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Full text of "Intermediate Algebra"

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68071 > 



OSMANIA UMVERSITY LIBRARY 

Call No. tt^^ Accession No. ig, . 

Author WkU- , to! \VMVW, L. 

Title 

This book should be returned on or before the date last marked below. 



INTERMEDIATE 



ALGEBRA 



FOR COLLEGES 



BY WILLIAM L. HART 

PROFESSOR OF MATHEMATICS 
UNIVERSITY OF MINNESOTA 



D. G. HEATH AND COMPANY 



Copyright 1948, by D. C. HEATH AND COMPANY 

No part of the material covered by this 
copyright may be reproduced in any form 
without written permission of the publisher. 

PRINTED IN THE UNITED STATES OF AMERICA (408) 



PREFA Cf 



HIS BOOK offers a collegiate substitute for third semester high 
school algebra. The text was designed for a college student who will 
study it either (1) as a preliminary to taking college algebra, or (2) as 
terminal work hi algebra which is intended as a prerequisite for ele- 
mentary courses in various fields of natural or social science, or in 
business administration. A suitable selection of content from the 
book would provide a satisfactory algebraic foundation for a first 
course in trigonometry or in the mathematics of investment. In 
the case of a student of the assumed preparation, the text provides 
sufficient material for a substantial course utilizing from 40 to 60 
class hours. 

The plan of the text was based on the assumption that the typical 
student involved is of a mature age but studied his elementary 
algebra so long ago that practically all fundamentals must be taught 
as if they were relatively new material for him. Hence, the early 
chapters of the book present a mature but frankly elementary treat- 
ment of the foundations of algebraic technique with a generous amount 
of discussion and problem material. Also, appropriate refresher 
work on arithmetic is provided incidentally hi the algebraic problems 
and explicitly in an early optional chapter devoted to computation. 
The tempo of the discussion in the text is gradually increased until, 
in the later chapters, distinctly collegiate speed is attained so that the 
student will find it easy to make the transition into a substantial 
second course devoted to college algebra. The text makes no attempt 
to present material which custom dictates as primarily within the 
sphere of college algebra, although such material frequently may 
enter the most substantial courses in third semester algebra at the 
secondary level. However, in the interest of efficiency and math- 
ematical simplicity, the terminology and general viewpoint of the 
text is distinctly collegiate. Emphasis is laid on the logical sequence 
of topics, accuracy of definitions, and the completeness of proofs. 



vi PREFACE 

SPECIAL FEATURES 

Adult nature of the presentation. The discussion in the text is 
couched at a level suitable to the maturity of college students. 
Hence, the available space and assumed class time are utilized mainly 
to explain and illustrate the mathematical principles involved and 
only the necessary minimum attention is devoted to artificial mo- 
tivation of the type which might properly be expanded for younger 
students. 

Terminology. Particular emphasis is given to the language con- 
cerning variables, functions, equations, and the most elementary 
aspects of analytic geometry because of the importance of this vo- 
cabulary in fields of application which the students will enter in 
college. The technical vocabulary of the algebraic content is limited 
by excluding terms which are of small or doubtful utility. 

Illustrative material. Extensive use is made of illustrative ex- 
amples to introduce new theory, to recall previous knowledge, and to 
furnish models for the student's solutions of problems. 

Emphasis on development of skill in computation. The viewpoint 
is adopted that the student needs refresher training in the operations 
of arithmetic, as well as* new mature appreciation of various features 
of computation. Hence, work with fractions and decimals is intro- 
duced quickly, and substantial early sections are devoted to a dis- 
cussion of approximate computation. Also, the exercises and applica- 
tions continue to demand computing skill throughout the text, and 
the chapter on logarithmic computation is made very complete. 

Supplementary content. A small amount of material not essential 
in the typical course is segregated into obviously independent sec- 
tions labeled with a black star, *. Also, the teacher will understand 
that several of the later chapters are optional and that their omission 
in whole or in part will not interfere with the continuity of other 
chapters. The book was planned to eliminate the necessity for fre- 
quent omissions by the teacher. 

The extent and grading of the exercises. The problem material 
is so abundant that, in many exercises, either the odd-numbered or 
the even-numbered examples alone will be found sufficient for the 
student's outside assignments, and the balance may be reserved for 
work in the classroom. In each exercise, the problems are arranged 



PREFACE vii 

approximately in order of increasing difficulty. Examples stated in 
words are emphasized at the appropriate places to avoid the develop- 
ment of an inarticulate form of algebraic skill. However, the text 
does not spend valuable time in specialized training to develop 
problem solving skills devoted to artificial or unimportant types of 
problems. 

Answers. The answers to odd-numbered problems are provided 
in the text, and answers for even-numbered problems are furnished 
free in a separate pamphlet at the instructor's request. 

Flexibility. The grading of the exercises, various features in the 
arrangement of theoretical discussions, and the location of certain 
chapters are designed to aid the teacher in adapting the text to the 
specific needs of his class. 

Composition and appearance. The absence of excessively small 
type, the generous spacing on the pages, and the special care taken 
in the arrangement of the content into pages create a favorable setting 
for the use of the book by both the teacher and the student. 



University of Minnesota WILLIAM L. HART 



CONTENTS 



CHAPTER 

1. THE FUNDAMENTAL OPERATIONS 1 

2. INTRODUCTION TO FRACTIONS AND EXPONENTS 22 

Review of Chapters 1 and 2 46 

3. DECIMALS AND ELEMENTS OF COMPUTATION 48 

4. LINEAR EQUATIONS IN ONE UNKNOWN 58 

5. SPECIAL PRODUCTS AND FACTORING 82 

6. ADVANCED TOPICS IN FRACTIONS 104 

Review of Chapters 4, 5> and 6 118 

' 7. RECTANGULAR COORDINATES AND GRAPHS 119 

8. SYSTEMS OF LINEAR EQUATIONS 131 

9. EXPONENTS AND RADICALS 142 

10. ELEMENTS OF QUADRATIC EQUATIONS 170 

11. ADVANCED TOPICS IN QUADRATIC EQUATIONS 186 
r I2. THE BINOMIAL THEOREM 204 
^. RATIO, PROPORTION, AND VARIATION 210 

14. PROGRESSIONS* ~~" "~~~ 221 

15. LOGARITHMS ._. 240 

1 6. SYSTEMS INVOLVING QUADRATICS 262 

APPENDIX 277 
TABLES 

I POWERS AND ROOTS 283 

II FOUR-PLACE LOGARITHMS OF NUMBERS 284 

in COMPOUND AMOUNT: (1-K*)* 287 

IV PRESENT VALUE OF $1 DUE AFTER 

k PERIODS: (1 -M) ~* 88 

ANSWERS TO EXERCISES 289 

INDEX 315 



CHAPTER 



THE FUNDAMENTAL OPERATIONS 



1 . Explicit and literal numbers 

In algebra, not only do we employ explicit numbers like 2, 5, 0, 
etc., but, as a characteristic feature of the subject, we also use letters 
or other symbols to represent numbers with variable or undesignated 
values. For contrast with explicit numbers, we agree that number 
symbols such as a, b, x, and y will be called literal numbers. In this 
book, as a rule, any single letter introduced without a qualifying 
description will represent a number. 

2. Signed Numbers * 

The numbers used in the elementary stages of algebra are called 
real numbers. They are classed as positive, negative, or zero, 0, which 
is considered neither positive nor negative. The word real is used 
with reference to these numbers in order to permit contrast with a 
type of number called imaginary, which will be introduced at a later 
stage. 

ILLUSTRATION 1. 17, f , and are real numbers. 

In arithmetic, the numbers employed consist of zero, the integers 
or whole numbers 1, 2, 3, , and other unsigned numbers which we 
express by means of fractions or the decimal notation. These num- 
bers, except for zero, will hereafter be called positive numbers. When 
we choose, we shall think of each positive number as having a plus 
sign, +-, attached at the left. 

ILLUSTRATION 2. The positive number 7 may be written -f 7 for emphasis. 

* For a logical foundation for algebra, see pages 1-78 in College Algebra, by 
H. B. FINE; GINN AND COMPANY, publishers. 



2 THE FUNDAMENTAL OPERATIONS 

In a later section, we shall formally define the negative numbers, 
which will be described as the "negatives" of the positive numbers. 

ILLUSTRATION 3. Corresponding to + 6 we shall introduce the negative 
number 6. 

Positive and negative numbers may be contrasted concretely hi 
assigning values to quantities which are known to be of one or other 
of two opposite types. In such a case, we conveniently think of any 
positive number P and the corresponding negative number P as 
being opposiies. With this hi mind, we frequently refer to the signs 
"+" and " "as being opposite signs. 



ILLUSTRATION 4. In bookkeeping, if a gain of $5000 is assigned the value 
+ $6000, a loss of $3000 could be given the value - $3000. 

We shall desire the results of operations which we shall define for 
signed numbers to correspond with our intuitions when the numbers 
are interpreted concretely. 

ILLUSTRATION 5. Let t indicate an increase of temperature when t is 
positive and a decrease when t is negative. Let time be considered positive 
in the future and negative in the past. Then, the following concrete state- 
ments should correspond to the indicated addition or multiplication. 

A decrease of 20 and then a rise of 8 creates a decrease of 12; or, 

(- 20) + 8 = - 12. 

A decrease of 10 per hour in temperature for the next 3 hours mil create a 
decrease of 30 ; or, 

(+ 3) X (- 10) = - 30. 

// the temperature has decreased 10 per hour for the preceding 4 hours, 
the temperature 4 hours ago was 40 higher than now; or, 

(- 4) X (- 10) = + 40. 

EXERCISE 1 

Under the specified condition, what meaning would be appropriate for the 
indicated quantity with opposite signf 

1. For 5 miles, if + 5 miles means 5 miles north. 

2. For + $10, if - $15 means $15 lost. 

3. For + 8, if 3 means a fall of 3 in temperature. 



THE FUNDAMENTAL OPERATIONS 3 

4. For - 14 latitude, if + 7 latitude means 7 north latitude. 

5. For - 170' altitude, if -f 20' altitude means 20' above sea level. 

6. For -f 30 longitude, if - 20 longitude means 20 west of Greenwich. 

7. For - 5', if + 5' means 5' to the right. 

Introduce your own agreements about signed values and express each of the 
following facts by adding or multiplying signed numbers. 

8. A gain of $3000 followed by a loss of $9000 creates a loss of $6000. 

9. A fall of 40 in temperature followed by a rise of 23 creates a fall of 17. 

10. Thirty-five steps backward and then 15 steps forward bring a person 
to a point 20 steps backward. 

11. If you have been walking forward at a rate of 25 steps per minute, 
then 6 minutes ago you were 150 steps back from your present position. 

12. If the water level of a river is rising 4 inches per hour, then (a) the 
level will be 24 inches higher at the end of 6 hours; (6) the level was 36 niches 
lower 9 hours ago. 

3. Extension of the number system 

At this point, let us start with the understanding that we have at 
our disposal only the positive numbers and zero. Then, we shall 
extend this number system to include negative numbers, properly 
defined, and shall introduce the operations of algebra for the whole 
new number system. Hereafter, when we refer to any number, or 
use a literal number without limiting its value, we shall mean that it 
is any number of the final number system we plan to develop. 

4. Algebraic operations 

The fundamental operations of algebra are addition, subtraction, 
multiplication, and division. Whenever these operations are intro- 
duced, the results in applying them will be the same as in arithmetic 
when only positive numbers and zero are involved. 

5. Multiplication 

The result of multiplying two or more numbers is called their 
product and each of the given numbers is called a factor of their 
product. To indicate multiplication, we use a cross X or a high dot 
between the numbers, or, in the case of literal numbers, merely 
write them side by side without any algebraic sign between them. 



4 THE FUNDAMENTAL OPERATIONS 

We separate the factors by parentheses if the dot or cross is omitted 
between factors which are explicit numbers, or when a factor not at 
the left end of a product has a plus or a minus sign attached. 

ILLUSTRATION 1. 6-3 = 6X3 = 6(3) = 18. We read 6X3, 6-3, 
or 6(3) as "six times three." 

ILLUSTRATION 2. 4ab means 4 X a X b and is read "four a, b" If 
a = 2 and b = 5, then 4ab = 4(2) (5) = 40. 

If N is any number, we agree that 

(+ 1) X # = N; N X = 0. (1) 

6. Negative numbers 

Let 1 be a new number symbol, called "minus 1," to which we 
immediately assign the following property: 

(-Ijx (-1) = +1. (1) 

By our standard agreement about multiplication by + 1, 

(4-1) x (-1) = -1. (2) 

If P is any positive number, we introduce P as a new number 
symbol, called "minus P," to represent ( 1) X P. That is, 

- P = (- 1) x P. (3) 

We call Pa negative number. Our number system now consists of 
the positive numbers, zero, and the negative numbers. 

ILLUSTRATION 1. Corresponding to + 6, we have the negative number 
6, defined as ( 1) X 6. In concrete applications, corresponding to 
each positive number, we may think of multiplication by 1 as having 
the property of producing a number of opposite type, called negative. 

7. Absolute value 

The absolute value of a positive number or zero is defined as the 
number itself. The absolute value of a negative number is the given 
number with its sign changed from minus to plus. The absolute 
value of a number N is frequently represented by the symbol | N |. 

ILLUSTRATION 1. The absolute value of + 5 is + 5. The absolute value 
of - 5 is also + 5. We read | - 3 | as "the absolute value of - 3." We 
have | - 3 | = 3 and = 0. 



THE FUNDAMENTAL OPERATIONS 5 

8. Inserting signs before numbers 

If we insert a plus or a minus sign before (to the left of) a number, 
this is understood to be equivalent to multiplying it by + 1 or 1, 
respectively. 

ILLUSTEATION 1. -f 5 = (+ 1) X 5 = 5. - 16 = (- 1) X 16. 
+ a = (+ 1) X a = a. - a = (- 1) X a. 

Hereafter, we shall act as if each explicit number or literal number 
expression has a sign attached, at the left. If no sign is visible, it 
can be assumed to be a plus sign because, for every number N, we 
have N = H- N. 

9. Properties of multiplication 

We agree that the following postulates * are satisfied. 

I. Multiplication is commutative, or the product of two numbers is 
the same in whatever order they are multiplied. 

ILLUSTRATION 1. 7 X 3 = 3 X 7 = 21. ab = ba. 

ILLUSTRATION 2. (- 1) X (+ 1) = (+ 1) X (- 1) = - 1. 

II. Multiplication is associative, or the product of three or more num- 
bers is the same in whatever order they are grouped in multiplying. 

ILLUSTRATION 3. 5X7X6 = 5X(7X6) = 7X(5X6)= 210. 

abc = a(bc) = b(ac) = (a6)c. 

We read this " o, b, c, equals a times b, c, equals b times a, c, etc." 

ILLUSTRATION 4. The .product of three or more numbers is the same 
in whatever order they are multiplied: 

abc --= a(bc) (bc)a = bca = (ac)b = acb, etc. 

10. Computation of products 

To compute a product of two numbers, find the product of their 
absolute values, and then 

I. give the result a plus sign if the numbers have like signs; 

II. give the result a minus sign if the numbers have unlike signs. 

* A postulate is a property which is specified to be true as a part of the defi- 
nition of the process. 



6 THE FUNDAMENTAL OPERATIONS 

The preceding facts about a product are arrived at naturally in any 
example by recalling the multiplication properties of 1. State- 
ments I and II are called the laws of signs for multiplication. 

ILLUSTRATION 1. ( 5)( 7) = 4- 35 because 

(- 1) X 5 X (- 1) X 7 - (- 1)(- 1)(5)(7) = (+ 1)(36). 
(- 4) X (+ 7) - (- 1) X 4 X 7 - - 28. 

In a product, the result is positive if an even number (2, 4, 6, ) 
of factors are negative, and the product is negative if an odd number 
(1, 3, 5, ) of factors are negative. 

ILLUSTRATION 2. - 3(- 2)(- 5) (4- 6)(- 5) * - 30. 

ILLUSTRATION 3. 
(- 1)(- 1)(- 1)(- 1) - [(- 1)(- !)][(- 1)(- 1)] - (+ !)(+ 1) - 1. 

11. Division 

To divide a by 6, where b is not zero, means to find the number x 
such that a = bx. We call a the dividend, b the divisor, and x the 

quotient. We denote the quotient by a -5- 6, or r> or a/6. The fraction 

a/6 is read "a divided by 6," or "a over 6." In a/6, we call a the 
numerator and 6 the denominator; also, a and 6 are sometimes called 
the terms of the fraction. The fraction a/6, or a -s- 6, is frequently 
referred to as the ratio of a to 6. 

ILLUSTRATION 1. 36 * 9 = 4 because 4 X 9 = 36. 

The absolute value and sign of any quotient are a consequence of 
the absolute value and sign of a corresponding product. 

To compute a quotient of two numbers, first find the quotient of their 
absolute values, and then apply the laws of signs as stated for products. 

40 
ILLUSTRATION 2. XTo = ~ 4 because 10 X ( 4) = 40. 

i" *" 9 

__ 40 

=4-5 because 5 X (- 8) - 40. 

o 

Note 1. Division is referred to as the inverse of multiplication. Thus, 
if 7 is first multiplied by 5 and if the result, 35, is then divided by 5, we obtain 

7 unchanged. Or, division by 5 undoes the effect of multiplication by 5. 
Equally well, multiplication is the inverse of division. i 



THE FUNDAMENTAL OPERATIONS 7 

EXERCISE 2 

Read each product or quotient and give its value. 
1. 7 X 8. 2. (- 3) X (- 5). 3, (- 2) X (6). 

4. 8 X (- 3). 5. (- 9) X (- 4). 6. (- 3) X 5. 

7. 4 X 0. 8. (+ 5) X (- 3). 9. (- 2) X (+ 4) 

10. (+ 4)(+ 6). 11. (- 7)(- 8). 12. X (- 3). 

13. (- 1)(- 5). 14. - (- 4). 15. - (+ 8). 

16. + (- 7). 17. + (+ 4). 18. - (+ 1). 

19. - (- 3). 20. - (- 1). 21. + (- 9). 

22. (- 7)(- 4)(6). 23. (- 2)(- 7)(- 3)(4). 

24. 5(- 2)(7)(- 3)(- 4). 26. (- 5)(- 3)(- 4)(- 2). 

26. - 6(- 4)(- 3). 27. - 3(5) (2) (- 4). 

28. - (- 7)(- 4). 29. - 4(- 5)(6)(- 3). 

+ 16 . 31 ~ 16 39 -11. 

-+8 31 * 8 3 -3 12 



-42 -36 -28 +39 

^-i8' 36 'T7' 



38. (- 1)(- 1)(- 1)(- 1)(- 1). 39. (- 1)(- 1)(- 1)(- 5). 

State the absolute value of each number. 
40. 16. 41. - 52. 42. - 33. 43. - }. 44. -f 14.2. 

45. Find the product of 3, 5, and 4. 

46. Find the product of 5.3, 4, and + 2. 

47. Find the product of 1.8, 2, and 4. 

Read each symbol and specify its value. 

48. | 7 |. 49. | + 4 |. 50. | - 6 |. 61. | - 31 . 52. - 1.7 . 



53. Compute 4o6c ifa= 3, 6= 4 and, c = 2. 

54. Compute 3xyz if x = 2, y 10, and 2 = 5. 

65. Compute 2abxy if a = 3, b 4, x 3, and y =* 5. 
56. Compute 5hkwz if h 3, k = 2, w = 5, and z 2. 



8 THE FUNDAMENTAL OPERATIONS 

12. Addition 

The result of adding two or more numbers is called their sum. 
Usually, to indicate the sum, we take each of the numbers with its 
attached sign, supplying a plus sign where none is written, and then 
write these signed numbers in a line. Each number, with its sign, is 
called a term of the sum. We usually omit any plus sign at the left 
end of a sum. 

ILLUSTRATION 1. The sum of 15 and 17 is written 15 + 17; the sum 
is 32, as in arithmetic. 

We can state that a plies sign between two numbers indicates that 
they are to be added, because we could write a sum by inserting a 
plus sign before each term and then writing the numbers in a line. 
However, this might introduce unnecessary plus signs. The most 
useful statement is that, when numbers are written in a line, con- 
nected by their signs, plus or minus, this indicates that the numbers 
are to be added. 

ILLUSTRATION 2. The sum of 17 and 12 is represented by 17 12. 
Later, we will justify saying that this equals 5, the value of the expres- 
sion in arithmetic. By using a needless plus sign we could have written 
17 + (- 12) for the sum. 

We specify that the number 1 has the new property that the 
sum of 1 and + 1 is zero. That is, 

-1 + 1 = 0. (1) 

Also, for any number N, we agree that N + = N. 
The operation of addition satisfies the following postulates. 

I. Addition is commutative, or the sum of two numbers is the same 
in whatever order they are added. 

ILLUSTRATION 3. 5 + 3 = 3 + 5 = 8. a + 6 = 6 + o. 

ILLUSTRATION 4. 0= 1 + 1 = + 1 1. 

II. Addition is associative, or the sum of three or more numbers is the 
same in whatever order they are grouped in adding. 

ILLUSTRATION 5. 3 + 5 + 7 = 3 + (5 + 7) - 5 + (3 + 7) = 15. 



THE FUNDAMENTAL OPERATIONS 

ILLUSTRATION 6. 



= c -+ (a + &) = c + a -{- b, etc. 

Thus, the sum of three or more numbers is the same in whatever order 
they are added. 

Addition and multiplication satisfy the following postulate. 
III. Multiplication is distributive with respect to addition, or * 

a(6 + c) = ab + ac. 
ILLUSTRATION 7. 8 X (5 + 7) = (8 X 5) + (8 X 7) = 40 -f 56 = 96. 

% 

1 3. Introduction of the negative of a number 

The negative of a number N is defined as the result of multiplying 
N by 1, so that the negative of N is N. The negative of a 
positive number is the corresponding negative number. The negative 
of a negative number is the corresponding positive number. 

ILLUSTRATION 1. The negative of + 5 is 5. The negative of 5 is 
+ 5 because 

- (- 5) = (- 1) X (- 5) = + 5. 

Thus, we notice that, if~one number is the negative of another, then the 
second number is the negative of the first. 

We observe that the sum of any number and its negative is 0. 

ILLUSTRATION 2. By Postulate III of Section 12, 

- 5 + 5 = [(- 1) X 5] + [(+ 1) X 5] = 5 X (- 1 + 1) = 5 X = 0. 

- a + a = [(- 1) X o] + [(+ 1) X a] = a X (- 1 + 1) = a X = 0. 

14. Subtraction 

To subtract b from a will mean" to find the number x which when 
added to 6 will yield a. This also is the definition used for subtrac- 
tion in arithmetic. Hence, the result of subtracting a positive num- 
ber from one which is no larger f will be the same in algebra as in 
arithmetic. 

- We read a(b + c) as "a times the quantity b + c." 

t This is the only case of subtraction which occurs in arithmetic because 
negative numbers are not used in that field. Thus, (15 25) has no meaning 
until negative numbers are introduced. 



?0 THE FUNDAMENTAL OPERATIONS 

% 

ILLUSTRATION 1. The result of subtracting 5 from 17 is 12 because 
12 + 5 - 17. 

If x is the result of subtracting b from a, then by definition 

a = 6 + x. (1) 

On recalling arithmetic, we would immediately like to write x = a b, 
which means the sum of a and b. To prove that x = a b, we add 
b to a, as given in equation 1 : 

_&=-6 + a = -&+(& + z) = (-6 + 6) + z = z. (2) 

T, 

In (2), we proved that the result of subtracting b from a is obtained 
by adding b to a. Thus, 

to subtract a number, add its negative. (3) 

We define the difference of two numbers a and b as the result of 
subtracting the second number from the first. If x is this difference, 
we proved in (2) that 

x = a - b. (4) 

Thus, we can say that the minus sign in (4) indicates subtraction, 
just as in arithmetic. However, it is equally important to realize 
that (a b) means the sum of its two terms a and b. 

ILLUSTRATION 2. The difference of 17 and 5 is (17 5). The difference 
(17 5) represents the sum of 17 and 5. Also, (17 5) represents 
the result of subtracting 5 from 17, which is 12. 

Note 1. In a difference a 6, the number b which is subtracted is called 
the subtrahend, and the number a, from which b is subtracted, is called 
the minuend. These names will not be mentioned very often. 

1 5. Computation of a sum 

The computation of a sum of two signed numbers or, as a special 
case, the subtraction of one number from another, always will lead 
to the use of one of the following rules. 

I. To add two numbers with like signs, add their absolute values and 
attach their common sign. 

II. To add two numbers with unlike signs, subtract the smaller ab- 
solute value from the larger and prefix the sign of the number having 
the larger absolute value. 



THE FUNDAMENTAL OPERATIONS 11 

ILLUSTRATION 1. 7 + 15 = 22, just as in arithmetic. 

EXAMPLE 1. Add 5 and 17. 

SOLUTION. The sum is - 5 - 17 = - (5 + 17) = - 22 by Rule I. 
To verify this, we use Postulate III of Section 12: 

- 5 - 17 = [(- 1) X 5] + [(- 1) X 17] = (- 1) X (54-17) - - 22. 

ILLUSTRATION 2. The sum of 6 and 6 is zero or, as in arithmetic, the 
result of subtracting 6 from 6 is zero: 6 6 = 0. 

ILLUSTRATION 3. The sum of 20 and 8, which is the same as the result 
of subtracting 8 from 20, is 20 - 8 = 12. 

ILLUSTRATION 4. The sum of 20 and 8 is, by Rule II, 

- 20 + 8 = - (20 - 8) - - 12. 
To verify this, we recall that 20 = 12 8. Hence, 

-20 + 8= -12-8 + 8= -12 + 0= -12. 
Essentially, + 8 cancels 8 of 20 and leaves 12. 

EXAMPLE 2. Subtract 15 from 6. 

FIRST SOLUTION. Change the sign of 15 and add: 

result'^ - 6 + 15 = 9. 

SECOND SOLUTION. Write the difference of 6 and 15: 
result = - 6 - (- 15) = - 6 + 15 = 9. 

CHECK. Add - 15 and 9: - 15 + 9 = - 6. 

16. Algebraic sums 

An expression like 

c - 3 - 5a + 76 (1) 

is referred to as a sum, or sometimes as an algebraic sum to emphasize 
that minus signs appear. Expression 1 is the sum of the terms 
c, 3, 5a and 76. In connection with any term whose sign is 
minus, we could describe the effect of the term in the language of 
subtraction instead of addition, but frequently this is not desirable. 
To compute a sum of explicit numbers, first eliminate parentheses 
by performing any operations indicated by the signs. Then, some- 
time^, it is desirable to add all positive and all negative numbers 
separately before combining them. 



12 THE FUNDAMENTAL OPERATIONS 

ILLUSTKATION 1. Since + ( 12) = 12 and ( 7) = + 7, 
- 16 - (- 7) + (- 12) + 14 = - 16 + 7 - 12 + 14 = - 28 + 21 = - 7. 
Or, we could compute mentally from left to right: 

-16 + 7w-9; -12 is -21; + 14 is - 7. 

Note 1. It is undesirable to use unnecessary plus signs. Thus, in place 
of + (- 3) + (+ 8) we write -3 + 8. 

EXAMPLE 1. Compute (5z 3o6) if x 3, a = 4, and 6 = 7. 
SOLUTION. 5x - 3ab = [5(- 3)] - [3(- 4) (7)] = - 15 + 84 = 69. 

Comment. Notice the convenience of brackets to show that the multi- 
plications above should be done before computing the sum. 

1 7. Summary concerning the number zero 

We have mentioned that the operation a -*- 6 is not defined when 
b = 0. That is, division by zero is not allowed. However, no ex- 
ception has arisen in multiplying by zero, adding or subtracting 0, 
or in dividing zero by some other number. Thus, if N is any number, 

AT + = AT; N-Q = N; # X = 0. 

If N is not zero, then TT = because = X N. 

r 

ILLUSTRATION 1. 6 + = 0. 3X0 = 0. = = 0. 

5 

Note 1. Contradictions arise when an attempt is made to define division 
by zero. Thus, if we were to define 5/0 to be the number x which, when 
multiplied by zero, will yield 5, then we would obtain 5 = Q-x, or 5= 0, 
which is contradictory. 

EXERCISE 3 

Compute each sum mentally. 
1. 20 + 7. 2. - 28 - 9. 3. - 13 - 5. 

4. 36-4. 6. 28 - 15. 6. 12 - 17. 

7. - 16. 8. 6 - 25. 9. - 13 - 19. 

10. -35 + 6. 11. 7 - 42. 12. - 25. 

13. - 6 + 6. 14. -18 + 3. 15. -13 + 0. 

16. 16.8 - 15.2. 17. - 13.7 - 4.5. 18. - 5 + 3.4. 



THE FUNDAMENTAL OPERATIONS 13 

What is the negative of the number? 

19. 7. 20. - 4. 21. - 13. 22. 0. 

Without writing the numbers in a line, (a) add the two numbers; (b) sub- 
tract the lower one from the upper one. 

23. 45 24. 36 26. - 13 26. - 12 27. - 53 

16 - 19 - 17 + 18 4- 17 



28. 


- 13 
-24 


29. 



17 


30. 



- 15 


31. - 


-4.3 
7.6 


32. 


-.87 
1.35 



Find the value of the expression in simple form. 
33. + (- 5). 34. - (- 8). 35. - (+ 7). 36. + (- 6). 

Compute each sum. 
37. - (- 3) + 7. 38. 4- (4- 4) - (- 8). 39. - 6 + (- 4). 

40. - (- 4) 4- (- 9). 41. 16 - (- 6). 42. - 15 - (4- 12). 

43. 12 - 7 - 5 4- 3. 44. - 16 - 3 + 5 - 7 4- 6. 

46. - 10 4- 17 - 8 4- 14. 46. 43 - 25 - 6 4- 8 4- 12. 

47. - 16 4- 14 4- 36 4- 8. 48. 3 - 16 4- 17 - 8 - 9. 

49. - 3 4- (- 5) - (- 16) - (- 4). 
60. 5 - 7 4- (- 3) - (4- 16) - 3.7. 

Find (a) the sum and (b) the difference of the two numbers. 
51. 16 and 12. 62. 15 and - 3. 63. - 33 arid - 7. 

64. - 14 and - 5. 66. 6 and 38. 66. 15 and 67. 

Compute the expression when the literal numbers have the given values. 

67. 16 4- Bab] when a = 4 and b = 7. 

68. 2a 3cd; when a = 2, c = 3, and d 5. 



69. 26 7 4- 4oc; when a = 5, b = 7, and c = 3. 

60. x 2yz 3; when x = 4, y = 2, and z 7. 

61. 2a 4- 56 - 16.3; when a = 5 and 6 = - 6. 

Find (a) the sum of the numbers; (b) their difference; (c) their product; 
(d) the quotient of the first divided by the second. 

62. - 60 and - 15. 63. and - 14. 64. - 12 and 4. 
66. 6 and - 3. 66. - 52 and - 13. 67. and 23. 



14 THE FUNDAMENTAL OPERATIONS 

1 8. Real number scale 

On the horizontal line in Figure 1, we select a point 0, called the 
origin, and we decide to let this point represent the number 0. We 
select a unit of length for measuring distances on the line. Then, 
if P is a positive number, we let it be represented by the point on 

i i i i ixj i i i i i i i i i i i ix i i i 

7 6 -543-2-1 1 2 3 4 56 

Fig. 1 

the line which is at P units of distance from to the right. The 
negative number P is represented by the point which is at P units 
of distance from to the left. The points at whole units of distance 
from represent the positive and negative integers. The other 
points on the line represent the numbers which are not integers. 
Thus, all real numbers are identified with points on the real number 
scale hi Figure 1. If M is any real number, it can be thought of as 
a measure of the directed distance from to M on the scale, where 
OM is considered positive when the direction from to M is to the 
right and negative when the direction from to M is to the left. 

19. The less than and greater than relationships 

All real numbers can be represented in order from left to right on the 
scale in Figure 1. We then say that one number * M is less than a 
second number N, or that N is greater than M, in case M is to the 
left of N on the number scale. We use the inequality signs < and > 
to represent less than and greater than, respectively. This definition 
includes as a special case the similar notion used in arithmetic for 
positive numbers. The present definition applies to all real numbers, 
positive, negative, or zero. If a 7* 6,f then either a < b or a > b. 

ILLUSTRATION 1. In each of the following inequalities, we verify the 
result by placing the numbers on the scale in Figure 1. Thus, 7 < 3 
because 7 is situated to the left of 3 in Figure 1. We read this inequality 
as " 7 is less than 3." 

4<6; 0<8; - 4 < 0; - 6 < 5. 

To say that P > is equivalent to saying that P is positive, be- 
cause the numbers to the right of in Figure 1 are positive. To say 
that M < is equivalent to saying that M is a negative number. 



* Until otherwise indicated, the word number will refer to any real number. 
t We read 7* as "not equal." 



THE FUNDAMENTAL OPERATIONS 15 

20. Numerical inequality 

We say that one number 6 is numerically less than a second number 
c in case the absolute value of bis less than the absolute value of c. To 
distinguish this relation from ordinary inequality, we sometimes place 
the word algebraically before greater than or less than when they are 
used in the ordinary sense. 

, 

ILLUSTRATION 1. 5 is numerically less than 9 because | 5 1 < 1 9 1. It 
is also true that 5 is algebraically less than 9, because 5 < 9. 

ILLUSTRATION 2. We see that 3 is numerically less than 7 because 
| - 3 | = 3 and | - 7 | = 7, and 3 < 7. On the other hand, - 7 < - 3. 
Thus, 3 is algebraically greater than 7 but numerically less than 7. 

In Illustration 2, we observe a special case of the fact that, if one 
negative number b is algebraically less than a second negative num- 
ber c, then b is numerically greater than c. 

EXERCISE 4 

Construct a real number scale 10 inches long, and mark the locations of the 
positive and negative integers from 10 to + 10, inclusive. Then, read each 
inequality and verify it by marking the two numbers on your scale. 

1. 5 < 9. 2. < 7. 3. - 3 <0. 

4. - 3 < 8. 5. - 5 < 2. 6. - 7 < - 4. 

7. 8 > - 9. 8. 5 > - 3. 9. - 3 > - 10. 

Mark the numbers in each probkm on your scale and decide which sign, 
< or > , should be placed between the numbers. 

10. 7 and 9. 11. - 2 and 5. 12. and 8. 

13. - 6 and 0. 14. - 3 and 3. 15. - 2 and - 7. 

16. - 9 and 10. 17. 7 and 5. 18. 8 and - 3. 

19. - 6 and - 3. 20. - 7 and - 9. 21. 8 and - 10. 

Read the inequalities and verify that they are correct. 
22. |-7| <|8|and - 7 < 8. 23. | 4| < | 9| and 4 < 9. 

24. | - 3 1 > 1 1 but - 3 < 0. 26. | - 8 | > 3 but - 8 < 3. 

26. | -5| <|"-7|but -5> -7. 

27. ] - 6 i < ] - 9 1 but - 6 > - 9. 



76 THE FUNDAMENTAL OPERATIONS 

Stale which number (o) is algebraically greater than the other; (6) is nu- 
merically greater than the other. 

28. - 8; 6. 29. - 5; - 3. 30. 7; 4. 

31. 0; - 3. 32. 5; 0. 33. - 2; 7. 

34. - 3; - 7. 36. 2; - 6. 36. - 8; - 3. 

21 . Signs of grouping 

Parentheses, ( ), brackets, [ ], braces, { }, and the vinculum, 

, are symbols of grouping used to indicate terms whose sum is to 

be treated as a single number expression. Any general remark about 
parentheses which follows will be understood to apply as well to any 
other symbol of grouping. 

Parentheses are useful for enclosing and separating algebraic ex- 
pressions written side by side as an indication that they are to be 
multiplied. 

ILLUSTRATION 1. 3( 5) means 3 times 5, or -f 15. 

In reading algebraic expressions, the student may use the words 
"the quantity 1 ' as he comes to the left marker for any symbol of 
grouping enclosing more than one term. 

ILLUSTRATION 2. (3 5o)(2 + 6a) is read "the quantity 3 5a times 
the quantity 2 + 6a." If a = 4, 

(3 - 5o)(2 -f 60) - (3 - 20) (2 + 24) = (- 17) (26) = - 442. 

Parentheses can be employed to avoid ambiguity hi regard to the 
order of application of the fundamental operations. 

ILLUSTRATION 3. Doubt arises as to the meaning of (9 6 -f- 3). Does 
it mean (9 6) -s- 3, which equals 1, or does it mean 9 (6 -5- 3), which 
is (9 2) or 7? The two possible meanings were written without ambiguity. 

ILLUSTRATION 4. If a = 2, b = 3, and c = 5, then 

5-3o& + 2oc = 5- [3(2)(- 3)] + [2(2) (5)] 

= 5 - (- 18) + 20 = 5 + 18 + 20 = 43. 

A factor multiplying a sum within parentheses should be used to 
multiply each term of the sum. 

ILLUSTRATION 5. - 3(2a) = (- 3) (2) (a) = - 6a. 



THE FUNDAMENTAL OPERATIONS 17 



ILLUSTRATION 6. - 4(a - 26 + 7) - - 4a - 4(- 2fe) - 4(7) (1) 

- - 4a + 86 - 28. (2) 

The student should write (2) without writing the right-hand side of (1). 



ILLUSTRATION 7. - ( 5 + 3z y) = (' 1)(- 5 + 3x y) 

= 5 - Zx + y. 

The presence of a plus sign or a minus sign before any algebraic 
expression indicates that it is to be multiplied by + 1 or 1, re- 
spectively. Multiplication by + 1 would leave a sum unchanged, and 
multiplication by 1 would change the signs of all its terms. These 
remarks justify the following rules. 

I. To remove or to insert parentheses preceded by a plus sign, rewrite 
the included terms unchanged. 

II. To remove or to insert parentheses preceded by a minus sign, re- 
write the included terms with their signs changed. 



ILLUSTRATION 8. + (3r 7 -f %) = &c 7 -f 

- (- 2 + 5x - 7y) = (- 1)(- 2 + 5x - 7y) = 2 - 5x + 7y. 



ILLUSTRATION 9. In the sum on the left-hand side below, we enclose all 
terms after the first in parentheses preceded by a minus sign. 

5a - 2c + 3d - 8 = 5o - (2c - M + 8). 

The signs of the terms enclosed were changed in order that, if the parentheses 
were removed, the original terms would be obtained. 

In performing an operation which removes parentheses, if only 
explicit numbers are involved, it is best to compute each sum within 
parentheses before they are removed. 

ILLUSTRATION 10. - 3(2 - 5 + 7) = - 3(4) = - 12. 

EXERCISE 5 

f 

Compute the expression. 
1. (- 2)(- 5). 2. 7(- 3). 3. (- 6)(4). 

4. (0)(- 3). 5. - (- 5). 6. + (- 4). 

7. 2(- 5 + 9). 8. - 3(5 - 14). 9. - 4(- 5 - 6). 

10. (- 8 + 6)(- 5 -f 12). 11. (7 - 3) (16 - 5 - 2). 



J8 THE FUNDAMENTAL OPERATIONS 

12. Compute (36 4oc 7) if 6 2, a 3, and c = 5. 

13. Compute (- 6 + 3oc - 6c) if a - 3, b - 4, and c = - 2. 

14. Compute (3 - 5a)(- 2 + 60) when a = 4. 

Compute (c 2d)(4d 3c) w#A <Ae pfoen vaJwes for c and d. 
16. c = 4; d = 3. 16. c = - 2; d = 5. 17. c = - 2; d = - 3. 

Rewrite, by performing any indicated multiplication and removing paren- 
theses. Evaluate, if no letters are present. 
18. - (17 - 8 - 3). 19. - (2 - 5 + 16). 20. + (3 - 6 + 15). 

21. - (2a - 56 + c). 22. + (- 3 + 7a - b). 23. + (31 - 5a + y). 
24. - (- 3 - 5z + 4y). 26. - (80 - 36 - c). 26. - (2x - 5y - 9). 
27. 3(5a). 28. - 2(4c). 29. - 3(- 5o). 30. 5(- 3x). 
31. 4(2o - 3). 32. - 2(x - y + 8). 33. - 5(3 - a - 6c). 
34. 2(- 5 + 7a - 46c). 36. 3(6 - 4a + 56). 36. - 4(- 5 + 66). 

Rewrite, enclosing the three terms at the right in parentheses preceded by a 
minus sign. 
37. _ 5 + 7 a _ 46. 38. - 60 -f 46 - c. 39. 6 - 3x - 4y. 

40. 2a - 3 + 56 - c. 41. 16 - 4a - 6 -f 3c. 

42. - 13 + 5 - c -f 06. 43. 2ac -f- 3 - 5a + 4c. 

44. Compute (5 - 17) -5- 3; 5 - (17 -4- 3). 

22. Similar terms 

Two terms such as 5a6c-and 7o6c, where the literal parts are the 
same, are called similar terms or like terms. In a term such as 5a6c, 
the explicit number which is a factor is called the numerical coefficient 
of the term, or, for short, the coefficient. 

ILLUSTRATION 1. The numerical coefficient of 5o6c is 5; of 7o6c is 
7; of abc is 1. We never write the coefficient when it is 1. Thus, we 
would never write Io6c for abc. 

A sum of similar terms can be collected (added) into a single term, 
by use of the distributive property of multiplication. 

To collect a sum of similar terms, add their numerical coefficients 
and multiply the result by the common literal part. 



THE FUNDAMENTAL OPERATIONS 19 

ILLUSTRATION 2. 5ab + lab - ab(5 -f- 7) 12o6. 

If we think of "06" as a concrete object, the result here is obvious: 5 of 
the objects plus 7 of them equals 12 of them. 

ILLUSTRATION 3. 9o6 + 4o6 = ab( 9 + 4) = o6( 5) = 606. 

In finding a sum of algebraic expressions, we collect similar terms. 
A direction to collect terms means to collect similar terms. 

EXAMPLE 1. Find the sum and the difference of 

3a 5y - 8 and 3y 2a 6. 

FIRST SOLUTION. 1. The sum is (3a 5y 8) + (3y 2o 6) 
= 3a - 2a - 5y + 3y - 8 - 6 = a - 2y - 14. 

The student should learn to omit the intermediate details hi such a solution. 
2. The difference is 

(3a 5y 8) (3y 2a 6) = 3a - 5y - 8 - 3y -f 2o -f 6 
= 3a + 2a - 5y - 3y - 8 -f- 6 = 60 - Sy - 2. 

SECOND SOLUTION. To find the sum, arrange the given expressions with 
like terms in separate columns, and add. To find the difference, change 
the signs in (3y 2a 6), or take its negative, and add similarly. 

A 1 1 i wv vi/ <-* jiii *^O """ *^2/ *"" o 

^Lrta: < . ^oa: < i 

^<z ot/ + o 



r 3a _ 5y .. 8 r 

\ - 2a + 3y- 6 AflW< \ 



= a 2y 14. Difference = 5a Sy 2. 

In finding the difference, we could have changed the signs of ( 2o + 3y 6) 
mentally without rewriting at the right. Thus, from the columns for the 
sum: subtracting 2a from 3a gives 5a; etc. 

EXAMPLE 2. Perform indicated multiplications, remove parentheses, and 
collect terms: 

3(4a - 3xy - 56 - 2) - 2(- 5 + 3o -f- 56 - fay). (1) 

SOLUTION. The sum equals 

12o - 9xy - 156 - 6 + 10 - 60 - 106 + 12*y 

= 6a -f 3xy - 256 + 4. (2) 

Comment. We can obtain a relatively certain check on the work by sub- 
stituting values for the letters in (1) and (2). The results should be equal. 
In checking by this method, avoid substituting 1 for any letter. 

CHECK. Substitute a = 2, b = 4, x = 3, and y = 4: 
In (1): 3(8 - 36 - 20 - 2) - 2(- 5 + 6 + 20 - 72) = - 48. 

In (2): 12 -f 36 - 100 + 4 = 52 - 100 - - 48, This checks. 



20 THE FUNDAMENTAL OPERATIONS 

23. Nests of grouping signs 

If a symbol of grouping encloses one or more other symbols of 
grouping, remove them by removing the innermost symbol first, and so 
on until the last one is removed. Usually, we enclose parentheses 
within brackets, and brackets within braces. . 

ILLUSTRATION 1. [3t/ (2x 5 4- 2)] 

= - [3^ - 2x -f 5 - z] = - 3y + 2x - 5 + z. 



ILLUSTRATION 2. - [16 - (2 - 7) - (- 3 + 5)] 

= - [16 - (- 5) - (2)] = - [16 + 5 - 2] = - 19. 

EXERCISE 6 

Collect similar terms. 
1. 3o + 80. 2. - 56 + 76. 3. - 13z - 5x. 

4. 19o6 -f 5o6. 6. - 2cd + 8cd. 6. 5xy 5xy. 

7. 3x - a + 2z - 5a. 8. - 3a + 46 -f 7a - 96. 

9. - 2o - 3c + 5a - 8c. 10. 6z - 4y + 12z + 3y. 

11. lie - 5cd + 8c - 13crf. 12. Qh - 5kw - llh + 7kw. 

(a) Add the two expressions; (6) subtract the lower one from the upper one. 
After each operation, check by substituting convenient values for the letters in 
the given expressions and the final result. 

13. - 4a + 76 - 3 14. 3a - 56 + 5 

2a - 96 + 7 - 9a - 46 - 7 

16. 3x - 5a6 - c 16. - 2a - 46c - 2d 

6a6 - 3c 5a - 66c -f d 



17. 3m - 5k - h 18. - 7r -f 3* - 6 

- 6m + 4fc - 5h 8r - 5s -f 9 



Find J/ie sum of the three expressions separated by semicolons. First arrange 
the expressions with like terms in separate columns. 

19. 3x - 2y - 5; - fix + 7y - 8; - 12z - 8y -h 13. 

20. - 3 + o - 6c; 26c - 3a + 5; 4a - 56c - 9. 

21. 2oc - 5xz/ - 86; 4cy 3ac -h 56; 76 - 6xy - 4ac. 

22. 7ad - 56 - 4a; 6a - 3ad + 56; 3o -h 5arf - 96. 



THE FUNDAMENTAL OPERATIONS 21 

Remove parentheses and collect similar terms. 

23. 2(a - 36) - 5(6 - 2a) + 3(- a - 36). 

24. 5(x - 3y + 5) - 2(* + 2? - 4) - 3(- 4z - y + 6). 
26. 6(2a - h - k) - 3 (a - 4A + 5Jfc) + 2(4A - 3a - &). 

By w$e of parentheses, write an expression for the quantity described; then 
remove parentheses and collect terms. 

26. Subtract 3a - 26 and 2a + 56 from - a 56. 

HINT. The result equals (- a - 56) - (3a - 26) - (2a + 56). 

27. Subtract 2a - 3y - 5 from 5a + 7y - 8. 

28. Subtract 3a 7h and 5A 6a from the sum of 2o 3h and 5a -f 



29. Multiply 3o - 5y - 3 by 2, and - 5o -f 7y - 5 by - 3, and then 
add the results. 

Remove all signs of grouping and collect terms. 

30. - [4o - (2a + 3) J 31. - [2* + (3 - 40 J 

32. - (a - 2) - [2o - (a - 3)]. 33. a + [6 + (a - 6)]. 

34. 9 - - (6 - 2)]. 36. 2r + [> - ( + 4r)J 

36. - [> - (y + 3)]. 37. 2i/ - 5 + [3 - 2(y - 2) J 

38. 3o - [3a - 4(5 - a)]. 39. 3 - [2x - 2[> - (2x - 5)]). 

40. - {a - [a - (2a - 7)]}. 41. - {26 - [6 - (36 - 4)]}. 

42. - 3(2z - 3y) + 2[2 - (5x - y)] - (* + 7y - 8). 

43. 2(3/i - A; - 5) - 3[A - 2(* - 3)] - 4(A - 2k -h 2). 



CHAPTER 



2 



INTRODUCTION TO FRACTIONS AND 
EXPONENTS 



24. Fractions in lowest terms 

The basic properties of fractions as met in arithmetic are primarily 
consequences of the definition of division. These properties extend 
immediately to fractions as met in algebra, where the only essential 
new feature is the introdmction of negative numbers hi the fractions. 
We shall recall and use the properties of fractions without rehearsing 
the sequence of definitions and proofs which, logically, would be 
necessary hi building a foundation for the use of fractions hi algebra. 

FUNDAMENTAL PRINCIPLE. The value of a fraction is not altered if 
both numerator and denominator are multiplied, or divided, by the 
same number, not zero. 

ILLUSTRATION 1. 



5 


5X3 15 


a ac 


7 
36 


7X3 21 
36 -M2 3 


b~ be 


84, 


84-5-12 7 



ILLUSTRATION 2. On multiplying numerator and denominator by 1 
below, we obtain 

- 3 (- 1)(- 3) 3 



-4 (- 1)(- 4) 4 

We say that a fraction is in lowest terms if its numerator and de- 
nominator have no common factor except + 1 and 1. 

To reduce a fraction to lowest terms, divide numerator and denomina- 
tor by all their common factors. 

ILLUSTRATION 3. = = =- (Divide out ac) 

7acy 7y 



INTRODUCTION TO FRACTIONS AND EXPONENTS 23 

210 0X3X7X2 7X2 14 



ILLUSTRATION 4. 



135 3X3X3X3 3X3 9 



In the preceding line we divided out the factor 5 and one 3 from numerator 
and denominator. 

25. Change in sign for a fraction 

If the numerator or denominator of a fraction is multiplied by 1, 
the sign before the fraction must be changed. These actions are 
equivalent to multiplying the fraction by two factors 1, whose 
product is H- 1. This keeps the value of the fraction unaltered. 

T - 0-3 (- l)(o - 3) 3-Q 
ILLUSTRATION 1. = = - ~ = = 

26. Multiplication and division of fractions 

To multiply one fraction by another, multiply the numerators for a 
new numerator and multiply the denominators for a new denominator. 

3 ^ 6 18 a c ac 

ILLUSTRATION 1. = X = = ^=- i* j ** cr 

5 7 oo o a oa 

To divide one fraction by another, invert the divisor and multiply the 
dividend by this inverted divisor. 

T o 4 . 3 4 7 28 

ILLUSTRATION 2. F^JT^K'Q 7c' 

.) i O o ID 

a 

a _._ c _b _ a d _ ad 
b ' d c be be 

d 

It is frequently useful to recall that any number can be expressed 
as a fraction whose denominator is 1. By use of this fact we verify 
the following results. 

To multiply a fraction by a number, multiply the numerator by the 
number. 

To divide a fraction by a number, multiply the denominator by the 

number. 

7 

_/5\ 75 35 7. 571 7 

ILLUSTRATION^ 7( g ) = rg =* T 5 "" 4 = 4 = 5*4 " 20* 

1 



24 INTRODUCTION TO FRACTIONS AND EXPONENTS 

T A , . 6 5 . 6 5 7 35 

ILLUSTRATION 4. 5 = = T ^ = 7 = -- 

7 1 / 1 o o 

/a\ c a ac 

ILLUSTRATION 5. c (r) = T ' I = T" 

\o/ 16 o 

f? _i. ?^_ ? I 
b ~ b ' I 6 c be 



EXERCISE 7 

Express the result as a fraction in lowest terms without a minus sign in 
numerator or denominator. 



1 18 

10 30' 


00 J.K 

2 3 . 

72 21 


A 66 
4 '7f 


6 '35' 


fi 21 


OfJ K7 

7 . 8 


9 63 - 


10 42 - 


6 ' 14* 


120 38 


9 ' 81 


10 ' 63 


11 ~ 5 


>fl O 


Id ~ ^. 


15 78 - 


11. 3 . 


l2 ' - 7 13 * - 2 


14 ' 35 


15 ' - 26 


16 M 


n^ 


10 3% 


19 27a . 


16 ' 16c* 


OQ / >* / i / 


lo. ~ 

6?/ 


18 ' 6a6 


he 

*>- 


21 4a 


- 2 


^^ 5ad! 

oo . . 


^ 1 * Q^ 

da 


22> 66 


Z3% - 3d 


u. 3 4 

VL*L 

^ 5 7 


*{! 


9R 15 21 
26. y- T - 


27 5.?. 
4 '' 2 rf 


28. H- 

2i O 


29 n -^- 
29. y . 5 


30 15 J- 5 - 


31. ^^l 4 

38 57 



3 2 .5(|). 33. 21 (?). 34. g). 36. eg). 

36. 5(4). 37. (6). 38. 1 + 2. 39. | + 6. 

40. y + 15. 41. y - 6. 42. || -*- 2a. 43. y * d. 

.,325 .. 3o 56 2c ... x 2c 3d 

' 



l_ 8 12a 

48. - 49. 60. - 



5c 15 



INTRODUCTION TO FRACTIONS AND EXPONENTS 25 



15 


14 


3A 


4w 




51. 1- 


62 - fo' 


M k 

53 -T 


^ yet- 

54. TT-- 

2w 




65. 6 -*- 1- 


66. 5 * ~ 


67. 5 + 4' 


68. 5d -T- 


3d 


2 


7 


1U 




c 


69 4- 


* 


61. - < 

c 


AO 5w? 
62. 7= 

15w> 





7 9 d 8 

-Ml -H)H)H)H)- -(-SX-SX-8- 

27. Positive integral exponents 

We write a 2 to abbreviate a -a, and a 8 for a -a' a. We call a 2 the 
square of a and a 3 the cube of a. 

ILLUSTRATION 1. 5 2 = 5-5 = 25. 5 3 = 5* 5' 5 = 125. 

If m is any positive integer,* we define a m by the equation 

9 

a m = a-a-a- -a. (m factors a) (1) 

We call a m the mth power of the base a and call m the exponent of 
this power. The exponent tells how many times the base is used as 
a factor. 

ILLUSTRATION 2. 3 4 = 3-3-3-3 = 81. (- 4) 2 = (- 4)(- 4)'= + 16. 
(_ 4)3 = (- 4)(- 4)(- 4) = - 64. (?)' = . f.f = |. 



We notice that, by the laws of signs, an even power of a negative 
number is positive and an odd power is negative. 

By definition, b 1 = b. Hence, when the exponent is 1 we usually 
omit it. Thus, 5 means 5 1 and y means y l . 

28. Index lows 

The following laws for the use of exponents are called index laws. 
At present we will illustrate them, and verify their truth in special 
cases. The proofs of the laws will be given later. 

* Until otherwise specified, any literal number used in an exponent will 
represent a positive integer. 



26 INTRODUCTION TO FRACTIONS AND EXPONENTS 

I. In multiplying two powers of the same base, add the exponents 

a m <i n = c m+n . 

ILLUSTRATION 1. a 3 o 2 = a 3+2 = a 6 , because 

a 3 o 2 = (aa-o)(a-o) = o0'O-a-a = a 5 . 
szV = x^x 4 = z 1+2+4 = a: 7 . 

II. 7n obtaining a power of a power, multiply the exponents: 

(<z m ) n = fl 17 " 1 . 

ILLUSTBATION 2. (a 3 ) 2 = a (3X2) = a 6 , because 

(a*) 2 = a 3 -a 3 = (a-a-a)(a-a-a) = a-a-a-a-a-a = a 6 . 

III. To obtain a power of a product, raise each factor of the produc 
to the specified power and multiply: 

(abc) n = a n 6 n c n . 

ILLUSTRATION 3. (o6) 3 = o 3 ^ 3 , because 

(oft) 3 = ab'ab'ab = (a-0'o)(6-6-6) 



IV. To obtain a power of a fraction, raise the numerator and denom 
inator to the specified power and divide: 

fa\ n _ a? 
W " &"' 

T A fa\* X 4 u /^\ 4 X X X X X 4 

ILLUSTRATION 4. (-) =-7 because (-1 = ------- ~i* 

\yl y* \yi y y y y y 4 

T c / 2 \ 4 24 16 /T 

ILLUSTRATION 5. (^) = ^ = o7* (Law IV 

(-!)'-[<-(!)]'= -"(1)'- -I- -fi- 



If we recall that an odd power of a negative number is negative, we ma; 
abbreviate the preceding line by omitting the first two steps. 

ILLUSTRATION 6. (2o 2 6) 3 = 2 3 (a 2 ) 3 6 3 = 8a 6 6 3 . 

ILLUSTRATION 7. 


<IAWS 



\ W / (A') 4 A" A 12 



,, 



INTRODUCTION TO FRACTIONS AND EXPONENTS 27 

EXERCISE 8 

Compute by the definition of an exponent. 
1. 2 4 . 2. 5 3 . 3. 10 2 . 4. 10 s . 5. 10 4 . 

6. 10 5 . 7. (- I) 2 . 8. (- I) 3 . 9. (- I) 4 . 10. (- 5)*. 

11. (- 3) 3 . 12. (- 2) 6 . 13. (- 5) 3 . 14. 6 2 . 15. (- 3)*. 

16. 10*. 17. (- 10) 3 . 18. () 4 . 19. (f) 3 . 20. ($). 

21. (- |) 3 . 22. to) 4 . 23. - 2 4 . 24. - 3 3 . 25. - 6 s . 

26. 2(3 4 ). 27. 3(- 5 2 ). 28. 6(10*). 29. 5(4 3 ). 30. - 2(- 5). 

31. For what values of the positive integer n will ( l) n be positive 
and when will it be negative? 

32. Compute 3a 2 6 if a = - 2 and b = 4. 

33. Compute 5xy* if x = 3 and y 2. 

34. Compute - 4W if h = - 3 and k = 2. 

35. If x is positive or negative, what is the nature of x 2 , positive or nega- 
tive? If x is negative, what is the nature of x 3 and of x 4 ? 

Perform the indicated operation by use of the laws of exponents. 



36. a b a 4 . 


37. 3V. 


38. xx 3 . 


39. y*y 7 . 


40. 10H0 6 . 


41. xx*x*. 


4$ T yy*y 7 . 


43. 6*6*6. 


44. 3 3 3 B . 


45. a h a k . 


AA /.i/2fi 
vv. XX. 


47. (a*) 4 . 


48. (2 2 ) 3 . 


49. (xyY. 


50. (cd) 3 . 


61. (3x) 2 . 


62. (c 3 ) 6 . 


63. (A 3 ) 2 . 


64. (ox 2 ) 3 . 


56. (36) 4 . 


/c\ 3 

' w 


67 - 4 - 


58. (?) 3 - 
\a/ 


. 


60. (?) 4 . 


61. (f). 


62. (- f) 4 . 


83. (- J) 


64. I-}*- 


/2/i\ 4 
66. I ; ) 


J% 4 / O*C \ 

66. ( 


67. (- S 



68. (xV) 4 - 69. (3C 2 ) 3 . 70. (2o 2 6 4 ) 4 . 71. (3xu;) 4 . 

72. (- 2A 2 ) 3 . 73. (- 3x 2 ) 4 . 74. (- 5xt/ 2 ) 3 . 75. (- 

(O/7\4 /^ 2 rr\' / 3 \* / 

S)- 77 -(^)- 78 -(-^)- ro -(- 

80. Find the value of x 3 - 3x 2 -h 4x - 7 when (a) x = 3; (6) x * - 2. 



28 INTRODUCTION TO FRACTIONS AND EXPONENTS 

29. Integral rational terms 

At present, in any sum, the typical term will be either an explicit 
number or the product of an explicit number and powers of literal 
numbers, where the exponents are positive integers. The explicit 
number is called the numerical coefficient, or for short the coefficient 
of the term. A term of this variety, or a sum of such terms, is said 
to* be integral and rational * in the literal numbers. 

An algebraic sum is called a monomial f if there is just one term, 
a binomial if there are just two terms, and a trinomial if there are 
just three terms. Any sum with more than one term also is called a 
polynomial. 

ILLUSTRATION 1. 3z + 7ab is a binomial. 

ILLUSTRATION 2. In the trinomial 8 + x 3a6 2 , the terms are 8, 
x, and 3a& 2 , whose numerical coefficients are 8, 1, and 3, respectively. 

ILLUSTRATION 3. 5z 2 x 7 is an integral rational polynomial in x. 

To multiply two integral rational terms, multiply their numerical 
coefficients and simplify the product of the literal parts by use of the 
law of exponents for multiplication. 

ILLUSTRATION 4. 6a6 2 (3a 2 6 4 ) = (6)(3)(aa 2 6 2 6 4 ) = 18a 3 6 6 . 



To multiply a polynomial by a single term, multiply each term of the 
polynomial by the single term and form the sum of the results. 



ILLUSTRATION 5. 5(3z 2 - 2x - 5) = 15z 2 - lOz - 25. 



EXERCISE 9 

Perform the indicated multiplications and simplify the results by use of the 
law of exponents for multiplication. 

2. 32/(22/ 6 ). 3. ab(3a). 4. 



5. - 5(32 7 ). 6. - 2a 2 (3a 9 ). 7. cd^cP). 8. x(- 

* The word integral refers to the fact that the exponents are -integers. The 
force of the word rational will be pointed out later. 

t This name need not be used very often because the simple word term is 
just as desirable. 



INTRODUCTION TO FRACTIONS AND EXPONENTS 29 

9. - 4fl(4). 10. 3c(- c 3 ). 11. - 2z(+ xy*). 12. - **(- 2z 2 ). 

13. - ax*(- 2o 2 x). 14. 5oy(- 2xy 2 ). 15. 2a6(- 4a 8 6 2 ). 

16. - 8m 3 (- 2m). 17. - 4r z h(- 6rA 4 ). 18. - Gc 2 ^- 3cd). 

19. 3(4z - y). 20. - S(5x - 2z 2 ). 21. - 5(3 - 4a). 

22. 4(2a - 56). 23. - 3(- 5x - 4y). 24. x(2x 2 - 3x). 

25. 2x(- 3z - 5z 3 ). 26. o(a 2 - a 3 ). 27. 2* 2 (3 - z 4 ). 

28. - 4a(l - 2a6). 29. - 3fc(fc - M). 30. ah\a - 
31. - 5w(2 - 3u> 4- 4i^). 32. - 
33. 3a6 2 (2a 2 6 3 )(a6 4 ). 34. - 
35. - 4m 3 n(- 3m)(2mn 3 ). 36. 4?/0 2 ( 

37. 2aj3a- 38. 2a^- 3a 2 6 3 . 39. - 



Multiply the polynomial by the term which is beneath it. 

40. 5z 3 - 3x 2 - 2x - 5 41. 6 - 

3x 



42. a 2 - 2a6 - 6 2 43. 3 - Gab - 5a 2 6 2 - a 3 6 3 

- 2ab - Sab 



44. 15(2z 2 - J + f). 46. 8(| - iy + Jif). 

46. 12(f - f* + 2 2 )- 47. 16(Ja 2 - fa - J). 

48. (12 - 6a + 24o 2 ). 49. J(- 16 -f 8x - 

60. 



30. Multiplication of polynomials 

To form the product of two polynomials, multiply one of them by 
each term of the other and collect similar terms. 

ILLUSTRATION 1. (2x 3y)(x 2 xy) = 2x(x 2 xy) 3y(x 2 xy) 

= 2s 3 - 



ILLUSTRATION 2. (x + 5) 2 = (x -f 5) (a: + 5) 

= x(x + 5) + 5(* + 5) = x 2 + 5x -f 5x + 25 = a? -f 10x + 25. 

Before multiplying, if many terms are involved, arrange the poly- 
nomials in ascending (or descending) powers of one letter. 



30 



INTRODUCTION TO FRACTIONS AND EXPONENTS 



ILLUSTRATION 3. To multiply (x 2 + 3x* - x - 2) (2s + 3) : 

3x + x 2 - x-2 
(Multiply) _ 2x + 3 

6s 4 + 2x - 2x 2 - 4x (Multiplying by 2x) 

Ox 3 + 3x 2 - 3x - 6 (Multiplying by 3) 
(Add) 6x* + llx 3 + a* - 7x - 6 = product. 



EXAMPLE 1. Multiply 

SOLUTION. 
x 2 + 



- 2x*y - 



-f 
-f 



x* y 2 + 2xy by y 2 + 2 



CHECK. Place x = 2 and y = 3. 

= 4 - 12 - 9 = - 17. 

= 4 + 12 + 9 = 25. 
Hence, the product should equal 
25- (- 17) = -425 

when x = 2 and y = 3. 

= 16 - 144 - 216 - 81 = - 425. 



Comment. The preceding numerical check does not absolutely verify the 
result, but almost any error would cause the check to fail. In checking, if 
1 were substituted for a letter, we could not detect an error in any of its 
exponents because every power of 1 is 1. Hence, in numerical checks, 
avoid substituting 1 for any letter. 

