x when y = 7.
First Solution. Since xocy, the yariation is direct.
Hence,
y:yi^x:xi,
Taking y,=6, a?i=5, y=7,
7:6=:aj:5.
Hence,
a5 = 5f.
Second SoltUion. Since
««y,
we have
o^sA;^.
To determine the constant k, use the given simultaneous
values of the variables x and y, viz., x = 5 when y = 6. We
^^*a^ 5 = A;. 6.
Hence, ^ = f .
Substituting the value of /c, xss^y.
When y = 7, we have a? = ^ x 7 = &J.
The student should master both solutions.
2. If x varies inversdy as y, and a; = 5 when y s= 6, find a;
when y = 7.
i^ir«< Solution. Since a? oc , we obtain the proportion
7 : 6 : 5 : a;.
x==^.
Second Solution. Since a; x  ,
a; = — •
Taking a? = 5, y = 6, ^ "^ fi'
A; = 30.
30
Substituting the value of A;, a; = — •
y
When y = 7, a; = 8^ = 4^.
PROPORTION, VARIATION, FUNCTION 95
3. If z varies jointly as x and y, and 2 = 5, when xssS and
y = 4, find z when a? =a 7 and y = 6.
First Solution, By proportion, z:zi = xy: x^yi.
Take ajj = 3, yi = 4, «i = 5, a; = 7, y = 6.
Then, « : 5 = 42 : 12,
2; = 17f
Second Solution, Since zccxy^
z = kxy.
To determine A;, substitute the simultaneous values,
2 = 5, a? = 3, y = 4,
5 = A: X 3 X 4,
I. — . 6
a;— ly.
Hence, « = ji^ ajy.
When aj= 7, y = 6, 2 = ^ x 42 = 17f
4. If z varies directly as x and inversely as y, and 2 = 5,
when a? = 3, y = 4, find 2; when a; = 7 and y = 6.
Solution. Here 2 oc , hence
z = k>.
y
Substitute the simultaneous values 2 = 5, a; = 3, y = 4,
5 = A;.f.
Hence, ^ = ^,
A 20 a?
ana 2 = — •  •
3 y
Whena? = 7,y = 6, 2; = ^.J = 7f
Let the student give the solution by proportion.
6. The distance described by a body falling from rest varies
as the square of the time. If in 2 seconds it falls 64.4 feet,
find the distance it falls in 6^ seconds.
Solution* The distance (d) varies directly as the square of the time {t).
Hence, d = kfi.
96 ELEMENTARY ALGEBRA
When « = 2, d = 64.4, hence, 64.4 = ifc . 2^,
A: = 16.1,
and d = 16.1 <2.
When t = 6i seconds, d = 680.2 + ft.
6. The brightness or intensity of lig:ht varies inversely as
the square of the distance from the source of the light. How
many times brighter is the page of a book illuminated at a
distance of 3 ft. from the source than at a distance of 5 ft. ?
Solution. The intensity (i) varies inversely as the square of the dis
tance (d).
That is, t = — , or i.ii = di^.d^,
d^
Since we do not have sufficient data to determine the numerical value
of k, it is easier to work the example by proportion.
We obtain i :ii = ^: 3«.
Or 1 = 25 = 24.
ii 9
At a distance of 3 ft. the page is 2} times brighter than at 5 ft.
7. If a: varies as y, and if x is 144 when y is 3, find the value
of y, when x is 360.
8. If X varies as the square of y, and if x is 144 when y is
3, find the value of y when x is 360.
9. If x varies jointly as y and z, and op = 10 when y = 15,
and z = 18, find x when y =5 and z = 24.
10. If oj varies inversely as the cube of y, and a; = 54 when
y = 3, find x when y = .1.
11. The areas of similar triangles vary as the squares of
homologous sides. If homologous sides of two such triangles
are 7 and 10, and the area of the larger triangle is 13.5, what
is the area of the smaller triangle ?
12. The areas of two similar triangles are 36 and 43 ; one
side of the smaller triangle is 5. Find the homologous side of
the larger triangle, correct to two decimal places.
PROPORTION, VARIATION, FUNCTION 97
13. Two pieces of round timber (right cylinders) have the
same altitude ; their girths are 3.4 ft. and 3.1 ft. Find the
ratio of their volumes.
14. If a carriage wheel 4 ft. 2 in. in diameter makes 240
revolutions in going a certain distance, how many revolutions
will a wheel 4 ft. 8 in. in diameter make in going the same
distance ?
15. The weight of a sphere of given material varies as the
cube of the radius. If a sphere having a diameter of 3 in.
weighs 4 lb., find the weight of a sphere of the same material,
but with a radius of 2 in.
16. The rents of an estate should be divided between A and
B in the ratio of 4:5. However, A is paid $ 250, and B is
paid $ 275. Which has been overpaid, and how much ?
17. If a train, whose speed is 55 miles per hour, makes a
certain trip in 3^ hr., how long will it take a slower train to
travel  of that distance, its speed being to that of the fast
train, as 4 is to 9 ?
18. If the pressure p exerted upon the air in a bicycle pump
and the volume of the air v obey Boyle's law, accordiug to
which pv = constant, and if the pressure is 25 lb. per square
LQch when the volume is 27 cubic inches, what will the pres
sure be when the volume is reduced to 20 cubic inches ?
19. Part of the expenses of a certain school was fixed ; part
of them varied with the number of pupils. If the yearly ex
penses were $ 75,000 when the number of pupils was 710, and
$ 85,000 when the number of pupils had increased to 900,
find the yearly expense when the number of pupils was 800.
20. If the ratio ofSx + ytoAx — Sy equals the ratio of 21
to 2, what is the ratio oix to y?
21. At the end of 12 days, 15 men finish one fourth of a
piece pf work. How many additional men must be engaged
in order to complete the remaining work lq 18 days ?
H
98
ELEMENTARY ALGEBRA
22. If a body falls from rest through a distance of 16.1 feet
during one second, how far will it fall in 8.3 seconds ?
23. Compare the intensity of illumination of the page of a
book at a distance of 6.3 feet from a source of light with the
intensity at a distance of 4.9 feet.
24. If the surface area of a certain sphere is 35 sq. in., what
is the surface area of a sphere having a radius 4.7 times longer ?
26. If the volume of a certain sphere is 500 cu. in., what is
the volume of a sphere whose radius is 2.9 times longer ?
26. If the base of a triangle is increased 10fold and its alti
tude is reduced to j^ its original length, find the ratio between
the initial and final areas.
27. Two casks of similar shape have their homologous linear
dimensions in the ratio of 2 : 3.4. Find the ratio of their
capacities.
28. The radii of the bases of two cylindrical tanks are 1 ft.
5 in., and 1 ft. 11 in., respectively ; the heights of the tanks are
6 ft. and 5 ft., respectively. Compare the capacities of the
tanks.
29. The area of a circle is found by the formula A = wR*,
where v = 3.1416. Draw a graph by which the areas can be
determined by inspection.
B
A
Point
i
.79
A
1
3.14
B
li
4.91
C
H
7.07
D
U
9.62
E
2
12.56
F
3
28.27
Q
4
60.27
H
Assuming values of i?, as shown in the
table, compute the values of the area A,
Explain the construction of Fig. 13.
J
PROPORTION, VARIATION, FUNCTION
99
30. In Fig. 13,
find by inspection
the areas of circles
having radii of 2J,
2^, 2f , 3j, ^, ^.
31. If a garrison
of 2000 men is pro
visioned for 40 days,
how long will their
provisions last if
they are joined by
375 other men ?
32. The surface
of a sphere varies
directly as the
square of the diam
eter. If the diam
eter is 2 inches, the
surface is 12.666
square inches. Find
the surface when the diameter is 7 inches.
33. The value of a piece of land has risen in ten years from
$ 1200 to $ 3200. How would you calculate the rise in value of
another similarly situated piece which was worth $1750 ten
years ago ?
34. The rent of a house in a town has gone from $ 85 to $ 70
per month during the last 15 years. If the value of other houses
in that neighborhood has decreased in the same ratio, what was
the decrease in the rent of a house now rented at $ 42 ?
36. The diagonal of a cube varies directly as the length of
an edge. When the edge measures 3 inches, the diagonal
measures 5.196 inches. What does the diagonal measure when
the edge is 3.8 inches ?
1,
{
lO
I
1
r
>
/v
/
*r
/
/
—
)>
Z'
f?
>
'
—
■i\
t\
.y^
/
^}
G
I
J
f,
f
■S
J
I
lA
/
~{
sU
J
/
/
Fj
/
1
E
/
J
T
)/
'
c
/
B>f
A
/^
^
2C
_,
2 S
Radius of Circle
Fia. 13.
100 ELEMENTARY ALGEBRA
36. The weight of a copper coin varies directly as its thick
ness and also as the square of its diameter. A coin weighing
11 grams has a diameter of 3 cm. and a thickness of .2 cm.
What is the weight when the diameter is 2.5 cm. and the
thickness .3 cm. ?
37. A's rate of working is to B's as 3 : 4 ; B's rate of work
ing is to C's as 6 ; 5. How long will it take C to do work
which A can do in 8 hours ?
38. The distance through which a body falls from rest varies
as the square of the time of falling. If a body falls 257.6 ft.
in 4 seconds, how far will it fall in 5^ seconds ?
39. If a stone weighing 5 pounds falls from rest through a
distance of 144.9 ft. in 3 seconds, how far will a stone fall from
rest in 3 seconds, if its weight is 500 pounds ? See Ex. 38.
40. The tractive force or pull necessary to move a vehicle at
a uniform rate (say 3 miles an hour) varies directly as the
pressure (weight of vehicle and load) and inversely as the
square root of the average radius of the wheels. If on level,
paved, or macadam roads a pull of 57 pounds per ton of pres
sure is necessary when the front and rear wheels average 50
inches in diameter, what pull is necessary to move 3500 pounds
when the wheels of the vehicle average 38 in. in diameter ?
41. If a tractive force of 75 lb. is necessary to keep a load
of 1 ton in motion in a vehicle whose wheels average 38 in. in
diameter, on a dry and hard earth road, what is the tractive
force for 5 tons when the wheels average only 26 in. in
diameter ? See Ex. 40.
42. If on an earth road, in sticky mud \ in. deep, the trac
tive force per ton is 119 lb. when wheels 38 in. in diameter are
used, what tractive force is required to move 4 tons when the
diameters are 50 in. ?
43. The tractive force per ton over dry, cloddy plowed ground
is 252 lb. for wheels 50 in. in diameter ; what is the tractive
force for J of a ton when the wheels are 26 in. in diameter ?
PROPORTION. VARIATIpkf rtkc'TtON^ '' \ IQl
GRAPHS EXHIBITINO EkplRICAL 'iJATA
106. 1. By a certain plan of life inaurance a single premium
is paid in order that the insured may receive $100 when he ia
60 years old or that Ms beneficiaries may receive $100 if he
dies before that age. The premium depends upon the age at
which the i:
Age next birthday :
15 20 25 30 36 40 45 60
Premium: $41.20 $46.25$49.30 953.60 $60.10 $65.10 $72.05 $80.20
Draw a graph and
use it to find the pre
mium one would pay
at the age of 18, 23,
32, 38, 43. 2
It U convenient to g
mark oB along the zaxie J
the ages above 16 ; along a
the {(axis the premiums g
aftoiie $40. By this <le J
vice a much smaller sheet **
of Bquare paper can be
used. See Fig. 14.
2. In a city in the
northern part of the
United States the
times of sunrise on
certain days in May, June, and July
Miv
;20
.04
Jf:
i^es above 16 years
Fig. 14.
as follows ;
46
Draw a graph showing the hour of sunrise from May 10 to
July 30.
Mark oS on one axis the number of days after May 10, od the other
axis the number of minuteB after 3 : 44 a.h.
• • • * • .
• • 
«
CHAPTER IV
LOGARITHMS
107. In an equation 10^ = N
the exponent L is called the logarithm of the number N, to the
base 10.
For example,
102 = 100, hence 2 is the logarithm of 100, to the base 10.
We write, 2 = log 100.
10> = 1000, hence 3 is the logarithm of 1000, to the base 10.
We write, 8 = log 1000.
10^ = 3.16227+ (verify this), hence .6 is the logarithm of 8.16227+, to
the base 10. ^^ ^^ite, .6 = log 3.16227+.
10* = 10i(10)i = 81.6227+, hence 1.5 is the logarithm of 81.6227+, to
the base 10. ^^ ^^te, 1.6 = log 31.6227+
As in this chapter all logarithms are taken to the same base 10, no con
fusion will arise from the omission, hereafter, of the phrase *' to the base
10."
Verify the following statements :
10» =1000 10* = 10(10)* = 31.6227+
10* =100 IQi =3.16227+
10^ =10 1
lfti = — =.316227+
10^ =1 ^^ lOi
101 = .1 .1
10. = .01 ioi = ^i=. 0316227^
10» = .001 1
lOf = A = .00316227+
10* = 100(10)* = 316.227+ 10*
102
LOGARITHMS 103
Hence, by the definition of logarithms,
3 = log 1000 ^.5 = log 316.227+
2 = log 100 1.5 = log 31.6227+
1 = log 10 .5= log 3.16227+
O = logl .6= log ,316227+
 1 =. log .1  1.5 = log ^0316227+
 2 = log .01  2^5 = log .00316227+
 3 = log .001
LOGARITHMIC CUBVB
108. On a piece of square paper as small as that in Fig. 15, it is
not convenient to draw a curve that will exhihit the logarithms
of numbers ranging from ,001, the smallest, to 1000, the largest
number considered above. The curve shows the logarithms of
positive numbers between .1 and 50. The numbers are laid off
on the as&xiB, their respective logarithms on the yaxis.
Fig. 15.
— 1 = log .1 looatea the point A
= log 1 locates the point B
.6 = log 3.16+ IncatPB the point C
1 = log 10 locates the point D
1.6 = log 31.6^ locates the point E
curve, some additional values were used.
104 ELEMENTARY ALGEBRA
ORAL BXBBCISBS ON THB LOGARITHMIC CXJBVB
109. 1. Where does the curve cut the ataxia ? How much
is log 1 ?
2. What is the algebraic sign of the logarithms of all num
bers larger than 1 ?
3. For what range of numbers are the logarithms negative ?
4. Does the logarithmic curve extend to the left of the yaxis ?
6. Do negative numbers have real logarithms ?
6. Does the logarithm increase as a variable number in
creases ?
By inspection of Fig. 15, find approximately the logarithms
of the following numbers :
7. 5. 10. 35. 13. 2.
8. 15. 11. 45. 14. 4.
9. 25. 12. 50. 16. 6.
By inspection of Fig. 15, find approximately the numbers
corresponding to the following logarithms :
16. 1.6. 20. 1.2. 24. .8.
17. 1.5. 21. 1.1. 26. .3.
18. 1.4. '22. 1. 26. 0.
19. 1.3. 23. .7. 27. —.5.
Find by inspection, how many times greater
28. log 4 is than log 2. 33. log 27 is than log 3.
29. log 8 is than log 2. 34. log 16 is than log 4.
30. log 16 is than log 2. 36. log 25 is than log 5.
31. log 32 is than log 2. 36. log 36 is than log 6.
32. log 9 is than log 3.
1
LOGARITHMS 105
Show by Fig. 15 that
37. log 50 = log 5 4 log 10. 40. log 48 — log 8 = log 6.
38. log 40 = log 5 + log 8. 41. log 24  log 4 = log 6.
39. log 42 = log 6 4 log 7. 42. log 32 — log 4 = log 8.
ASSUMPTIONS
•
110. In drawing the logarithmic curve in Fig. 15 we located
a few points and then drew a smooth curve through those points.
This process is based on the following assumptions which
admit of proof : (1) All positive numbers, whether rational or
irrational, have logarithms ; (2) In all cases, the logarithm of a
number increases when the number itself increases.
FUNDAMENTAL THEOREM
111. In studying the logarithmic curve we noticed that
log 50 = log 5 4 log 10.
This illustrates an important theorem.
Let N and Ni be any two positive numbers.
Also, let JVr=10%
and Ni = 10^.
Then their product is, N Ni = lO^+^i.
By definition of logarithms, L = logarithm of N,
and Zi = logarithm of Ni,
X + Zi = logarithm of N • ^i.
Hence, the theorem.
The logarithm of the product of two positive numbers is equal
to the sum of the logarithms of the numbers.
112. The integral part of a logarithm is called its charac
teristiCy and the decimal part is called its mantissa.
For example, log 31.6227+ = 1.5.
The characteristic of log 31.6227+ is 1.
The mantissa of log 31.6227+ is .6.
106 ELEMENTARY ALGEBRA
The tables of logarithms give only the mantissa ; the charao
teristic can be supplied by two easy rules.
It has been found convenient to take the mantissas of all
logarithms positive. The characteristic is positive in the
logarithms of numbers larger than 1, and negative in the
logarithms of numbers smaller than 1. That is, the char
acteristic of log, 100 is positive; the characteristic of log .01
is negative. If the characteristic is negative, the negative
sign is placed over the figure, as a reminder that the — does
not apply to the mantissa. Thus the characteristic of log .06
is 2.
For the purpose of deriving the rule for determining the
characteristic of the logarithm of a number, we restate some
of the relations in § 107 :
1000 = 10* 1 =10°
100 = 10^ .1 =10^
10 = 10^ .01 =102
1 = 10° .001 = io»
Since 569.5 lies between 100 and 1000, its logarithm lies between 2 and 8.
That is, log 569.6 = 2 + a mantissa.
Since 86.6 lies between 10 and 100, log 86.6 = 1 + a mantissa.
Since 7.03 lies between 1 and 10, log 7.03 = + a mantissa.
Since .673 lies between .1 and 1, log .673 = ~ 1 + a mantissa.
Since .046 lies between .01 and .1, log .045 = ^ 2 + a mantissa.
Since .0078 lies between .001 and .01, log .0078 = — 8 + a mantissa.
By inspection of these relations we obtain the rule :
1. If the first Bigni&CAnt figure of a number is n places to the
' left 1
/ ' of units' place, the characteristic of the logarithm is
I vOflv
( ■ ^
I 4 w 1
— n
LOGARITHMS 107
The following illustrations will make this rule plainer :
J
S
1^ . Characteristic
6 6 9.6 2 .
8 6.6 1
7.03.
0.673 1
0.046 2
0.0078 3
It will be seen that when a number is 10 or larger than 10, the first
significant figure is to the left of units ^ place ; if the number is less than
1, the first significant figure is to the rig?U of units' place. This is shown
by the dots placed above the digits ; the number of dots placed over a
number indicates the number of units in the characteilstic.
BXBBCISBS
113. By inspection, tell the characteristics of the logarithms
of the following numbers :
1. 123. 4. 17.754. 7. .091. 10. 2000.
2. 97. 6. .7894. 8. .00034. 11. .00005.
3. 2.345. 6. .005. 9. .0034. 12. 13.764.
114. The following is an important property possessed by
logarithms constructed to the base 10 :
If two numbers contain the same figures in the same order , hut
differ in the position of the decimxxl point, their logarithms to the
base 10 have the same mantissa.
Suppose the two numbers are 7896.1 and 7.8961.
We have 7896.1 = 7.8961 x 1000.
By §111, log 7896.1 = log 7.8961 + log 1000.
But log 1000 = 3.
Hence, log 7896.1 = log 7.8961 + 3.
108 ELEMENTARY ALGEBRA
Since the two logarithms differ by the integer 3, the decimal parts (the
mantissas) of the logarithms must be the same.
For example, log 7896.1 = 3.8974.
log 7.8961 = 0.8974.
How the mantissa is found from the tables will be explained next.
FINDING LOGARITHMS
115. 1. Find the logarithm of 789.
The characteristic is 2.
The mantissa is taken from the table, p. 110. The column on thcr ex
treme left contains the first two significant figures of the number whose loga
rithm is sought. In this case we look for the figures 78. In the same row
with 78, and in the column headed * * 9 ' ^ ( 9 being the third digit in 789) , is the
number 8971 ; this is the mantissa sought, expressed to four decimal places.