EXERCISE 10 

Multiply and collect similar terms. Check by substituting values for the 
letters, when directed by the instructor. 



1. (x - 3)(x + 4). 

3. (x - 5)(2x + 7). 

5. (5a - 7)(4a - 3). 

7. (2h - ZK)(2h + 3k). 

9. (a + 36)(2a - 56). 

11. (3r - 5s) (3r + 5). 

13. (06 - 3)(2a6 -I- 5). 

15. (cd - x)(cd +'x). 

17. (a -f 3)*. 18. (2x - t/) 2 . 

21. (3a - 26) 2 . 22. (2x + 4/0 2 . 

26. (y - 2)(y -f- 5). 



2. (x + 9)(x + 10). 

4. (x - 3) (7 - x). 

6. (3y - 2) (2 - By). 

8. (3w - 4r)(2w + 3r). 
10. (x + y)(x - y). 
12. (a 2 - 2)(2a 2 - 5). 
14. (3 - 5x')(2 + 3x 2 ). 
16. (ay - 2z)(2ay + 3). 

19. (h - 4fc) 2 . 20. (4 -f 
23. (ax - 6) 2 . 24. (5 - x 2 y) 2 . 
26. (c 2 - 2a 2 )(c 2 + 3a 2 ). 



INTRODUCTION TO FRACTIONS AND EXPONENTS 



3? 



27. (a s - 6 2 )(3a 3 + 46 2 ). 

29. (y - 2)(i/ 2 + 2y + 4). 

31. (x - 3)(x 2 + 2x - 5). 

33. (2 + x)(3 - 4x - x 2 ). 

35. (1 - 2x)(2x - x 2 + 5). 

37. (5 - 6) (4 - 26 - 6 2 ). 

39. 



40. 
41. 
42. 

43. (5 -h 3x 2 - 

44. (x 2 -+ 3x + 
45. (5z - 2y + 3) 2 . 

47. (a + 6)(a 2 - db -f fe 2 ). 
49. (a; + 3)(* - l)(2a? - 5). 
61. (2a - 3)(3a + 5)(a - 2). 



28. (x 2 - 
30. (4a 2 - 2a + l)(2a + 1). 
32. (a - 4)(3a 2 - 2a + 1). 
34. (c + 2)(2c - 5 - 3c 2 ). 
36. (h - 4)(2A 2 - 1 + A). 
38. (2 - y)(y + 5 - 
- 2x 2 + 5x - 7)(2x - 1). 

3). 



3). 

3 + x). 
a: 2 + 3 - 
46. (x + 

48. (x - 2y)(x* + 2xy + 
50. (3x - y)(2x + y)(2x - y). 
62. (x - a*)(x 2 + a*x + a 2 *). 



63. (x n - 

64. (x 4 + T/ 4 - x?y + X 2 y 2 )(x 2 - 2xi/ - 



31 . Exponents in division 

By the definition of a power, 

a 6 a'CL'd'd' 
= 

a 3 



a-a-a 
a-a-a 



a-a 

T" 

1 



a 



a-a-a-a-a a-a 



= a 2 ; (Divide out a-a-a) 
= -= (Divide out a a a) 



The preceding results illustrate the following index law. 

In dividing one power of a specified base by another power of the 6oae, 
subtract the exponents: 



(ifm>n); = 



(ifm<n). (1) 



32 INTRODUCTION TO FRACTIONS AND EXPONENTS 

In addition to formulas 1, we note that 

a n 

= 1- (2) 

a 5 x* I 1 h? 

ILLUSTRATION 1. . = 1. -^ = -^-^ -: 77 

a 6 x 10 x ia ~ 4 x 6 h* 



In dividing one integral rational term by another, we use formulas 
1 and 2 to reduce the fraction to lowest terms. 

ILLUSTRATION 2. In the following simplification, we can think of dividing 
numerator and denominator by 5ox B . Or, we can think of dividing numerator 
and denominator by 5 and, also, of applying formulas 1 to the powers of 
a and of x, separately: 

- 15ax 6 3 a 3 x 6 Sa 3 " 1 3a 2 



lOax 9 2 a x 9 2x 9 ~ 5 2x 4 

16a 3 x 2 16 a 3 x 2 8a 2 

ILLUSTRATIONS. 



ILLUSTRATION 4. 



/ 1\ a w 6 
= { ^) -r -= = 
\ 2/ a 4 w> 2 



2a 3 
where the intermediate details should be performed mentally. 

- 6ay 6 a 3 y* 2 

ILLUSTRATION 5. f = ^= - ~ = 

15 a d 



32. Division by a single term 

To divide a polynomial by a single term, divide each term of the 
polynomial by the single term and combine the results. 



6x 2 - 9x 3x 4 . 6x 2 9x 

ILLUSTRATION 1. 



= x 8 - 2x -f 3. 
EXAMPLE 1. Perform the division: (4o 2 6 4 - 8a 2 6 - 26 2 ) ^ (- 2O6 8 ). 

SOLUTION. Write the quotient as a fraction: 

4a 2 6 4 8a 2 6 



- 2a6 3 - 206 s 



INTRODUCTION TO FRACTIONS AND EXPONENTS 33 

EXERCISE 11 

Perform the division, expressing the result as a fraction or sum of fractions 
in lowest terms by use of the law of exponents for division. Express each final 
fraction without a minus sign in numerator or denominator. 

y "a 2 x 8 h 9 x 2 



6 - ;r 7 - 5r 8 

a o 
-<t Sw 3 .. 15a6 


J.X* J.4. 

4t/ 5a 

91 /t2 Q/r^ 

IK ^rltt 1fi OX 


96 


15 ' 3a 16 ' 12X 3 ' 

1ft . . 9A 


/i*4/)/2 nrfi 
oq * 4/ . Qd 


276C 3 

/y3 J)2 
^ ft/ 


^* " c ^* flt 

xy 6 a 6 o 

^^.^ jbOo 6 ^%^^ "" ^ A 
O>7 . OQ 



3x 



5a 7 

^21.5 

^^ 26. 

/ 



on 36m 

*'*' To 

18mn 
32. ~ 9A3t * 



35. (8x 3 ?/) -* (- 24X 3 ?/). 36. (- 49rV) * (- Trs 4 ). 

37. (- 80) + (24a6 2 ). 38. (4ac) -s- (72a 3 c 2 ). 

39. (48xV) ^- (- 8x2/ 2 ). 40- (- 36a6 2 ) -i- (- 4a6). 



. 6a H- 206 . 5/i - 25A; . 4a + 166 

41. - -- 42. 



5 4 



7x 2 - 5x ._ 3a 3 - 2a 4 

-- 45. - 

x a 2 



- 15o 4 



_ 3a2 

60. (8a 3 - 6a 2 - 4a) + (~ 2). 61. (6x - 3x 2 - 9) 4- (- 3). 

62. (x 4 - 3x 3 - 5x 2 ) H- x 2 . 63. (y* - 5y - y 3 ) -5- (- y). 

54. (_ 36 + 126 - 96 2 ) ^ 3. 66. (x 4 - 3x 2 + 5x) - 

66. (32a 2 6 4 + 48aV) ^- (16a 2 ). 67. (21a 2 6 2 - 



34 INTRODUCTION TO FRACTIONS AND EXPONENTS 

_ - 02? - Sax 4- 5o RA 7x* - 4x* - 3x + 2 

DO. 





ox 2x 

- 7a6 2 c 3 + 6c* c . 

61 * 



M - 276" co a&d 3 - a 2 6d - a 3 

62 ' - -- 63 ' 



33. Fundamental equation of division 

In the following long division, we use the customary terminology 
of arithmetic: 

15 (Quotient) 



(Divisor) 17 259 (Dividend) 



- 

89 
85 
4 (Remainder) 



259 = (17 X 15) + 4; = 15^. (2) 



In Section 11, when we met the notation a -5- 6, we called its complete 
value the quotient of a divided by 6. In (1), the complete value of 
the quotient is 15^. Hence, if there were danger of lack of clearness, 
in (2) we would call 15 the partial quotient. Frequently, the word 
quotient refers to a partial quotient. When appropriate, we shall use 
the qualifying word partial to prevent ambiguity. 

After any step in a division process similar to that met in (1), the 
remainder and partial quotient satisfy the equation 

dividend ,. . , remainder /ON 

. -7T-7 - = quotient H 37-7 - (3) 

divisor H divisor v ' 

Or, dividend = (quotient) (divisor) + remainder. (4) 

Equation 4 is frequently called the fundamental equation of division. 
The first equation in (2) is an illustration of (4). 

34. Division by a polynomial 

The long division process for polynomials is similar to long division 
in arithmetic. We say that the division is exact if the final remainder 
is zero. 



INTRODUCTION TO FRACTIONS AND EXPONENTS 35 

EXAMPLE 1. Divide: (x 2 + 3x - 40) -5- (x - 5). 

SOLUTION, (x 2 -f- x) x; this is the first term of the quotient. Then, 
x(x 5) = x 2 5x; we subtract this from the dividend, obtaining 8x 40. 

2. (8x -T- x) = 8; this is the second term of the quotient. Then, 
S(x 5) = 8x 40. We subtract this, obtaining zero. 

x + 8 (Quotient) 

(Divisor) x - 5 | a 2 + 3x - 40 (Dividend) 
(Subtract) x* - 5x 

8x -40 
(Subtract) Sx - 40 

(Remainder) 

CHECK. Since the division is exact, we obtain a check by verifying that 

(quotient) (divisor) = dividend. 

We find (x -f 8)(x - 5) = x* + 3x - 40, which checks. 

EXAMPLE 2. Divide: (4x 3 - 9x - 8x 2 + 7) -r (2x - 3). 

SOLUTION. 1. We first arrange the dividend in descending powers of a?, 
obtaining . 

4x 3 - &c 2 - 9z + 7. 



2. Since (4x 8 -s- 2x) = 2z 2 , this is the first term of the quotient; etc., 
as follows. 

2x 2 - x 6 (Quotient) _ 
(Divisor) 2x - 3 j 4r* - Sx 2 - 9x -f 7 (Dividend) 
2x*(2x - 3) -> (Subtract) 4s 3 - 6x 2 



- 2x 2 ) + 2x] = - x. - 2x 2 - 9x 

x(2x - 3) - (Subtract) - 2x 2 + 3x 



[(- 12x) -r 2x] = - 6. - 12x + 7 

- 6(2x - 3) - (Subtract) - 12x + 18 

11 (Remainder) 

Conclusion. From equation 3, Section 33. 

4x* - 8x 2 - 9x + 7 _ _ _ _ 11 
2x-3 "^ X 2x-3 

CHECK. Substitute x = 3 in the dividend, divisor, and quotient: 
dividend = 16; divisor = 3; quotient = 9; remainder = 11. 

Refer to equation 4, Section 33: does 

16 = 3(9) - 11 - 27 - 11 - 16? 
Since this equality is satisfied, we conclude .that the solution is correct. 



36 INTRODUCTION TO FRACTIONS AND EXPONENTS 

SUMMABT. To divide one polynomial by another: 

1. Arrange each of them in either ascending or descending powers 
of some common letter. 

2. Divide the first term of the dividend by the first term of the divisor 
and write the result as the first term of the quotient. 

3. Multiply the whole divisor by the first term of the quotient and 
subtract the product from the dividend. 

4. Consider the remainder obtained in Step 3 as a new dividend and 
repeat Steps 2 and 3; etc. 

Note 1. The numerical check in Example 2 does not constitute an ab- 
solute verification of the solution. To verify the result, we should (without 
substituting a special value for x) multiply the divisor by the quotient, add the 
remainder, and notice if we thus obtain the dividend. 



EXERCISE 12 

Divide and summarize as in the solution of Example 2, Section 34. // the 
division is exact, check by multiplying the divisor by the quotient. If the division 
is not exact, check by substituting convenient values for the literal numbers. 



1. (x* 4- 7z 4- 12) ^ (x + 3). 2. (</ 2 4- 15y + 36) + (y + 12). 

3. (c 2 - lOc 4- 21) -h (c - 7). 4. (d 2 - 12d + 35) -J- (d - 5). 

6. (s 2 4- 6s - 27) ^ (s 4- 9). 6. (2z 2 - 13x + 15) + (x - 5). 

7. (y 2 4- Qy - 40) 4- (y + 10). 8. (54 + 3z - x 2 ) -f- (6 + x). 
9. (4c 2 - 15) -h (2c + 3). 10. (6a 2 - ab - 6 2 ) ^ (3a + 6). 

11. (x 4 + 3x 2 - 4) ^ (x 2 - 1). 12. (h* + 3h* - 10) -h (ft 2 -f 5). 

13. (2z 2 -f 7z + 8) -i- (z -f 2). 14. (3a 2 - 7) + (a - 5). 

15. (2a 2 - ab - 66 2 ) -*- (2o + 6). 16. (6z 2 + zy - 2t/ 2 ) ^ (3z + 2y 

17. (4z 6 + Sz 3 - 6) -i- (4z 3 - 3). 18. (2a - a 8 - 15) -!- (2a 3 + 5). 

19. (5x + 3x 2 - 3) -f- (x - 2). 20. (a + 6a 2 + 3) -f- (2a - 1). 

21. (x 3 + 3x 2 - 2x - 6) -* (x 4- 3). 

22. (60 - 2a 2 4- 3a 3 - 26) -* (a - 2). 

23. (4x 3 - 9* - 8x 2 4- 7) + (2x - 3). 

24. (36y - 9 - 19y 2 - 15^) -5- (3y 2 4- 5y - 4). 



INTRODUCTION TO FRACTIONS AND EXPONENTS 37 

26. (Sy* - 1% 2 - 6 + lly) * (W - 3y + 2). 

26. (2^ - 5y 2 - 12 + llr/) * (2y - 3). 

27. (x 4 - 4z 3 + 3z 2 - 4x + 15) -J- (z - 3). 

28. (&c 2 - 3 - 5z 3 ) -^ (7* - 2 + 5z 2 ). 

29. (2z 3 + Wy + 12^ + 17zt/ 2 ) * (2x + 3t/). 

30. (a 3 - 3a 2 6 - 6 3 + Soft 2 ) 4- (a 2 - 2a6 + 6 2 ). 

31. (a: 3 + 27) (* + 3). 32. (a 3 + ft 3 ) -s- (o + 6). 
33. (x 3 - 2/ 3 ) --(- y). 34. (IGx 4 - y 4 ) * (2x - y). 

8^-27 



- 43x 2 - 9x - 5 



- 7 > 3*' + 22 
40. (6a 3A + ISo 2 * - 4a A - 15) -5- (2a* + 3). 



35. Fractions with a common denominator 

In a fraction, the bar should be thought of as a vinculum, a symbol 
of grouping, which encloses the numerator. We use this fact in 
adding fractions. 

2 _ a 

ILLUSTRATION 1. In the fraction -- = > the bar encloses 3 a and 

5 

the fraction equals (3 a) * 5. 

SUMMARY. To express a sum of fractions with a common denominator 
as a single fraction: 

1. Form the sum of the numerators, where each is enclosed in paren- 
theses and is given the sign of its fraction. 

2. Divide by the common denominator. 

8 3,9 8-3 + 9 14 

ILLUSTRATION 2. = -= + = = - = - = -=- 

5 o 5 o o 

T o^ 5 - x , 3 - 2x 
ILLUSTRATION 3. 7= --- r= -- =q 

lla lla lla 

6 - (5 - x) + (3 - 2x) = 6-5 + s + 3-2s = 4 - x 
lla lla lla ' 



38 INTRODUCTION JO FRACTIONS AND EXPONENTS 

36. Alteration of a denominator 

To change a fraction to an equal one having an added factor in 
the denominator, we must multiply both numerator and denominator 
by this factor, in order to leave the value of the fraction unaltered. 



ILLUSTRATION 1. To change $ to 14ths, we multiply numerator and de- 
nominator by 2 because ^ = 2: 

3 = 3 X2 6 

7 7X2 14' 

ILLUSTRATION 2. To change the following fraction to one where the de- 
nominator is 6o 3 6, we multiply numerator and denominator by 2a 2 6, because 
-i- 3a = 



5 - x 2a?b(5 - x) 10a 2 6 - 2a?bx 
3a 2a 2 6(3o) 6a 3 6 

37. Prime integers 

An integer is said to be prime if it has no factors except itself, and 
-f- 1 and 1, which are factors of any algebraic expression. Two 
factors are considered essentially the same if they differ only in sign, 
and then their product can be expressed as a power of either one 
of them. To factor an integer will mean to express it as a product of 
powers of distinct prime factors. 

ILLUSTRATION 1. 2, 3, 5, 7, 11, etc., are prime numbers. 
ILLUSTRATION 2. When factored, 200 = 2-2-2-5-5 = 25 2 . 

38. Lowest common multiple 

At present, when we refer to a monomial, or single term, we shall 
mean an integral rational term whose numerical coefficient is an 
integer. 

The lowest common multiple, LCM, of two or more monomials is 
defined as the 'term with smallest positive coefficient, and smallest 
exponents for the literal numbers, which has the given term as a 
factor. As a special case, the LCM of two or more integers is the 
smallest positive integer having each given integer as a factor. 

ILLUSTRATION 1. The LCM of 3, 5, and 10 is 30. 
ILLUSTRATION 2. The LCM of 30& 8 and 5a 2 6 is 



INTRODUCTION TO FRACTIONS AND EXPONENTS 39 

SUMMARY. To find the LCM of two or more terms: 

1. Express each term with its coefficient factored. 

2. Form the product of all letters in the terms and all the different 
prime factors of the coefficients, giving to each letter or integral factor 
the highest exponent it possesses in the given terms. 

EXAMPLE 1. Find the LCM of 20a 2 6 s and 70o 4 6. 

SOLUTION 1. In factored form, 20a 2 6 8 = 2 2 -5a 2 6 3 ; 70a 4 6 = 2-5-7a 4 6. 

2. Hence, LCM - 2 2 -5-7a 4 6 3 = 



Note 1. In brief, the LCM of two or more terms equals the product of the 
LCM of their coefficients and the highest powers of the Utters seen in the terms. 

EXERCISE 13 

Express the sum of fractions as a single fraction in lowest terms. 
,3,79 ft 2 18 ,6 

L i+i-r 2 -5-y + 5* 

2,6 __a 4 c _d _ 7 

3 *3+3 3* 888' 

.3 5,7-6 A 11 5 , 

5. ---- -- o. ---- 



a a a z z z 

_ 3 2a - 56 5 6 - 3a 

7 '7 -- 7 -- 8 '3~~T~* 



^ 40 - 3 5 2y - 5 

n . _ --- _ . _. 12. - - - . 

* 3 -a; 



14 
a a a 

- 



206 3 

Write the missing numerator or denominator to create equality. 

<i ^ 1ft 5 _ 10 3 _ 6 

17. - - 18. - - 19. 



40 INTRODUCTION TO FRACTIONS AND EXPONENTS 

A. - __ 24 -^- = __ 26 

~ 



36x 7 



Express the fraction as an equal one having the specified denominator. 

2 5 

26. ; denom., 21. 27. 5 ; denom., 32. 

7 o 

28. ; denom., 40. 29. |; denom., 35. 

5 

30. *j\ denom., be. 31. =-; denom., Qy. 

32. -= -', denom., lOxy 3 . 33. ,5-5,; denom., 18xV- 

5x?/ 2 3x 2 ?/ 4 

Q 1 

34. STTL; denom., 20ft 2 fc 4 . 35. ; denom., 20a 5 6 3 . 



Find the LCM o/ /ie given terms. As a partial check, verify that each term 
is a factor of the LCM. 
36. 5; 4; 10. 37. 16; 24; 48. 38. 12; 54; 30. 

39. 15; 12; 75. 40. 200; 36; 28. 41. 300; 27; 21. 

42. e^; 9x 2 t/ 2 ; 15xy*. 43. 8a 2 6 6 ; 4a 3 6 4 ; 6a6. 

44. 2o6; 14a 2 6 3 ; 66 4 . 45. 6a 2 x; 4a'x 2 ; 9ax 4 . 

46. 3zV; 12xi/ 3 ; 20x. 47. 5W; 10^; 16^. 

48. Change , ^, and ^ to the denominator 42, and then add the fractions. 



39. Addition of fractions * 

To combine a sum of fractions into a single fraction, the given 
fractions must be changed to new forms having a common denomina- 
tor. We define the lowest common denominator! LCD, of a set of 
fractions as the LCM of their denominators. 



ILLUSTRATION 1. The LCD of J, f , and is the LCM of 6, 5, and 8 or 
3-5'8 or 120. Hence, to add the fractions, we. change them to new fractions 
whose denominator is 120: 

1 4_ ? 4- 1 = 2Q . 3-24 7-15 = 20 + 72 + 105 197 
6 + 5 + 8 6-20 + 5-24" l "8.15 120 120* 

* To simplify the present chapter, we will deal only with the case where 
each denominator is an integral rational term. 



INTRODUCTION TO FRACTIONS AND EXPONENTS 41 

SUMMARY. To express a sum of fractions as a single fraction: 

1. Find the LCD of the given fractions. 

2. For each fraction, divide the LCD by the denominator and then 
multiply numerator and denominator by the resulting quotient, to 
change to an equal fraction having the LCD. 

3. Form the sum of the new numerators, where each one is enclosed in 
parentheses and is given the sign of its fraction. 

4. Place the result of Step 3 over the LCD, remove parentheses in the 
numerator, and reduce the fraction to lowest terms. 

ILLUSTRATION 2. In the following sum the LCD is 20; we think of 2 as 
2/1. We observe that ^ = 5 and f = 2. Hence, 

2-20 5(3s) 2(7 - 2x) 



o _ _ 

4 10 20 20 20 

40 - 15x - 2(7 - 2x) 40 - 15s - 14 + 4x 26 - llx 
20 20 20 

3w 7 
EXAMPLE 1. Express as a single fraction: 



Sax* 
SOLUTION. 1. The LCD is 15a 3 z 2 . We have 

15a 3 z 2 Ida's* r 

, . = 3z; -r r = 5a 2 . 
5a 3 x 3ax 2 

2. Hence, we multiply by 3x and 5o 2 in the given fractions to change to 
the LCD: 



_ 7(5a 2 ) = 9xy - 35a 2 
3aa: 2 (5a 2 ) 



EXERCISE 14 

Combine into a single fraction in lowest terms. 



1. 


5 

8" 


-! 


2. 


f- 


^fr 


3. 


5 

6 


1 

3* 


4. 


3 
5 


9 
10 


5. 


13 
16 


3 

8* 


6. 


1_ 


4 


7. 


2 

7 


5 
"*"2l' 


8. 


1 

6 


3 

5* 


9. 


3~ 


* 


10. 


2c 
5 


d 
2* 


,11. 


3 


k 
4* 


12. 


^H 
2 ^ 


~5 



14. i - - 1 15. ? _ 2 + 

83 7 ^ 21 



42 INTRODUCTION TO FRACTIONS AND EXPONENTS 

1A 12 4 , 17 3 74 lg 0,1,4 

16. y-5 + 3. 17 ' To-30 + 5' 18 ' " 2 + 6 + 15 

- 3 3 3o-6 7 



a-3 . 64-2 2a; - 3 , 3 - 



3a a + 7 2 x - 3 g - 7 

~ 2 ~ 



25 10 18 

v\ b c <u h w 

- Ta ~ 6^' ^ S "" 25* . S ~ 33k 



4 OB A- 1. 34 -I- A. 

36' 12y 2y Gx Sxy 



36 - - 37 _ 

36> 37 ' 



39 JL - ^ " 1 40 

3a 2 5a6 

_ 62^ 



4 10 9 54 

2X 



JIt a: - 1 x - 3 2o~3 5o + 3 

47 - 



y-2 



Ki - 2 - 4* 4a - 5 3a + 2 

51. g -- 1 - -2J 52. ? 2 + 

2-x 4 +. 



/NTRODUCT/ON TO FRACT/ONS AND EXPONENTS 43 

40. Mixed expressions 

A sum of an integral rational part and of one or more fractions is 
referred to as a mixed expression. If a mixed expression occurs as a 
factor in a product or as the numerator or denominator of a fraction, 
it is usually desirable to combine the mixed expression into a single 
fraction before performing other operations. 

ILLUSTRATION 1. We refer to a number such as 5| as a mixed number. 

23 17 23(17) 391 



T o 

ILLUSTRATION 2. 






ILLUSTRATION 3. 3z 2 -h - is a mixed expression. 



ILLUSTRATION 4. (2 + |j(3 | 



15 - a 



= (6 + a) (15 ~ a) 90 + 9a - a 2 
15 15 

'41 . Complex fractions * 

A simple fraction is one without any fraction in its numerator or 
denominator. A complex fraction is one having one or more fractions 
in the numerator and denominator. 

SUMMARY. To reduce a complex fraction to a simpty fraction: 

1. Express the numerator and denominator as simple fractions. 

2. Form the quotient of the simple fractions and reduce the result to 



lowest terms. 



- , 3 5+3 8 

I _X^ __ 
A -t- 



T t 5 5 5 8 3 12 

ILLUSTRATION 1. - = g ^ g - - = 

333 

Sometimes it may be convenient to reduce a complex fraction to 
a simple fraction by the single operation of multiplying both numerator 
and denominator of the complex fraction by the LCD of all simple 
fractions involved. 

* In this chapter, the problems will avoid questions of factoring which will 
be met in Chapter 5. 



44 INTRODUCTION TO FRACTIONS AND EXPONENTS 

ILLUSTRATION 2. To reduce the given complex fraction of Illustration 1 to 
a simple fraction, we observe that the LCD of 3/5 and 4/3 is 15. Hence, 
we multiply both numerator and denominator by 15, observing that 
15$) - 9 and 15(j) = 20: 

1 + 5 15 + 9 24 12 



4 30-20 10 5 



, , 2 15o6 
3a + yr 



T Q 56 56 

ILLUSTRATION 3. = = 



6o6-7 
36 

15o6 + 2 36 45a6 



56 606 - 7 30a6 - 35 
where we divided out 6 from numerator and denominator. 

26-5 26-5 

36 36 26 - 5 1 26-5 

ILLUSTRATION 4. 



4-6 4-6 36 4-6 126 - 36* 

1 

DEFINITION I. The reciprocal of a number H is = 
ILLUSTRATION 5. The reciprocal of 3 is J. 

r 

ILLUSTRATION 6. The reciprocal ofvis? = Y'o =:: o' 
The reciprocal of T is 



a 1 a a 
Thus, the reciprocal of a fraction is the fraction inverted. 

EXERCISE 15 

Express the fraction or product as a simple fraction in lowest terms. 
1. (4 + f)(5). 2. (3 - f)(4f). 3. (2f)(3f). 

4. (6 -f |) (6 - f). 6. (f - 36) (f - 46). 



INTRODUCTION TO FRACTIONS AND EXPONENTS 45 

A 7 15 ~* 

6 - T+T 7 - T+T 

+ -IS 



7 2a 2 6 






10. 

12. f-r-^- 13. g^-4- 14. 



* i + 5 "* 1 - f "'5 

5c~f 
J-2c' 

2_ 3 4_ 3 5_ 3 

16. 2 16. - r- 17. T- 

5 . . e , o . , o 

- + 4 6 +: 4 + ^ 

X* Fj tif^ 

\Jb \J l/Vx 



J. , 2 _ _ + 

1ft 2x y 19 2 

TIT' 

a: ^ 
01 _ 



2 3 


5x 


5 


x" h 4y 


3 


+ 3z 


3 5 


5 


3 


a 2 26 


w\ 


2 y 


2 3 


23 ' 2 


1 


a 4a6 


32/ 5 


' 2x 


5 3 


w; 2 


1 2 


2xw 


9A 3 





27 . . 28. . 29 . 

2 + ? ' + ^ 

3 2a 56 

30. !i. 31. 1- 32. 



. 

- 2 a: -f 3y 

Find the reciprocal of the expression, and express the result as a simple 
fraction in lowest terms. 
33. 75. 34. 17. 36. . 36. f 37. - . 

38. 10. 39. 12? . 40. 2J. 41. - 5J. 42. a. 

44. . 45. 



46. By writing a fraction, show that to divide a number N by a number H 
means the same as to multiply N by the reciprocal of H. 



46 INTRODUCTION TO FRACTIONS AND EXPONENTS 

EXERCISE 16 
Review of Chapters 1 and 2 

Compute each expression, leaving any fraction in lowest terms. 
1. (- 3)(- 4)(- 5). 2. - (- 2)(- 5). 3. - (- 3)(- 4)(0). 

. (-7X-3) (- 2) (5) ( 



* -14 " (-3X-6) ~ (-2X-4) 

7. - 7 + 19 - 16. 8. 3 - {- 4) - 7. 9. - 2 -f (- 3) + 6. 



Find the absolute value of the expression. 
10. - 8. 11. (- 3)(- 2). 12. | - 14 |. 13. | 17 |. 

(a) Add the two numbers; (6) subtract the lower one from the upper number. 

14. 17 16. - 23 16. - 15 17. 17 

- 5 - 9 _ 29 25 

18. Read the expressions 4 < 2 and 17 > 0, and verify their truth 
by marking the numbers on a real number scale. 

Insert the proper sign, < or > , between the numbers. 
19. 11 and 19. 20. - 15 and - 27. 21. and - 6. 

22. Which one of the numbers 15 and 7 is less than the other? Which 
one is numerically less than the other? 

Perform any indicated operation, removing any parentheses, reducing any 
final fraction to lowest terms, and employing the laws of exponents to simplify 
expressions. 

23. - (3a - 26 - c). 24. 2(3 - 5a - c). 26. - a 2 (3a 8 6 - aft*). 



26. 5(- 3*). 27. 3*y(2* - Wy). 28. - 2*y(- 3* - 5j/). 

29. 3(2a - h + A) - 2(3a + 4h - 3k). 

30. - x*(3xy* - 2xy + 3) + 4i/(2xV - 3x* + 5). 

31. 3o - [2a - 3(5 - a)]. 32. - (6a - 6) - 2[3 - (2o - 46)]. 

- 15 4 14 3 6 



37. 17. 38. | -4- 6. 39. 7 + |- 40. ~ 



41. - 2AW(3iUPX- 4A 2 fc). 42. - 4/*V(2% 2 - 3/fy - 5A). 



INTRODUCTION TO FRACTIONS AND EXPONENTS 47 

43. (Si 8 !/) 4 . 44. (- 2oV). 46. (- 

46. (-2)'. 47. (I)'- .(-?)'. .(*)' 

' ' -(I)- 

66. (2* + 3) (2s - 7). 66. (3x* - 7x)(2x 

57. (2z - 2/)(3z - %). 68. (2a - 36) 2 . 

69. (3x* - 5x - 2r> + 3)(4z + 5). 



64. (6o 3 - 19o 2 + 21a - 9) ^- (2o - 3). 
66. (18 + 4x 3 - 26a; - 2x 2 ) ^- (2x - 5). 

A7 4y + 8 4x - 7s* 

67 ' 



Express as a single fraction in lowest terms. 

2 7 . 10 - ft 2h-3 4A-7 

_ -4- / II. ' ti . i 

33^3 5 10 

-_2. 72 j! _ 4. A. 

3 ' " 12 " 2xy 3y 2 4x 2 

2 - 3x 5 - By -. 2a 3 3a - 6 

iO ^ * rfc I4 Q * AV ^ y 



5 .!__:>. 35 

76. -J. 76. | |- 77. -J | 

5 a 56 



78. ^-~- 79. 

9 J_ _ _. 

a x y 



5 
-L^. _ 3 



3 1 

81. Find the reciprocal of -=; of (a 5). 

o 6 



CHAPTER 



3 



DECIMALS AND ELEMENTS OF 
COMPUTATION 



42. Decimal notation 

The decimal notation * for writing numbers is called a place sys- 
tem, for which the base is 10. In this notation, each number is an 
abbreviation for a sum involving units and powers of ten, written by 
use of the Hindu-Arabic digits or figures (0, 1, 2, 3, 4, 5, 6, 7, 8, 9). 

ILLUSTRATION 1. 3456 = 3(1000) + 4(100) + 5(10) + 6 

= 3(10 3 ) + 4(10 2 ) + 5(10) + 6. 

We have 3456 as the sum of 6 units, or ones, 5 tens, 4 hundreds, and 3 thousands. 
ILLUSTRATION 2. 23.572- = 2(10) + 3 + ~ + ^ + 2 



10 ' 100 ' 1000 



10 ' 10 2 ' 10 3 

Counting to the left of the decimal point in a number, the places 
are named the units' place or ones' place; tens' place; hundreds' place; 
thousands' place; ten thousands' place; etc. Counting to the right -of 
the decimal point, we have the tenths' place; hundredths' place; 
thousandths' place; etc. The name of any place is the value of a 
unit located there. These values should be remembered in terms of 
powers of ten. 

ILLUSTRATION 3. 1000 = 10 3 ; .001 = rrr = -; .01 = T- 



ILLUSTRATION 4. We read .5073 as tc point, 5, oh, 7, 3," or as "5073 
ten-thousandths" 

* In this chapter, unless otherwise specified, any number referred to will 
be positive. 



DECIMALS AND ELEMENTS OF COMPUTATION 49 

In this chapter we will think of each number in its decimal form. 
The part to the right of the decimal point is called the decimal part 
of the number. We consider each number as having a decimal part 
even when the decimal part is zero and the number is then an integer. 
The decimal places of a number are those places to the right of the 
decimal point in which digits are actually written in the number. 

ILLUSTRATION 5. 23.507 has three decimal places and its decimal part 
is .507. 

In any number, observing its digits from left to right, we may 
visualize an endless sequence of zeros at the right of the last digit 
not zero, if there is such a last digit. A number of this character is 
called a terminating decimal. An endless decimal is one in which, 
however far we proceed to the right, we never reach a digit beyond 
which all digits are zeros. Hereafter, unless otherwise stated, any 
number mentioned will be a terminating decimal. 

ILLUSTRATION 6. 35.675, or 35.67500 , is a terminating decimal. The 
familiar number TT = 3.14159 is an endless but not a repeating decimal. 
The fraction J is equal to the endless repeating decimal .333 



If a unit in any place in the decimal notation is multiplied by 10, 
we obtain a unit in the next place to the left. If we divide a unit in 
any place by 10, we obtain a unit in the next place to the right. The 
preceding remarks justify the following convenient rules. 

I. To multiply a number by 10, move the decimal point one place to 
the right in the number. 

II. To divide a number by 10, move the decimal point one place to the 
left in the number. 



ILLUSTRATION 7. 10(315.67) = 3156.7. ~~ = 31.567. 

To divide by 1000, which is 10(10) (10), we move the decimal point three 
places to the left: 21.327 * 1000 = .021327. 

43. Addition of decimals 

In finding a sum of decimals, after they have been arranged with 
the decimal points in a column, it is desirable to add once going up- 
ward and then downward to check in each column. 



50 DECIMALS AND ELEMENTS OF COMPUTATION 

ILLUSTRATION 1. To find the sum of 31.457, 
2.6, 3.15, and 101.41, we annex zeros to extend each 

number to the 3d decimal place, and then add. 

3.150 



EXAMPLE 1. Add: 



1.573 + 3.671 - 1.157 - 4.321 + 10.319. Sum 138 ' 617 



31.457 



101.410 



SOLUTION. The sum of the positive terms is 15.563; of the negative terms 
is - 5.478. 

(15 563 1 
5 478 f sum - 10.085. 

EXERCISE 17 

Write the number in decimal form. 
1. 3 and 25 hundredths. 2. Point, oh, 1, 5, 3, 9, oh, 4. 

3. 10*. 4. 10'. 6. ~ 6. i- 7. i- 

Write each number as a sum involving powers of 10, with one term corre- 
sponding to each digit (not zero) of the number. 
8. 567. 9. 3149. 10. 16,342. 11. .319. 12. 27.0457. 

Write the number in decimal notation equal to the sum. 
13. 5(100) + 3(10) +6 + A + A 



14. 2(1000) -f 4(10) + 3 + 1 + JL + -J- 

15. 5(10*) + 7(10') + 3(10) + 5 + + 



Compute the indicated sum. 

16. 2.057 + 3.11 + 4.985 + 3.05 + 1.5 + 2.177 + .459. 

17. 3.193 + .098 -f 1.567 -f 2.457 + 3.167 + 2.13 -f- 1.5072. 

18. 21.675 - 14.521. ' 19. .0938 - .0257. 20. 1.721 - 2.468. 
(a) Add the numbers. (6) Subtract the lower number from the upper one. 

21. 5.26 22. .357 23. - 43.8468 24. .02438 

1.38 .2983 - 59.923 - .5729 

25. Compute 2.156 - 3.149 + 4.183 + 2.147 - 4.159. 



DECIMALS AND ELEMENTS OF COMPUTATION 51 

26. Compute 13.083 + 2.148 - 41.397 - 2.453 + 12.938. 

Perform the multiplication or division mentally. 

27. 10(.532). 28.1000(1.0219). 29. 10*(32.653). 30. 100(.00415). 

.0317 13.257 57.38 .0498 



oo ' wo IS" ISSo* 

44. Multiplication of decimals 

Any number whose decimal part is not zero can be written as a 
fraction whose numerator is an integer and denominator is a power of 
10. The exponent of 10 hi the denominator can be taken equal to the 
number of decimal places in the given number. 

ILLUSTRATION 1. 1.21(.205) = 

_ 121 205 (121) (205) 24,805 
10 2 ' 10 s 10 2+ * 10 6 

We conclude that the five decimal places in the result are a consequence 
of the law of exponents for multiplication, because the factors had two and 
three decimal places. This result is a special case of the following rule. 

SUMMARY. To multiply two decimals: 

1. Find the digits of the product by multiplying the factors with their 
decimal points disregarded (or even removed). 

2. Add the numbers of decimal places in the factors to find the number 
of decimal places in the product, and insert the decimal point. 



ILLUSTRATION 2. To multiply .0238 X 112.75, we find 
the digits of the product, at the right, and then point off 
(2 -f 4) or 6 decimal places to obtain 2.683450. Notice 
that the final zero had to be written and counted in fixing the 
decimal point. In stating the final result, we could then 
omit the zero. 



11275 
(X)238 

90200 
33825 
22550 
2683450 



The digits in the product of two decimals depend only on the 
digits of the factors. If the decimal points are moved hi the factors, 
this only alters the position of the decimal point in the result. 

ILLUSTRATION 3. In Illustration 2, .0238(112.75) = 2.683450. 
Hence, 2.38(1127.5) - 2683.450. 



52 DECIMALS AND ELEMENTS OF COMPUTATION 

EXERCISE 18 

Perform the following multiplications. 
1. 3.51(1.4). 2. .46(.107). 3. 13(.461). 

4. .0142(3.6). 5. 21.38(.024). 6. 156.1(1.38). 

7. 398.4(.0342). 8. .00175(.2147). 9. .0346(.00157). 

10. 85.2(1.356). 11. 9.137(.2346). 12. 74.308(.00259). 

45. Significant digits 

In any number N, let us read its digits from left to right. Then, 
by definition, the significant digits or figures of N are its digits, in 
sequence, starting with the first one not zero and ending with the 
last one definitely specified. Notice that this definition does not 
involve any reference to the position of the decimal point in N. 
Usually we do not mention final zeros at the right in referring to the 
significant digits of N, except when it is an approximate value. 

ILLUSTRATION 1. The significant digits of 410.58 or of .0041058 are 
(4, 1, 0, 5, 8). 

46. Approximate values 

If T is the true value and A is an approximate value of a quantity, 
we agree to call A T the error of A. 

ILLUSTRATION 1. If T = 35.62, and if A = 35.60 is an approximation to 
T, then the error of A is 35.60 - 35.62, or - .02. 

The significant digits in an approximate value A should indicate 
the maximum possible error of A. This error is understood to be 
at most one half of a unit in the last significant place in A, or, which is 
the same, not more than 5 units in the next place to the right. 

ILLUSTRATION 2. If a surveyor measures a distance as 256.8 yards, he 
should mean that the error is at most .05 yard and that the true result lies 
between 256.75 and 256.85, since the error (plus or minus) might be =fc .05. 

In referring to the significant digits of an approximate value A, 
it is essential to mention all final zeros designated in A. 

ILLUSTRATION 3. To state that a measured weight is 35.60 pounds should 
mean that the true weight differs from 35.60 pounds by at most .005 pound. 
To state that the weight is 35.6 pounds should mean that the true weight 



DECIMALS AND ELEMENTS OF COMPUTATION 53 

differs from this by at most .05 pound. Thus, there is a great distinction 
between 35.6 and 35.60 as approximate values although there is no difference 
between 35.6 and 35.60 as abstract numbers. 

47. Rounding off a number 

In referring to a place in a number, we shall mean any place where 
a significant digit stands. In referring to a decimal place, the word 
decimal will be explicitly used. 

To round off N to k figures, or to write a fc-place approximation 
for N, means to write an approximate value with k significant digits 
so that the error of this value is not more than one half of a unit 
in the kth place, or 5 units in the first neglected place. This condition 
on the approximate value of N leads us to the following method. 

SUMMARY. To round off a number N to k places, drop off the part of 
N beyond the kth place and then proceed as follows: 

1. // the omitted part of N is less than 5 units in the (k -f- l)th place, 
leave the digit in the kth place unchanged. 

2. // the omitted part of N is more than 5 units in the (k -f l)th place, 
increase the digit in the kth place by 1. 

3.* // the omitted part of N is exactly equal to 5 units in the (k + l)th 
place, increase the digit in the kth place or leave it unchanged, with 
the object of making the final choice an even digit. 

ILLUSTRATION 1. The seven-place approximation to TT is 3.141593. On 
rounding off to five places (or four decimal pldles) we obtain 3.1416. We 
changed 5 to 6 because .000093 > .00005. On rounding off TT to three 
places we obtain 3.14. 

ILLUSTRATION 2. In rounding off 315.475 to five figures, with equal 
justification we could specify either 315.47 or 315.48 as the result. In 
accordance with Item 3 of the Summary, we choose 315.48. 

48. A notation for large numbers 

For abbreviation, or to indicate how many digits in a large number 
are significant, it is sometimes convenient to write a number AT as the 
product of an integral power of 10 and a number equal to or greater 
than 1 but less than 10, with as many significant digits as are justified 
by the data. 

* Item 3 could be replaced by various similar and equally justified agreements. 



54 DECIMALS AND ELEMENTS OF COMPUTATION 

ILLUSTRATION 1. If 5,630,000 is an approximate value, its appearance 
fails to show how many zeros are significant. If just five digits are significant 
we write 5.6300C10 6 ), and, if just three are significant, 5.63C10 6 ). 

49. Accuracy of computation 

By illustrations, we can verify that the following rules do not 
underestimate the accuracy of computation with approximate values. 
On the other hand, we must admit that the rules sometimes over- 
estimate the accuracy. However, we shall assume that a result ob- 
tained by these rules will have a negligible error in the last significant 
place which is specified. 

I. In adding approximate values, round off the result in the first 
place where the last significant digit of any given value is found. 

II. In multiplying or dividing approximate values, round off the 
result to the smallest number of significant figures found in any given 
value. 

ILLUSTRATION 1. Let a = 35.64, 6 = 342.72, and c = .03147 be approxi- 
mate values. Then, a + b -\- c is not reliable beyond the second decimal 
place because both a and b are subject to an unknown error which may be 
as large as 5 units in the third decimal place. Hence, we write 

a + b + c = 378.39147 = 378.39, approximately. 

ILLUSTRATION 2. If x = 31.27 and y = .021 are approximate values, 
then,*by Rule II, we take xy = .66, because y has only two significant digits: 

xy = 31.27(.0ll) = .65667 = .66, approximately. 

ILLUSTRATION 3. If a surveyor measures a rectangular field as 385.6' by 
432.4', it would give an unjustified appearance of accuracy to write that 
the area is (385.6) (432.4) = 166,733.44 square feet. For, an error of .05 foot 
in either dimension would cause an error of about 20 square feet in the area. 
By Rule II, a reasonably justified result would be that the area is 166,700 
square feet, to the nearest 100 square feet, or 1.667(10*) square feet. 

In problems where approximate values enter, or where approxi- 
mate results are obtained from exact data, the results should be 
rounded off so as to avoid giving a false appearance of accuracy. 
No hard and fast rules for such rounding off should be adopted, 
and the final decision as to the accuracy of a result should be made 
only after a careful examination of the details of the solution. 



DECIMALS AND ELEMENTS OF COMPUTAT/ON 55 

EXERCISE 19 

Round off, first to five and then to three significant digits. 
1. 15.32573. 2. .00132146. 3. .3148638. 4. 5.62653. 

6. 195.635. 6. .00128558. ' 7. .0345645. 8. 392.462. 

Tell between what two values the exact length of a line lies if its measured 
length in feet is as follows. 

9. 567. 10. 567.0. 11. 567.4. 12. 35.18. 

Assuming that the numbers represent approximate data,* find their sum 
and product and state the results without false accuracy. 

13. 31.52 and .0186. 14. .023424 and 1.14. 15. .0047(11.3987126). 

HINT for Problem 15. In proceeding to multiply or divide approximate 
values, there is no advantage in keeping many significant digits in one 
value when other values have relatively few significant digits. A conservative 
rule is that, before multiplying or dividing, we may round off any given 
value to two more significant digits than appear in the least accurate of the 
given values.^ 

Write the number in ordinary decimal form. 
16. 100(3.856). 17. 27.38(10 2 ). 18. 1.935(10<). 19. 2.056(10). 

Write as the product of a power of 10 and a number between 1 and 10. 
20. 38.075. 21. 675.38. 22. 153,720,000. 23. 45,726. 

Given that the number is an approximate value, write it as the product of an 
integral power of 10 and a number between 1 and 10, assuming, first, that there 
are five significant digits and, second, that there are three significant digits. 

24. 9,330,000. 25. 453,120. 26. 23,523,416. 27. 72,200,000. 

In the following problems, state each result without false accuracy. 

28. The measured dimensions of a rectangular field are 469.2 feet and 
57.3 feet. Find the perimeter (sum of lengths of sides) and area of the field. 

29. The measured dimensions of a rectangular box are 20.4 feet, 16.5 feet, 
and 7.8 feet. Find the volume of the box in cubic feet. 

30. Given that one cubic foot contains approximately 7.5 gallons, how 
many gallons are contained by 2.576 cubic feet? 

* In this book, unless otherwise stated, the numerical data in any problem 
should be assumed to be exact. Results obtained from exact data may some- 
times be rounded off. 

t See Note 3 in the Appendix for a convenient abridged method for multiplyic 
two numbers with many significant digits. 



56 DECIMALS AND ELEMENTS OF COMPUTATION 

50. Division of decimals 

When one decimal is divided by another, the division process some- 
times gives a zero remainder if the work is carried to a sufficient 
number of decimal places. Usually, however, we may expect a 
remainder not zero however far we continue the division. We can 
always arrange the details so that the actual work amounts to 
division by an integer. This arrangement is useful in locating the 
decimal point in the result. 

ILLUSTRATION 1. To compute 372.173 -5- 1.25 with accuracy to two 
decimal places, we first indicate the division as a fraction, and then multiply 
its numerator and denominator by 100, to obtain an integer as the divisor: 



372.173 = 372.173(100) 37,217.3 
1.25 1.25(100) ~ 125 



(1) 



1.25, 



297.738 (Quotient} 
372.17*300 (Dividend) 
250 



122 
112 



67 
75 



At the right, we insert the original 
decimal points in dividend and di- 
visor and mark with "A" the new lo- 
cations of the decimal points observed 
in the final fraction in (1). The mul- 
tiplication by 100 in (1) is equivalent 
to the action of moving the decimal 
point two places to the right in both 
dividend and divisor. In the process 
of division, the integral part of the 
quotient ends when we begin using 
digits of the dividend at the right of 
the new decimal point , " A ." If we place 
each digit of the quotient directly 
above the last digit being used at that 
stage from the dividend, the decimal 

point in the quotient occurs exactly above the altered decimal point, A , in 
the dividend. We find the quotient to three decimal 
then round it off to 297.74, which we say is correct to 
To check, we compute 

1.25(297.74) = 372.175. 



92 3 

87 5 
4 80 
3 75 
1 050 
1J300 
50 



places, 297.738, and 
two decimal places. 



We do not obtain exactly 372.173 because 297.74 is not the exact quotient. 

Note 1. In any division, estimate the result before dividing, to check the 
location of the decimal point in the quotient. Thus, in Illustration 1, a 
convenient estimate would be 375 -5- 1.25, or 300. Then we are sure that 
the actual answer should have three places to the left of the decimal point. 



DECIMALS AND ELEMENTS OF COMPUTATION 57 

Note 2. In dividing approximate values, obtain the quotient to one more 
figure than is specified as reliable by Rule II, Section 49. Then, round off 
the quotient in the preceding place. 

Any terminating decimal can be expressed as a fraction, which then 
may be reduced to lowest terms. Conversely, we can express any 
common fraction as a decimal by carrying out the division indicated 
by the fraction, to obtain either a terminating or an endless decimal. 

ILLUSTRATION 2. On carrying out the division, we obtain J = .875, 
exactly. 



3125 = 25 
1000 8 



ILLUSTRATION 3. 3.125 = 



ILLUSTRATION 4. To express ii as a decimal 
we divide, at the right. After reaching .458 in the 
quotient, we meet 8 each time as the only digit in 
the remainder. Hence, 3 will be obtained endlessly 
in the quotient. The result is the endless repeating 
decimal .4583, where the dot above 3 means that 
the digit repeats endlessly. 



24 



.45833 



11.00000 
96 



140 
120 
200 
192 



80 
72 



80 etc. 



EXERCISE 20 

(a) Write each expression as a fraction, and then alter it to make the divisor 
an integer. (6) Divide until the remainder is zero. 

1. 3.562 -T- 2.6. 2. 2.849 -f- .74. 3. 140.14 -^ 2.45. 

Obtain the result of the division correct to three decimal places. 
4. 381.32 -5- 58. 5. .083172 -5- .316. 6. .5734 *- 12.8. 

Assume that the numbers in the following problems are approximate values. 
Carry out the division and state the result without false accuracy, according 
to Rule II, Section 49. 
7. 573.2 * 3.83. 8. 19.438 -s- 2.21. 9. .09734 -f- 3.265. 

10. 98.3 -s- 21.473. 11. .003972 4- .0139. 12. .01793 ^ .5634. 

Change the decimal to a fraction in lowest terms. 
13. 2.75. 14. .0125. 16. 2.375. 16. .0175. 17. .0325. 

Obtain the decimal equal to the given fraction. If the decimal repeats endlessly, 
carry the division far enough to justify this conclusion. 
18. f . 19. f . 20. A. 21. A- 22. ft. 23. 



CHAPTER 



4 



LINEAR EQUATIONS IN ONE UNKNOWN 



51 . Terminology about equations 

An equation is a statement that two number expressions are equal. 
The two expressions are called the sides or members of the equation. 
An equation in which the members are equal for all permissible 
values of the letters involved is called an identical equation, or, for 
short, an identity. An equation whose members are not equal for all 
permissible values oi the letters is called a conditional equation. 

ILLUSTRATION 1. In the following equation, by carrying out the multipli- 
cation on the left-hand side, we verify that the product is the same as the 
right-hand side. This is true regardless of the values of a and b. Hence the 
equation is an identity. 

(a + 26) (a + 36) = a 2 + 5a6 + 66 2 . 

ILLUSTRATION 2. The equation x 2 = is a conditional equation be- 
cause the two members are equal only when x =-2. 

The word "equation" by itself will be used in referring to both 
identities and conditional equations, except where such usage would 
cause confusion. Usually, however, the word "equation" refers to 
a conditional equation. At times, to emphasize that some equation 
is an identity, we shall use " s " instead of " = " between the members. 

A. conditional equation may be thought of as presenting a ques- 
tion: the equation asks for the values which certain letters should 



. , 

values are requested, are called unknowns. Some of the letters in 
an equation may represent known numbers. 

ILLUSTRATION 3. x* + 3x - 5 = is an equation in the unknown x. 



LINEAR EQUATIONS IN ONE UNKNOWN 59 

52. Solution of an equation 

An equation is said to be satisfied by a set of values of the unknowns 
if the equation becomes an identity when these values are substituted 
for the unknowns. A solution of an equation is a set of values of the 
unknowns which satisfies the equation. A solution of an equation in 
a single unknown is also called a root of the equation. To solve an 
equation in a single unknown means to find all the solutions of the 
equation. 

ILLUSTRATION 1. 4 is a root of the equation 2x 3 = 5, because when 
x = 4 the equation becomes [2(4) 3] = 5, which is true. 

53. Equivalent equations 

Two equations are said to be equivalent if they have the same 
solutions. 

EXAMPLE 1. Solve: 3 - 3x = - 7 - 5x. (1) 

SOLUTION. 1. Add 5x to both members: 

1 - 3x -V 5x = - 7 - 5x -V 5x, or 3 + 2x = - 7. (2) 

2. Subtract 3 from both members, or, which is the same, add 3 to 
both sides: 

- 3 + 3 -f- 2x = - 7 - 3, or 2x = - 10. (3) 

3. On dividing both sides of (3) by 2 we obtain 

x = - 5, (4) 

and conclude that this is the only solution- of (1). 

CHECK. Substitute x = 5 in (1). 
Left-hand side: 3 - 3(- 5) = 3 + 15 = 18. 

Right-hand side: - 7 - 5(- 5) = - 7 + 25 = 18, which checks. 

Comment. Each equation obtained from (1) was equivalent to it and this 
would justify us in saying that (1) has just one root, x = 5, without any 
necessity for the check. 

The equivalence of (1), (2), (3), and (4) is a consequence of the 

following familiar facts: if equal numbers are added to or subtracted 
from equal numbers the results are equal; if equal numbers are multiplied 
or divided by equal numbers the results are equal. 



60 LINEAR EQUATIONS IN ONE UNKNOWN 

ILLUSTRATION 1. Any value of x satisfying (1) will also satisfy (2), be- 
cause we pass from (1) to (2) by adding equals, 5x and 5x, to the equal sides 
of (1). And, any value of x satisfying (2) will satisfy (1) because we can 
pass from (2) back to (1) by the inverse operation of subtracting 5x from both 
sides of (2). Hence, (1) and (2) are satisfied by the same values of x, and 
thus are equivalent. 

In solving an equation in a single unknown, by use of the following 
operations we pass from the original equation to progressively simpler 
equivalent equations, which finally yield the desired roots. 

SUMMARY. Operations on an equation yielding equivalent equations. 

1. Addition of the same number to both members. 

2. Subtraction of the same number from both members. 

3. Multiplication (or division) of both members by the same num- 
ber, provided that it is not zero and does not involve the unknowns. 

Note 1. We observe that Operation 2 is a special case of Operation 1 
because subtraction of a number N is equivalent to addition of N. The 
division part of Operation 3 is a special case of the multiplication part, 
because division by a number N is equivalent to multiplication by 1/N. 

Convenient mechanical processes, and corresponding terminology, 
grow out of Operations 1, 2, and 3. 

A term appearing on both sides of an equation can be canceled, by 
subtracting the term from both sides. 

ILLUSTRATION 2. Given: x -j- 3 = f -f 3. 

Subtract 3 from both sides: x = f. 

A term can be transposed from one member to the other, with the sign 
of the term changed, by subtracting it from both members. 

ILLUSTRATION 3. Given: x + 5 = 7. 

Subtract 5 from both sides, or transpose 5: 

x 7 5, or x = 2. 

ILLUSTRATION 4. Given: x 4 = 9. 

Transpose 4: x 9 + 4, or x = 13. 

The signs of all terms on both sides may be changed, by multiplying 
both sides by 1. 



LINEAR EQUATIONS IN ONE UNKNOWN 61 

ILLUSTRATION 5. Given: 3x 6 = 5 ox. 

Change all signs (multiply both sides by 1) : 

6 = 5 H- ax. 



54. Degree of a term 

The degree of an integral rational term in a certain literal number 
is the exponent of the power of that number which is a factor of the 
term. If the term does not involve the number, the degree of the 
term is said to be zero. The degree of a term in two or more letters 
together is the sum of their degrees in the term. The degree of a 
polynomial is defined as the degree of its term of highest degree. 

ILLUSTRATION 1. With x as the literal number involved, the degree of Sx 3 is 
3. The degree of 2x is 1 because x x l . The degree of (5x 3 3z 2 -f 2x 7) 
is 3 

ILLUSTRATION 2. The degree of 3#V in x is 2, in y is 3, and in x and y 
together is (3 -f- 2) or 5. 

ILLUSTRATION 3. A polynomial of the first degree in x is called a linear 
polynomial in x and is of the form ax + b, where a and b do not involve x. 

55. Linear equations 

An integral rational equation is one in which each member is an 
integral rational polynomial in the unknowns. A linear equation, or 
an equation of the first degree, is an integral rational equation in which 
the terms involving the unknowns are of the first degree. 

ILLUSTRATION 1. The equation 3x 7 = 5 is a linear equation in x. 

SUMMARY. To solve a linear equation in one unknown: 

1. If fractions appear, place parentheses around each numerator and 
clear of fractions by multiplying both members by the LCD of the 
fractions; then, remove parentheses and combine terms. 

2. Transpose all terms involving the unknown to one member and all 
other terms to- the other member. Combine terms in the unknown, 
exhibiting it as a factor. 

3. Divide both sides by the coefficient of the unknown. 

4. To check the solution, substitute the result in the original equation. 



62 LINEAR EQUATIONS IN ONE UNKNOWN 

r. *oi x x 3 3 + x n 

EXAMPLE 1. Solve: --- ~ = r -- 2. 



SOLUTION. 1. The LCD is 30. Hence, multiply Ijoth sides by 30, ob- 
serving that 



30 -p = 10(z-4); 30-- = 15(s - 3); 30 -_ = 3(3 + 

10(x - 4) - 15(x - 3) = 3(3 + x) - 60; 
10* - 40 - 15* + 45 = 9 + 3x - 60; 

- 5x + 5 = 3x - 51. 
/ 

2. Subtract 3z and 5 from both sides: 

- 5x - 3x = - 51 - 5; - 8x = - 56. 

3. Divide both sides by 8: x = 7. 

CHECK. Substitute x 7 in the original equation. 

r * 1 j ; 7 ~ 4 7 - 3 3 4 1 o 1 

Left-hand side: ^ --- j; = ___ == 1_2= 1. 

o A o 

34-7 10 

Right-hand side: ^. -- 2 = -r 2=1 2= 1. This checks. 

In the case of a linear equation in a. single unknown x, if the un- 
known remains in the equation after Step 2 of the standard method 
of solution is applied, the equation is then of the form ex = 6, where 
c j* 0, and b and c do not involve x. On dividing both sides of ex b 
by c we obtain x = b/c. That is, a linear equation in a single un- 
known has just one root. 

Note 1. In directions for solving an equation, in this book, a specification 
to add, subtract, multiply, or divide will mean to perform the operation on 
both sides of the preceding equation. 



EXERCISE 21 

Solve the equation for the literal number in it. 
1. 5x - 3 = x + 7. 2. 3z - 6 = 18 - x. 



3. 5 - 2y = 2 - 3y. 4. 3 - 3x = - 7 - 5x. 

5. 2 + 5 - 4(2 - ). 6. 2y - 4 = 1 - 4y. 

7. 5 - 5y = 5 - 4y. B. 2(4 - x} = 8 - 3x. 

9. Sy + 3 - 5y - 4. 10. 7 - x = 2(1 - 



LINEAR EQUATIONS IN ONE UNKNOWN 



63 



11. 2(7 + a) - 1 - 7x. 
13. 9 - 7h = 14A - 12. 

16. 52 - 11 + 32 = 2 - 3. 

17. 5x -- 1 - 4* -f 2. 
19. 4* + * - 3* - }. 
21. to - f - 3s + ft. 

, 3 h 



25 -4- _ 
25 . - - 4 - 3 



27. _ _ _ 

10 2 6 2 



31. 



__ 3 
" T "" 5* 

- 7 _ 4 + y 



_ 

35. 



5 - 



25 5x - 3 



3 -a: _ 5 a; - 2 
~6~ ~ 6 " ~2~ 



6 



3x - 2 



a; - 5 
T- + 
43. .26 - z = .98 - 3. 

46. .26a: - .2 = .53* - .38. 

47. 2.5* - 3.7 = 13.5 - 1.8x. 
49. .19* - .358 = .032 - .07*. 



12. llh - 5 - 8fc - 4. 
14. 7k + 12 = 2k - 7. 
16. 11 - 6w> - 34w - 9. 
18. 42 - i = 32 + f . 



22. 5* + Y = 17* - 
5* 3* 3* 



-------. 

* 15 3 ~ 5 5 



28. - = - - 
10 3 2 12 



3 
5 



13 
TO 




32. 



3 - 2x 9 x - 3 



37 ^ 2a; + 7 
TO + ~~5~~ 



- 5 



8 13 x - 2 



- 3 



2s - 1 



16 



44. 3x - .55 = .33 - 
46. 2.3z - 2.4 = 1.6 - 1.7*. 
48. .21* - .46 = .79 + .96*. 
60. 4.088 + .03* - 3* - .07. 



64 LINEAR EQUATIONS IN ONE UNKNOWN 

61. 2.035 - Mix = - .212* - .215. 

62. 3(5z + 2) - 12 = 25(2z + 1) - 3. 

63. 2y -I - 10(y + 1) = 2(2 - 3t/) - 1. 

64. 82 - 2(32 - 1) = 7z - 3(z - 1). 
66. 2x - 6(z + 1) = 1 - 3(3* - 1). 

66. S(w + 2) - 5(2w - 1) = 6(w - 2) + 3. 



68. 4r - 9 = 7(2 - r) - 6(r + 1). 

Solve for P, or A, correct to two decimal places, by first clearing of fractions. 
69. 300 = P[l + f (.07)]. 60. 250 = A[l - J(.05)]. 

61. 500 = A[l - A(.06) J 62. 750 = P[l + tt(.07) J 

56. Simple Factoring of polynomials 

If all terms of a polynomial contain a common factor, we may 
express the polynomial as a product of this factor and a second factor. 
The second factor is the sum obtained by dividing each term of the 
polynomial by the common factor of the terms. 

ILLUSTRATION 1. 3x + ax -f bx = z(3 + a + 6). 

+ 2xy + 4zy 3 = xy(3y + 2 



At present, in solving equations in an unknown x, we will be in- 
terested in factoring polynomials only where a: is a common factor 
of the terms. If we express such a polynomial as the product of x 
and another factor, we then refer to this factor as the coefficient of x. 

ILLUSTRATION 2. On factoring, we obtain 

2x + 4ox + %cx = 2x(l -f 2a + c). (1) 

In (1), we say that the coefficient of a: is 2(1 -f- 2a -f c). 

57. Constants and variables 

In a given problem, a constant is a number symbol whose value does 
not change during the discussion involved. A variable is a number 
symbol which may take on different values. Any explicit number, 
such as 7, automatically is a constant wherever it appears. 



LINEAR EQUATIONS IN ONE UNKNOWN 65 

ILLUSTRATION 1. The volume V of a sphere is given by the formula 
V = Jur 3 where r is the radius. In considering all possible spheres, r and V 
are variables but TT is a constant, approximately 3.1416. 

58. Literal equations 

Sometimes an equation in an unknown x may involve other literal 
numbers besides x. In such a case, during the process of solution 
for x, we assume that the other literal numbers are constants. The 
summary of Section 55 still specifies our method of solution. 

EXAMPLE 1. Solve for x: 3bx 2 = 2cx + 

SOLUTION. 1. Transpose 2 and 2cx: 

36z - 2cx = 2 + a. (1) 

2. Factor on the left in (1) : 

x(3b - 2c) = 2 + a. (2) 

3. Divide both sides of (2) by (36 2c), the coefficient of x: 

_ 2 + a 
X ~ W^2c' 

2x 3 x 

EXAMPLE 2. Solve for x: ^r -- = 

ao a 2a 

SOLUTION. 1. The LCD is 2ab. Hence, multiply both sides by 2ab, 
noticing that 

2a6(|) = 4*; 2o6(|) - 66; 2^) - te. 

We obtain 4z 66 = bx. 