Hence we have log 789 = 2.8971.
2. Find the logarithm of .02738.
The characteristic is 2. The significant figures are 2738. On p. 109
look in the column on the extreme left for 27 ; in the column headed
** 3 '* and in the same row with 27 is the mantissa .4362.
Hence log .0273 = 2.4362. But we want log .02738. This is not found
in this table and must be determined by a process called *^ interpolation.'^
The table is so constructed that interpolation is made easy. In the
column on the extreme right, headed **789,*' we find, under the 8, and
in the same row with 4362, the number 13.
Add 13 to 4362. We obtain 4376. This is the mantissa, obtained by
interpolation.
Hence, log .02738 = 2.4376.
116. It may be added that logarithmic computation in gen
eral is only approximate. With a fourplace table like the
one in this book, logarithms of numbers are found to four
places only. Absolute accuracy is, therefore, out of the ques
tion. This fact does not materially diminish the value of
logarithms, for the reason that in practical problems, results
correct to three, four, or five decimal places are satisfactory.
For instance, in the computation of the interest on a given
sum, a result correct to the nearest cent is all we want.
LOGARITHMS
109
117.
Logarithms
No.
10
11
11
IS
14
16
16
17
18
19
1
S
S
0128
0681
0S99
1239
1668
4
0170
0569
0984
1271
1684
1875
2148
2405
2648
2878
6
0212
0607
0969
1308
1614
6
T
0294
0682
1088
1867
1678
8
9
1 S S
4 6 6
T 8 9
0000
0414
0792
1189
1461
1761
2041
2804
2653
2788
0048
0458
0828
1178
1492
1790
2068
2880
2577
2810
8082
3248
8444
3686
8820
0086
0492
0864
1206
1528
1818
2096
2856
2601
2888
8064
8268
8464
8666
8888
4014
4188
4846
4502
4664
0268
0646
1004
1886
1644
0884
0719
1072
1899
1708
1987
2268
2604
2742
2967
0874
0765
1106
1480
1782
4 812
4 811
8 710
8 610
8 6 9
17 2126
1619 28
14 17 21
18 16 19
12 15 18
29 88 87
26 80 84
24 28 81
28 26 29
2124 27
1847
2122
2880
2625
2866
1908
2176
2480
2672
2900
1981
2201
2455
2695
2928
1950
2227
2480
2718
2945
2014
2279
2629
2766
2989
8201
8404
8698
8784
8962
8 6 8
8 5 8
2 6 7
2 6 7
2 4 7
11 14 17
11 18 16
10 12 15
9 12 14
9 1118
20 22 26
18 2124
17 20 22
1619 21
16 18 20
80
SI
SS
SS
S4
8010
8222
8424
8617
8802
8075
8284
8488
8674
8856
8096
8804
8502
8692
8874
8118
8711
8892
4065
4282
4898
4548
4698
8189
8846
8641
8729
8909
8160
8865
8660
3747
8927
4099
4265
4426
4579
4728
4871
5011
6146
6276
6408
8181
8386
8679
8766
8946
2 4 6
2 4 6
2 4 6
2 4 6
2 4 5
81118
81012
81012
7 911
7 9 11
15 17 19
14 16 18
14 15 17
18 15 17
12 14 16
86
S6
ST
SS
S9
SO
S4
S6
S6
ST
S8
S9
40
41
4S
4S
M
46
46
4T
48
49
60
61
6S
6S
64
8979
4160
4814
4472
4624
8997
4166
4880
4487
4689
4081
4200
4862
4518
4669
4048
4216
4878
4688
4688
4829
4969
6105
5287
6866
4082
4249
4409
4664
4718
4867
4997
5182
5268
6891
4116
4281
4440
4594
4742
4188
4298
4456
4609
4767
2 8 6
2 8 6
2 8 6
2 8 6
18 4
7 910
7 810
6 8 9
6 8 9
6 7 9
12 14 15
1118 16
11 18 14
11 12 14
10 12 18
4771
4914
5061
5186
6816
5441
5668
6682
6798
6911
6021
6128
6282
6386
6185
6582
6628
6721
6812
6902
4786
4928
5065
6198
6828
5468
6576
6694
6809
6922
6081
6188
6248
6845
6444
6642
6687
67aio
6821
6911
4800
4942
6079
5211
5840
4814
4955
6092
6224
5858
4848
4988
5119
6260
5878
4886
6024
6169
6289
6416
4900
5088
5172
6802
6428
18 4
1 8 4
18 4
18 4
18 4
6 7 9
6 7 8
6 7 8
6 6 8
6 6 8
10 11 18
10 11 12
9 1112
9 10 12
9 1011
6465
5587
6706
6821
6988
6042
6149
6268
6855
6464
6478
6599
5717
6882
6944
6068
6160
6268
6865
6464
6561
6666
6749
6889
6928
7016
7101
7186
7267
7848
5490
5611
5729
6848
5956
6602
6628
5740
6866
6966
6076
6180
6284
68S5
6484
6580
6676
6767
6857
6946
7088
7118
7202
7284
7864
6514
6685
6762
6866
6977
6086
6191
6294
6895
6493
6527
5647
5768
5877
5988
5589
5668
5776
6888
6999
6107
6212
6814
6416
6618
6609
6702
6794
6884
6972
5661
6670
6786
5899
6010
1 2 4
1 2 4
1 2 3
1 2 8
12 8
6 6 7
6 6 7
6 6 7
6 6 7
4 5 7
91011
81011
8 9 10
8 910
8 9 10
6064
6170
6274
6876
6474
6671
6665
6768
6848
6987
7024
7110
7198
7276
7866
6096
6201
6804
6405
6503
6599
6698
6T85
6875
6964
7050
7185
7218
7800
7880
6117
6222
6826
6425
6522
1 2 8
12 8
12 8
12 3
1 2 8
4 6 6
4 6 6
4 6 6
4 6 6
4 5 6
8 9 10
7 8 9
7 8 9
7 8 9
7 8 9
6661
6646
6789
6880
6820
7007
7098
7177
7269
7840
6690
6684
6776
6866
6955
7042
7126
7210
7292
7872
6618
6712
6808
6898
6981
7067
7162
7285
7816
7396
12 8
12 8
12 8
12 3
12 8
4 6 6
4 5 6
4 6 5
4 4 5
4 4 5
7 8 9
7 7 8
6 7 8
6 7 8
6 7 8
6990
7076
7160
7248
7824
6998
7084
7168
7251
7882
7060
7148
7226
7808
7388
12 8
1 2 8
12 2
12 2
12 2
8 4 6
3 4 6
8 4 6
8 4 5
8 4 5
7 8
6 7 8
6 7 7
6 6 7
6 6 7
110
ELEMENTARY ALGEBRA
Logarithms
No.
7404
1
6
7419
8
4
6
7448
6
7451
7
7469
8
7466
9
1 6
8
4
6
6
7 8 9
7412
7427
7485
7474
1 2
2
8
4
ft
5 6 7
M
7482
7490
7497
7505
7618
7520
7528
7686
7548
7661
1 2
2
8
4
5
5 6 7
67
7069
7566
7674
7682
7589
7697
7604
7612
7619
7667
1 2
2
8
4
5
5 6 7
M
7634
7642
7649
7657
7664
7672
7679
7686
7694
7701
2
8
4
5 6 7
M
7709
7716
7728
7781
7808
7788
7810
7746
7752
7760
7767
7889
7774
2
8
4
5 6 7
60
7782
7769
7796
7818
7825
7882
7846
2
8
4
5 6 6
61
78m
7860
7868
7876
7882
7889
7896
7908
7910
7917
2
8
4
5 6 6
6S
7924
7981
7988
7945
7962
7969
7966
7978
7980
7987
2
8
8
5 6 6
66
7998
8000
8007
8014
8021
8028
8065
8041
8048
8066
2
8
8
5 5 6
64
66
8062
8069
8075
8082
8089
8096
8102
8109
8116
8122
2
8
8
6 5 6
8129
8186
8142
8149
8166
8162
8169
8176
8182
8189
2
8
8
5 5 6
66
8195
8202
8209
8215
8222
8228
8285
8241
8248
8264
2
8
8
5 6 6
67
8261
8267
8274
8280
8287
8293
8299
8806
8812
8819
2
8
8
6 5 6
66
8825
8881
8888
8844
8861
8867
8868
8870
8876
8882
2
8
8
4 5 6
66
8888
8895
8401
8468
8407
8470
8414
8476
8420
8482
8426
8482
8494
8489
8446
8606
2
2
8
4 6 6
70
8451
8457
OvOO
8600
2
2
8
4 5 6
71
8618
8619
fmt>
8681
86B7
8648
8549
8666
8661
8667
2
2
8
4 5 5
76
8678
8679
8686
8691
8697
8606
8609
8616
8621
8627
2
2
8
4 5 5
76
8688
8689
8646
8651
8657
8668
8669
8676
8681
8686
2
2
8
4 5 5
74
8692
8696
8766
8704
8762
8710
8768
8716
8774
8722
8779
8727
8788
8789
8746
8802
2
2
8
4 5 6
76
8751
8785
8791
8797
2
2
8
8
4 5 5
76
8806
8814
882(r
1B825
8831
8887
8842
8848
8864
8869
2
2
8
8
4 5 5
77
8866
8871
8876
8882
8887
8898
8899
8904
8910
8916
2
2
8
8
4 4 5
78
8921
8927
8982
8988
8948
8949
8954
8960
8965
8971
2
2
8
8
4 4 6
76
80
8976
9081
8982
9086
8987
9042
8998
8998
9004
9009
9015
9020
9074
9026
9079
2
2
8
8
4 4 5
9047
9058
9058
9068
9069
2
2
8'
8
4 4 5
81
9085
9090
9096
9101
9106
9112
9117
9122
9128
9188
2
2
8
8
4 4 5
86
9188
9148
9149
9164
9160
9166
9170
9175
9180
9186
2
2
8
3
4 4 5
66
9191
9196
9201
92()6
9212
9217
9222
9227
9232
9288
2
2
8
8
4 4 5
64
9248
9248
9299
9258
9258
9268
9269
9820
9274
9279
9880
9284
9885
9289
9840
2
2
8
8
4 4 5
86
9294
9804
9809
9815
9825
2
2
8
8
4 4 5
66
9845
9850
9856
9860
9865
9870
9876
9380
9886
9890
2
2
8
3
4 4 5
87
9895
9400
9405
9410
9415
9420
9425
9480
9486
9440
1
2
2
3
3 4 4
86
9445
9450
9465
9460
9465
9469
9474
9479
9484
9489
1
2
2
8
8 4 4
66
9494
9499
9604
9652
9609
9518
9618
9623
9671
9528
9576
9688
9638
1
2
2
8
8 4 4
90
9542
9547
9567
9562
9566
9581
9586
1
2
2
8
8 4 4
91
9590
9596
9600
9605
9609
9614
9619
9624
9628
9683
1
2
2
3
8 4 4
96
9688
9648
9647
9662
9657
9661
9666
9671
9676
9680
1
2
2
3
8 4 4
98
9685
9689
9694
9699
9708
9708
9718
9717
9722
9727
1
2
2
8
8 4 4
94
9731
9786
9741
9745
9791
9760
9764
9769
9806
9768
9768
9814
9773
9818
1
2
2
8
3 4 4
96
9777
9782
9786
9796
9800
9809
1
2
2
8
3 4 4
96
9828
9827
9882
9886
9841
9846
9860
9854
9869
9868
1
2
2
3
3 4 4
97
9868
9872
9877
9881
9886
9890
9894
9899
9903
9908
1
2
2
3
8 4 4
96
9912
9917
9921
9926
9980
9984
9989
9948
9948
9962
1
2
2
3
8 4 4
99
9956
9961
9966
9969
9974
9978
9983
9987
9991
9996
1
2
2
3
3 8 4
LOGARITHMS 111
BXBBCISBS
118. Find the logarithms of :
1. 870. 4. .01235. 7. 13.8964.
2. 63. 6. .003986. 8. 1089.
3. 708. 6. .7943. 9. 1008.
FINDING ANTILOGARITHMS
119. 1. Given log n = 1.2355, find the antilogarithm n.
The number n, frequently called the arUilogarithm, may be found by
steps which are the reverse of those taken in finding the logarithm.
At first consider only the mantissa 2366. On page 109 find the mantissa
2366. It occurs in the same row as the number 17 on the extreme left,
and in the column headed ** 2.^* ^ The order of the figures in the answer
is 172. '
Where should the decimal point be ? That is determined with the aid
of the characteristic. All we need to do is to apply the rule relating to
characteristics. Place the decimal point so that the characteristic of the
resulting number is 1.
That number is 17.2. Thus, n = 17.2.
2. Given log n = 2.8764, find the antilogarithm n.
Look in the table for the mantissa 8764. It cannot be found.
Look then for the next smaller mantissa. It is 8762 ; the order of
figures in the antilogarithm of 2.8762 is 762. But we want the antiloga
rithm of 2.8764. This is found approximately by the following ** inter
polation.*'
Find the difference between 8762 and the given mantissa. This differ
ence is 2.
Look in the same row with 8762, and in one of the righthand columns
for the number 2. You see it in the column headed ** 3 '* and also in the
column headed "4."
Take either 3 or 4 as the figure found by interpolation. Which of these
is the more accurate, we cannot tell from these tables. Take 3.
Write 3 after 752, and we have the required order of figures ; namely,
7623.
Place the decimal point so that, by the rule, the characteristic of the
resulting number is 2. We obtain n = .07623, nearly.
112 ELEMENTARY ALGEBRA
EXEBCISBS
120. Find the antilogarithms of the following :
1. 1.7931. 4. 1.7605. 7. 3.4567. 10. 3.6709.
2. 2.3181. ' 6. 0.9064. 8. 0.7064. 11. 3.5374.
3. 3.9876. 6. 2.1907. 9. 2.3706. 12. 5.9860.
121. 1. Using logarithms, find the product n = 7893 x 879G.
By § 111, the logarithm of a product is the sum of the logarithms c
the factors. We find , „^^„ ., „^«„
log 7893 = 3.8973
log 8796 = 3.9448
log n = 7.8416
We find the antilogarithm n = 69430000.
Observe that this answer is only approximate. The true answer is
69426828. Closer approximations may be obtained by logarithms, by
using a table containing figures to 6, 6, 7 or more places.
2. Find the product x = .9062 x .007362.
The characteristics of the factors are I and 3. When logarithms are
added or subtracted, it is easier to write a negative characteristic as the
difference of two positive integers, thus : 1 = 9 — 10, 3 = 7 — 10.
Accordingly, we get log .9062 = 9.9572 — 10
log .007362 = 7.867010
Add log X = 17.8242  20
Or loga:= 3.8242
X = .006671, nearly.
BXEBCISBS
122. Using logarithms, compute the following products :
1. 123 X 978. 4. 37.61 x 3.945. 7. 9999 x 7777.
2. 12.34 X 75.34. 6. 1479 x 6984. 8. 19283 X 87045.
3. .0009638 X 7894. 6. 2673 x 7654. 9 93069 x 80493.
LOGARITHMS 113
FUNDAMENTAL THEOREMS
123. In § 111, we proved the important theorem that
Tlie logarithm of a product is equal to the sum of the logarithms
of the factors.
We proceed to establish two other theorems that are no less
fundamental.
The logarithm of a quotient is equal to the logarithm of the
dividend minus the logarithm of the divisor.
The proof is similar to that of the first theorem.
Let N and Ny^ be any two positive numbers.
Let also JV^=10^
and ^1 = 10a.
Divide N by iVi, — = lO^^i.
By definition of logarithms,
. L = logarithm of Ny
Li = logarithm of Ni,
and L — Li= logarithm of —
Hence the theorem is proved.
The logarithm of a positive number with the eocponerU n is n
times the logarithm of the number.
In this theorem n may be a positive integer or a positive
fraction.
Let ^be the number, and let
JV^=1(P.
Raise both sides to the nth power,
N^ = (1(F)».
Simplify N^ = lO"^,
114 ELEMENTARY ALGEBRA
By the definition of logarithms,
L = the logarithm of N
and nL = the logarithm of N\
Hence the theorem is proved.
If w =— , then this theorem gives the logarithm of y/W.
m
124. 1. Using logarithms, find the quotient a?= ^.
We find log 9876 = 3.9946
log 6987 = 3.8443
log a; = 0.1603
The autilogarithm x = 1.414, nearly.
2. Find a? = (8.786)*.
We find log 8. 786 = 0. 9438
Multiply by 6, 6
loga; = 4.7190
X = 52360, approximately.
3. Compute aj =(.01237)*.
The characteristic of .01237 is 2 ; we write it 8 — 10.
Hence, log .01237= 8.092310
Multiply by 7, 7
66.6461  70
Divide by 3. Since 70 » 3 introduces a fraction, it is more convenient
to write the characteristic, 66 — 70, in the form 16 — 30.
The logarithm is then 16.6461  30.
Dividing by 3, log a; = 6. 5487  10.
X = .00003537, nearly.
BXBBCISES
jl25. Evaluate the following by means of the fourplace table
of logarithms :
1. 6.823 X 2.315. 4. 2293 x 4489 ^ 7895.
2. 46420 X 27.49. 6. .943 x 9855 ^ .0896.
3. 32.25^6.923. 6. (2.519)*.
LOGARITHMS 115
7. (.007668)'. ^^ 98.6 x 7.639
8. a/4978000000. ^^
9. V752 X V3450. .. (273)«(743)«
10. (1.63)*. ' (897)V
11. (3.197)1 ^^ V3:598 ' ^479 ,
12. V;03276. * V933
PROBLEMS
126. In the following problems let w = 3.1416, B = radios.
1. Find the length of a circle (i.e. the circumference) whose
radius is 3.17 ft.
2. Find the area of a circle whose radius is 9.795 in. (Area
3. Find the area of a circle whose radius is 7.891 cm.
4. Compute the curved surface of a right circular cylinder
whose height is 7.9 in. and whose base has a diameter of
79.86 in.
6. Calculate the volume of a sphere whose radius is 2.97 yd.
Volume of a sphere = .
3
6. What is the area of the surface of a sphere when R
= 0.7 in. . Area of spherical surface = 4 wlfi,
7. Compute the leg of a right triangle when
the hypotenuse is 25.97 in. and the other leg is
17.8 in.
In Fig. 16, X = Vh^  63 = V(h + 6) (A  6).
8. Compute the area of a triangle whose sides are, respec
tively, 17.89, 20.96, 19.78.
Area of triangle =>/«(« — a) (« — 6) (« — c) , where a, &, e are the sides
of the triangle and s is half the sum of the sides.
116 ELEMENTARY ALGEBRA
9. In a right triangle the hypotenuse is 9.675, one leg is
7.98 ; find the length of the other leg.
10. Find the area of the surface of a hemispherical dome,
the diameter of which is 40 feet.
11. A triangular piece of ground measures along the edges
315 yd., 541 yd., 479 yd. Find the area.
12. The volume of a pyramid is one third the product of
its base and altitude. What is the volume of a pyramid whose
height is 7.82 ft. and the sides of whose triangular base are
6.92 ft., 5.83 ft., and 4.91 ft. ?
EXPONENTIAL EQUATIONS
127. An exponential equation is one in which the unknown
occurs in an exponent.
Thus, 12* =15, a**+'=6 are exponential equations which can
be solved with the aid of logarithms.
Since log 12* = x log 12,
we obtain from 12* = 15, ' x log 12 = log 16.
Divide both sides by log 12, x = l^i? .
log 12
Finding the logarithms, we obtain x = — .
1 i7fii
is computed like any other quotient.
1.0792 y J ^
log 1.1761 = 0.0705
log 1.0792 = 0.0.382
Subtracting, log x = 0.0373
X = 1.090, nearly.
Note that in computations that are only approximate the answer 1.090
receives a different interpretation than the answer 1.09. The difference
is this : the answer 1.090 shows that an effort was made to ascertain the
third decimal figure and that it was found to be 0. On the other hand,
1.09 means that no effort was made to ascertain the value of the third
figure. Consequently 1.090 signifies an answer carried to 3 decimals,
while 1.09 signifies an answer carried to but 2 decimals.
LOGARITHMS 117
EXERCISES
128. Solve for a?;
1. 7* = 17. 3. 9»+^ = 87. 6. 25*"+» = 795.
2. 126* = 98. 4. IS*' = 25. 6. 9* = 13.
PROBLEMS
129. 1. In what time will $ 1 double itself at 4 %, interest
compounded ajinually ?
At the end of the first year the interest is $ .04, the amount is $ 1.04.
At the end of the second year the interest is $ (1.04)(.04), the amount
is $1.04 + (1.04) (.04) = I (1.04)2.
At the end of the third year the interest is $ (1.04)2 (.04), the amount
is $(1.04)8.
At the end of x years the amount will be $ (1.04)'.
We have the equation, (1.04)* = 2.
x= 17.7 nearly.
2. In what time will $ 1 treble itself at 4 %, interest com
pounded annually ?
3. A sum of $ 1000 bears 5 % interest compounded annually.
What is the amount after 20 years ?
4. Compute the compound interest on $ 879 at 4 % for 13
years.
HISTORICAL NOTE
130. Logarithms constitute one of the most useful inventions in
mathematics. A great French astroaomer once said that logarithms ^* by
shortening the labors, doubled the life of the astronomer.*' Logarithms
were invented by John Napier, Baron of Merchiston, in Scotland, who in
1(314 published a table of logarithms and described their use. An inde
pendent inventor of logarithms was the Swiss watchmaker and astron
omer, Joost Burgi, who published a table in 1620, at a time when Napier's
logarithms were already known and admired throughout Europe. Napier's
and BtLrgi's logarithms differed somewhat from each other and from the
logarithms described in this book. Our system of logarithms to the base
118 ELEMENTARY ALGEBRA
10 was designed by John Napier and his friend Henry Briggs, conjointly,
before the year 1617. At one time Napier lived in a beautiful castle on
the banks of the Endrick. Tet even in such surroundings he was not
free from annoyances which hindered intellectual work. On the opposite
side of the river was a lint mill, and its clack greatly disturbed Napier.
He sometimes desired the miller to stop the mill so that the train of his
ideas might not be Interrupted. Much of Napier^s time was taken up in
the management of his estate. Joost BUrgi was not rich, like Napier ;
the necessity of earning a livelihood greatly hampered his scientific work.
But both Bttrgi and Napier were men who loved mathematics and were
able to achieve great results in spite of many hindrances.
y
7
iti
CHAPTER V
QUADRATIC EQUATIONS AND THBIK PKOPBKTIXS
REVIEW
131. In solving a quadratic equation of the form (j^ ^hx
h c = by " completing the square," the first step in our ex
planation was to divide both sides of the equation by the coeffi
cient of aj*. There are other methods of procedure which possess
certain advantages ; fractions having large denominators may
be avoided, or, fractions may be avoided altogether. These
methods will be explained in § 191.
In the following exercises solve by any method. To find
the square roots of numbers, use the tables in § 197.
1. 6aj2f20a; = 17. 6. 35 a?, 3a^= 11.
2. aaj» I afta; f 4 aft = 0. 6. 3aa^f 6 a; — 5a* = 0.
3. 5aj«h21a;=30. 7. 40a?3a^ = 12.
4. oo* f o&c t a« = 6». 8. (a+6)«*2(a— 6)aJfa=0.
EQUATIONS QUADRATIC IN FORM
132. Equations like o^ f osff" f 6 = are said to be quad
ratic in form, because the unknown x occurs only with the
exponents n and 2n, where one exponent is twice the other.
Equations of this form can be solved by the methods used in
solving ordinary quadratic equations.
1. Solve a;*7aj*f10 = 0.
Factoring, (a;«  6) {x^  2) = 0.
a;3  6 = 0, «a  2 = 0.
From the tables in § 107 we obtain x = ± 2.236, x =± 1.414, approxi
mately.
119
120 ELEMENTARY ALGEBRA
2. Solve aJ»f3aj»4l=0.
Complete the square, a^ + S a^ + { ^^ i 4.  = .
x=y_,±:^.
As there are in general three cube roots to a number, and
there are in this ease two different numbers under the radical
sign, it is seen that there are six roots of the g^ven equation.
Of these, four are imaginary. As we wrote the answer, only
the two principal roots are indicated.
BXBBOISBS
133. Solve:
1. ic* 4 6 a^  16 = 0. 6. 6 m«  23 m» 4 20 = 0.
2. a«2aj*4l = 0. 7. y»2y»143 = 0.
3. aJ»h3aj»hl = 0. 8. 2;* = 266.
4. a?«8aj*20 = 0. 9. y«"52r + 6 = 0.
6. 2a?«7a^30 = 0.
RELATIONS BETWEEN ROOTS AND COEFFICIENTS
134. From the solution of aac*  to  c = 0, we obtain
— 6 ± V6» — 4 ac
X =! •
2a
If we designate the two roots by x^ and ajj, we have
2a
V6*
4ac
2
a
V62
4ac
and Am = —
^ 2a 2a
QUADRATIC EQUATIONS 121
The sum of the roots is
_6
a
The product of the roots is
^^ / 6 + V6»  4 acY^ b  V624 ac\
^^ =(,^^ — 2^ — ^;(, — Ui — )
6* — (6* — 4ac) 4ac c
4 a* 4 a' a
If we divide both sides of ax^ f 6aj + c = by a, we obtain
a a
in which the coefficient of a^ is unity. We see that, in the last
equation,
(1) The sum of the roots is the coefficient of x with the sign
changed; and
(2) The product of the roots is the absolute term.
To illustrate: Solve ac^ + 6« + 6 = 0.
The roots are — 2 and — 8. Their sum is — 6, the coefficient of x with
its sign changed ; their product is + 0) the absolute term.
From this relation between the coefficients and the roots, we
may form an equation, if the roots are known.
Form the equation whose roots are — 1 and 6.
We assume that, in the required equation, the coefficient of x^ is 1.
Then, xi + X2 = 6f the coefficient of x with its sign changed ; xix% = — 6,
the absolute term.
Hence, the equation isx^ — 6x— 6=0.
Or, we might write the equation thus :
xa(l + 6)x+(l6)=0.
Whence, ac* — 6 a; — 6 = 0.
122 ELEMENTARY ALGEBRA
135. Form the equation whose roots are :
1. + 5, 4 2. 6. 2^, 3. 9. V2,  V3.
2. 3,8. 6. i, f 10. I4V2, 1V2.
3. 4,6. 7. —a, —6. 11. — V5, — VS.
4. 7, 0. 8. m2,  3 m\ 12. V2,  1 V2.
If you know one root of a given quadratic equation, how
can you calculate the other: (1) from the coefficient of x,
(2) from the absolute term, (3) by the Factor Theorem ? Use
each of these methods in turn, to find the second root in each
of the following equations :
13. aj2 — 8.7 05 I 17.6 = 0, one root 5.5.
14. aj2 — 2.1 a; + .9 = 0, one root 1.5.
16. a;2 — 1.7 a; — 4.8 = 0, one root 3.2.
16. a;2 4 11.8 X + 34.17 = 0, one root — 6.1.
17. aj2 f 3.6 a:  25,92 = 0, one root  7.2.
18. a;2 — .085 x + .00175 = 0, one root .05.
NATURE OF THE ROOTS
136. We shall discuss the nature of the roots of the quad
ratic equation asc* + 6» f c = 0, in which the letters a, 6, c are
assumed to be real and rational numbers* When some of
these letters are imaginary or irrational, the conclusions which
we shall draw do not necessarily hold. The solution of
ax2 j 6aj 4 c = is .
— b± Vb^ — 4 ac
x = •
2a
The nature of the roots depends upon b^ — 4:acy which is
called the discriminant.
I. When b^—4:ac=0, the roots are real and equal. Explain.
II. When 62 _ 4 etc > 0, the roots are real and uneqtuil. If
62 — 4ac is a perfect square, both roots are rational; if
QUADRATIC EQUATIONS 123
6* — 4ac is not a perfect square, both roots are irrational.
Why?
III. When 6* — 4 ac < 0, both roots are imaginary. Explain.
To iUustrate: 1. In as^ lOx + 26 = 0, 6^ _ 4 ^j^ = 100  100 = 0.
Hence the roots are real, rational, and equal. The roots are 6, 6.
2. In flr2  Sjc  28 = 0, 62 «. 4 ^  9 _. 112 = 121. Hence, the roots
are real and unequal; since 121 is a perfect square, both roots are
rational.
3. Inx«2a: + 6 = 0, 63 — 4ac = 4 — 24 = 20. Hence both roots
are imaginary,
EXEBOISBS
137. Determine, without solving, the nature of the roots of :
1. aj2aj12 = 0. 6. 7a;2 12ajH4 = 0.
2. iB24.8a;16 = 0. 7. 4 «2  12 «'=  9.
3. a.2__8ic25=:0. 8. 4 aj2 12 a? 55 = 0.
4. 5aj2f.7a; + l = 0. 9. a?2_2a;H3 = 0.
6. 2aj2 43a; = 8. 10. 3aj2f.l3a:fl = 0.
For what values of a are the roots of the following equa
tions equal ? Real and unequal ? Imaginary ?
11. aaj24.4a; = — 1. 13. 3a«h3aj + a = 0.
12. aj^^ cue 4 5 = 0. 14. ax'^ { ax \ 2 = 0.
GRAPH OF THE QUADRATIC EQUATION, ax^^bx^c = y
138. The value of the expression, ooj* ^ bx\c, for given
values of a, 6, and c, depends upon the value of the variable x.
For that reason the expression is called a function of x. If we
write aoi^ f fta? + c = y, then for every value of x there is a
corresponding value of y. "^
The values of x and the corresponding values of y may be
taken as the coordinates of points on the graph of the equation
aa? h 6a5 4 c = y.
124 ELEMENTAHY ALGEBRA
1, Draw the graph of a!* — 4 a! — 5 = y.
"
V
Point
16
A
7
B
C
6
D
8
B
9
F
8
6
S
1
I
2
7
J
8
16
. K
ThuB it will be seen (Fig. 17) that the graph
crones the xazis at the points (5, 0) and
(1,0). Upon solving ib«4i6^0, we
find tbat X = + 6 and — 1, and these an th«
values of x which make ^ = 0.
QUADRATIC EQUATIONS
2. Draw the graph of a? 

f
Poinl
6
A
6
i
B
C
3
D
2
E
1
F
In rig. 18 It is seen that tbe grapb touches
the zaxis, but does not ctobs it. The graph
tumB and proceede upward.
Instead of two points of intersection with
the xasis, as in Fig. IT, there is la Fig. 18
one point o( contact; the two points have
UDlted in a single point. Evidently there are
two roots at that point. Upon solving the
equation we find that z = 3, S ; 3 is called a
doable root.
BLBMBIfTARY ALOBBBA
3. Diaw the graph of a^ — 2x+ 3 = y.
=«
*
Point
6
18
A
4
11
B
3
6
C
3
8
D
1
2
E
8
F
_ 1
6
2
11
H
3
18
I
Fig. 19 UluHtratea the fact that not eyeiy
gnph cati the zaxis.
K we solve z>  a x 4 8= 0, we find that
K = 1 ± v'^. Now V^^ l8 an imaginary
Dumber ; therefore, there U no real valae of
z which will make x>  2 x + 8 equal to 0.
The grapb makes this fact evident at a
glance.
We see that the two niota of the eqoaUon
ax* + bz 1 c = may be real and unequal, as
in Fig. IT, leal and equal, as in Tig. 18, oi
imaginary, as In Fig. 19,
CHAPTER VI
SYSTEMS OF EQUATIONS SOLVABLE BY QUADRATICS
I. ONE EQUATION IN THE SYSTEM IS LINEAR
139. Solve
aj« + 3/2 = 26.
«  3/ = 1,
(1)
(2)
Let us solve this system, first graphically, then alge
braically.
The equation a; — y = 1 is linear and is represented by a
straight line. The graph of aj* f y* = 25 is a circle.
X
y =
1
X
y
Point
1
1
A
D
x^
+ y2 =
= 26
X
y
Point
6
C
4
±3
D,E
3
±4
F, G
±6
H,I
3
±4
K,L
4
i3
M, N
6
Fig. 20 shows the graph of x^ + y'^ = 26, (1), and «  y = 1, (2). The
line (2) intersects the circle in two points (—3, — 4) and (4, 3).
The coordinates of these points satisfy both equations and are there
fore the values sought ; namely,
05 = 4 and y = 3 ; x = — 8 and y = — 4.
127
ELEMENTARY ALGEBRA
140. Observe that x and y, when subject to the one condi
tion a? + t^ = 25, are variables, because they are capable of
taking on successively a never ending aeries of different values.
For the same reason x and y are also variables when subject to
the one condition x~y = l. But x and g can no longer aa
same successively a never ending series of different values,
■when they are subject at the same time to both conditions^
('' ~ ( . Hence x and y are now no longer variables,
x^y=\ f
but constants. It is the purpose of the solution to find the
values of these constants.
In drawing graphs we are dealing with variables; in
finding the points of intersection of these graphs, we are
finding the values of constants which, at the outset, are
unknown.
SYSTEMS OF EQUATIONS 129
141. The algebraic solution of
aj« 4. y« = 26 (1)
xy = l (2)
can be effected easiest by the method of suhstitutixm.
From (2), x = 1 + y.
Substitute in (1), 1 + 2 y + y2 __ ^a = 26.
Whence, y^ ^ y _ 12 = 0.
(y + 4)(y3)=0.
y = — 4, or + 3.
Substitute in (2), x =  3, or + 4.
The two sets of roots are, ac = — 3, y = — 4, and as = 4, y = + 3.
In checking, substitute the values of x and y in both of the given
equations ; the answers might be wrong, yet might satisfy one of the two
equations. For instance, x = 3, y = 2 will satisfy (2), but not (1).
If in Fig. 20 the line should move, parallel to its present position, until
it were tangent to the cirele, we would have just one set of values satis
fying the pair of equations. Show that this is the case, if the equations
are 052 ^ y2 ::= 26 and ac — y = 6 V2.
If the line should not intersect the cirele, nor be tangent to it, we
would have a set of imaginary values satisfying both equations. Show
that this is the case, if the equations are x^ + y^ = 25, x — y = 10.
142. Instead of elimination by substitution, some special
device is frequently used.
1. Solve ic«y» = 5, (1)
xy = b. (2)
Divide (1) by (2), x + y =  1 (3)
xy=6 (2)
Add (2) and (3), 2 x =  6
X =3
Subtract (2) from (3), 2 y = 4
y = 2. Ans. x=— 3, y = 2.
ELEMENTARY ALGEBRA
'
v=
6
j?_^ =
S
«
V
Po[nt

V
Points
5
6
A
B
3
8
«
4
±a
±2
±3.3*
±8.3*
A,Ai
S,Bi
C, C,
E,E,
Ezplaiu bow Fig.
21 exLibita the values
of X and y which
nitisEy eqnations (1)
and (2).
Square (1),
(2) 4,
Add (8) and (4),
Extract square root,
Add (6) and (1),
Snbtroct (1) from (6),
The MlB of roots are
iEys= — 5,
ixv + y^ = 36.
x' + 2aK + v^ = 16.
x + y =
a)
(2)
(S)
(<)
(5)
(«)
(1)
SYSTEMS OF EQUATIONS 131
8. Solve a' + y» = 20, (1)
xy = 8. (2)
MulUpl? (2) by 2, 2 a^, = 16. (3)
Add (1) and (8), i" + 2 37 + k^ = 36. (4)
Subtract (1) and (S), a?  2 xy + v* = 4. (6)
Extract square root of (4) itnd (6),
Whence, 2a:=±8, ±4,
a: =±4, ±2.
2y=±i, ±8,
If = ± 2, ±4.
The sets of rootfl are, i =± 4, v = ± 2; a: =± 2, v = ±*
The graphic soIuUon is exhibited in Fig. 22.
132
ELEMENTARY ALGEBRA
X^
' + y« = 20
X
y
Point
±4.47
A,Ai
±4.47
B,Bi
2
±4
C, Ci
2
±4
A A
3
±3.3+
^, ^1
3
±3.3+
F,Fi
4
±2
a, Gi
4 ±2
H,Hi
jcy = 8
X
y
Point
±2
±4
a, A
±3
±2}
/, /i
±4
±2
^,^1
±5
±^
A A
4. Solve
aj«y» = 91,
a? — 3/ = 7.
t
Divide (1) by (2),
x^^xy\y^ = 13.
Square (2),
052  2 ay ± y2  49.
Subtract (4) from (3),
3 xy =  36,
3cy=12.
Add (5) and (3),
a;2±2xy±ya = l.
Extract square root
a5 + y = ± 1
05 — y = 7.
Whence
2 X = 8, 6,
X = 4, 3.
And
2y=6,.
8,
y=3, .
4.
The sets of roots are
aj = 4,y=3;
x = 3,y
6. Solve
aj*h 2^ = 97,
a; 4 y = 5.
=4.
Let x = t« + r; y = w — t>.
Substitute in (1) and (2),
«* + 4 m8» + 6 mV 4.4Mt>8 + v*±t«*4M8v±6 m%2_ 4 u«« + «*
Whence, 2 u* + 12 u2»2 + 2 1;* = 97,
and u±t> + M — 1> = 6.
Whence, 2 u = 6,
(1)
(2)
(8)
(4)
(6)
(«)
(7)
(2)
(1)
(2)
97. (3)
(4)
(6)
SYSTEMS OF EQUATIONS 133
Substitute u = J in (4), aA + 75 »2 + 2 r* = 97. (6)
Whence, 16 r* + 600 »2 = 151.
Completing the square, 16 «* + 600 1?2 4 5625 = 5776,
4v« + 75=±76,
4 1?2 = 1,  151,
t^ = i H^»
Hence, « = 3, 2, J ± J V 161,
The sets of roots are
y = 2, 3, J q: i V 151.
05 = 3, y = 2; x = 2, y = 3; a; = J±J V 151, y = J ^ } V  151.
Check each set of real roots.
The check for the imaginary roots is as follows :
(1) 3ff.^ ii V 161  u^fiA T ^i^ V 151 + ^¥<P + W
:p ijA VIST _ xi2A ± ij4 V 161 + »3^ = 97,
iJi = 97,
97 = 97.
(2) jijVirirHf TiV^=T5l = 5,
5 = 5.
143. The equations aj*  ^ = 97 and x\y = 5 are called
symmetrical equations, because they remain unaltered when x
is written for y and y is written for x.
An expression is symmetrical with respect to two or more
letters, if it remains unaltered when the letters are inter
cttanged.
Every system of two equations that are symmetrical, or sym
metrical concept for the signs, can be solved by assuming x^u^v
and y = u — V, Instances of equations symmetrical except
for the signs are a* — y* = a, a — y = 6.
134 ELEMENTARY ALGEBRA
II. A SYSTEM OF TWO EQUATIONS IN X AND F, BOTH
QUADRATIC
144. Any tioo equations of the form aa^ ■i'by^=c,or aa? + hxy = c,
or oo* + bxy \ cy^ == d, can be solved by assuming y = vx, and
determining the constant v.
It is seen that in these equations all the terms which contain x and y
are of the second degree in z and y. Some authors call these equations
homogeneous. We avoid this term for the reason that the word homoge
neous is more commonly used to designate equations like ax^'^hxy\'Cy^=0,
in which every term of the equation contains x and y^ and is of the same
degree in x and y. See § 8.
Solve a^\xy = — l, (1)
y'2xy = S. (2)
Let y = vx.