2. Transpose terms, and solve for x : 

fiA 

4x - bx = 66; x(4 - 6) = 66; x = ^y 

37 5 

EXAMPLE 3. Solve for x: s -- =- = 7- (4) 

4 



SOLUTION. The LCD is 12z. Multiply both sides by 12x: 

18 - 28 = - 15*; - 10 = - ISz; x = . (5) 

CHECK. Substitute x = in (4) : 

3 797 5 



____ _ , 

5 
Right-hand side: T- This checks. 



66 LINEAR EQUATIONS IN ONE UNKNOWN 

Comment. In this chapter, the unknown will occur in a denominator 
only under the most simple conditions. In solving (4), an incorrect choice 
of the LCD, containing an unnecessary factor, might have prevented the 
equations in (5) from being linear in x. 

59. Formulas 

Frequently, the data in a problem come to us as the values of a 
set of variables, which we represent as literal numbers. Sometimes 
we are able to write a mathematical expression for one of the variables 
in terms of the others. The resulting equation is referred to then as 
a formula for the first variable. An algebraic formula is one involv- 
ing only the operations of algebra. 

ILLUSTRATION 1. The Fahrenheit reading, F, and centigrade reading, (7, 
in degrees for a given temperature are related by the equation 5F = 9(7 -f 160. 
On solving for F, we obtain a formula for F in terms of C: 

F = f C + 32. (1) 

To find F corresponding to 36 centigrade, substitute C = 36 in (1) : 

F = f (36) + 32 = 64.8 + 32 = 96.8. 

EXERCISE 22 

Solve for x, or y, or z, whichever appears. Reduce any final fraction to 
lowest terms. 

1. bx = 3 H- c. 2. 16z - 4 = h. 3. ex 5a = 3h. 



4. ay by = 5. 6. 3x = ax 4- 26. 6. 2z = bz a. 

7. 2ay 5c = 3by -f 4a. 8. 7x d = Sax + 8. 

9. ax - Sax = 5c - bx. 10. 2dx - 3 = d?x + b. 



11. = a. 12. 3x = ~ 13. = d. 

b be 

- ax c A ... 26 3x n .,_ a?x , 

14. -T j = 0. 15. --- = 0. 16. -r- = 2a 3 . 

b d c a 3 



17 _ _ _ 18 _ = o 19 - - 

JL f 3T7" mo U. AO rt i \J AU - 



2c 6 "' *" 4 ~ BC 



a x x a 

2 24.--- = -1 

o 3 2c c 2 1% 



LINEAR EQUATIONS IN ONE UNKNOWN 67 

26. 3a(2x - 36) - 5(cz - 3) = 26. 26. 4c(ax - 6c) - 2a(bx + a 2 ) = 0. 
97 4 2 1 5 29 3 

V/ - - = - Ho* ' - == - - * 

3z 15 3z 24 4x 



29 - = + . 

10* 12 ^ 15z 662 15az 

31. In the Fahrenheit-centigrade equation, 5F = 9C + 160, solve for C 
in terms of F. Then, use the resulting formula to find the centigrade tem- 
perature correct to tenths of a degree corresponding to the following Fahren- 
heit temperatures: (a) 32; (6) 212; (c) 80; (d) 50. 

32. Let an object be shot vertically upward from the surface of the earth 
with an initial velocity of v feet per second, and let us neglect air resistance 
and other disturbing features. Then, it is proved in physics that s = vt %gP, 
where s feet is the height of the object above the surface at the end of t 
seconds and g = 32, approximately, (a) Find s if v 100 and t = 6. 
(6) Solve for v to obtain a formula for v in terms of s and t. (c) From (6), 
compute the velocity with which the object must be shot to attain a height 
of 1000 feet in 5 seconds. 

33. Solve / = ma for a. 34. Solve 8 = k -{- vt for v. 
36. Solve I = a + (n l)d for a; for ; for d. 



I M 
36. Solve I = for M; for t\ for J. 



37. Solve S - - r- for a. 

r 1 

38. Solve S = - r for I: for a. 

r 1 

Each of the following problems states a rule for computing values of a certain 
variable in terms of others. State the rule by means of a formula. 

39. The average, A, of three numbers M, N, and P is one third of their 
sum. 

40. The cost of electricity for a house in a certain city is 7jf per kilowatt- 
hour for the first 10 kilowatt-hours, 5^ per kilowatt-hour for the next 20 kilo- 
watt-hours, and 2%i per kilowatt-hour for all over 30 kilowatt-hours. Write 
an expression for the total cost, C, of (a) 60 kilowatt-hours; (6) n kilowatt- 
hours where n > 30. 

41. On any order for more than 200 units of a certain manufactured 
product, the cost is 15^ per unit for 200 units and 12^f per unit for the re- 
main(}er of the order. Write a formula for the cost, C, in dollars, of n units 
if n > 200. 



63 LINEAR EQUATIONS IN ONE UNKNOWN 

60. Algebraic translation 

In applying equations in the solution of problems stated in words, 
we translate word descriptions into algebraic expressions. 

ILLUSTRATION 1. If x is the length of one side of a rectangle and if the 
other dimension is 3 units less than twice as long, then the other dimension 
is (2x 3) ; the perimeter (sum of lengths of sides) is 

2z + 2(2z - 3), or 6z - 6, 
and the area is x(2x 3). 

ILLUSTRATION 2. If x, y, and z are, respectively, the units', tens', and 
hundreds' digits of a positive integer with three digits, the value of the 
integer is x + Wy + lOOz. 

SUMMARY. To solve an applied problem by use of equations: 

1. Introduce one or more letters to represent the unknowns and give 
a description of each one in words. 

2. Translate the given facts into one or more equations involving the 
unknowns, and solve for their values. 

3. Check the solution by substituting the results in the written state- 
ment of the problem. 

EXAMPLE 1. $350 is to be divided between Jones and Smith so that 
Jones will receive $25 more than Smith. How much does each receive? 

SOLUTION. 1. Let x be the number of dollars which Smith receives. 
Then, Jones receives (x + 25) dollars. 

2. The sum of the amounts received is $350, or 

x + (x + 25) = 350. .. (1) 

On solving (1), we obtain x = 162.50. Hence, Smith receives $162.50 and 
Jones receives $162.50 + $25 or $187.50. These results check. 

EXAMPLE 2. Find two consecutive even integers such that the square 
of the larger is 44 greater than the square of the smaller integer. 

SOLUTION. 1. Let x represent the smaller integer. Then, the larger 
integer is x + 2. Their squares are x* and (x -f 2) 2 . 

2. From the data, (x + 2) 2 - x 2 = 44. (2) 
Expanding: x* + 4z + 4 x* = 44; 

4z = 40; x = 10. 

3. The integers are 10 and 12. 



LINEAR EQUATIONS IN ONE UNKNOWN 69 

CHECK. 10 2 - 100; 12 2 = 144; 144 - 100 = 44. 

EXAMPLE 3. How long will it take Jones and Smith, working together, 
to plow a field which Jones can plow alone in 5 days and Smith, alone, 
in 8 days? 

SOLUTION. 1. Let x days be the time required by Jones and Smith, 
working together. 

2. In 1 day, Jones can plow J and Smith J of the field. Hence, in x 

X 3/ 

days Jones can plow ^ and Smith can plow ^ of the field. 

5 o 

3. Since the whole field is plowed in x days, the sum of the fractional 
parts plowed by the men in x days is 1 : 

x = 3^ days. 



EXERCISE 23 

Solve by use of an equation in just one unknown. 

1. A line 68 inches long is divided into two parts where one is 3 inches 
longer than the other. Find their lengths. 

2. A rope 36 inches Sng is cut into two pieces such that one part is 
4 inches less than twj 'as long as the other part. Find the lengths of 
the parts. ;he 

3. Find the dinu J P| ( jns of a rectangle whose perimeter is 55 feet, if 
the altitude is f of the base. 

4. One dimension of a rectangle is the other. Find the dimensions 
of the rectangle if its perimeter becomes 130 feet when each dimension 
is increased by 5 feet. 

5. What number should be subtracted from the numerator of fj to 
cause the fraction to equal f ? 

6. Find two consecutive positive integers whose squares differ by 27. 

7. Find three consecutive integers whose sum is 48. 

8. Find the angles of a triangle where one angle is three times a second 
angle and six times the third angle. 

9. A rectangle and a triangle have equal bases. The altitude of the 
rectangle is 25 feet and of the triangle is 20 feet. The combined area of the 
triangle and the rectangle is 280 square feet. Find the length of the base. 

10. J?ind two consecutive positive odd integers whose squares differ 
by 32,* 



70 LINEAR EQUATIONS IN ONE UNKNOWN 

11. The length of a rectangular lot is three times its width. If the length 
is decreased by 20 feet and the width is increased by 10 feet, the area is 
increased by 200 square feet. Find the original dimensions. 

12. A triangle and a rectangle have the same base. The altitude of 
the rectangle is 4 feet longer, and the altitude of the triangle is 5 feet shorter 
than the base. The area of the rectangle is 90 square feet greater than 
twice the area of the triangle. Find the length of the base. 

13. A sum of money amounting to $13.55 consists of nickels, dimes, and 
quarters. There are three times as many dimes as nickels and three less 
quarters than dimes. How many of each coin are there? 

14. If (59 3z) is divided by the integer x, the quotient is 5 and the 
remainder is 3. Find the value of x. 

HINT. Recall: dividend = (quotient) (divisor) + remainder. 

16. A peddler sold 7 bushels more than f of his load of apples and then 
had 9 bushels less than f of the load remaining. Find his original load. 

16. In 1 hour, Jones can plow J of a field and Roberts -fa of it. If they 
work together, how long will it take them to plow the field? 

17. A room can be painted in 21 hours by Smith and in 14 hours by 
Johnson. How long will it take them to paint the room working together? 

18. How long will it take two mechanical d f .chdiggers to excavate a 
ditch which the first machine, alone, could con lete in 8 days and the 
second, alone, in 11 days? Jo* 

* 

19. How long will it take workers A and B, together, to complete a 
job which could be done by A alone in 7 days, and by B alone in 9 days? 

20. How long will it take to fill a reservoir with intake pipes A, B, and 
C open, if the reservoir could be filled through A alone in 6 days, B alone 
in 8 days, and C alone in 5 days? 

2JL. If 1000 articles of a given type can be turned out by a first machine 
in 9 hours, by a second in 6 hours, and by a third in 12 hours, how long 
will it take to turn out the articles if all machines work together? 

22. An integer between 10 and 100 ends in 5. By writing an equation, 
find the integer if it is 5 times the sum of its digits. 

61. Percentage 

The words per cent are abbreviated by the symbol % and mean 
hundredths. That is, if r is the value of h%, then 

h % = 156 = ' 



LINEAR EQUATIONS IN ONE UNKNOWN 7? 

fi 4 7*i 

ILLUSTRATION 1. 6% = = .06. 4J% = = .0475. 



From equation 1, we obtain h = lOOr; hence, to change a number 
to per cent form, we multiply r by 100 and add the % symbol. 

ILLUSTRATION 2. If r = .0175, then lOOr = 1.75 and .0175 = 1.75%. 

18 9 
ILLUSTRATION 3. cnn = xrv\ = - 0225 = 2 - 25 %- 

oLMJ 



The description of a ratio M/N in per cent form is the background 
for the following terminology. 

If M is described by the relation M = Nr, where r is the ratio 
of M to N, we sometimes say that M is expressed as a percentage 
of N, with r as the rate and AT as the base for the percentage: 

M = Nr, or percentage = (base) (rate) ; (2) 

M . percentage /0 , 

r = -TT* or rate = * r - (3) 

W base v ' 

ILLUSTRATION 4. To express 375 as a percentage of 500, we compute the 
rate r = fj$ = .75. Hence, 375 = .75(500), or 375 is 75% of 500. 



EXAMPLE 1. Find the number of residents of a city where 13% of the 
population, or 962 people, had influenza. 

SOLUTION. Let P be the number of residents: 



.UP - 962; P - , = 7400. 

.lo lo 

Note 1. In the formation of a mixture of different ingredients, we shall 
assume that there is no change in volume. Actually, a slight gain or loss of 
volume might occur, for instance, as a result of chemical action. 

In the typical mixture problem, where one special ingredient is 
involved, the equation for solving the problem frequently can be 
obtained by writing, in algebraic form, the statement that 

( the sum of the amounts of the] _ ( amount of the ingredient } , . 
\ ingredient in the parts / \ in the final mixture, j 

If the price of a mixture is the fundamental feature, the equation 
may be obtainable by using equation 4 with the costs of tHe parts 
thought of in place of the ingredients. 



72 LINEAR EQUATIONS IN ONE UNKNOWN 

EXAMPLE 2. How many gallons of a mixture containing 80% alcohol 
should be added to 5 gallons of a 20% solution to give a 30% solution? 

SOLUTION. 1. Let x be the number of gal. added. In x gal., 80% pure 
alcohol, there are .80x gal. of alcohol. 

2. In 5 gal., 20% pure, there are .20(5) gal. of alcohol. 

3. In (5 4- x) gal., 30% pure, there are .30(z + 5) gal. of alcohol. 

4. The alcohol in the final mixture of (5 -f- x) gal. is the sum of the alcohol 
in the x gal. and in the 5 gal. Or, 

.30(z + 5) = .80* + .20(5); x = 1. 

EXAMPLE 3. What percentage of a 20% solution of hydrochloric acid 
should be drawn off and replaced by water to give a 15% solution? 

SOLUTION. 1. Think of the solution as consisting of 100 units of volume; 
then the solution contains 20 units of acid. 

2. Let x% be the rate for the percentage which should be drawn off. 
Then, from the 100 units we should draw off x% of 100, or x units. 

3. In x units there are .20# units of acid. There remain (20 .20z) 
units in the final solution of 100 units, after water is added. Therefore, 



.15 = 2 ; 15 = 20 - .20z; x = 25. 



Or, we should draw off 25% of the original solution. 

EXERCISE 24 

Change to decimal form. 
1. 5%. 2. 4J%. 3. 3|%. 4. 45%. 5. 126.3%. 6. 

Change to per cent form. 
7. .07. 8. .0925. 9. .025. 10. .0575. 11. 1.35. 

Compute each quantity. 
12. 6% of $300. 13. 3^% of 256. 14. 110% of 1250. 

Express the first number as a percentage of the second. 
15. 75, of 200. 16. 400, of 640. 17. 350, of 200. 

18. The average price of copper per pound in the United States was 
approximately $.138 in 1926, $.081 in 1931, and $.215 in late 1947. Find 
the per cent of change in the price from 1926 to 1931; from 1931 to 1947. 

Solve each problem by using an equation in just one unknown. 

19. If 385 is 85% of x, find x. 20. If 268 is 24% less than y, find y. 



LINEAR EQUATIONS IN ONE UNKNOWN 73 

21. After selling 85% of a stock of dresses, a merchant finds that he 
has 84 dresses left. What was his original stock? 

22. A merchant buys 100 dozen shirts at $13.20 per dozen. He sells 
90 dozen at a markup of 30% over the purchase price. At what price per 
shirt could he afford to sell the remaining 10 dozen to clear his stock if he 
desires his total receipts from the shirts to be 25% greater than the cost? 

23. $3000 of Smith's income is not taxed by the state where he lives. 
All of his income over $3000 is taxed 2% and all over $8000 is taxed 3% in 
addition. If he pays a total tax of $800, what is his income? 

24. Under the taxes of Problem 23, with an additional surtax of 5% 
on all income over $20,000, Johnson's tax is $1400. Find his income. 

25. A merchant has some coffee worth 70^f per pound and some worth 
50^. How many pounds of each are used in forming 100 pounds of a 
mixture worth 65^ per pound? 

26. How many gallons of a mixture of water and alcohol containing 
60% alcohol should be added to 9 gallons of a 20% solution to give a 30% 
solution? 

27. How many gallons of a solution of glycerine and water containing 
55% glycerine should be added to 15 gallons of a 20% solution to give a 
40% solution? 

28. How many ounces of pure silver must be added to 150 ounces, 45% 
pure, to give a mixture containing 60% silver? 

29. A feed merchant wishes to form 200 bushels of a mixture of wheat 
at $1.25 per bushel and wheat at $.80 per bushel, so that the mixture will 
be worth $1.00 per bushel. How much of each kind should he use? 

30. How many pounds of cream containing 35% butterfat should be 
added to 800 pounds of milk containing 3% butterfat to give milk con- 
taining 3.5% butterfat? 

31. An automobile radiator holds 8 gallons of a solution containing 
40% glycerine. How much of the solution should be drawn off and replaced 
by water to give a solution with 25% glycerine? 

32. What percentage of a 30% solution of sulphuric acid should be 
drawn off and replaced by water to give a 20% solution? 

33. What percentage of a 40% solution of alcohol and water should 
be replaced by pure alcohol, to give a 75% solution? 

34. What percentage of a mixture of sand, gravel, and cement, con- 
taining 30% cement, should be replaced by pure cement in order to give 
a mixture containing 40% cement? 



74 



LINEAR EQUATIONS IN ONE UNKNOWN 



62. Lever problems 

A lever consists of a rigid rod with one point of support called the 
fulcrum. A familiar instance of a lever is a teeterboard. If a weight 
w is attached to a lever at a certain point, the distance h of w from 
the fulcrum is called the lever arm of w, and the product hw is called 
the moment of w about the fulcrum. The following statement is 
demonstrated in physics. 

LEVER PRINCIPLE. // two or more weights are placed along a lever 
in such a way that the lever is in a position of equilibrium, then, if each 
weight is multiplied by its lever arm, the sum of these products for all 
weights on one side of the fulcrum equals the sum of the products for all 
weights on the other side. In other words, the sum of the moments of 
the weights about the fulcrum is the same on both sides. 

Note 1. In all lever problems in this book, it will be assumed that the 
weight of the lever is negligible for the purpose in view. 



ILLUSTRATION 1. In Figure 2, the sum 
of the moments for the weights at the left 
is (5-80 -h 4-250) or 1400, and for those 
at the right is 

(4- 100 + 5-200), or 1400. 
Hence, this lever is balanced. 




MIAMI 





Fig. 2 



> x- 



EXAMPLE 1. Two girls, weighing 75 pounds and 90 pounds, respectively, 
sit at the ends of a teeterboard 15 feet long. Where should the fulcrum be 
placed to balance the board? 

SOLUTION. 1. Let x feet be the distance from the fulcrum to the lighter 
girl. Then, the lever arm for the other 
girl is (15 x) feet. 

2. Hence, from Figure 3, 

75* = (15 - z)90; 
x = 83^ feet. 

EXERCISE 25 

1. A weight of 300 pounds is placed on a lever 20 feet from the fulcrum. 
Where should a weight of 275 pounds be placed to balance the lever? 

2. A weight of 60 pounds is placed on a lever 8 feet from the fulcrum, 
How heavy a weight should be placed 12 feet from the fulcrum on the 
other side to give equilibrium? 




LINEAR EQUATIONS IN ONE UNKNOWN 75 

3. A teeterboard is balanced when one girl weighing 80 pounds sits 4 feet 
from the fulcrum, another girl weighing 100 pounds sits 7 feet from the 
fulcrum on the other side, and a third girl sits 6 feet from the fulcrum. 
How much does the third girl weigh? 

4. Jones and Smith together weigh 340 pounds. Find their weights 
if they balance a lever when Jones sits 5 feet from its fulcrum on one side 
and Smith sits 6 feet from the fulcrum on the other side. 

5. A 40-pound weight is placed 6 feet from the fulcrum on a lever, and 
a 60-pound weight 8 feet from the fulcrum on the other side. Where should 
a 30-pound weight be placed to give equilibrium? 

6. How heavy a weight can a man lift with a lever 9 feet long if the 
fulcrum is 2 feet from the end under the weight and if the man exerts a 
force of 140 pounds on the other end? 

7. How many pounds of force must a man exert on one end of an 8-foot 
lever to lift a 300-pound rock on the other end if the fulcrum is 1J feet 
from the rock? 



63. Uniform motion 

When we say that a body is moving in a path at constant speed, 
we mean that the body passes over equal distances in any two equal 
intervals of time. Such motion is referred to as uniform motion in 
the path. The velocity * or speed or rate of the body in its path is 
defined as the distance traveled in one unit of time. If v is the veloc- 
ity, and d is the distance traveled in t units of time, 

d = vt. (1) 

j 
Since v = - , the velocity is referred to as the rate of change of the 

t 

distance with respect to the time. In stating a velocity, the units for 
the measurement of time and of distance must always be mentioned. 

ILLUSTRATION 1. If an airplane flies 1250 miles in 5 hours at uniform 
speed, the speed is 

d 1250 _ n ., , 

v = - = > or 250 miles per hour. 
t o 

The speed of the airplane per minute is 2 - or 4J miles. 

* In physics, velocity is denned as a vector quantity, possessing both magnitude 
and direction. In this text, wherever the word velocity is used, it will refer to the 
magnitude (positive) of the velocity vector. 



76 LINEAR EQUATIONS IN ONE UNKNOWN 

Note 1. All motion considered in this book will be uniform motion. If 
the velocity of a moving body is variable, a discussion of the motion must 
bring in more advanced notions met in physics and calculus. 

EXAMPLE 1. A messenger, traveling at a speed of 65 miles per hour, pur- 
sues a truck which has a start of 2 hours and overtakes the truck in 3 hours. 
Find the speed of the truck. 

SOLUTION. Let x miles per hour be the truck's speed. Then, 
3. (65) = (3 + 2)z; 195 = 5z; x = 39 miles per hour. 

The equation d = rt applies in the discussion of any variable 
quantity d which changes uniformly at a specified rate r with respect 
to change in the time t. Thus, we may refer to a rate of increase 
or a rate of decrease under various conditions. 

EXAMPLE 2. A motorboat went 70 miles in 4 hours when traveling at 
full speed upstream on a river whose current flows at the rate of 6 miles 
per hour. How fast can the boat travel in still water? 

SOLUTION. 1. Let x miles per hour be the speed of the boat in still water. 
In travel upstream, the rate of the current reduces the speed of the boat to 
(x 6) miles per hour. 

2. From d = vt, with t = 4, v = x 6, and d = 70, 

70 = 4(s - 6). (2) 

On solving (2) we obtain 4x = 94; x 23.5. 

Hence, the boat can travel 23.5 miles per hour in still water. 

64. Radius of action of an airplane 

Suppose that an airplane flies with a groundspeed * of Gi miles f 
per hour in a particular direction from a base B and then back along 
this path at a groundspeed of (? 2 miles per hour, when the engines are 
working at full power, and while the wind maintains a constant 
direction and speed. G\ and <? 2 in general would be different because 
of the effects of the wind. Suppose that the gasoline tanks of the 
airplane permit it to operate at full power for only T hours after 
leaving B. We refer to T as the available fuel hours. Then, it is of 
interest to consider the maximum length of time, h hours, during 
which the airplane may fly out if it is to return to B by the end of 

* Speed with respect to the ground as contrasted to the airspeed, or speed with 
respect to the air, which itself may be in motion because of a wind. 

t The subscript 1 on GI is just a tag. We read "Gi" as "G sub 1 " or "G, 1," 



LINEAR EQUATIONS IN ONE UNKNOWN 77 

T hours. The distance R which the airplane flies out from B in 
h hours is called the radius of action of the airplane in the specified 
direction, with the given wind. It can be proved * that 



EXAMPLE 1. How many fuel hours must be available in order to have 
900 miles as the radius of action in a direction where the groundspeeds 
out and back are, respectively, 300 and 200 miles per hour? 

SOLUTION. Substitute R = 900, Gi = 300, and G z = 200 in the second 
equation in (1) : 

_ 7X300X200) 
900 " 100 + 200 ' r 



Hence, T = = 7J hr. 



EXERCISE 26 

1. At what rate does an automobile travel if it goes 450 miles in 9 hours? 

2. Jones and Smith travel toward each other from points 500 miles 
apart, Jones at the rate of 60 miles per hour and Smith at 50 miles per hour. 
When will they meet if they start at the same instant? 

3. Two men start at 7 A.M. from the same place, in opposite directions, 
at speeds of 36 miles and 48 miles per hour, respectively. When will they 
be 600 miles apart? 

4. At 6 A.M., a motorcycle messenger starts from a city at a speed of 
45 miles per hour to meet a regiment which is 120 miles away and is ap- 
proaching at a speed of 5 miles per hour. When will the messenger meet 
the regiment? 

6. One man can run 400 meters in 54 seconds and a second man can run 
the distance in 60 seconds. How long will it take the faster man to gain 
a lead of 12 meters on the slower man if they start together in a 400-meter 
race? 

6. An airplane leaves the deck of a battleship and travels south at the 
rate of 230 miles per hour. The battleship travels south at the rate of 20 
miles per hour. If the wireless set on the airplane has a range of 800 miles, 
when will the airplane pass out of wireless communication with the ship? 

* See page 51, in WILLIAM L. HART'S College Algebra, 3d Edition, D. C. HEATH 
AND COMPANY. 



78 LINEAR EQUATIONS IN ONE UNKNOWN 

7. How many seconds will it take for a man to travel y miles if he travels 
x miles in t seconds? 

8. In an 800-meter race between two men, the winner's time is 2 minutes, 
and his lead is 40 meters. How many seconds would it take the loser to 
run 800 meters? 

9. An airplane flew 850 miles in 2J hours against a head wind blowing 
30 miles per hour. How fast could the plane fly in still air? 

10. Two men start together in a race around a 300-yard oval track, one 
man at a speed of 9 yards per second and the other man at 7J yards per 
second. When will the faster man be exactly one lap ahead? 

11. When will Johnson be twice as wealthy as Smith if each has $4000 
now and if their estates are increasing at the annual rates of $400 for Smith 
and $1200 for Johnson? 

12. Jones can run around a 400-meter track in 65 seconds. How long 
does Smith take to run the 400 meters if he meets Jones in 35 seconds after 
they start together in a race around the track in opposite directions? 

In each problem, (a) find the radius of action for a flight by an airplane 
in a direction where the groundspeeds have the indicated values; (b) find the 
number of hours flown on the maximum outward trip. 

13. Sixteen fuel hours available; groundspeed out is 240 miles and back is 
210 miles per hour. 

14. Twelve fuel hours available; groundspeed out is 190 miles and back is 
225 miles per hour. 

16. Fourteen fuel hours available; groundspeed out is 175 miles and back 
is 200 miles per hour. 

// an airplane is to have the specified radius of action in a direction corre- 
sponding to the given groundspeeds, find the number of fuel hours which must 
be available. 

16. Radius of action is 1350 miles; groundspeed out is 200 miles and back 
is 225 miles per hour. 

17. Radius of action is 750 miles; groundspeed out is 195 miles and back 
is 175 miles per hour. 

18. How many fuel hours must be available to permit an airplane flight 
out from a field for 5J hours in a direction such that the groundspeed out is 
200 miles and back is 185 miles per hour? 

19. At how many minutes after 2 P.M. will the minute hand of a clock 
overtake the hour hand? 

20. After 10 P.M., when will the hands of a clock first form a straight line? 



LINEAR EQUATIONS IN ONE UNKNOWN 79 

65. Interest 

Interest is income received from invested capital. The capital 
originally invested is called the principal. At any time a'fter the 
investment of the principal, the sum of the principal and the interest 
due is called the amount. 

The rate of interest is the ratio of the interest earned in one year 
to the principal. If r is the rate and P is the principal, then 



_ interest per year _ 

T - J 



interest for one year = Pr. (2) 

Thus, the interest is a percentage of the principal, with r as the rate. 

ILLUSTRATION 1. If $1000 earns $36.60 interest in one year, then, from 
equation 1, r =^^ = .0366, or r = 3.66%. 



66. Simple interest 

If interest is computed on the original investment during the whole 
life of a transaction, the interest earned is called simple interest. 

Suppose that P is invested at simple interest for t years at the 
rate r. Let / be the interest and F be the final amount at the end 
of the t years. Then, the interest for one year is Pr and, by defini- 
tion, the simple interest for t years is t(Pr) or Prt', that is, 

/ = Prt. (1) 

Since amount equals principal plus interest, 

F = P + /. (2) 

From (1), 

P + / - P + Prt = P(l + rt). 

Hence, from (2), 

F = P(l 4- rt). (3) 

In equations 1 and 3, t represents the time expressed in years. 
If the time is described in months, we express it in years assuming a 
year to contain 12 equal months. If the time is given in days, there 
are two varieties of interest used, ordinary and exact simple interest 
In computing ordinary interest we assume a year to contain 360 days, 
and, for exact interest, we assume a year to contain 365 days. 



80 LINEAR EQUATIONS IN ONE UNKNOWN 

To find the amount F when P, r, and t are given, first find the 
interest from 7 = Prt and then compute P + / to find F. 

Note l'. Unless otherwise specified, the word "interest" in this book will 
refer to simple interest. 

ILLUSTRATION 1. If $5000 is invested for 59 days at 5%, 
the ordinary interest due is 5000(.05)^y = $40.97; 

the final amount due is 5000 + 40.97 = $5040.97. 

EXAMPLE 1. If $1000 accumulates to $1250 when invested at simple 
interest for 3 years, find the interest rate. 

SOLUTION. 1. We have P = $1000; F = $1250; I = 1250 - 1000 = $250. 

2. From 7 = Prt with t = 3, 

OKA 
250 = 1000(r)(3); 250 = 3000r; r = = .08J = 



In F = P(l + rt), the principal P is frequently called the present 
value of the amount F because, if P is invested today at the rate r, 
the accumulated amount at the end of t years will be F. 

EXAMPLE 2. Find the present value of $1100 which is due at the end of 
21 years, if money can be invested at 4%. 



SOLUTION. 1. We have F = $1100, r = .04, and t = 
2. Hence, from (3), 

1100 = P[l + f(.04)]; 1100 = P(l + .10); 



1.1P = 1100; P = ~ = $1000. 

JL .L 

CHECK. 7 = 1000(.04)(f) = $100; F = 1000 + 100 = $1100. 



EXERCISE 27 

Find the ordinary interest and the final amount. 
1. On $5000 at 6% for 216 days. 2. On $8000 at 4J% for 93 days. 

Find the exact interest and the final amount. 
3. On $3000 at 4% for 146 days. 4. On $2500 at 51% for 27 days. 

6. Find the amount due at the end of 8 months if $150 is invested 
at 9%. 

With the given data, solve F = P(l + rt) for P, to the nearest cent. 
6. F = $1000; r = .03; t - f 7. F = $3000; r = .05; t = 



LINEAR EQUATIONS IN ONE UNKNOWN B1 

8. At what rate will $750 be the interest for 5 years on $6000? 

9. Find the invested principal if it earns $375 interest in 3 months when 



the interest rate is 

10. Find the principal if it earns $150 interest in J year at 8%. 

11. (a) Find the principal which will amount to $1300 by the end of 
6 years when invested at 5%. (6) Verify the result by computing interest 
on it for 6 years. 

12. Find the present value of $1888 which is due at the end of 4 years, 
if the interest rate is 4%. 

13. Jones agreed to pay Smith $6000 at the end of 5 years. What should 
Jones pay immediately to cancel his debt if Smith agrees that he can invest 
money at 4%? 

14. Roberts buys a bill of goods from a merchant who asks $2000 at 
the end of 2 months. If Roberts y wishes to pay immediately, what should 
the seller be willing to accept if he is able to invest his money at 8%? 

15. A debtor owes $1100 due at the end of 2 years and he requests the 
privilege of paying an equivalent smaller sum immediately. At what 
simple interest rate would the creditor prefer to compute the present pay- 
ment, at 5% or at 6%, and how much would he gain by the best choice? 

16. How long will it take a given principal to double itself if invested 
at 5% simple interest? 

17. A man invests $7000, one part of it at 5% and the balance at 4%. 
If the total annual interest is $320, how much is invested at each rate? 



CHAPTER 



5 



SPECIAL PRODUCTS AND FACTORING 



67. Square root 

If R 2 A, we call R a square root of A. 

ILLUSTRATION 1. 4 is a square root of 16 because 4 2 = 16. 

Every positive number A has two square roots, one positive and 
one negative, with equal absolute values. The positive square root 
is denoted by H- VZ, or simple VA, and the negative square root 
b VZ. We call VZ a radical and A its radicand. Unless other- 
wise stated, the square root of A will mean its positive square root. 
By the definition of a square root, 

If x is positive or zero, 

~ - x. (2) 



ILLUSTRATION 2. V9 = 3 because 3 2 = 9. The two square roots of 9 
are V9 and V, or V9, or 3. We read " rfc" as "plus or minus." 

ILLUSTRATION 3. VJ J because (|) 2 = 4- 

68. Perfect squares 

In the square of an integral rational term, each exponent will be an 
even integer because, in squaring, each original exponent is multiplied 
by 2. 

ILLUSTRATION 1. (3xV) 2 = 3 2 (x 2 ) 2 (i/ s ) 2 



An integer is said to be a perfect square if it is the square of an 
integer. The student should learn the most common perfect square 
integers, with the aid of Table I, page 283. 



SPECIAL PRODUCTS AND FACTORING S3 

ILLUSTRATION 2. The perfect square integers are 1, 4, 9, 16, 25, 36, etc. 
Their square roots are, respectively, 1, 2, 3, 4, 5, 6, etc. Thus, \/25 = 5 
because 5 2 = 25. 

An integral rational term is said to be a perfect square if it is the 
square of some other term of the same variety. Hence, in a perfect 
square, each exponent is an even integer. 

ILLUSTRATION 3. 25a 2 6 4 is a perfect square because 25a 2 6 4 = (Safe 2 ) 2 . 

I. To find the square root of a perfect square monomial: 

1. Rewrite the literal part with each exponent divided by 2. 

2. Multiply by the square root of the numerical coefficient of the 
given term. 

ILLUSTRATION 4. Vl6x*y* = VT6Vx*y* = 4#y, because 



II. To find the square root of a fraction, find the square root of the 
numerator and of the denominator and divide: 




/ - v^4 2 

ILLUSTRATION 5. \/T^ = 7= = ^* 

V25 5 





IWQa 6 \/100a 6 10o 3 
ILLUSTRATION 6. 




Note 1 . For the present, we shall consider \/A only where A is a perfect 
square monomial, or where A is a fraction whose numerator and denominator 
are perfect square monomials. All literal numbers in A will be supposed 
positive or zero. 

EXERCISE 28 

Find the two square roots of the number and check by squaring the results. 
1. 25. 2. 49. 3. 121. 4. 64. 6. fc. 6. 



Find each square root and check by squaring the result. Inspect Table I if 
necessary. 
7. V9. 8. VIOO. 9. \/81. 10. \/l44. 



84 SPECIAL PRODUCTS AND FACTORING 



11. Vl96. tf. Vf. 13. \/if. 14. 

15. V. 16. Vg. 17. VS- 18. 

19. Vz 4 . 20. V. 21. Vo 2 ". 22. 



23. Vtf*. 24. V^ 5 . 25. Viol 26. 

27. \/4^. 28. \/16^. 29. \/49?. 30. 



31. \64a. 32. fo 2 . 33. \49w*c 4 . 34. 







39. tv 40. t/=?. 41. \(1- 










each quantity. 
51. (\/37) 2 . 52. (\/142^) 2 . 63. (V^) 2 . 64. (>/659)2. 



55. A negative number can have neither a positive nor a negative square 
'root. Why is this true? 

69. Products of binomials 

When desired, the product of two binomials may be found by 
longhand methods. 

ILLUSTRATION 1. (3s - 5y)(2x - 7y) = 3x(2x - ly) - 5y(2x 7y) 
= 6x 2 - 21xy - Wxy + 35t/ 2 = 6x 2 - Zlxy + 35y 2 . 

ILLUSTRATION 2. (x -f y)(x y) = x(x y) + y(x y) 

x z xy + xy y z = x 2 y 2 . 

ILLUSTRATION 3. (a + 6) 2 = (a -f fe)(a + 6) 

= a(o + 6) + 6(a + 6) = a 2 + 2ab + ft 2 . 

70. Special products 

The student should be able to dispense with the longhand methods 
of the preceding section and should form the product of two bi- 
nomials mentally. Products of the following types occur frequently. 
The student should verify each right-hand member. 



SPECIAL PRODUCTS AND FACTORING 85 



I. a(x + y) = ax -f 

II. (*4-y)(*-if) = x 2 - 

III. (a + b)* = a 2 + 2a& + 

IV. (a - &) 2 = a 2 - 2ab + 

V. (x + a)(x + 6) = x 2 + (ax -f 6x) -f ab. 

VI. (ax + b)(cx + d) = acx 2 -f (aa*x + bcx) + 6<f. 

It proves convenient to memorize Types II, III, and IV as formulas 
and also in words. 

ILLUSTRATION 1. Type II states that the product of the sum and the differ- 
ence of two numbers is the difference of their squares. 

ILLUSTRATION 2. Type III states that the square of the sum of two numbers 
equals the square of the first, plus twice the product of the numbers, plus the 
square of the second number. 



ILLUSTRATION 3. (c - 2d)(c + 2d) = c 2 - (2d) 2 = c 2 - 4#. (Type II) 

ILLUSTRATION 4. From Type III with a = 3x and 6 =* 2y, 

= 9x 2 



The right members of Types V and VI should not be committed to 
memory. However, the nature of the right members should be 
memorized, with (ax 4- bx) in Type V and (adx + bcx) in Type VI 
remembered as the sum of the cross products. 

ILLUSTRATION 5. To obtain (2# 5) (3x + 7) : 

f 2* -5 ' 

j^XT I ( Product = 6z 2 x 35. 
3z -F7 



Sum of the cross products: I4x 15x = x. 

The diagram and auxiliary computation of the sum of the cross products 
should be omitted and replaced by mental computation as in the next il- 
lustration. 



ILLUSTRATION 6. (2x - 7fc)(3z -f 2h) = 6x 2 - I7hx - 

because the sum of cross products is 2lhx + 4hx, or I7hx. 

ILLUSTRATION 7. (x 2 - 3y*) 2 - (z 2 ) 2 - 2(a; 2 )(3^) + W) 2 (Type IV) 



86 SPECIAL PRODUCTS AND FACTORING 

ILLUSTRATION 8. (x 2 - 2y)(x* + 2y)(x* + 4t/ 2 ) (Type II) 

/)](z< + 4s/ 2 ) = (x* - 4i/ 2 )(z< -f 4s/ 2 ) 
- (4i/ 2 ) 2 - x 8 - 



ILLUSTRATION 9. (- 3x - 4)(- 3s + 4) = - (4 + 3z)(4 - 3x) 

= - (16 - 9z 2 ) = - 16 + 9z 2 . 

EXERCISE 29 

Expand and collect terms, performing as miich of the work as possible mentally. 
1. 5(3a - 4t>). 2. 3c(2 - 6c). 3. ab(4x - ax). 

4. - 5x(2y - 3z). 6. (c - d)(c + d). . 6. (h - 2k)(h + 2k}. 

7. (a + y)\ 8. (c - 3x)(c + 3x). 9. (4 - y)(4 + y). 

10. (5 - 2y)(5 + 2y). 11. (3 + 2r)(3 - 2r). 

12. (3x - 4) (3 + 42). 13. (a 2 - 36) (a 2 -f 36). 

14. (a6 - 2) (06 + 2). 16. (a - 2) (a - 4). 

16. (c + 3). 17. (x + 5) 2 . 18. (y - 4) 2 . 

19. (2a - 5) 2 . 20. (3x - 2) 2 . 21. (2z - 

22. (3x - 4y) 2 . 23. (2a + 6) 2 . 24. (x - 

26. (3 -h *)(2 4- x). 26. (2x + 5y)(- 2x - 5y). 

27. (x - 5)(* -f 9). 28. (x + 13)(* - 4). 
29. (a + 26) (a + 36). 30. (w - 2z}(w -f &). 
31. (2x + 3)(3x + 4). 32. (3x - 5)(2x - 3). 
33. (4y - 3x)(2y - x). 34. (2^/ + w)(y + 5w). 

36. (2y - 3)(3y -h 5). 36. (y - 3)(2y + 7). 

37. (3u> -f 5)(7w ~ 2). 38. (3 - 4z)(2 + 5x). 
39. (4x - 3y)(2x + 3y). 40. (3u - 5w)(2w + 3io). 
41. (6 - 5x)(~ 2 -h x). 42. (- 3 - 2x)(2 - x). 
43. (3* - 4)(- x + 5). 44. (- y - 3)(fy + 4). 
45. (x 2 -h 2) 2 . 46. (4 + 36 2 ) 2 . 47. (2xy - 

48. (4a^ - 2/) 2 . 49. (3 + 46x) 2 . 60. (x - 2wxr 2 ) 2 . 



SPECIAL PRODUCTS AND FACTORING 87 

61. (x - J) 2 . 62. O/ + i) 2 . 63. (J - 2*)'. 
64. (we 2 - 2a)(wx* 4- 2o). 66. (cd - 3^)(cd + 3x). 
66. (x + .l)(z + .5). 67. (a; - .2)(x -f .5). 

68. (.3 + z)(.2 - *). 69. (3 - .2z)(2 -f .5x). 

60. (a - 6)(o + b). 61. (Jo - #>)(Ja + J6). 

62. (& + fe)(fz - /). 63. (.4* - .3)(.2x - .5). 
64. (x - y)(x 4- 2/X* 2 + ^ 2 ). 66. (2 - a; 2 )(2 + z')(4 



66. (w - 3)(w 4- 3)(w> 2 + 9). 67. - 7x(2ax - 3x 2 - 

68. - 3yz(2y* - 3yz + 2 2 ). 69. (- 3 - 4)(- 2 -f fa). 

70. (- x - t/) 2 . 71. (- 2s - 3y) 2 . 72. (- 3 - 

73. [2(* - ?/)] 2 . 74. [3(o + 6)] 2 . 76. [5(2c 

76. (3o: 2 - 8)(x* + 2). 77. (4z 2 - 3)(3x 2 + 2). 

78. (2x 2 - 3i/ 2 )(x 2 -f 4?/ 2 ). 79. (2a 2 + 56 2 )(a 2 - 36 s ). 

80. (z 3 + 3)(3x 3 - 4). 81. (3a - 26 s ) (7a 3 + 66 s ). 

82. (2u 4 - 3v 2 )(3w 4 + 2v 2 ). 83. (4x - 3^)(3x + 2y*). 

84. (2x 2 - 5i/ 2 )(2z 2 -f 5y 2 ). 86. (3w 2 - 7v 2 )(3w 2 -f 2V 2 ). 

86. (2a6c 2 - 



71. Grouping in multiplication 

The method of the following illustrations is particularly useful in 
applications of Types II, III, and IV of Section 70. 

ILLUSTRATION 1. (c -f 2d - lla)(c -f 2d -f Ho) 

= [(c + 2d) - llo][(c + 2d) -f- llo] (Type II) 

= (c + 2d) 2 - (Ho) 2 = c 2 + 4cd + 4cP - 121o 2 . 

ILLUSTRATION 2. (2x + y - 3z) 2 = [(2x + y) - 3] 2 (Type IV) 

= (2x + t/) 2 - 2(32) (2* -f y) + (3s) 2 
= 4s 2 . + 4x + t/ 2 - 12x2 - 6z + 9s 2 . 



ILLUSTRATION 3. (a + 6 + c + 2) = [(a + 6) + (c + 2)? (Type III) 
- (a + &) 2 + 2(o + 6)(c 4- 2) + (c + 2) 2 
= a 2 4- 2o6 4- 1 2 4- 2oc 4- 4o 4- 26c 4- 46 4- c 2 4- 4c 4- 4. 



88 SPECIAL PRODUCTS AND FACTORING 

ILLUSTRATION 4. (2x - 3 + 2y)(2x + 3 - 2y) 
= [2* - (3 - 23,)] O + (3 - 2y)] 

- (3 - 2y) 2 = 4x* - 9 4- I2y - 



EXERCISE 30 

Expand and collect terms by use of preliminary grouping. 
1. [(* 4- y) 4- 2] 2 . 2. [(o - 6) + 5J. 3. [3 - (2x - y)J. 

4. (2 + a 4- w)\ 6. (3s + y + 5) 2 . 6. (* - 2y - 3) 2 . 

7. (4a - b - c) 2 . 8. (- 2 + a + 6) 2 . 9. [2x - 3 (a - fc 2 )] 2 . 

10. (2x - 3* 2 + 3?/) 2 . 11. [(ar + y) - 3][> + y) + 3J 

12. [(c + 2x) - 2][(c + 2x) + 2], 13. [4 - (2o + 6)] [4 + (2o + 6)1 
14. (a + w -f 4) (a + w - 4). 15. (a + 6 - x)(a + b + x). 

16. (3* + y - 2) (3* + y + 2). 17. (3a - y + 4)(3o - y - 4). 

18. (a - fc 2 + )(a -f t 2 4- 2). 19. (z 2 - y + 2)(a; 2 + y - ). 

20. [(a 4. 6) + (c - 3)] 2 . ' 21. [2s + y + o - 3] 2 . 

22. (a 4- c 4- 6 - 5) 2 . 23. (2x - z + y - 2) 2 . 

24. (o 4- b 4- c 4- d)(a + 6 - c - d). 
26. (2x + y-z + 3)(2x + 2 

26. (a + 3y 5)(a 4- 3y - 

27. (c - 2d - a - x)(c 2d + a + x). 

28. Expand (x -\- y + z) 2 and state the result in words. 

Use the formula of Problem 28 to expand each square. 

29. (2a 4- 36 H- 4c) 2 . 30. (3a - 26 4- 3c) 2 . 

31. (w - to 4- 3a) 2 . 32. (42 s - 3*i/ 2 - 5X 8 ) 2 . 

72. Terminology about factoring 

In our discussion of factoring, unless otherwise stated, the coeffi- 
cients will be integers in any polynomial referred to. Such an expres- 
sion will be called prime if it has no integral rational factors except 
itself, or its negative, or 1. No simple rule can be stated for determin- 
ing whether or not an expression is prime. 



SPECIAL PRODUCTS AND FACTORING 89 

ILLUSTRATION (. We shall say that (x y) is prime although 

x - y = (Vx -f Vy)(Vx - Vy), 



because these facltors are not integral and rational. Other prime expressions 

are (x + y), (x* --f- y 2 ), (z 2 + xy + ?/ 2 ), and (x* xy + 



To factor a polynomial will mean to express it as a product of 
positive integral powers of distinct prime factors. 

ILLUSTRATION 2. To factor 4c 4 46 2 z 2 , we write 



After an expression has been factored, the factors should always 
be verified toy multiplying them to obtain the given expression. 

73. Factoring by inspection 

Each ty/pe formula of Section 70 becomes a formula for factoring 
when read! from right to left. 

I. ax + ay + az = a(x -f y -f z). 

iLLUS-niiATiON 1. by 2 + 3y + % 2 = t/(6?/ + 3 + Sy). 

ILLUSTRATION 2. If a factor 2x z y 3 is removed from the term 
the remaining factor can be verified by division: 



Hence, 14xV = 2xV(7x?/ 2 ). 

ILLUSTRATION 3. In the following factoring, we remove the common 
factor hxy 2 from each term : 

3 - 



II. T/ie difference of two squares equals the product of the sum and 
the difference of their square roots: 



ILLUSTRATION 4. # 2 9 = (x 3)(z -f 3). 

ILLUSTRATION 5. To factor 25s 2 Oy 4 , we observe that 25a? = (6x) 2 
and Oy 4 = (32/ 2 ) 2 . Hence, 

H- 



ILLUSTRATION 6. a 4 IGt/ 4 = (a 2 - 4y 2 )(a 2 4- 

- (a - 



90 SPECIAL PRODUCTS AND FACTORING 

74. Perfect square trinomials * j 

An integral rational polynomial with three tei;ms is called a 
trinomial. The square of any binomial is a perfect s}quare trinomial. 
A trinomial of this type can be recognized and factored by the 
formulas of Types III and IV of Section 70. 

Perfect square trinomials: 

III. a 1 + 2ab + fc* = (a + &); 

IV. a 1 - 2a& + b* = (a - &). 

In a perfect square trinomial, we notice that 

1. two terms are perfect squares, and 

2. the third term is plus (or minus) twice the product of the square 
roots of the other terms. 

To verify that a trinomial is a perfect square, take the square roots 
of the terms which are perfect squares, compute the third tlerm which 
should be present, and check by inspection. \ 

ILLUSTRATION 1. To factor 4z 2 2Qxy -f 25y 2 , we observe perfect squares 
4z 2 and 25y 2 , whose square roots are 2x and 5y. Hence the third term should 
be 2(2x)(5y) = 2Qxy, which checks, and gives 

- 2Qxy + 25y* = (2x - 



ILLUSTRATION 2. 162 4 + 242^ + 9w 2 = (4Z 2 -f 3w)*. ' 

EXERCISE 31 

Factor by use of Types I and II. // fractions occur, leave the factors in 
the form which arises most naturally by standard methods. Check by multiplying 
the factors. \ 

bx. 2. 2cx + 4<fc. 3. 6zy 2 -f 2ax. 



4. bx -4- x -f- c 2 ^. 5. 2cy 

6. 3o& + 2a - 5a 2 . 7. - ac + 3bx + ex 

8. - 5y* - 3y -f ay 2 . 9. - 4o< -f < 2 - cf 3 . 

10. 46V -f 6te 2 + 86cx 2 . 11. 3aV - 2o?/ 2 + ay 

* See Note 5 in the Appendix for an explanation of the process for finding the 
square root of a number expressed in decimal notation, by means of pure arith- 
metic. This process is intimately related to the formula of Type III. 



SPECIAL PRODUCTS AND FACTORING 91 

12. 6aV - 3<w 2 + 40^. 13. 2t0 4 z - 6w*c 2 + 5w*r*. 

14. x 2 - a 2 . 15. w> 2 - s 2 . 16. y* - 25. 

17. 64 - x*y\ 18. 36 - 2 s . 19. 4x* - y\ 

20. 9x 2 - 25s 2 . 21. 36# - 121. 22. 9s* - 1. 

23. 4a 2 - 9ft 2 . 24. 1 - 25x 2 . 25. 256a* - 1. 

26. 9s 2 - J. 27. J - w*. 28. 9a 2 6*u> - 16s*u>. 

29. 25W? 2 - c 2 ^. 30. 49u 2 - 16v^. 31. 36a 2 6 s - 64x 8 . 

32. 9 2 - 144a 2 6 2 . 33. ax* - ay*. 34. 



Which trinomials are not perfect squares? 
36. x 2 + 3x + 4. 36. a 2 + a + 1. 



37. 4z 2 + 6z + 9. 38. 9x 4 - 6x 2 

39. 3z 2 + fay + 4y*. 40. 4z 2 - 



Insert any missing term to complete a perfect square. Then factor by use 
of Types III and IV. 

41. x 2 -f 2bx + b*. 42. rf 2 + 2<fy + y 2 . 



43. a 2 - 2a + 1. 44. x 2 -f ( ) + 16. 

46. u? - ( ) -h 36. 46. 4z 2 - 20xz + 25 8 . 

47. x 2 -h 81 - l&c. 48. x 2 + x + J. 
49. 49x 2 + 14ox + a 2 . 60. 1 + z 2 z 2 - 2x2. 

51. 64 - 16a6 + a 2 6 2 . 62. 9a 2 + ( ) + 256. 

53. 4x 2 + ( ) + 9 2 . 64. 166 2 - ( ) + 49x 2 . 

66. 4c 2 (P - ( ) + 25a 2 . 56. 9x 2 - ( ) + hW. 

67. - 30xy + 9x 2 + 25y 2 . 68. 24ax + 9x 2 + 16a 2 . 
69. 4x 4 - 28x 2 + 49. 60. 25 - 30x 2 + 9x 4 . 
61. 4a 4 - 12a*b* + 96 4 . 62. 



Factor. 
63. 49 2 - 4ft 2 . 64. 75o 2 - 3a 2 6 2 . 65. 



. 25s 2 - 30^2 + 9u 2 . 67. x 2 -f 25^ - 

68. 9X 4 + 49s/ 4 - 42xV- 60. 4a 2 x - 4ax -f x. 



92 SPECIAL PRODUCTS AND FACTORING 

70. x* - 9y*> 71. 98u 4 - 50. Y2. 9oz 2 - lay*. 

73. 25x 2 - 1006 4 . 74. 3a 2 x 2 - 5ay. 76. 16x 4 - 

76. 25wV - TOu 2 ^ 2 + 49t^. 77. 18u 2 - 60wt> 

78. 2w 6 - 12t*V + 18y 8 . 79. 147x 2 - 



First factor and then compute. Check by expanding the original expression. 
80. 23 2 - 17 2 . 81. 52 2 - 48 2 . 82. 27 2 - 23 2 . 

83. 104 2 - 96 2 . 84. 45 2 - 55 2 . 85. 37 2 - 33 2 . 

75. Factoring trinomials by a trial and error method 

We recall the formulas of Types V and VI of Section 70. 

V. x* + (a + b)x + ab = (x + a)(x + b). 

VI. flex* -f (ad + bc)x + bd = (ax + b)(cx + d). 

Certain trinomials of the form * gx* -f hx + k can be factored by a 
trial and error method suggested by the preceding formulas. 

\ 

EXAMPLE 1. Factor: x 2 2x 8. 

SOLUTION. 1. We wish to find a and 6 so that 

(x + a)(x + 6) = x* + (a + 6)a? + ab = x 2 - 2x - 8. 

2. Hence, ab = 8; thus a and 6 have opposite signs and are factors 
of 8. Since the sum of the cross products is 2x, we guess that a = 4 
and 6 = 2. This is correct because 

(x - 4)(* + 2) = x* - 2x - 8. 

EXAMPLE 2. Factor: 15x 2 + 2z 8. 

SOLUTION. 1. We wish to find a, 6, c, and d so that 

(ax + 6) (ex -f d) = oca; 2 + (ad + bc)x + bd = 15z 2 + 2z - 8. 

Hence, oc = 15, 6d = 8, and the sum of the cross products is 2x. 

2. First trial. Since oc = 15, choose a = 15 and c = 1 ; since bd = 8, 
choose 6 = 2 and d = 4. This selection is wrong because 

(15z + 2)(z - 4) = 15s 2 - 5&c - 8. 

3. Second trial. Choose a 3, c 5, 6 = 2, and d = 4. This selection 
is correct because 

(3x - 2) (fir + 4) = 15x 2 -f- 2z - 8. 



* If g, h, and k were chosen at random, without a common factor, the trinomial 
would probably be prime. Later, we shall discuss a condition which g, h, and A; 
satisfy when and only when the trinomial is not prime. 



SPECIAL PRODUCTS AND FACTORING 93 

If one prime factor is merely the negative of another, we do not 
consider them as distinct prime factors; we combine their powers 
into a single power of one of them. 

ILLUSTRATION 1. In ( x 2}(x -f- 2) = x* 4x 4, we notice that 
( x 2) = (x + 2). Hence, we write 

- x z - 4x - 4 = - (x + 2)(z + 2) = - (x + 2) 2 . 



Note 1. The preceding factoring methods apply to polynomials in which 
the coefficients are any real numbers, not merely integers as in the illus- 
trations. The nature of the coefficients which we agree to allow in a poly- 
nomial and its factors affects our definition of a prime expression but not our 
general factoring procedure. 

EXAMPLE 3. Factor: 6z 4 x* 15. 

SOLUTION. By trial and error, 6s 4 - z 2 - 15 = (3z 2 - 5)(2z 2 + 3). 

EXERCISE 32 

Factor by trial and error methods. 

1. x* + Sx + 15. 2. z 2 -f lOz + 21. 

3. a 2 - 8a + 12. 4. y* - 1y + 12. 

6. x 2 - Sx + 15. 6. 2 2 - 52 - 6. 

7. < 2 + 4* - 21. 8. w* - 5w - 24. 
9. z 2 - 3x - 18. 10. a 2 + 6a - 16. 

11. w* + 2w - 48. 12. 4 - 3y - ^ 2 . 

13. 15 - 2w - w> 2 . 14. 8 - 7a - a 2 . 

16. 24 + 2w - w 2 . 16. 6 2 + 36 - 28. 

17. 32 - 4y - y\ 18. 27 + Qw - w? 2 . 
19. 54 - 3& - fc 2 . 20. 36 + 5h - K. 
21. x 2 - 6z - 72. 22. '2o: 2 + 7z + 3. 
23. 5a 2 + 12a + 7. 24. 3o 2 + 8a 4- 5. 
25. 10z 2 - llz + 3. 26. 3a 2 - lOa + 7. 
27. 8x 4 - 10z 3 + 3z 2 . 28. 2x* - x* - 

29. 3 + 2 2 - 5. 30. 



94 SPECIAL PRODUCTS AND FACTORING 

31. 3x* + x 9 - 10. 32. 15j/ 2 4- 4y - 4. 

33. 8u* 6I0 8 9. 34. 5 4- 3x 4- 2z 2 . 

35. 150 4 - a 2 - 28. 36. 8 4- 2y 2 - 15^. 

37. 7 - 19z - 6z 2 . 38. - 12fc 2 - 8fc 4- 15. 

39. 27z 2 4- 3x + 2. 40. 5a 2 4- J2a& 4- 7b*. 

41. 3s 2 4- 5xy 4- 2y 2 . 42. 3x 2 4- 7ax 6a 2 . 

43. Sw 2 4- 14u 15z 2 . 44. ISwr 4 4- 9t^ 20. 

45. 5w 2 28wu> 4- 12^. 46. 45fc 2 Sxy 4t/ 2 . 
Factor by the appropriate method. 



47. 6a 2 - 13o6 4- S6 2 . 48. 4a; 2 - 7xy + 3t/ 2 . 

49. lOOa 2 - x*. 60. 49 2 - 46 2 . 61. 

52. Tc 2 + 19cd - 6tf. 63. 64a 2 - 48ac + 9c 2 . 

64. - 2z 2 + 15 4- x. 65. - 6z 2 + 20 - 7*. 

66. 9 4- 250 2 io 2 30tw. 57. 2x? 

58. 8a 2 c - 18c. 59. W + 

60. 3a 4- 13a6 4- 10a6 2 . 61. 25x 2 - 1006 4 . 

62. 75ccP + 30c 2 rf + 3cX 63. 2r - llhr 4- ISAV. 

64. .Sic 4 - .16d 4 . 65. ^ - 16y*. 

66. Sx 4 - 16x 2 H- 3. 67. 31x - 5x 2 - 3. 

68. xV 4- 9xy 52. 69. 3s 4 7z 2 20. 

70. 6s - 1 9s*. 71. 

72. 9s 2 * - 4. 73. 

74. - 4x 2 4- 12s - 9. 76. - 9a 2 4- 30a6 - 2S6 2 . 

76. 3s 4 - 17x 2 -f 10. 77. 2X 4 + x 2 - 15. 

78. 3s 4 - 5xV - 2y*. 79. 3O 4 



76. Factoring by use of grouping 

The following methods make frequent use of the fact that an ex- 
pression enclosed within parentheses should be treated as a single 
number expression. 



- SPECIAL PRODUCTS AND FACTORING . 95 

ILLUSTRATION 1. 5(x a) 3(x a) *= 2(x a). 

ILLUSTRATION 2. To factor the following expression, we observe the 
common factor (a 6), and remove it from each term: 

2c(o - b) 4- d(a - b) - (a - 6)(2c + d). 

ILLUSTRATION 3. After grouping, we observe a common binomial factor, 
and then complete the factoring: 

bx + by 4- 2&c 4- 2hy = (bx + by) + (2kc + 2%) 

(* 4- y) = (6 + 2A)(* + y). 



ILLUSTRATION 4. The second term below was altered by changing signs 
(or, multiplying by 1) both within and without the parentheses in order 
to exhibit the same binomial factor as the first term: 

a - 6) + 40(6 - 2a) - 3x(2a - b) - 4y(2a - b) 

4t/)(2a - 6). 



ILLUSTRATION 5. In order to factor below, we group two terms within 
parentheses preceded by a minus sign, and hence change the signs of the 
terms, in order to exhibit the same factor as observed in the other terms: 

xz kx 4- kw wz (xz wz) (kx kw) 
z(x w) k(x w) = (z k)(x w). 

ILLUSTRATION 6. 6 - 3s 2 - 8x + 4z 3 = (6 - Sx) - (3z 2 - 
= 2(3 - 40) - x*(3 - 4c) = (3 - 40) (2 - &) 



ILLUSTRATION 7. We factor below as the difference of two squares: 
(c - 2z) 2 - (6 - a) 2 = [(c - 2x) - (b - a)][(c - 2) + (6 - a)] 

= (c - 2x - 6 + a)(c - 2x + 6 - a). 

ILLUSTRATION 8. a 2 - c 2 + 6 2 d 2 - 2db - 2cd 

= (a 2 - 2a6 -|- 6 s ) - (c 2 + 2cd + <P) = (a - 6) 2 - (c -f d) 2 
= [(a - 6) - (c + d)][(a - 6) + (c + 
= (a 6 c d)(o 6 + c H 



EXERCISE 33 

Factor. 



- 5(* 4- 2|/). 2. 4(3A + *) - 9(3A 4- 

3. c(x 4- y) 4- <*(* 4- J/). 4- 5a(c - 3d) ~ 26(c - 3d). 

6. 2A(m - 2) - 3fc(m - 2). 6. 2c(x 4- 40) 



96 SPECIAL PRODUCTS AND FACTORING 

7. - 5c(r + ) + 2d(r + s). 8. - 2x(a + h) - 3y(a + h), 

9. 3A(w ) (> *) 10. 2x(h - 2fc) + 3% - 6ky. 

11. 3a(w - 2k) + 26u> - 46fc. 12. fee + fa/ -h 2hx + 2%. 

13. 3ac + 3bc + ad + bd. 14. 2a + 2ay + bx + by. 

15. cr cs + 3dr 3ds. 16. 4&c 46A + bcx 56c. 

17. 2cx + cy 2dx %. 18. box + 26z lOod 4bd. 

19. 4Ax 4bh 8cx + 8bc. 20. 36tt> 360 4aw -f 4az. 

21. (x 3 - 2a; 2 ) - (x - 2). 22. (ax 3 + fee 2 ) - 4(ax + b). 

23. x* + 2x* + x + 2. 24. ax 2 + 6z 2 -h ad 2 + b&. 

26. x 8 - 3x 2 + x - 3. 26. 2x 2 - 4x + 1 - 8z 3 . 

27. a 3 - 3a 2 - 3 H- a. 28. 2 + 4x - lOx 4 
29. 3X 3 - 2x 2 + 6x - 4. 30. 4 - &c 2 - 5x + 
31. 2(r s) x(s r). 32. a(x y) + 6(?/ 
33. x 2 - (s H- 3) 2 . 34. (w - I) 2 - 16& 2 . 

35. (2z + wY - y*. 36. (4a - fe) 2 - (2x - ?/) 2 . 

37. (c - 3d) 2 - (2x + y) 2 . 38. (4x - 3y) 2 - 25. 

39. z 2 + 20 H- 1 - 9* 2 . 40. 4i^ + 20w + 25 - 81 2 . 

41. 2/ 2 + 2y + 2 2 - 4x 2 . 42. 9w> 2 - 4a 2 - 4ab 6 2 . 

43. 4a 2 - 92 2 - 62 - 1. 44. 16i/ 2 - a 2 -f 2ab - fe 2 . 

46. 9x 2 2/ 2 + 2y z 2 . 46. t^ 2 4x 2 y* 4xy. 

47. 16a 2 - 1 - 9x 2 H- 6x. 48. a 2 c - a 2 d 

49. bx 4 -by 4 + ex* - cy 4 . 60. a 2 - ft 2 - a + 

61. 2* - w 2 + t*> - 2 2 . 62. ch -f 6dfc 

- 53. r 2 + 6r< + 9/ 2 - a 2 - 2a6 - fe 2 . 

64. 4z 2 + 4x^ -f 2/ 2 - 9a 2 - I2at - 4< 2 . 

66. c 2 + 4c 4- 4 - 9eP - 6dh - ft 2 . 

66. 16x 2 - 24xj/ H- 9j/ 2 - 9a 2 - 12a - 4. 

67. 9x 2 - Qxy + 2/ 2 - 25a 2 + 10a6 - 6. 



SPECIAL PRODUCTS AND FACTORING 97 

68. 4z 2 - 4xy -f y* - 9s 2 + 6w? - w 2 . 

59. 6 2 - 9z 2 + 2ab + a*. 

60. 4# - 1610 2 - 4cd + c 2 . 

61. 4o 2 + 96 2 - 4z 2 - y 2 - 4zy - 12o6. 

62. a 2 - 96 s - d 2 - 2o - 1 - 66d 

63. 16s 4 - Sly 4 + 4z 2 - 9y\ 
'64 cV - 81c 2 + 324 - 4s 4 . 