Then x^ + rx* =  1, (3)
and fAc^^2vx^ = 8. (4)
From (3), (1 + v)x^=h and «» =^J. (6)
1 + 1?
From (4), («2  2 v)x^ = 8, and x^ = — ^ (6)
•• i + t,t^2i? . ^^
Whence,  «« + 2i? = 8 + 8», (8)
and t^ + 6i? + 8 = 0, (9)
and (v + 4)(i? + 2) = 0.
Then v =  4,  2.
Substitute in (6), x^=^ = l, and a;2 = — = 1,
K=±jV3, ±1,
y=zvx = Tiy/S, T2.
The two sets of roots are a; = ± J VS, y = =F J\/3 ; x =±1, y = T2,
To find the value of y care must be taken to use the value of v with
that value of x which was obtained by substituting the value of v.
When — 4 was substituted for v in (5), yielding x = ±}V3, the — 4
must be multiplied by ± J\/3 (since y = vx), to obtain the correspond
ing value of y.
Hence the values of x and y must be carefully paired as shown above.
When ± and =F are used in the answers, the upper signs go together
and the lower signs go together.
SYSTEMS OF EQUATIONS 135
145. The method just explained possesses the great advan
tage of always yielding results. Very often, however, much
shorter solutions can be given by special devices. Frequently
a third equation can be derived from the two given equations
which is simpler than one or both of those given. The solu
tion is then obtained by the use of this simpler equation along
with one of the original ones.
For example, solve a* — 3 ajy =
143,
(1)
2^ + ajy =
168.
(2)
By adding (1) and (2) the shnpler equation is obtained,
x^ 2icyhy2=26.
Extract the square root of both sides,
« — y = ± 6.
(8)
(3) is a linear equation. It gives
X = ± 6 + y.
Substitute in (2),
y2±5y4y2=168,
2y^±6y = l6S,
Taking the upper sign in ±6y,
_ 6 ± V1369
Taking the \ower sign in ± 6y,
y = ^'r''
Hence,
y = 8, 8,V. ¥•
Substitute the values of y in (2),
a: = 18, 13, Y, ^.
The sets of roots are a; = ±13, y = ±8; x = ±^,y=±^.
POSSIBILITY OF SOLUTION BY QUADRATICS
146. It is readily seen that when, in a system of two equa
tions, one equation is of the second degree and the other is of
the first degree or linear, a solution may always be obtained
by quadratics.
If, however, both equations are quadratics, this is usually
not the case. Only special types of quadratic equations, such
as have been studied in this chapter, and others of similar
136 ELEMENTARY ALGEBRA
character, admit of being solved by quadratics. The general
case is far more complicated. Given
'^ aal^ + bxy \ q/^ \ dx \ ey \f=
and Oiic* + b^xy + c^ + d^x + e^y +/i = 0,
to find X and y. If y is eliminated by substitution, the result
is a quartic equation, which is of the fourth degree and cannot
be solved by quadratics, except in special cases. It is shown in
more advanced algebras that the algebraic solution of a quartic
equation depends in general upon the solution of a certain
cubic equation or equation of the third degree. The solution
of cubic and quartic equations is not explained in this book.
The given equations of the second degree in x and y can be solved by
quadratics whenever the auxiliary cubic equation here mentioned possesses
a rational root. This rational root can be found by the factor theorem
(see § 78) ; the other two roots of the cubic can then be found by quad
ratics, as can also the four sets of roots of the given equations. (See
F. Cajori, Theory of Equations, New York, 1914, pp. 7173.)
BXBBCISBS
147. Solve
1. 2 ic* — 3 a?y = 5, 6. ic* + 3 a?^ = — 5,
oj — y = 2. 2xy + y^ = 'l^,
2. a?y = l, 7. a + 2/ = a,
iK* + y2 = 85. 4:xy — a^=z — b^,
X y 15' x^ — y^ = 4: mn.
1_^1^_34 9. aj*y* = 2401,
a^ f 225' 352^2/2^49.
4. aj» H y» = 28, lo. ar' + 7 a?y =  104,
a + y = 2. 5xy'y^=:129.
6. xy = 45, 11. a5 + y* = 33,
aj + y = 14. x + y = 3.
SYSTEMS OF EQUATIONS 137
12. xy = ly 15. aj3 I y3 «. 109^
a , y _. 2 J 05*^ + ojy* = — 36.
13. L3y = 0, "• fy'=10.
7a^ + 3a,4.V = 43. 2a + 6;.^ + 33^' = 14.
14. a:^'y^ = m^y 17. 3aj232/2 = 9,
18. a^—7x + y^—7yy
3(a + 2/) = 2icy.
PROBLEMS
148. 1. Prices of two kinds of bicycles are such that 7
of one kind and 12 of the other kind can be obtained for
$640. Three more of the latter can be purchased for $180
than can be purchased of the former for $120. Find the
price of each.
2. A person lends $ 2500 in two separate sums at the same
rate of interest. At the end of one year the first sum with
interest amounts to $997.50; at the end of two years the
second sum with (simple) interest amounts to $1705. Find
the two separate sums and the rate of interest.
3. Prove that if the sum of two real numbers is multiplied
by the sum of their reciprocals, the product cannot be less
than 4.
Let X and y be the two real numbers, p the product considered ;
solve for the unknown ratio .
y
4. A traveler starts from A toward B, another traveler
starts at the same time from B toward A. In two hours they
meet 20 miles from A. When the second traveler arrives at
A, the first is still 13^ miles from B. Find the distance be
tween A and B.
138 ELEMENTARY ALGEBRA
6. If in aoc^ + to  c = 0, the coefficients are related to each
other in such a way that a h 6 = ^ and a = 2 c, what must be
the value of a, so that 8 will be a root of the given quadratic
equation?
6. The difference between two numbers is 12, and the
difference between their cubes is 7488. What are the two
numbers ?
7. A sum of money at simple interest for four years
amounts to $2240. Had the rate of interest been 1 % higher,
the sum would have amounted to $80 less than this in two
years. Find the capital and the rate.
8. The sum of the areas of two circles is 694.2936 sq. in. ;
the sum of their radii is 21. Find their radii.
9. A circular track is constructed so that the width of the
track is ^ of the inside diameter. The area of the track is
2600 sq. yd. What are the inside and outside lengths of the
track?
10. Find two numbers such that their sum is equal to their
product and also to the difference of their squares.
11. Find two numbers such that the sum of the numbers is
equal to the sum of their squares and also to twice their
product.
12. Find two numbers whose sum is 8 and whose product
is 26.
13. Find three numbers such that the sum of the squares of
the first two is equal to three times the square of the first minus
the square of the second, and is also equal to the first minus the
second plus twice the third, and also to the sum of the three
numbers.
CHAPTER VII
EXPONENTS, RADICALS, IMA6INASIES
MEANINGS OF DIFFERENT KINDS OF EXPONENTS
149. The different kinds of exponents which have been stud
ied thus far have been interpreted in the following manner, m
and n being taken to be positive iategers :
a" = a • a • a ••• (to n factors),
1  1
a~* = ^i or more generally, a " = — ;;;, where a ^ 0,
a •»
a° = 1, where a =^ 0,
While a number a has in general n different nth roots, it is agreed that
1
only one of them shall be represented by a** or Va, namely, the socalled
principal root This restriction was made in order to avoid unnecessary
complication and confusion in the interpretation of expressions and equa
tions involving radicals. Accordingly, Vi = + 2, — Vi = — 2, \^ = + 2,
\/8=2, \/^^ = 2, v^^=42, v^=+2, etc.
The exponents considered above are all rational numbers. But other
exponents have been brought to our attention. In the study of logarithms,
mention was made of the fact that logarithms are often irrational numbers.
Since logarithms are realjy exponents, it follows that exponents may be
irrational. In more advanced books still another kind of exponent is
considered, namely, the exponent that is an imaginary number.*
* For a fuller treatment of the theory of exponents, consult H. B. Fine,
College Algebra^ 1904, p. 376.
139
140
ELEMENTARY ALGEBRA
OPERATIONS WITH EXPONENTS
150. Operations involving exponents are subject to the fol
lowing laws :
1. Law of multiplication : a"^ a'* = a"*"*^.
2. Law of division : a"^ ^ a"" = a"*"".
3. Law of involution : (a*)* = a"»*.
1 m
4. Law of evolution : (a") ** = a" .
DIFFERENT KINDS OF NUMBERS
151. In the study of arithmetic and algebra several different
kinds of numbers have come to our notice. These may be
classified as follows :
Numbers
Real
(Classified according to signs)
(Classified according to express
ibility in HinduArabic nu
merals)
Positive
Negative
Rational
Irrational, as
V6, ir =
3.14169...
Imaginary
Pure Imaginary, as hV— 1, where &^0
Complex Numbers, as a\hy/— 1, where 6^0
Arithmetic deals with real numbers, all positive, but some of
them irrational.
The irrational numbers of arithmetic arise in the process of
finding roots.
In algebra use is made of the numbers encountered in arith
metic, but convenience forces upon us the need of considering
also negative numbers for the purpose of indicating relations of
opposition, as temperatures above or below a certain fixed point,
debts or assets, distances to the right or left, etc. In the solution
of quadratic equations we meet a still different type. If we try
to solve ic2 . 1 = 0, we are confronted with the symbol V— 1.
EXPONENTS, RADICALS, IMAGINARIES 141
This result greatly embarrassed mathematicians of the eight
eenth and of earlier centuries ; no satisfactory explanation of
it could be made at first. But now the symbols are recognized
as constituting a new type of number, the socalled imaginary
number, which deserves a legitimate place in algebra and is of
great service in certain advanced developments of algebra, that
are useful in the study of polyphasy electric currents and of
other advanced topics in mathematical physics. Just as nega
tive numbers have been found truly useful in elementary
algebra, so imaginary numbers have been found useful in ad
vanced algebra. The symbol V— 1, or this multiplied by
any real number 6, such as 6V— 1, is called a pure imagi
nary number.
Expressions of the type a ± 6V— 1, which are the sum
or difference of a number a and of a pure imaginary
b^/— 1 (a and b being any real numbers, but b ^ 0), are
called GfmSplex numbers. The term " complex " was intro
duced because the parts of the expression are partly real
and partly imaginary.
SIMPLIFYING RADICALS
152. A radical is said to be in its simplest form :
(a) When the index of the root is as small as possible,
(b) When the expression under the radical sign, called the
radicandy is integral,
(c) When the radicand contains no factor with a negative ex
ponent, or raised to a power equal to or greater than the index
of the root.
\/25 is not in its simplest form, because the index 4 of the root is not
as small as possible ; we have y/2b = 6* = 5* = V5.
^\ is not ill its simplest form, because the radicand, \, is fractional.
y/c^ is not in its simplest form, because the exponent 4 is greater than
8, the index of the root ; we have ■\/a*6 = ay/ab.
\/3 ab^ is in its simplest form, since it fulfills all three conditions.
142 ELEMENTARY ALGEBRA
1. Simplify J^^.
3 gi gj 8 3 *
Notice that the denominator was rationalized by multiplying numer
ator and denominator by 8^.
2. Simplify </3 a*b^c.
6* ft ft
3. Other illustrations of simplifying are :
'^7a« ^ 3^ g* _ 3^ ■ q ■ aM _ a j/Wab^
V72  36 VS = 36^(2  v^)* = 6 V2  V3.
BXBBOISBS
153. Remove all negative exponents, compute the values, and
simplify the expressions :
1. 27*. 4. 16"*. 7. 64"*. ^^' (A)"'
2. 16* 5. 64i. 8. (!)». 11. (.16)*.
3. 27*. 6. 64^. 9. 8"*. 12. (.125)*.
13. (_ + 6)0. 16. (100«+10)^ ^g gmn ^ gmft
5«25\
14. 70x(.26)*. 17 — 1 20. ^""^
16. (.49/'. 18. g'^g'". 21. «:l±«^+l.
EXPONENTS, RADICALS, IMAGINARIES 143
22. V60. 25. V2000. 28. VST. 31. V 800.
23. Vm. 26. 7V20. 29. \/48. 32. v^729.
24. V98. 27. 12 V8. 30. ^"08. 33. \/2450.
34. 2V6x5>/3. 38. 3Vlbx24V6fVl5.
36. 5V2x7V6. 39. 3 ViO X 2 Vl5 4 3 V2.
36. 2Vl6 X 5V2 X 2V3. 40. 5Vl6 X 3V6 4 2V5.
37. 2Vl5j5V3x2V2. 41. (6.25)*.
42. (2.25)1 46. (.216)1 60. * (^)*.
43. (.25)"*. 47. (.0625)*. 61. ~ (6.25)i
44. (.125)"*. 48. (.125)*. 62. (a*6*)«.
46. (.008)*. ^49. (fj)*. 63. (a26»)i
64. (a«6»)"'^. S6. c^^c^xcK
66. a^^a^y. a*. 67. 2* x 64* 4 729*.
Eationalize the denominators of the following fractions and
then find their values to three figures :
58. i. Hint. J=i..^ = ^. 69. ^
V2 V2 V2 V2 2 2V3
60. 4= 61. 4=r 62. V. 63. ^^^.
Vl4 V2.3 V:22
Calculate the values of the following expressions when
m = V2, n = V3, p = V5, using the tables in § 197 :
64. 7 m* — n*. 67. phi^ — nha\ 70. (n+J>)w.
66. 1©* — wiV. 68. — ^ 71. ^
^^ 2p pn
« • . « s «A wi — w »»« ^* — ^**
66. mH^\rM. 69. ^ ^  72.
5jp* — mnp
144 ELEMENTARY ALGEBRA
78. Which of the expressions is the larger, V5 or v/lT ?
V6 = 5* = 5* = 126^
\^ = 11*=11* = 121*.
But 126^ > 12li Hence V5 is the larger.
Arrange in the order of magnitude in each example :
74. \/ii, </2l. 77. 2, Vily v^.
76. V2, v^^. 78. •^, ^/Vf, VO.
76. VO, \/9. 79. \/3, 5^, vT7.
80. Using logarithms, compute the value of y/7943.
Log 7943 = 3.9000.
Log \/7943 = 1.8000.
v'mS = 19.96, nearly.
81. Using logarithms, compute the value of x =(59.45)"^.
(59.46)"* = — ^
(69.46)
I
Log 1 = 0.0000
Log 69.46 = 1.7742.
Log (59.46)' = j(1.7742) = 1.1828
Log X = 2.8172
X = .00664, nearly.
Compute the values of the following radicals :
82. ^/96S, 84. ^790. 86. (154)"*.
83. VTOSS. 85. (12)A 87. (376)J.
ADDITION AND SUBTRACTION OF RADICALS
164. The indicated additions and subtractions of radicals
can be reduced to one term, only wh^n the terms in the ex
pression are similar after they are simplified.
Thus, \/l8 + V60 = SV^ + 6 V2 = 8 V2;
EXPONENTS, RADICALS, IMAGINARIES 145
EXBBCISBS
155. Perform the indicated operations :
1. V63+\/784. 3. A/24 + </8i+\/375.
2. V294+v^576. 4. 50\/^^+<^7000^56.
6. Vr^  V9a^  V16 a^dP.
6. 6V3+8VfV^.
7. 4V^15v^60V.
10. w3\/«S^+wV\/.
MULTIPLICATION OF RADICALS
\
156. E.eal radicals of the same index are multiplied to
gether according to the formula
a y/b X c y/d = ocVftd.
When the real radicals have different indices^ it is easiest to
change to the exponential notation, then perform the multipli
cation and, after reducing fractional exponents to a common
denominator, return to the radical notation. This may be
shown as follows :
m n
Special care must be exercised when the expressions to be
multiplied involve imaginary numbers. It is customary, for
brevity, to represent the pure imaginary V— 1 by the single
146 ELEMENTARY ALGEBRA
letter. t. Hence the complex number a+V— 16 may be
written a f ib.
The following has been adopted as a definition of V— 1 or f :
7%e number V— 1 is one whose square is — 1.
Accordingly, (V— 1)* = i* = — 1.
All multiplications involving imaginary numbers must be
made to conform with this fundamental definition.
Particular care must be exercised to avoid the following procedure:
V^n. times \/^n[=V(l)(l) = >/TT= 1.
This procedure is in violation of the definition of V— 1 given above,
and must therefore be avoided. Always take V— 1 x V— 1 = — 1, or
The multiplication of expressions involving imaginaries can
be performed easiest^ if we take the precaution to vrrite a radical
V— a in the form V— 1 Va or i Va.
For example,
V— a X V— 6 = i Va X I* V^ = i^ Voft = — VoS.
1. Multiply 2 Va  3 V26 + 4 V3 a6 by 5 Va6.
2 V53\/26 + 4 V3a6
10aV6166 V2a + 20a6V3
2. Multiply Va + 2 V 6  3 c by Va6.
Va + 2f>/63c
tVa6
ai
iVb^2hy/aSciy/ab
EXPONENTS, RADICALS, IMAGINARIES 147
BXBBOISBS
157. Perform the indicated operations and simplify the
result :
1. V5.V125. 18. V2^^V2T«.
2. \/3\/9. 14. ^^r^.vsTft.
8. SVa.SVoft. 15. (V2a+V36)».
4. ^cC^iy/a^xVaK le. (a+V6)».
6. ^ ! Va X \/a». 17. (a + V^^)*.
6. aV^x26V^=n:. 18. Vab^Va.
7. 3i.4iV5. 19. </^^Va6.
8. 4iVc"j2t. 20. SVx^^'^V23r^.
9. 5V^V3x6V^^. 21. V^^^a64V^=^
10.  M2xV. 22. iaXi6V2xic.
11. (^abVcy. 23. V^Ta . v"=^ . V^=^.
12. — V2a' X V2a&. 24. t V« • t V^ • * Vy2.
26. ^J^^g X \/^^ X ^v^^=^.
26. (2Vi — 3Vy+V2«)5Vc^.
27. 6^^(2V«+Vy«).
28. (2V23V^^H^v/"=^2</2)\/2.
29. (2V5</62V^2.
80. (VssV^^)*.
81. Change to an equivalent fraction having a rational
V3+V2.
denominator,
V3V2
If >/3 >/2 is multiplied by VS + V2, the product is (>/3)2(V2)2
or 3 — 2. Hence, multiply both numerator and denominator of the given
fraction by VS + V2. We obtain
(V3+V2)2 ^ 8 + 2V6h2 ^ 6 + 2v^ ^g ^ g^/g
(V3)2_(v^)2 32 1
148 ELEMENTARY ALGEBRA
Change to equivalent expressions having a rational denomi
nator:
82. ' ^ . 34. ^^A 86. t±^.
V3VO VaVft 2hV3
33. — 1 36. ^■^^. 37. ^
V64.V5 V2V6 2V53V3
38. Is — 1 h V5 a root of the equation aj2 42aj — 4 = 0?
Write  1 + V6 for X,
(_ 1 4. V6)a + 2( HV6) 4 = 1  2V6 + 6  2 + 2V6  4 = 0.
Hence, — 1 + V6 is a root of the equation.
Is the binomial written after each equation a root of that
equation ?
39. aj»4ajl = 0, 2hV3.
40. aj2a;3 = 0, (lVi3).
41. a;2h3a;hl = 0, 3hV5.
42. 2aj2aj2 = 0, i(lVl7).
THE SQUARE ROOT OF a±2y/b
158. Since ( V3 ± V2)«= 3 ± 2 V6 + 2 = 5 ± 2 V6, it follows
that V6±2V6 = V3± V2.
We notice that 5 is the sum of 2 and 3 ; 6 their product.
This suggests the following rule for finding the square root of
ah2V6:
Find two numbers whose sum is a and whose product is b.
Write down the square root of the larger number^ plus (or minus)
the square root of the smaller.
This rule is of practical use only when the two numbers
sought are rational. In other cases the resulting expression is
more complicated than the given expression; hence to be
rejected.
EXPONENTS, RADICALS, IMAGINARIES 149
1. Find the square root of 8 + V^O.
Writing the expression in the form a ± 2Vbj 8 + 2\/i6.
The two numbers whose sum is 8 and product 16 are 6 and 8.
Hence, V8+V60 =Vb + VS.
2. Find Vg  4 V2.
Changing to the form a ± 2 V6, 6 — 2v^.
The two numbers whose sum is 6 and product 8 are 4 and 2.
Hence, Vo 4V2 = Vi  v^ = 2  V2.
BXBBCISBS
159. Find the square root of :
1. 3h2V2. 7. 9h6v^.
2. 74V3. 8. IV2.
3. 9h2Vi4. 9. t^VJ.
4. 9 42V20. 10. X — Vaj»  1.
6. 16 46V6. 11. (a2h4 6)4aV6.
6. 8h2V7. 12. x^2y + 2V2xy.
IRRATIONAL EQUATIONS
160. An irrational or radical equation is an equation in which
the unknown number x and expressions containing it occur
under radical signs or with fractional exponents.
For example, 2 — V« = Va? — 1, aj^  5 «» = 6.
It is agreed among mathematicians that in equations of this
sort, only the principal roots ot^/x, Va; — 1, x^ are to be taken.
The solution of irrational equations may be explained by
examples.
150 ELEMENTARY ALGEBRA
1. Solve 2  Vic  Va?  1 = 0.
Solution. Transpose one of the radicals, 2 ^ Vx = y/x— 1.
Square both sides, 4 — 4 V« + x = x — 1.
Simplify and transpose to the right all
terms except 4 Vie, 4>/x = 5.
Sqoare both sides again and solve for x, ^ = il'
Substitute in the given equation, 2 — Vf  = V{J — 1,
Hence x = f  is the value sought.
»=t
2. Solve 1 h Va; h 3 = V«.
Square both sides, 1 + 2Vx + 3 + x + 3 = x.
Simplify and isolate the radical, Vx + 8 = ^ 2.
Square both sides again, x + 3 = 4«
x=l._
Substitute x in the given equation, 1 + \/4 = VI,
1 + 2 = 1.
This is absurd ; x = 1 is not a root of the given equation. This is an
example of a radical equation which cannot be satisfied by any value of x.
The equation is impossible. It is therefore not true of radical equations
that values of x satisfying them always exist.
It will be instructive to examine just how the false value x = 1 was
obtained. When both sides of an equation involving x are squared, new
values of x may be introduced, giving rise to extraneous roots. By sub
stitution we can ascertain in which place the extraneous root came in.
It is found that x = 1 does not satisfy equations (1) and (2), but it does
satisfy 3. Hence, the extraneous root x = 1 was introduced when both
sides of the equation Vx + 3 = — 2 were squared.
Extraneous roots arise here in the same way as in the solution of frac
tional equations. One simple way of determining whether a value of x
obtained by squaring both sides is an extraneous root or not, is by sub
stitution in the original equation ; if it is not satisfied, then the root is
extraneous.
From the explanation given above it is evident that, though all opera
tions in finding x may be performed without error, the answer may never
theless be false. It is therefore necessary to substitute the value of x in
the original irratiohdl equation, to ascertain whether that value is true or
false.
EXPONENTS, RADICALS, IMAGINARIES 151
While x = l IB not a root of 1 + Vx + 3 = Vx, it will be noticed that
it 18 a root of 1 — Vx + 3 = — Vx. The solution of the latter equation
gives rise to equation (3), and to x = 1.
3. Solve Vx'^ h V3a;h3 « 3. (1)
Transpose so that the most complicated radical stands alone on one
side of the equation, , ,
' Va;23=V3a; + 3. (2)
Square both sides, a; — 2 — 6y/x — 2+9 = 3a; + 3. (3)
Simplify, keep only the radical on left side, — 6 Vx — 2 = 2 x — 4. (4)
Divide both sides by 2 — 3 Vx2 = x  2. (6)
Square both sides, 9x— 18=x3— 4x+4. (6)
Solve the quadratic x^  13 x + 22 = 0.
(x2)(xll)
X = 2, X = 11.
Substitute x = 2 in (1), y/2^^ + V9 = 3.
3 = 3.
Hence X = 2 is a root of (1).
Substitute x = 11 in (1), Vll2 + V33 + 3=3.
3 + 6 = 3.
Hence x = 11 is not a root of (1). Nor is it a root of (2), (3), (4), and
(6) . But it is a root of (6). It follows that x = 11 is an extraneous root,
obtained in squaring (6).
4. Solve V6 — a; h V« + 7 = 6.
Place one radical on one side by itself, V6x = 6— Vx+7.
Square both sides, 6 — x = 26 — 10 Vx + 7 + x + 7.
Simplify and keep only the radical on left side.
10Vx+7 = 2x + 26.
Divide both sides by 2, 6 Vx+7 = x + 13.
Square both sides, 26 x + 176 = x^ + 26 x + 169.
Solve the quadratic x* + x — 6 = 0.
(x + 3)(x2) =0.
X =  3, X = 2.
152 ELEMENTARY ALGEBRA
Substitute x = ^ 3 in the given equation,
>/9 + \/4 = 6,
3 + 2 = 6.
Hence z = — 3 is a root of the given equation.
Substitute x = 2 in the given equation,
V4+\/9 = 6,
2 + 3 = 6.
Hence x = 2 is also a root of the given equation.
BXBROISBS
161. Solve the following irrational equations, testing care
fully each value of x that may arise :
1. Va? + 4 = 4— Vof — 4.
2. Va?2+Vaj3 = 6.
3. Va;— l4Vah6 = V4aj9.
4. Vic+Vx+l = V2a; 1,
6. Vy^^ = VyhV2.
6. vVhlVy+l = 0.
7. 4V5aj* 5a: 19 + 3 = 31.
8. Vaj* + 21+a=21.
IRRATIONAL EQUATIONS QUADRATIC IN FORM
162. The equation a;« + 5 «» + 6 = is said to be quadratic
in fomiy because it contains only two different powers of x, one
exponent of x being double the other exponent.
If we put x^ = y, then %' = y^ and we obtain the quadratic
y^ + 5y + Q = 0.
Solving this we obtain y = — 2 and 2( = — 3.
Hence, x ■ = — 2 and a;» = — 3.
Cubing both members, x=—S and a; = — 27.
EXPONENTS, RADICALS, IMAGINARIES 153
Substituting x = — 8 in the given equation,
(_8)*+5(8)* + 6 = 0.
Simplify, (2)2 + 6(2)+6=0,
4  10 + 6 = 0.
Hence, x = — 8 is a root of the given equation.
Substituting x = — 27 in the given equation^
(27)*H6(27)t + 6 = 0,
(3)2+6(3)+6 = 0,
916 + 6 = 0.
Hence, x = — 27 is a root of the given equation.
Solve a:*  7 «* = 8.
The exponent  is double the exponent } ; hence the equation is quad
ratic in form.
Let X* = y, then x* = ^,
Solving, y2 _ 7 y _ 8^
y = 8, y = l.
We obtain x' = 8 and x^ = — 1. (1)
The exponent f in x* signifies the third power of the fourth root of x ;
we wish to find x itself.
To find the value of x^, we must extract the cube root of both sides of (1).
We obtain x* = + 2 and x^ = — 1. (2)
To find the value of x, we must raise both sides of (2) to the fourth
power.
We obtain x = 16 and x = 1.
Substitute x = 16 in the given equation, (16)^ — 7(16)^ = 8,
(4)3_7(2)« = 8,
64  66= 8.
Hence, x = 16 is a root of the given equation.
Substitute x = 1 in the given equation, (1)^ — 7(1)* = 8,
17 = 8.
Hence, x = 1 is not a root of a given equation. Where did this false
value enter ? We see that x = 1 satisfies neither x* = — 1 nor x* = — 1 ;
hence this extraneous root appeared when the sides of x^ = — 1 were
raised to the fourth power.
154 ELEMENTARY ALGEBRA
In solving (1), we might have reversed the two operations, by first rais
ing both sides to the fourth power and then extracting the cube root. But
the operations would have involved larger numbers iu the case of %* = 8.
Be careful to avoid mistakes in solving equations like x* = 8.
A frequent error is to conclude from x* = S
that X = 8*.
As a matter of fact, x = 8>.
The safer way is to proceed by two steps as is done above :
The first step (extraction of the cube root) changes the exponent of x
from i to }.
The second step (raising to the fourth power) changes the exponent
of x from ^ to 1.
*
BXEBOI8B8
163. Solve:
1. aj*7xl410 = 0.
2. 8ajl66ajl48 = 0.
3. 2* 32* + 2 = 0.
4. (a? h 1)*  ll(a; + l)i 4 30 = 0.
6. a;*a;i + 132 = 0.
6. y* h 19 yA 216 = 0.
7. (a; h 1)»  60(a; + l)i = 256.
8. a;*h6a?*— 55 = 0.
GRAPHIC REPRESENTATION OF COMPLEX NUMBERS
164* The graphic representation of ordinary positive and negative
numbers is familiar to all students of algebra. On a straight line,
usually drawn horizontally, a point is chosen as the starting point or
origin ; positive numbers are shown by distances to the right, negative
numbers by distances to the left.
The graphic representation of complex numbers like a + 15, where
i = \/— 1, is not so obvious. In the seventeenth century John Wallis of
EXPONENTS, RADICALS, IMAGINARIES 155
the University of Oxford made attempts to find a geometrical interpre
tation of such numbers, but was not able to devise a general and con
sistent scheme. Not till the close of the eighteenth century did a
satisfactory plan suggest itself to mathematicians. In 1797 a Norwegian
surveyor, by the name of Caspar Wessel, found a graphic representation
which agrees with that now adopted. But his writings failed to attract
the notice of mathematicians. A little later a Frenchman by the name
of Argand had a similar experience. More successful in reaching the ear
of the mathematical public on this matter was the German mathematician,
Carl Friedrich Gauss. He placed the theory of complex numbers on a
firm basis and, by his reputation, induced others to enter upon a more
careful study of this subject and to adopt the geometrical interpretation
that is given now in works on algebra.
4, 3,2, i,
■I i 1 H
+1, +2, +3, +4, +6,
■H 1 1 1 H
•2/1
Fio. 23.
165. This interpretation of imaginary or complex numbers
is really quite simple. Two axes are drawn perpendicular to
each other, one axis horizontal, the other vertical, as in Fig. 23.
Along the horizontal axis real numbers are marked off, positive
156
ELEMENTARY ALGEBRA
numbers to the right and negative numbers to the left of the
origin. The pure imaginary numbers V— 1, 2V— 1,
f 3V— 1, ••• are marked off along the perpendicular axis, from
the origin up. The pure imaginary numbers — V— 1,
— 2V~ 1, — 3 V— 1, ••• are marked off along this same
perpendicular, but from the origin douni.
Complex numbers are pictured in this diagram by points in
the plane. In case of 3 f 1 4, measure off three unit distances
on the horizontal axis to
the right, and 4 unit dis
tances up. This locates the
point A in the plane (Fig.
24). The point ^ in the
plane is accordingly the
graphic representation of
3 + *4.
To find the geometric
representation of 3 — 1 4,
pass from the origin 3 units
to the right, as before, and
4 units down; the point B
in Fig. 24 represents 3— i 4.
Similarly, C represents
— 3 — i 4, and D represents
3 4i4.
The representation is similar for other complex numbers.
It is seen that every complex number can be represented in this
way, and that every point in the plane stands for some number.
All real numbers are confined to the horizontal axis ; all pure
imaginary numbers are confined to the vertical axis ; all other
numbers a ± ih lie in one or another of the four quadrants.
The axis for the pure imaginaries is used in much the same
way as the yaxis when the graph of an equation containing
the variables x and y is drawn. Only now the vertical axis
carries imaginary numbers, while before the yaxis carried real
values of the variable y.
"
""
1
■\ ,
A
J
J ^
^
A
f^ 1
, T
J<
'
>u
Fig. 24.
EXPONENTS, RADICALS, IMAGINARIES 157
BXBBCISBS
166. Locate the' points whicli represent the following com
plex numbers :
1. 5 4* 4. 4. 5 — 16. 7. 3_4V31.
2. _5_i6. 6. 1h*. 8. 445y^^.
3. 5 4* 6. 6. 6 4t6. 9. V3 + V^.
10. (2ht3)h(l4i2).
IHint. (2 + i 3) + (1 + i 2) = 3 + <6 ; locate 3 + » 5.]
11. (4 + t)(2 + i2).
12. (5i2)4(3t4).
13. (6t5)'(6i7),
CHAPTER VIII
SBRISS AND LIMITS
ARITHMETICAL SERIES
167. The numbers 6, 9, 13, 17, 21 appear upon examination
to have been selected according to some law and arranged in a
definite order. Each number after the first is greater than the
one immediately preceding by 4. Such a regulated succession
of numbers is called series, , When, as here, the increase is the
same throughout, the series receives the special name of
arithmetical series or arithmetical progression.
An arithmetical series is a succession of numbers in which
each number after the first minus the preceding one always
gives the same difference.
This difference is called the common difference. Instead of
increasing, the numbers in the series may decrease, so that the
first number is the largest and the common difference is
negative.
Arithmetical series are frequently encountered in the study
of mathematics, hence it is desirable to develop certain formulas
relating to such series.
BXBBCISBS
168. State which of the following series are arithmetical
series :
1. 10,8,6,4,2. 4. a, ahd, a2d, a43d.
^ 9 7 5 3 1
2. i, 1, H, 2, 2^, 3. 5. _,_,_,__.
3. 10, 9, 7, 6, 4. 6,x,x — y,x—2y,x^S y,
158
SERIES AND LIMITS 159
THE LAST TERM
169. If a stands for the first number or term of an arith
metical series, and d for its common difference^ then the series
may be written in general terms thus,
a, a h d, a h 2 d, a + 3 d, a h 4 d, etc.
Let I stand for the last term of the arithmetical series. The
value of the last term evidently depends upon three things ;
namely, the value of a, the value of d, and also the number of
terms in the series. Denote the number of terms by n.
It is to be observed that the coefficient of d in the second
term is 1, in the third term is 2, in the fourth term is 3, in the
fifth term is 4. If n denotes the number of terms, what must
be the coefficient of d in the last term ? From what we have
observed it must be one less than the number of the term, that is,
n — 1. Hence we have the formula for the nth term,
I = a +(n  l)d. (A)
BXBBCISBS
170. 1. Find the 10th term in the arithmetical series, 2, 7,
12, 17, . . .
Here a = 2, d = 6, w = 10.
Substitute these values in (A), Z = 2 +(10  1)6,
« = 47.
Hence the 10th term is 47, as may be verified by writing down all the
terms to the 10th term.
2. Find the 16th term of the series —2, —4, —6, — 8, • • •.
3. Find the 12th term of the series 1, 1^, 2, 2^, 3, • • ..
4. Find the 20th term of the series 5, 5 \2x, 5 4 4 «,
6. Find the 24th term of the series V2, V2 4. 1, V2 h 2,
V2+3, ....
160 . ELEMENTARY ALGEBRA
6. Find the (n — l)th term of the series 3, 6, 9, 12, 15, • • •.
Find the expression for any given term (the nth term) of
the following series of numbers ; .
7. 3, 6, 7, 9, . . .. 10. I, V^, IVV, • • •.
8. 7, 12, 17, 22, . . .. 11. 66, 59, 62, • • ..
9. 13, 17, 22, 26^, . . .. 12. 2.4, 2.1, 1.8, • • ..
13. Find the (n — 2)th term of the series 5, 1, — 3, — 7, • • .
14. A bullet is fired vertically upward so that at the end of
the first second it has a velocity of 200 ft. per sec, at the end
of the second second a velocity of 168 ft. per sec, at the end
of the third second a velocity of 136 ft. per. sec, and so on.
Compute the velocity at the end of the sixth second, at the end
of the tenth second. Interpret the second answer.
15. A body falling from rest falls 16 ft. during the first
second, 48 ft. during the second, 80 ft. during the third, 112
ft. during the fourth, and so on. How far will it fall during
the ninth second ?
ARITHMETICAL MEANS
171. The arithmetical means between two numbers are num
bers which, together with the two given numbers as first and
last terms, form an arithmetical series.
If the two given numbers are 5 and 50, then 14, 23, 32, 41
are four arithmetical means, because 5, 14, 23, 32, 41, 50 is an
arithmetical series.
EXBSBCI8BS
172. 1. Insert six arithmetical means between 7 and 63.
The six terms to be found, and the given numbers 7 and 63, will make
8 terms.
We have n = 8, a = 7, Z = 63. Substitute in (4),
63 = 7+(8l)d,
63=7 + 7d,
d=8.
Hence the required series is 7, 15, 23, 31, 39, 47, 55, 63.
SERIES AND LIMITS 161
2. Insert 5 arithmetical means between 6 and 72.
3. Insert 4 arithmetical means between 7 and — 23.
4.' Insert 8 arithmetical means between 7 and S^,
5. Insert 3 arithmetical means between x and y,
6. Insert 7 arithmetical means between V7 and 10V7.
7. Insert an arithmetic mean between 100 and 133. *
SUM OF AN ARITHMETICAL SERIES
173. The sum of the terms of an arithmetical series can
always be found by writing down all the terms, and adding
them. But, if the number of terms is great, this operation is
quite laborious. We proceed to derive a formula by which the
sum of a large number of terms may be computed with less
labor.
Observing that, if the last term is I, the term immediately
preceding may be written I — d, the term before this I — 2dy
and so on, we may indicate the sum of the series thus,
^ = a+(a + d) + (ah2d)+ ... +(Z2d)h(Z d)+L (1)
Reverse the order of the terms in the right side of (1),
^=:Z+(Zd)4.(?2d)h ... h(a + 2d)+(a + (f)+a. (2)
Adding (1) and (2),
In (3) there are as many parentheses (a + 1) as there are
terms in the series ; hence.
162 ELEMENTARY ALGEBRA
BXBBCI8BS
174. 1. Find the sum of 20 terms of the series 99, 103, 107,
111, ....
Hero n = 20, a = 09, d = 4.
Henoe, I = a + (fi  1)<2 = 99 + 19 x 4 = 175,
and ^ = (a+Z) = y(99+176) = 10x274=2740.
The roqaired sum is 2740.
2. Find the sum of 25 terms of the series — 60, — 46,
 40, • . . .
3. Find the sum of 18 terms of the series 11, 8, 6, 2, • • • .
4. Find the sum of the first 200 integers, 1, 2, 3, • • • .
6. Find the sum of all the integers between 60 and 76,
excluding —50 and 75.
6. Find the sum of 15 terms in 1^, If, 2, 2f, • • •.
7. In formula (J3) substitute for I its value as given in
formula {A) and derive a formula for S which does not con
tain L
8. Find the sum of the first one hundred odd numbers.
9. Find the sum of the first one hundred even numbers.
10. How many of the integers 1, 2, 3, • • • must be added to
yield the sum 55 ?
Use the formula ^9 =  (2 a +(n  l)(l), derived in Ex. 7. Do both
answers satisfy the conditions of the problem ?
11. How many terms of the series 5, 4, 3, • • . are necessary
to yield a sum 9 ?
Do both answers satisfy the conditions of the problem ?
12. The second term of an arithmetical series is 11, the
fifth term is 20 ; find the 14th term.
13. The first week a store was opened the expenses ex
ceeded the income by % 52.25. The second week the loss was
SERIES AND LIMITS 163
$41.75. If the improvement in the trade continued at the
same rate, how much profit was made in 24 weeks ?
14. A man enters an office at a salary of $ 1200, which is
increased annually $ 75. How much will the firm pay him
during 18 years ?
15. The three formulas found below give the salaries offered
by three companies to men entering their employ. S is the
monthly salary in dollars earned after a given number of
years (n). Calculate which company after 20 years' employ
will give the highest salary. How much does a man earn in
20 years in each case ?
(a) 5'=100 4w, (b) /S' = 95 44n, (c) S=SS+^n.