77. Cube of a binomial 

We verify that 

(x + y)* =(x + y) 2 (x + y) - (z 2 + 2^ 4- 2/ 2 )(z 4- y) 
= x 3 -t- 2x 2 !/ + xt/ 2 + x?y + 2^!/ 2 + y 3 . 

On collecting terms we obtain (1) and, similarly, we could verify (2) : 

(x + y) 8 = y* + 3x*y + 3xy* + y 8 ; (1) 

(x - y) 8 = x 8 - 3x a # + 3xy a - y 8 . (2) 

The student should memorize these formulas. 

ILLUSTRATION 1. From formula 1, with x = 2a and y = 6, 
(2a + 6) 3 = (2a) 3 + 3(2a) 2 (6) + 3(2a)(6 2 ) + 6 s 

12a 2 6 



ILLUSTRATION 2. From formula 2, 

(4 - xY = 64 - 



78. Sum and difference of two cubes 

By long division we could verify that 



a 



- a 2 



Hence, we have the following formulas, useful for factoring when 
read from left to right, and useful in multiplication when read from 
right to left. 



98 SPECIAL PRODUCTS AND FACTORING 

ILLUSTRATION 1. By use of (1), read from right to left, with 6 3, 
(a - 3)(a 2 4- 3a + 9) - a 3 - 3* - a 8 - 27. 

ILLUSTRATION 2. From formula 2 with a = 3s and b 
27s 8 + 8s/ 3 = (3*)' 
= (3* 4- 
4- 



ILLUSTRATION 3. 1 - 64s 8 - I 8 - (4z) 8 = (1 - 4z)(l + 4z 4- 16z 2 ). 
EXAMPLE 1. Factor: j/ 6 - 19?/ 8 - 216. 

SOLUTION. tf - lty - 216 = (y 3 - 27) (^ + 8) 

4). 



EXERCISE 34 

Divide by long division and check by use of Section 78. 



a 3 -A 8 o + 276 3 . 8s 3 

O. 



*' x + y ~ a-h ~ a 4- 36 * 2x - 3y 

Multiply by inspection. 
5. (c 4- w?)(c 2 - cw 4- w*). 6. (u v)(w 2 4- uv 4- t> 2 ). 

7. (3a - c)(9o 2 4- 3ac 4- c 2 ). 8. (1 - w)(l 4- w 

9. (1 3x)(l 4- 3x 4- 9x 2 ). 10. (2 3u)(4 4- 

11. (6 2x)(6 2 4- 2bx 4- 4x 2 ). 12. (4# 4- l)(16y 2 4y 4- 1). 

Factor. 
13. d 3 - ^. 14. A 3 4- 2*. 15. y 3 - 27. 16. u 3 4- 1. 

17. 1 - v 9 . 18. 8 - x 8 . 19. s 3 4- 1000. 20. 64 - t^. 

21. 1 - 27s 3 . 22. 125 4- 8^. 23. s 3 - SwA 24. 8 - 27x 3 . 

25. 216s 3 - yV. 26. x 8 - 64^. 27. 343a 3 - Sz 3 * 3 . 

Expand each cube by use of the formulas of Section 77. 
28. (c 4- d)*. 29. (h - A;) 3 . 30. (2 4- y) 3 . 31. (u 4- 3) 3 . 

32. (5 - y) 8 . . 33. (2x 4- w) 8 . 34. (y - 3x) 8 . 35. (4x 4- y) s . 

36. (a - ft 2 ) 8 . 37. (a 2 - 2x) 8 . 38. (x 2 - i/ 2 ) 8 . 39. (c - 2ft 2 ) 8 . 

40. (o - 2s 8 ) 8 , 41. (2c* - 3s) 8 . 42. (.1 - 2x) 8 . 



SPECIAL PRODUCTS AND FACTORING 99 

Factor. 

43. y* + 7z* - 8. 44. 276 + 26fc - 1. 

46. 8z - 19xV - 27y. 46. 64a - 16a*6 + 6 6 . 

47. a - 3o 2 + 3a - 1. 48. s 3 + 62? + 12s + 8. 
49. w - 9w*x + 27w*c 2 - 27s 8 . 60. 125u s - 75u 2 -f 
51. (c - d) 3 - a s . - 62. (h - x)* - (y - 



*79. Trinomials equal to differences of squares 

An expression of the form z 4 -f kx 2 y 2 + y 4 can be written as the dif- 
ference of two squares if the expression becomes a perfect square 
after the addition of a perfect square multiple of x 2 y 2 . 

EXAMPLE 1. Factor: 64o* - 64a 2 6 2 4- 256*. 

SOLUTION. 1. A perfect square involving 640 4 aqd 25ft 4 is 

(8a 2 -f 56 2 ) 2 = 640* + 80a 2 6 2 -f 256 4 . 

2. Hence, 64a 4 64a 2 6 2 + 256 4 becomes a perfect square if we add 144a 2 6 2 . 
Therefore, we add 144o 2 6 2 and, to compensate for this, also subtract 144a 2 6 2 : 

64a* - 64o 2 fe 2 + 256 4 = (640 4 - 64O 2 ?) 2 -f- 256 4 -h 144a 2 6 2 ) - 144a 2 6 2 
= (64a 4 + 80a 2 6 2 + 2S6 4 ) - 144a 2 6 2 = (8a 2 + S6 2 ) 2 - 144a 2 6 2 
= (8a 2 + 5ft 2 - 12a6)(8a 2 H- S6 2 + 12a&). 

EXAMPLE 2. Factor: . Qx 4 - I6x*y* + 4y*. (1) 

SOLUTION. 1. The perfect squares involving 9s 4 and 4y* are 

= 9s 4 db 



In order to obtain + 12a:V from (1), we would have to add 2&rV> but 
this is not a perfect square. To obtain 12#V we must add 4afy 2 , which 
is a perfect square. 



2. Add, and also subtract, 4zy in (1) : 

Ox 4 - 16zy + 4^ = (9s 4 - 16*y + 40* 

2xy). 



*EXERCISE 35 

Factor by reducing to a difference of two squares. 
1. a* + 2 + L 2. ^ - 3 2 + 1. 3. Qo 4 + 2o l -f 1. 



4. ftc 4 + lla; 2 + 4. 5. s 4 + A 2 ^ + h*. 6. 9s 4 - 10*' -h 1. 



700 SPECIAL PRODUCTS AND FACTORING 

7. 4t0* + Sa 2 ^ -h 9a 4 . 8. a< - 9oV + Ify/ 4 . 9. 25a 4 - 5aW -f 46*. 

10. 4# + 4d?h* + 25A 4 . 11. s* + 4. 12. w^ + 4s 4 . 

13. s 4 -h 64A 4 . 14. 625Z 4 + 4w*. 16. 81* 4 + 64s 4 . 

16. x* - 12aW + 160 4 . 17. Oa 4 - 16a 2 c 2 + 4e*. 

18. 19aV + 4x* -h 490 4 . 19. 25O 4 + 9^ - 34ay. 

20. 4a^ - 24 + 25. 21. 



*80. Perfect powers 

An integral rational term is said to be a perfect nth power if it is 
the nth power of an integral rational term. 



ILLUSTRATION 1. IGa 4 ^ 8 is a perfect 4th power because 16o 4 6 8 = (2O6 2 ) 4 . 

ILLUSTRATION 2. 8o 6 6 6 is a perfect cube because 8o 6 6 6 = (2a 2 6 2 ) 3 . The 
original exponents have 3 as a factor. 

In a perfect nth power, each exponent has n as a factor because in 
raising a term to the nth power we multiply each of the original ex- 
ponents by n. 

ILLUSTRATION 3. (2 3 a 2 6 4 ) n = 2 3n a 2n 6 4n . 

*81 . Special cases of sum or difference of perfect powers 

I. // n is even, commence factoring (a n b n ) by recognizing it as the 
difference of two squares. 



ILLUSTRATION 1. x 6 - t/ 6 = (x 3 ) 2 - (y 3 ) 2 = (a: 3 - ^)(x 3 + y*) 

= (x - t/)(x 2 + xy + y*)(x + y)(x* - xy + y*). 



We could have commenced by factoring (x 8 t/ 6 ) as the difference of two 
cubes, but this would have been an inefficient method. 

ILLUSTRATION 2. To factor 16o 4 6 4 81, we observe that each term is a 
perfect square. Hence, 

16a 4 & 4 - 81 = (4a 2 6 2 - 9)(4a 2 6 2 + 9) 
= (2afe - 3)(2afe + 3)(4a 2 6 2 + 9),' 

where the final factor is a prime sum of perfect squares. 

II. // n is odd and has 3 as a factor, we can commence factoring 
(a n 6 n ) by recognizing it as the sum or difference of two cubes. 



SPECIAL PRODUCTS AND FACTORING 707 

ILLUSTRATION 3. x 9 + y 9 = (z 3 ) 3 + (y 3 ) 3 

= (x 3 + 2/*)(z 6 x 3 !/ 3 + |/*) (Using Section 78) 



*EXERCISE 36 

Express the perfect power as the 3d or 4th power of some other term, which- 
ever is the case. 

1. 8a 3 6 3 . 2. 27ay. 3. 16a 4 6 4 . 4. 81xy. 

5. 125xy. 6. 64xy. 7. 256W 8 !; 12 . 8. 625ay. 

Factor each expression which is not prime. 

9. a 4 - x 4 . 10. y 4 - 81. 11. 16 - w 4 . 12. Six 4 - y 4 . 

13. x 8 - y 8 . 14. x 4 -f y 4 . 15. 81 - 16x 4 . 16. y - x*. 

17. w - 1. 18. a 6 - 64. 19. z - 64/. 20. a 6 + 64. 

21. x 6 + 1. 22. 729 - a 6 . 23. 729 + x 6 . 24. 125 - <z fl . 

25. 256 - a 8 . 26. h 9 - k*. 27. a 9 + 6 9 . 28. a 8 + 6 s . 

29. Six 8 - y 4 . 30. 16x 4 - Sly 8 . 31. 625 - 16X 8 . 32. x 8 - w 8 * 8 . 

33. a 6 - 646. 34. 64 + xy . 35. 8a 3 - 27x 6 . 36. x u - y u . 



*82. Properties of factors of a" db b n 

We have verified special cases of the following results, where n 
represents a positive integer. Any special case of the results can be 
checked * by long division. 

I. For every value of n, (a n b n ) has (a b) as a factor; in other 
words, (a n b n ) is exactly divisible by (a b). 



ILLUSTRATION 1. a 3 b 3 (a 6) (a 2 + ah -f 

a* - b 4 = (a - 6)(a 3 + o 2 6 + 
a 4 - 16 = a 4 - 2 4 = (a - 2)(a 3 + 2a 2 + 4a + 8). 

II. // n is even, (a n b n ) has (a -{-b) as a factor. 

ILLUSTRATION 2. a 2 6 2 = (o 6) (a + b). 

4 _ 54 = ( a + 5)( s _ 2& 4. a &2 _ 



* A convenient method for giving a general proof of the results is met in a 
more advanced section of algebra. 



J02 SPECIAL PRODUCTS AND FACTORING 

III. Ifnis odd, (a n + b n ) has (a -f 6) as a factor. 



ILLUSTRATION 3. a 8 -f 6 s = (a + 6) (a 2 ab -f 

a 7 + & 7 - (a + 6)(a - a 6 6 + o 4 ^ - a 8 6 + a'6 4 - 



IV. If n is even, (a n 4- 6 n ) does not have either (0 6) or (a + 6) 
as a factor. 

ILLUSTRATION 4. (a 2 + fe 2 ) and (a* + o 4 ) are prime, (a 6 H- 6*) is not 
prime but it does not have either (o + 6) or (a 6) as a factor: 

6 + 6 ( a 2 + &)(a4 _ 2&2 _|_ J|) f 

where each factor is prime. 

Special cases of the following general properties were exhibited 
by the second factors in Illustrations 1, 2, and 3. 

A. When (a* b n ) is divided by (a 6), all coefficients in the quo- 
tient are -f 1. 

B. When (a n + b n ) or (a n b n ) is divided by (a + 6), the coefficients 
in the quotient are alternately -f 1 and 1. 

Factors obtained by reference to (I), (II), and (III) are not always 
prime. Also, as seen in Illustration 4 and Section 81, an expression 
of the type a n 4- & n , with n even, may be factorable although (IV) 
is true. In finding the prime factors of a n b n , first use the methods 
of Section 81 if possible, before employing (I), (II), and (III). 



ILLUSTRATION 5. z 6 + 64 = (x 2 ) 8 + 4 3 = (z 2 + 4)(x* - 4s 2 + 16). 



ILLUSTRATION 6. x 9 - y 9 = (z 8 ) 8 - (y 8 ) 8 = (z 8 - ^(x 6 + xV -f 

= (x - 



ILLUSTRATION 7. x 4 y 4 = (x 2 ) 2 (y 2 ) 2 = (x 2 # 2 )(x 2 + y 2 ) 

- (* - y)(* + y)(* 2 + 2/ 2 ). (1) 

By use of (I), 

x 4 - ^ - (x - 2/)(x 8 + x*y + xy 2 + y 8 ). (2) 

Equation 1 shows that the second factor in (2) is not prime; this factor 
could be factored J>y grouping: 

= &(x -f y) + j/ 2 (x + y) 



Thus, we finally arrive at the factors obtained in (1) but by a much less 
desirable process. 



SPECIAL PRODUCTS AND FACTORING f03 

*EXERCISE 37 

Find the quotient by long division, and the remainder if the division is inexact. 

x 4 + 16 



. . - 

s + y s + 2y z + 2 



eocft reswft without using long division t by use of properties A and B 
o/ Section 82, and cAeefc 6y multiplication. 



~ 1 u * - 16 



13. ( - y>) + (a + y ). 14. (i6x* - a 4 ) -*- (2a? + a). 

15. (a 8 - 8) * (a - 2). 16. (243* 6 - 1) * (3x - 1). 

8 J?. 19 8 " 

y 19. - sr - 

^ oo s" 

" 



26 ' 



a;* - 2a 
166* 



Factor eacA expression which is not prime. 

28. a 5 - c 6 . 29. a 4 - w 4 . 30. u 7 - v 7 . 31. u* + t^ 6 

32. 32 + x 6 . 33. 1 - ^. 34. a* - 256s/ 8 . 35. u - . 

36. t> 6 - 32u 8 . 37. 32a 6 - 1. 38. x n + y l \ 39. 128 + 

40. x 6 - 243j/*. 41. a 8 - 27z. 42. 16x 4 + 

43. 4x* + 1. 44. lutx 4 + 810 4 . 45. w 3 * + y 

46. 32x w + y*. 47. x 16 + y". 48. u ffi - 

49. 512 - x 9 . 50. z + 512a'. 51. u 



CHAPTER 



6 



ADVANCED TOPICS IN FRACTIONS 



83. Reduction of fractions to lowest terms 

Whenever we make a reference to factoring in a fraction, it will be 
assumed that the numerator and denominator are integral rational 
polynomials with integral coefficients. In the final result of any 
operation on fractions, we agree to leave any expression in a factored 
form if it arises naturally. 

SUMMARY. To reduce a fraction to lowest terms: 

1. Factor the numerator and denominator. 

2. Divide both numerator and denominator by all their common 
factors. 

ILLUSTRATION 1. In the following fraction, we divide both numerator 
and denominator by 3z 4y and indicate this by cancellation. 



3x 2 + 2xy - Sy z &xr=^Sy)(x + 2y) x + 2y 

ILLUSTRATION 2. In reducing the following fraction to lowest terms, we 
first notice that one factor in the numerator is merely the negative of a fac- 
tor of the denominator. 

x 2 -9 (x-3)(x 



12 + 2x - 2x 2 2(3 - z)(2 + x) 

_. 
x - 3) 



In the preceding line, we obtained (x 3) in the denominator by multiplying 
(3 x) by 1, and hence it was necessary to change the sign before the 
fraction to keep its value unaltered. 



ADVANCED TOPICS IN FRACTIONS 105 

EXERCISE 38 



Reduce to lowest terms. 





1 , wy a*b(x - 2y) 6c -f 6d 

irv n * r ni A/ r\ \ * 3 



3c + 3d 
c 2 d(o 4- 36) c 4at/ 26y 



cz(z + y) cd*(a + 36) * 2ac 6s 

z 2 - y 2 ax - ex A 



~ a 2 -c 2 
1ft ^ ^. t1 4o 2 - 96 2 

10. : 11. ^ ^* 12. 



cx + cy ' 2ax 36x * 4ax 26x 

m 2 -m-42 a 2 + 2a - 15 

13 ' m 2 - 3m - 28 14 - a 2 + a - 



_ w f 13a; - 10 -A 3x 2 - 7ax 4- 4a 2 

lo. 



17. 



4 3a: 2 4" 2ac 8a 2 

- 12a 2 -A a 2 - 4aa; 

' 2a 2 - 9ax - 



i i t a IW ; ; 

4#2/ 4" ty x* 4" bxy -\- 

ax + ay 



. U rt , 

2a 26 



22. 



Reduce to lowest terms with as few minus signs as possible remaining in 
the numerator and denominator. 

-3 OK -2z-2y Ofl 2a + 2d 



27. 



a o 4- 6 

ax 5a 



y) 2 5 - z 

a; 2 - Ox 4- 9 



* 3v - 3w 18 - 2z 2 (y - 2z) 2 

h 3x - 9 . 9 - 15z 4- 4a; 2 

94* * 

- 15 2a: 2 15ca; + 20dx 9c 



- 27 10x 2 + 29x - 21 

~ 5 8a + 276 s 



39. 



26x 2 - 26 3 - x 3 4- 6^ x* - 



?0<$ ADVANCED TOPICS IN WACT/ONS 

84. Lowest common multiple of polynomials 

The LCM of two or more integral rational polynomials is defined 
as the polynomial of lowest degree in all the literal numbers, with 
smallest integral coefficients, which has each given polynomial as a 
factor. Two results for a LCM which differ only in sign will be 
considered essentially identical because usually the sign of a LCM is 
of no importance. To find a LCM, first factor the polynomials. 

\ 

ILLUSTRATION 1. The LCM of 

2(3 - z)(3 + a), 4(z - 3)(s - 1), and 3(z - 3)' 

is 4-3(x - 3) 2 (z + 3)(z - 1). We did not consider (3 - x) and (x - 3) 
as distinct factors because 3 x (x 3). 

The LCD of two or more fractions is the LCM of their denomina- 
tors. We shall deal with the notion of a LCM only where it is a LCD. 

Note 1. The highest common factor (HCF) of two or more integral 
rational expressions is the expression of highest degree, with largest integral 
coefficients, which is a factor of each of the given expressions. Thus, the 
HCF of 6sV and 4xy* is 2xy 8 . We shall not find it essential to use the 
HCF terminology. 

85. Addition of fractions with polynomial denominators 

SUMMARY. To express a sum of fractions as a single fraction: 

1. Find the LCD; that is, factor each denominator and form the 
product of all different prime factors, giving to each factor the highest 
exponent with which it appears in any denominator. 

2. For each fraction, divide the LCD by the denominator and then 
multiply both numerator and denominator by the resulting quotient, 
to express the fraction as an equal one having the LCD. 

3. Combine the new numerators just obtained, with each numerator 
placed in parentheses preceded by the sign of its fraction, and divide 
by the LCD. 

Note 1. To check the addition of fractions, substitute explicit values 
for the literal numbers in the given sum and the final result. 

EXAMPLE 1. Express as a single fraction: 

4x 



-9 x* + x-6 ^ 



ADVANCED TOP/CS IN FRACT/ONS 107 

SOLUTION. 1. Factor the denominators: 

* 2 - 9 - (* - 3)(x + 3); z 2 + x - 6 - (x + 3)(* - 2). 

Hence, LCD (a - 3) (a; -f 3) (a? - 2). 

2. In the 1st fraction, LCD * (x 2 - 9) = x - 2. 

3. In the 2d fraction, LCD -*- (x 2 + x - 6) = * - 3. 

4. We multiply numerator and denominator by x 2 in the 1st fraction, 
and by x 3 in the 2d fraction: 



x* - 9 z 2 + x - 6 
* - 2) 3z(a; - 3) 



(* - 3)(* + 3)(x - 2) (as - 3)(x + 3)(* - 2) 

4sQc - 2) - 3a(:c - 3) = x* + x 

(* - 3)(* -f 3) (a? - 2) (a; - 3)(x + 3)(a; - 2) 

CHECK. When x - 4, we obtain: 

16 -12 16 6 10 



(2) 
(3) 



In (1) : 



16-9 16 + 4-6 7 7 7 



For the result in (3) : j-. ow , , ow , ^r -TT which checks. 

(4 - 3)(4 + 3)(4 - 2) 7 

Comment. With practice, the student should be able to omit details 
such as those on the right in (2). 

ILLUSTRATION 1. In the following addition of fractions, we change signs 
in the second denominator in order to exhibit the identical nature of two 
factors in the denominators: 

5 7 5757 

X g\ 1 I A J f* .. 



3(3c-2d) 6c - 4d 3(3c - 2d) 2(3c - 

(5-2) - (7-3) _ 11 
3-2(3c-2c*) 6(3c-2d)' 



EXERCISE 39 

Change the fraction to an equal one with the specified denominator. 

1. 3x/(x 2); new denominator, (x + $)(x 2). 

2. 2y/(y 4); new denominator, (y - 4)(3y 1). 

3. 3z/(2z 3) ; new denominator, 4x 2 - 9. f 

4. 2/(a -f 2) ; new denominator, 2o 2 -f 6. 

5. (3 a)/(2 a); new denominator, 2a 4. 



708 ADVANCED TOPICS IN FRACTIONS 

Combine into a single fraction in lowest terms. Where letters are involved, 
check by substitution when directed by the instructor. 

6 _Z. _ 4-- 7 5? _ 8 ? 3a 4- 2 

10 30 + 5* v 2a 46* 3 x - 5 ' 

12 32 

" nf -i\ ^p - / iv * XU. _ . 

3(a - 6) 5(a - 6) 7x 4- 



00 
29. 



11. s-- - v--- 12. 



3x 3y 5x 5y a 6 a 4- 6 



3 2 - 3s 
15. s - FJ + ^j - 16. 



2c - 6d ' 3d - c 2a - 46 66 - 3a 

4 2 



- 1 ^ 6x - 3 3x - 2 2x 



20. 



6x 4- 6 9a 2 d 2 6a 

a s a K 

21. 



* 2x 



23. 3a-+l. 24. !- 

- 3 



OK 4 5 10 

26. _ , ~ + - - -; 26. 



-f 2y ' a: 2 - y 2 4c - z 2 ' 3a; 2 - 48 



07 2fl ~ n i 3a - 4n a - 4 2 - llo 

2a - 2n "*" 6n - 6a* 2a - 4 "*" 2 - a ' 

4- 2x - 1 2x4-1 



x 4- 4 x 2 4- x - 12 x 2 4- 4x - 60 x - 6 

" 3n - 3 ~~ n 2 4- 3n - 4* 32> o 2 - 16 ~ a 2 4- 8a 4- 16 



33 2c ~ 3 I 4 34 g + 5x2 I 3 

***** no 10 I o > i 1 ^ i ? *" ^ ^ .2 i^ 



- 18 ' 3C 2 - lie 4- 6 & - ^ ' 2x - 2y 

5 



- 27 4x 2 - 12x 4- 9 x* + 8 x 2 - 2* 4- 4 

37 3s* 5s 2 - 3 2s* - 3 s 4- 3 

" t " 



:c 4 -4""2x 4 4-x 2 -6 2^4-3^-2 x 6 - 4 



ADVANCED TOPICS IN FRACTIONS J09 



39 3x-2 2s + 5 

2z 2 - x - 3 3s 2 + 6* + 3 ~ r 

x - 2 3s + 5 , 2a; - 



+ x - 6 2z 2 - x - 6 4* 2 - 9 
7 3 - 



< 

** 



49 

** & _o _r ro i 



8s 2 - 18 ' 2x* - toe + 9 2z 2 - 3z - 9 

a + 36 a - 26 



6a 2 - 06 - fe 2 ' 3o 2 + 7a6 + 26 2 2a 2 
r - o 2r - a 



___ 
r 2 - 6ar + 9a 2 r 2 - 9a 2 



q 



86. Factoring in multiplication or division of fractions 

To multiply or divide fractions involving polynomials, factor the 
numerators and denominators and divide out all common factors 
from the numerator and denominator of the final result. 



7x - 15 2x 2 - 19z + 42 
ILLUSTRATION 1. ^ _ ^ _ u -- ^ _ 12 

(2x - 3) (a; + 5) (2s -'7) (a - 6) = (x + 6)(s - 6) 
(2x - 7)(x + 2) ' 4(2z - 3) 4(x + 2) ' 

where we divided both numerator and denominator by (2x 3)(2z 7). 

xy* - y 8 

T n x* + x z y xy* w 3 x* xy 2t/ 2 

ILLUSTRATION 2. -5 ^ 75 = -f-j ^ -5 ** 

x* 2xy + y z x* + x*y x 2 



xy 



= y 2 (a? - y) (a; - 
' 



(^ + y) (x - y) 2 a: 2 (x - 

where we divided both numerator and denominator by (x y)(x 



T - 4 . t ON 2a; - 4 . x -2 

ILLUSTRATION 3. -T - =- - (x 2) = -^ - = -. -- : 

z 2 5 v a; 2 5 1 

= 2(g - 2) 1 2 

a; 2 - 5 'x - 2 x 2 - 5* 

where we divided out (re 2). 

Whenever a mixed expression is involved in the numerator or de- 
nominator of a fraction, or as a factor in a product, it is advisable 
to change the mixed expression to a single fraction as the first step 
hi simplification. 



710 ADVANCED 7OP/CS IN FRACTIONS 

At 

EXAMPLE 1. Reduce to a simple fraction: 



u 



SOLUTION. Express the numerator and denominator of the complex 
fraction as simple fractions and divide: 

25 



u 



2 _ 



9 9 9u-25 



3u - 5 9 3u - 5 



5) 



9 (3t< - 5) 3 

Comment. A somewhat shorter solution is obtained if, as the first step, 
we multiply both numerator and denominator by the LCD, 9, of the fractions 
involved in them. 



ILLUSTRATION 4. (1 -f ~ J * (2 =-) 



2 
Zx 2 1 



2 + 3s . 4 - 9s* 

*"^ *""' 



2 (2 -f- 3z)(2 - 3z) 2 - 3s 

EXERCISE 40 

Perform the indicated operation and reduce to a simple fraction in lowest 
terms. Check by substituting values for the letters, where directed by the in- 
structor. 

3a - 36 a + 26 2c-4da6-3a 



a-&' 6-3 *6c-26c 



4 tf- 



hx-hy cw-bw, - 

ab-ac 3x-Zy ch - ex 



K "" ^2 IA\ A 

5. -= - r- (a; 2 lo). o. 



-= - r- . . = ^ - r 

a; 2 4a; 5w aw 5k ok 



- 1 . 6y - 2 2s - 2y . (a? - 

- 16 ' / 2 * 



A / K o *N IA 9 

9. (5x 3x 2 ) 4- - r- =- 10. = - =- ? -- - 

v ' x + 3 3s 3y 

6 2 



- 9 a* + 06 4a* - 96 



14. 



3x-l 



ADVANCED TOPICS IN FRACTIONS 

_ ox-f 6x 



17. 



20. 



23. 



26. 



a 



IU 



29 . 



9x 2 - 1 


o 2 a 2 


XO* ." ' " 


4x + 5 

ill. 

-4-3 


3x 
18. LI-. 

?_ 

-* 

z ^ 
2o 


a 2 

._! 
19. ^ j- 

6 a 
2 


l-i 

1 erf 
1 


I-* 

OK W . 


5y 


i T * 

24_ fl ' +6 


r.~* 


a+1"" r 
a 1 


i - ? 

u 
2x 2 -f 5x - 12 


x a 


AI 2 2o 

Q7 IMMiMMWMW 


Oft xa + 2x + 1 


A 




^5x 2 4- x 6 


X CL 

/, , 4 M 


4 4a 

* * \ L , 5 


r/if\ /Olf A'l'X 

'* ju / y ^ * C L 



-I- 2 



32. (l - 

\ o 2 



1- 



33. 



2s* + 5z - 12 



35. 



nx + 



34. 



on at; 



x 

ex ac). 
y 



36. 



*~ 2? 



on- 1 

37. _i-- 



_ 



112 ADVANCED TOPICS IN FRACTIONS 

41 / a 4 - 816 4 __ a + 36 \ _._ a 2 + 606 + 
U 2 c - 3o6c + 96^ ' a 2 - 6a6 + 96 2 / * a 3 + 276 8 

42 c 4 ~ 2c 3 d + 4c 2 <P . (c*d - c* c*'+ 8d \ 
ac + 6W - 2ad - 36c T \c 2 - 4cP ' a 8 - 276 3 / 






44 



3o H- 46 5oc H- 



5o 

t~ y """ ~~~* 
a 



16o 2 __ 

z o 2 ^ a 



47. 



4- 7^ 2 



Find <Ae reciprocal of the expression and reduce the result to a simple fraction 
in lowest terms. 

/3a:+l g . 3 \ 
l2^^2 - 5 + JTT/' 

Reduce to a simple fraction in lowest terms. 

a* x 4 + 4x 2 + 8 



6* x* - 4 

51. _L_ -.. 62. * . 4 



a 4 x + 2 

.. A-2 2_V_? L_\ 

V a + 3/U + 2 3 -a/ 



64. ? 66. 



A 2a 
x r-7 4o 



x 1 + 2a 



66. _ T _. 67. 



4o - 1 4o - 



a- 1 



ADVANCED TOPICS IN FRACTIONS 773 

87. Equations involving fractions 

To solve an equation involving fractions, we proceed as follows. 

1. Factor all denominators and form the LCD in factored form. 

2. Enclose each numerator in parentheses and multiply both sides of 
the equation by the LCD to clear the equation of fractions. 

3. Remove parentheses and solve. 



G, i i 2x 2x + 28 , 

EXAMPLE 1. Solve: 3 ~ 2^3 = (1) 



SOLUTION. 1. The LCD is (2z + 3)(2z - 3), or (4z 2 - 9), 

2. Multiply both sides by the LCD: 

2x(2x - 3) - 2x(2x + 3) = 2x + 28, (2) 

Or 

because (2x + 3)(2z - 3) = 2x(2x + 3); etc. 



3. Expand in (2) and collect terms: 

4x* - Qx - 4x 2 - Gx = 2x + 28; - 28 = 14x; x = - 2. 

The student should check by substituting x 2 in (1). 

88. Operations leading to extraneous roots 

A. // both members of an equation are divided by an expression 
involving the unknowns, the new equation may have fewer roots than 
the original equation. 

ILLUSTRATION 1. By substitution, we verify that x 1 and x = 2 are 
roots of x z 3x H- 2 = 0. On dividing both sides by (x 2) we obtain 



x* - 3x + 2 A (x - A - A 

- ^ = 0. or - - ^^ - - = 0, or x 1 = 0. 
. x 2 x 2 

The final equation has just one root, x = 1. The root x 2 was lost by 
the division. 

In solving algebraic equations, we usually avoid operations of 
Type A in order that roots may not be lost.* 



B. // both members of an equation are multiplied by an expression 
involving the unknowns, the new equation thus obtained may have 
more solutions than the original equation. 

* See Note 4 in the Appendix for a "proof" that 2 = 1, in which the fallacy 
involves an operation of Type A which conceals a division by zero. 



114 ADVANCED TOPICS IN FRACTIONS 

ILLUSTRATION 2. The equation x 3 * has just one root, x 3. If 
both sides of x 3 are multiplied by (x -f 2) we obtain 

(x + 2)(x - 3) - 0, or x* - x - 6 = 0. 

By substitution, we verify that this equation has two roots, x = 3 and 
x 2. The root 2 was introduced by the multiplication. 

A value of the unknown, such as x * 2 in Illustratioij 2, which 
satisfies a derived equation but does not satisfy the original equation, 
is called an extraneous root. 

Whenever an operation of Type B is employed, test att values 
obtained to reject extraneous roots, if any. 

EXAMPLE 1. Solve: -5 = ; = H 17 = 0. 

S)*Z MMB I 1l** ^" I O* ^M I 

v A *C/ X C/ |^ JL 

SOLUTION. The LCD is x 2 1; multiply both sides by x 2 1: 
s - 1 + 2(x - 1) = 0; 3z - 3; qr a: - 1. 

TEST. Since x 1 makes z 2 1 = in the denominators of the given 
equation, 1 cannot be accepted as a root because division by zero is not ad- 
missible. Hence, 1 is an extraneous root and therefore the given equation 
has no root. 

EXAMPLE 2. On a river whose current flows at the rate of 3 miles per 
hour, a motorboat takes as long to travel 12 miles downstream as to travel 
8 miles upstream. At what rate could the boat travel hi still water? 

SOLUTION. 1. Let x miles per hour be the rate of the boat in still water. 
Then the rate of the boat in miles per hour going upstream is (x 3) and 
downstream is (x + 3). 

2. From the standard equation d = vt of uniform motion, we obtain 
t d/v. Hence, the time in hours for traveling 

g 
8 miles upstream is =; 

X u 

12 
12 miles downstream is r-x 



3. Hence, j- = --. (1) 

4. Multiply both, sides of (1) by (x - 3)(s + 3): 

S(x + 3) = 12(x -3); 8x + 24 = I2x - 36; 

4x . 60; x = 15. (2) 

Thus, the boat travels 15 miles per hour in still water. 



ADVANCED TOPICS IN FRACTIONS 115 

EXERCISE 41 

Each equation will reduce to a linear equation if cleared of fractions properly. 
This reduction may be prevented and extraneous roots may be introduced i] 
unnecessary factors are employed in the LCD. Solve each equation and check. 

* 61 . ^ x 3 .75 



*" x - 2 ~ 2 '* x~=3 " 2 

A 7 3 A K I 7 1 

4. !SS 4. 6. 7: 

If* <7* J9 9 

mLf ml/ mm& 4mr 



2* - 2 w ~ + 1 3< + 4 

.a 10. 2 5 



12. 



* + 2 2x + 2 
a; x - 3 



14 2h + 1 3x - 3 3* - 6 

2 _ 1 14< *-3 m ** + 20- 



- * x 2 - 1 "I - 1 t* + t-2 

-* + 14 






17. j^j - 1 - g^' W. 



2 ^ 6* 2 -h6 3 ~ 3:C 4.2. 



2^ . O A^r2 .. /r _ 9 IT 9 

" j \JJb ^^ J(/ ^^ mi IJvU ^^ mm 

\ - - 3 "" x 02 4x 2 + 3 = 1 

} .3 + * 8x 3 + 1 2x + 1 

* J ' ft 2* + 3 3 - * 2 

* - 5 * 2 - 6* + 5* 

3 3* 



2 + 3* 4 - 9* 2 
4* + 3 7 - 2* 2 



26. 



' * - 2 2* - 3 2x 2 - 7x + 6 

-7 



- 14 



w 6 5 _ 5-4to ^ _3 2 2s-l 

27 : ^ TT T"3 ii ?r^* * ^ i_ yi ^. c **" 

X + 4 JC 



176 ADVANCED TOPICS IN FRACTIONS 

29. In a certain fraction, the denominator exceeds the numerator by 5. 
If the numerator and denominator are both increased by 3, the fraction 
equals f . Find the original fraction. 

30. On six quizzes in mathematics, a student has obtained 70% as his 
average score. How many scores of 86% each must be obtained to bring 
his average score up to 80%? 

31. In one hour, Jones can plow J of a field. If Jones and Smith both 
work, they can plow the field in 2 hours and 24 minutes. How many hours 
would it take Smith alone to plow the field? 

32. When the wind velocity is 40 miles per hour, it takes a certain airplane 
as long to travel 320 miles against the wind as 480 miles with it. How fast 
can the airplane travel in still air? 

33. When the wind velocity is 20 miles per hour, it takes a certain air- 
plane 90% as long to travel with the wind to any destination as it would 
to return to the starting place against the wind. How fast can the airplane 
travel in still air? 

34. A fuel tank has one intake pipe which fills it hi 8 hours. A second 
intake pipe is installed and it is found that, when both are in use, they 
fill the tank hi 2 hours. How long would it take the second pipe alone to 
fill the tank? 

35. On a river whose current flows at the rate of 3 miles per hour, a 
motorboat takes as long to travel 12 miles downstream as to travel 8 miles 
upstream. At what rate could the boat travel in still water? 

36. Two rivers flow at the rates of 3 miles and 4 miles per hour, re- 
spectively. It takes a man as long to row 13 miles downstream on the 
slower river as to row 15 miles downstream on the faster river. At what 
rate can he row in still water? 

89. Solution of literal equations involving factoring 

In solving a linear equation in a single unknown x, when other 
literal numbers occur in the equation, it may be necessary to factor 
either in clearing of fractions or in simplifying the final result. 

EXAMPLE 1. Solve for x: b(b + x) = o 2 ax. 

SOLUTION. 1. Expand: fc 2 + bx = o 2 ox. 

2. Add ox; subtract 6 2 : o& -f 6x = o 2 6 2 . 

3. Factor: x(o + 6) = (o - 6) (a + 6). 

4. Divide by (a + 6) : x = a b. 

Comment. To check, substitute x = a 6 in the original equation. 



ADVANCED TOPICS IN FRACTIONS 117 



EXAMPLE 2. Solve for *: ? + L+* = !L. 

6 2 a 2 a 6 a + 6 

SOLUTION. The LCD is (6 a) (6 + a). We first rewrite the equation 
to change the sign in the denominator (a 6) and then clear of fractions 
by multiplying by the LCD : 

aw + 6* w -\- b w + a 
6 2 a 2 6 a a + 6 

Multiply by (6 2 - a 2 ) : 

010 + 6 2 - (w + 6) (6 + a) - (w + a)(6 - a). (1) 

The student should expand in (1) and solve, to obtain w = o. 

EXERCISE 42 

otoe for x or # or is, whichever appears. 
1. ex 3o = 2/i. 2. 7x a = 3ox 5. 



3. 3o* - bz = 9a 2 - 6 2 . 4. 36x - 96 2 = 2ax - 4o 2 . 

6. 4a3 a 2 = 4z 1. 6. mnx a anx m. 

7. x 2 - 3n 2 = (3n - z) 2 . 8. 6(6 - x) = a 2 + ax. 
9. a6x a 2 = 6 2 o6x. 10. bx bd ad ax. 

11. 6(6* - a) = a*x + 6 2 . 12. hz - A 2 = kz - k*. 

13. 2bx + 6a 2 = 3ac + 4a6. 14. ax - ab - a 2 = b(x - 26). 
15. acx + adz + d 2 = c 2 6cx 6ax. 

x(a - 46) - 6 2 x + 6 _ x + a 
b ' a 2 -6 2 ~^a- b*~*a + b 

17 c = d 18 

'* * 



4A 
19. 



- d 2x - c x + 26 x - 2a 

x 26 



2x + h 4x 2 - W 2x - h * 2oc + x a - b 

x a 2 x + 2a 



21. 



26 2 + 06 - a 2 6 + a 26 s + 4a6 + 2a 2 



22. Solve C = r- for 6. 23. Solve S - - ^ for r. 

6 a r 1 

24. Solve s = r for 6. 

c b 



118 ADVANCED TOPICS IN FRACTIONS 

EXERCISE 43 
Review of Chapters 4, 5, and 6 

Perform the indicated operation and collect terms. 
1. (3x - 5j/)(3x + 5y). 2. (4x 2 - 3yz)(4x* + 82/2). 

3. (2* + 3) 2 . 4. (x - 2*)*. 5. V - 3w>) 2 . 6. (2a + 56) 2 . 

7. (a - 4)(a 2 + 4a 4- 16). 8. (2x - 3s)(4x 2 + 6x -f 92 s ). 

Factor. 
9. j/ 2 - 25s*. 10. 4z* - 9W. 11. * 

12. M 4 - Slj/ 4 * 4 . 13. o* - 276. 14. 8w + 

15. % 2 + 122/2 2 + 4z*. 16. 2/ 2 + y - 12. 

17. 8 + 42 - 21. 18. 6x 2 + x - 15. 

19. 2 - 12x 2 + 5x. 20. 4A 2 - 28/wo + 49t^. 

21. S* 2 - 30; -h 45u>*. 22. 06 + 26c + 3ad H- 6cd. 

23. 2a 8 -h 4a 2 - 2o - 4. 24. (m + w?) 2 -f 4m + 4w; -f- 4. 

25. x* - a 2 - 6a6 - 96 2 . 26. x 2 + 4x -h 4 - 9a 2 . 



3 4- 5 1 2 

-- j- . j^ 



Reduce to a simple fraction in lowest terms. 

l-~ 
. 5 Q * T 29 - 

O j * 

_ _ (j -~ . $ -f- _ 

x y 9 ox 

3x - 1 2x + 3 M 2a - 6 , 56 



3y-2 2y-5 3c - 6 36 



or x and 

3x 2 + x . 12 4 H- 2x 



' 3+lx 
3 . 1 + 4x 



2 

2 ~- 



" 7 - 2x 4x 2 - 16x + 7 
a - 6 a 6 . A 3X 2 - 2x -h 9 3 + 2x 






CHAPTER 



7 



RECTANGULAR COORDINATES AND GRAPHS 



II 



I I I I I I 



4 
--3 
- - 2 



90. Rectangular coordinates 

On each of the perpendicular axes OX and OF in Figure 4, we lay 
off a scale with as the zero point on both scales. In the plane of 
OX and Y we shall measure vertical Y 

distances in terms of the unit on OF 
and horizontal distances in terms of 
the unit on OX. We agree that 
horizontal distances will be consid- 
ered positive if measured to the right 
and negative if to the left; vertical dis- 
tances will be considered positive if 
measured upward and negative if 
downward. Let P be any point in the 
plane. 

The horizontal coordinate, or the 
abscissa of P, is the perpendicular 
distance, x, from OY to P; this di- 



~ 1 







-6-5-4-3-2-1 

1 + 



III 



2 



3 



-4.. 



123456 
I I I I I. I 



V 



IV 



Fig. 4 



rected distance is positive if P is to the right of OF and negative if P 
is to the left of OF. 

The vertical coordinate, or the ordinate of P, is the perpendicular 
distance, y, from OX to P; this directed distance is positive if P is 
above OX and negative if P is below OX. 

Each of the lines OX and OF is called a coordinate axis, and the 
abscissa and ordinate of P together are called the rectangular co- 
ordinates of P. The point at which the axes intersect is called 
the origin of the coordinate system. When the axes are labeled 
OX and OF as in Figure 4, we sometimes refer to the abscissa as the 
z-coordinate and to the ordinate as the y-coordinate. 



720 RECTANGULAR COORDINATES AND GRAPHS 

Notice that there is no necessity for using the same unit of length 
for the scales on OX and OF. 

ILLUSTRATION 1. In Figure 4, the coordinates of P are x == 5J and 
y = 2. The coordinates of a point are usually written together within pa- 
rentheses with the abscissa first. Thus, we say that P is the point (5J, 2). 
In Figure 4, R is the point ( 3, 4). 

Note 1. The coordinate axes divide the plane into four parts called 
quadrants, which we number I, II, III, and IV, counterclockwise. 

To plot a point, whose coordinates are given, means to locate the 
point and to mark it with a dot or a cross. 

EXAMPLE 1. Plot the point ( 3, 4). 

FIRST SOLUTION. At 3 on OX, erect a perpendicular to OX. Go up 
4 vertical units on this perpendicular to reach the point R in quadrant II 
which is ( 3, 4). 

SECOND SOLUTION. At + 4 on OF, erect a perpendicular to OF. Go to 
the left 3 units on this perpendicular to reach ( 3, 4). 

Note 2. The word line in this book will refer to a straight line unless 
otherwise specified. 

EXERCISE 44 

Plot the following points on a coordinate system on cross-section paper. 
1. (3, 4). 2. (3, 0). 3. (1, - 2). 4. (- 3, - 5). 

6. (0, - 2). 6. (- 5, 0). 7. (0, 7). 8. (- 3, 4). 

9. (- 2, - 3). 10. (- 3, 5). 11. (4, - 4). 12. (- 2, 1). 

13. Three corners of a rectangle are (3, 4), ( 5, 4), and (3, 1). Find 
the coordinates of the 4th corner and the area of the rectangle. 

Find the area of a triangk with the given vertices. 
14. (4, 3); (4, 7); (- 2, 3). 15. (0, - 4); (3, - 4); (3, 2). 

16. (- 2, 1); (3, 1); (5, 5). 17. (0, 0); (5, 3); (5, 7). 

18. A square, with its sides parallel to the coordinate axes, has one 
corner at ( 3, 2) and lies above and to the left of ( 3, 2). If the units 
of length on the axes are the same and if each side of the square is 4 units 
long, find the coordinates of the other corners. 



RECTANGULAR COORDINATES AND GRAPHS 

In which quadrant does a point lie under the specified condition? 

19. Both coordinates are negative. 

20. The abscissa is negative and the ordinate is positive. 

21. The abscissa is positive and the ordinate is negative. 

22. A line is parallel to OX and passes through the point where y = 3 
on OF. What is true about the ordinates of points on the given line? 

23. A line is perpendicular to OX at the point where x 2. What 
can be stated about the abscissas of points on the given line? 

24. How far apart are the lines on which the abscissas of all points are 

3 and 4, respectively? 

25. How far apart are the lines on which the ordinates of all points are 

7 and 3, respectively? 

91 . The function concept 

We recall that, in a given problem, a constant is a number symbol 
whose value is not subject to change during the course of the dis- 
cussion, and a variable is a number symbol which may take on 
different values. When desirable, we may think of a constant as a 
variable which can assume only one value. 

If a first variable, x, and a second variable, y, are so related that, 
whenever a value is assigned to x, a corresponding value (or corre- 
sponding values) of y can be determined, we say that y is a function 
of x. Then x is called the independent variable and the second variable, 
y, which is a function of x, is called the dependent variable. To say 
that y is a function of x means that the value of y depends on the 
value of x. 

ILLUSTRATION 1. In the formula A = Trr 2 for the area of a circle, if r 
is a variable then A is a variable and A is a function of r. 

Any formula in a variable x represents a function of x; the values 
of the function can be computed from its formula. 

ILLUSTRATION 2. (3z 2 H- 7x -f- 5) is a function of x. If x = 2, the value 
of the function is (12 +14 + 5) or 31. 

Note 1. If just one value of y corresponds to each value of x, we say that 
y is a single-valued function of a;; if just two values of y correspond to each 
value of x, then y is a two-valued function of x; etc. 



722 



RECTANGULAR COORDINATES AND GRAPHS 



92. Graph of a (unction 

Let y represent any function of x. Then, each pair of corresponding 
values of x and y can be taken as the coordinates of a point in an 
(x, y) coordinate system. This leads us to adopt the following 
terminology. 

DEFINITION I. The graph of a function, y, of x is the set of all points 
(or the locus of points) whose coordinates form pairs of corresponding 
values of x and y. 

To graph a function will mean to draw its graph. In graphing 
a function, we usually plot the values of the independent variable on 
the horizontal axis of the coordinate system. 

A linear function of x is a polynomial of the first degree in x and 
hence has the form ax 4- 6, where a and 6 are constants. In Illustra- 
tion 1 below we meet a special case of the fact that the graph of a 
linear function of x is a straight 'line. This fact, whose proof we 
shall omit, is the basis for the name linear function of x. 

ILLUSTRATION 1. If x is the independent variable, in order to graph the 
function (J:c 3), we introduce y to represent it. That is, we let y \x 3. 
If x = 5, then y = f( 5) 3 = 6. Hence,, one point on the graph 
is (5, 6). Similarly, we let x 0, 2, etc., and compute the 
corresponding values of y given in the 
following table. We plot (- 5, - 6), 
( 2, 4J), etc., in Figure 5 and join them 
by a straight line, which is the graph of the 
function. From the graph, we read that 
the value of the function is zero (the graph 
crosses the z-axis) when x = 5. The func- 
tion equals 2 when x = 1, approximately. 



X = 


- 5 


-2 





3 


6 


y = 


-6 


-4i 


-3 


-li 


f 



o 



Fig. 5 



If y is a linear function of x, we need only two pairs of values of x 
and y to obtain the graph, because a straight line is definitely located 
if we know two points on it. However, in graphing any linear func- 
tion, we shall compute three values of the function hi order to check 
the arithmetic involved. If the corresponding three points do not lie 
on a line, an error is indicated. 



RECTANGULAR COORDINATES AND GRAPHS 



123 



Note 1. In graphing, do not choose the position of the origin or the scales 
on the coordinate axes until after a reasonably complete table of values has 
been prepared. Then, make the appropriate selections of origin and scales 
so that as large a graph as possible may be placed on the available paper. 

If a function of x is defined by a formula, in general its graph 
is a smooth curve* To graph such a function, we introduce some 
letter, such as y, to represent the function, compute a table of cor- 
responding values of x and y, and draw a smooth curve through the 
corresponding points on a coordinate system. 



ILLUSTRATION 2. To graph x 2 4x -f 6, we 
let y represent the function, 

y = x* - 4x + 6, 

compute the following table of values, and plot 
the points. The graph, in Figure 6, is a curve 
called a parabola. 



x = 


- 1 


1 


2 


3 


5 


y = 


11 


3 


2 

' 


3 


11 




Fig. 6 



93. Functions not defined by formulas 

Functions not defined by formulas arise frequently. Sometimes 
the only information concerning a function consists of a table of 
corresponding values of the function and the independent variable, 
where the table may be obtainable by experimental means or obser- 
vation. In drawing the graph of such a function, sketch a smooth 
curve through the points obtained from the given values, unless 
otherwise directed. Instead of drawing a smooth curve through the 
points, it is sometimes desirable to connect them by segments of 
straight lines and thus to obtain a broken-line graph. 

Note 1 . The intersection of the'coordinate axes may be selected to repre- 
sent any convenient value, not necessarily zero, on either scale. 

ILLUSTRATION 1. The second row of the following table gives the general 
wholesale price index number of the United States Department of Labor 
for the critical depression months from June, 1930, to June, 1931. A value 
like 86.8 means 86.8% of the average level in 1926. To graph the index 

* Or, m some cases, two or more disconnected smooth curves. 



124 



RECTANGULAR COORDINATES AND GRAPHS 



number as a function of the time, in Figure 7, we choose coordinate axes, 
with time plotted horizontally and index number vertically. We take 1 month 
to be the unit of tune. We let the intersection (origin) of the axes represent 
June, 1930, on the axis of abscissas and 60 on the vertical axis, and assign 
units on the axes to suit the size of the figure. Then, for December, 1930, 
we plot the point (6, 78.4), etc. We join the plotted points by a reasonably 
smooth curve, which is the graph of the function. From the graph, extended 
as a guess to July, 1930, we estimate that the index number then was 68.6. 



JUNK '30 


Jui/r 


AUG. 


SEPT. 


OCT. 


Nov. 


DEC. 


JAN. '31 


FEB. 


MAR. 


APR. 


MAY 


JUNE 


86.8 


84.0 


84.0 


84.2 


82.6 


80.4 


78.4 


77.0 


75.5 


74.5 


73.3 


71.3 


70.0 



f=0is June, 1930 
1 is July, 1930 



8 9 10 11 12 13 




. 7 



EXERCISE 45 

The letter x represents the independent variable in all problems where it 
appears. Clearly indicate the scale on each coordinate axis employed. 

1. Graph the function (2x -\- 3). From the graph, (a) read the values of 
the function when x = 2J and x = 3J; (6) read the values of x corre- 
sponding to which the values of the function are 2, 0, and 3. 

Graph the function of x and, from the graph, read the value of x for which 
the function equals zero. 

2. 3x + 5. 3. 3 - 4z. 4. - 2 - 5x. 5. 



6. - 2x. 
10. 7. 



7. -2 
11. -4. 



3*. 



8. 4 - 2x. 
12. 0. 



9. - 3 - 2x. 
13. x. 



HINT for Problem 10. Any constant can be considered as a function of 
any variable x, with just one value for the function. The graph is horizontal. 



RECTANGULAR COORD/NATES AND GRAPHS 



125 



14. Graph the function of y defined by (3y 4), with the y-axis horizontal 
and with z used as a label for the function. 



15. Graph (x* 6z + 7) by computing its values for the following values 
of x: 1, 0, 2, 3, 4, 6, and 7. From the graph, (a) read the values of the 
function when x = 5 and x 1; (6) read the values of x for which the func- 
tion equals or 10. 

16. Graph ( x 2 4x + 6) by computing its values for the following 
values of x: 6, 5, 4, 3, 2, 1, 0, 1, and 2. From the graph, 
read the values of x for which the function (a) equals 0; (6) equals 3. 

17. The table gives the total mileage of hard-surfaced roads forming 
parts of state highway systems in the United States at the ends of various 
years. Graph the mileage as a function of the time. 



YEAR 


1926 


1929 


1931 


1934 


1939 


1940 


1942 


MILEAGE 


54,000 


75,000 


96,000 


110,000 


120,000 


122,000 


130,000 



18. The table gives the time it takes money to double itself if invested 
at certain rates of interest, compounded semiannually. Graph the time 
as a function of the rate. From the graph, find the time for money to 
double at 3i%. 



TIME, YEARS 



46J 



34f 



28 



23J 



14 



ill 



RATE 



1% 



4% 



19. The velocity of sound in air depends on the temperature of the air. 
By use of the following data, graph the velocity as a function of the 
temperature. From the graph, read the velocity if the temperature is 
35; 8.5; 120. 



VELOCITY, FT. PER SEC. 


1030 


1040 


1060 


1080 


1110 


1140 


1170 


TEMP. (FAHRENHEIT) 


-30 


-20 





20 Q 


50 


80 


110 



HINT. Let the origin represent 1000 feet on the vertical axis. 

20. The table gives the number of divorces per 1000 marriages in various 
years in continental United States. Graph the number of divorces as a 
function of the time. 



YEAR 


1890 


1900 


1916 


1922 


1930 


1934 


1937 


1940 


DIVORCES 


62 


81 


108 


131 


170 


157 


173 


169 



726 



RECTANGULAR COORDINATES AND GRAPHS 



21. The weight of a cubic foot of dry air at an atmospheric pressure of 
29.92 inches of mercury, under various temperatures, is given in the follow- 
ing table, where weight is in pounds, and temperature is hi degrees Fahren- 
heit. Graph the weight of air as a function of the temperature. 



TEMP. 





12 


32 


52 


82 


112 


152 


192 


212 


WEIGHT 


.0864 


.0842 


.0807 


.0776 


.0733 


.0694 


.0646 


.0609 


.0591 



22. The following table gives the "thinking distance" t, and the "braking 
distance" b involved when a motorist, traveling at s miles per hour, decides 
to stop his car. The value of t is the distance traveled by the car in } second, 
the interval which elapses between the instant an average driver sees danger 
and the instant he applies his brakes. The sum d t + b is the total 
distance the car will travel before stopping after danger is seen. On one 
coordinate system, draw graphs of t as a function of s and d! as a function of 8. 



8 (mph) 


20 


30 


40 


50 


60 


70 


t (feet) 


22 


33 


44 


55 


66 


77 


b (feet) 


21 


46 


82 


128 


185 


251 



94. Functional notation 

Sometimes we represent functions by symbols like /(#), H(x), K(s), 
etc. The letter in parentheses tells what the independent variable is. 
The letter to the left is merely a convenient name for the function. 

ILLUSTRATION 1. We read "f(x)" as " the /-function of x," or for short 
"/ of x." We may represent 3s 2 5 by f(x) and write f(x) = Zx 2 5; we 
read this "/ of x is 3x 2 5." H(y) would represent a function of y. For 
instance, we may let H(y) 7y* + 6. 

If F(x) is any function of x and a is any value of x, then 
F(d) represents the value of F(x) when x = a. 

ILLTTSTRATION 2. "F(a) " is read F of a." If F(x) - 3s 2 - 5 - x, 
F(3) - 3-3 2 - 5 - 3 - 19; 
- 3) - 3(- 3) 2 - 5 + 3 = 25; 
;- 6 s ) = 3(- fc 2 ) 2 - 5 - (- 6 2 ) - 36 4 - 5 -f 6*; 
[F(~ 2)] 2 - (12 - 5 + 2) 2 = 81; 
5F(2) - 5(12 - 5 - 2) - 25. 



RECTANGULAR COORDINATES AND GRAPHS 127 

A variable z is said to be a function of two variables x and y in 
case a value of z can be determined corresponding to each pair of 
values of x and y. Similarly, we may speak of a function of three 
variables, or of any number of variables. The functional notation 
just introduced for functions of a single variable is extended to func- 
tions of more than one variable. 

ILLUSTRATION 3. F(x, y) would be read "F of x and y" and would repre- 
sent a function of the independent variables x and y. Thus, we may let 



2. 
Then, F(2, 1) - 2 + 3 + 2 - 7. 

EXERCISE 46 

Vf( x ) = 2z + 3, find the value of each symbol. 



6. /(- f). 

// G(z) = 2z 3 2 , find the valve of the symbol or an expression for it. 
1. 0(- 3). 8. 0(6). 9. 0(J). 10. 0(o). 11. 0(2c). 12. 0(3z). 

13. If F(x) = *' - x + 3, find F(- 2); F(6); F(c 2 ); F(* - 2). 



14. If 0(ii = , find 0(2); 30(1); [0(3)?; 



15. If KM - , find K(2); 2K(4); [JC(3)? ; 



16. If F(x, y) - ac + 2y, find F(2, - 3); F(- 1, 4); F(o, 6). 

17. If F(x t y) - x 2 + 3xt/, find F(- 1, 2); F(- 3, - 2); F(c, 26). 



18. If F(x) = x* - ar, find F: 

19. If /(x) =x*-4x + 5, graph /(a?) by use of /(- 1), /(O), /(I), /(2), 



20. If /(x) = ^ - 12x + 3, graph /(x) by use of /(- 4), /(- 3), /(- 2), 
/(- 1), /(O), /(I), /(2), /(3), and /(4). 

t 

95. Functions defined by equations 

A solution of an equation in two variables x and y is & pair of cor- 
responding values of x and y which satisfy the equation. Usually, 
an equation in two variables has infinitely many solutions. 



728 RECTANGULAR COORDINATES AND GRAPHS 

ILLUSTRATION 1. Consider 3z 5y = 15. If x = 3, then 9 5y = 15, 
or y = f . Hence, (x = 3, y = f ) is a solution of the given equation. 
If y 0, then 3x = 15 or x 5; hence (x = 5, y, = 0) is another solution. 
Thus, by substituting values for either variable and computing values of the 
other variable, we could find as many solutions as we might desire. 

In case x and y are related by an equation, then usually we may 
think of y as a function of x and, likewise, of as a function of y. 
This is true because, in general, for each value of either variable we 
can find corresponding values of the other variable by use of the 
equation. In particular, a linear equation in x and y defines either 
variable as a linear function of the other variable. 

ILLUSTRATION 2. From 3z 5y = 15, on solving for x we obtain 

x = 5 + fe/; 

on solving for y we obtain 

y - & - 3. 

Hence a; is a linear function of y and, equally well, y is a linear function of x. 

96. Graphical representation of an equation 

The graph, or the locus, of an equation in two variables x and y 
is the locus of all points whose coordinates (x, y) form solutions of the 
equation. If we think of a: as an independent variable, the graph of 
the equation is identical with the graph of the function, y, of x, defined 
by the equation. In particular, if a, 6, and c are constants, the graph 
of the linear equation ax + by = c ISQ. straight line. For, the graph 
of this equation is the graph of the linear function of x, or of the 
linear function of y, defined by the equation. 

ILLUSTRATION 1. From Zx 5y 15, we obtain y = far 3. The graph 
of 3x 5y = 15 is the graph of the linear function fx 3; this graph is 
found in Figure 5, page 122. 

The abscissa of any point where a graph on an (#, y) coordinate 
system meets the a>axis is called an x-intercept of the graph. The 
ordinate of any. point where the graph meets the 2/-axis is called a 
y-intercept of the graph. To find the x-intercept (or intercepts) of 
the graph of an equation hi x and y, place y - in the equation and 
solve for x; to find the y intercept (or intercepts), place x = and 
solve for y. 



RECTANGULAR COORDINATES AND GRAPHS 



129 



SUMMARY. To graph a linear equation in x and y: 

1. Place x and compute y, to find the y-intercept. 

2. Place y = and compute x, to find the x-intercept. 

3. Find any other solution of the equation and draw the line through 
the points determined on plotting the solutions obtained. 



ILLUSTRATION 2. To graph 3x % = 15, first let x and obtain 
5y = 15, or y 3; hence, (0, 3) is a 
point on the graph. If y = then 3x = 15, 
or x 5; the x-intercept is 5, or (5, 0) is a point 
on the graph. The graph is shown in Figure 5, 
page 122. 



ILLUSTRATION 3. The graph of the equation 
x 5 = consists of all points (x, y) in the 
coordinate plane for which x = 5, and the value 
of y is of no importance because it does not 
occur hi the equation. Hence, the graph of 
x 5 = is the line perpendicular to the x-axis 
at the point where x = 5, as shown in Figure 8. 



-H 



1 



o - 



44 



44 



x 



Fig. 8 



97. Equation of a line 

An equation of a curve on an (x, y) coordinate plane is an equation 
in the variables x and y whose graph is the given curve. ILtwo equa- 
tions have the same graph, in general the equations differ only in 
nonessential features. Hence, although a given curve may have 
infinitely many different equations, we shall refer to any one of these 
as the equation of the curve. 

ILLUSTRATION 1. 3x + 2y = 7 is the equation of a certain straight line. 
This line also is the graph of 6x 4 4y = 14 because these two equations 
have the same solutions. 

Frequently we refer to a function of a variable x, or to an equation 
in x and y, by giving the function or equation the name of its graph. 

ILLUSTRATION 2. Thus, we may refer to the line 3x + 2y = 7, or to the 
parabola y = x 2 4x + 6 (see Figure 6, page 123). 

We shall assume without proof the fact that the equation of any 
straight line on an (x, y) coordinate plane is of the form ax 4 by = c 
where a, 6, and c are constants. The equation of a line is a linear 



130 



RECTANGULAR COORDINATES AND GRAPHS 



relation between x and y which is true when and only when the point 
(x, y) is on the line. F 

ILLUSTRATION 3. The equation of the vertical 
line 3 units to the left of the y-axis is x 3. 

ILLUSTRATION 4. Let P, with coordinates 
(x, y), be any point on the line through (0, 0) 
and (1, 2). Then, from similar right triangles 
hi Figure 9, 



V % 
- = 



or y 



x I 
this is the equation of the line. 




Graph each equation. 
1. 3x + 2y - 6. 

4. 2x + 7y - 0. 
7. 4x - 5w - 20. 



EXERCISE 47 

2. 3y - 4s - 12 0. 
6. 3x - y - 9. 

Q O/M __ 

O* <6C 



10. s-7. 
14. y - 0. 



11. y - 5. 
15. 



12. a; = - 3. 
16. &c -f 9 = 0. 



3. &c - 5y - 0. 
6. 3x - 15 + 6 
9. 5j/ 4- s = 10. 
13. y - - 4. 
17. 3y + 4 - 0. 



Give fte equation of the line satisfying the given condition. 

18. The horizontal line (a) 6 units above OX"; (6) 4 units below OX. 

19. The vertical line (a) 5 units to the right of OF; (6) 4 units to the left 
of OF. 

20. The line on which the ordinate of each point is (a) the same as its 
abscissa; (6) the negative of its abscissa. 

Without graphing, find the coordinates of the points where the graph of the 
equation cuts the axes. 
21. Zx + 5y = 15. 22. 2x - 5y = 10. 23. - 3x + 2y - 5 = 0. 

24. Find an expression for the linear function of y defined by the equation 
2x + 7y = 9. 

25. Find an expression for the linear function of x defined by the equation 
3a? - 5w - 11. 



CHAPTER 



8 



SYSTEMS OF UNEAR EQUATIONS 



98. Graphical solution of a system of two equations 

A solution of a system of two equations in two unknowns, x and 
2/, is a pair of corresponding values of x and y which satisfy both 
equations. If a system has a solution, the equations are called 
simultaneous. 

/ x - y = 5, (1) 

1 x + 2y - 2. (2) 



EXAMPLE 1. Solve graphically: 



SOLUTION. 1. In Figure 10, AB is the 
graph of (1) and CD is the graph of (2). 
AB consists of all points whose coordinates 
satisfy (1) and CD consists of all points 
whose coordinates satisfy (2). Hence, the 
point of intersection, E, of AB and CD is the 
only point whose coordinates satisfy both 
equations. 