GEOMETRICAL SERIES
176. The series 3, 6, 12, 24, is not arithmetical ; the dif
ference between successive terms is not the same. The
successive terms are formed in accordance with a diflerent law.
It is readily seen that any term after the first is derived from
the preceding one by multiplying by 2. Such a series is called
a geometHcal series or a geometrical progression.
A geometric series is a succession of numbers in which each
number after the first, when divided by the preceding number,
always gives the same quotient.
The quotient is called the common ratio.
EXERCISES
176. Which of the following series are geometrical and
which arithmetical?
1. 7, 35, 175, .... 5. 2,  4, 8,  16, • . ..
2. 6, 12, 18, .... 6. 2, 1, 0,  1, • • •.
3. 2, 8, 32, .... 7. V2, 1, V, . • .
4. 2, 8, 32, 40, .... 8. 2i, 2i, 2f , . . ..
164 ELEMENTARY ALGEBRA
LAST TERM OF A GEOMETRICAL SERIES
177. If the first term of a geometrical series is a, the com
mon ratio is r, the number of terms n, then an expression for
/, the last term, may be obtained by inspecting the terms in
the general geometrical series,
a, ar, ar^, ar*, ar^^ • • ••
We observe that in the second term, the exponent of r is 1, in
the third term it is 2, in the fourth term it is 3, in the fifth term
it is 4. Evidently, in the nth term, the exponent of r is n— 1.
Hence we have the formula for the last term of a geomet
rical progression, ^ ^ ^^i ^q^
BXBBOISBS
178. 1. Find the seventh term in the geometrical series,
16, 32, 64, . . ..
Here a = 16, r = 2, n = 7.
Hence, I = ar^^ = 16(2)^1 = 16 x 64 = 1024.
The seventh term is 1024. This result may be verified easily by writ
ing down the first seven terms of the series.
2. Find the 8th term of the series 6, 2^, 1 J, • • •.
3. Find the 6th term of the series 130, 390, 1170, • • ..
4. Find the 9th term of 3,  6, 12,  24, . • ..
5. Find the 8th term of 3, — 1, i, — , • • •.
6. Indicate the 15th term of the series — y 1, — , •••.
V2 6
GEOMETRICAL MEANS
179. The geometrical means between two numbers are
numbers which, together with the two given numbers as first
and last terms, form a geometrical series.
If the two given numbers are 8 and ^, then 4, 2, 1 are three
geometrical means, because 8, 4, 2, 1, ^ is a geometric series.
SERIES AND LIMITS 165
BXBBCISBS
180. 1. Insert four geometrical means between 5 and 160.
The two given numbers and the four means make together 6 terms.
We have a = 6, 1 = 160, n = 6. We must find r.
By(C), l = ar^\
160 = 6 »*.
r = 2.
Hence the required geometrical series is 5, 10, 20, 40, 80, 160.
2. Insert six geometrical means between 10 and — 1280.
Solve r^ =  128 by trial.
3. Insert four geometrical means between 3 a and 96 a*.
4. Insert one geometrical mean between 133 and 1197.
What is the geometrical mean between two numbers ? What
is the arithmetical mean between two numbers ?
6. Two numbers differ by 6, and their arithmetical mean
exceeds their geometric mean by 1. Find the numbers.
SUM OF A GEOMETRICAL SERIES
181. The sum of the tirst n terms of a geometrical series
may be indicated thus,
S = a^ar^ar^\ f aV"^ f ar^"^ + ar^K (1)
Multiply both sides of (1) by r,
r/S = ar 4 ar* 4 a^^ 4 • • • 4 a^*"* 4 ar""* 4 ar*'. (2)
Subtract (1) from (2) and observe that all the terms dis
appear in subtraction, except a and ai**. We obtain,
rS — S = ar^ — a
S(r  1) = a(r 1)
r — 1
166 ELEMENTARY ALGEBRA
BXBBOISB8
182. 1. Find the sum of six terms of the geometrical series
11, 22, 44, ... .
Here a = 11, n = 6, r = 2.
Hence. /? = <'!^rL!l = yC^.rJl = ll(?lrill= n x 63 = 603.
r1 21 1
2. Find the sum of seven terms of 1, ^, ^, • . • .
3. Find the sum of six terms of 1, — V2, 2, — 2 V2, • • • .
4. Find the sum of five terms of a?, — ay', ajy*, ....
6. Show that formula ((7) may be written 8 = ^\ "~ — ^ •
1 — r
For what values of r is this form more convenient than (C) ?
CL ^— 1*1
6. Show that (0) may be written S =
1 — r
7. What will $1000 amount to in four years, interest 3 %,
compounded annually ?
8. What will $ 600 amount to in two years at 4 ^ annual
interest, compounded semiannually ?
9. What will $ 700 amount to in eight years at 4 % annual
interest, compounded semiannually ?
We simplify thia computation by the use of logarithms.
The amount is given by the expression x = ^700 (1.02)".
log 1.02 = 0.0086,
log (1.02)w = 16 X 0.0086 = 0.1376
log 700 = 2.8451
logaj = 2.9827
flc = f 961.
This answer is only approximate. Had we used a sixplace table of
logarithms, instead of the fourplace table, the answer would have come
out 9 960.06, correct to the nearest cent.
SERIES AND LIMITS 167
10. What sum of money, at 3 % interest, compounded
annually, will amount to $ 1000 in 20 years ?
Let Jjjl X be the sum, in 20 years this amounts to $x(1.03)*>.
Hence the equation, x(l.OS)^ = 1000
1000
(1.03)20
Log 1000
= 3.0000
Log 1.03
= 0.0128.
Log (1.03)«>
= 20 X 0.0128
= 0.2660
loga; = 2.7440,
x = $664.60.
The answer,
correct to the nearest cent, is $ 663.68.
11. Wliat sum of money, at 4 % interest, compounded
annually, will amount to $ 695 in 10 years ?
12. A sum of $ 497 draws compound interest at 3^ % for 7
years. Find its amount.
13. If $ 1 could have drawn 6 % compound interest during
the past 500 years, what would be its amount now ?
14. Owing to the introduction of automobiles the number of
horses in a town decreased in five years from 560 to 336.
Assuming a constant annual rate of decrease, find the number
after four more years.
15. Determine the present value of $ 1000 due in 5 years,
on the supposition that money can be invested so as to yield
3 % interest when compounded annually.
16. A man undertook to pay $ 100 to a charity one year,
$ 90 the next year, ^ of the latter sum the third year, and so
on, until his death. He died after making 21 payments.
What was the total of his gifts ?
Use logarithms in the computation.
17. Another man pledged $ 175 and promised to increase
the amount by ^ every year. What is the total of his dona
tions after 15 years ?
168 ELEMENTARY ALGEBRA
18. A man saved $ 250 every year and invested it in a busi
ness that brought him 4 % per annum compound interest. He
made these investments f6r six consecutive years. What was
the total sum standing to his credit immediately after he paid
in his last $250?
19. A man's business is increasing at a uniform rate, and so
rapidly that his income is doubled in two years. What is
the rate of increase or growthfactor f
If at starting his business was 9 a, it was 9 or at the end of the first
year, and 9 ar^ at the end of the second. We have then aH = 2 a, or
r = \/2. The growthfactor = >/2.
20. What is the annual rate of increase, or annual growth
factor, if income increased uniformly and doubled in the course
of four years ?
21. In 4 years the population of a town increased from
2000 to 2662. Determine the growthfactor on the supposition
that the rate of increase was uniform.
22. A young tree grew in 6 years from a height of 32 in. to
a height of 20 ft. 3 in. What constant annual growthfactor
would account for this increase ?
23. The population of a mining town, now 10,000, is falling
off at the rate of 1 % per annum. What will the population
be 2 years from now ?
24. Between the ages 9^ and 14^ the average height of boys
increases from 60 to 60 inches. Find how tall a boy would be
at 10, 11, 12^, and 13 years of age, on the supposition that
the growthfactor was the same for these years. Compare
your results with the statistical averages which give the
heights as 51.9, 53.6, 55.4, 57.5 inches, respectively.
Use logarithms.
SERIES AND LIMITS 169
INFINITE GEOMETRICAL SERIES
183. An important kind of series arises when the number
of terms in it is no longer restricted to a fixed finite number,
but is permitted to increase without end.
Consider the series 1, ^, J, \, ^.
It has five terms. Imagine now that the number of terms
is much larger, and that this number is steadily growing. We
obtain a series which may be written,
1» h h h^y^fhy^ infinity. (1)
The number of terms is taken larger than any large finite
number that we may name.
We say that the series is infinite. The characteristic prop
erty of it is that it has no last term. Whenever we try to
seize upon any one term, say ■— , as being the last one, we can
always go one step farther and write down the term — ^ which
comes after it. Considerations of this sort compel us to accept
the conclusion that there is no last term in such a series.
Another noticeable feature about the series (1) is that its
terms decrease toward the right; that is, the terms farther
from the starting point are always the smaller. The nth term
may be written
By taking n sufficiently large, we may make this nth term
smaller than any small number which we may name. For
instance, the nth. term will become smaller than jJ^, if we
1 11
take n = 8. In this case —  becomes — = — , which is less
2»i 27 128'
thauy^.
If you name some smaller number, say y^T^, then again a
value of n can be chosen, so that the nth term becomes smaller
than it. Take, n = 11, then h. = hxh = :^y^l= ^
2w 2' 2» 128 8 1024'
170 ELEMENTARY ALGEBRA
which is less than y^Vir* ^^^ ^^"^7 name a still smaller fraction,
and again a value can be assigned to n in the expression for the
nth term, which makes that nth term less than that small fraction.
No matter how small a fraction is written down, we can always
find some term in series (1) which is smaller than that fraction.
In such a case we say that the term — ^ approojohea zero
as a limitf as n increases wUhoiU end.
SUM OF AN INFINITE GEOMETRICAL SERIES
184. We proceed now to find the sum of the infinite series
IH 1 1 [•••H : h • • • to infinity. Examine formula
(C) ; we see that
o __ a(r* — 1) _ a(l — r *) __ a — ar^ __ a^ ar*
r — 1 1 — r 1 — r 1 — r 1 — r
We know that this series gives the sum correctly for any
finite and fixed number of terms. What is the sum when the
series becomes infinite ?
In the infinite series 1, J., J, i., . . . we have r = J^, a = 1.
Other geometrical series may be written, but in all of them
that we shall consider now, we shall assume that the ratio r is
numerically less than 1. In that case the power r* becomes
smaller for larger values ofn and approaches the limit zero as n
increases without end.
That this is so, may become clearer by a second illustration.
Let r = I, then r^ = ^, r" = ^^y, r* = \^, r* = ^^^, and so on.
The diminution in the value of the power is not as rapid now
as in the case, r = ^ ; nevertheless, a value of the integer n can
always be found, such that r" becomes less than any small frac
tion previously named. To make r" less than y^itj ^® need
only take n = 12. Since
jj 64 .f 12 «8 ^ 64x64 4096
r = , we get r^^ = r^ *t^ =
729' ° 729x729 531441
SERIES AND LIMITS 171
In this fraction the numerator is seen to be less than the
onehundredth part of the denominator.
185. What happens to the fraction , when the integer
1 — r
n LQcreases without end, and r is numerically less than 1 ?
Bear in mind that a, the first term of the series, is a fixed
number ; r is also a fixed number, hence the denominator 1 — r
is a fixed number. Only the factor r" changes, as n increases
without end. And we have seen that, under these circum
stances, r* approaches the limit zero. Hence the numerator
ar** also approaches the limit zero. Hence, the entire fraction,
, approaches the limit zero as n increases without end.
1r
The series,
S = a \' ar \ aj^ { ...4 ar"""^ + . . .
which for a fixed, finite number of terms n has the sum
a ar^
aS =
1 — r 1 —r
has the sum, 8 =  , (D)
1 — r
when the series becomes infinite, so that n increases without
end, provided that r is numerically less than 1.
From what we have said it appears that by the " sum " of
an infinite series we mean the value which the sum of the first
n terms of the series approaches as a limit, when n increases
without end.
When the common ratio r of the geometrical series is 7iot
numerically less than 1, the above reasoning does not apply.
In that case r" does not approach the limit as w increases
without end. In the series 1, 2, 4, 8, • • • r = 2, and ?•* = 2" ; 2"
becomes greater and greater as n increases. Thus 2* = 4,
2* = 8, 2* = 16, etc. It is seen that 2» increases without limit
172 ELEMENTARY ALGEBRA
when n increases. Hence the fraction does not ap
1 — r
proach the limit zero.
When in an infinite geometrical series the ratio r is numeri
cally less than 1, so that the sum of the first n terms ap
proaches or converges to as a limit, the series is said to
1 — r
be convergent.
Convergent aeries play a very important part, both in theo
retical and in applied mathematics.
BXBROISBS
186. 1. Find the sum of the infinite geometrical series,
^2 4 8 ^2"^
Here a = 1, r = J.
Hence ,8^ = _?_ = J_ = 2.
1r 1J
The Bum of the series is 2.
To convince himself of the correctness of this result the student may
find the sum of 8, 4, 6, or more terms and see that the smn gets very
close to 2, the closer when more terms are added.
Let 8n mean the sum of the first n terms, then we see that 82 = 1.5,
8z = 1.76, 84 = 1.876, 8s = 1.9376, 8^ = 1.96875. By actual addition we
can approach to the limit 2 closer and closer, but we cannot reach it,
because we can perform this addition for only a finite number of terms.
2. Find the sum of the infinite geometrical series 1, ^, ^,
ift^? • • ••
3. Find the sum of the infinite geometrical series 1, ^,
4. Find the sum of the infinite geometrical series 2, — J,
h — T^7) • • ••
6. Find the sum of the infinite geometrical series 1, — , ,
11 2* 2
^'2^'
SERIES AND LIMITS 173
6. Find a common fraction which expresses the exact
value of the repeating decimal, .252525 • • •.
The repeating fraction may be written as an infinite geometrical
series thus,
.2625 ... = tW + tAW + Tirt^inj + •
Here o = fW* »• = nflftnr ■*• ^ = rir
Hence ^ = _«_ = .JL= 25,^^ ^26 ^^
l_r 1tJ^ 100 100 99'
Find the common fraction which is the exact value of each
of the following repeating decimals :
7. .333.... 8. .151515.... 9. 7.363636....
10. .555... 11. .325325. ... 12. 7.8212121....
13. A man undertook to pay $500 to a charity one year,
$ 450 the next year, ^ of the latter sum the third year, and
so on, until his death. What was the outside limit of the
expectation of this charity? If the man died after making
15 donations, how much did his total payments fall short of
that sum ?
Use logarithms in performing the computation.
THEORY OF LIMITS
187. In deriving the sum of an infinite converging geomet
rical series we touched upon the theory of limits which plays
a fundamental r51e in higher mathematics. The ideas which
we brought out suggest the following definition of the limit
of a varying number.
2%e limit of a variable is a fixed number which the variable
approaches in siich a way that the difference between the variable
and its limit becomes and remains numericaUy less than any num
ber, however smallj but never becomes zero.
Thus, in fi; = 1+1 + 1+...+ ^
2 4 ^2*»i'
the variable is 8n^ which increases when the integer n increases.
174 ELEMENTARY ALGEBRA
The limit of 8^ is the number 2.
As n increases without end, the variable Sn approaches 2 and may be
made to approach it so closely that 2 — i9» is less than any previously
assigned small number different from zero, such as ^ithm ^^ looiooc
A variable of some importance is the fraction , when n
n
takes in succession the values 1, 2, 3, 4, • • •. It is seen that
the variable  becomes smaller and smaller and approaches
n
the limit zero. Under the same conditions the fraction ,
n
where a is any fixed number, will likewise approach the limit
as the integer n increases without end.
Quite different is the behavior of , when a is a fixed num
X
ber different from zero, and x takes in succession the values
1, ^, i, i, • • •. For simplicity, let a = 1.
Then, for a; = 1,  = 1, for a; =i,  = 8,
X S X
fora: = i, ^ = 2, for a: = 1, ^= 16,
2 a? 16 a?
for a? =  ,  = 4, for a; = —  ,  = 32,
4' a? ' 32' a? '
and so on. The smaUer the denominator, the larger the value
of the fraction.
We let the variable x approach the limit zero in such a way
that it does not reach zero. We stipulate that x shall not be
come zero for the reason that we cannot divide by zero, as was
explained in a previous chapter. If, then, x is permitted to
approach as a limit, without reaching its limit, then the
fraction  increases in value without limit. That is, we can
x 1
select a value for a;, such that  will be greater than any pre
viously assigned large number.
SERIES AND LIMITS 175
For instaoice, select the number 1000 ; we can take x so small
that  is greater than 1000. Let « = —  =
X ° 210 2*^.2^ 32.32
= , then  = 1024, which is more than 1000. In the same
1024' X '
way, X can be so chosen, that  exceeds 10,000, or any other
X
fixed number. Hence the theorem,
As the variable x is made to approach the limit zero, without
rea>ching zero,  increases so as to exceed any large number that
X
may be previously assigned.
This theorem is sometimes condensed in this manner :
As X approaches zero,  approaches infinity.
X
A still further compression of the theorem consists in the
use of the symbolism
This symbolism is objectionable, because it may convey the
idea that division by is permissible. The reader must re
member that these syjnbols are intended merely as a con
venient abbreviation for the theorem given above.
EXEBCISBS
a I 35
188. 1. What value does the fraction — ^—  approach, when
b — X
a and b are fixed numbers (b ^ 0) and cc is a variable approach
ing the limit zero ?
Under these conditions the numerator a + ^ approaches a, the denomi
nator h — x approaches 6 .
Hence the fraction approaches •
h
176 ELEMENTARY ALGEBRA
2. What value does the fraction — approach, when x in
creases without limit ? a + x
The reasoning is easier, if we first divide numerator and denominator
by X, so that x occurs in the denominators of the minor fractions.
«l
We obtain,
a — x^x
a^x ^ A.I
X
Since a is a fixed number, and x increases without limit, the fraction
 approaclies the limit zero, hence, the numerator  ~ 1 approaches — 1
X X
as a limit ; the denominator  f 1 approaches + 1 as a limit ; the com
ae
plex fraction approaches ^^— or — 1 as a limit.
What is the limiting value of the following fractions when
X approaches zero as a limit ?
' 4 — 6aj' ' 1— «* * 2 — aj*'
*• ~ • o. — . o.
aj46 Shaj* aj*f2a:*i3
What is the limiting value of the following fractions when
the variable x increases without limit ?
X l2aj a — ar
10. _?_. 12. «±^. 14. ^+£±«.
x + 6 c + dx x^ — xib
SUPPLEMENT 177
SUPPLEMSNT
THE HIGHEST COMMON FACTOR BY THE METHOD OF
DIVISION
189. The method of finding the h. c. f. of two polynomials
not easily factored is similar to the method for finding the h. e. f .
of two numbers in arithmetic, neither of which can be easily
factored. The method was first suggested by Euclid (300 B.C.).
BXAMPIiB
Find the h. c. f . of 1343 and 3002.
Id43)3002[2
268.6
316) 1348 [4
1264
79)316 [4
316
Process : Divide 3002 by 1343.
Take the remainder 316 as a new divisor, and 1348 as a new dividend.
Take the next remainder 79 as a new divisor, and 316 as a new dividend.
The next remainder is zero.
The last divisor 79 is the h. c. f . of 1343 and 3002.
Proof: Let ^ = 1343, i? = 316.
B = 3002, Q = 2.
We have, as always.
Dividend = Divisor x Quotient + Remainder.
B = AxQ +B,
or B = BAQ
From the last equation we see that any factor of both A and B
must be also a factor of B — AQ^ and therefore a factor of the re
mainder B. Hence, all common factors of A and B are common factors
of A and B,
Instead of finding the h. c. f. of A and B, we may therefore find the
h. c. f . of A and B ; the latter process involves smaller numbers.
Repeating the above process, divide A by B. Let Q' be the new quo
tient 4, and B' be the new remainder 79. As before, all common factors
of A and B are factors of B' ; for, A = Q'B + B' and B' = A— Q'B.
178 ELEMENTARY ALGEBRA
Hence, the problem reduces itself farther, to finding the h.c. f. of B
and B' ; that is, to finding the h. c. f. of 816 and 79. As 816 = 79 x 4, it
follows that 79 is the h. c. f . required.
This process may be used in finding the h. c.f. of poly
nomials which cannot be easily factored. Numerical factors
common to the polynomials are easily detected by inspec
tion. Such numerical common factors, as well as other
common factors that can be found by inspection, should be
removed at once.
BXAMPLB
Find the h.c.f. of 2iB»22aj12 and 4iB»16aj«16rc12.
Process : 2 x«  22 «  12 = 2(a*  11 a: 6).
4 x» + 16 a;a + 16 a + 12 = 4(x» + 4 a;2 + 4 x + 8).
The numerical factor 2 is common to the polynomials.
x»llx6)a* + 4a;2+ 4x + 8Ll
z« 11 a; 6
4a:«16x + 9)4x»44a;24 a;16
16x2
63x
24
4
60x2«
212 X
96
60x2
225 X
136
18 13xh39
X + 3)4 x2 + 16 X + 9 4x48
4x2+ 12x
3x + 9
3x + 9
The h.c.f. =2(x + 8).
To avoid fractions any expression may be multiplied or divided by
any number which is not a factor of the other. In the example above,
when 4 x2 in the first remainder will not be contained in x^ a whole num
ber of times, the expression x'— llx — 6 is multiplied by 4. Again
when 4 x2 is not contained in — 16 x2 a whole number of times, the
SUPPLEMENT 179
expression — 15 x^ — 63 2c — 24 is multiplied by 4. In the remainder
13 X + 30, the factor 13 may be discarded because the expression
4 a;2 __ 15 x j does not have a factor 13. It would be wrong to multiply
4 x^ + 16 X + 9 by 13, because 13 is a factor of 13 x + 30.
Continue to divide until the remainder is of a lower degree than the
divisor. Then take the remainder for a new divisor and the previous
divisor for a new dividend and proceed as before.
The 1. c. m. of two expressions not easily factored may be found by
first finding their h. c. f . The 1. c. m. required will then be the product
of either expression and the quotient obtained by dividing the other ex
pression by the h.c. f. It is seldom that resort to this long process be
comes necessary.
EXERCISES
190. Find h. c. f. of :
1. 2a2 + a — 3and4a'8a2 — a — 6.
2. 2a^ — 5a;2 — 2aj + 2and2aj» — 7a;2 + 9aj — 3.
4. a* — a' + 2 a2 — a 4 1 and a* + a'  2 a2 + a + 1.
5. 8 am«24 am^} 22 am\6 a and 6 m'+13 m2 + 8 m + 1.
6. 2a:310aj214aj + 70anda^3aj27a;15.
7. S xlh/—10 a^y\7 xh/—2 xy and 6 a^y 11 oiih/ ^8 ah/— 2 xh^.
8. 4a»  4a2  5a + 3 and 10a2 19a + 6.
The h. c. f . of three or more expressions may be found as follows : Find
the h. c. f. of two of them ; then find the h. c. f. of this result and another
of the given expressions, and so on. The last h. c. f . is the one required.
9. 2a^2aj2_2aj4, 3iB»6a;2+9aj18,and8aj»12a;2
4aj8.
10. a^x — 6 a2aj f 11 ox — 6 aj, cj^x — 9 a2a;  26 ox — 24 aj, and
a«aj2 — 8 a2aj2 4. 19 aaj2 — 12 x^.
180 ELEMENTARY ALGEBRA
SOLUTIONS OF QUADRATIC EQUATIONS
191. In previous solutions of the quadratic equation ax^ 4
6a; 4 c = 0, obtained by completing the square, the first step
was to divide both sides by a. There are other methods of
procedure which possess certain advantages. If a, 6, c are
integers and it is desired to avoid fractions in the process of
completing the square, the Hindu method of completing the
square may be used. It is as follows :
Given ox* + ftx = — c.
Multiply both sides by /our times the coefficient ofx^^ or 4 a.
We obtain 4 a^c^ \ 4 abx =  4 ac.
Divide 4 abx by twice the square root of 4 a'x' ; this gives 5.
Add the square of 6 to both sides,
4a^x^ + 4abx + fe^ = 52 _ 4 <c.
Take the square root of both sides,
2ax + b = ± y/h^  4 oc.
Transpose h and divide by 2 a, x ^  & J:>/"&^  4ac ^
2a
Sometimes fractions may be avoided by the use of a smaller
factor than 4 a. Consider the equation,
3a? h 4a; = 8.
Make the first term a perfect square by multiplying by 3,
9a;2+12x = 24.
Divide 12 x by twice the square root of 9 x^ ; this gives 2.
Add the square of 2 to both sides,
9«3 + i2a;4 = 28. _
Then 8a; + 2 =±^28, _
2 J:>/28
"" 3 '
2±2V7.
3
Solve by completing the square without introducing frac
tions :
1. 2a« 13a; 410 = 0. 3. 4a;* + 6aj 41 =0.
2. 5a;2h7a;13 = 0. 4. 7a;2  14a;  3 = 0.
SUPPLEMENT 181
MATHEMATICAL INDUCTION AND PROOF OF THE
BINOMIAL THEOREM
192. There is an important method of reasoning in mathe
matics, called mathematieal induction, which we shall use in
proving the binomial formula. The method will be grasped
more readily, if we give a nonmathematical illustration of it.
Non 'mathematical Illustration,
For the sake of argument, suppose it established as true
(1) That there was once an Indian named Hiawatha*
(2) That every Indian named Hiawatha (if ever such an Indian ex
isted) had a grown son named Hiawatha^
From these two propositions certain conclusions can be drawn.
By (1) we know that an Indian named Hiawatha once existed.
By (2) we know that he had a grown son named Hiawatha. Desig
nate him Hiawatha II.
By (2) we know that Hiawatha II had himself a grown son named
Hiawatha. Designate him Hiawatha III.
By (2) we know that Hiawatha III had himself a grown son named
Hiawatha. And so on.
Thus the inference is drawn from propositions (1) and (2) that there
existed an unbroken, neverending line of descent of Indians named
Hiawatha.
Mathematical Illustration
It is to be proved that tf^e sum of the first n positive integers i« ^ (n h 1).
I. We see by trial that 1 + 2 = J(2 + 1),
1 h 2 + 3 = i(3 h 1).
This shows that the theorem is correct for the particular cases n = 2
and n = 8.
II. We establish the conditional proposition, that if the theorem is
true for some value of n, say n = m, then the theorem must be true for
n = m + l.
If the theorem is true for » = m, we have
l2 + 3 + ..+m = ^(w+l).
Add (m I 1) to both sides,
1 + 2 + 31 ... +m+(w+l) = ^(m+l)+(w + l).
182 ELEMENTARY ALGEBRA
Combine terms on the right side,
1 + 2 + 3 + . . . + m + (m + 1)= 5L±i([w, + 1]+ 1).
As this result is the original theorem for n = m + 1, we have shown
that the theorem is true forn = m + 1, provided it is true for n = m.
Next we apply the reasoning called mathematical induction :
By I, the formula
1 + 2 + 3 + . .. + n = (n + l)
is true for n = 3.
By II it follows that, being true for n=3, it must be true also for n=4.
By II it follows that, being true for n = 4, it must be true also for
n = 6. And so on for n = 6, 7, 8, • • •. Consequently, the theorem is
true for any positive integral value of n, no matter how great it may be.
Thus, the theorem is established.
Note. Very often a beginner fails to i)erceive the need of such an
involved argument. If a theorem is true for certain special cases, he may
conclude at once, without further argument, that it is true for all cases.
But such jumping at conclusions is dangerous. The relation a^ = 2 a
holds for a = 0, also for a = 2 ; is it true generally ? Is it true for any
other value of a ?
PROOF OF THE BINOMIAL THEOREM WHEN THE EXPO
NENT OF THE BINOMIAL IS A POSITIVE INTEGER
193. By actual multiplication it is found that
(a ± 6)*= a?±2db\h\
{a ± by= a« ± 3a«6 + 3a6» ± b\
(a ± by= a* ± 4 a'6 4 6 a'fe* ± 4a&» + b\
By inspection we find that these products follow the follow
ing laws :
I. The first term is a raised to the same power as that of
the binomial. In each succeeding term the exponent of a
decreases by 1.
SUPPLEMENT 183
II. The factor b does not appear in the first term ; the ex
ponent of 6 in the second term is 1 and increases by 1 in each
succeeding term.
III. The coefficient of any term after the first is found by
multiplying the coefficient of the preceding term by the ex
ponent of a in that term, and divided by one more than the
exponent of 6. ,
IV. If the binomial is a + b, the signs of the product are
all plus ; if the binomial is a — &, the signs are alternately f
and — .
V. The number of terms is one more than the exponent of
the binomial.
VI. Each term is of the same degree as the binomial.
If we use the exponent n, these six laws are embodied in the
following formula : *
(a + by = a» h na^'^b + ^l^^^lDa^^^^
_^ n(«l)(«2) ^^y + . • . + naly^+ b". (1)
Thus far, this formula has been shown to be true only for the
particular cases n = 2, 3, or 4. (Substitute 4 for n and show
that the product of (a + by is obtained.) Is formula (1) true
for all positive integral values of n ? Were we to attempt to
prove its generality by actual multiplication, we would soon
weary ; the task would be impossible. By mathematical in
duction the proof is short.
Let us establish the conditional proposition that if formula
(1) holds for, say n = m, it must hold also for m +1. In (1)
write m in place of n. Then multiply both sides of the re
sulting equation by (a + b). We obtain,
* By the notation 2 1 we mean 2 x 1 » by 3 ! we mean 3x2x1; generally,
nl=»n(nl)(n — 2) ...3.2.1. WecaUn! "factorial n."
184 ELEMENTARY ALGEBRA
(a + 6)« = a* + ma^'^b+ 5*l5Lziiia«»a6a + • • • + mab"^^ + ft*, (2)
(a + 6)= a} h
(a + 6)"+i
2J
^1 « I
+ (m+l)a6« + &«+i. (8)
The operation given above should be studied very carefully. Several
steps require explanation. We have multiplied both sides of equation
(2) by a + 2). Examine every step carefully. In the right sides we first
multiplied by a, then by b. The dots stand for terms that could not all
be written down, because the number of terms is m + 1, and m is not
given in HinduArabic numerals. The term *»(t»— l) q,>25> which is
2!
obtained when we multiply by 5, is not written down and is among the
terms represented by dots. The term ^(^~ ^/ qaftmi ig obtained by
multiplying by b the term ^(^~ ^^ q2fc«>a which in (2) lies among the
terms represented by dots.
Be sure to verify the addition of the partial products.
We observe that (3) is the same in form as (1) ; that is, if
in (1) we write m f 1 for w, we obtain (3).
By the above multiplication we have proved that the for
mula (1) is true for n = m f 1, provided it is true for n = m.
Now we proceed to the argument by mathematical induction :
Formula (1) is known from actual multiplication to be true
for n = 4. But we have shown that if (1) is true for w = 4, it
must be true for n = 5.
Again, if (1) is true for n = 5, it must be true for n:=6.
And so on, for n = 6, 7, 8, • • • .
Hence formula (1) is true for any positive integral value of
the exponent.
SUPPLEMENT 185
BXEBCISBS
194. 1. Expand (a + 3 y)\
In formula (1) substitute a = a;, 6 = (3y), n = 6. We obtain,
Simplifying,
(«+3y)«= a^ + 18a^y+186a:*yH540a:»y8+1215a;2y*+1468icye+729 y«.
2. Expand (2p — qy.
In formula (1) substitute a = 2p, 6 =(— g), n = 6.
(2i> g)6 =(2p)6 + 5(2p)*( g) + ^ (2p)»(g)«
Simplifying,
(2p  g)6 = 32 j)6  SOp^g + 80i)«g«  40p2ga + I0)gf4  g6.
Expand :
3. (a26)». 7. (2i> 43^)5. 11. (a + 6)'.
*• («iy)* 8. (V2a6y. 12. (a + V^6)».
6. (3aj2/. _ / ,
/I Y 9. (V2a;+^/2y)«. 13. (a V 16)*.
\m "" **/ ' 10. (a  h)\ 14. ( V^ «+ V^ y)*.
195. It can be proved that, under certain limitations, tlie
Binomial Formula is applicable to cases in which the exponent
w of (a + 6)** is not a positive integer, but is a negative integer,
or a positive or negative fraction. In such cases the binomial
expansion becomes an infinite series. Such an infinite series
can be used for purposes of computation only when it is con
vergent; that is, only when the sum of the first r terms of the
series approaches a finite constant as a limit, as r increases
without end. When a > 6, the expansion, expressed in ascend
ing powers of 6, is always a convergent series. Omitting all
proofs, we proceed to applications.
186 ELEMENTARY ALGEBRA
1. Find approximatiiely the cube root of 27^.
This can be obtained by the regular process of extracting the cube root,
or by the use of logarithms. A third method is by means of the binomial
theorem.
We notice that (27.2)> is a little over 8. Take 272 = 27 + .2, where
27 is a perfect cube. Then by the binomial formula, taking
a = 27, 6 = .2, n = 1,
(27 + .2)* =(27)1 + l.(27)*(.2) + Ki.Ilil(27)**(.2)2 + . . .
Here .2 is small when compared with 27. Hence the series is convergent.
In the expansion, .2 occurs to the first power in the second term, to
the second power in the third term, and to higher powers in the terms
foUowing which are not put down.
The simplified value of the second term is + ^\^ = .007407, of the
third term is — ^^^ = — .000018.
If we take only the first two terms in the expansion, and neglect all the
rest, we obtain an approximate value for the cube root, 8.007407.
If we take the first three terms, the more accurate value, 3.007889, is
obtained.
If in a calculation, account is taken of a number h which is small com
pared with the other numbers involved, but the square and higher powers
of h are discarded, then the calculation is said to be carried to the fir^t
approximation.
If both the first and second powers of b are used, but no higher powers of
&, then the calculation is carried to the second approximation. And so on.
In the above example, where h = .2, the first approximation to the
cube root of 27} is 8.007407 ; the second approximation is 8.007889.
EXBBCISBS
196. Find to a first approximation the values of the follow
ing expressions in which the second number is much smaller
than the first :
1. (32^)*. 4. (a* 4 6)*.
2. (63)* =(641)*. 6. Va« + 2 c.
3. V99.=V1001. 6. ^/IpTq,
7. (a«36)*. 8. (9m«n)^. 9. ■\/a^—\/a.
SUPPLEMENT 187
10. A brass cube 1 inch each way is heated until its edges
are 1.003 inches long. Compute to the first approximation the
area of each face and the volume of the cube.
11. A metal cube 2 inches each way is cooled so that its
volume is reduced to 7.998 cubic inches. What is now the
length of an edge ?
12. Compute to a second approximation the amount after
10 years, of $ 575 at 2 %, interest compounded annually.
Evaluate 676 (1 h «02)^<> by retaining the first and second powers of
.02 in the binomial expansion, but discarding higher powers. Does the
first approximation give results differing from those due to simple
interest ?
13. In measuring the diameter of a sphere whose diameter
is actually 5", an error of 1 % is probable. What is the probable
error, computed to a first approximation, of the volume derived
from the inaccurate diameter ?
The volume of a sphere =  irt*.
Take r» = \(6 + .06)».
14. How great an error is made by assuming
1a^
1 + a
= 1 — a + a* — a' + a*.
when a = ^ ? What is the ratio of the error to the whole of
the fraction ?
16. Proceed on the assumption that the earth is a sphere,
the radius of which is 4000 miles. On the supposition that its
radius was at one time 10 miles greater than at present, calcu
late to the first approximation, and also to the second approxi
mation, the amount that it has lost during the contraction,
(1) in superficial area, (2) in volume.
[The superficial area = 4 irr^.]
188
ELEMENTARY ALGEBRA
197. A Table of Squares and Cubes, Square Boots and Cube
Roots of Numbers from 1 to 200,
Square
•
Cubs
Sqvars
Cubs
8QVAXB8
CVBSt
No.
— * —
1
K0OT8
Roots
ft
SQVAKn
CVBM
No.
61
Roots
Roots
1
1
1.000
1.000
2,601
132,651
7.141
3.708
4
8
2
1.414
1.260
2,704
140,608
62
7.211
3.733
9
27
8
,M^.?32
1.442
2,809
148,877
68
7.280
3.756
16
64
4
2.000
1.587
2,916
167,464
64
7.348
3.780
26
125
6
2.236
1.710
3,026
166,376
66
7.416
3.803
36
216
6
2.449
1.817
3,136
176,616
66
7.483
3.826
49
343
7
2.646
1.913
3,249
186,193
67
7.660
3.849
64
612
8
2.828
2.000
3,364
195,112
68
7.616
3.871
81
729
9
3.000
2.080
3,481
206,379
69
7.681
3.893
100
1,000
10
3.162
2.164
3,600
216,000
60
7.746
3.915
121
1,331
11
3.317
2.224
3,721
226,981
61
7.810
3.936
144
1,728
12
3.464
2.289
f 3,844
238,328
62
7.874
3.958
169
2,197
18
3.606 <
3.351
• 3,969
260,047
68
7.937
3.979
196
2,744
14
3.742
2.410
4,096
262,144
64
8.000
4.000
225
3,376
16
3.873
2.466
4,226
274,625
66
8.062
4.021
266
4,096
16
4.000
2.620
4,356
287,496
66
8.124
4.041
289
4,913
17
4.123
2.671
4,489
300,763
67
8.185
4.061
324
6,832
18
4.243
2.621
4,624
314,432
68
8.246
4.062
361
6,859
19
4.359
2.668
4,761
328,609
69
8.307
4.102
400
8,000
20
4.472
2.714
4,900
343,000
70
8.367
4.121
441
9,261 ' 21
4.583
2.769
6,041
357,911
71
8.426
4.141
484
10,648 22
4.690
2.802
6,184
373,248
72
8.486
4.160
629
12,167
28
4.796
2.844
5,329
389,017
78
8.644
4.179
676
13,824
24
4.899
2.884
6,476
405,224
74
8.602
4.198
625
15,625
26
6.000
2.924
5,626
421,876
76
8.660
4217
676
17,676
26
6.099
2.962
5,776
438,976
76
8.718
4.236
729
19,683
27
6.196
3.000
6,929
466,533
77
8.776
4.254
784
21,952 28
6.292
3.037
6.084
474,562
78
8.832
4.273
841
24,389 ; 29
6.385
3.072
6,241
493,039
79
8.888
4.291
900
27,000 80
6.477
3.107
6,400
612,000
80
0.9S%
4.309
961
29,791 81
5.568
3.141
6,561
631,441
81
9.000
4.327
1,024
32,768 82
5.657
3.175
6,724
651,368
82
9.056
4.344
1,089
35,937 88
6.746
3.208
6,889
671,787
88
9.110
4.362
1,156
39,304
84
6.831
3.240
7,056
692,704
84
9.165
4.380
1,225
42,875
86
6.916
3.271
7,226
614,125
86
9.219
4.397
1,296
46,656 861
6.000
3.302
7,396
636,056
86
9.274
4.414
1,369
50,653
87
6.083
3.332
7,669
658,503
87
9.327
4.431
1,444
54,872
88
6.164
3.362
7,744
681,472
88
9.381
4.448
1,621
59,319
89
6.245
3.391
7,921
704,969
89
9.434
4.465
1,600
64,000
40
6.325
3.420
8,100
729,000
90
9.487
4.481
1,681
68,921
41
6.403
3.448
8,281
753,571
91
9.539
4.498
1,764
74,088
42
6.481
3.476
8,464
778,688
92
9.691
4.614
1,849
79,507
48
6.657
3.503
8,649
804,367
98
8.644
4.631
1,936
85,184
44
6.633
3.530
8,836
830,584
94
9.695
4.647
2,025
91,125
46
6.708
3.557
9,025
857,375
96
9.747
4.563
2,116
97,336 , 46
6.782
3.683
9,216
884,736
96
9.798
4.679
2,209
103,823
47
6.856
3.609
9,409
912,673
97
9.849
4.595
2,304
110,692
48
6.928
3.634
9,604
941,192
98
9.900
4.610
2,401
117,649 49
7.000
3.659
9,801
970,299
90
9.960
4.626
2,500
125,000 60
7.071
3.684
10,000
1,000,000
100
10.000
4.642
SUPPLEMENT
S4IM1B
Cdbh
Ko.