2. We observe that E has the coordinates 
(4, 1). Hence, (x = 4, y = 1) is the only 
solution of the system. These values check 
when substituted in (1) and (2). 



O 




B 



D 



Fi 9 . 10 



SUMMARY. To solve a system of two equations in two unknowns 
graphically: 

1. Draw the graphs of the equations on one coordinate system. 

2. Measure the coordinates of any point of intersection of the graphs; 
these coordinates form a solution of the system. 

Usually a system of two linear equations in two unknowns has 
just one solution, as was the case in Example 1, but the following 
special cases may occur. 



132 SYSTEMS OF LINEAR EQUATIONS 

A. // the graphs of the equations are parallel lines, the system has 
no solution and the equations are called inconsistent equations. 

B. // the graphs of the equations are the same line, each solution of 
either equation is also a solution of the other and hence the system has 
infinitely many solutions. In this case the equations are said to be 
dependent equations. 

Note 1. Usually a graphical solution gives only approximate results, 
because in obtaining them we estimate certain coordinates visually. 

EXERCISE 48 

Solve graphically. If there is no solution, or if there are infinitely many, 
state this fact with the appropriate reason. 



, 2 , , 

\ y + 2x = - 3. \ 2y - x = - 5. \3y + 4z = 23. 

f 2y - Zx = 0, f 3x + 8 = 0, f 5y - 3 = 0, 

\ 4y + 3z = - 18. \ 6x + 7y = 5. \ Wy + 3z = 4. 



/ 2y - 5x = 10, R f 2z - 3y = 0, f 3z + 5y = 2, 
\ 2y - 2a; = 3. \ to + 7y = 0. \ 2x - 3y = 5. 



, n , 
' - y = 6. \ 2y - 4x = 5. \ 4x - 6 = 



2a; + 2y = 7. \ 6* - % = 3. \ 10y - 2a; + 4 = 0. 

16. (a) Graph x + 3y = 5. (6) Multiply both sides of the equation by 
2 and graph the new equation, (c) By inference, state how two linear 
equations are related if they have the same graph. 

99. Elimination by addition or subtraction 

T, 1 a i r j f 4* + 5y = 6, (1) 

EXAMPLE 1. Solve for a; and y: < , / 

4. 



SOLUTION. 1. Multiply (1) by 3: 12x + 15^ = 18. (3) 

2. Multiply (2) by 5: 10z + I5y = 20. (4) 

3. Subtract, (3) - (4): 2x = - 2; x = - 1. (5) 

In obtaining equation 5, we have eliminated y by subtraction. 

4. On substituting x = 1 in (2) we obtain 3y = 4 -f 2 or y = 2. 

5. The solution of the system is (x = 1, y = 2). The student should 
check this solution by substitution in (1) and (2). 



SYSTEMS OF LINEAR EQUATIONS 



133 



SUMMABY. To solve a system of two linear equations by elimination 
by addition or subtraction: 

1. In each equation, multiply both members, if necessary, by a properly 
chosen number to obtain two equations in which the coefficients of one 
unknown have the same absolute value. 

2. Add, or subtract, corresponding sides of the two equations obtained 
in Step I so as to eliminate one unknown. 

3. Solve the equation found in Step 2 for the unknown in it, and sub- 
stitute the result in one of the given equations to find the other un- 
known. 

If two linear equations in x and y are inconsistent or dependent, 
then, in eliminating one unknown, the other will also be eliminated. 
If the equations are dependent, an identity = results from this 
elimination. If the equations are inconsistent, a contradictory 
equation such as = 36 is obtained. We shall omit proving these 
facts but shall exhibit special cases of them. 

Note 1. Hereafter, to solve a system of equations will mean to solve 
algebraically, unless otherwise stated. 

If the given equations involve fractions, clear of fractions before 
applying the preceding method. 



EXAMPLE 2. Solve for x and y: 

\. 

SOLUTION. 1. Multiply (6) by 2: 

6z + 4t/ = 12. (8) 

2. Subtract, (7) - (8) : 

= 12. (9) 

Hence, the given equations are inconsistent 
because a contradictory statement, 12 = 0, 
results from the assumption that a pair of 
values of x and y exists which satisfies (6) 
and (7). 

COMMENT. In Figure 11, AB is the graph of 
(6) and CD is the graph of (7). It is observed 
that these lines are parallel and hence do not 



3x + 2y = 6, 
+ 4y = 24. 

D Y 



(6) 
(7) 




Fig. 11 



intersect, which agrees with the preceding algebraic proof that (6) and (7) 
have no solution. 



134 



SYSTEMS OF LINEAR EQUATIONS 



1. 

4. 



EXERCISE 49 

Solve by elimination by addition or subtraction and check. 

Zx - y = 7, 
2x + 3w = 12. 



- 2* = 0, 
-f 2y = 0. 

3s - 2y = 2, 

- 3z = 2. 



2. 
5. 
8. 



-3 = 7: 



_ _ 
322' 



10. 



10 /3z + 5y = 9, . 
13 ' \102,-7*=-8. 



11. 



2x = 3y + 12, 

8 = 0. 

9, 
H- Jy - 

x 1 



6. 
9. 



10, 
= 6z 4- 14. 



-I- 2x - - 3, 
-f 2s - - f 



12. 



2z 

x 
2" 



14. 



= - 3, 
11* + 5y = - 15. 



15. 



7 
3 



- y = 8, 
7x + 4y = 43. 



Proceed with the solution until you recognize that the equations are incon- 
sistent or dependent. Then check by graphing the equations. 






2* - 



- 10. 



* 



- 7 = 



19. < 

1 00. Elimination by substitution 

EXAMPLE 1. Solve for x and y: 
SOLUTION. 1. Solve (2) for x: 



4x 



- 6, 
= 4. 



2. Substitute * - J(4 - 3y) in (1): 

4(4 - 



- 6. 



- 2y - 3 = 0, 
6. 



(1) 
(2) 

(3) 
(4) 



In obtaining equation 4, we have eliminated x by substituting for x from 
one given equation into the other. 

3. Solve (4) for y: 



4. Substitute^/ = 2 in (3): 

* - 1(4 - 6) - - 1. 
Hence, the solution of the system is (x = 1, y - 2). 



SYSTEMS OF LINEAR EQUATIONS 



135 



SUMMARY. To solve a system of two linear equations by elimination 
by substitution: 

1. Solve one equation for one unknown in terms of the other and sub- 
stitute the result in the other equation. 

2. Solve the equation obtained in Step 1 for the second unknown. 

3. Substitute the value of the second unknown in any equation in- 
volving both unknowns and find the value of the first unknown. 



1. 



EXERCISE 50 

Solve by elimination by substitution and check. 
x = 3y - 1, ^ / 2x + y = - 3, 

- y - 15. 



- 3 = 4. 



fw = 2t; + 4, 
\ 100 - 2u - 1. 



I i 

L I 

f 
\ 



u w + 1, 
f 2w> - 0. 



5. 
8. 



- x = 0, 
4.y = 0. 



= -3, 
5x-2y = 21. 



6. 



9. 



-f 3x + 4 = 0, 
x + 2y = 0. 



y = 14. 



3z + 5y = 0. 
10-15. Solve Problems 1-6 of Exercise 49 by substitution. 

Clear of fractions if necessary and solve by any method. Do not restrict 
your choice to just one of the two available methods. 

<^Js m ^" tJ 1/ " ^\ ^^ I \JJb mm ^ m 9J u m **^ ^J * ^ ^*. I is i "*T^ JL^Xo r * ^*!^^t 

i^wv \J ^^ f ^ ^y I ^*~*r ij ^^ f * ^f J j 

2x - 7y = 3. * \ 4y - 9x = 5. " \ 6r + 21s = - 7. 

- 4y = .5, 

== .O. 



- 1.5. 



~.6a; = 3.45, 

~~"~ "~ O I 



22. 



25. 



04 



Q ^7 2 I 

= i 



26 



27. 



- 2y 4- 3 



- J = 0, 



' 



-? = o. 



* H- Jy = A- 

2x + 5y 



29. 



*-2 2-f y 



' 



4 



IZ + i 

^2 



3a? + y + 4 . 3 _ 
10 "^5 



- 1 - 



1 _ 



8 - y - te 



- 2 



3x 



- 3 



136 SYSTEMS OF LINEAR EQUATIONS 

101. Systems involving literal coefficients 

If a system involves other letters than the unknowns, it is usually 
best to solve by finding each unknown in turn by elimination through 
addition or subtraction. 

T> i c i / j ( ax + by = e t (I) 

EXAMPLE I. Solve for x and y: < . ' , ;; 

\cx-\-dy = f. (2) 

SOLUTION. I. Multiply (I) by d: adx + My = de. (3) 

2. Multiply (2) by b: bcx + bdy = of. (4) 

3. Subtract, (3) (4) v x(ad be) = de bf. (5) 

4. Suppose that ad be 7* and _ de bf 
divide by ad be in (5). ad be 

5. By similar steps [multiplying (I) _ of ce 
by c and (2) by a] we find y. ad be 

1 02. Systems linear in the reciprocals of the unknowns 

- + - = 17, (1) 

x y 

EXAMPLE 1. Solve: { 

- + - = 2. . (2) 
c y 

10 ^ 
SOLUTION. 1. Multiply (2) by 5: + - = 10. (3) 

x y 

2. Subtract, (3) - (1): - = - 7; 7 = - 7x; x = - 1. 

X 

3. Substitute x = - 1 in (1) : - 3 + - = 17; y = ~ 
The solution of the system is (x 1, y = i). 

Comment. Equation 1 is said to be linear in l/x and l/y because, if 
we let u = l/x and v = l/y, equation 1 becomes 3u -j- 50 = 17. Similarly, 
equation 2 becomes 2u H- t> 2. In place of the preceding solution, we 
could first solve for u and v; then their reciprocals would give x and y. 



EXERCISE 51 

Solve for the literal numbers without first clearing of fractions. 



1. 



3 - 5 - 17 
- 17 ' 



x 



y 



3. 



y 



2 4 

_ 4- - 

5w ^ y 

3 4- 5 

u + 2i 



SYSTEMS OF LINEAR EQUATIONS 



137 



x y 



6. 



(x y 
Solve for x and y,orforw and z. 



M-H, 



-4. 



6. 



10 J_ 9 

-- h - = 
u v 



u 



ax - 2y = 2 + 6, 
= 2 - 26. 



7. 



= a -j- 6, 
*' 1 2abx - aby = a? - 6 2 . 

11. / ??* + ? = * 12. 



8. 
10. 



2cz dy = c 2 + d 2 , 



-f y = 2c. 

610 as 6 s = 0, 
aw; 4- bz bw = a 2 . 



ax + by = 3, 



14. 



2aw 



= 4a 2 + 6 2 , 



16. 



\ w> - 22: = 2a - 6. 



= 2a_ 6 

26 a 6 a* 



f 2aw + 62 = 06, 
oi/ = 3. *"" \ w - bz 3a6 + 26. 

aw + bz = a 2 + 6 s , 



15. 



6w> - a = a 2 + 6 2 . 



17. 



_ 

2a 



1. 



2m 


n 








, 


2n + 3j 


1 m x 




m 


2n 




n + x 


2m - 3y 


. 



103. Solution of a system of three linear equations 

A system of three linear equations in three unknowns usually has 
one and only one solution. In special cases, however, such a sys- 
tem may have no solution, in which case the equations are called in- 
consistent, or infinitely many solutions, in which case the equations are 
called dependent. Such cases will not be considered in this book.* 



EXAMPLE 1. Solve for x, y, and z: 



3z + y - z = 11, 
x + 3y - z = 13, 
x + y - 82 = 11. 



2x - 2y = - 2. 
+ 9t/ - 82 = 39. 

2x -f 8y = 28. 



(1) 

(2) 
(3) 

(4) 
(5) 
(6) 



= 30; y = 3. 



SOLUTION. 1. Subtract, (1) (2): 
Multiply (2) by 3: 
Subtract, (5) - (3) : 

2. Solve (4) and (6) for x and y: 
Subtract, (6) - (4) : 

Substitute y = 3 in (4): 2x - 6 = - 2; x = 2. 

3. Substitute (x = 2, y = 3) in (1) : 6 + 3 - z = 11; z = - 2. 
The solution of the given system is (x = 2, ?/ = 3, z = 2). 

* For a more complete treatment, see College Algebra, Third Edition, by 
WILLIAM L. HART; D. C. HEATH AND COMPANY. 



738 



SYSTEMS OF LINEAR EQUATIONS 



SUMMARY. To solve a system of three linear equations in three un- 
knowns: 

1. From one pair of the equations, eliminate one of the unknowns; 
eliminate this unknown from another pair of the original equations. 

2. Solve the resulting equations for the two unknowns in them. 

3. Substitute the values of the unknowns found in Step 2 in the 
simplest of the given equations and solve for the third unknown. 

irNote 1. To solve a system of four linear equations in four unknowns, 
first we would obtain three equations in three of the unknowns by eliminating 
the other unknown from three different pairs of the original equations. 
Then, we would solve the new system of three equations and, later, obtain 
the value of the fourth unknown. A similar but more complicated method 
would apply to systems in five or more unknowns. A more elegant method 
is presented in advanced college algebra. 



1. 



EXERCISE 52 

Solve. Do not commence by clearing of fractions. 

' 3y - 5x = 1, [ 2x + y = 2, 

2. \ 2y - 5z = 7, 
Qx + 2z = 1. 



10. 



3a? + - 1, 



3. 



' 12x - 



3, 



x y 22 = 1, 
- 2 = 0. 



' x - y + fa = 7, 
6. | 2x + 3y + 62 * 0, 
^ + 22/ + 92 = 3. 

' 2s - y + 2 = 2, 
8. < 12x -f y - 3 = 3, 
k 6x - y + 62 = 12. 



7 - 



+ 



= 3, 
= 2. 



5. 






x y z 

60 o 
l & m 

- H = 5. 

L* y z 



HINT for Problem 10. Let - 

x 



#> + 6c = 14, 
4c - 6 = 2 + 3a, 
14c - lOa - 96 = 10. 



7. { 2A + 3J5 - 2C + 1 = 0, 
.A+ 3B + 2C= 1. 



9. 



3e = 3, 
- 6z - 62 + 3 = 0, 

, 2 - y - x - 2. 



11. 



x y 2 

5+1--; 



* 



M, - = v, and - ~ w in all equations. 
y z 



SYSTEMS OF UNEAR EQUATIONS 139 



*12. 



5x y = 6, 

4y + * = 10, 



-f 62 - 3w = - 3. 
4s - y = - 2. 



*13. 



x + 2y + w = 4, 



2 - 3u = 6, 
- x 6 = - 2, 
+ - = - 2. 





104. Applications of systems of linear equations 

EXAMPLE 1. The sum of the digits of a certain two-digit number is 9. 
If the digits are reversed, the new number is 9 less than 3 times the original 
number. Find the given number. 

SOLUTION. 1. Let t be the tens' digit and u be the units' digit of the 
number. Then, the number is 10J -j- u. 

2. If the digits are reversed, u becomes the tens' digit and t the units' digit. 
The new number is IQu -f t. 



3. From the problem, \ 1n w ~ ' 

I lOtt 



t = 3(10< + u) - 9. 
We obtain t = 2 and u = 7. The original number is 27. 

EXAMPLE 2. Workmen A and B complete a job if A works 2 days and 
B works 3 days, or if both work 2f days. How long would it take each to 
do the job alone? 

SOLUTION. 1. Let x be the number of days it takes A to do the job 
alone, and y the number of days for B alone. 

2. The fractional part of the work which A does in one day is l/x and, 
for B, is 1/y. Since they do the whole job, under each set of given data, 

2,3 , , 12 1 , 12 1 t 

- + - = 1, and -= h -5 * 

x y o x 5 y 

On solving the system consisting of the preceding equations by the method 
of Section 102, we find x = 4 and y = 6. 

Any line on an (x, y) coordinate plane which is not parallel to the 
y-axis has an equation of the form y mx + 6, where m and 6 are 
constants. We can use this fact to obtain the equation of a line 
through two given points. 

ILLUSTRATION 1. To find the equation of the line through (4, 3) and (6, 9), 
we substitute each pair of coordinates for (x, y) in y = mx + 6: 

(when x 4 and y = 3) f 3 = 4m + 6, (1) 

(when x = 6 and y 9) \ 9 = 6m -f b. (2) 

The solution of [(1), (2)] is (m = 3, b = - 9). Hence, the equation of the 
desired line is y - 3x 9. It can be verified by graphing or by substitution 
that the given points lie on the graph of this equation. 



140 SYSTEMS OF LINEAR EQUATIONS 

EXERCISE 53 

Solve by introducing two or more unknowns. 

1. One angle of a triangle is 30 and a second angle is four times the 
third angle. Find the unknown angles of the triangle. 

2. The width of a rectangle exceeds its length by 5 feet and the per- 
imeter of the rectangle is 25 feet. Find the dimensions. 

3. Two angles are complementary and one exceeds the other by 7. 
Find the angles. 

4. A contractor has a daily payroll of $73 when he employs some men 
at $6 per day and the rest of his workers at $5 per day. If he should double 
the number receiving $5 and halve the number receiving $6 per day, his 
daily payroll would be $74. How many employees does he have? 

5. How much of a 20% solution of alcohol and how much of a 50% 
solution should be mixed to give 8 gallons of a 30% solution? 

6. How much milk containing 2% butterfat and how much contain- 
ing 6% butterfat should be mixed to form 100 gallons of milk containing 
3% butterfat? 

7. If each dimension of a rectangle were increased by 5 feet, the area 
would be increased by 95 square feet and one dimension would become 
twice the other. Find the original dimensions. 

8. If a two-digit number is divided by its units' digit, the result is 16. 
If the digits of the given number are reversed, the new number is 18 less 
than the original one. Find this number. 

9. A weight of 5 pounds is 6 feet from the fulcrum on the right-hand 
side of a lever. It is balanced if we place a first weight 4 feet from the 
fulcrum on the right and a second weight 7 feet from the fulcrum on the 
left, or if we place the first weight 8 feet to the right and the second weight 
9 feet to the left of the fulcrum. Find the unknown weights. 

10. If we seat a boy at 5 feet and a girl at 8 feet from the fulcrum on 
one side of a teeterboard, they balance a man weighing 160 pounds who 
is seated 6 feet from the fulcrum on the other side. Balance is maintained 
if the boy moves to 8J fe6t and the girl to 4 feet from the fulcrum o'n their 
side. Find their weights. 

11. When we divide a certain two-digit number by its tens' digit, the 
result is 13. If we reverse digits in the number and then divide by the 
original number, the result is 31/13. Find the original number. 

12. The sum of the reciprocals of two numbers is 1/2 and the difference 
of the reciprocals is 1/6, Find the numbers. 



SYSTEMS OF LINEAR EQUATIONS 

13. How much silver and lead should be added to 100 pounds of a mix- 
ture containing 15% silver and 30% lead to obtain an alloy containing 25% 
silver and 50% lead? 

14. How much chromium and nickel should be added to 100 pounds 
of an alloy containing 5% chromium and 40% nickel to give an alloy con- 
taining 15% chromium and 50% nickel? 

15. A man divides $10,000 among three investments, at 3%, 4%, and 
6% per annum, respectively. His annual income from the first two in- 
vestments is $80 less than his income from the third investment and his 
total income is $460 per year. Find the amount invested at each rate. 

16. Workmen A and B complete a certain job if they work together 
for 6 days or if A alone works for 3 days and B alone works for 10 days. 
How long does it take each man to complete the job alone? 

17. In a three-digit number which is 31 times the sum of the digits, 
the units' digit is one half the sum of the other digits. If the digits are 
reversed, the new number obtained is 99 greater than the original number. 
Find its digits. 

Find the equation of the line through the given points on an (x, y) coordinate 
system by solving a pair of equations for two unknowns, or by inspection. 

18. (2, - 3); (4, 3). 19. (- 3, 1); (- 2, - 3). 
20. (3, - 2); (- 3, - 12). 21. (- 4, 5); (8, - 4). 
22. (- 3, 5); (- 3, - 2). 23. (4, - 2); (9, - 2). 

24. An airplane, flying with the wind, took 2.5 hours for a 625-mile 
run and took 4 hours and 10 minutes to return against the same wind. 
Find the velocity of the wind and the speed of the airplane in calm air. 

25. An army messenger will travel at a speed of 60 miles per hour on 
land and in a motorboat whose speed is 20 miles per hour in still water. 
In delivering a message he will go by land to a dock on a river and then 
on the river against a current of 4 miles per hour. If he reaches his des- 
tination in 4J hours and then returns to his starting point in 3J hours, 
how far did he travel by land and how far by water? 



CHAPTER 



9 



EXPONENTS AND RADICALS 



105. Proofs of the index laws 

We have already employed the following results, called index laws, 
which govern the use of positive integral exponents. 

1. Law of exponents for multiplication: a m a n = a m+n . 

Proof. 1. By definition, a m a- a a; (ra factors) 

a n = a a a a. (n factors) 

2. Hence, a m a n = (a-o a)(a*a-a a) = a m+n . 

\_(m 4- n) factors a] 

II. Law for finding a power of a power: (a m ) n = a mn . 

Proof. 1. (a m ) n = a m -a m a m ; (n factors a m ) 

(By Law I) = a m+m+ ' * +m . (n terms m) 

2. Since (m + m 4- + m) to n terms equals mn, (a m ) n = a mn . 

III. Laws of exponents for division: 



= a "(ffm>n); _ = _ 



ILLUSTRATION 1. -5 = a 6 . 15 " ~V 

d Or a 

Proof, for the case n > m. By the definition of a m and a n , 

a m (m factors) 



o" ** a -a- -a-fil'ji- $' (n factors) 
1 1 



_____ 

n m) factors a] a*a a a 



n-OT 



EXPONENTS AND RADICALS 143 

IV. Law for finding a power of a product: (ab) n = a n b". 
Proof. (ab) n = ab-ab afc; (n factors ab) 

(n factors a, and b) = (a a - a) (b b b) ~ a n b n . 

Law IV extends to products of any number of factors. Thus, 

(abc) n = a n b n c n . 

ILLUSTRATION 2. (4a 2 6 4 ) 3 - 4 3 (a 2 ) 3 (6 4 ) 3 = Q4a*V*. 

(a\ n a n 
h) ^ M* 

T /a\ 4 a 4 /a 2 \ 2 (a 2 ) 2 a 4 

ILLUSTRATION 3. = . - 



D r /^ n a a a ( , a\ 

Proo/. ^) ^6T"6 ; (n factors^ 

(n factors a) _ a-a - a __ a^ 

(n factors 6) 6 6 6 b n 

Note 1. The determination of powers of numbers is called involution. 



EXERCISE 54 

Find each power by use of the definition of an exponent. 
1. 2 6 . 2. (- 5) 4 . 3. (- 3)*. 4. (.I) 4 . 6. (j) 2 . 6. (- ). 

7. What is the sign' of an odd power of a negative number? 

Perform the operations by use of the index laws. 
8. a 6 a 8 . 9. zV. 10. 2 3 2. 11. (x 3 ) 5 . 

12. (3z) 4 . 13. (2a 2 ) 6 . ' 14. (46) 8 . 15. (5afy) 4 . 

16. ddW. 17. (- 2z J ). 18. (- fy 4 )'. ' 19. (- 2a) 4 . 

20. (b*)\ 21. (a*) 2 . 22. (c*) 3 . 23. (d 2 )**. 

24. (o6 2 ). 26. (cd)*. 26. (- .2a 2 6) 3 . 27. (- . 

- 



er- 



144 EXPONENTS AND RADICALS 



- (' 

_ '. 62. (^'?. 63. 

\- yz 



64. Prove that part of Law III which applies if m > n. 

66. (a) Compute ( 2) 4 and 2 4 . (6) Under what condition on the 
positive integer n will ( 3) n 3 n ? 

106. Imaginary numbers 

We have called R a square root of A if R 2 A. If A is positive, 
it has exactly two square roots, one positive and one negative, denoted 
by =t VA. 

ILLUSTRATION 1. The square roots of 4 are db Vi or =t 2. 

If a negative number P has R as a square root, then R 2 P. 
But, if R is either positive or negative then R z is positive and thus 
cannot equal P. Hence, P has no positive or negative square root. 
Therefore, in order that P may have square roots, we define the 
symbol V P as a new variety of number, called an imaginary num- 
ber, with the property that 

(V^P)* - - P and (- V^) 2 = - P. 

Thus, P has the two imaginary numbers V P as square roots. 
As an immediate extension of the preceding terminology, we agree 
that, if M is a real number, each of the expressions (M + V P) 
and (M V P) will be called an imaginary number. 

ILLUSTRATION 2. The square roots of the negative number 5 are the 
imaginary numbers =fc V 5. (7 + V 18) is an imaginary number. 

Unless otherwise stated, any literal number in this book will 
represent a real number. Imaginary numbers will not enter actively 
into our discussions until we meet them in the solution of equations 
in a later chapter. 



EXPONENTS AND RADICALS 145 

Note 1. The somewhat unfortunate name imaginary number is inherited 
from a time when mathematicians actually considered such a number to 
be imaginary in the colloquial sense. In a similar fashion, our common 
negative numbers, at their first introduction into mathematics, were also 
called illusory or fictitious. The student will soon appreciate that imaginary 
numbers deserve consideration on an equal footing with real numbers. 
Imaginary numbers are indispensable not only in pure mathematics but 
also in important fields where mathematics is applied. Imaginary numbers 
will be studied in more detail in a later chapter. 

107. Roots 

We call R a square root of A if R 2 = A and a cube root of A if 
R z = A . If n is any positive integer we say that 

R is an nth root of A if R" = A. (1) 

ILLUSTRATION 1. The only nth root of is 0. 2 is a 5th root of 32 because 
2 5 = 32. - 3 is a cube root of - 27. 

The following facts are proved in college algebra when imaginary 
numbers are treated in a complete fashion. 

1. Every number A, not zero, has just n distinct nth roots, some or all 
of which may be imaginary numbers. 

2. If n is even, every positive number A has just two real nth roots , 
one positive and one negative, with equal absolute values. 

3. Ifn is odd, every real number A has just one real nth root, which is 
positive when A is positive and negative when A is negative. 

4. // n is even and A is negative, all nth roots of A are imaginary 
numbers. 

If A is positive, its positive nth root is called the principal nth root 
of A . If A is negative and n is odd, the negative nth root of A is called 
its principal nth root. 

ILLUSTRATION 2. The real 4th roots of 81 are =fc 3 and + 3 is the principal 
4th root. The principal cube root of + 125 is + 5 and of 125 is 5. 
All 4th roots of 16 are imaginary numbers. 

ILLUSTRATION 3. The real cube root of 8 is 2. Also, by advanced methods 
it can be shown that 8 has the two imaginary cube roots ( 1 + V 3) and 



146 EXPONENTS AND RADICALS 

108. Radicals 



The radical A, which we read the nth root of A , is used to denote 
the principal nth root of A when it has a real nth root, and to denote 
any convenient * nth root of A if all nth roots are imaginary. In 
v^L, the positive integer n is called the index or order of the radical, 
and A is called its radicand. When n 2, we omit writing the index 
and use VA instead of \^A for the square root of A. 



I. A is positive if A is positive. 

II. ^A is negative if A is negative and n is odd. 

III. v~A is imaginary if A is negative and n is even. 

ILLUSTRATION 1. ^'sl = 3. ^32 = 2. v / 8 is imaginary. 



ILLUSTRATION 2. y^y = ^ because ^) = 3* = gj' 

By the definition of an nth root, 

(VA) n = A. (1) 

ILLUSTRATION 3. (V3)* = 3. v / (169) 7 = 169. (^ctf) 8 - 2cd*. 

(v'c 2 + cd + d 2 ) 4 = c 2 + cd + #. 



Note 1 . To avoid ambiguous signs and imaginary numbers in elementary 
problems, the following agreement will hold in this book unless otherwise 
specified. If the index of a radical is an even integer, all literal numbers in 
the radicand not used as exponents represent positive numbers, and are such 
that the radicand is positive. 



By the definition of A as a principal nth root, it follows that, 
under the agreement f of Note 1, 



= a. (2) 

ILLUSTRATION 4. ^5 = x. v'S 4 = 5. 



* This matter of convenience is discussed in more advanced treatments of 
imaginary numbers. If all nth roots of A are imaginary, it is not usual to call 
any particular one of them the principal nth root. 

f If a is negative and n is even, then a n is positive and the positive nth root 

of a n is a, or "^a* a. This case is ruled out by Note 1 and does not 
come under formula 2. For instance, if a is negative, Vo* = a. 



EXPONENTS AND RADICALS 747 

EXERCISE 55 

State the two square roots of each number. 
1. 64. 2. 49. 3. 81. 4. 121. 5. J. 6. &. 7. .01. 

State the principal square root of each number. 
8. 16. 9. 144. 10. 100. 11. &. 12. &. 13. 

State the principal cube root of each number. 
14. 8. 16. - 27. 16. 27. 17. 125. 18. - 1. 19. - 216. 20. - 

State the two real fourth roots of each number. 

21. 81. 22. 16. 23. 625. 24. 10,000. 26. A. 26. .0001. 



Find the specified power of the radical, or the indicated root. 
27. Vtf. 

31. V&. 
35. (v^) 6 . 
39. v^=~8. 
43. ^64. 
47. v^~T. 
51. V400. 
65. v'J. 
59. VX)1. 60. 

109. Rational and irrational numbers 

A real number which can be expressed as a fraction m/n, where 
the numerator and denominator are integers, is called a rational 
number. All integers are included among the rational numbers be- 
cause, if m is any integer, then m can be expressed as the fraction m/1. 
A real number which is not a rational number is called an irrational 
number. 

ILLUSTRATION 1. 7, 0, and f are rational numbers. Any terminating 
decimal fraction is a rational number. Thus, 3.017, or 3017/1000, is a 
rational number. IT and V% are irrational numbers. A proof of the irration- 
ality of V2 is given in the Appendix, Note 1. Any irrational number can be 
expressed as an endless but not a repeating decimal. Thus, v = 3.14159 
and V% - 1.414 are endless but not repeating decimals. 




148 EXPONENTS AND RADICALS 

If A is not the nth power of a rational number, and v^A is real, 
then VA is irrational and is called a surd of the nth order. 



ILLUSTRATION 2. V3 is a surd. v^64 is not a surd because v'd? 2. A 
surd of the second order is sometimes called a quadratic surd and one of the 
third order a cubic surd. The values of various quadratic and cubic surds 
are given to a limited number of decimal places hi Table I. 

110. Rational and irrational expressions 

An algebraic expression is said to be rational in certain letters if it 
can be expressed as a fraction whose numerator and denominator 
are integral rational polynomials in the letters. An algebraic expres- 
sion which is not rational in the letters is said to be irrational in them. 



T . - . ,. . . , 

ILLUSTRATION 1. ~ - = - is rational in a and x. 

oa 



ILLUSTRATION 2. V3x -h y is not rational in x and y. 

ILLUSTRATION 3. The expression x\ /r 2 3x 2 \/5 is an integral rational 

polynomial in x. The presence of irrational explicit numbers, A/2 and V5, 
is of no concern. 

Hereafter, unless otherwise stated, in any integral rational poly- 
nomial we shall assume that the numerical coefficients are rational 
numbers. 

111. Perfect powers of rational functions 

A rational expression is called a perfect nth power if it is the nth 
power of some rational expression. Also, we say that a rational 
number is a perfect nth power if it is the nth power of some rational 
number. 

If an integral rational term is a perfect nth power, the numerical 
coefficient separately is a perfect nth power. Also, each exponent in 
the term has n as a factor, because in obtaining the nth power of a 
term each exponent is multiplied by n. In verifying that a term is a 
perfect nth power, first factor the coefficient. 

ILLUSTRATION 1. 32y 15 is a perfect 5th power: 32y 15 = 2V 5 = (2I/ 3 ) 6 . 

X* 

ILLUSTRATION 2. ^r* is a perfect square: 

e ^ 



X 2 



^ 

9o 6 \3o 3 / 



EXPONENTS AND RADICALS 149 

112. Elementary properties of radicals 

The following Properties I and II have already been met in Sec- 
tion 108 as direct consequences of the definition of an nth root. 



II. Va n = a. (If a is positive when n is even) 

III. Vob = vWb. 

ILLUSTRATION 1. V&x? = VsVtf = 3z, because (3x) 2 = 9z 2 . 

ILLUSTRATION 2. \fab = ^fc^fb because 

(vWfc) 3 = (^a) 3 ^) 8 = 06. 

IV " 

1V * 




T o 4/81 81 3 , /3\ 4 3 4 81 

ILLUSTRATION 3. ^ = = -, because (- j = ^ = 



T , 4 u a 

ILLUSTRATION 4. = -r-> because -= = .' = _ 

6 



n / * 

V. // m/n is an integer, va m = a n . 

ILLUSTRATION 5. v^o^ = a 1 * 1 = a 4 , because (o 4 ) s * a u . 



i . The following proofs are complete if a and 6 are positive, be- 
cause then all principal roots involved are positive. The interested student 
may consider the possibility of negative values for a and 6. To complete 
the proofs, it should be demonstrated that in all cases the two sides of each 
formula in (III), (IV), and (V) are either both positive or both negative and 
hence are equal. 



Proof of (III). Raise (v) to the nth power: 

(V~ a ^b) n = (^a) n (^6) n = ab. 
Hence, by the definition of an nth root, ^a^fb is an nth root of ab, or 

n/- n/r n/ r 

Vav 6 = v ab. 



Proof of (V). By Law II, page 142, (a n ) n a"' n - a m . Hence, 

in m 

a" is an nth root of a m , or a" = Va"*. 



150 EXPONENTS AND RADICALS 

We observe that Property II is a special case of Property V, with 
m = n. However, we dignify Property II by special attention be- 
cause, if we are able to express A as a perfect nth power, Property II 
gives VA by mere inspection. 



ILLUSTRATION 6. V&xfiy 9 - ^(2xV) 8 = 2zV . (Property II) 

Or, by Properties III and V, 



EXERCISE 56 

Each radicand is a perfect power. Find the specified root. 
1. VP. 2. 3d*. 3. Vol. 4. Va*. 5. 



6. Vy. 7. . 8. **. 9. . 10. 

11. <&. 12. v^io. 13. ^x 5 . 14. Vy*. 16. 

16. V*2. 17. -^. 18. -^7?. 19. \/A. 20. 

21. tff. 22. ^Jf. 23. ^3fc. 24. ^p. 25. 

26. Vm*. .27. V/8E 5 . 28. ^^. 29. 

30. V^. 31. tfrV. 32. \/9^. 33. 

34. v/*. 36. ^- .001. 36. v 5 . 37. 



38. ^SE 1 " 2 . 39. ^- 32 10 . 40. ^0625. 41. 

42. ^I6a. 43. VSS 5 . 44. ^06. 46. 





113. Fractional powers 

We have previously defined a? only when p is a positive integer. 
We shall now introduce other types of powers in such a way that all 
the types, together, will obey laws of the same forms as those for 
positive integral exponents. 

If fractional exponents are to obey the law of exponents for multi- 
plication, then, for example, 

akcfr o^ a, or (afy a; 

thus, a& must be a square root of a* Similarly, we should have 

- a 6 , or (a$) 2 a 6 , 



EXPONENTS AND RADICALS 151 

so that a$ should be a square root of a 5 . Accordingly, if m and n are 

m 

any positive integers, we define a n to be the principal nth root of a m : 

m 

a" = -v/a 1 "; (1) 



i 



[when m = 1 in (1)] a n = v^a. ^ (2) 

Thus, we may use fractional exponents instead of radicals to denote 
principal roots. The defining equation 1 is consistent with Property V 
of page 149, which was proved for the case where m/n was an integer. 

ILLUSTRATION 1. 8* = v^ = 2. (- 8)* = v^"8 = - 2. 



ILLUSTRATION 2. x = v. 8* = \& = te = 4. 

ILLUSTRATION 3. (- 8)* = v / (- 8) 2 = vlJ4 = 4. 



m 



In (1), we defined a n to be the nth root of a m . It is very useful to 

m 

prove the theorem that, also, a n is the mth power of the nth root of a, or 

m 

a" = (v^)". (3) 

To prove (3) we must show that the right members of (1) and (3) are 
identical, or that 

. (4) 



ILLUSTRATION 4. To show that \/aJ = ('^a) 4 , raise the right member 
to the 3d power and use the laws for integral exponents : 

[(>^)4]3 = (^)i2 = [t^) 3 ] 4 = (a) 4 - a 4 . 

Hence, (^a) 4 is a cube root of a 4 . If we assume that a is positive, then this 
cube root is positive and hence is identical with the principal cube root 
represented by v^o 4 . 



In accordance with Note 1 on page 146, we agree not to deal with 

m 

the symbol a" if n is evenjand a < 0. With this case eliminated, we 
prove (4) by raising the right-hand side to the nth power, showing 
that we obtain a m , and thus demonstrating that (\^a) m is an nth 
root of a m : 

~~ - a*. 



152 EXPONENTS AND RADICALS 

ILLUSTRATION 5. From (3), since \/64 = 2, 

64$ = (^64) 6 = 2 6 = 32. 

Notice the relative inconvenience of the following evaluation by use of (1) 

64* = ^oT 6 = &(* = v^ = 2 6 = 32. 



if Note 1 . The difficulties^ met if a n is used when a is negative and n is eve 
are illustrated below where a contradiction " + 1 = 1 " results from reel 
less use of the symbol ( 



- 1 = V^I = (- l)i = (- 1)* = v'Pl)' 2 = V 



114. Zero as an exponent 

If operations with a are to obey the law of exponents for multiplies 
tion, then we should define a so that 

a n 
aa n = a+ n = a n , or aa n = a n , or a = = 1. 

' ' a n 

Hence, if a ^ 0, we define a by the equation a = 1. 
ILLUSTRATION 1. By definition, 5 = 1. (17z) = 1. 

115. Negative exponents 

If a negative exponent is to obey the law of exponents for mul 
tiplication, then, for instance, we should have a 3 cr 3 = a 3 " 3 = a = 1 
Hence, if p is any positive exponent of the types previously intro 
duced, we define or* by the equation a p a~ p = 1, or 



ILLUSTEATION 1. -- Br - I _ 1. 

In a fraction, any power which is a factor of one term (numerate 
or denominator) may be removed if the factor, with the sign of it 
exponent changed, is written as a factor of the other term. That is 

a _ ax~ n 
bx ~ b 

n f a a I a _^ ax"" 
Proof. 7 T == r ar* = r 
' bx n b x n b b 



EXPONENTS AND RADICALS 153 

We may use negative exponents in avoiding fractions. 

ILLUSTRATION 2. -rj- = 17o TJ = 17o6~ 2 . 

ILLUSTRATION 3. To express the following fraction wityi positive ex- 
ponents, we may use (2) mechanically: 

3a" 2 6 3 _ 3c*b 3 _ Sc 3 ^ 3 fQ . 

IF 3 ^ 4 " = ~oJoJ = ~HT' ( ' 

In more detailed fashion we obtain (3) as follows: 

Q/7~ 27)3 /I \ / 1 \ O7)3 ^3 2J)3/^ 

Otv C/ / ^ JL i I / j -*- 1 OC/ O Ot/ O x *v 

Pw "r*-*')-'-^-?)--*-;?--*-- (4) 

The student should act as hi (3) but should also appreciate (4). 

EXERCISE 57 

Find the value of the symbol by changing from a fractional exponent to a 
radical or from a negative to a positive exponent if necessary. 

6. 4- 1 . 
10. 81*. 
16. (i)*. 
20. 16-*. 
25. (.36)*. 
'. 30. (- S)- 4 . 

31. (- 2)~ 6 . 32. (- 5)- 3 . 33. (- 1)*. 34. (- 8)*. 35. (- 8)-*. 
36. (- 27)-*. 37. (- 125)*. 38. (.0081)*. 39. (.0001)-*. 

Find the value of the symbol by use of formula 3, page 151. 
40. 8*. 41. 25*. 42. 4*. 43. 36*. 44. 81*. 

45. 125*. 46. ft)t. 47. (^)*. 48. (- 27)*. 49. (- 64)'*. 

Express with positive exponents. 



1. 9*. 


2. 25*. 


3. 8*. 


4. 27*. 


6. 2~ J . 


7. 35- 1 . 


8. 16. 


9. 4*. 


11. 3-2. 


12. 5~ 3 . 


13. S- 4 . 


14. 16*. 


16. (*)*. 


17. GM*. 


18. (- 8)*. 


19. 7. 


21. 9~*. 


22. 27-*. 


23. ft)-*. 


24. ()-' 


26. (.09)*. 


27. (- S)- 2 . 


28. 725. 


29. (- 5 



60. a~ 3 . 


51. b~*. 


62. c-. 


KS f t r ^ / ii 

i/W ftls t^ 


54. a~ 3 6 2 . 


66. c 2 d~ 8 . 


66. 62T 2 . 


67. 3A~ 4 . 


68. 2o6- 3 . 


69. 4x~*y. 


60. arV 5 ' 


61. 2oy~ 5 . 


62. Zax-*y-*. 


63. 4- 1 aar 3 . 


64. 5~ 2 ca;- 3 . 


66 


_. 3 

miljm 


67 


a 3 


69 


* b~~* 


f? 


a~ 8 


rf-2* 


' 5 



754 EXPONENTS AND RADICALS 



6a~ 8 
75.^- 76. ^r- 77. ^br 78. 



Write without a denominator by use of negative exponents. 

80. i- 81. i- 82. - 83. - 4 - 84. ^ 

y* & x* y 4 a* 

4a M 2a ftft &c* 

OO. -.i oft O v 



90. /- rvo\K* 8L 



(L03)" 8 (1.04)" "* (1.05) 8 

Rewrite, expressing each fractional power as a radical and each radical as a 
fractional power. 

94. x*. 95. z*. 96. a*. 97. &i 98. 3ai 

99. 5c*. 100. ax*. 101. fcci 102. v/^ 3 . 103. v/6. 

104. (56)*. 105. (6c)*. 106. Vy>*. 107. (2^)*. 108. (4C 3 ) 9 , 

109. (7a')i 110. (2a^)t. 111. <Q>. 112. ^fe 11 . 113. >/5?. 

114. -C^HM. 116. Vo 2 - 36. 116. #(a 4- 6) 2 . 117. ^(c - 3d) 3 . 

118. Va 2 -f 6 2 . 119. ^a 8 - fe 8 . 120. ^ST^ 3 . 121. vT^ll 
122. Compute (- i)~ 8 ; (- J)-; (- .04)~. 

116. Extension of the index laws 

We have defined a p if p is any rational number but we have proved 
the index laws only for positive integral exponents. A detailed 
discussion, which we shall omit (see Appendix, Note 2), shows that 
the formulas of Laws I to V of pages 142 and 143 apply if the ex- 
ponents are any rational numbers. 

Hereafter, unless otherwise specified, to simplify an expression 
involving exponents will mean to perform indicated operations as far 
as possible by use of Laws I to V, and to express the result without 
zero or negative exponents. Moreover, unless specifically requested, 
we shall not introduce radicals for powers having fractional exponents. 

In operations involving exponents, it is frequently convenient to 
express the numerical coefficients of terms as products of powers of 
prime factors. 



EXPONENTS AND RADICALS 755 

ILLUSTRATION 1. (x*)$ - aH = x 4 . 



/21te\* /8-27afa\* = /23 8 x\* 2*3*3* 
\125arV V 125 / V 5 8 / " 5* "" 25 



" 1 i 1 !"* 1 1 



/ 1 Y" 17 iV 

(-125) -Lrs/ 

EXAMPLE 1. Simplify: 



.-a 

^f 

FIRST SOLUTION. Change to positive exponents: 

J_ J_ 

1 q*y* 



1/ 2 



a* y 2 a V 

SECOND SOLUTION. Multiply both numerator and denominator by 
to eliminate negative exponents : 

(<rV*)(oV) 



(a- 2 + if*) 

1 



+ ay t/ 2 + a 2 

We may use the special products of Chapter 5 in multiplying or 
factoring polynomials involving negative or fractional exponents. 



ILLUSTRATION 2. (ar 2 y$)(x~* + y$) (Type II, page 85) 



ILLUSTRATION 3. (z* - 2?/- 1 ) 2 (Type IV, page 85) 



ILLUSTRATION 4. 2x- 2 + z" 1 - 6 = (2s;- 1 - 3)(xr l + 2). 



EXERCISE 58 

Simplify and, if no letters are involved, evaluate. 

1. a&e 2 . 2. a&c*. 3. aAr 4 . 4. a s a. 5. (a*) 4 . 

6. (3 4 )t. 7. (2)1 8. (x*) 4 . 9. (4z 2 )*. 10. (3x~^. 

11. (5a~ 2 ) 8 . 12. (cr 1 * 2 ) 8 . 13. (5)i 14. (*t)t. 16. (o*)*. 

16. (ax- 1 ) 4 . 17. (a*)- 1 . 18. (a 2 )" 2 . 19. (&)-*. 20. (a")-. 

21. (2X- 2 ) 8 . 22. (5a-) 2 . 23. (oV)-*. 24. 



?56 EXPONENTS AND RADICALS 

26. (3x-y) 2 . 26. (5arV 3 ) 2 . 27. (Gar 1 !/- 3 )- 2 . 28. 

29. 32i 80. 271 31. 125*. 32. 216i 

33. (4-V)t. 34. (a:- 4 !/ 2 )*. 35. (27a- 8 a*)i 36. (25ar 2 )-*. 



2-5 a \l n 1 

VT . Qfl - QQ q 

of. r* oo. TT: w rr 1 



f ^ ;j~ 44. ^- 46. -^. 46. 



47.^. 48.^. 49. ^S. 60. 



5L -^f-s- 62. i~ 63. 



67. PT-r - 68. 



69. (27w)i 60. (32a 6 6 6 )*. 61. (125x 6 )^. 62. (216ar 6 )i 



Simplify to a single fraction in lowest terms without negative exponents. 
63. a- 1 + b~ l . 64. 3a~ 2 + 6. 66. a^ - Zr 3 . 66. 5a~ l + &-' 

A7 a ~ lb Aft ^-^ AQ < !r V 2 7n S- 2 -4-2 

67 ' a -l + ' ** /,-2 J.A* 69 ' x^J_,r-2* 70 - 



rt -l J_ A-l 4-1 _ a - 

71. " + L- 72. T^ fl _.. 73. " " 74. 



a -2 _ 5-2 4-2 _ a -2 a - 8 _ 6 -- 3-2 + ^-2" 

76. (Sat)- 1 . 76. (c + M)' 1 - 77. (cr 1 + b^}~ 1 . 78. (4a~ 8 - &)->. 



Expand and simplify, without eliminating negative exponents. 
79. (ar 1 - j/- 1 )^- 1 + 2/" 1 ). 80. (3a - 6' 



81. (4x - y)(4a; + y). 82. (jg - 

83. (xi + y*)(a:* - y*). 84. (2xi - 3)(3x* + 1). 

86. (5a* - 2)(3ai + 4). 86. (a* - 26*)(a* + 36*). 

87. (<r + 6) 2 . 88. (or 1 + 3) 2 . 89. (a* + Z>*j 2 . 
90. (2a* + 36ty. 91. (a' 1 + y 2 ) 2 . 92. (x 4 - 2?/- 1 ) 2 . 
93. (a- 1 + 6) 3 . 94. (2 + ar 2 ) 3 . 96. (3 - y~ 1 ) 3 . 

2 + 4). 97. (3 - 6arH2 



EXPONENTS AND RADICALS . 157 

98. (z 3 V*)i 99. (3a;*) n . 100. (3*a*6 n )*. 101. (4a~*6 n ) m . 



/ a 6 * \* * / 4z 4 * \* /.125;r 9 \3 

-- 



/. 

102 - to 103 - 455= l04 - (* 



106. (a - 6*) (a 2 + a6* + 6). 106. (c* + <*)(< - <*d + 

if Factor into two factors involving fractional or negative exponents. When 
possible, factor as the difference of two squares or as a perfect square. 



ILLUSTRATION 1. x y = 

Or, x - y = (a*) 3 - (yty = (x* - y*)(x* + zfyi + 

107. a; 2 - jf. 108. a~ 2 - 6 4 . 109. Oar 2 - 

110. ^ - x*. 111. 4x^ - jft. 112. 9a* - 

113. 9x* - 25^. 114. 16a* - 49y*. 116. 4o - 96. 

116. x* - Zxy- 1 + tr 2 . 117. 2 - Gzor 1 + 9x~ 2 . 

118. 4a~ 2 - 4a~ 1 6~ 1 + b~\ 119. 9o~ 2 - tor*-* -f b~*. 

120. 4ot - 20ai6V + 256*. 121. 

122. a:- 2 - 4X- 1 - 5. 123. 

124. - tyf . 125. Sa + 276. 126. e - By. 127. 216 - 

if Find the quotient by use of factoring. 

128 . - r> 129 . uc!. 130 . 

- - 



117. Simplification of a radicand 

Although it is possible to express a radical as a power with a frac- 
tional exponent, in some operations it is convenient to retain the 
radical form. This remark applies in particular to the following 
operation. 

SUMMARY. To remove factors from the radicand in a radical of 
order n: 

1. Separate the radicand into factors of which as many as possible are 
perfect nth powers. 

2. Find the nth root of each perfect nth power and express the final 
result by use of the property 'Vab = v^v^S. 



158 EXPONENTS AND RADICALS 

ILLUSTRATION 1. A/147 = A/49-3 = A/49A/3 = 7A/3; 

A/147 = 7(1.732) = 12.124. (Table I) 



T o 3 /o , 5 3/3az 3 

ILLUSTRATION 2. \/3a + -7 = \; - 

\ x 3 \ x 



Hereafter, unless otherwise specified, if a radicand involves frac- 
tions, reduce it to a single fraction. If a radical is of order n, simplify 
the radicand by removing from it any factor which is a perfect nth 
power. Also, in a radical of odd order, change the radicand to a 
form where all signs are " + " if possible. 



ILLUSTRATION 3. v / ^"2 = v / ^~Iv / 2 = - 1 

v'- fl - 26 = v'- (a + 26) = v'^Tv'a + 26 = - v'a + 26. 

In a sum, two or more terms involving the same radical as a factor 
may be combined by factoring. 

ILLUSTRATION 4. 5V/3 + 26\/3 = (5 + 26)\/3. 

V20 -f 2\/45 = V5\/5 + 2V9A/5 = 2\/5 + 6^5 = 8>/5. 
(\/3 V7 + 3V^5) cannot be simplified in form. 

ILLUSTRATION 5. 7\/5 - 3V5 + 6\/5 = V5(7 - 3 + 6) = 10\/5. 
(Using V5 = 2.236 from Table I) = 10(2.236) = 22.36. 

EXERCISE 59 

Simplify the radical and then compute by use of Table I. 
1. A/18. 2. A/75. 3. V20. 4. A/24. 6. A/200. 

6. A/500. 7. A/27. 8. A/108. 9. A/72. 10. 



11. A/5. 12. A24. 13. A6. 14. A54. 16. v108. 

16. AV3T3. 17. xVZTs. 13. A^~l6. 19. A^^54 20. 

Simplify by removing perfect powers from the radicand. 
21. Vx\ 22. Vo*. 23. \Vy*. 24. \V5. 26. v^io. 2 6. 

27. A/9o*. 28. A/8^. 29. v^. 30. 



32. A20o*. 33. V2. 34. 



EXPONENTS AND RADICALS 



?59 



36. 4 
39. 
43. <> 




47. v^ 



61. 



* ^V12V 
69. V9 + 92/ 2 . 
62. 
66. 



36. V75?. 
40. V- 27a 8 . 

44. VsiaV 1 ' 
48. VxJla 5 . 

62. 




37. 
41. 
46. 
49. 

63. 




38. V375J 5 . 
42. ^ - 128x. 
46. V/320V 5 . 
60. 

64. 




66 - V- |i- 

60. V4 - 4o 2 . 



--S- 68. 




63. 



61. 
64. 



+ 5a 2 6. 



66. 



67. 



68. 



2d , d 

' ** * 

ab ^ a 2 



Simplify and then collect terms, exhibiting any common radical factor. 
71. 5A/2 + 3A/2. 72. 3\/3 - a\/3. 73. 2\/l8 + V50. 

74. Vl2 + V75. 76. Vl47 - VS. 76. ^24 -h 2V/8I. 

77. aV2 - 56V2. 78. VS + V25a. 79. 

80. ^Sa 4 + *VVa. 81. ^48^ - V48i/. 82. 



118. Products and quotients of radicals 

The product or the quotient of two radicals of the same order can 
be expressed as a single radical by use of the following properties 
of radicals. 




Also, we recall that, by the definition of a root, (\^J) n A. 

ILLUSTRATION 1. (2V3)(5V6) = loV3\/6 

= 10Vl8 - 30V2. 

ILLUSTRATION 2. (5^3) - 5(^3) 8 - 125(3) - 375. 

. 

ILLUSTRATION^ 



V S ^J/oS /^6 /a 1 G 
^ - Vg' ^p ~ W " V^ " b\b' 



760 EXPONENTS AND RADICALS 

ILLUSTRATION 4. 



EXAMPLE 1. Multiply (2\/3 + 3V2x)(3V3 - \/2z). 

SOLUTION. The product is 

6(V3) 2 - 2\/3\/2z -f 
= 18 - 2Vfo -f 9\/6i - 3(2x) = 18 - 

Comment. The student may prefer an expanded solution: 
2V3 + 3\/2z 

- V2s (multiply) 

4- 



18 + 7\/3\/2x - 3(2z) = 18 - Qx + 7\/6x. 

If we remove a positive factor P multiplying a radical \^A we 
must multiply by P n under the radical sign because 



ILLUSTRATION 5. 3\/6 = 



EXERCISE 60 

Express by means of a single radical, and then compute by use of Table I 
if necessary. 

2. V2\/3. 3. \/5vTo. 4. \/3Vl5. 



6. vWl2. 6. (-^2) 8 . 7. (2^6). 8. (3>/5) 2 . 

9. 3V3(2\/6). 10. 5V6(2V2l). 11. V30>/35. 

12. (2V3V5) 2 . 13. (3^2) 3 . 14. 

16. 3>? / 36(v / 45). 16. ^^l^ls. 17. v^ 



VT/s VT4 v'To \/QQ 

18. -^- 19. -^ 20. ^?- 21. 4^- 22. 



Express as a single radical and then simplify. 



23. -j^- 24. ^- 26. -^- 26. -^=- 27. 

- /rfc V4C ^ 7 ^^ a/^^r 



Multiply, simplify, and collect terms. 
28. V3aVl5a. 29. ZVxVSx*. 30. 



EXPONENTS AND RADICALS 76? 



31. zxfa*. ' 32. VfoV25xy. 33. 

34. (3VS). 36. 5\/3a~'3. 36. 2v / 4^ 4 . 37. 



38. (5i~+i;) 2 . 39. (6V?~+~1) 2 . 40. (- 

41. (2 + 3V6)(3 - V5). 42. (2\/2 - 3)(5\/2 + 2). 

43. (\/3 - \/2)(\/3 + V5). 44. (2\/3 - V5)(3V3 -f V5). 

46. (3\/2 + V3)(V2 + 4V). 46. (Vg - \/5)(V6 + Vg). 

47. (2V3 - V7)(2V3 + V?). 48. (5\/2 - 3\/3)(5V2 -f 3\/3). 
49. (2\/3 + V5)(V2 + 4). 60. (3^2 + V3)(\/2 + V6). 

61. (v^ + 5) 2 . 62. (3 + 5\/2)2. 63. (\/2 - 2\/3) 2 . 

64. (V - 2Vi)(Vx + 2V2). 65. 

56. (2#x - 3)(5v^ + 2). 67. 

68. (Vx - 5V^)2. 69. (Va + 6v^) 2 . 60. (aVy - 

61. \^^^- x- 62. v^- 



Replace the coefficient by an equivalent factor under the radical sign. 
63. 3V2a. 64. sVfo. 65. aVbx. 66. 

67. S^P. 68. 2^a. 69. 2V/36. 70. 



119. Rationalization of denominators 

To rationalize a denominator in a radical of order n, after the 
radicand has been expressed as a simple fraction, multiply both nu- 
merator and denominator of the radicand by the simplest expression 
which will make the denominator a perfect nth power. In particular, 
if the radical is a square root, we make the denominator a perfect 
square; if a cube root, we make the denominator a perfect cube. 

T 1 /3 /S 7 ? VSI 4.583 A , K frr , , T , 

ILLUSTRATION 1. 1/7 = \~7T = ~~T~~ ~ 7 = -655. (Table I) 

Notice the inconvenience of the following attempt at computation of 
without rationalization of the denominator; a long division is required. 

1<732 ' (Table 



T 

ILLUSTRATION 2. 




762 



EXPONENTS AND RADICALS 



ILLUSTRATION 3. 




3 /64ar4 __ 3/64~ _ 8/43.32? _ "^4 3 -3a; 2 _ 4^33* 
V 9 "" \9x* ~ \9z 4 -3z 2 "* ^oO = 3a; 2 




+ <* 



a 2 6 



oW 



EXERCISE 61 

Compute by use of Table I a/ter rationalizing the denominator. 
1. V}. 2. V|. 3. Vf . 4. V|. 5. Vf. 

6. VJV. 7. ^}. 8. ^J. 9. ^S. 10. ^5. 

^~dW 12. ^pj. 13. V^. 14. ^. 

15. ^T03. 16. Vl)07. 17. \/^12. 18. ^- .128. 



Eliminate any negative exponents, rationalize the denominators, and col' 
lect terms involving a common radical factor. 



19. \K- 



24. 
29. 
34. 






. 



26. i/T- 



26. 







S^- 31 



"V5S?" 





s=s- 36. 




22. \ F?> 



27. 



32 

32 - w * 



37. 





28. 



33. 







8/6 




9o 4 






tf-1 




44. v/ 

47. VF 5 . 48. VF. 
5 



41. 
46. 




42. 



46. 



49. 



60. v^a~ 7 6~ 2 . 



c 3 ^ 

V^Tft' 

61. 







66. V J + ar'. 67. 

60. ^^ 

63. 



68. v^| + or*. 69. v'a -f 
61. 5VJ -h V45. 62. 10V| 

$4. ^ -h <^. 66. ^F 1 



EXPONENTS AND RADICALS 163 

120. Additional devices for rationalizing denominators 

The method of the following illustration is frequently equivalent to 
the procedure of the preceding section. 

ILLUSTRATION 1. The denominator below is multiplied by ^2 in order 
to make the new radicand a perfect cube: 



T V3 V3 \/5 Vl5 

ILLUSTRATION 2. 7= = = = = = 

V5 \/5 V5 5 



If a denominator has the form aV& c V5, we can rationalize it 
by multiplying by aVb + c Vd because 

(aVb - cVd)(aVb + cVd) - (a\/6) 2 - (c\/5) 2 = a'6 - 

3V^ - \/3 3\/2 ~ V3 2V2 + \/3 
ILLUSTRATION 3. p - 7= = 7= - 7= 7= - 

2V2 - \/3 2\/2 - \/3 2>/2 + \/3 

6(\/2) 2 + (3 - 2) V5 - (V3) 2 9 + V6 9 + 2.449 _ 
- (V3) 2 8 ~ 3 5 



In finding the quotient of two radicals, it may be desirable to 
write the expression as a single radical before rationalizing. 




ILLUSTRATION 4. 



EXERCISE 62 

Rationalize the denominator and, if no letters are involved, compute by use 
of Table I. Collect terms in any polynomial. 



1 1 . 

JL y * 


2 -L. 

4B - 

*v 5 


< 6 

s 'vl' 


1 3 

* rn* 

V7 


6. A. 

V3 


fi 3 

D. . 

vU 


, V5 
7 ' Vl' 


8.^- 


*' vf' 


ia^. 

2V5 


11 


lg , i 


13 2 ~ 


V3 


14.; 


2V3 + 5 


15. 


1 


JLU* 

3 + 


V3 


V3-4 


3 + 2V2 


1ft. 


3 


17 . 




1ft 


V^-2\/3 



2>/2 + V3 V2 + V3 



164 EXPONENTS AND RADICALS 



V2 + 3V3 V2 + V5 



* y - 7=" 4v - - * . - - 

2V3 - V5 2V3 + 3 V2 3V5 - V2 



22. . 23. - 24. 

2\/3 - 3\/7 2V6 + V7 2 V? + \/3 

25. (3 - V2) -7- (2 + V2). 26. (\/3 + VTI) + (1 - VII). 



27- : S 28. -|L 29.-,^=. 30. -^=. 31. 



32. ~^> 33. -^:. 34. ~ 



36. -^^. 37. l . 38. 

X K Ol O */ *-v r" - 



40. 41. o/ ^ y . 42. = 

"v 16a 3 6 2 v27xy* 2\fc \^2a 



43. -^p=. 44. ^g- 1 . 46. 8 - + ^ 

3 + SVa; + 2 Vo + 6 - V2o 



3 2 

46. -= -= 7=' 47. =: p: >. = =.< 

V3-V2 + V5 V3-V6 + V5 V^-VJ 



49. vS- 50. v fc - 61. hfW - m ^ m 





1 21 . Changing from fractional exponents to radicals 

To change a product of powers involving fractional exponents to 
radical form, first change the fractional parts of the exponents to 
fractions with their lowest common denominator. 



ILLUSTRATION 1. 

1 22. Radical operations performed by using fractional exponents 

In this section the results will be desired in radical form. 

I. To find a power or a root of a radical: 

1. Express each radical operation by use of a fractional exponent 
applied to the radicand. 

2. Simplify the indicated power of the radicand and express the result 
as a radical. 



EXPONENTS AND RADICALS 165 

ILLUSTRATION 1. 



ILLUSTRATION 2. (2v / 5x) 2 = 4[(5z)*] 2 

We recognize that, in simple cases, it may be unnecessary to in- 
troduce fractional exponents in an operation of Type I. Also, with 
experience, one observes simple rules such as 

"the wth root of the nth root of A is the mnth root of A." (1) 
ILLUSTRATION 3. v^ 3 = 3. A^Va = v'a. 



II. T'o ^/ki the product or the quotient of two radicals of different 
orders: 

1. Express each radical as a fractional power of its radicand, and 
change the resulting fractional exponents to their LCD. 

2. Rewrite in radical form and combine into a single radical. 



ILLUSTRATION 4. V^3\/2 = 3*2* = 
ILLUSTRATION 5. 



(3a6)t 




27a 3 3a 



III. To reduce the order of a radical, when possible: 

1. Change to fractional exponents in lowest possible terms with a 
common denominator. 

2. Rewrite finally in radical form. 
ILLUSTRATION 6. ^625 = 



In reducing the order of a radical, it is convenient to commence 
by expressing the radicand as a power of some expression. 

ILLUSTRATION 7. \/16z 2 = v"(4z) 2 = 



166 EXPONENTS AND RADICALS 

1 23. Simplest radical form 

As far as problems in this text are concerned, we agree that an 
expression is in its simplest radical form if all possible operations of 
the following varieties have been performed, with any negative ex- 
ponents eliminated. 

SUMMARY. To reduce a radical expression to simplest form: 



1. .Express any power or root of a radical, or product of radicals, as a 
single radical. 

2. Reduce each radicand to a simple fraction in lowest terms. 

3. Rationalize all denominators. 

4. Remove from each radicand all factors which are perfect nth powers, 
where n is the order of the radical. 

5. Reduce each radical to the lowest possible order. 

6. Combine any terms with a common radical factor. 

It must not be inferred that the preceding operations need be 
performed in the specified order. 

To simplify a radical expression will mean to reduce it to simplest 
radical form. 

ILLUSTRATION 1. To simplify the following radical we rationalize the 
denominator, and finally notice that the order of the radical can be reduced. 



ef^~ = 6/q2.4c 2 = v 
\16c l \16-4c 12 



2c 2 2c 2 2c 2 



EXERCISE 63 

Change to simplest radical form. 

1. atb*. 2. x*y*. 3. 5ai 4. 2z*. 6. 

6. a$bt. 7. a$b%. 8. xly$. 9. xty$. 10. 

Reduce to a radical of lower order. 

11. \^. 12. ^wT 2 . 13. v^. 14. Vy\ 15. 

16. v^. 17. 1??. 18. Vtf. 19. #9. 20. 

21. ^27. 22. 4T25. 23. ^36. 24. ^49. 25. 

26. ^8l. 27. J/W. 28. v/iF 2 . 29. V%a*. 30. 



EXPONENTS AND RADICALS 

Change to simplest radical form. 
31. (V&. 32. VS. 33. V5. 34. 



167 



47. (2v / 3) 4 . 48. (v^ 2 ) 6 . 49. 



36. (v'a)'. 37. (\/3) 4 . 38. (V2)'. 

i > 

41. (V5). 42. (\/6) 4 . 43. 

4 fAj/0^4 
IBV* \ V ^t/y 

61. V^. 52. V\/5. 63. 

66. -V^a;. . 67. 

61. 

66. 

69. v^ -5- Va. 

72. V2 -5- -^2. 

75. v^ 4- \/2. 

78. ^ 

81. >^S. 82. 



39. (v"5) 4 . 
44. 



54. A/V1. 



35. 
40. 
45. 

50. 
66. 



62. 

66. ^2^2. 
70. v 
73. 
76. v 
79. v 
83. 



59. 

63. ^yv 
67. V 



84. 



86. vx^. 
90. 



94. 



87. 3 

91. (v^x 2 ) 6 . 

95. 



88. v 
92. 




99. 




100. 



101. 




46 4 



60. 
64. 
68. 

71. v^-r- 
74. ^3 -i- 
77. V6~ 
80. %Vy -. 
~ 85. 

89. 

93. 

97. Vx~* + 

102. 




103. (v^) 7 . 104. (2v / 5) 4 . 
107. Wjty 2 . 108. 
111. (cv/4) 3 . 112. v 



105. 

109. (av^) 6 . 
113. v 



106. 

110. (6V/3) 6 
114. V3a 






m . 



119. - 



+ a 



118. - 



120. -j 



122. 4 



+ 9y~ 2 . 



123. 



Va + b - \/3a 

;+ 



EXPONENTS AND RADICALS 



EXERCISE 64 
Chapter Review 

Find the value of each symbol, using Table I if necessary. 



16. 



21. 



26. 



6~ 3 . 


2. (- 2)~ 4 . 


3. (- 15). 


4. 125i 


5. 27*. 


4*. 




7. 25*. 


8. 9~*. 


9. (- 8)-i 


10. Vf 


Vf. 


12. (VIM) 2 . 


13. (V239) 3 . 


14. Vg. 


15. V^. 


y*- 


17. V^. 


18. V^^. 


19. VV|. 


20. ^Vlf , 


2V27 


22. 3VI * 


23 V ^ 


4% 4 ^^ O W 


9K 


V6 




Vl25a 
27. ^~ ^ 


-. 5! 


4O. j-j- 
-\/ JQ 

O "^ O * 
., 



y - y - 

V3 + V2 



2\/2 - 



Write without fractions by use of negative exponents. 

06' 



Ovr 



OJL* 



' a + 26 



Express without radicals, or negative signs or zero in the exponents, and 
simplify by use of the laws of exponents. 

33. &?. 34. <Vf. 35. V9. 36. Vi^. 37. 

38. v^p. 39. v^. 40. ty&o}. 41. ^C^. 42. 

43. &r 8 . 44. W~ 5 . 46. (16z)*. 46. (a6-)*. 47. 

K1 

51. 4 



62. (2o + 6- 2 )- 1 . 63. S^- 1 + 2y)~ l . 64. 5(a~ 2 + 36- 1 )' 2 . 

65. State the principal 4th root of 256 and principal cube root of 27. 
Change to simplest radical form. 

68. ^iaF 3 . 69. v'- 

62. #&. 
66. (V2a). 



66. V96V. 

60. 

64. aW. 

68. (2V3X) 3 . 

72. -C^^. 

76. V5V5. 



67. ^32^. 

61. 

65. 



73. 

77. \/2lfyr\ 



70. 

74. v/49. 
78. 



63. \/J 4- Vj. 
67. 

71. 

76. 
79. 



EXPONENTS AND RADICALS 

m 



84. 3\/ + 
86. (V2- 
89. 

91. 



+ &V8. 



Va 



. 
6 - Va -b 



93. 



96. 



96. V(a 



82 - 



86. 



87. - . 



5VI2 - a 

88. (Vo + 



90. 
92. 



276 3 -i- 



36. 



3 v 2 
94. x 



-J- x^a 2 - 6 2 . 



-f- 



97. V6- 1 - a~ 2 ^- (V a - 



169 



CHAPTER 



10 



ELEMENTS OF QUADRATIC EQUATIONS 



1 24. Terminology and foundation (or imaginary numbers 

By definition, R is a square root of 1 in case # 2 = 1. But, if R 
is either positive or negative, R 2 is positive and hence cannot equal 
1. Obviously R = is not a square root of 1. Hence, no real 
number is a square root of 1. Similarly, if P is positive, any square 
root R of the negative number P would satisfy the equation 
R z = P. But, R 1 is positive or zero for all real values of R and 
hence P has no real number R as a square root. Therefore, in 
order that negative numbers may have square roots, we proceed to 
define numbers of a new type, to be called imaginary numbers. 

Let the symbol V 1 be introduced as a new variety of number, 
called an imaginary number, with the property that * 



- 1 V - 1 = - 1. 




For convenience, we let i = V 1. Then, by definition i*i 1 
or i 8 = 1. We agree f that the operations of addition, subtraction, 
and multiplication will be applied to combinations of i and real num- 
bers as if i were an ordinary real literal number, with i 2 =A1 Then, 
in particular, 

(- ,') = ,-=_!, (2) 

so that i, as well as -h i, is a square root of 1. Any positive 
integral power of i can be easily computed by recalling that i 2 1 
and hence 

< -(*)(*) -(-1)(-1)-1. (3) 

* Refer to the introduction of negative numbers in Section 6 and observe the 
similarity of the present discussion, 

t This procedure can be arrived at logically by a more advanced discussion. 



ELEMENTS OF QUADRATIC EQUATIONS 171 

ILLUSTRATION 1. t 3 = i(i 2 ) t( 1) = i. 

t 13 = W - 



ILLUSTRATION 2. ' (3 + 5i)(4 + t) = 12 + 23t 4- 5t 2 

- 12 + 23i - 5 = 7 4- 23t. 

If P is any positive number, we verify that 

- - P; (- VP) = t^P = - P. 



Hence, the negative number P has the two square roots db t' 
Hereafter, we agree that the symbol V P or ( P)^ represents the 
particular square root t'VP. Then, P has the two square roots 
db V P = iVP. This agreement about the meaning of V P 
is equivalent to saying that we should proceed as follows in dealing 
with the square root of a negative number: 



(4) 
ILLUSTRATION 3. The square roots of 5 are 

= 




ILLUSTRATION 4. V- 4V- 9 = (iV4)(i\/9) = 6i 2 = - 6. 

]Vofe /. The formula VaVfe = Vo6 was proved only for the case where 
Va and VJU are real. We can verify that the formula does not hold if a 
and 6 are negative. Thus, by the formula, , 

V^I\A=~9 = V(- 4)(- 9) = V36 = 6, 
which is wrong, because the correct result is 6 (in Illustration 4). 

If a and b are real numbers, we call (a + bi) a complex number, 
whose real part is a and imaginary part is bi. If b 7* 0, we call (a 4- W) 
an imaginary number. A pure imaginary number is one whose real 
part is zero; that is, (a -f bi) is a pure imaginary if a = and 6^0. 
Any real number a is thought of as a complex number in which the 
coefficient of the imaginary part is zero; that is, a - a -f Oi. In 
particular, means (0 4- Oi). 



772 



ELEMENTS OF QUADRATIC EQUATIONS 



ILLUSTRATION 5. (2 3i) is an imaginary number. The real number 6 
can be thought of as (6 + Oi). 

Note 2. In this book, unless otherwise stated, all literal numbers repre- 
sent real numbers, except that hereafter i will always represent V 1. 
Any literal number in a radical of even order will be supposed positive, if 
this is possible and adds to our convenience. 

itNote 8. The student has seen that we introduce imaginary numbers in 
order to provide square roots for negative numbers. It might then be in- 
ferred, incorrectly, that still other varieties of numbers would have to be 
introduced to provide cube roots, fourth roots, etc., of positive and negative 
numbers and also roots of all orders of imaginary numbers. An extremely 
interesting theorem is that the real numbers and the imaginary numbers, 
as just introduced, provide all the numbers we need in order to have at our 
disposal roots of all orders of any one of these numbers. Explicitly, in more 
advanced algebra,* it is proved that, if k is any positive integer, then any 
complex number N has just k distinct kih roots, which are also complex 
numbers (including real and imaginary numbers as special cases). 



irNote 4- In the theory of electricity, it is customary to use j for V 1 
because the letter i is reserved for a different purpose. 

EXERCISE 65 

Express by use of the imaginary unit i and simplify the remaining radical. 
1. V^~9. 

6. V^ 



11. \ 
16. 




17. v^^OO. 



21. V-^36. 22. V-- 



27. V^ 






29. V- a 2 ** 2 . 



30. V- Ifo*. 



31. v 



32. V 



33. V- 



34. V- 12t0. 
o 2 ^ 



38.^ 



36. V 

A / cW 
89 - V- 26 ' 



36. V- 4ay. 37. V-27F. 



4L V- 5- 



46. -63. 46. -f 

* See De Moivre's Theorem and related topics in any college algebra. 



State the two square roots of each number. 
42. - 81. 43. - M. 44. - 



ELEMENTS OF QUADRATIC EQUATIONS 173 

Perform the indicated operation and simplify by use of i 1 1 until i 
does not occur with an exponent greater than 1. 

47. i 6 . 48. i 7 . 49. %. 60. t 8 . 61. t 18 . 62. i. 

63. (3 - t)(3 + t). 64. (3i + 5) (4 - 3i). 66. (3 + 2i)(3 - ft). 
66. (2i + 3) 2 . 67. (3i - 2)(5i + 7). 68. (4t + 3)(- ft + 5). 
69. (5 - 2i) 2 . 60. (3i - 4)*. 61. (4t + 5) 2 . 

62. (2i 2 - 3t -f 5) (ft - 3). 63. (i 3 - 2i 2 + 3)(3i 2 - 5). 

64. (4t - 7)(ft -f 5i 2 ). 66. (ft + 4i - t 2 )(2 + 3t). 

66. V^~2V^~8. 67. V^V- 75. 68. V- 27\^ : ~3. 

69. V^2(3 - 5V^1). 70. V^~3(5 - V^27). 71. (5 - V^~8) 2 . 

72. Substitute x =* 3 + 6i in (a; 2 - 60; + 34). 

73. If /(*) = 3a: 2 + 2x - 7, find /(ft); /(- 3i); /(2 - 5i). 

74. If /(x) = a* - a? - 3, find /(ft); /(I + i). 
Obtain an expression of the form a + bifor the fraction. 



3i-4 
HINT for Problem 75. Multiply numerator and denominator by 5 4i. 

125. Equations of the second degree 

A quadratic equation, or an equation of the second degree, is an 
integral rational equation in which, after like terms are collected, 
the terms of highest degree in the variables are of the second degree. 
A quadratic equation in a variable x can be reduced to the standard 

form 

ax a + bx + c = 0, (1) 

where a, b, and c are constants and a 7* 0. A complete quadratic 
equation in x is one for which 6 7* 0, and a pure quadratic equation is 
one for which & = 0. 

ILLUSTRATION 1. 3z 2 5x + 7 = is a complete quadratic equation 
and 5x 2 7 = is a pure quadratic in x. 

1 26. Pure quadratic equations 

To solve a pure quadratic equation hi x, solve the equation for x 2 
and extract square roots. 



174 ELEMENTS OF QUADRATIC EQUATIONS 

EXAMPLE 1. Solve: 7y* = 18 + 3y 2 . 
SOLUTION. 1. ly 1 - 3y* - 18; 4y 2 = 18. 

2. Divide by 4: y* = f . 

3. Hence, j/ must be a square root of j. On extracting square roots 
and using Table I, we obtain 

- 2.121. 




EXAMPLE 2. Solve: 2j/ 2 + 35 = - 5y*. 

SOLUTION. 7y 2 = 35; 2/ 2 5. Hence, y = =b V 5 = iV5. 

EXAMPLE 3. Solve for x: a*x* + ft 2 = o&c 2 + a 2 . 

SOLUTION. 1. o 2 ^ 2 a&c 2 = a 2 ft 2 . 

2. Factor: (a 2 - ab)x* = a 2 - 6 2 . 

3. Divide by (a 2 06) and reduce to lowest terms: 

2 . a?-& = (a ~ b)(a + 6) 
X tf-ab a(a - 6) '* 

. + 6 

or. a^ = ! 

a 

4. Extract square roots and rationalize the denominator: 



K a a 

1. If the coefficients in a quadratic equation are explicit numbers 
and if a radical occurs in any solution which is a real number, always com- 
pute the decimal value of the solution by use of Table I. If it is desired to 
check such a solution, substitute the radical form instead of the decimal 
value, unless otherwise directed by the instructor. The approximate decimal 
value, as a rule, could not lead to an absolute check. 

EXERCISE 66 

Solve for x, or otherwise for the letter in the problem. 
1. 5x 2 - 125. 2. 3x 2 = 12. 3. x 2 = - 9. 4. 4x 2 - - 9. 

6. 9x 2 - - 25. 6. 2x 2 - 3. 7. 5x 2 - 7. 8. 3x 2 - 11. 

9 A 7*2 == / in QT^ Jt 11 ^/vr2 = h 19 9hr% 1 ft 

t JU - G. XV* 7* / XX* OCftv " fii% JLm% &WU - 1Q. 

13. 15 - 162 2 - 4. 14. 9 - 7* * 6. 15. Jx 2 - 1 Jx 2 . 

16. 9x 2 + 49 - 0. 17. 7X 2 - 5 - 3s 2 . 18. Jx 2 - } * fcc 2 . 



ELEMENTS OF QUADRATIC EQUATIONS 175 

19. l&r 2 + 64 - 0. 20. f z* + 4 = a*. 21. fcc - | - 

22. 4oz 2 - c - rf. 23. 4a + 2cz 2 - 4d. 24. 9ac* - 46 

25. 4z 2 - 25a = 4&r*. 26. 9az* - 46 + 9cz*. 

27. 4x 2 + 25a = 25 + 4a 2 z*. 28. 2cx* + 4d - c* - 4<fc*. 

29. Solve /r = rov 2 for v. 30. Solve S J#* for . 

31. Solve A = ?rr 2 for r. 32. Solve A Jir 2 A for . 

or x. 



07 ^ 2 + 2 _5 04^.5 x 49 

33. -g- - g. 34. j - 35. j - - - 0. 



8 



4 2* + 3 2 as + 6 

127. Solution of an equation by factoring 

We know that a product of two or more numbers equals zero when 
and only when at least one of the factors is zero* This fact is the 
basis for the f ollowing method, which applies to integral rational equa- 
tions of any degree. 

SUMMARY. To solve an equation in x by use of factoring: 

1. Clear the equation of fractions if necessary by multiplying both 
sides by the LCD of all fractions involved. 

2. Transpose all terms to one member and thus obtain zero as the other 
member. Factor the first member if possible. 

3. Place each factor equal to zero and solve for x. 

EXAMPLE 1. Solve: 6 - 5z - 6x 2 - 0. 

SOLUTION. 1. Multiply both sides by 1, to obtain convenience in 
factoring with a positive coefficient for # 2 . 

6* 2 + 5x - 6 = 0. 

2. Factor: (3z - 2) (2* + 3) - 0. 

3. The equation is satisfied if 

3x - 2 or if 2x + 3 = 0. 

4. If 3x 2 = 0, then 3x 2; x = is one solution. 

5. If 2x 4- 3 = 0, then 2x = 3; x f is a second solution. 
* This fact holds for a product of complex numbers. 



776 ELEMENTS OF QUADRATIC EQUATIONS 

EXAMPLE 2. Solve: 4x* + 20s -f 25 0. 

SOLUTION. 1. Factor: 

(2* + 5) 2 - 0; or (2z + 5)(2* + 5) - 0. 

2. If 2z -f 5 = 0, then x = \ . Since each factor gives the same value 
for x, we agree to say that the equation has two equal roots. 

In solving an equation, if both sides are divided by an expression 
involving the unknowns, solutions may be lost. 

EXAMPLE 3. Solve: 5z 2 &t. 

SOLUTION. 1. Transpose 8z: 

5z 2 - Sx = 0; x(5x - 8) = 0. 

2. Hence, x or 5x 8 = 0; the solutions are and f . 

INCORRECT SOLUTION. Divide both sides of 5z 2 = &c by x: 

5x = 8. 

Then, incorrectly, we obtain x = f as the only solution. In this incorrect 
solution, the root x = was lost on dividing by x. 

Some literal quadratic equations can be solved by factoring. 
EXAMPLE 4. Solve for x: 2a?x* + 3a6x - 26 2 = 0. 

SOLUTION. 1. Factor: (2ax ft) (ax + 26) = 0. 

2. If 2ax - b = 0, then 2ax = 6; x = H-- 

3. If ax -f 26 = 0, then ax = - 26; x = --- 

a 

4. The equation has the two solutions 5- and --- 

^ 2a a 

EXERCISE 67 

Solve 6y factoring. 
1. x 2 - 3z = 10. 2. y 2 - 5y = 14. 3. s 2 -f * = 12. 



4. z 2 -f 3z - 28. 6. 21x = 14z 2 . 6. W - 144 = 0. 

7. 3x 2 - 7x - 0. 8. 6s 2 = Ifo. 9. 5z 2 - 9z - 0. 

10. x 2 + 8 - 6x. 11. 4x 2 - 25 = 0. 12. x 2 + 15 - 8x. 

13. 2s 3 + 5z - 3. 14. 3z 2 - 2z - 5. 15. 8z 2 -f 3 = 



ELEMENTS OF QUADRATIC EQUATIONS 177 

16. 16z 2 - 24x - 9. 17. 25y 2 - 20y - 4. 18. z 2 + 6* - - 9. 

19. 4y* + 40 = - 1. 20. 3z 2 + 2 = - 7*. 21. 2z 2 + 7x - - 6. 

22. lOz + 3 -f &c 2 - 0. 23. 12 - 5z 2 - 17z = 0. 

24. 6z 2 - 19* + 15 = 0. 25. 16z 2 + 40z + 25 = 0. 

26. 8 - 22z + 15z 2 = 0. 27. 15 - 7w - 4w 2 = 0. 

28. 8z 2 + 2x - 15 = 0. 29. 7z 2 + 9z - 10 = 0. 

30. 49x 2 + 28x = - 4. 31. 4 + 5x - 9x 2 = 0. 

32. 8 - 2x - x 2 = 0. 33. 6 + 5x - 6x 2 = 0. 

Solve for x or w or z. 

34. 3fo 2 + ex = 0. 35. 2or 2 - 3dx = 0. 



36. x 2 + or - 6a 2 = 0. 37. x 2 + 56x + 66 2 = 0. 

38. 2z 2 + bx - 36 2 = 0. 39. 3w; 2 - bw - 46 2 = 0. 

40. 46 2 z 2 -h 4a6x + a 2 = 0. 41. 66 2 z 2 - 7bx - 3 = 0. 

42. 2aV - abx - 36 2 = 0. 43. 9aV + 12a6x + 46 2 = 0. 



3 , 7 5 46 8 


2 + 3w? 


1 


" 4*" ' 8z 2 " - " 3i H 


h 3 ' 3^ + 1 


1 


6^ 9 
4. A7 


3 


5 


* # -f- 4 2 ' ** 


1 2x + 5 


3 


2111 1 r 
2 l li l 40 


2x 


1 


* 2 - 2 12 22-2 s + 


1 1 -4x 


x + 1 


10u> 4w+l 4w>-7 2x4 


-11 3x - 1 


n 



- 2 2w - 1 w 2x + 8 * - 1 
62. (x + 3)(2z - 5)(3z + 7) = 0. 63. 6z 8 -f ar 2 - 15* = 0. 
64. (2x - 3)(3x -f 5) = 2x + 7. 65. (3z - l)(2x + 5) = 3z + 19. 



128. Completing a square 

A binomial x* + pa: becomes a perfect square if we add the square of 
one half of the coefficient of x. That is, we complete a square if we 

add g)' or f : 



178 ELEMENTS OF QUADRATIC EQUATIONS 

ILLUSTRATION 1. x 2 fa becomes a perfect square if we add the square 
ofi(6), or 3: 

x 2 - Ox + 9 - (x - 3) 2 . 

ILLUSTRATION 2. To make z 2 7x a perfect square, we add (J) 2 or 

a* - 7z + ^ - (x - I) 2 . 



SUMMARY. To sofoe o quadratic equation in x by computing a square: 

1. Transpose all terms involving x to the left side and all other terms 
to the right member and collect terms. 

2. Divide both members by the coefficient of x 2 . 

3. Complete a square on the left by adding the square of one half 
of the absolute value of the coefficient of x to both sides. 

4. Rewrite the left member as the square of a binomial. 

5. Extract square roots t using the double sign on the right. 

EXAMPLE 1. Solve: a; 2 + 4x + 1 = 0. (1) 

SOLUTION. 1. Subtract 1: x 2 + 4x 1. (2) 

2. Since 4 -*- 2 = 2, add 2? or 4, to complete a square on the left: 

* 2 + 4z + 4 = 4- 1; (3) 

(x + 2) 2 - 3. (4) 

3. Extract square roots: x + 2 = V3 t or 

x - - 2 db V3. (5) 

Thus, the roots are irrational numbers. To compute approximate values 
for the roots to three decimal places, we obtain V3 from Table I : 

x * - 2 + 1.732 = - .268 x = - 2 - 1.732 = - 3.732. 

% 

EXAMPLE 2. Solve: 3x* - &e + 2 * 0. 

SOLUTION. 1. 3x* - &c ** - 2. 

8 2 

2. Divide by 3: s a 53 = ~- 



3. Since * 2 J, add (|)* or ^ to complete a square. 

8 



tr 
Hence, 



/4\* 16 2 16 6 

I _ I - .. . _ . ^ .. ^ . 

\3/ 9 39 9 
4 * 10 



ELEMENTS OF QUADRATIC EQUATIONS 179 



4. Extract square roots: 

4 



From Table I, VIo = 3.162. Hence, 



3.162 OOQ>y . 4-3.162 O . n 

SB 2.387, and x - = - .279. 



., 
o o 

EXAMPLE 3. Solve by completing a square: x* -f 4s + 7 0. 
SOLUTION. 1. x 2 4- 4x 7. 

2. Since (4 4- 2) * 2, we add 2* or 4 to both sides: 

& + 4* + 4 4 - 7; or (x + 2)' - - 3. 

3. Hence, x + 2 - =t V^IJ - * fx/3; 

x - - 2 *V3. 

EXAMPLE 4. Solve for x: ox* -f bx + c * 0. 

SOLUTION. 1. Subtract c: ax* + fez c. 

6 c 

2. Divide by a: x 1 -f- - --- 

a a 

O AJJ/^V 6* 1 , & , /&\* &* C 

3. Add (TT-) , or -:-=: rr* + -x + (^-) - ^-r --- 

\2a/ ' 4a 2 a \2a/ 4o a o 

c . ... / , 6 \ 8 6 - 4ac 

Simplify: (x + ^j -TT"' 

4. Extract square roots: 

6 



2o 

_ , , . 6 fe Vfe 2 4oc fe =fc V6* 4oc 

5. Subtract JT-: a; = =- =b ~ = s 

2a 2a 2a 2a 

Note 1 . An equality A 2 = B 1 is satisfied if A 5, that is, 

if A=B arif A = -B. (7) 

Equally well, A 2 = B 2 if - A 5, that is, 

t/ - A - J? or t/ - A * - B. (8) 

If both sides of each equation hi (8) are multiplied by 1, we obtain equa- 
tions (7). Therefore, on extracting the square roots of both sides of A* = B 1 , 
we obtain all possible information by writing A = B, instead of writing 
A = B, where we read d= as " + or ." That is, it is necessary to 
use the double sign on just one side if the square roots of both sides of an 
equation are extracted in solving it. 



780 ELEMENTS OF QUADRATIC EQUATIONS 

EXERCISE 68 

Find what must be added to the expression to make it a perfect square, and 
then write this square. 

1. x* - Sx. 2. x* + 10*. 3. * 2 - 2cx. 4. * a + 4dx. 



6. * 2 - J*. 6. x 2 + f*. 7. * 2 + j*. 8. x* - J*. 

Sofoe fo/ completing a square. 
9. a; 2 + Ox = 7. 10. a: 2 4- 10s -f 24 = 0. 11. * 2 + 4* = 21. 

12. a; 2 + 9 = 6*. 13. * 2 + 4* + 4 = 0. 14. 2w^ + 3 = 8w;. 

15. 2y* 4- 4y = 5. 16. x 2 + 13 = 6*. 17. x 2 + 5 = 4z. 

18. 6x 2 - 2 = 4z. 19. 9x 2 + 1 = 12*. 20. 9s 2 + 6* = 1. 

21. 4* 2 + 13 - 12*. 22. 4z 2 4- 4z - 3 = 0. 23. 3z 2 -h 8z = 1. 

24. 16x 2 + 9 = 24*. 26. 9* 2 - 12* = 21. 26. 4* 2 - 12* = 5. 
Verify the statement by substitution. 

27. * = (- 2 + V2) and * = (- 2 - \/2) satisfy * 2 + 4* + 2 = 0. 

28. * = (2 3i) are solutions of * 2 4* + 13 = 0. 

Solve for x by completing a square. 

29. * 2 - 2o* - 15a 2 . 30. * 2 + bx = 66 2 . 

31. 2* 2 - 56* = 36 2 . 32. 6* 2 - 4* - c = 0. 

33. 3* 2 -f 2o* - b = 0. 34. .3* 2 - .06* - .144 = 0. 

36. o* 2 + 4* - c = 0. 36. 2* 2 + bx + c = 0. 

37. Hx* + Kx + P - 0. 38. A* 2 + 2B* + C = 0. 



1 29. The quadratic formula 

In Example 4 on page 179, the quadratic equation 

ax a 4- bx + c = (1) 

was solved by the method of completing a square; the solutions were 
found to be 

x = -n. 

We call (2) the quadratic formula. In (2), it is permissible for a, 6, 
and c to have any values, with a j* 0. 



ELEMENTS OF QUADRATIC EQUATIONS 181 

SUMMARY. To solve a quadratic equation in x by use of the quadratic 
formula: 

1. Clear the equation of fractions and reduce it to the standard form 
ax* + bx + c = 0. 

2. List the values of the coefficients a, b, and c. 

3. Substitute the values of a, b, and c in the formula. 

ILLUSTRATION 1. To solve 3z 2 62 2 = 0, we observe that a = 3, 
6 = 6, and c = 2. Hence, from the quadratic formula, 

_ - (- 6) V(- 6) 2 - 4-3- (- 2) _ 6 2VT5 _ 3 3.873 
X ~ 6 6 ~ 3 ; 

x = 3 + 3 ' 873 = 2.291, and x = 3 " 3 ' 873 = - .291. 
o o 

ILLUSTRATION 2. To solve 2z 2 4x + 5 = 0, we notice that = 2, 
fe = 4, and c = 5. Hence, from the quadratic formula, 

4 db Vl6 - 40 4 V- 24 4 2t>/6 2 



4 ~4~4~2 

EXAMPLE 1. Solve for x: x* 3ex -f 5dx I5de = 0. 

SOLUTION. 1. Group terms in a:: z 2 4- x( 3e -f 5d) 15de = 0. 

2. In the standard notation, a = l, 6= 3e + 5rf, and c = 
From the formula, 

- (- 3e 4 



The radicand becomes 

9e 2 + 60efe = 25(^4- 30de + 9e* 



/ox 
(3) 



Hence, from (3), 

- (- 3e + &0 (5d + 3e) 

o* o*? .1 i .. . i i ............... * 

x 2 ' 

3e - 5d 4- 5d + 3e 3e - 5d - 5d 



x = 2 

The solutions are 3e and 



R . 

x = - - ** " 



. In deriving the quadratic formula, we showed that an integral 
rational equation of the 2d degree in x has just two roots (which sometimes 
are identical). This result is a special case of a general theorem that, if 
the degree of the equation is n, the equation has just n roots (with repetitions 
of values possible among them). 



782 ELEMENTS OF QUADRATIC EQUATIONS 

EXERCISE 69 

Solve by use of the quadratic formula.* Check if directed by the instructor. 
1. 6z a + x - 2 - 0. 2. 3y + 2y - 5 0. 3. 6y 2 - 7y - 3 =* 0. 



4. 7j/ 2 - 8y_ = 12. 6. j/ 2 - 2y 4- 10 - 0. 6. z 2 + 13 - 4z. 

7. 4* 2 + 9 - 12*. 8. 16z 2 - 25 - 0. 9. 4z 2 - &c + 1 - 0. 

10. 9* 2 + &c + 1 - 0. 11. 36z 2 - 49 = 0. 12. By - 18 - Ify 8 . 

13. 9z 2 + te - 1. 14. 4 + 4x - 6z 2 . 15. 2z 2 + 3 - 8z. 

16. 2* 2 - 2s - 7. 17. 4z* 4- 3 - 2z. 18. 9s 2 + 6z = 7. 

19. 9s 2 + 16 - 0. 20. 4z* + 13 - 40. 21. 3s 2 = 4z - 8. 

22. 4s* + 5 - 82. 23. a* + .15 = .8z. 24. x* + 6 = 6z. 

25. 4x 2 -f 53 = 4x. 26. 3x 2 -f 2x = 9. 27. 4x 2 + Sx = - 9. 

28. 9z 2 - 27 - &c. 29. 4x 2 -f 13 = 12x. 30. 18z 2 + 33x = 40. 

31. 25z 2 -f 4 - 20c. 32. 21x 2 + 19a; - 12. 33. 9y z + 23 = 3(ty. 

34. 24y 2 + 2y = 15. 36. 4r 2 -h 29 = - 8x. 36. 16a; 2 + 34x = 15. 

Solve for x or y by use of the quadratic formula. 
37. &c 2 - 5dx - 6tf - 0. 38. 2z 2 + hx - 16A 2 = 0. 

39. ox 2 - dx + 3c - 0. 40. 2oc 2 -f 36x - c = 0. 

41. 5&2/ 2 - 3Jty + 6 - 0. 42. 2/uc 2 -f 3x - 5A = 0. 

43. y 2 + 2cy + dy -f 2crf * 0. 44. y* - 46y + Say - 12o& = 0. 

45. 5fcc* - 6 + lOfcc - 3a: - 0. 46. 8% 2 4- 12y - 15 - 10% = 0. 

47. 6% 2 - 4hy -f 10 - 15y - 0. 48. 2x 2 - 3hx + A* - x = 1. 

49. 3z 2 -f 3&c 2 - 6x + 5/uc - 10 + 5x = 0. 



50. Check the solutions ( 6 V6 J 4oc) -i- 2a by substitution in 
the equation ax* + 6x -f c = 0. 

51. Solve for y in terms of a: a; 2 2y 2 2 + xy + a; + 5y = 0. 

52. Solve for x in terms of y: 2y 2 + 15z 2 - 2 - x + 3y IZxy = 0. 

63. Solve for a; hi terms of y in Problem 51. 

64. Solve for y in terms of in Problem 52. 

* Table I is useful in detecting perfect square numbers. Thus, if we meet 
V1764, by reference to Table I we observe that 1764 * (42)*. 



ELEMENTS OF QUADRATIC EQUATIONS 



183 



130. Outline for solution of quadratic equations 

A pure quadratic equation should be solved by merely extracting 
square roots, as in Section 126. Any other quadratic equation should 
be solved by factoring if factors can be easily recognized. In all 
other cases, solve by use of the quadratic formula, unless otherwise 
specified. The method of completing the square is not recommended 
in any problem unless specifically requested; this method was in- 
troduced mainly as a means for deriving the quadratic formula. 

1 31 . Applications of quadratic equations 

From geometry, we recall the Pythagorean 
theorem, which we associate with the triangle 5 

in Figure 12. . R 9 . 

// a and b are the lengths of the perpendicular sides and c is the 
length of the hypotenuse of a right angled triangle, then a 2 -f fc 2 * c 2 . 

Applications of the preceding theorem frequently introduce quad- 
ratic equations. 

EXAMPLE 1. Find the length of a side of an equilat- 
eral triangle whose altitude is 3 feet shorter than a side. 

SOLUTION. 1. In Figure 13, let ABC represent 
the triangle. Let x feet be the length of a side of 
AABC. Then, the lengths of AD and DC are, re- 
spectively, \x and (x 3). 

2. From the Pythagorean theorem, 

AC 2 = Iff + 




or 



(X ~ 



(1) 



3. Simplify in (1) : x* = ^ -f z 2 - fa + 9; 

x 2 - 24* + 36 - 0. 

4. Solve (2) by the quadratic formula: 

24 



(2) 



X 



rf 171 * 8 - 12*6^8; 



= 22.39 and 1.61. 



(Using Table I) 



The smallest root has no significance in the problem because (1.61 3) 
is negative. Hence, the side of the specified triangle is 22.39 feet long. 



184 ELEMENTS OF QUADRATIC EQUATIONS 

EXAMPLE 2. An airplane flies 560 miles against a head wind of 40 miles 
per hour. The plane took 28 minutes longer for this flight than would have 
been the case in still air. How fast could the airplane travel in still air? 

INCOMPLETE SOLUTION. 1. Let x miles per hour be the speed of the air- 
plane in still air. In flying against the wind, the speed is (x 40) miles 
per hour. From the equation distance = (rate) (time), the flight times for 
a distance of 560 miles against the wind and in still air are, respectively, 

560 , 560 

and 



x _ 40 x 

2. From the statement of the problem 

560 560 28 

" /JA " 



x - 40 x '60 
The student should clear of fractions and solve for x. 

MISCELLANEOUS EXERCISE 70 

Solve each equation by three methods, (a) by factoring, (b) by completing a 
square, and (c) by use of the quadratic formula. 

1. 2* 2 + 5* = 3. 2. 3* 2 + 5 = 16*. 3. 12* 2 + 11* = 15. 

Solve for x or y or z by the most convenient method. 
4. 2/ 2 - 33 = ty. 6. x* - 4x = 45. 6. 16</ 2 + 9 = 24p. 

7. 14* 2 - x - 3. 8. 3 2 - 6 - 2*. 9. 49* 2 + 4 = 14*. 

10. 4y* + 17 = I2y. 11. 1 + 25* 2 = 10*. 12. 6* 2 + 5* = 56. 

13. 25* 2 - 20* = 1. 14. 16 - 5* 2 = 0. 16. 6* 2 = 7*. 

16. 6* 2 + 5 = 0. 17. 9 - II* 2 - 0. 18. 4* 2 = 49*. 

19. 5y 2 + 36 - 0. 20. 16y 2 + 1 = &y. 21. 20* 2 + 13* = 21. 

2- * =3. 



. 

3 x 5 + x 



3* o OB 7 



* - 2 * 2 - 4 

X 2 1 5 ~* 3-2* 



- 
4 "" 2*T~I ~ 6T= 2 - * 2* " l 

28. \g& + a* 3S. 29. 4* 2 -f 26* + b - 1. 

30. &* 2 - 2fcc + 2 - *. 31. 3* 2 4- hx + 3fc* -f W? 0. 

32. a* 2 - 26* = 2* + 3. 33. c* 2 + 2hx = 5 + 4kx. 



ELEMENTS OF QUADRATIC EQUATIONS 185 

34. Solve for x by completing a square: hx* + 2fcc m 0. 

35. Solve for x by completing a square: dx* 3cx + h = 0. 

eocA problem by introducing only one unknown number. 

36. Divide 45 into two parts whose product is 434. 

37. The area of a rectangle is 221 square feet and one side is 4 feet 
longer than the other. Find the dimensions. 

38. Find two consecutive integers whose product is 306. 

39. Find a number which is ^^ less than its reciprocal. 

40. Find the length of a side of a square where a diagonal is 6 feet longer 
than a side. 



41. Find the length of a side of an equilateral triangle whose altitude 
is 2 feet shorter than a side. 

42. After plowing a uniform border inside a rectangular field 50 rods long 
by 40 rods wide, a farmer finds that he has plowed 60% of the field. Find 
the width of the border. 

43. The diameter of a circular field is 40 yards. What increase in the 
diameter will increase the area by 440 square yards. (Use IT = 3^.) 

44. A circular field is surrounded by a cinder track whose width is 
20 feet and area is J of the area of the field. Find the radius of the field. 

46. An airplane flew 660 miles in the direction of a wind and then took 
40 minutes longer than on the outward trip to fly back against the same 
wind. If the plane flies at the rate of 200 miles per hour in still air, how 
fast was the wind blowing? 

46. Jones travels 4 miles per hour faster than Smith and covers 224 
miles in one hour less time than Smith. How fast does each man travel? 

47. A motorboat takes 2 hours to travel 8 miles downstream and 4 miles 
back on a river which flows at the rate of 2 miles per hour. Find the rate 
at which the motorboat would travel in still water. 

48. If- A is the measured cross-section area of a chimney, its so-called effec- 
tive area E is the smallest root of the equation E* 2AE + A* .36A = 0. 
Solve for E in terms of A and, from the result, find E if A =20 square feet. 

49. If an object is shot vertically from the surface of the earth with 
an initial velocity of v feet per second, and if air resistance and other dis- 
turbing factors are neglected, it is proved hi physics that $ = vt %g&, 
where s feet is the height of the object above the surface at the end of t 
seconds and g 32, approximately, (a) Solve for t in terms of s. (6) If 
v = 200 feet, use Part a to find when = 500 feet and feet. 



CHAPTER 




ADVANCED TOPICS IN QUADRATIC 
EQUATIONS 



1 32. Graph of a quadratic function 

A quadratic function of x is a polynomial of the second degree in 
x and hence has the form 

ax* -f bx + c, 

where a, b, and c are constants and a ^ 0. 
EXAMPLE 1. Graph the function z 2 2x 3. 

SOLUTION. Let y = x z 2x 3. We select values for x and compute 
the corresponding values for y. In Figure v 

14, we plot the points (- 3, 12), (- 2, 5), 
etc. In the table of values, we arrange 
the values of x in their natural order as 
they appear on the z-axis, because then the 
corresponding points on the graph are met 
hi their natural order as we draw the curve. 
The curve through the plotted points is 
the graph of the function and is called a 



2? == 


-3 


- 2 





1 


2 


4 


5 


y = 


12 


5 


-3 


4 


-3 


5 


12 




Fig. 14 

parabola. The point V at the rounded end is called the vertex of the 
parabola. Since V is the lowest point of the graph, the ordinate of V, 
or 4, is the smallest or minimum value of the function, and we call V 
the minimum point of the graph. The vertical line through V is called 
the axis of the parabola. The part of the curve to the right of this axis 
has exactly the same shape as the part to the left. That is, the parabola is 
symmetrical with respect to its axis. The equation of the axis of the parab- 
ola y = x 2 2x 3 shown in Figure 14 is x 1. 



ADVANCED TOPICS IN QUADRATIC EQUATIONS 187 

If a parabola is concave downward (open downward), instead of 
concave upward as shown in Figure 14, then the vertex of the parab- 
ola is its highest point and is called the maximum point of the curve. 

At a more advanced stage, we meet proofs of the following facts: 

I. The graph of ax* + bx -f c is a parabola with its axis perpendicular 
to the x-axis; this parabola is concave upward if a is positive and 
concave downward if a is negative. 

II. The abscissa of the vertex of the parabola is x = 5-; when x 

has this value, the function has its minimum or its maximum value 
according as a is positive or negative. 

2 

ILLUSTRATION 1. In Figure 14, at V, x = 7:-^- = 1. 

A- \ 

SUMMARY. To form a table of values in graphing a quadratic func- 
tion f(x): 

1. Find the coordinates of the vertex of the graph. 

2. Choose pairs of values of x where, in each pair, the values are equi- 
distant from the vertex, one value on each side; the values off(x) cor- 
responding to each pair will be equal. 

ILLUSTRATION 2. In Example 1, the abscissa of V is x 1. Then, we 
selected z = 1 db 1, orz = 2 and z = 0; x I 3, or z = 4 and x = 2; 
etc. The corresponding pairs of values of y are equal. 

*EXAMPLE 2. Divide 50 into two parts such that their product will be 
a maximum. 

SOLUTION. 1. Let x be one part; (50 x) is the other part. 

2. Let/Or) represent the product #(50 x), or f(x) = 50z x 2 . 

3. The maximum of f(x) is attained at the vertex of the parabola which is 
the graph off(x), or when x = [50 * ( 2)] = 25. Hence, the product 
of the parts of 50 will be greatest when they are equal, each being 25. The 
corresponding largest product is 625. 

Note 1. After having formed a table of values for graphing a quadratic 
function of x, select the scales on the coordinate axes with care. Choose 
the unit for distance on the z-axis large enough to spread out the parabola 
in order to make it generously open. Choose the vertical unit independently 
of the previous choice of the z-unit in order to be able to plot all points from 
the table of values on the available part of the cross-section sheet. 



188 ADVANCED TOPICS IN QUADRATIC EQUATIONS 

EXERCISE 71 

For each function, (a) find the coordinates of the vertex of the graph and the 
equation of its axis; (b) graph the function, with values of x extending at least 
4 units on each side of the vertex; (c) state the maximum or minimum value 
of each function. 

1. x*. 2. 4z 2 . 3. - x\ 4. - 6z 2 . 5. * 2 + 5. 



6. x* - 4. 7. a* + 6z + 5- 8. z 2 - 4x + 7. 

9. - 2z 2 + 4* + 3. 10. - 3z 2 + I2x. 11. 2z 2 -f 8z + 3. 

12. - 3s 2 + 6z - 5. 13. 4z 2 - 12*. 14. 2z 2 - 20z + 4. 

State whether the function has a maximum or a minimum value, and obtain 
this value without graphing by finding the coordinates of the vertex. 

16. 4z 2 - IQx + 3. 16. - 3z 2 + 24z - 7. 17. - 6z 2 + 8. 

18. If an object is shot vertically upward from the earth's surface with 
an initial velocity of 96 feet per second, (a) draw a graph of the distance s 
as a function of t\ (b) from the graph, find when the object commences 
to fall, the maximum height which it reaches, and when it hits the sur- 
face. (Recall the formula of Problem 49, page 185.) 

19. If an object is shot vertically upward from the earth 's surface with 
an initial velocity of 80 feet per second, find when the object reaches its 
maximum elevation, without graphing. (See Problem 18.) 

20. Graph the function z 3 I2x + 3 by use of the integral values of 
x from 4 to 4 inclusive. 

ifGraph each of the following functions, with enough computed points to 
obtain a graceful curve. 

21. x 3 . 22. x 4 . 23. - x*. 24. - z 3 . 

26. x s + 2x* - 4* + 3. 26. - 3x* - 4z 3 + 12* 2 + 6. 



if Solve each problem by introducing just one unknown x and then finding 
the maximum of a quadratic function of x, without graphing. 

27. Divide 60 into two parts whose product is a maximum. 

28. Find the dimensions of the rectangular field of largest area which 
can be inclosed with 600 feet of wire fence. 

29. In forming a trough with a rectangular cross section and open top, 
a long sheet of tin is bent upward on each long side. If the sheet is 30 
inches wide, find the dimensions of the cross section with the largest possible 
area. 

30. Divide H into two parts whose product is a maximum. 



ADVANCED TOPICS IN QUADRATIC EQUATIONS 789 

1 33. Graphical solution of an equation 

If x has a value for which the graph of f(x) meets the x-axis, then 
with this value of x we have/(x) = 0. Hence we are led to the follow- 
ing procedure. 

SUMMARY. To find approximate values of the real roots of an equa- 
tion in x graphically: 

1. Simplify and transpose all terms to one member to obtain an equa- 
tion of the form f(x) = 0. 

2. Graph the function f(x) and measure the abscissas of the points 
where the graph meets the x-axis; each of these abscissas satisfies the 
equation f(x) = 0. 

EXAMPLE 1. Solve z 2 2x 3 = graphically. 

SOLUTION. 1. Let y = x* 2x 3 and consider its graph in Figure 14, 
page 186. The graph crosses the x-axis at x = 3 and x = 1. 

2. Since y = when x = 3 and when x = 1, these are values of x for 
which z 2 - 2x 3 = 0. That is, 3 and 1 are roots of the equation. 

If the roots of f(x) = are imaginary) this would be indicated by 
the fact that the graph of f(x) would not meet the x-axis. 

1 34. Graphical solution of a quadratic equation 

In order to solve the equation 

ax 2 + bx + c = (1) 

graphically, we construct the graph of the quadratic function 

ajc 2 + bx + c. (2) 



The parabola, which is the graph of this function, 

I. cuts the x-axis in two points when and only when equation 1 has 
unequal real roots; 

II. touches the x-axis in just one pointy or is tangent to the x-axis, when 
and only when the roots are equal; 

III. does not meet the x-axis when and only when the roots are 
imaginary. 

Note 1. A parabola can be defined geometrically as the curve of inter- 
section when a right circular cone is cut by a plane which is parallel to a 
straight line on the cone through its apex. 



790 



ADVANCED TOP/CS IN QUADRATIC EQUATIONS 



ILLUSTRATION 1. In Figure 15, parabolas 
I, II, and III are, respectively, the graphs of 
the functions in the left members of the fol- 
lowing equations. 

(I) z 2 - 2x - 8 - 0; 

(II) z 2 - 2x + 1 = 0; 

(III) *'-2z + 5 = 0. 

From the graphs, we see that (I) has the 
roots x - 4 and x - - 2; (II) has equal 
roots, x = 1; (III) has imaginary roots. 

In graphing a specified quadratic func- 
tion, we have no license to simplify its 
form by multiplication or by division. 
But, before solving a quadratic equation graphically, we may (1) clear 
the equation of fractions; (2) divide out any common constant factor 
from all terms; (3) make the coefficient of z 2 positive. Operation 3 
would cause the corresponding graph to open upward. 




Fig. 15 



EXERCISE 72 

Find the real roots of the equation graphically. 
1. 2x - 5 = 0. 2. x* + 2x - 8 = 0. 



4. 

7. 

10. 



+ 4x + 7 

- 13 - 

- 12x - 9. 



5. x* - 2x + 3 = 0. 
8. 4c - 2z 2 - 5. 
11. 2z 2 = 4x + 3. 



3. x 2 + 6# + 9 = 0. 
0. 5. x* - 2x + 3 = 0. 6. f z = 2 - z 2 . 

12. 3z 2 = 6x - 5. 

grrap/& o/ x 2 4- 4z -f 4 and then t in Problems 13 and 14, find 
specified results by use of the graph. 

13. Find the values of x for which the value of the function is 1. 

14. Solve re* + 4x -f 4 = 6 by inspection of the graph. 

15. By use of a single graph, solve each of the following equations graph- 
ically: 2s 2 - 5x = 0; 2s 2 - 6x + 3 = 0; 2z 2 - 5x - 7 = 0. 

1 35. Character of the roots 

Let r and s represent the roots of ax 2 -f- foe + c 0. Then, from 
the quadratic formula, 



ADVANCED TOPICS IN QUADRATIC EQUATIONS 



197 



- b + V6 2 - 4ac 
2a ; 



8 



b - Vb 2 - 4oc 
2a 



(1) 



We assume that a, 6, and c are real numbers and that a j^ 0. 
Then, the roots are imaginary when and only when b 2 4oc is 
negative; if one root is imaginary, the other is also. 

If ft 2 4oc 0, then r = s = 6/2a. Moreover, if r , on sub- 
tracting the expressions in (1) we obtain 



= r s = 



2Vb 2 - 4oc 
2a S 



a 



0; V6 2 -4oc - 0. 



Hence, if r s then fc 2 4ac = 0. 

From the preceding remarks and Section 134, we see that the 
items in any row of the following summary hold simultaneously. 



THE ROOTS OF 
ax 2 + bx + c = 



real and unequal 

real and equal 

imaginary 



THE VALUE OP 



6 2 - 4ac > 
6 s - 4oc = 
6 2 - 4ac < 



THE GRAPH OF 
ax z + bx + c 



cuts x-axis in two points 

is tangent to x-axis 

does not touch x-axis 



If a, 6, and c are rational numbers, the roots are rational when 
and only when V& 2 4oc is real and is a rational number. That is, 
the roots are rational numbers when and only when fe 2 4oc is a perfect 
square. 

We call 6 2 4oc the discriminant of the quadratic equation 
ax* + bx + c = 0, or of the quadratic function ax 2 -f bx 4- c, because, 
as soon as we know the value of 6 2 4oc, we can tell the general 
character of the roots of the equation without solving it, and the 
general nature of the graph of the function without graphing it. 

ILLUSTRATIONS OF THE USE OF THE DISCRIMINANT 



EQUATION 


DISCRIMINANT 


HENCE, THE ROOTS ARE 


4z 2 - 3z + 5 = 


(-3)2 _ 4-4-5 = - 71 


imaginary numbers 


4z 2 - 4x + 1 = 


(- 4) 2 -4-4 = 


real; equal; rational 

% * 4 


4z 2 - 3x - 5 


(-3) 2 + 4-4-5 - 89 


real; unequal; irrational 




a 2 - 2x - 3 = 


(- 2) 2 - 4(- 3) = 16 = 4 2 


real; unequal; rational 



192 ADVANCED TOPICS /N QUADRATIC EQUATIONS 

Before computing the discriminant in any equation, it should be 
simplified by clearing of fractions and combining terms. 

EXAMPLE 1. State what you can learn about the graph of the quadratic 
function 3a 2 -\- 5x 6 without graphing. 

SOLUTION. 1. The discriminant of the function is 25 72 = 47. 

2. Hence, the graph would not touch the z-axis. Since the coefficient 
of z 2 is r- 3, the graph is concave downward and therefore must lie wholly 
below the a>axis. 

136. Conjugate imaginarics 

If two imaginary numbers differ only in the signs of the coefficients 
of fyeir imaginary parts, then either of the given numbers is called 
the conjugate of the other. 

ILLUSTRATION 1. The conjugate of (3 + 5i) is (3 5i). The conjugate 
of (a + bi) is (a bi). 

When the roots of a quadratic equation are imaginary, these roots 
are conjugate imaginary numbers, because the imaginary parts come 
from V6 2 4oc in the quadratic formula. 

ILLUSTRATION 2. The roots of x* -f- 4z + 5 = are 



- 4 =t v 16 - 20 . . , . . . 
x = ~ = 2 i, conjugate imaginanes. 



EXERCISE 73 

Compute the discriminant and tell the character of the roots, without solving. 
1. y* - 1y + 10 = 0. 2. y* - 4i/ - 21 = 0. 3. z 2 + 2z - 2 - 0. 



4. 3s 2 - 5z + 7 = 0. 6. 9z 2 + 12z + 4 = 0. 6. 4x* + 4z = 3. 

7. 30 - % 2 = 2%. 8. 3z - 2 = 5z 2 . 9. 25 + 4z 2 = 20z. 

10. 2x - 3 - fo 2 . 11. 5x 2 + 1 - 2z. 12. 25z 2 + 1 

13. 8a? - 7 - 0. 14. 5z 2 - 3x = 0. 15. 1 - 2x - 

16. 3 + 5z 2 = 0. 17. Qx = ftc 2 + 4. 18. z 2 -f .4a; 4- .3 = 0. 

Solve graphically; check the graph by computing the discriminant and thus 
determining the character of the roots. 

19. x 2 - 4x = 6. 20. z 2 + 7 - 4x. 21. 4z 2 + 4z = 1. 



ADVANCED TOPICS IN QUADRATIC EQUATIONS 193 

Compute the discriminant of the function and, without graphing, state all 
facts which you can learn about its graph. 

22. 4x* - I2x + 9. 23. 2z 2 - 3z - 5. 24. 3** - 4x. 



25. - 3z 2 + 5z - 7. 26. 4z 2 + 5x + 7. 27. - 3s 2 - 2x + 4. 

Specify the conjugate number for the imaginary number. 
28. 3 + 7t. 29. - 2 - W. 30. - 2 + V^~9. 31. 6\^"T. 

1 37. Sum and product of the roots 
By use of 



- 6 + V6 2 - 4oc , - 6 - Vfe 2 - 4oc 

r = - o - and s = - ~ - ' 
2a 2a 

- 26 6 



, . . 
we obtain r 



rs = 



2a a' 

- b + Vb 2 - 4oc - 6 - Vb 2 - 4oc 
2a ' 2a 



= - (b 2 - 4oc) ^ 4oc = c 

4a 2 4a 2 a 



Hence, for the equation ox 2 -f 6x + c = 0, 

swm o/ the roots eqvals -- : r-fs= -- (1) 

c c 

product of the roots equals -: rs = (2) 

ILLUSTRATION 1. For 3z 2 5x + 7 = 0, we find r + s = and rs = J. 

1 38. Factored form of a quadratic function 

THEOREM I. If r and s are the roots of ax 2 + bx + c = 0, 

ax* + frt + c = a(x - r)(x - 5). (1) 

Proof. 1. We can write 

ox 2 + bx + c = at x 2 4- -a? 4- - 1 

\ o a/ 

5 c 

2. From Section 137, - = (r + s) and - = rs. Hence, 

(Z d 

4- bx + c = a[x 2 (r + s)x 4- rs] = a(x r)(x s). 



194 ADVANCED TOPICS IN QUADRATIC EQUATIONS 

ILLUSTRATION 1. A quadratic equation whose roots are 5 and 3 is 

(x + 3)(z - 5) - 0, or s 2 - 2x - 15 - 0. [a - 1 in (1)] 

ILLUSTRATION 2. A quadratic equation whose roots are J(2 =b 3t) is 

> - i(2 - 30] - 0. 



To eliminate fractions we use a = 4 = 2-2, and then group within paren- 
theses to exhibit the sum and difference of two quantities, as an aid in 
multiplying: 



0; 



[(2z - 2) - 3t][(2s - 2) + 3t] - 0, or (2s - 2) - 9i - 0. 

Since i 2 = 1, we have 

4z 2 - &c + 4 -h 9 - 0, or 4s - &c + 13 - 0. 



Under certain circumstances, we have seen how to solve the equa- 
tion ax 2 -f bx + c = by first factoring the function ax 2 + bx 4- c. 
Formula 1 permits us to reverse this process and to /actor the funo 
tion by first solving the equation (of course, not using factoring in 
the solution). 

EXAMPLE 1. Factor 6x 2 23a? -f- 20 by first solving an equation. 
SOLUTION. 1. Solve 6z 2 23z + 20 0, by the quadratic formula: 

23 V49 23 db 7 
* " 12 - l2- ; 

5 A 4 

a? jr ana a; 

A O 

2. From formula 1 : 

62* - 23* + 20 = 6(z - $)(* - j) - (2* - 5)(3* - 4). 



Formula 1 states that any quadratic function of x can be expressed 
as a product of factors which are linear in x. However, these factors 
involve rational, irrational, or imaginary coefficients depending on 
the nature of the roots r and s. In particular, from the facts about 
rational roots, on page 191, we draw the following conclusion: 

// a, b, and c are rational numbers, ax* 4- bx + c can be expressed 
as a product of real linear factors with rational coefficients when and 
only when the discriminant ft 2 4ac is a perfect square. 



ADVANCED TOPICS IN QUADRATIC EQUATIONS 795 

EXERCISE 74 

Find the sum and the product of the roots of each equation, in the unknown x, 
without solving for x. 

1. x 2 + 5x - 3 = 0. 2. 2x - 5x* + 7. 3. 4z* = 3z - 6. 



4. 7 &c - 2z 2 . 5. 7 - 3x = 4z*. 6. 5s 2 - 17 = 0. 

7. 18 = 5x 2 . 8. 12z 2 + 3 = 0. 9. 5 - 9z* = 7s. 

10. 2z 2 - 3 - 5z. 11. 5 = 7z 2 - 4x. 12. 4z 2 - 12s - 9. 

13. ax* + dx = h. 14. cz 2 = 3z 6. 15. 4z* = ox + c. 

16. 2z 2 + 3z + as + c - 0. 17. 5z* + <w 2 + 3x + d = 0. 

18. 2z 2 + 3a; + 2a + d = 0. 10. & + ex* - 2x + ex - d - 5. 

Compute the indicated product. 



20. 4(* - f)(x + f). 21. 6(x - J)(* + 

22. (a; - 2 + 3i)(x - 2 - 3i). 23. (x + 1 - 2V2)(z + 1 + 2\/2). 

Form a quadratic equation with integral coefficients having the given numbers 
as roots. 
24. 3; - 7. 26. - 2; - 3. 26. J; - f . 27. J; 2. 

28. -f; -f. 29. 2; -. 30. - f; - f . 31. \/2. 

32. }V3. 33. 3V2. 34. 2i. 35. t. 

36. 1 db V2. 37. - 2 =t V5. 38. 3 2\/2. 

39. J j}V3. 40. - J =fc JV2. 41. 3 5t. 

42. - 2 3i. 43. 4 2i. 44. - Ji. 

45. 2 2iV5. 46. i f tV3. 47. - Jt V2. 

Factor, after first solving a related quadratic equation by use of the quadratic 
formula. 

48. 12z' + llz - 36. 49. 27z* + 2Lc - 40. 50. 27x 2 - fay - 16j/ 2 . 
51. 24x 2 - 13* - 60. 62. 48x 2 + 50s - 75. 53. 27x 2 + 12z - 32. 

Without factoring or solving any equation, determine whether or not the ex- 
pression has real linear factors with rational coefficients. 

64. 8z* + 7* - 2. 56. llz 2 + 12* - 5. 56. 6z* + 25xy + 26j/ J . 

if Factor, perhaps by use of imaginary or irrational numbers. 
57. a? + Ox + 10. 58. 4s* - 12x + 7. 59. 2z* - 2x + 6. 



796 ADVANCED TOPICS IN QUADRATIC EQUATIONS 

1 39. Equations in quadratic form 

EXAMPLE 1. Solve: x* - 5z 2 + 6 = 0. (1) 

SOLUTION. 1. Factor: (z 2 - 3)(z 2 - 2) = 0. 

2. If x* - 3 = 0, then x = \/3; if z 2 - 2 = 0, then x = V2. 

The given equation has four solutions, Vjji and =t V2. 

SECOND SOLUTION. 1. Let y x 2 ; then i/ 2 = z 4 and, from (1), 

2/ 2 - % + 6 = 0. 

2. Solve for y: 

(y - 3)(y - 2) - 0; 
hence, y = 3 and ?/ = 2. 

3. If y = 2, then z 2 = 2 and z = A/2. 

4. If y = 3, then x 2 = 3 and x = \/3. 

Comment. The given equation is said to be in the quadratic form in z 2 
because we obtain a quadratic in y on substituting y x*. 

EXAMPLE 2. Solve: 2ar 4 - z~ 2 - 3 = 0. 

SOLUTION. 1. Let y = or 2 ; then 2/ 2 = ar 4 and 2# 2 # 3 0. 

2. Solve for y: (2y - 3)ft/ + 1) = 0; 

hence, y = 1 and y = f . 

o ^1^9 1 

3 Tf 11 - thpn r~ 2 = - = - r 2 *= - rr = -4- -Vfi 
^ "~ 2 2' a: 2 2' 3* 3 



v 



4. If 2/ = - 1, then or 2 = - 1; = - 1; x* = - 1; x = t. 

5. The solutions are =fc t and 

EXAMPLE 3. Solve: (x* + 3z) 2 - 3x 2 - ftc - 4 = 0. 
INCOMPLETE SOLUTION. 

1. Group terms: (z 2 + 3z) 2 - 3(x 2 + 3x) - 4 = 0. 

2. Let y = x* + 3z; then y 2 3y 4 = 0; hence, y = 4, and y = 1. 



We should then solve 

z 2 3x = 4 and x 2 + 3z = - 1. 



1 . In solving an equation of the form x k = A where A; is a positive 
integer greater than 2, we agree for the present that we desire only real 
solutions unless otherwise specified. The real solutions, if any, of x k = A are 
the real A?th roots of A. Thus, x 4 = 8 has no real solutions while z 6 = 64 
has the real solutions x = 'v/ci = 2. 



ADVANCED 7OP/CS IN QUADRATIC EQUATIONS 197 

EXAMPLE 4. Obtain all roots by use of factoring: 8z 3 + 125 = 0. 
SOLUTION. 1. Factor: (2x + 5)(4z 2 - lOx + 25) = 0. 

2. Hence, 2x + 5 = 0, or 4z 2 lOz + 25 = 0. 

3. The solutions are 



x = - J and x = i(10 VlOO - 400) = 

EXAMPLE 5. Find the four 4th roots of 625. 

SOLUTION. 1. If z is any 4th root of 625, then x* 625. 

2. Solve for x: x 4 - 625 = 0; 

(z 2 - 25)(z 2 + 25) = 0; z 2 = 25 or x* = - 25.' 
Hence, x 5 and x = 5i are the desired 4th roots of 625. 

Note 2. In this section the student has met further illustrations of the 
truth of the theorem that an integral rational equation of degree n in a single 
variable x has exactly n roots (we admit the possibility that some of the roots 
may be equal). Also, we have seen illustrations of the related fact that, if 
n is a positive integer, every number H has exactly n distinct nth roots, 
some or all of which may be imaginary. 

EXERCISE 75 

Solve by the method of page 196, without first clearing of fractions when 
they occur. Results may be left in simplest radical form. 

1. x* - 5z 2 + 4 = 0. 2. & - 10z 2 + 9 = 0. 3. x* - 8x* + 16 = 0. 

4. 9z 4 + 4 = 13z 2 . 5. 4z 4 + 15z 2 = 4. 6. y 4 - ?/ = 2. 

7. y 4 4- 7y z = 18. 8. x 4 - 9 = 0. 9. Sly 4 - 16 = 0. 

10. x 6 - 8 = 7r*. 11. 27z + 1 = 28X 3 . 12. 8/ + 39y 3 = 5. 

13. 4ar 4 - liar 2 -3 = 0. 14. 36X" 4 - 13ar 2 + 1=0. 

i 

16. 250T 4 - 26x~ 2 + 1 = 0. 16. 2 + 17ar 2 - 9Z- 4 = 0. 

17. 8z 6 + 35z 3 + 27 = 0. 18. 1 - 2ar 2 - 3or 4 = 0. 

19. (x 2 - z) 2 - (&c 2 - &c) + 12 = 0. 

20. (x 2 + 3z) 2 - 3z 2 - 9z - 4 = 0. 

21. 2(2z 2 - x) 2 - 6z 2 + 3z - 9 = 0. 

22. z 2 + 4z 2 - 17z 2 - 60 - 68z - 0. 



798 ADVANCED TOPICS IN QUADRATIC EQUATIONS 



26 
25 ' 



x* + 3* 



29. a; 4 + 2x* + a; 2 - (14z 2 + 14z) + 24 = 0. 

30. 4x* - 4s 3 + * 2 + 4z 2 - 2x - 15 = 0. 

31. z w - 30z 5 = 64. 32. 6x 6 + 7s = 20. 



35 



2 T 
HINT for Problem 33. Let y = (2 - x)/x 2 . 

36. 2x* - llax 2 + 12a 2 = 0. 36. (2z 2 - 3az) 2 - 2o 2 a; 2 + 3a 3 x = 2a 4 . 

if Find all roots by first using factoring. 

37. 27z* - 8 - 0. 38. z 8 + 8 = 0. 39. x 3 - 27 = 0. 40. 16z 4 - 81. 

41. 81 - 625z 4 = 0. 42. Sy 3 - 125 - 0. 43. 125z + 27 = 0. 

if Find the three cube roots of each number. 
44. - 27. 46. 64. 46. 1. 47. - 1. 48. 8. 49. J. 60. ^y. 

if Find the four 4th roots of each number. 
61. 1. 62. 16. 63. 81. 64. 625. 66. 16. 66. 256. 67. Jf . 

1 40. An operation sometimes leading to extraneous solutions 

Let M N represent any equation. On squaring both sides, we 
obtain M 2 = N 2 , which is satisfied if3f = ATorifM = N. Hence, 
the solutions of M 2 =* N* consist of all solutions of M = N together 
with those of M = N. 

ILLUSTRATION 1. x - 5 is the only root of x 3 2. (1) 

On squaring both sides, we obtain (x 3) a * 4. (2) 

On solving (2) for x we find x 3 = 2; hence, x = 5 or x 1. 
Therefore, (2) has the root x = 1 besides the root x = 5 of (1). 