Root*
Etwm
IKIFIBES
ClTBia
No.
8OT.«
Roots
Roan
10,201
1 " 01
10,050
*.667
1 1
12.288
O.320
10,<HH
1 OS
10.100
4.672
1 2
12.329
6.337
10.609
1 27
10.149
4.688
S
12.369
0.348
10,816
1 64
10.198
4.703
4
12.410
6.360
11,025
1 25
10.247
4.718
ISS
12.500
5.372
11236
1 16
10.296
4.733
1 B
12.490
6.383
lt.449
1 43
10.344
4.747
1 7
12.530
5.395
i: a
1 12
10.392
4.763
1 8
12.670
6.406
I )i
I 39
'
10.440
4.77V
I 8
12.610
5.418
i: »
1 00
10
10.488
4.791
160 12.649
6.429
i: a
1, , 31
1
10.536
4.806
1 1 . 12.689
6.440
11 H
1,404,928
1
10.683
4.820
2
12.728
5.461
i: 19
1,442,897
1
:o.&to
4.830
3
12.767
5.483
l,4Ht,544
1
10.677
4.849
84
a
5.474
i: »
1,020,870
1
10,724
es
5.486
i; 16
1,660,896
1
10.770
4^877
66
12.884
5.496
i: »
1,601,613
1
10.817
4.891
7
12.923
6J107
i; 14
1,643,032
1
10.863
8
12.961
5.018
H 11
1.685,169
1
10.909
4^919
eg
13.000
6.029
u «
1,728,000
£0
10.954
4.932
13.038
5.040
U 11
1,771,061
2
IIJWO
4.M6
71
13.077
6.660
1B1B848
11.IM0
4.960
7S
13.116
5.661
U 19
1,W«B«T
8
11.091
4.973
73
13.103
6.572
11 '6
1,! 4
24
11.136
4.987
4
13.191
5.583
U »
i; s
U
11.180
6.000
7fi
13229
5.593
11 '6
2,1 6
s
11.220
0.013
6
13.266
5.604
If !9
2/ 3
2
11.269
6.027
7
13.304
5.615
1« H
2,< a
8
11.314
6.040
78
13,342
0.820
It H
2, 8
29
11.368
0.053
79
13.379
6.636
1( 10
2
S
11.402
6.066
80
13,416
5.646
1^ 11
2 1
S
11.446
81
13454
5.607
n »
2.1 8
s
11.489
6:092
82
13.491
5.667
n 19
2r 7
3
11.533
0.104
183
6.677
n i6
2,' 4
8
11.576
6.117
184
13!s6fl
0.688
If !9
Z 5
8
11.619
5.130
186
13.601
0.698
It e
2J 6
S
11.662
6.143
186
6.708
It 19
3.. 8
3
11.705
0.155
187
13,870
0.71S
11 4
2,1 2
S
11.747
6.168
188
13.711
6.729
1! 11
2) g
3
11.71«
0.180
89
13.748
0.739
11 10
2.'
4
11,832
5.192
™...0
190
13.784
5.749
li a
2.: ;l
4
11.874
0.206
36,481
191
13,820
5.769
a M
2; e
4
11.916
6.217
36,864
1 2
13.866
6.769
20,419
2,' 1
4
11.968
0.229
37.249
193
13.892
5.779
20,736
2; 4
4
12.000
5.241
37,636
194
13.928
5.789
21,026
3.1
4
12.042
5.2H
38,025
9S
13.964
21,316
3 4i
4
12.083
5.266
38.416
96
14.000
5.809
21800
3; 3
4
12.124
0.278
38,809
97
14.036
5.819
21,904
3,; S
4
12.1H6
5.290
39,204
98
14.071
0.828
22.201
3,. 9
4
12.207
5.301
30,601
9
14.107
5.838
22,600
3;.
S
12.247
5.313
40;000
800
14.1*2
6.848
REVIEW EXERCISES
(Sblbotbd from Collbob Entbanob Examinations)
FACTORING
198. 1. Factor 6 a* + 10a6 — 4&«, aj» + 125, 1 — a? — a^ + a^,
a^ + a?* + 1. Univ, of State of N.Y.
2. Find the prime factors of
(a) (a? ««)» + («  !)• + (1  x)K
(6) (2aj + a — ft)*  (aj  a + by. Mass. InstUute of Tech.
8. Factor a? — 2 mas  wi* — n',
a^^^ — 11 aj»+^"» h 18 ajy»* Univ. of Pa.
4. (a) Resolve the following into their prime factors :
(1) (aj«y«)«3^.
(2) 10aj»7aj6.
(b) Find the h. c. f . and the 1. c. m. of
aj»3aj*4a53,
aj' — Sa^ — a5f3. Colunibia.
6. Find the h. c. f . : »* — y*,
a?* h 2«y — 3 y*. Cornell.
6. Factor the following expressions :
(a) ai — 6i,
(5) aj*yV — aJ*2 — yH; h 1,
(c) 16(a; f y)* — (2 a; — y)* Jfoun^ Holyoke.
190
REVIEW EXERCISES
191
FRACTIONS
a
199. 1. Simplify tj
a
fe' +
cb
b
Harvard.
2, Simplify the expression
ichy —
ajhy
xy
x\yl
a^ — y^
Cornell.
3. Divide /^g^ziJ^^ ^4y Wa^ + y^^^±y.\
\a?y^ Q^ — xyJ Xx — y xy — yy
Sheffield.
4. Simplify the expression
A y\A ab¥ \ a* ^ ab
\ ay\ a* Ja^ + b^ ' a^^b^'
Mass. Inst. Tech,
6. Find the product of
Vasaar.
12 3
6. If m = T » n = , p = — ;—p^f what is the value of
m
+
a + l' a + 2'^ a43
+
1 — m 1 — w 1 —p
Univ. of Pa.
EQUATIONS
200. 1. Solve ^ + ^±^ = ?.
aj 4 X { a 2
Tale.
2, Solve for x : — z=. H — ^=3 = Va — b. Univ. of Pa.
Va — X y/x — b
192 ELEMENTARY ALGEBRA
3. Solve for a; : ?i^?6=s Cornell
x1 a?2aj3
4. (a) Solve for a? : V2 a?  3 a + V3 a? — 2 a = 3Va.
(b) Solve for m:
1 = — —.4 ^" . Univ. of Pa.
6. Solve H = 0. Princeton.
a? 4 1 05 — 1 a? — 3 a? — 5
6. Solve the equation V2 x } 5 ^ V6 — as = 1.
Uwiv. of Pa.
7. Solve 3 aJ'  11 aj = 70. CTniv. o/ /Seoie o/ N. T.
8. Solve the quadratic equation
aj» — 1.6 a? 4 0.3 = 0. Harvard.
9. If d is one of the roots of the equation aa? + 6aj  c = 0,
find the other root. Univ. of Pa.
10. Solve — ^ 4  = 3. Univ. of State ofN. T.
x\ 1 X
11. ■ Solve Vajf a f Vi 4 y/x — a = 0.
Uhry. o/ State of N. Y.
12. (i) Solve for t: {t 2)(t 4 3) = (3 « 4 4)(2  t).
(ii) Solve for x :
■^ H ^ = . . Univ. of Pa.
V^T2 V3aj2 V3a:»44a;4*
REVIEW EXERCISES
193
RADICALS
201. 1. Find the square root of
af^^2ix^ — x^\Sa^\2x + l. Univ, of State of N.T.
2.
3.
Univ, of State of N, Y.
Solve for x and y the equations
Vx^y = a H 6,
* — y = (« — ft) Va? + y.
C/niv. of Pa,
4. Simplify (a? f 1)' — xy/x H 1, V— m'n • V— mn',
i 1
6*^
C/niv. o/ Pa,
5. Simplify the product of
(ayx~^)^y (bxy~^y, and (y*a"'6~*)*. Princeton.
6. Solve the simultaneous equations
xih2yi^i,
7. Simplify (a) Ve  V20. (6) i^v^Tl + o:^
8. Find the value of a^6^, if
a = a?^y » and 6 = ^ «"^y*,
FoZe.
Cornell,
and reduce the result to a form having only positive exponents.
Haroard.
9. Simplify — — ^ — ^— , and compute the value of the
fraction to two decimal places. Yale.
o
194
ELEMENTARY ALGEBRA
SYSTEMS OF LINEAR EQUATIONS
. 1. Solve = a,  +  = 0, = c.
X y y ^ z X
Univ. of State of N.T.
2. Solve the equations, 2 a + 5 y = 85,
2^452 = 103,
2 2 I 5 a? = 57. Vassar.
3. Solve the following set of equations :
« H y =  1.
ajh3y + 22 = 4.
05 — y + 42 = 5.
Cornell.
4. Solve the equations :
7a?h6 ig^ 5g13 8yx
11 "^^ 2 5 '
3(3iB + 4) = 10y15.
Mass, Inst. Tech.
6. Draw the lines represented by the equations
3 a— 2y= 13 and 2a;5y= — 4,
and find by algebra the coordinates of the point where they in
tersect. Univ. of Col.
6. Solve the simultaneous equations :
a + a
and verify your results. Harvard.
SYSTEM OF EQUATIONS, ONE OR BOTH QUADRATIC
203. 1. Solve the system :
aJ* + y' + «y = 54^,
2. Solve
a? — y = 2,
a^ — a^ = 30.
Univ. of Pa.
Univ. of State of N. T.
REVIEW EXERCISES 195
3. Solve for t and u:  H — = 74, = 2. Univ. of Pa.
4. Solve the equations
2 a? — 3 y = 0. Columbia.
5. Plot the following two equations, and find from the graphs
the approximate value of their common solutions :
a^ f y2 = 26.
4 a;* f 9 y* = 144. ColunMa.
6. Solve the following pair of equations for x and y :
a' + y' = 4,
a? = (1 H V2)y  2. Oomc/Z.
7. Solve
a; — y = 2,
and sketch the graphs.
Mass. Inst. Tech.
BINOMIAL THEOREM
204. 1. Write in the simplest form the last three terms of
the expansion of (4 a' — o^x^y. Tale.
2. Expand (1 — V2)*^ by the Binomial Theorem.
Univ. of Pa.
3. Write out by the binomial theorem the first ^ve terms of
/^2 a  — y . Univ. of State of N. T.
4. Find the seventh term of f a +  ) • Columbia.
\ ay
6. In the expansion of ( 2a?f^] the ratio of the fourth
term to the fifth is 2 : 1. Find a?, Princeton,
196 ELEMENTARY ALGEBRA
PROBLEMS
205. 1. Two cars of equal speed leave A and B, 20 mi.
apart, at different times. Just as the cars pass each other an
accident reduces the power and their speed is decreased 10 mi.
per hour. One car makes the journey from A to B in 56 min.,
and the other from B to A in 72 min. What is their common
speed ? Tale.
2. The sum of two numbers is 13, and the sum of their
cubes is 910. Find the smaller number, correct to the second
decimal place. Vassar.
3. A man arranges to pay a debt of $ 3600 in 40 monthly
payments which form an A. P. After paying 30 of them he
still owes ^ of his debt. What was his first payment ?
Princeton.
4. A page is to have a margin of 1 inch, and is to contain
35 square inches of printing. How large must the page be, if
the length is to exceed the width by 2 inches ? Mount Holyoke.
5. Insert 10 arithmetical means between 10 and 61, and
find the sum of the entire series. Univ. of Pa.
6. A man bought a number of cattle which cost him in all
$672. If each head had cost him $4 less, he would have
been able to buy 3 more. How many did he buy and at what
price? Univ. of Pa.
7. A boy is 5 years older than his sister and ^ as old as
his father ; the sum of the ages of all three is 51. Find the
age of the father. Univ, of State of N. T.
8. Find two consecutive numbers whose product is 306.
Univ. of State of N. T.
9. A man engaged to work a days on these conditions : for
each day he worked he was to receive b cents, and for each day
he was idle he was to forfeit c cents. At the end of a days he
received d cents. How many days was he idle ? Univ. of Pa.
REVIEW EXERCISES 197
10. For what values of m will the roots of the equation
(m h )a?2 — 2(m + l)aj f 2 = be equal ?
Form the equation whose roots are  and ^. Univ. of Pa.
11. If a number of two digits be divided by the product of
its digits, the quotient will be 6. If 9 be added to the number,
the sum will be equal to the number obtained by interchanging
the digits. What is the number ? Univ. of Pa.
12. A and B each shoot thirty arrows at a target. B makes
twice as many hits as A, and A makes three times as many
misses as B. Find the number of hits and misses of each.
Univ. of Cat.
13. The sides of a triangle are a, h, c. Calculate the radii
of the three circles having the vertices as centers, each being
tangent externally to the other two. Harvard.
14. The force P necessary to lift a weight IT by means of a
certain machine is given by the formula
P=ah6Tr,
where a and h are constants depending on the amount of fric
tion in the machine. If a force of 7 pounds will raise a weight
of 20 pounds, and a force of 13 pounds will raise a weight of
60 pounds, what force is necessary to raise a weight of 40
pounds ?
(First determine the constants a and h.) Harvard.
16. Find the sum of 8 terms of the progression
6 + 3^ h 2 f .... Halyard.
16. How many terms must be taken in the series 2, 5, 8, 11,
• • ., so that the sum shall be 345 ? Mass. Inst. Tech.
INDEX
Numbers refer to sections.
Absolute, term, 70.
value of a number, 1.
Addition, 2.
checking, 2.
fractions, 68.
radicals, 154.
Algebraic expression, 1.
Antecedent, 98.
Antilogarithm, 119.
Approximations, 195.
Arithmetical mean, 171.
Arithmetical operations, order of,
Arithmetical series, 167.
Associative laws,' 76.
Base, 107.
Binomial, 1.
surds, 158.
theorem, 84, 85, 192, 193.
Braces, 6.
Brackets, 6.
Characteristic, 112.
Checking, addition, 2.
division, 9.
equation, 20.
multiplication, 8.
subtraction, 3.
Coefficient, 1.
detached, 8.
Commutative laws, 76.
Complex numbers, 151, 164.
Consequent, 98.
Constants, 140.
Contact, point of, 138.
Convergent series, 195.
Coordinates, 24.
Degree, of equation, 19.
of term, 8.
Determinants, 87.
Discriminant, 136.
Dissimilar terms, 1.
Division, 9.
by zero, 94.
checking, 9.
of fractions, 62.
of radicals, 157.
Double root, 138.
Elimination, by addition, 29.
by comparison, 29.
5. by substitution, 29.
Equation, 19.
axioms used in, 20.
complete quadratic, 70.
containing fractions, 95.
cubic, 19.
degree of, 19.
double root, 138.
exponential, 127.
general quadratic, 70.
graph of linear, 22.
graph of quadratic, 138,
impossible, 160.
incomplete quadratic, 70.
indeterminate, 22.
involving parentheses, 21.
irrational, 160.
linear, 19, 22.
of condition, 19.
of identity, 19.
quadratic, 19, 70, 131.
quadratic form, 132, 162.
quartic, 19.
radical, 160, 162.
root of, 19, 146.
Equations, equivalent, 26, 27.
homogeneous, 144.
inconsistent, 27.
independent, 24, 27.
linear, 87.
199
200
INDEX
Equations — continued
one linear, one quadratic, 139.
simultaneous linear, 22.
simultaneous quadratic, 139.
solution by quadratics, 146.
symmetrical, 143.
Exponents, 1, 149, 150.
fractional, 35.
negative, 37.
positive integral, 32.
sero, 36.
Extraneous roots, 95, 160.
Extremes, 98.
Factor, 1.
H. C. F., 66, 189.
imaginary, 82.
irrational, 82.
literal, 1.
prime, 45.
rational, 45.
rationaUxing, 152, 157.
theorem, 78.
Factoring, 45, 79.
Formulas, 31, 75, 97. 100, 101, 102,
105, 126.
Fractions, 58.
addition, 68.
complex, 64.
division, 62.
multiplication, 62.
reduction of, 58.
subtraction, 68.
value of, 60.
Fu{iction, 100, 101, 103, 138.
Geometrical mean, 179.
Geometrical series, 175.
infinite, 183, 184, 185, 195.
Graphs, 103, 106.
identical, 26.
of complex numbers, 164.
of linear equation, 22.
of quadratic equation, 138.
of simultaneous linear equations,
24.
parallel, 27.
variation shown by, 103, 105.
H. C. F.. 66, 189.
Hindu method, 191.
Historical Notes, 77, 130, 164.
Homogeneous expressions, 8, 144.
Imaginary, factors, 82.
numbers, 138, 151, 152, 156.
Inconsistent equations, 27.
Integral expression, 45.
Interpolation, 119.
Intersection, 138.
Involution, 39.
Is not equal to, 76, note.
Laws of Algebra, 76.
L. C D., 68.
L. C. M., 66.
Limits. 167, 185, 187.
Logarithmic, curve, 108, 110.
table, 117.
Logarithms, 107, 117, 123.
Mantissa, 112.
Mathematical Induction, 192.
Mean proportional, 98.
Means, 98.
Monomial, 1.
Multiplication, 8.
by detached coefficients, 8.
checking, 8.
fractions, 62.
radicals, 156.
type forms, 17.
Number, absolute value, 1.
Numbers, complex, 151, li4.
imaginary, 138. 151, 166.
irrational, 39, 151.
negative,. 151.
positive, 151.
rational, 45, 151.
real, 151, 165.
Order of fundamental operations, 5.
Origin, 164.
Parentheses, 6, 12.
equations, 21.
insertion of, 14.
removal of, 12.
Polynomial, 1.
INDEX
201
Portraits, de Morgan, 77.
Wallis, frontispiece.
Power, 1, 39.
law of signs, 33.
Principal roots, 38, 39, 160.
Problems, practical, 31, 93, 97, 104,
105, 126. 129.
Products, special, 17.
Proportion, 98.
Quadratic^ equation, 19, 70.
complete, 70, 73.
general, 70.
graph of, 138.
incomplete, 70.
solved by completing square, 73,
191.
solved by factoring, 73.
solved by formula, 73.
nature of roots, 136.
relation of roots and coefficients,
134.
Quaternions, 76.
Radicals, 152.
addition, 154.
division, 152^
multiplication, 156.
rationalizing, 152, 153.
simplifjring, 152.
subtraction, 154.
Radicand, 152.
Ratio, 98.
Rational integral expression, 45, 78.
Remainder theorem, 78.
Review exercises, 198.
Root, double, 138.
of equation, 19.
square, 41, 43, 158.
Roots, 39.
extraneous, 95, 160.
nature of, 136.
principal, 38, 39, 160.
relation between roots and coeffi
cient, 134.
Series, arithmetical, 167.
convergent, 185, 195.
geometrical, 195.
infinite, 183, 195.
Similar terms, 1.
Simultaneous, equations, 22, 139.
homogeneous, 144.
sjrmmetrical, 143.
Square root,_41.
of a+2V6, 158.
of fractions, 43.
of numbers, 43.
tables of, 197.
Subtraction, 3.
checking, 38.
fractions, 68.
radicals, 154.
Tables of square roots, 197.
Terms, dissimilar, 1.
similar, 1.
Trinomial, 1.
Variables, 101, 138, 140.
Variation, 102, 103.
Vinculum, 6.
Zero, division by, 94.
exponent, 36,
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