If an operation on an equation in x produces a new equation 
which is satisfied by values of x which are not roots of the given 
equation, we have agreed to call such values extraneous roots. From 



ADVANCED TOPICS IN QUADRATIC EQUATIONS 



199 



the preceding discussion, we observe that, if both members of an 
equation are squared* extraneous roots may be introduced. 

ILLUSTRATION 2. In Illustration 1, z = 1 is an extraneous root. 

141. Irrational equations 

An irrational equation is one hi which the variables occur under 
radical signs or hi expressions with fractional exponents. 

EXAMPLE 1. Solve for x in the folio whig equations (a) and (6). 



(a), 2* -2 = 



4. 



(6) 2x - 2 = - 2 z* + 4. 



SOLUTION. 1. Square both sides: 

2. 4x* - 8x .+ 4 = 2x* + 4. 

3. 2z 2 - Sx = 0; 2z(z - 4) = 0. 

4. x = and # = 4. 

TEST. Substitute x = in (a) : 

Does - 2 = V4? Or, does 
-2 = 2? No. 

Substitute x 4 in (a) : 

Does 8 - 2 = V36? Yes. 

x = is not, and = 4 is a root. 



SOLUTION. 1. Square both sides: 

2. 4z 2 - Sx + 4 = 2x 2 + 4. 

3. 2z 2 - &c = 0; 2z(z - 4) = 0. 

4. x = and x = 4. 

TEST. Substitute x = in (6): 
Does - 2 = - VI? Yes. 
Substitute x *= 4 in (6) : 
Does 8 - 2 = - V36? Or, does 

6 - 6? No. 
z = 4 is not, and x = is a root. 



Comment. We met the extraneous roots x = in solving (a) and x = 4 
hi solving (6). The test of the values obtained in Step 4 in either solution 
was necessary in order to reject these extraneous roots. The necessity for 
the test is also shown by the fact that, although (a) and (b) are different 
equations, all distinction between them is lost after squaring. 

SUMMARY. To solve an equation involving radicals: 

1. Transpose the most complicated radical to one member and all 
other terms to the other side. 

2. // the most complicated radical is a square root, square both mem- 
bers; if a cube root, cube both members; etc. 

3. Repeat Steps 1 and 2 with the effort to eliminate all radicals in- 
volving the unknowns. Then, solve the resulting equation. 

4. Test each value obtained in Step 3 by substitution in the given 
equation to determine which values are roots. 

* Also true if both sides are raised to any positive integral power. 



200 ADVANCED 7OP/CS IN QUADRATIC EQUATIONS 



Note 1. Recall that, if A is positive, VA, or A$, represents the positive 
square root of A and that VA represents only the principal nth root of A. 

m 

Also, in testing for extraneous roots, remember that we are using a to 
represent only the principal nth root of a m . 



EXAMPLE 2. Solve: (x - 2)* - v2* + 5 = 3. 



SOLUTION. 1. Vx - 2 = 3 -f V2z + 5. 



2. Square: * - 2 = 9 + 6v2*45 + 2x + 5. 

3. Simplify: - * - 16 = 6V2* 4 5. 

4. Square: * 2 + 32* + 256 = 36(2* + 5) ; 

x*- 40* + 76 = 0; (x - 38)(* - 2) = 0. 
Possible roots of the given equation are x = 38 and x = 2. 

TEST. Substitute x = 2 and * = 38 in the original equation: 



* = 2: does V2 - 2 - vT+5 = 3, or does -3 = 3? No. 



* = 38: does V38 -2 - V76 + 5 = 3, or does 6-9 = 3? No. 

Hence, neither x = 2 nor * = 38 is a root. Therefore there are no solutions 
for the given equation. 

EXERCISE 76 

Solve for x or y or z. 



2. V3 -z = 5. 



1. Va+2 = 3. 



8. v'J+l = 1. 



11. 3* = 5\/2. 



10. (3 + z 



14. 2V3 5s = 0. 



13. (2* + 3)* = - 5. 




3. V2 - 7z = - 4. 



6. v"6* - 2 = 4. 
9. (2 -f z)* = 4. 
12. (2 - 2)* = - 2. 
15. Vy = 6 - y. 

= 3-2*. 17. 3V* + 9 = 2*. 18. 4* 2 + zV3 = 0. 

- 2* -2 = 0. 20. v / * ? ^"24* -3 = 0. 21. 2i/ 2 - 3j/\/5 = 0. 

4- 1 = 1 - Vi. 
26. V2* 4 4 4- V^ = 1. 

^2 - V2*4-3 = 2. 27. V? -2* - V3 - * = 1. 

28. V3 -2* + V242* = 3. 29. V2 - 4* + 2\/l - 3* = 2. 



30, g*4 2* = 2. 31. 2*45- = 2. 



ADVANCED 7OP/CS IN QUADRATIC EQUATIONS 201 

32. V3 + x - (3 - x)* = x*. 33. 2Vx 2 + x - 2 - x = x 2 - 2. 

34. Solve for x: V3x + a - 3Vx + Va = 0. 

35. Solve for z: Vz - a + V2z + 3a = V5a. 

36. V3-f 3x = 2V3x - 2 - \/3 - x. 

37. Solve for x: Vx" + V3x + 46 = 2V2x -h b. 

38. Solve v = V20s, (a) for s; (6) for 0. 

39. Solve t = 7r\/-> (a) for Z; (6) for g. 

*40 Solve: 4x* + 7x - 2 = 0. 
SOLUTION. 1. Let y = xi; then y 2 = x$ and 4g/ 2 + 7y 2 = 0. 

2. Solve for y: (4y - l)(y + 2) = 0; y = - 2 and y = J. 

3. If y = 2, then xi = 2 or 'v'x = 2; hence, x = 8. 

4. If y = J, then x^ = J or -Six - J; hence, x = (J) 3 = 

irSolve by reducing to a quadratic in some new variable. 
41. 52 + 3Vz = 2. 42. 2x* + 9xi = 5. 43. 3X" 1 + 5 = 

44. 3x + 7xi = 6. 45. 2x* = 6 + a. 46. 2x -1 + x~* = 6. 

47. 3xi = 8xi - 4. 48. 4x^ = 7x* + 2. 49. 4x~ l + 3x~* = 1. 

60. x 2 + 2x - Vz 2 + 2x - 6 = 12. 61. 2x 2 + 3V2x 2 + 3 = 7. 

if Find all real roots. 

62. #* = 8. 53. x* = 32. 64. 2$ = 16. 65. x* = - 8. 

66. y* = 4. 67. x* = 9. 68. & = - 243. 69. x* = - 27. 

60. (2x + 1)* = 4. 61. (5 - 3x)* = 27. 62. (2 + 3x)* = 8. 

63. 2x~ 3 + 15x-* -8 = 0. 64. 3X 3 + 26x* -9 = 0. 

*1 42. Miscellaneous problems about the roots 

EXAMPLE 1. If c is a constant and 2 is one root of the equation 

3x 2 - 7x + c = 0, 
find the other root. 

SOLUTION. Let the roots be r and s, with r 2. Then, from page 193, 

r + 8 = : or, 2 + s - *; * - 



202 ADVANCED TOPICS IN QUADRATIC EQUATIONS 

EXAMPLE 2. Find the constant h if one root of the following equation 
exceeds the other by 5: x 2 x Zh = 0. 

SOLUTION. 1. Given condition: r s = 5. (1) 

2. Sum of the roots: r + s 1. (2) 

3. Product of the roots: rs 2h. (3) 

4. Solve (1) and (2) for r and s: r = 3; 8 = - 2. (4) 

5. Substitute (4) in (3): - 6 = - 2A; h = 3. (5) 

EXAMPLE 3. Find the values of k for which the following equation in 

x has equal roots: 

kx* + 2s 2 - 3kx + fc = 0. 

SOLUTION. 1. Group the terms in standard form: 

(k + 2)z 2 - 3kx + fc = 0. 

Hence, the standard coefficients are a = k + 2, 6 = 3k, and c = &. 

2. If the roots are equal, the discriminant 6 2 4oc is zero: 
discriminant = (- 3&) 2 - 4(& + 2)(fc) =0; or 5k* - Sk = 0. 

3. Hence, k(5k 8) = 0; or k = and k = f . 



THEOREM I. // one root of ox 2 + bx + c = is 2fte negative of the 
other root, then 6 = 0. 

Proof. 1. If r and s are the roots, then r = s, or r + s = 0. 

2. Hence, r + s = = 0, or 6 = 0. Therefore 6 = 0. 

' a ' 

THEOREM II. 7/6 = 0, then one root of ax 2 + bx + c is the 
negative of the other root. 

Proof. Since 6 = 0, then =0 = r + s. Hence, r = s. 



. Note 1. Theorem II is the converse of Theorem I. Theorem II could 
have been abbreviated by adding the words and conversely at the end of 
Theorem I. Or, both theorems are included in the following statement: 
" One root of ax* + bx + c = is the negative of the other root when and only 
when 6 = 0." In this statement we justify the phrase only when by 
Theorem I, and the word when by Theorem II. 

EXAMPLE 4. Find the values of the constant h so that the equation 
+ 9A 2 a? 3 -\- x will have one root the negative of the other. 

SOLUTION. 1. Write in standard form: hx* -f x(W 1) 3 0. 
2. From Theorem II: W - 1 = 0; h - d= . 



ADVANCED TOPICS IN QUADRATIC EQUATIONS 203 

*EXERCISE 77 

By use of the discriminant, find the values of the constant k for which the 
equation will have equal roots for the unknown x. 

1. 4x* - Zkx + 1 - 0. 2. 4z 2 + 5kx + 4 = 0. 3. 2fer 2 + 9 - \2x. 



4 fcc 2 + 3fo + 5 = 0. 5. x 2 - fcr 2 - 

6. 5s 2 - 2fcc - k = 0. 7. & 2 :t 2 - fcc - a; 2 - x = 3. 

8. x 2 kx + a; k - 0. 9. fee + x 2 + Axe 2 - 2x = 4. 



Find <Ae values of the constant k for which the graph of the function of x 
will be tangent to the x-axis. 

10. 5z 2 - 2kx + k. 11. 2kx* - 3fcc + 5. 12. s 2 - 3x - k - fcc. 



13. 2z 2 + 2z - 3fc - 2A;x. 14. 2x 2 - 2Jb 2 - 5kx + 5. 

In all probkms, x is the unknown and all other letters are constants. 

15. If one root is 3, find the other root: 2z 2 5x + d = 0. 

16. If one root is 2, find the other root: 3x 2 -f dx + 5 = 0. 

17. If one root is 5, find the other^oot: 2z 2 + bx 3 = 0. 

18. If one root is J, find the other root: 3z 2 + 7x + h = 0. 

Find the value of the constant h under the given condition. 

19. The sum of the roots is 5: 3hx 2 4x 5hx + 6 = 0. 



20. The sum of tl|e roots is 7: * 5x 2 - hx* -f Mhx + 4 = 0. 

21. The product of the roots is 9: 2x 2 3/u: 2 - 6z + 4/i = 0. 



22. The product of the roots is - 6: 3/w; 2 + 5x + h - 1 = 0. 

23. One root exceeds the other by 2: 2x 2 4A + 5x = 0. 

24. One root exceeds the other by 3: 3z 2 5x + 3h 6 = 0. 
26. One root is four times the other: 2x 2 + 20z + A 2 = 13. 

One root is the negative of the other; find h. 
26. hx* - Ox + Zhx - 5 = 0. 27. 2to 2 - ihx - 5h?x + 6 = 0. 



n 

28. x 2 + 12z - 3A 2 x + h = 0. 29. h*x* +^h*x + 5hx - 4 = 0. 



30. hx* + Wx = 3 + x. 31. x 2 - 3A 2 z = A - 2x. 

32. 3z 2 + 5h?x = 2 + x. 33. x 2 - hx - /i 2 a; + 2x = 0. 

34. Prove that, if ox 2 + bx + c = has one root zero, then c = 0, and 
conversely. 



CHAPTER 




THE BINOMIAL THEOREM 



143. Expansion of a positive integral power of a binomial 

By multiplication, we obtain the following results: 
(x + y) 1 = x + y\ 



(x + yY = & + 3x*y 4- 

(x + y) 4 = x* + 4x*y 4- 6z 2 2| 4- 4XI/ 3 4- 

(re 4- y) 6 = z 5 4- 5z 4 i/ + lO^y + lOz 2 */ 3 4-.5Z2/ 4 + y*. 



We see that, if n = 1, 2, 3, 4, or 5, the expansion of (x + y) n con- 
tains (n 4- 1) terms with the following properties: 

I. In any term the sum of the exponents of x and y i& n. 

II. The first term. is x n , and in each other term the exponent of x is I 
less than in the preceding term. T r 

III. The second term is nx n ~ l y, and in each succeeding term the ex- 
ponent of y is 1 more than in the preceding term. 

IV. // the coefficient of any term is multiplied by the exponent of x in 
that term and if the product is divided by the number of that term, the 
quotient obtained is the coefficient of the next term. * 

ILLUSTRATION 1. In (at^ y} 4 , the third term is 6zV* By Property IV, 
we obtain (6-2) -5- 3, Or 4, amhe'^pefficient of the fourth term. In (x + y) B , 
the fourth term is lQx*y*; m^rqjperty IV, we obtain (10-2) -s- 4, or 5, as 
the coefficient of the fifth term.* s 

V. The coefficients of terms equidistant from the ends are the same. 

ILLUSTRATION 2. The coefficient of the second term equals that of the 
next to the last term, etc. 



THE BINOMIAL THEOREM 205 

We shall assume that Properties I to V are true if n is any positive 
integer, although we have merely verified their truth when n - 1, 2, 3, 
4, and 5. The theorem which justifies this assumption is called the 
binomial theorem, which we shall accept without proof in this text. 

EXAMPLE 1. Expand (c -f w) 7 . 

SOLUTION. 1. By use of Properties I, II, and III, we obtain 
(c + w) 7 = c 7 + 7cfiw + c 5 2 -f cV + eW + c 2 ^ 6 + cufl + w 7 , 
where spaces are left for the unknown coefficients. 

2. By Property IV, the coefficient of the third term is (7-6) 4- 2, or 21; 
that of the fourth term is (21 -5) -5- 3, or 35. 

3. By Property V, we obtain the other coefficients; hence, 
(c + w>) 7 = c 7 + 7c*w + 21cV + 35cV -f 



(w\ 6 
2a -r 1 

SOLUTION. 1. (2a - |) = [(2a) + (- |) ]' 



2. We use Properties I to V with x 2a and y = ^ and keep the terms 

o 

of the binomial within parentheses in finding the coefficients: 

_ 6 = (2a) J ' 



- I or 

80 . - 160 ... 20 



ATote 1. The following array of numbers is called Pascal's Triangk. The 
successive rows give the coefficients in the successive positive integral powers 
of x -f y. To form any row after 
the second, we first place 1 at the 
left; the 2d number is the sum of 
the 1st and 2d numbers in the pre- 
ceding row; the 3d number in the 
new row is the sum of the 2d and 
3d numbers in the preceding row; 
etc. This triangle was known to Chinese mathematicians in the early 
fourteenth century, and it appeared in print in Europe for the first time 
in 1527. 



1 
1 1 

121 
1331 
14641 
1 5 10 10 5 



206 THE BINOMIAL THEOREM 

The preceding diagram exhibits the fact that the largest coefficient 
hi any power of x + y is the coefficient of the central term oMerms. 

We observe that the signs are alternately plus and minus hi the 
expansion of a power of a binomial where one term bears a plus sign 
and the other term bears a minus sign. 

144. The factorial symbol 

The symbol n\ is read "n factorial," and is an abbreviation for the 
product of all integers from I ton inclusive, where n is a positive integer. 

ILLUSTRATION 1. 5! = 1-2-3-4-5 = 120. 31 = 1-2-3 = 6. 

1L = 1-2-3-4-5-6-7 1 J_ 

10! 1-2-3-4-5-6-7-8-9-10 8-9-10 720* 

EXERCISE 78 

Expand each power by use of Properties ItoV. 
1. (a + 6) 6 . 2. (c - d) 6 . 3. (x - y) 9 . 4. (c + 3) 6 . 

6. (2 + a) 4 . 6. (x - 2o) 7 . 7. (36 - y). 8. (2c + 3d) 8 . 

9. (a + 6 2 ) 8 . 10. (c 8 - 3d} 4 . 11. (a 2 - 6 2 ). 12. (c - x 3 ) 6 . 

13. (x - i) 6 . 14. (1 - a) 9 . 16. (Vz - v^). 16. (z* + a) 6 . 

17. (- a + JT 2 ) 4 . 18. (2T 8 - x) 5 . 19. (z* - 2a~ 1 ) 4 . 



20. + . 2 1 



1.(?-36>V 
, \a / 



Find only the first three terms of the expansion. 
23. (a + 12) 16 . 24. (c - 3) 25 . 26. (a 2 + 6 3 ) 20 . 26. (1 + 2a) 10 . 

27. (1 - .I) 22 . 28. (1 + .2) 12 . 29. (1 - V2) 12 . 30. (1 - 3s 3 ) 18 . 
31. (2x - a 2 ) 80 , 32. (z* + &)". 33. (a" 1 + 3) M . 34. (x - a" 2 ) 11 . 

36. (* - y) n . 36. (a + z)*. 37. (z 2 - y) m . 38. (w; 2 .-h )*. 

Compute each factorial expression. 

71 ot 

39. 6! 40. 8! 41. 11! 42. ^ 43. ^fVi 

3! 5! 4! 

1 45. General term of the binomial expansion 

By use of Properties I to IV of Section 143, we obtain 



THE BINOMIAL THEOREM 207 

, , v. n , i , n(n - 1) 
(z T y) % + nx n ~ l y H ~-jr ^ 

, n(n- 



(1) 



In (1) we read the dots " " as "and so forth." In the terms 
in y, y* t and y 8 , we observe special cases of the following facts, which 
we shall accept without proof. 

SUMMARY. Description of the term involving y r , in (x + y) n : 

A. The exponent of x is n r. 

B. The denominator is the product of all integers from I tor inclusive; 
that is, the denominator is rl 

C. The numerator of the coefficient has r factors, the first being n and 
each other being 1 less than the preceding factor. The last factor is 
n r + 1. 

f- 

When (A), (B), and (C) are combined, they state that 

. . t . r n(n 1) (n r 4* 1) . r /\ . 
the term involving y r is - p x n r y r . (2) 



By use of formula 2, we may write 

(x + y) n = *" + nx~*y + n(n x n ~*y* -f 



, n(n - 1) * (n - r + 1) vn _ r , ir 
H -- -? - AC" r y r 



. . . J_ l*n 

r! - ir i-y- 



(3) 



We refer to (3) as the binomial formula. By use of (2), we can 
write any term of (3) without writing the other terms. Hence, we 
refer to (2) as the general term of the expansion of (x + y) n . 

ILLUSTRATION 1. The term involving y 4 in the expansion of (x + y) 7 is 

7.5.5.4 

jj x*y* or 35zV. 

EXAMPLE 1. Find the 8th term of (3o* - 6) 11 . 

SOLUTION. The 8th term will involve the 7th power of the 2d term of 
the binomial. Hence, use (2) with r = 7, x - 3a*, and y - b: 



8th term is '' <3*ty<- W - - 26,730a'&'. 



208 THE BINOMIAL THEOREM 

Note 1. To derive a formula for the rth term in (3), we notice that this 
term will contain y*" 1 as a factor. Hence, we substitute (r 1) for r in 
(2) and find that 

*i ~*u t n(n 1) (n r 4 2) , 

tte rth term is ' ^. ixn-r+i^r-i. (4) 

We may call (4), as well as (2), the general term. Example 1 could have been 
solved by use of (4) with r = 8. 

EXAMPLE 2. Find the term involving z lz in the expansion of (v z 3 ) 7 . 
SOLUTION. Since 2" = (z 3 ) 4 , we use formula 2 with n = 7, r = 4, x v, 

and y z 3 : the term is T~o~q~l t;3 (~~ 2;3 ) 4 > or 35V 3 ;? 12 . 

EXAMPLE 3. Compute (1.01) 6 correct to 3 decimal places. 

SOLUTION. (1.01) 6 = (1 4 .Ol) 6 

= I 6 + 6(1) 6 (.01) 4 15(1) 4 (.01) 2 4 

= 14 6(.01) + 15(.01) 2 -f 20(.01) 3 4 

= 1 -f .06 + .0015 + .000020 H (negligible terms) 



= 1.06152 = 1.062, approximately. 

EXERCISE 79 

Find only the specified term. 

1. Term involving y 6 in the expansion of (a 4 y) 9 . 

2. Term involving x 6 in the expansion of (z + x) 10 . 

3. Term involving y* in the expansion of (x y) 7 . 

4. Term involving z 6 in the expansion of (x 4 3y) 8 . 
5. 4th term of (a x) 9 . 6. 3d term of (w z) 11 . 

7. 6th term of (a 2 4 x) 7 . 8. 10th term of (x 2 4 y 3 ) 10 . 

9. 4th term of (x - 5y) 7 . 10. 5th term of (1 - .02) 7 . 

11. 6th term of (1 4 .I) 8 . 12. 4th term of (Jx - 

13. Term involving 2 s in the expansion of (x 2 2 ) 6 . 

14. Term involving w 10 in the expansion of (w* 4 I/ 8 ) 8 . 

15. Term involving y 8 in the expansion of (x y) n . 



THE BINOMIAL THEOREM 209 

16. Term involving z 3 in the expansion of (o a 

17. Term involving x% in the expansion of (y + 

18. Middle term of (x* - y) w . 19. Middle term of (a + 3* 2 ) 8 . 

20. Middle terms of (a* + 2/) 7 . 21. Middle terms of (2s* + j/*). 

22. Term involving -j in the expansion of ( r ) 

23. Term involving -g in the expansion of (sp -\ ) 

(y x\* 
o) ' 

Find the term or terms with the largest coefficient in the expansion of the 
power. 

25. (a + x)*. 26. (c - w>) 10 . 27. (a 2 + 6) 9 . 28. (c - rf 3 )". 

Compute by use of the binomial theorem. In any problem involving decimals t 
use only enough terms to obtain the result correct to three decimal places. 

29. (10 - a) 4 . 30. (100 - 2) 3 . 31. 99 4 . 32. 39 4 . 

33. 51 3 . 34. (1.01) 7 . 35. (1.01) 12 . 36. (1.02). 

37. (1.03) 7 . 38. (.99) 6 . 39. (.98) 6 . 40. (1.04) 10 . 

41. C1.02) 11 . 42. (.52) 8 . 43. (.49) 9 . 44. 101 B . 45. 62*. 



CHAPTER 



13 



RATIO, PROPORTION, AND VARIATION 



146. Ratio 

The ratio of one number a to a second number 6 is the quotient 
a/6. The ratio of a to 6 is sometimes written a:b. A ratio is a frac- 
tion, and any fraction can be described as a ratio: 

a:b = |- (1) 

The ratio of two concrete quantities has meaning only if they are 
of the same kind. Their ratio is the quotient of their measures in 
terms of the same unit. 

ILLUSTRATION 1. The ratio of 3 feet to 5 inches is ^. 

147. Proportion 

A proportion is a statement that two ratios are equal. That is, a 
proportion is merely a statement that two fractions are equal. The 
proportion * 

a:b = c:d means that -r = 3- (1) 

o a 

The proportion a: 6 = c:d is read "a is to 6 as c is to d." We say 
that the four numbers a, 6, c, and d form a proportion. 

'In a proportion a: 6 = c:d, the first and fourth numbers, a and d, 
are called the extremes, and the second and third, b and c, are called 
the means of the proportion. 

ILLUSTRATION 1. To solve the proportion a;: (25 x) 3:7, we first 
change it to fractional form, and then solve the resulting equation: 

x 3 

Z * *'* J x = 75 3z; ICte 75; hence, x - 7.5. 

*"~ X f * 



RAT/0, PROPORTION, AND VARIATION 211 

EXAMPLE 1. Divide 36 into two parts with the ratio 3:7. 

SOLUTION. 1. Let x and y be the parts; then x + y " 36. (2) 

x 3 

2. Also, x\y 3:7, or - =* = Hence, 7x 3y. (3) 

3. On solving the system [(2), (3)] we obtain (x - 10.8, y = 25.2). 

ATote Jf . If two triangles (or polygons of any number of sides) are similar, 
then (a) the ratio of any two sides of one triangle equals the ratio of the corre- 
sponding sides of the other triangle, and (6) the area of one triangle is to the 
area of the other as the square of any side of the first triangle is to the square of 
the corresponding side of the other triangle. 

EXAMPLE 2. The sides of a triangle are 12, 8, and 15 inches long. In a 
similar triangle, the longest side is 40 inches long. Find the other sides. 

SOLUTION. 1. Let x and y be the lengths in inches of the sides of the 
similar triangle corresponding to those sides which are 8 and 12 inches 
long in the first triangle. Then, 



y:12 = 40:15 or - - ; (4) 

r 4ft 

z:8 = 40:15 or | - ~ (5) 

o JLO 

2. Solving (4) and (5) we find y = 32 feet and x - 21J feet. 



EXERCISE 80 

Express each ratio as a fraction and simplify. 

2. J:. 3. 5:7. 4. x:- 6. 



Find the ratio of the given quantities. 
6. 75 pounds to 160 ounces. 7. 27 days to 156 hours. 

8. 51 pints to 17 quarts. 9. 25 miles to 3175 yards. 

10. 72 cubic feet to 1320 cubic inches. 

Change to fractional form and solve. 
11. 3:(20 - 2x) = 5:2. 12. (2 - 3y):(4 + 5y) = 3:2. 

13. x:(x - 25) - 6. 14. (2 - x):(3 + x) - (4 - *):(2 -f x). 

15. 2*:(5 - 3z) * 2. 16. (4 + x):(3 + x} - - 2:5*. 

17. A line 18 niches long is divided into two parts whose lengths have 
the ratio 5: 4. Find the lengths. 



272 RATIO, PROPORTION, AND VARIATION 

Solve by introducing one or more unknowns. 

18. Divide 45 into two parts whose ratio is 4: 11. 

19. Divide 90 into two parts such that the ratio of one part decreased 
by 5 to the other part decreased by 10 is 1:4. 

20. Find two numbers whose difference is 18 and whose ratio is 4:3. 

21. The sides of a triangle are 12, 8, and 18 inches long. In a similar 
triangle, the shortest side is 40 inches long. Find the other sides. 

22. The sides of a polygon are 10, 7, 4, and 8 inches long. If the long- 
est side is lengthened by 2 feet, by how much should the other sides bo 
lengthened to obtain a similar polygon? 

23. A triangle whose base is 15 inches long has an area of 220 square 
niches. Find the area of a similar triangle whose base is 6J feet long. 

24. The area of a quadrilateral is 49 square feet and its longest side 
is 12 feet long. Find the area of a similar quadrilateral whose longest 
side is 15 feet long. 

25. The area of a triangle is 150 square feet and its shortest side is 
12 feet long. Find the shortest side of a similar triangle whose area is 
30 square feet. 

26. A man 6 feet tall stands at the foot of a tower and casts a shadow 
10 feet long. How high is the tower if its shadow is 69 feet long? 

27. A man 5J feet tall stands 40 feet from a street light and casts a shadow 
9J feet long. How high is the light? 

28. Solve the proportion 3: a; = z:27 for #. 

29. Solve the proportion, a:x = x:b for x. 

If a:x x: 6, then x is called a mean proportional between a and b. 
If a:x = x:b, then x* = ab or x = =t Vo&; or, if neither a nor b is zero, 
there are two mean proportionals between a and b. Find the mean proportionals 
between each of the following pairs of numbers. 

30. 64 and 4. 31. - 4 and - J. 32. 2 and 8. 

33. 25 and 25. 34. - 2 and 8. 36. - 3 and 27. 

36. 2o 3 and 4o. 37. y* and x~ 4 . 38. zy and 

-- ,.-,,-12. _ * + 2z + 4 

' 



4 y 2-t/ ' 2-2 

*// a:b = c:x, then x is called the fourth proportional to a, b, and c. 
Find the fourth proportional to each set of numbers: 

41. 2, - 5, and 14. 42. 5, 4, and 7. 43. 3, 6, and a'6. 



RAT/O, PROPORTION, AND VARIATION 213 

*// a : 6 = 6 : x, then x is called the third proportional to a and b. Find the 
third proportional to each pair of numbers: 

44. 18; 50. 45. 2J; . 46. 2xy, y. 47. 5m*n; 3m 3 . 

*// a:b = c:d and if no denominator involved is zero, prove the following 
properties of the proportion. 

48. PROPERTY I. ad = 6c; or, in any proportion, the product of the means 
equals the product of the extremes. 

49. PROPERTY II. - == -?; or, the means may be interchanged without de- 

C tt 

stroying the proportion. (The resulting proportion is said to be obtained 
from a:b = c:d by alternation.) 

50. PROPERTY III. - = - (The resulting proportion is said to be ob- 
tained from a:b = c:d by inversion.) 

51. PROPERTY IV. T = -7 (Said to be obtained by composi- 
tion. To prove, add 1 to both sides of a: b = c:d.) 

52. PROPERTY V. r = -7 (Said to be obtained by division.) 

148. Direct variation 

Let x and y be related variables. Then, we say that 

y is proportional to x, or 

y varies directly as x, or 

y is directly proportional to x y or 

y varies as x } 

in case there exists a constant k such that, for every value of x, there 
is a corresponding value of y given by 

y = kx. (1) 

We call k the constant of proportionality or the constant of variation. 

ILLUSTRATION 1. The circumference C of a circle varies directly as the 
radius r because C = 2irr. The constant of proportionality is 2ir. 

11 
From y = kx, we obtain k = - Hence, if y is proportional to x, 

X 

the ratio of corresponding values of y and z is a constant. Con- 



RATIO, PROPORTION, AND VARIATION 
versely, if the ratio of corresponding values of two variables y and x 

y 

is a constant, then y is proportional to x, because the equation - * k 

X 

leads to y = kx. 

ILLUSTRATION 2. If y is a function of x and if it is known that - = 4, 

X 

then y = 4x, and y is proportional to x. 

If y is proportional to x, then x is proportional to y. In other words, 
the proportionality relationship is a reciprocal property. This is true 
because, if y kx, then 

* - T y. (2) 

Hence, if y varies as x, with fc as the constant of proportionality, tnen 

x varies as y, with l/k as the constant of 

proportionality. 

If y varies directly as x, so that equation 
1 is true, then the graph of the relationship 
is a straight line, because (1) is linear in 
x and y. We observe that, for any value 
of k, the graph of (1) passes through the 
origin, because (x 0, y = 0) is a solution 
of (1). 

ILLUSTRATION 3. If y is proportional to x, 
with 3 as the constant of proportionality, then 
y = 3x. The graph of this equation is given 
in Figure 16. 

1 49. Inverse variation Fis. 16 

We say that 

y is inversely proportional to x, or 

y varies inversely as x, 

in case there exists a constant k such that, for every value of x, there 
is a corresponding value of y given by 

(i) 




k 

y = 
* x 



From this equation, k - xy, or the product of corresponding values 
of x and y is a constant. 



RATIO, PROPORTION, AND VARIATION 



215 



ILLUSTRATION 1. The time t necessary for a train to go a given distance 
d varies inversely as the speed s of the train because t =* d/8. The constant 
of proportionality here is d. 

If y varies inversely as x, with k as the constant of proportionality, 
then likewise x varies inversely as y, because the equation A; xy, 
which comes from (1), leads to 
both of the equations 



y 



* K 

- and = - (2) 

x y 




ILLUSTRATION 2. If y varies in- 
versely as x, with 4 as the con- 
stant of proportionality, then 
y = 4/x or xy = 4. The graph of 
y as a function of x is the graph of 
the equation xy = 4. This graph 
has no points for which x = 
or y = because in such cases 
xy = 0, and hence xy j* 4. We 
make up the following table of val- 
ues by substituting the values of x 
in y = 4/z. The graph, in Figure 17, extends beyond all limits upward 
and downward, approaching the y-axis as shown. Similarly, as x grows 
numerically large without bound, either through positive or through negative 
values, y approaches zero and the curve approaches the z-axis. The curve 
in Figure 17 is an illustration of a hyperbola.* 



Fis. 17 



X 


-8 


4 


- 2 


1 


- i 




* 


i 

16 


1 


2 


4 


8 


y 


-i 


j 


-2 


4 


- 16 


4 


2 


1 


i 



1 50. Joint variation 

We say that 

z varies jointly as x and y, or 

z is directly proportional to x and y, or 

z is proportional to x and y, or 

z varies as x and y, 

in case z is proportional to the product xy, or 

z'= hxy, 
* Graphs of hyperbolas are considered in detail in Chapter 16. 



d) 



(2) 



216 RATIO, PROPORTION, AND VARIATION 

where k is a constant of proportionality. Notice that the significance 
of the word and hi each statement hi (1) is that x and y are multiplied 
in (2). Any of the various types of variation may be combined. 

ILLUSTRATION 1. To say that z varies directly as x and y and inversely 
as uP means that z kxy/vP. 



ILLUSTRATION 2. If P = lOafy/s 3 , then P varies directly as # 2 and y, 
and inversely as z 3 . 

1 51 . Applications of variation equations 

Suppose it is known that certain variables are related by a variation 
equation, with an unknown constant of proportionality, k. Then, 
if one set of corresponding values of the variables is given, we can 
find A; by substituting the values in the variation equation. 

EXAMPLE 1. If y is proportional to x and w 2 , and if y = 36 when x = 2 
and w ~ 3, find y when x = 3 and w = 4. 

SOLUTION. 1. We are given that y - ku?x, where k is an unknown 
constant. 

2. To find k, substitute (y = 36, x = 2, w = 3) in y = kw*x: 

36 = &(3 2 )(2); 36 = 18fc or k = 2. (1) 

3. From (1), y = 2w*x. (2) 

4. Substitute (a? - 3, w - 4) in (2): y - 2- 16-3 = 96. 

Notice that the following steps were taken hi Example 1. 

1. The variation statement was translated into an equation involving 
an unknown constant of proportionality. 

* 

2. The unknown constant was found by substituting given data. 

3. The value of the constant of proportionality was substituted in the 
equation of variation, and this equation was used to obtain the value 
of one variable by use of given values of the other variables. 

Useful information can be obtained by use of an equation of varia- 
tion on many occasions when the data are not sufficient to enable us 
to find the value of the constant of proportionality. 

EXAMPLE 2. The kinetic energy of a moving body is proportional to the 
square of its velocity. Find the ratio of fye kinetic energy of an automobile 
traveling at 50 miles per hour to the kinetic energy of the same automobile 
traveling at 20 miles per hour. 



RATIO, PROPORTION, AND VARIATION 217 

SOLUTION. 1. Let E be the energy, and v the velocity in miles per hour. 
Then, E = kv 2 , Where A; is a constant of proportionality. (The data do not 
permit us to find the value of &.) 

2. Let EI be the energy at 20 miles per hour, and E 2 the energy at 50 miles 
per hour. Then, 

#1 - *(20) 2 or E t - 400fc; (3) 

# 2 = &(50) 2 or # 2 = 2500&. (4) 

3. From (3) and (4), 

E 1 _ 2500fe _ 26 ^ 
Ei ~ 400& ~ 4 " **' 

Thus, E 2 is 6J times as large as EI. 

Note 1. In applications of an equation of variation, the constant of pro- 
portionality will depend on the units in terms of which the variables in the 
problem are measured. Hence, if the constant is determined for one set of 
units, care must be exercised to employ the same units whenever this value 
of the constant is used. 

EXERCISE 81 

Introduce Utters if necessary and express the relation by an equation involv- 
ing an unknoum constant of proportionality. 

1. H varies directly as x and inversely as wP. 

2. B is proportional to x 2 and inversely proportional to z. 

3. Z is proportional to \ /r x and varies inversely as y*. 

4. K is proportional to z and w 2 and inversely proportional to xy. 
6. (x + 2) is inversely proportional to (y + 3). 

6. The area of a triangle is proportional to its altitude. 

7. The volume of a sphere is proportional to the cube of its radius. 

8. The volume of a specified quantity of gas varies inversely as the 
pressure applied to it, if the temperature remains unchanged. 

9. The weight of a body above the surface of the earth varies inversely 
as the square of the distance of the body from the earth's center. 

10. The power available in a jet of water varies jointly as the weight of 
the water per cubic foot, the cube of the water's velocity, and the cross- 
section area of the jet. 

11. The maximum horsepower of the boiler which can be served by a 
chimney of given cross-section area is proportional to the square root of 
the height of the chimney. 



218 RATIO, PROPORTION, AND VARIATION 

For each formula, give a statement about the variable on the left side in the 
language of variation. All letters except the constant k represent variables. 

12. y - 7w. 13. z 3x\ 14. z = 5xy*. 16. u = 



4* 4 - <> rt 

16. w = -T- 17. 10 r- 18. u = 19. w - 

y # y 2 z 

By employing all data, obtain an equation relating the variables. 

20. P is directly proportional to x* and P 18 if x = 4. 

21. R is inversely proportional to x and directly proportional to y, while 
R = 4 when x 3 and y 5. 

22. U varies directly as 3 and y, and inversely as z 2 ; C7 = 15 when a; = 5, 
y = 2, and 2 = 3. 

23. # varies jointly as x and y and inversely as Vz; H = 6 when z = 2, 
y 3, and 2 = 9. 

24. If w is proportional to z and if w = 5 when x = 7, find w when 
* - - 6. 

25. If y is inversely proportional to x and if y 5 when x = 20, find y 
when x 15. 



26. If H is proportional to x and inversely proportional to y, and if 
H = 3 when 2 and y = 4, find H when y = 9 and x = 5. 

27. The distance fallen by a body, starting from a position of rest in a 
vacuum near the earth's surface, is proportional to the square of the num- 
ber of seconds occupied in falling. If a body falls 256 feet in 4 seconds, 
how far will it fall in 7 seconds? 

28. The kinetic energy E, of a mass of m pounds moving with a velocity 
v, is proportional to mv 1 . If E = 2500 foot-pounds when a body weighing 
64 pounds is moving at a velocity of 50 feet per second, find the kinetic 
energy of a body weighing 30 pounds whose velocity is 2400 feet per minute. 

29. If one body is sliding on another, the force of sliding friction is pro- 
portional to the normal pressure between the bodies (if this pressure is 
moderate). If the sliding friction between two cast-iron plates is 60 pounds 
when the normal pressure is 270 pounds, find the normal pressure when the 
sliding friction is 600 pounds. 

30. The maximum safe load of a horizontal beam supported at its ends 
varies directly as its breadth and the square of its depth and inversely as 
the distance between the supports. If the maximum is 2400 pounds for a 
beam 4 inches wide and 10 inches deep, with supports 15 feet apart, find 
the maximum load for a beam of the same material which is 3 inches wide 
and 5 inches deep, with supports 25 feet apart. 



RAT/0, PROPORTION, AND VARIATION 219 

31. How far apart may the supports be placed if a beam 5 inches wide 
and 8 inches deep, like those in Problem 30, supports 6000 pounds? 

32. A beam like those in Problem 30 is 6 inches wide and the supports are 
12 feet apart. How deep must the beam be to support 3500 pounds? 

33. The approximate amount of steam per second which will flow through 
a hole varies jointly as the steam pressure and the area of a cross section 
of the hole. If 40 pounds of steam per second at a pressure of 200 pounds 
per square inch flows through a hole whose area is 14 square niches, (a) how 
much steam at a pressure of 260 pounds per square inch will flow through 
a hole whose area is 20 square inches; (6) what is the area of a hole which 
allows 30 pounds of steam to flow through it when the pressure is 300 pounds 
per square inch? 

34. The electrical resistance of .a wire varies as its length and inversely as 
the square of its diameter. If a wire 350 feet long and 3 millimeters in di- 
ameter has a resistance of 1.08 ohms, find the length of a wire of the same 
material whose resistance is .81 ohm and diameter is 2 millimeters. 



36. If y is proportional to x and if y = 16 when x = 4, graph the relation 
between x and y. Make a statement about the change in the value of y t 
(a) if x varies from any given value to a value three times as large; (6) if 
x increases by 25% from a given value. 

36. Repeat (a) and (b) of Problem 35 in case y is inversely proportional 
to x and y = 16 when x = J. 

37. The approximate velocity of a stream of water, necessary to move a 
round object, is proportional to the product of the square roots of the ob- 
ject's diameter and its specific gravity. If a velocity of 11.34 feet per second 
is needed to move a stone whose diameter is 1 foot and specific gravity is 4, 
how large a stone with specific gravity 3 can be moved by a stream whose 
velocity is 22.68 feet per second? 

38. Read Example 2 in Section 151. Find the ratio of the kinetic energy 
of a skater whose speed is 20 miles per hour to his energy when his speed 
is 15 miles per hour. 

39. The horsepower that can be safely transmitted by a solid circular 
steel shaft varies jointly as the cube of its diameter and the number of revolu- 
tions it makes per minute. If a shaft 1.5* in diameter rotating at 1520 revolu- 
tions per minute can' transmit 135 horsepower, find the speed at which the 
shaft could transmit 162 horsepower. 

40. The illumination received from a source of light varies inversely as 
the square of the distance from the source, and directly as its candle power. 
At what distance from a 50 candle power light would the illumination be 
one half that received at 30 feet from a 40 candle power light? 



220 RATIO, PROPORTION, AND VARIATION 

41. Newton's Law of Gravitation states that the force with which each of 
two masses of m pounds and M pounds attracts the other varies directly 
as the product of the masses and inversely as the square of the distance 
between the masses. Find the ratio of the force of attraction when two 
masses are 8000 miles apart to the force when they are 2000 miles apart. 

42. As a first approximation, it is found that the wind pressure on a 
surface at right angles to the direction of the wind varies jointly as the area 
of the surface and the square of the wind velocity. What wind velocity 
would be necessary to cause the pressure on 40 square feet of surface to 
be double the pressure exerted on 10 square feet by a wind velocity of 30 miles 
per hour? 

43. The current in an electric circuit varies directly as the electromotive 
force and inversely as the resistance. In a certain circuit, the electromotive 
force is A volts, -the resistance is b ohms, and the current is c amperes. If 
the resistance is increased by 20 ; %, what per cent of increase must occur 
in the voltage to increase the current by 30%? 

Note 1. The statement x is to y is to z as r is to s is to t, or x, y, and z 
are proportional to r, s, and t is abbreviated by x: y:z = r:s:t, and means 
that there exists a number k 9* such that x =? AT, y = ks, and z kt. 

if Find x, y, and z under the given conditions. 

44. x:y:z = 4: 2:5, and x + %y + z = 40. 



HINT, x 4k; y = 2&; z = 5k. Substitute in the given equation 
and find the value of k. 

46. x:y:z = 5: 3:2, and x y z = 12. 

46. x:y:z = 2:5:1, and x* + y 2 + 2 = 120. 

47. x:y:z = 3: - 1:2, and x 2 + y 2 + *> = 56. 

48. Divide 2800 into four parts proportional to 5 : 3 : 4 : 2. 

49. Divide 1250 into four parts proportional to 3:5:11:6. 



CHAPTER 



14 



PROGRESSIONS 



1 52. Arithmetic progressions 

A seqwnce of things is a set of things arranged in a definite order. 
An arithmetic progression (abbreviated A.P.) is a sequence of numbers 
called terms, each of which, after the first, is derived from the pre- 
ceding one by adding to it a fixed number called the common differ- 
ence. The common difference can be found by subtracting any term 
from the one following it. 

ILLUSTRATION 1. In the arithmetic progression 9, 6, 3, 0, 3, , the 
common difference is 3. The 6th term would be 6. 

153. The nth term in an arithmetic progression 

Let a be the first term and d be the common difference. Then, the 
second term is a + d; the third term is a -f 2d; the fourth term is 
a -f 3d. In each of these terms, the coefficient of d is 1 less than the 
number of the term. Similarly, the tenth term is a + 9d. The nth 
term is the (n l)th after the first term, and is obtained after d has 
been added (n 1) times, in succession. Hence, if I represents the 
nth term, 

/ = a + (n - l)d. (1) 



ILLUSTRATION 1. If a = 3 and d 4, the 18th term is 3 -f- 17(4) = 71. 

1 54. Sum of an arithmetic progression 

Let S be the sum of the first n terms of an A.P. The first term is a; 
the common difference is d', the last term is I; the next to the last 
term is I d, etc. On writing the sum of the n terms, forward and 
backward, we obtain 



222 PROGRESS/ONS 



l; (1) 
S~l + (l -d) + (l -2d) + --- + (a + 2d) + (a + d) + a. (2) 

On adding corresponding sides of (1) and (2) we obtain 



where there are n terms (a + 1). Hence, 2S = n(a + 0, or 

S = 5 ( a + I). (3) 

EXAMPLE 1. Find the sum of the A.P. 8 + 5 4- 2 4- to twelve terms. 
SOLUTION. 1. First obtain I from I = a + (n l}d. We have 

a 8, d - - 3, and n = 12: I = 8 + 11(- 3) = - 25. 

2. From (3), S = 6(8 - 25) = - 102. 

If we rewrite (3) in the form 

S = np-J)> (4) 

we observe that the sum of an A.P. of n terms equals n times the 
average of the first and last terms. 
On substituting I * a + (n l)dm (3), we obtain 

S = % [2a 4- (n - l)d]. (5) 

& 

The quantities a, d, I, n, and S are called the elements of the general 
arithmetic progression. When three of the elements are given, we 
may obtain the other two by use of I = a 4- (n l)d and formulas 
3 and 5. 

EXAMPLE 2. Find the remaining elements in an A.P. for which a = 2, 
I 402, and n = 26. 

SOLUTION. 1. We wish to find d and S. From (3), S = 13(404) = 5252. 
2.' From I = a + (n - l)d, 402 = 2 + 25d; hence, rf = 16. 

If a sequence of three numbers a, 6, and c forms an A.P., then 
b a c 6, because each side of this equation is equal to the 
common difference. 

EXAMPLE 3. Find the value of k if (17, k, 29) form an A.P. 
SOLUTION, k - 17 = 29 - k', 2k = 46; hence, k = 23. 



PROGRESSIONS 223 

t 

EXAMPLE 4. Find the sum of the A.P. 6 + 9 + 12 H h 171. 

SOLUTION. 1. We have given a - 6, d 3, and I = 171. 

2. To find n, use Z = a + (n - l)d: 

171 = 6 + 3(n- 1); 171 = 6 + 3n-3; n = 56. 

KA 

3. To find 5, use (3) : 8 = ^ (6 + 171) - 4956. 

& 

EXAMPLE 5. Find the 3&th term in an A.P. where the 4th term is 8 
and the common difference is 5. 

SOLUTION. 1. Think of a new A.P. where 8 is the 1st term; the former 
39th term is the 36th term of the new progression. 

2. Use I = a + (n - l)d with a = - 8, d = 5, and n = 36: 
desired 39th term = - 8 + 35(5) - 167. 

EXERCISE 82 

Write the first six terms of an A.P. from the given data. 
1. a = 15; d = 3. 2. a = 17; d = - 3. 

3. a = 18; d = 2. v / 4. o = &; d = A. 

Which sequences do not form arithmetic progressions? 
6. 3, 7, 11, 15. 6. 15, 17, 20, 22. 7. 23, 20, 17. 8. 35, 32, 30, 28. 

Find the value of b for which the sequence forms an A.P. 
9. 3, 8, b. 10. 25, 21, b. 11. 15, b, 13. * 12. b, 17, 23. 

Find the specified term of the A.P. by use of a formula. 

13. Given terms: 4, 7, 10; find the 50th term. 

14. Given terms: 5, 8, 11; find the 29th term. 

15. Given terms: 4, 2, 0; find the 41st term. 

16. Given terms: 3, 3J, 3J; find the 83d term. 

17. Given terms: 2.4, 2.6, 2.8; find the 39th term. 

18. Given terms: 3, 2.95, 2.9; find the 201st term. 

Find the last term and the sum of the A.P. by use of formulas. 

19. 8, 13, 18, to 15 terms. 20. 3, 5, 7, to 41 terms. 

21. 9, 6, 3, to 28 terms. 22. 13, 8, 3, to 17 terms. 

t 

28. 2.06, 2.02, 1.98, to 33 terms. 24. 5, 4i, 4, to 81 terms. 



224 PROGRESSIONS 

In each problem, certain of the elements a, d, I, n, and S are given. Find 
the missing elements. 

25. a = 10, 1 = 410, n - 26. 26. a = 4, 1 - 72, n = 18. 

27. a - 17, 1 - 381, d - 4. 28. i - 53, d = 4, n = 19. 

29. I - 87, d - - 3, n - 18. 30. a = 27, Z = 11, d = - J. 

31. a - 60, Z = 0, d = - f. 32. S = - 2496, n = 52, a = 3. 

33. S = 2337, n - 38, d - J. 34. n = 26, S = 5278, d = 16. 

Ftnd the value of k for which the sequence of three terms forms an A.P. 

36. (3 - 2*); (2 - *); (4 + 3*). 36. (2 + *); (2 + 4fc); (6fc - 1). 

37. Find the 45th term in an A.P. where the 3d term is 7 and the common 

difference is J. 

t 

38. Find the 59th term in an A.P. where the 4th term is 9 and the common 
difference is .4. 

39. In the A.P. .97, 1.00, 1.03, -, which term is 5.02? 

40. In the A.P. 16, 13.5, 11, , which term is - 129? 

41. Find the common difference of an A.P. whose 6th term is 9 and 
37th term is 54. 

155. Arithmetic means 

The first term, 'a, and the last term, l } in an arithmetic progression 
are called the extremes of the progression. The other terms are called 
arithmetic means between a and /. To insert k arithmetic means be- 
tween two numbers, a and J, means to find a sequence of k numbers 
which, when placed between a and Z, give rise to an A.P. with a and / 
as its extremes. 

EXAMPLE 1. Insert five arithmetic means between 13 and 11. 

SOLUTION. 1. After the means are inserted, they will complete an A.P. 
of seven terms, with a = 13 and I = 11. We shall find d for the progres- 
sion and then form the terms. 

2. From I = a + (n - l)d, 

- 11 - 13 + 6d; d - - 4. 

3. Hence, the missing terms are (13 4), or 9; (9 4), or 5; etc. 
The A.P. is (13, 9, 5, 1, 3, 7, 11). Therefore the arithmetic means 
are (9, 5, 1, - 3, - 7). 



PROGRESSIONS 225 

When a single arithmetic mean is inserted between two numbers, 
it is called the arithmetic mean of the numbers. Thus, if (6, A, c) 
form an A.P., then A is called the arithmetic mean of 6 and c. Then, 
A - b = c - A or 2 1 A = c + 6. Hence, 



x _ 

A 



or the arithmetic mean of two numbers is one half of their sum. Thus, 
the arithmetic mean of b and c is the number which is frequently 
called the average of 6 and c. 

ILLUSTRATION 1. The arithmetic mean of 7 and 15 is J(7 + 15) = 11. 

Note 1. The average of k numbers is denned as their sum divided by k. 
As a generalization of equation 1, the average of k numbers is frequently 
called the arithmetic mean of the numbers. Unless we are dealing with just 
two numbers, so that k 2, the arithmetic mean of k numbers has no con- 
nection with the notion of arithmetic means as they occur in arithmetic pro- 
gressions. 

1 56. Applications of arithmetic progressions 

In a problem dealing with an A.P., write down the first few terms 
of the progression and describe them in the language of the problem. 
Then, decide which elements are known and which you wish to find. 

EXAMPLE 1. A man invests $1000 at the end of each year for 30 years at 
6% simple interest. Find the accumulated value of his investments at the 
end of 30 years, if no interest is withdrawn until then. 

SOLUTION. 1. The first $1000 invested will draw interest at 6% for 

29 years, or a total of $1740 interest; the resulting amount at the end of 

30 years is $2740. 

2. The second $1000 invested will draw interest for 28 years; the re- 
sulting interest is $1680 and the amount at the end of 30 years is $2680. 

3. Etc.; the $1000 invested at the end of 29 years will draw interest for 
just one year; the resulting amount is $1060. The last $1000 is invested at 
the end of 30 years and receives no interest. 

4. The total amount at the end of 30 years is 



274Q + 2680 + 2620 + + 1060 + 1000. 
We wish S for an A.P. in which a = 2740, n = 30, and d = - 60. 

5. From S - 5 (a + !), S =.^(2740 + 1000) * $56,100. 



226 PROGRESSIONS 

EXAMPLE 2. A contractor has agreed to pay a penalty if he uses more 
than a speckled length of time to finish a certain job. The penalties for 
excess time are $25 for the 1st day and, thereafter, $5 more for each day 
than for the preceding day. If he pays a total penalty of $4050, how many 
excess days did he need to finish the work? 

SOLUTION. 1. The penalties are $25, $30, $35, , which form an A.P. 
where a 25, d = 5, and S = 4050. We wish to find the number of terms, n. 

2. From S = Jn[2a + (n - 1)<T|, 

4050 - -[50 + 5(w - 1) J (1) 

* / 

3. To solve (1), multiply both sides by 2: 

8100 - 50n + 5n 2 - 5n; 
w 2 + 9n - 1620 - 0. (2) 

On solving (2) by factoring, or the quadratic formula, we find n = 36 and 
n = 45. The negative root has no application in the problem. Hence, 
there are 36 excess days. 

CHECK. The student should compute the sum of 

25 + 30 + 35 H to 36 terms. 



EXERCISE 83 

1. Insert four arithmetic means between 2 and 17. 

2. Insert five arithmetic means between 2 and 40. 

3. Insert five arithmetic means between 7 and 17. 

4. Insert four arithmetic means between 19 and 12. 

6. Insert six arithmetic means between 15 and 16.5. 

6. Insert seven arithmetic means between f and 7. 

7. Insert five arithmetic means between f and 6. 

/ m 

Find the arithmetic mean of the numbers. 
8. 6; 38. 9. 15; 37. 10. - 13; 27. 11. - 15; - 23. 12. x; y. 

13. Find the sum of all even integers from 10 to 380 inclusive. 

14. Find the sum of all odd integers from 15 to 361 inclusive. 

15. Find the sum of the first 38 positive integral multiples of 3. 

16. Find the sum of all positive integral multiples of 5 which are less 
than 498, 



PKOGKESS/ONS 227 

17. There are 16 rows of billiard balls in a symmetrical triangular 
arrangement on a table, with 46 balls in the first row and 3 less balls in 
each other row than hi the one preceding it. How many balls are on the 
table? 

18. Find the sum of all positive and negative integral multiples of 6 
between 55 and 357. 

19. The horizontal base of a right triangle is 15 feet long and the side 
perpendicular to this base is 45 feet long. At intervals of 1 foot on the 
base, a perpendicular is drawn to the base and reaches to the hypotenuse. 
Find the sum of the lengths of all perpendiculars, including the vertical 
leg of the triangle. 

20. A man invests $1000 at the end of each year for 12 years at 6% 
simple interest. What is the accumulated value of his investments at 
the end of 12 years? 

Find the total sum of money paid by the debtor in discharging his debt. 

21. Debtor borrows $10,000. Agrees to pay: at the end of each year 
for 10 years, $1000 principal and simple interest at 3% on all principal 
outstanding during the year. 

22. Debtor borrows $20,000. Agrees to pay: at the end of each year 
for 20 years, $1000 principal and simple interest at 5% on all principal 
outstanding during the year. 

23. A man invests $1000 at the beginning of each year for 20 years at 
5% simple interest. Find the accumulated value of his investments at the 
end of 20 years. 

24. The 4th term of *an A.P. is 215 and the 44th term is 55. Find the 
sum of the first 20 terms. 

25. If y = 5x -j- 8, find the sum of the values of y corresponding to 
the successive integral values x = 1, 2, 3, , 30. 

26. The bottom rung of a ladder is 28 inches long and each other rung is 
one half inch shorter than the rung below it. If the ladder has 18 rungs, how 
many feet of wood were used in making the rungs? 

*. 

157. Geometric progressions 

A geometric progression (abbreviated G.P.) is a sequence of num- 
bers called terms, each of which, after the first, is obtained by multi- 
plying the preceding term by a fixed number called the common ratio. 
The common ratio equals the ratio of any term, after ihe first, to the one 
preceding it. 



223 PROGRESSIONS 

ILLUSTRATION 1. In the G.P. 16, 8, -f 4, 2, , the common ratio is 
- J; the 5th term would be (- J)(- 2) = + 1. 

To determine whether or not a sequence of numbers forms a 
geometric progression, we divide each number by the one which pre- 
cedes it. All of these ratios are equal if the terms form a G.P. 

8 x 64 

ILLUSTRATION 2. If 3, 8, and x form a G.P., then = or x = -=- 

O O O 

If the terms of a G.P. are reversed, the terms will form a G.P. whose 
common ratio is the reciprocal of the ratio for the given G.P. 

ILLUSTRATION 3. In the G.P. (4, 8, 16, 32), the common ratio is 2. When 
the terms are reversed, we have (32, 16, 8, 4), where the ratio is 



ILLUSTRATION 4. The G.P. (a, ar, ar 2 , ar 3 ) has the common ratio r whereas 
the G.P. (ar 8 , ar 2 , ar, a) has the common ratio arVar 8 or 1/r. 

1 58. The nth term of a geometric progression 

Let a be the first term and r be the common ratio. Then, the 
second term is ar; the third term is ar 2 . In each of these terms the 
exponent of r is 1 less than the number of the term. Similarly, the 
eighth term is or 7 . The nth term is the (n l)th after the 1st and 
hence is found by multiplying a by (n 1) factors r, or by r n-1 . 
Hence, if I represents the nth term, 

(1) 



ILLUSTRATION 1. If a = 3 and r = 2, the 7th term is 3(2 6 ) = 192. 

1 59. Sum of a geometric progression 

Let S be the sum of the first n terms of a G.P. The terms are 
(a, ar, ar 2 , *, ar n " 2 , ar"- 1 ), where ar n " 2 is the (n l)th term. 

Hence, S =* a + ar 4- ar 2 + + ar n ~* -f ar n-1 , (1) 

and Sr ar -f ar 2 + ar 8 -f -f- ar n-1 + ar n ; (2) 

in (2) we multiplied both sides of (1) by r. On subtracting each side 
of (2) from the corresponding side of (1), we obtain 

S - Sr = a - ar n , (3) 

because each term, except ar n , on the right in (2) cancels a correspond- 
ing term in (1). From (3), S(l r) - a ar n , or 



PROGRESSIONS 229 



Since I - ar n ~ l y then rl ar n . Hence, from (4), 



In using (4), it is sometimes convenient to rewrite it as 

1 - r n 
S = ai L. (6) 

EXAMPLE 1. Find the sum of the G.P. 2, 6, 18, to six terms. 

SOLUTION. n = 6; a = 2; r = 3. From (4), 

2 - 2-3 2 - 1458 79R 
S== 1-3 = -2 = 728 ' . 

Formula 5 is convenient when Z is explicitly given. 

EXAMPLE 2. Find the sum of the geometric progression 
(1.05) 2 + (1.05) 6 + (1.05) 8 + - + (1.05) 86 . 

SOLUTION, a = (1.05) 2 ; r = (1.05) 3 ; I = (1.05) 35 . From formula 5, 

o = (1.05) 2 - (1.05) 8 (1.05) 36 (L05) 2 - (1.05) 88 
6 1 - (1.05) 3 1 - (1.05) 3 

Note L When a sufficient number of the elements (a, r, n, Z, S) are given, 
we find the others by use of I = ar n-1 , (4), and (5). 

EXAMPLE 3. If S = 750, r = 2, and Z = 400, find n and a. 

a rl 



SOLUTION. 1. From S 



l 



t-rrr\ < K/ % 

750 = -= =~; hence, a = 50. 

J. ~~ & 

2. From I = ar n ~\ 400 = 50(2*-'); 2r~ l = = 8; 

2n-i _ 23; hence, n 1 = 3, or n - 4. 

If three numbers (a, 6, c) form a G.P.. then - = T- 

a o 

10 50 
ILLUSTRATION 1. If (a, 10, 50) form a G.P. then ~ Tn or a 



230 PROGRESSIONS 

1 60. Geometric means 

The first term, a, and the last term, I, in a G.P. are called the 
extremes of the progression. The other terms are called geometric 
means between a and I. To insert k geometric means between two 
numbers, a and I, means to find a sequence of k numbers which, 
when placed between a and Z, give rise to a G.P. with a and I as its 
extremes. In asking for geometric means, we shall desire only real 
valued means. 

EXAMPLE 1. Insert two geometric means between 6 and ^. 

SOLUTION. After the means are inserted, they will complete a G.P. of 
four terms with a = 6 and I = *-. We shall find the common ratio of the 
progression, and then its two middle terms. From I = ar n ~ l t with n = 4, 
we obtain 

i? = fir- rsA. 'OE..? 

9 or; 27 ; r ^27 3' 

The G.P. is (6, 4, f , *). The geometric means are 4 and f . 

EXERCISE 84 

Write the first four terms of a G.P. for the given data. 
1. a = 5; r = 3. 2. a = 16; r = J. 

3. a = 4; r - 2. 4. a = 27; r = - J. 

In case the numbers form a G.P., state its common ratio and write two 
more terms of the G.P. 

6. 4, - 12, 36, 108. 6. 10, 5, J, f . 7. 8, f , f , ft. 8. 4, 2, 1, 0. 

\ 

9. a, ax, ax 2 , ax 8 . 10. (1.02) 4 , (1.02), (1.02) 8 . 

11. (1.01)- 5 , (1.01)- 8 , (1.01)-*. 

Find the value of x so that the three numbers form a G.P. 
12. 5, 20, x. 13. x, 12, 36. 14. 4, x, 16. 15. x, - 4, ia 

16. If the G.P. (81, 27, 9, 3) is reversed, what fact do you observe? 

By use of I = ar""" 1 , find the specified term of the given G.P. without finding 
the intermediate terms. 

17. 3, 9, 27; find the 6th term. 18. 4, - 12, 36; find the 9th term. 
19. 12, 6, 3; find the 8th term. 20. 6, - 3, f ; find the 9th term. 



PROGRESSIONS 237 

Find the last term and the sum of the G.P. 
21. 4, 12, 36, to 7 terms. 22. 12, 6, 3, to 6 terms. 

23. 5, - 15, 45, to 6 terms. 24. 25, 2.5, .25, to 7 termu. 

25. 3, - 6, 12, to 7 terms. 26. jfc, - }, 1, to 6 terms. 

27. 3, 66, 126 2 , to 8 terms. 28. 4, 8z 2 , 16s 4 , to 7 terms. 

Employ formula 5 on page 229 to find the sum of the G.P. 
29. 4 + 2 + 4- ife. 30. 5 + 15 + + 3645. 

Find tfie missing elements of the G.P. 
31. a = 5; r - 2; Z = 640. 32. a 2; r = 3; Z = 486. 

33. r = 10; a - .001; I - 100. 34. 5 - 2186; I = 1458; a = 2. 

35. S = 275; r = - 2; Z = 400. 36. S = *#*; a - - f ; Z - 135. 
37. a = 256; r - i; 2 = J. 38. a = 1458; r = J; J - . 

Find Me specified term without finding the first term of the G.P. 

39. The 10th term, if the 6th term is 5 and common ratio is 2. 

40. The 12th term, if the 8th term is 25 and common ratio is .1. 

41. The 4th term, if the 8th term is 40 and common ratio is 2. 

42. The 5th term, if the 9th term is 80 and common ratio is J. 

* 

Insert the specified number of geometric means. 
43. Five, between 2 and 128. 44. Five, between 128 and 2. 

45. Three, between 4 and 324. 46. Four, between J and 81. 

j 

47. Six, between .1 and 1,000,000. 48. Three, between 16 and .0001. 

// x and y are of the same sign, and if a single geometric mean G of the 
same sign is inserted between x and y, then G is called the geometric mean 
of x and y; (x, G, y) form a G.P. Find the geometric mean of the numbers 
49. J; 16. 60. i; 36. 51. 4; 25. 52. - 9; 

53. Find the geometric mean of x and y. S*wte the result in words 

Find an expression for the sum and simplify by use of the laws of er 
but do not compute. Use formula 5 on page 229 when convenient. 

54. 1 + (1.03) + (1.03) 2 + - + (1.03). 

55. 1 -I- (1.05) + (LOS) 2 + + (1.05) 

56. (1.02)* + (1.02) + (1.02)< + + (1.02)". 



232 PROGRESSIONS 

67. (1.06) 4 -I- (1.06) 6 + (1.06) + - - - + (1.06); 

58. 1 + (1.02) 3 + (1.02) H ---- to 21 terms. 

69. (1.02)- 16 + (1.02)-" + (1.02)- 18 + + (1.02)-*. 

60. (1.03)-" + (1.03)-" + (1.03)~ 12 + + (1.03)- 4 . 

61. 1 + (1.02)* + (1-02) + (1.02)* + - - -f ( 



62. If the 7th term of a G.P. is 5 and the llth term is 3^, find the inter- 
mediate terms. 

63. For what values of k. do the three quantities (k + 3), (6k -f 3), 
and (20k + 5) form a G.P.? - . 

64. Find the sum of a G.P. of 7 terms whose 3d term is j and 6th is -fo. 

-66. How many ancestors have you had in the twelve preceding gener- 
ations if no ancestor appears in more than one line of descent? 



5. An investment paid a man, in each year after the first, twice as 
much as in the preceding year. If his investment paid him $13,500 in 
the first four years, how much did it pay the investor in the first and the 
fourth years? 

67. In a lottery, it is agreed that the first ticket drawn will pay its 
owner $.10 and each succeeding ticket twice as much as the preceding 
one. Find the total amount paid on the first 10 tickets drawn. 

68. Find the sum of the first 19 positive integral powers of 1.03, given 
that (1.03) 10 = 1.344. 



1 61 . Applications of geor #r. progressions 

>.. , .1; .. . i ./,.'.' 

*W^\en a proHe*^ is met where a sequence of terms is suspected of 

forming ^ A.P., generally it is an advantage to compute the explicit 

values of the first few terms in simplest form in order to verify the 

'istence of a common difference between the terms. On the other 

% d, if a sequence of terms is suspected of forming a G.P., it is best 

4te the first few terms, without actually computing them, in a 

r hich will exhibit clearly any constant factor which appears to 

T e powers. 

* 1. A rubber ball is dropped from a height of 100 feet. On 

i, the ball rises one half of the height from which it last fell. 

*, has the ball traveled up to the instant it hits the ground for 



PKOGRESS/ONS 233 

SOLUTION. 1. We list the first few distances traveled by the bouncing 
ball: 

1st fall = 100ft. 

1st rise = i(100)ft. \ 



=i(100)ft. =100 ft. 

2d rise = i(i)(100) ft. 4(100) ft. \ 
3d fall = i(100)ft. )-m-100)ft. 

3d me = i(i)(100) ft. 

- 10 ft. 

efc. 

2. The 1st fall brings in an unsymmetrical term. On neglecting it tem- 
porarily, the total distance, in feet, traveled otherwise up to the time of the 
12th fall is the sum of 

100, i(100), J(100), - . to eleven terms. (1) 

In (1) we have a G.P. with a = 100, r = J, and n = 11. The sum S is 
obtained from 

~ rn - ion 1 " (i) 11 _ ininsik! - 100(2047). 

~ ~ ~ " ~ 1024 ' 



_ 25(2047) _ 
256 ~ 

3. On adding the 1st fall to S, we find that the total distance traveled 
by the ball is 299|f J feet. 

MISCELLANEOUS EXERCISE 85 

Solve by methods involving progressions. 

1. If $500 is to be divided between 10 men so that the first one re- 
ceives $5 and each succeeding man obtains a fixed amount more than the 
preceding man, how much will the 10th man receive? 

2. At a bazaar, tickets are marked with the consecutive even integers 
2, 4, 6, and are drawn at random by those entering. If each person 
pays as many cents as the number on his ticket, how much money is received 
if 1000 tickets are sold? 

3. The path of each swing, after the first, of a pendulum bob is .9 as 
long as the preceding swing. If the first swing is 40 inches long, how far 
does the pendulum travel on the first 8 swings? 

4. A man piles 152 logs in layers so that the top layer contains 2 logs 
and each lower layer has one more log than the layer above. How many 
logs will be in the lowest layer? 



234 PROGRESSIONS 

6. In a professional golf tournament, the total prize money of $5187 
is divided among the six players with lowest scores, so that each man above 
the lowest receives as much as the man below him. How much does the 
man with the lowest score receive? 

6. A body, dropped from a position of rest in a vacuum near the earth's 
surface, will fall approximately 32 feet farther in each second, after the first, 
than in the preceding second. If a body -falls 10,000 feet in 25 seconds, 
how far does it fall in the first second? 

7. Find an expression for the sum of the first n positive integers. 

8. Find an expression for the sum of the first n positive even integers. 

9. At the beginning of each year, a man invests $300 at simple interest 
at the rate 7%. At the end of 15 years, what is the total value of his in- 
vestments if none of them have been disturbed, and if all required interest 
is paid on that date? 

10. The radiator of a motor truck contains 10 gallons of water. We draw 
off 1 gallon and replace it with alcohol; then, we draw off 1 gallon of the 
mixture and replace it by alcohol; etc., until 9 drawings and replacements 
have been made. How much alcohol is in the final mixture? 

11. Find the sum of the first 40 positive integral powers of x. 

12. In creating a vacuum in a container, a pump draws out J of the 
remaining air at each stroke. What part of the original- air has been re- 
moved by the end of the 7th stroke? - 

13. A pendulum bob moves over a path 15 inches long on its first swing. 
In each succeeding swing the bob travels fo^ur fifths of the distance of the 
preceding swing. How far does the bob travel during the first six swings? 



14. In a potato race, tw^njiy potatoes are placed at intervals of 5 feet 
in a line from the starting poih^with the nearest potato 25 feet away. 
A runner is required to bring the potatoes back to the starting place one 
at a time. How far would he run in bringing in all the potatoes? 

15. A speculator will make $1200 during the first month and, there- 
after, in each month, $100 less than in the preceding month. If his original 
capital is $2700, when will he become bankrupt? 

16. Two men start in a distance run. One man proceeds at a uniform 
speed of 300 yards per minute. The second man travels 435 yards hi the 
first minute, but, thereafter, in each minute he goes 30 yards less than 
in the preceding minute. When will the first man overtake the second? 

17. Prove that the squares of the terms of a G.P. also form a G.P. 
Then state a more general theorem of this nature. 

18. Prove that the reciprocals of the terms of a G.P. also form a G.P. 



PROGRESSIONS 235 

19. A rubber ball is dropped from a height of 300 feet. On each re- 
bound, the ball rises one third of the height from which it last fell. What dis- 
tance has the ball traveled up to the instant the ball hits the ground for 
the 7th tune? 

20. In a certain positive integral number of three digits, the digits form 
an A.P. and their sum is 15. If the digits are reversed, the new number is 
594 less than the original number. Find the original number. 

Note 1. If P is the value of a certain quantity now, and if its value in- 
creases at the rate i (expressed as a decimal) per year, then the new value 
at the end of one year is (P -f Pi), or P(l 4- i). That is, the value at the 
end of any year is (1 -f t) times the value at the end of the last year. The 
values at the ends of the years form a G.P. whose common ratio is (1 + i). 
If A represents the value at the end of n years, then 



This formula is referred to as the compound interest law because, if a 
principal P is invested now at the rate t, compounded annually, the amount 
A at the end of n years will be P(l + i) n > Applications involving compound 
interest will be treated later. In all of the following problems, it will be 
assumed that any rate is constant. 

21. If 300 units of a commodity are consumed in a first year, and if 
the annual rate of increase of consumption is 6%, (a) give an expression 
for the amount consumed in the 7th year; (6) find the total consumption 
in the first 12 years, given that (1.06) 12 = 2.012. 

22. A corporation will sell $1,000,000 worth of its products this year 
and the sales will increase at the rate of 5% per year. Find the total sales 
during the first 25 years, given that (1.05) 26 = 3.38635494. 

23. The population of a city increased from 131,220 to 200,000 in 4 years. 
Find the rate of increase per year. 

24. A piece of property was purchased 4 years ago for $4860 and its 
value now is $15,360. Find the annual rate at which the value increased. 

25. The value of a certain quantity decreases at the rate w (expressed 
as a decimal) per year. If // is the value now, and K is the value at the 
end of n years, prove that K = H(l w} n . (This formula is the basis 
for computing depreciation charges in business under the so-called constant- 
percentage .nethod.) 

26. A n-.otor truck was purchased for $2500, and its value 4 years later 
is $1024. .Find the rate, per year at which the value has depreciated. 

27. A tytel, purchased 3 years ago for $512,000, is sold for $343,000. 
Find the n te per year at which its value has depreciated. 



236 PROGRESSIONS 

*162. Harmonic progressions 

A sequence of numbers is said to form a harmonic * progression if 
their reciprocals form an arithmetic progression. 



ILLUSTRATION 1. The sequence (1, J, , ^, J) is a harmonic progression 
because the reciprocals (1, 3, 5, 7, 9) form an A.P. 

To insert k harmonic means between two numbers, we first insert 
k arithmetic means between the reciprocals of the numbers. The re- 
ciprocals of the arithmetic means are the harmonic means. 

EXAMPLE 1. Insert five harmonic means between 4 and 16. 

SOLUTION. 1. First, we insert 5 arithmetic means between J and $. 

2. From I = a -f (n l)d, with a = J, I = -j^, and n 7, we find 

A = 4 

3. Hence, the A.P. is (J, & 

4. The corresponding harmonic progression is (4, 3ft -^, 3?, 8, ^, 16). 

Hence, the harmonic means are (3ft ^ , ^, 8, 



*EXERCISE 86 

Insert the specified number of harmonic means. 
1. Four, between and ^. 2. Five, between J and 



3. Four, between & and . 4. Four, between 4 and 24. 

5. Five, between f and J. 6. Four, between J and 3. 

// (c, .ff, d) /orw a harmonic progression, then H is called the harmonic 
mean of c and d. Find the harmonic mean of the numbers. 

7. 4; J. 8. 9; 6. 9. 4; - 8. 10. 12; 36. 11. x and y. 

*163. Geometric progressions with infinitely many terms 

Let S n represent the sum of the progression o, ar, ar 2 , , ar"" 1 . 
Then, by (4), page 229, 

a + ar + ar 2 H ---- 

* Suppose that a set of strings of the same diameter and substance are 
stretched to uniform tension. If the lengths of the strings form a harmonic 
progression, a harmonious sound results if two or more strings are caused to 
vibrate at one time. This fact accounts for the name harmonic prop ession. 



PROGRESSIONS 237 

ILLUSTRATION 1. Consider the endless geometric progression 

1, - -> , =jj^j to infinitely many terms. . (2) 

In (2), r = -; the nth term is ^; 1 - r = ^; ar" = ^- 

By(l), i + l + 1 + 1 + ...+^ ^2--^. (3) 

If n grows larger, without limit, the term ^ grows smaller, and is as near 



to zero as we please, if n is sufficiently large. Thus, if n 65, 

J_ = JL _ 1 _ 
2 n-i 2 64 18,446,744,073,709,551,616* 

which is practically zero. Hence, in (3), 8 n will be as near to (2 0) as we 
please for all values of n which are sufficiently large. To summarize this 
statement we say that as n becomes infinite S n approaches the limit 2, and 
we call 2 the sum of the progression 1, 4, 4, 4, to infinitely many terms. 
We sometimes use "n >oo " to abbreviate "n becomes infinite." Then, 
our conclusion can be briefly written limit S n = 2. 



n 



Now, consider (a, ar, ar 2 , to infinitely many terms), under the 
condition that r is a number between 1 and -f 1. Then, as n > oo , 
the absolute value of the numerator ar n in (1) grows smaller, and is 
as near to zero as we please for all values of n sufficiently large. 
Hence, from (1) we see that, as n grows large without limit, the value 
of S n approaches 

/ a 0_\ a 

\r=r r^7/' or r^"r" 

That is, limit S n = =-2 (4) 

n too 1 r 



This limit of the sum of n terms, as n becomes infinite, is callejd 
the sum of the geometric progression with infinitely many terms. If 
S represents this sum, then 



- r 
Thus, if | r | < 1, 



(5) 



(a + or + <n* + to infinitely many terms) = * _ (6) 



238 PROGRESSIONS 

Note 1. Recognize that S in (5) is not a sum in the ordinary sense of 
the word, but is the limit of the sum of n terms as n grows large without bound. 

ILLUSTRATION 2. By use of (6), with a = 5 and r }, 



4- 4- T + 



to infinitely many terms] 7-377 = 10. 



Practically, this means that, if we should add a relatively large number of 
terms, we would obtain approximately 10, and that by adding enough terms 
we can obtain as close to 10 as we may desire. Thus, Su = 



The indicated sum of a sequence of numbers is frequently called 
a series. An expression of the form 

+ u% + u* + to infinitely many terms (7) 



is called an infinite series. Accordingly, the expression on the left in 
(6) is referred to as the infinite geometric series. 

EXAMPLE 1 . Find a rational number equal to the endless repeating decimal 
.5818181 

SOLUTION. Because of the meaning of the decimal notation, 

.5818181 - = .5 + .081 4- .00081 H ---- to infinitely many terms, 

where we notice that (.081 + .00081 + ) is an infinite geometric series 
with a .081, and r = .01. By (6), the sum of this series is 

.081 = .081 = 9 
1 - .01 .99 110* 

Hence, .5818181 ---- .5 + rf^ - & + rf * - f? - 



Comment. By use of the method of Example 1, we can show that any 
endless repeating decimal is a rational number. 

In any infinite series such as (7), let S n represent the sum of the 
first n terms. Then we say that the series has a sum S, and call 
the series a convergent infinite series which converges to S in case the 
limit of S n is S as n becomes infinite. If S n has no limit as n becomes 
infinite, we say that the infinite series is divergent, or diverges. 

Note 2. In this section we have proved that the infinite geometric series 
in parentheses in (6) has a sum, or converges, when | r \ < 1. When | r \ ^ 1, 
the series is divergent, or does not have a sum, because in this case S n in (1) 
does not approach a limit as n > oo . Thus, for the G.P. (1, 2, 4, ) where 
r = 2, we find that S n increases beyond all bounds as n - oo . 



PROGRESSIONS 239 

*EXERCISE 87 

Find the sum of each of the followng infinite geometric series by use of the 
established formula. 

I- 7 + J + I + ' ' 2. 12 + 3 + J + - -. 

3. 15 + 5 -f f + . . 4. 10 - 5 + f + 

6. 1 - J + t 6. 1 - J + J . 

7. 1 - .01 + .0001 . 8. .8 + .08 -f .008 H . 

Find a rational number equal to the given endless repeating decimal, where 
repeating parts are written three times. 

9. .333 . 10. .444 . 11. .666 . 

12. .0999 . 13. .8333 . 14. .1666 . 

15. .212121 . 16. .050505 . 17. .030303 . 

18. .838383 . 19. .454545 . 20. 4.222 .. 

21. .2111 . 22. .345345345 . 23. .210210210 - . 

24. 252.525.- -. 25. 16.7167167 - -. 26. 25.05050 -. 
27. .153846153846153846 . 28. .076923076923076923 . 

29. A pendulum is being brought to rest by air resistance. The path 
of each swing, after the first, of the pendulum bob is .98 as long as the 
path of the previous swing (from one side to the other). If the path of 
the first swing is 30 inches long, how- far does the bob travel in coming to 
a position of rest? 

30. A rubber ball is dropped from a height of 100 inches. On each 
rebound the ball rises to f of the height from which it last fell. Find the 
distance traveled by the ball in coming to rest. 

31. The side of a certain square is 10 inches long. A second square is 
drawn by connecting the mid-points of the sides of the 1st square; a 3d 
square is drawn by connecting the mid-points of the sides of the 2d square; 
etc., without end. Find the sum of the areas of all the squares. 

Note 1. If | r \ < 1, we know that S, of (5) on page 237, is approximately 
equal to S n if n is large, and our confidence in this approximation increases 
as n increases. When n is large, it is decidedly easier to compute S than 
/Sn, and hence it is convenient at tunes to use S in place of S n < 

32. (I) Find Sn for the G.P. 3, f , f , . (II) Find the sum of this pro- 
gression extended to infinitely many terms. 

33. Find &, approximately, for the G.P, 8 -f J + f H . 



CHAPTER 



15 



LOGARITHMS 



164. Logarithms 

Logarithms are auxiliary numbers which are exponents, and which 
permit us to simplify the operations of multiplication, division, rais- 
ing to powers, and extraction of roots, applied to explicit real numbers. 

Previously, we have introduced exponents only under the assump- 
tion that they are rational numbers. In connection with logarithms, 
however, when we mention an exponent, it may be any real number, 
rational or irrational. A logical foundation for the use of irrational 
exponents is beyond the scope of this text. Hence, without dis- 
cussion, we shall assume the fact that irrational powers have meaning 
and that the laws of exponents hold if the exponents involved are 
real numbers, either rational or irrational, provided that the base 
is positive. 

ILLUSTRATION 1. The student may safely use his intuition in connection 
with the symbol 10^ = 10 1 - 414 ---. Closer and closer approximations to 
10^ are obtained if the successive decimal approximations to v^ are used 
as exponents. 

165. Logarithms to any base 

In the following definition, b represents any positive number, not 
1, and N is any positive number. 

DEFINITION I. The logarithm of a number N to the base b is the 
exponent of the power to which b must be raised to obtain N. 

In other words, if 6* N then x is the logarithm of N to the 
base 6. To abbreviate "the logarithm of N to the base b," we write 
"log*, i/V." Then, by Definition I, the following equations state the 
same fact, the first equation in exponential form and the second in 



LOGARITHMS 247 

logarithmic form: 

N = b* and x = log* N. , (1) 

ILLUSTRATION 1. If N - 4 5 , then 5 is the logarithm of N to the base 4. 

ILLUSTRATION 2. "Iog 2 64" is read "the logarithm of 64 to the base 2": 

since 64 = 2*, hence Iog 2 64 = 6. 

ILLUSTRATION 3. Since v^5 = 5$, hence Iog 5 \^5 = j = .333 . 



ILLUSTRATION 4. To find Iog 2 J-, we express J as a power of 2: 
since 5 = r = 2~ 3 , Aence Iog 2 = 3. 

o & o 

ILLUSTRATION 5. If logb 16 = 4, then b 4 = 16; 6 = v'Te = 2. 

ILLUSTRATION 6. If log 2 = J, then cri = 2. Hence, 

1 



For any base b, we have 6 = 1 and 6 1 = b. Hence, 

log & 1 = 0; log b 6 = 1. (2) 

Note 1. We do not use b = 1 as a base for logarithms because every 
power of 1 is 1 and hence no number except 1 could have a logarithm to 
the base 1. 

'x 

In the definition of log& N, we stated that N was a positive number. 
That is, in this book, if we speak of the logarithm of a number N 
we shall mean a positive number N. Also, we stated that the base 
b is positive. These agreements were made to avoid meeting im- 
aginary numbers as logarithms. In advanced mathematics it is 
proved that, if N and b are positive, there exists just one real logarithm 
of N to the base b. 

EXERCISE 88 

1. Since 27 = 3 s , what is the logarithm of 27 to the base 3? 

2. Since 625 = 5 4 , what is the logarithm of 625 to the base 5? 

3. Since J = 3" 1 , what is the logarithm of J to the base 3? 

4. Since 1 = 6, what is the logarithm of 1 to the base 6? 



242 LOGARITHMS 

5. If the logarithm of N to the base 4 is 3, find N. 

6. What number has 2 as its logarithm to the base 4? 

Find the number N whose logarithm is given. 
7. log, N - 4. 8. log. N = 2. 9. logio N - 1. 

10. logio N - 2. * 11. Iog 7 N = 0. 12. log* JV - - 1. 

13. log w AT = - 1. 14. Iog 2 'AT - - 3. 16. Iog 4 AT = - 3. 

16. Iog 10 N - - 4. 17. Iog 4 AT = J. 18. Iog 26 N = J. 

19. logs AT - |. 20. logioo N - 2. 21. logioo AT = f . 

22. Find the logarithm of 125 to the base 5. 

23. Find the logarithm of 1,000,000 to the base 10. 

Find the specified logarithm. 
24. logs 25. 25. log* 16. 26. Iog 2 8. 27. log? 49. 

28. log, 27. 29. Iog 4 64. 30. logio 1000. 31. Iog 5 625. 

32. logu 5. 33. Iog 9 3. 34. logioo 10. 35. log^ 3. 

36. Iog 4 J. 37. loge J. 38. Iog 2 J. 39. log, &. 

kFind the value of the unknown letter in the problem. 
40. Iog6 16 - 2. 41. log* 125 - 3. 42. log* 625 = 4. 

43. Iog5 1000 = 3. 44. log, 9 - J. 46. Iog 4 = - 1. 

46. loga 7 - - 1. 47. log* J - - 2. 48. log y = 2. 

49. Iog6 2 - J. 60. logb -fs - 2. 61. Iog6 .0001 = - 2. 

62. logj 16 = x. 63. log^ N == - 3. 64. logs N = - f . 

166. Common logarithms 

Logarithms to the base 10 are called common logarithms and are 
the most useful variety for computational purposes. Hereafter, un- 
less otherwise stated, when we mention a logarithm we shall mean a 
common logarithm. For abbreviation, we shall write log N, instead 
3! logw N, for the common logarithm of N and read log N as the 
logarithm of N. Then, from the definition of a logarithm, the follow- 
ing equations are equivalent: 

N = 10 and x = log N. (1) 



LOGARITHMS 



243 



ILLUSTRATION 1. Since 10,000 - 10 4 , hence log 10,000 = 4. 

Since ^Io = 10*, hence log ^Io .333 
Since .01 = 1Q- 2 , hence log .01 - 2. 

We have seen that log N may be either positive, negative, or zero, 
depending on the value assigned to N. Also, we notice that log N 
is an integer when and only when N is an integral power of 10. 

ILLUSTRATION 2. For future reference, the student should verify the 
following logarithms by use of the definition of a logarithm. 



N = 


.0001 


.001 


.01 


.1 


i 


10 


100 


1000 


10,000 


100,000 


log AT- 


-4 


-3 


2 


i 





1 


2 


3 


4 


5 



167. Some properties of logarithms 

The following properties hold if the logarithms are taken to any 
base 6, but we shall write proofs only for the case where b = 10. 

1. The logarithm of a product equals the sum of the logarithms of its 
factors. For instance, 

log MN = log M + log N. (1) 

ILLUSTRATION 1. If M = 10 3 and N = 10 8 , then MN = 1010 = 10 8 . 
Also, by the definition of a logarithm, log M = 3, log N = 5, and log MN = 8. 
Hence, log MN log M + log N in this special case. 

Proof of Property I. 1. Let log M = x and log N y. Then, 
M = 10*, N - 10", and MN - 10*10* = 10*+. 

2. Since MN = 10*+", hence log MN - x + y - log Af -f log 



II. !T/ie logarithm of a quotient equals the logarithm of the dividend 
minus the logarithm of the divisor. That is, 



log 7 = log M - log N. 



(2) 



Proof. 1. Let log M x and log N * y. Then, 

(By a law of exponents) 



2. Hence, log -^ =* x - y - log Af - log AT. 



244 



LOGARITHMS 



ILLUSTRATION 2. If we are given log 3 .4771, then 

log 300 = log 3(100) - log 3 + log 100 = .4771 + 2 - 2.4771; 
log .003 = log Ttffof = log 3 - log 1000 = .4771 - 3 - 2.5229. 



EXERCISE 89 

Express each number as a product or quotient, and find its logarithm by use 
of the given logarithms and the logarithms of integral powers of 10. 

log 2 = .3010; log 3 - .4771; log 7 = .8451; log 17 = 1.2304. 
1. 6. 2. 21. 3. 34. 4. 51. 5. 30. 6. 70. 



7. 1700. 


8. 2000. 


9. 42. 


10. | . 


11. J. 


12. Jf 


13. 


14. f 


16. A. 


1& rbs- 


17. A- 


18. .17. 


19. .007. 


20. .0003. 


21. .017. 


22. .0119. 


23. .042. 


24. 10.2. 



168. Characteristic and mantissa 

Every number, and hence every logarithm, can be written as the 
sum of an integer and a decimal fraction which is positive or zero and 
less than 1. When log TV is written in this way, we call the integer 
the characteristic and the fraction the mantissa of log N. 

log N = (an integer) -f (a fraction, ^ 0, < 1) ; 

log N = characteristic + mantissa. (1) 

ILLUSTRATION 1. If log N = 4.6832 = 4 -f- .6832, then .6832 is the man- 
tissa and 4 is the characteristic of log N. 

ILLUSTRATION 2. If log N = 3.75, then log N lies between 4 and 
3. Hence, log N 4 -f- (a fraction). To find the fraction, subtract: 
4 - 3.75 = .25. Hence, log TV = - 3.75 = - 4 + .25. 

ILLUSTRATION 3. The following logarithms were obtained by later meth- 
ods. The student should verify the three columns at the right. 





LOGARITHM 


CHARACTERISTIC 


MANTISSA 


log 300 = 2.4771 


= 2 + .4771 


2 


.4771 


log 50 - 1.6990 


= 1 -f .6990 


1 


.6990 


log .001 - - 3 


= - 3 -f .0000 


-3 


.0000 


log 6.5 0.8129 


= + .8129 





.8129 


log .0385 - 1.4145 


= - 2 + -5855 


-2 . 


.5855 


log .005 = - 2.3010 


= - 3 + .6990 


Q 


.6990 



LOGARITHMS 245 

1 69. Properties of the characteristic and mantissa 

ILLUSTRATION 1. All numbers whose logarithms are given below have the 
same significant digits (3, 8, 0, 4). To obtain the logarithms, log 3.804 
was first found from a table to be discussed later; the other logarithms 
were then obtained by the use of Properties I and II. 

log 380.4 = log 100(3.804) = log 100 + log 3.804 - 2 + .5802; 
log 38.04' = log 10(3.804) - log 10 + log 3.804 - 1 + .5802; 

log 3.804 =.5802 - + .5802; 

Q 804 
log .3804 =log~ = log 3.804 - log 10 = - 1 + .5802; 



log .03804 = log - = log 3.804 - log 100 = - 2 + .5802. 



Similarly, if N is any number whose significant digits are (3, 8, 0, 4), then 
N equals 3.804 multiplied, or else divided, by a positive integral power of 
10; hence, it follows as before that .5802 is the mantissa of log N. 

In Illustration 1, the characteristic of log 380.4 is 2, of log 38.04 
is 1, etc. These facts could have been learned as follows. 

ILLUSTRATION 2. To find the characteristic of log 380.4, notice the two 
successive integral powers of 10 between which 380.4 lies: 

100 < 380.4 < 1000. 
Hence, - log 100 < log 380.4 < log 1000; or, 2 < log 380.4 < 3. 

Therefore, log 380.4 = 2 + (a fraction, > 0, < 1); or, by definition, the 
characteristic of log 380.4 is 2. 

\ 

In Illustration 1 we met special cases of the following theorems. 

THEOREM I. The mantissa of log N depends only on the sequence 
of significant digits in N. That is, if two numbers differ only in 
the position of the decimal point, their logarithms have the same 
mantissa. 

THEOREM II. When N > 1, the characteristic of log N is an integer , 
positive or zero, which is one less than the number of digits in N to the 
left of the decimal point. 

THEOREM III. If N < 1, the characteristic of log N is a negative 
integer; if the first significant digit of N is in the fcth decimal pkce, 
then 'k is the characteristic of log N. 



246 LOGARITHMS 

ILLUSTRATION 3. By use of Theorems II and III, we find the character- 
istic of log N by merely inspecting N. Thus, by Theorem III, the char- 
acteristic of log .00039 is 4 because "3" is in the 4th decimal place. 
By Theorem II, the characteristic of log 1578.6 is 3. 

1 70. Standard form for a negative logarithm 

Hereafter, for convenience in computation, if the characteristic of 
log N is negative, - ft, change it to the equivalent value 

[(10 - k) - 10], or [(20 - *) - 20], etc. 

ILLUSTRATION 1. Given that log .000843 = 4 + .9258, we write 
log .000843 - - 4 + .9258 = (6 - 10) + .9258 = 6.9258 - 10. 

The characteristics of the following logarithms are obtained by use of 
Theorem III; the mantissas are identical, by Theorem I. 



IST SIGNIP. DIGIT IN 



ILLUSTRATION 



Loo AT 



STANDARD FORM 



1st decimal place 
2d decimal place 
6th decimal place 



N = .843 
N - .0843 
N = .00000843 



- 1 + .9258 = 9.9258 - 10 

- 2 + .9258 - 8.9258 - 10 

- 6 + .9258 = 4.9258 - 10 



1 71 . Four-place table of logarithms 

Mantissas can be computed by use of advanced mathematics, and, 
except in special cases, are endless decimal fractions. Computed 
mantissas are found in tables of logarithms, also called tables of man- 
tissas. 

ILLUSTRATION 1. The mantissa for log 10705 is .029586671630457, to 
fifteen decimal places. 

Table II gives the mantissa of log N correct to four decimal places 
if N has at most three. significant digits; a decimal point is under- 
stood in front of each mantissa in the table. If N lies between 1 
and 10, the characteristic of log N is zero so that log N is the same 
as its mantissa. Hence, a four-place table of mantissas is also a 
table of the actual logarithms of all numbers with at most three sig- 
nificant digits, from N , 1.00 to N 10.00. In case N is less than 1 
or greater than 10, we must supply the characteristic of log N by 
use of Theorems II and III besides obtaining the mantissa of log N 
by use of Table II. 



LOGARITHMS 247 

EXAMPLE 1. Find log .0316 from Table II. 

SOLUTION. 1. To obtain the mantissa: find "31" in the column headed 
N in the table; in the row for "31," read the entry in the column headed 
"6." The mantissa is .4997. 

2. By Theorem III, the characteristic of log .0316 is - 2, or (8 - 10): 
log .0316 = - 2 + .4997 - 8.4997 - 10. 

ILLUSTRATION 2. From Table II and Theorem II, log 31,600 4.4997. 
EXAMPLE 2. Find N if log N = 7.6064 - 10. 

SOLUTION. 1. To find the significant digits of N: the mantissa of log N 
is .6064; this is found in Table II as the mantissa for the digits "404." 

2. To locate the decimal point in N: the characteristic of log N is (7 10) 
or - 3; hence, by Theorem III, N = .00404. 

ILLUSTRATION 3. If log N 3.6064, the characteristic is 3 and, by 
Theorem II, N has 4 figures to the left of the decimal point: the mantissa 
is the same as in Example 2. Hence, N 4040. 

DEFINITION I. A number N is catted the antilogarithm of L in case 
log N ~ L, and for abbreviation we write N = antilog L. 

ILLUSTRATION 4. Since log 1000 = 3, hence 1000 = antilog 3. 
ILLUSTRATION 5. In Example 2 we found antilog (7.6064 10). 

EXERCISE 90 

In Probkms 1 to 8, each number is the logarithm of some number N. State 
the characteristic and the mantissa of log N. 

1. 2.9356. 2. 15.2162. 3. - 1.300. 4. - 2 + .3561. 

6. 3.5473. 6. 7.2356 - 10. 7. - 5.675. 8. 5.1942 - 10. 

Write the fottowng negative logarithms in standard form. 
9. - 1 + .2562. 10. .3267 - 3. 11. .4932 - 6. 12. - 3.4675. 

State the characteristic of the logarithm of each number. 
13. 637,500. 14. 368. 16. .000673. 16. .00897. 17. .000007. 

Use Table II to find the four-place logarithm of the number. 
18. 65.4. 19. 43.2. 20. 178. 21. .00785. 22. .0346. 

23. 9.46. 24. 6530. 25. 17,800. 26. .00005. 27. .086. 

28. .000358. 29. 101,000. 30. .00089. 31. 157,000. 32. .0000002. 



248 LOGARIJHMS 

Find the antilogarithm of the given logarithm by use of Table II. 
83. 2.3856. 34. 3.3927. 36. 3.6684. 36. 1.8785. 37. 0.1553. 

38. 2.1461. 39. 1.8692. 40. 0.9727. 41. 2.4800. 42. 0.5611. 

43. 7.7701 - 10. 44. Q.8041 - 10. 46. 8.9823 - 10. 

46. 4.8915 - 10. 47. 4.9542 - 10. 48. 2.9340 - 10. 

49. 9.4216 - 10. 60. 8.7284 - 10. 61. - 2.3010. 

1 72. Interpolation for a mantissa 

Interpolation in a table of mantissas is based on the assumption 
that, for small changes in N, the corresponding changes in log N are 
proportional to the changes in N. This principle of proportional parts is 
merely an approximation to the truth but leads to results which are 
sufficiently accurate for our purposes. 

We agree that, whenever a mantissa is found by interpolation 
from a table, we shall express the result only to the number of decimal 
places given in table entries. Also, in finding N by interpolation in 
a table of mantissas when log N is given, we agree to specify just four 
or just five significant digits according as we are using a four-place 
or a five-place table. No greater refinement in the result is justified 
because the unavoidable error, which may arise, frequently will be 
as large as 1 unit in the last significant digit which we have agreed 
to specify, although the error is rarely larger. 

t 

EXAMPLE 1. Find log 13.86 by interpolation in Table II. 

SOLUTION. 1. We notice that 13.80 < 13.86 < 13.90. Hence, by the 
principle of proportional parts, we assume that, since 13.86 is 3% of the way 
from 13.80 to 13.90, 

log 13.86 & & of the way from log 13.80 to log 13.90, or 
log 13.86 log 13.80 + .6(log 13.90 - log 13.80). 

2. Each logarithm below has 1 for its characteristic, by Theorem II. 



From table: log 13.80 = 1.1399" 

log 13.86 = ? 
From table: log 13.90 = 1.1430J 



Tabular difference is 
31 .1430 - .1399 = .0031. 
.6(.0031) = .00186, or .0019. 



log 13.86 = 1.1399 + .6(.0031) = 1.1399 + .0019 = 1.1418. 



Comment. We found .6(31) = 18.6 by use of the table headed " 31 " under 
the column of proportional parts in Table II. 



LOGARITHMS 



249 



ILLUSTRATION 1. To find log .002914: 



10 



~ T2910: mantissa is -46391 ' 

L2914: mantissa is ? J 
_ 2920: mantissa is .4654 _ 



Tabular difference is 
15 .4654 - .4639 - .0015. 
x - .4(15) = 6. 



Hence, the mantissa for 2914 is .4639 4- .0006 = .4645. 



Hence, by Theorem III, log .002914 = - 3 + .4645 7.4645 - 10. 

EXAMPLE 2. Find N if log N = 1.6187. 

SOLUTION. 1. The mantissa .6187 is not in Table II but lies between the 
consecutive entries .6180 and .6191, the mantissas for 415 and 416. 

2. Since .6187 is -fa of the way from .6180 to .6191, we assume that N is 
of the way from 41.50 to 41.60. 



11 



[1.6180 = log 41.501 

Ll.6187 = log N J* 

. ^1.6191 = log 41.60 J 



41.60 - 41.50 = .10 
.10 x = A(-IO) = .064, or 

approximately .06. 



N = 41.50 + -&C10) = 41.50 + -06 - 41.56. 



ILLUSTRATION 2. To find N if log N = 6.1053 - 10: 



34 



15 



r.1038, mantissa for 1270H 
L.1053, mantissa for ? J 
.1072, mantissa for 1280 J 



10 



= .4. Hence, 
x = .4(10) = 4. 
1270 + 4 = 1274. 



Hence, .1053 is the mantissa for 1274 and N = .0001274. 



Comment. We obtain Jj = .4 by inspection of the tenths of 34 in the 
columns of proportional parts. We read 

13.6 = .4(34) or ^ = .4, and ^ = .5. 



Since 15 is nearer to 13.6 than to 17, hence J| is nearer to .4 than to .5. 

Note 1 . When interpolating in a table of mantissas, if there is equal reason 
for choosing either of two successive digits, for uniformity we agree to 
make that choice which gives an even digit in the last significant place of 
the final result of the interpolation. 

4 1 73. Scientific notation for a number 

Any positive number N can be written in the form 

. (1) 



250 LOGARITHMS 

where P is a number greater than or equal to 1 but less than 10, and 
fc is an integer, either zero or positive or negative. We refer to the 
right-hand side of (1) as the scientific notation for N. 

ILLUSTRATION 1. 5,832,900 = 5.8329(10*). 

.00000058329 = 5.8329(.0000001) - 5.8329(10~ 7 ). 

The scientific notation gives a brief and easily appreciated form 
for writing very large or very small numbers. 

ILLUSTRATION 2. The nucleus of an atom has a diameter which is esti- 
mated as less than 3(10~ 12 ) centimeters. The mean distance from the sun 
to the outermost planet Pluto is approximately 3.67 X 10 9 miles. One 
light-year, the distance which light will travel in one year hi interstellar 
space, is approximately 6 X 10 12 miles. 

In equation 1, N and P have the same significant digits because the 
factor 10* merely alters the position of the decimal point to change P 
into N. Hence, the scientific notation is very useful hi writing a 
number N, particularly if it is very large, when we wish to show how 
many digits in N are significant. This feature was referred to in 
Section 48, page 53, for the case where k of (1) is positive. 

ILLUSTRATION 3. If 68,820,000 is the approximate value of some quantity 
and if just five digits are significant, this is not indicated by the usual form 
of the number. We write it as 6.8820(10 7 ) to show that one of the zeros 
is significant. 

ILLUSTRATION 4. If N = 1.352(10*), then, by Property I, page 243, 

log N = log 1.352 + log 10 8 = 0.1309 + 8 = 8.1309. 
Thus, 8 is the characteristic and log 1.352 is the mantissa of log N. 

\ 

Consider any number N, where N = P(10*) where k is an integer 
and 1 ^ P < 10. Then k is the characteristic and log P is the 
mantissa of log N, because 

log N - log P-f log 10* - log P + k, 
where log P < 1, since 1 ^ P < 10. 

ILLUSTRATION 5. If log N = 9.7419, and if we use the form N => P(10*), 
we have k = 9 and log P = .7419: 

P - 5.520 and 2V * 5.520(10*). (Four digits significant.) 



LOGARITHMS 251 

Note 1. Besides common logarithms, the only other variety used ap- 
preciably is the system of natural, or Naperian logarithms, for which the 
base is a certain irrational number denoted by e where e = 2.71828 
Natural logarithms are useful for theoretical purposes. 

Note 2. logarithms were invented by a Scotchman, JOHN NAPIER, Baron 
of Merchiston (1550-1617). His original logarithms were not the same as 
those now called Naperian logarithms, in his honor. Common logarithms, 
also called Briggs logarithms,' were invented by an Englishman, HENRY 
BRIQQS (1556-1631), who was aided by Napier. 

EXERCISE 91 

^i 

Find the four-place logarithm of each number from Table II. 
1. 1826. 2. 25.63. 3. 532.2. 4. 12.67. 

5. 35.94. 6. 1.293. 7. .3013. 8. .4213. 

9. .5627. 10. .03147. 11. .01563. 12. .001139. 

13. 90,090. 14. 203,500. 15. .001439. 16. .05626. 

17. 1.233(10- 4 ). 18. 1.417(10'). 19. 3.126(10 3 ). 20. 2.438(10-*). 

Find the antilogarithm of each four-place logarithm from Table II. 
21. 3.2367. 22. 7.1247 - 10. 23. 6.1640. 24. 8.9935 - 10. 

26. 3.1395. 26. 2.9276. 27. 1.6016. 28. 0.4906. 

29. 6.3350 - 10. 30. 4.1436 - 10. 31. 9.6715 - 10. 32. 8.0255 - 10. 

33.- 8.8862. 34. 2.1952. 35. 0.0130. 36. 5.5511. 

37. 5.9885 - 10. 38. 8.3358 - 20. 39. 9.6270 - 10. 40. 6.4228. 

1 74. Computation of products and quotients 

Unless otherwise specified, we shall assume that the data of any 
given problem are exact. Under this assumption, the accuracy of a 
product, quotient, or power computed by use of logarithms depends 
on the number of places in the table being used. The result is fre- 
quently subject to an unavoidable error which usually is at most a 
few units in the last significant place given by interpolation. Hence, 
usually we must compute with at least five-place logarithms to obtain 
four-place .accuracy, and with at least four-place logarithms to obtain 
three-ylace accuracy. As a general custom, in any result we shall 
give all digits obtainable by interpolation in the specified table. 



252 LOGARITHMS 

EXAMPLE 1. Compute .0631(7.208) (.5127) by use of Table II. 

SOLUTION. Let P represent the product. By Property I, we obtain 
log P by adding the logarithms of the factors. We obtain the logarithms 
of the factors from Table II, add to obtain log P, and then finally obtain 
P from Tkble II. The computing form, given in blackface type, was made 
up completely as the first step in the solution. 

log .0631 = 8.8000 - 10 (Table II) 

log 7.208= 0.8578 (Table II) 

log .6127 = 9.7099 - 10 (Table 'II) 



(add) log P = 19.3677 - 20 - 9.3677 - 10. 
Hence, P - .2332. [ = antilog (9.3677 - 10), Table II] 

431.91 



r> or.* 

EXAMPLE 2. Compute q 



, 



SOLUTION. 1. By Property II, log q equals the logarithm of the numer- 
ator minus the logarithm of the denominator. 

2. Before computing, we round off each given number to four significant 
digits because we are using a four-place table. For instance, 15.6873 be- 
comes 15.69. 

log 431.9 - 2.6354 (Table II) 

_(-) log 15.69 = 1.1956 (Table II) 

log q = 1.4398. Hence, q = 27.53. (Table II) 

257 
EXAMPLE 3. Compute q = Sn^s' 

SOLUTION. We employ Property II. 

log 267 = 2.4099 = 12.4099 - 10 
(-) log 8966 - 3.9521 = 3.9521 

log q = ??? = 8.4578 - 10. Hence, q = .02869. 

Comment. When we first tried to subtract log 8956 from log 257, we 
saw that the result would be negative because log 8956 is greater than log 257. 
In order that log q should appear immediately in the standard form for a 
negative logarithm, we changed log 257 by adding 10 and then subtracting 
10 to compensate for the first change. Actually, 

log q = 2.4099 - 3.9521 - - 1.5422 = 8.4578 - 10. 

Whenever it is necessary to subtract a larger logarithm from a smaller 
one in computing a quotient, add 10 to the characteristic of the smaller 
logarithm and then subtract 10 to compensate for the change. 



LOGARITHMS 253 

A * * (4.803) (269.9) (1.636) 

EXAMPLE^ Compute g= v (7880)(253.6) ' 

INCOMPLETE SOLUTION. First make a computing form. 

(log 4.803 - . . flog 7880 = 

(+) log 269.9 = *" N \log 263.6 = 

[log 1.636 = log denom. = 

log numer. = 

( ) log denom. = 

log q = Hence, q = 

EXAMPLE 5. Compute the reciprocal of 189 by use of Table II. 

1 



SOLUTION. Let R = 



189 



log 1 = 0.0000 = 10.0000 - 10 
(-) log 189 - 2.2765 = 2.2765 

log R = ? = 7.7235 - 10. 

Hence, R = .005290. 

Comment. In writing any approximate value, it is essential to indicate 
all final zeros which are significant. Hence, in writing R = .005290 in 
Example 5, the final zero was essential. It would have been wrong to 
write R .00529, because this would not show that we had reasonably ac- 
curate information concerning the next digit, zero. 

Note 1. Before finding the four-pl&ce log NUN has more than four 
significant digits, round off AT to four significant digits. 

EXERCISE 92 

Compute by use of four-place logarithms. 
1. 31.57 X .789. 2. .8475 X .0937. 3. 925.618 X .000217. 

4. 925.6 X .137. 6. .0179 X .35641. 6. 3.41379 X .0142. 

7. (- 84.75) (.00368) (.02458). 8. (- 16.8) (136.943) (.00038). 

HINT. Only positive numbers have real logarithms. First compute as if 
all factors were positive; then determine the sign by inspection. 

675 728.72 .0894 

** 



13.21 895 .6358 325.932 

568.5 1A 753.166 1K .0421 1 

lv ^rfc ^ ^ * JL4 /\rti*n o * JLO 



23.14 9273.8 .53908 w 100,935 



254 LOGARITHMS 

16.083 X 256 9.32X531 



17 

JL f 



47 + .0158 .8319 X .5685 .53819 X .0673 
.42173 X .217 A 5.4171 X .429 



' .3852 X .956 18.1167 X 37 .00073 X .965 
(- .29)(.038)(- .0065) n . (5.6) (- 3.9078) (- .00031) 



(- 1006.332) (2.71) (132) (- 1.93) 

Compute the reciprocal of the number. 
25. 63283. 26. .00382. 27. .02567. 28. .0683(.52831). 

29. (a) Compute 652(735); (6) compute (log 652) (log 735). 

80. (a) Compute .351 * 625; (6) compute (log .351) -5- (log 625). 

*175. Cologarithms * 

The logarithm of the reciprocal of N is called the cologarithm of N 
and is written colog N. Since log 1=0, 

colog N = log - = - log N. (1) 



T i n i noi i ! log 1 = 10.0000 - 10 

ILLUSTRATION 1. Colog .031 - log -ggjl ( , )log .Q* 31= 8.4914-10 

colog .031 = 1.5086. 

The positive part of colog N can be quickly obtained by inspec- 
tion of log N: subtract each digit (except the last) in the positive part 
of log N from 9, and subtract the last digit from 10. 

,, , n * 16.083 X 256 , r . . ., 

EXAMPLE 1. Compute q = 47 v AICQ "7 use ^ coioganthms. 



SOLUTION. To divide by N is the same as to multiply by 1/N. Hence, 
instead of subtracting the logarithm of each factor of the denominator, we 
add the cologariihm of the factor: 

16.083 X 256 _ , . / 1 \ / 1 

" (16 ' 083 X 



47 X .0158 

log 16.08 = 1.2063 
log 266 = . 2.4082 

log 47 - 1.6721; hence, colog 47 - 8.3279-10 

log .0158 - 8.1987 - 10; hence, colog .0168 - 1.8013 

q = 5542. < - (add) log q - 13.7437 - 10 - 3.7437. 

* The instructor may wish to direct occasional use of cologarithms. 



LOGARITHMS 255 

1 76. Computation of powers and roots 

We establish the following property of logarithms as an aid to 
computing powers. 

III. The logarithm of the kth power of a number N equals k times 
the logarithm of N: 

log N* = k log N. (1) 

Proof. Let x log N. Then, by the definition of a logarithm, 
N - 10*. Hence, 

N k - (10*)* - 10**. (A law of exponents) 

i 

Therefore, by the definition of a logarithm, 

log N k kx - k log N. (Using x = log N) 

ILLUSTRATION 1. Log 7 6 = 5 log 7. Log N* = 3 log N. 

EXAMPLE 1. Compute (.3156) 4 . 

SOLUTION. By Property III, 

' log (.3156)* = 4 log (.3156) - 4(9.4991 - 10). 
log (.3166)* - 37.9964 - 40 = 7.9964 - 10. 
Therefore, (.3156) 4 = .009918. 

Recall that any root of a number is expressible as a fractional power. 
Hence, as a special case of Property III we obtain 



IV. log 

Proof. Since 3/N = W*, we use Property III with k T- 



log y/Jf log JV - T log N. 



ILLUSTRATION 2. Since VJ? = Ni and 



log VN = ilog N\ log */N = \ log N. 

2 3 



EXAMPLE 2. Compute ^.08351. 

SOLUTION. By Property IV, log 4/N * J log AT. Hence, 



256 



LOGARITHMS 



log v^.08361 = 



58.9218 - 60 
6 



8.9218 - 10 

_ * 

6 
= 9.8203 - 10. 



(2) 



Therefore, \/. 08361 .6611. 



Comment. Before dividing a negative logarithm by a positive integer, 
usually it is best to write the logarithm in such a way that the negative part 
after division, will be 10. Thus, in (2), we altered (8.9218 10) by 
subtracting 50 from 10 to make it 60, and by adding 50 to 8.9218 to 
compensate for the subtraction; the result after division by 6 is in the 
standard form for a negative logarithm. 

T7 on*/ 05831) 8 \* 
EXAMPLE 3. Compute q = [ ~= ] 

\65.3VT46/ 

SOLUTION. 1. Let F represent the fraction. Then log q = log F. 
2. Notice that log (.5831) 3 = 3 log .5831; log Vl46 - i log 146. 

log .6831 = 9.7658 - 10 

log 146 = 2.1644 

3 log .6831 = 9.2974 - 10 \ 
( -) log denom. = 2.8971 j 

log F = 6.4003 - 10; 21ogF - 2.8006 - 10 = 42.8006 - 50. 

, 21ogF 42.8006 - 50 c , Am 

log q = jp = g = 8.5601 10. 

Hence, q = .03632. 



/ log 66.3 = 1.8149 

1 j log 146 - 1.0822 

log denom. = 2.8971. 



EXERCISE 93 

Compute by use of four-place logarithms. 
1. (17.5) 8 . 2. (3.1279) 4 . 3. (.837)*. 



6. VT09. 

9. (1.04) 7 . 

13. (700,928)*. 

17. (- 1.03)*. 

21. (143.54)*. 



6. -v/2795. 
10. (10,000)*. 

18. (- 1796)*. 
22. 




4. (.0315) 8 . 
8. 



12. (.0138273)*. 
16. 



20. (- .00831) 8 . 



23. (- .0057)*. 24. (157)" 8 . 
HINT for Problem 24. Recall that (1$7)~ 8 - 1/(157 8 ). 

26. (13.67)*. 26. (3.035)-*. 27. (.98)-*. 28. (.831447)- 6 . 



LOGARITHMS 257 

29. (1.03)-. 30. (1.05)-. 31. (1.04) 100 . 32. (1.04)-. 



1. Given the seven-place log 1.04 = 0.0170333, compute Problems 
31 and 32 and compare with the less accurate former answers. 

Compute by use of four-place logarithms. 
33. .958(12.167) 2 . 34. 10'- 66 V265. 35. 10 2 W1J8147. 36. 



37. 5 3 * 4 ; 31 !.- 38. (25 ;I^ 2 >'. 39. - 0198 



21.4V521.923 * 1893.32 "" (3.82616) 



40 758>32 41 * / 89-1 A0 / 47.5317 

' (46.3) 3 " \163 X .62* ^ \.031 X .964* 



10-*V?78 10-i"v387 .. ^- 463.19 

(.983174)' ** (57)(8.64)' ' 



46 /(- 316)(.198) /54.2VT89\ 

* \ .756392 ' 47 ' V .157386 ) 



5731.84 



ATofe ^. Observe that no property of logarithms is available to simplify 
the computation of a sum. Use logarithms below wherever possible. 



49. . + 89.532 ^45 - 364.1 R1 

V57 + 2.513 ' U ' (.9873) 2 + 16.3* 5L "453 X .110173 



62 - + 1 B3 log 86 log 567 - 20 

" (1.03)* + 1 log 53.8* * log 235 

66. (2.67) 1 -* 2 . 66. (53.17)- 4 . 67. (60.2)--. 58. (.065) - M2 . 

HINT for Problem 57. - .43 log 59.2 = - .7621 = 9.2379 - 10. 
69. Compute (a) (antilog 2.6731) 2 ; (ft) [antilog (- 1.4973)] 2 . 

DEFINITION I. The geometric mean of n numbers is defined as the nth 
root of the product of the numbers. Thus, the geometric mean of M, N, P, Q, 
and R is ^MNPQR. 

In each problem, find the geometric mean of the given numbers. 
60. 138; 395; 426; 537; 612. 61. .00138; .19276; .08356; .0131. 

// a, 6, and c are the three sides of a triangle t it is proved in trigonometry 
that A, the area of the triangle, is given by 



A = VS(S - a)(8 - b)(8 - c), where S = J(a + 6 + c). 

Find the area of a triangle whose sides are as follows. 
62, 375.40; 141.37; 451.20. 63. .089312; .0739168; .024853. 



258 LOGARITHMS 

The time t in seconds for one oscillation of a simple pendulum whose length 
is I centimeters, is given byt wv-> where g = 980 and v = 3.1416. 

64. (a) Find the time for one oscillation of a simple pendulum .985 centi- 
meters long. (6) Find I if the time for one oscillation is 3.75 seconds. 

65. Let d be the diameter in inches of a short solid circular steel shaft 
which is designed to transmit safely H horse power when revolving at R 
revolutions per minute. A safe value for d is 




Find the number of horse power which can be safely transmitted at 1150 
revolutions per minute if d - 1.9834. 

66. The weight w, in pounds of steam per second, which will flow through 
a hole whose cross-section area is A square inches, if the steam approaches 
the hole under a pressure of P pounds per square inch, is approximately 
w .0165AP-* 7 . How much steam at a pressure of 83.85 pounds per square 
inch will flow through a hole 12.369 inches in diameter? 

*1 77. Exponential and logarithmic equations 

A logarithmic equation is one in which there appears the logarithm 
of some expression involving the unknown quantity. 



EXAMPLE 1. Solve for x: log x + log -^- 6. 

SOLUTION. By use of Properties I and II of page 243, 

log x + log 2 -f log x log 5 = 6. 
2 log x - 6 + log 5 - log 2 - 6.3980. (Table II) 

log x - 3.1990; x - antilog 3.1990 - 1581. (Table II) 

An equation where the unknown quantity appears in an exponent 
is called an exponential equation. Sometimes, an exponential equa- 
tion can be solved by taking the logarithm of each member. 

EXAMPLE 2. Solve 16* * 74. 

SOLUTION. Equate the logarithms of the two sides: x log 16 = log 74; 

log 1.869 - 0.2716 

log 74 1.8692 (-) log 1.204 0.0806 
* * log 16 " 1.2041* log x - 0.1910; hence x - 1.552, 



LOGARITHMS 259 

*1 78. Logarithms to various bases 

The base 10 is convenient for a system of logarithms when they 
are being used to simplify computation. The only base other than 
10 which is used appreciably is the irrational number e = 2.71828- , 
which is fully as important a constant hi mathematics as the familiar 
number IT. Logarithms to the base e are called natural, or Naperian, 
logarithms. Natural logarithms have many advantages over com- 
mon logarithms for advanced theoretical purposes. 

Recall that the equations N = 6* and x = log& N are equivalent. 
Hence, if N and b are given, we can find log$ N by solving the exponen- 
tial equation N b* by use of common logarithms. In particular, 
the natural logarithm of N can be found by solving N e* for x. 

EXAMPLE 1. Find log, 35. 

SOLUTION. Let x = log, 35; then, 35 = e*. On taking the common 
logarithm of both sides we obtain x logio e = logio 35. 

logio 35 1.5441. log 1.544 = 10.1886 - 10 

X logio e 0.4343' (-) log .4343 - 9.6378 - 10 * 

x . 3.555 = log. 35. log * - 0.5508. 

THEOREM I. If a and b are any two bases, then 

log N = (toga &) (log* N). (1) 

Proof. Let y = lo& N; then N - If. (2) 

Hence, logo N = logo If - y logo b (logo 6)(log& N). 

The number log a b is called the modulus of the system of base a 
with respect to the system of base b. Given a table of logarithms 
to the base 6, we could form a table of logarithms to the base a 
by multiplying each entry of the given table by logo 6. 

*EXERCISE 94 

Solve for x or for n, or compute the specified logarithm. 
1. 12' - 28. 2. 51* - 569. 3. &* - 28(2'). 4. 15 s * - 85(3*). 

6. .67* = 8. 6. .093* - 12. 7. (1.03)-" - .587. 8. (1.04)" - 1.562. 

9. (1 ' 05 iT e "" 1 - 6.3282. 10. log W + log - - 5.673. 

.UO * 

11. log. 75. 12. log. 1360. 13. log, 10. 14. logu 33. 15. log.s 23.8. 



260 



LOGARITHMS 



16. Find the natural logarithm of (a) 4368.1; (6) 4.3681. (Notice that 
the results do not differ by an integer, so that Theorem I of page 245 does 
not hold for natural logarithms.) 

*1 79. Graphs of logarithmic and exponential functions 

We recall that y = log a x and x = a v are equivalent relations. We 
call logo x a logarithmic func- 
tion of x and a v an exponential 
function of y. 



ILLUSTRATION 1. In Figure 18, 
we have the graph of y = log, x. 
For any base a > 1, the graph of 
logo x would be similar. This 
graph assists us hi remembering 
the following facts. 

I. ' If x is negative, Iog x is 
not defined. 

II. If < x < 1, loga x is negative, and log a 1 = 0. 

III. // x increases without limit, log a x increases without limit; 
if x approaches zero, log a x decreases without limit. 

Since y = log a x is equivalent to x = a v , these equations have the 
same graph. Thus, in Figure 18 we have a graph of x *= e v . 




*EXERCISE 95 

1. Graph y = logio x for < x f 30. From the graph, read the value of 
.3. 10 -.e. io.s. 

2. Graph y = 2* from x = 6 to x = 4. From the graph, read log z 6. 



*1 80. Applications to compound interest 

On page 235 we saw that, if $P is invested at the interest rate i com- 
pounded annually, the amount $A on hand at the end of n years is 
given by the formula 

A = P(l + i) n - 

ILLUSTRATION 1. If $1000 is invested for 20 years at 5% compounded 
annually, the amount at the end of the time is 

A = 1000(1.05)*> = 1000(2.653) - $2653. (From Table III) 



LOGARITHMS 261 

For unusual values of t, it is impossible to employ an interest 
table as in Illustration 1, but then logarithms can be used. 

EXAMPLE 1. Compute A = 2000(1. 036) 26 . 

SOLUTION. log 1.036 = 0.0153; 25 log 1.036 = 0.3825 

log 2000 = 3.3010 (+) 
Hence, A = $4826. log A = 3.6835. 

We claim only three significant digits in the result because accuracy is lost 
in the multiplication 25 log 1.036. The 4th digit in 4826 is unreliable. 

EXAMPLE 2. Solve for the rate i: 2500 = 2000(1 + i) 8 . 

SOLUTION. 1. (1 + t) = $$& or (1 + i) 8 = 1-250. 

2. Hence, 1 4- i = v' 1.250; we compute this root by use of Table II. 
log 1.250 = 0.0969; J log 1.250 = 0.0121; hence, v / L250 = 1.028. 

3. Therefore, 1 + i = 1.028; or, i = 1.028 - 1 = .028 = 2.8%. 

^EXERCISE 96 

By use of Table III, find the compound amount at the end of the time if 
the money is invested at the specified rate compounded annually. 

1. $2500; at 4%, for 16 years. 2. $1200; at 6%, for 13 years. 

3. $1600; at 3%, for 35 years. 4. $400; at 5%, for 42 years. 

From A = P(l + i) n , we obtain P = A(l + i)-*. Use this result 
and Table IV to solve Problems 5 and 6. 

5. What principal should be invested now at 5% compounded annually 
to create $2500 as the amount at the end of 15 years? 

6. What principal should be invested now at 6% compounded annually 
to create $1000 as the amount at the end of 26 years? 

7. At what interest rate compounded annually will a $2000 principal 
grow to the amount $3500 at the end of 10 years? 

8. At what interest rate compounded annually will a $3000 principal 
double itself by the end of 15 years? 

9. How long will it take $300 to grow to the amount $750 if invested 
at 5% compounded annually? (Recall Section 177.) 

10. How long does it take money to double if invested at 6% compounded 
annually? 

11. How long does it take money to double if invested at 6% simple 
interest? Compare with the result of Problem 10. 



CHAPTER 



16 



SYSTEMS INVOLVING QUADRATICS 



1 81 . Graph of a quadratic equation in two variables 

A solution of an equation in two variables x and y is a pair of values 
of the variables which satisfies the equation. The graph or locus of 
the equation is the set of all points whose coordinates, (z, y), form 
real-valued solutions of the equation. 

EXAMPLE 1. Graph: z 2 -f- j/ 2 = 25. 

SOLUTION. 1. When x = 0, y 2 = 25; y = 5. Two solutions of the 
equation are (0, 5) and (0, 5). 

2. When y - 0, x 2 = 25; x - 5. Two 
solutions of the equation are (5, 0) and 
(- 5, 0). 

3. We plot the four points just found, with 
the same unit on OX and OY in Figure 19, 
and verify an advance inference that the 
graph is a circle whose center is the origin 
and radius is 5. 

Comment. Let P, with the coordinates 
(x, y), be any point in the coordinate plane 
with origin at 0, in a system where the same Fig. 19 

unit is used in measuring all lengths. Then 

x* + y* - (OP) 2 . 

Hence, if x 2 + j/ 2 = 25, the point P must lie on a circle about as center 
with (OP) 2 = 25, or with radius 5. 




EXAMPLE 2. Graph: 9x* 4y* * 36. 

SOLUTION. 1. Solve for x in terms of y: 9z 2 ~ 36 + 



(1) 



SYSTEMS INVOLVING QUADRATICS 



263 



- 

x - 



g 



9 



x = 9+ 2 . 



(2) 



2. We assign values to y and compute the corresponding values of x. 
Thus, if y = 0, then z = =t \/9 = =t 2. If y = 3, then 

* = =fc vl8 - 2V2 = 2.8. 
We tabulate the corresponding values of x and y in the following table. 



(a) 


y = 


-6 


-3 





3 


6 


x - V9 + y 2 


x 


4.5 


2.8 


2 


2.8 


4.5 


(6) 


y = 


-6 


-3 





3 


6 


x = - V9 + y 2 


X = 


-4.5 


-2.8 


-2 


-2.8 


-4.5 



We plot the points given by the pairs of values of x and y in the table. In 
Figure 20, the points listed for (a) in the table give the open curve FDE; 
the points for (6) give HBO. These two open curves, together, are called 
a hyperbola, and it is the graph of equation 1. Each piece of the hyperbola 
is called a branch of it. 

Comment. The equation 9x 2 4y 2 = 36 
defines z as a tow-valued function of y, as 
shown in (2), or y as a tow-valued function 
of x. The graph of the equation consists 
of the graphs of the two single-valued irra- 
tional functions 



x = 



and x 




The graph of the first of these is the branch 

FDE and the graph of the second is HBO. 

The two branches together make up the graph 

of equation 1. The branches are symmetr- Fig. 20 

rical with respect to the y-axis. 

Note 1. To every hyperbola there correspond two characteristic lines, 
called asymptotes, which are indicated by dotted lines in Figure 20. As 
we recede out on any branch of the hyperbola, the curve approaches the 
corresponding asymptote but never reaches it. By moving far enough out 
on the branch, we may approach the asymptote as closely as we please. It 
is proved in analytic geometry that the equations of the asymptotes for equa- 
tion 1 are obtainable as follows: . 



264 



SYSTEMS INVOLVING QUADRATICS 



1. Replace the constant term in the equation by 0, and factor the left member: 

9x 2 - 4t/ 2 = 0; (3x - 2y)(3x + 2y) = 0. 

2. Equate each factor separately to zero: 

3x - 2y = and 3x + 2y = 0. 

These are the equations of the asymptotes. 

EXAMPLE 3. Graph: x 2 -f 4y 2 = 25. (3) 

SOLUTION. 1. Solve for y: 

y 2 = i(25 - x 2 ) ; y = JV25 - x 2 . (4) 

2. To obtain real values for y, the numerical value of x may not be allowed 
to exceed 5. Thus, if x = 8, V25 x 2 = V 39, which is imaginary. 

3. Place x = in (3) to determine the t/-intercepts: 



- 25; 



f . 



Hence, two points on the graph are (0, 
and (0, f ), labeled A and C in Figure 21. 

4. Place y in (3) to determine the 
x-intercepts: 

x 2 = 25; x = 5. 

Hence, two points on the graph are (5, 0) 
and (- 5, 0), labeled B and D in Figure 21. 




Fi 9 . 21 



5. As many more points as desired can be found by substituting values 
of x in (4) and computing the corresponding values of y. When the points 
are joined by a smooth curve we obtain the oval ABCD in Figure 21. The 
curve is called an ellipse. The graph of the positive valued function 
y = JV25 x 2 , from (4), is the half of the ellipse above the x-axis. The 
graph of y = JV25 x 2 is the lower half, which is symmetrical to the 
upper half. The whole ellipse is the graph of (3). 

The facts stated in the following summary are proved in more 
advanced mathematics. 

SUMMARY. The graph of , any quadratic equation in two variables 
x and y with real solutions is either an ellipse, a hyperbola, a circle, a 
parabola, a pair of straight lines, or a single point. 

1. J/fc is positive, the graph of jc a H- y a = c is a circle whose radius is 
\/c and center is the origin, provided that the same unit is used on 
the scales of the x-axis and y-axis. 



SYSTEMS INVOLVING QUADRATICS 265 



2. // a, 6, and c have the same sign, the graph of ax 9 + &y* = c is an 
ellipse, with center at the origin; if a b, the ellipse is a circle, pro- 
vided that the same unit is used on the scales of the x-axis and y-axis. 

3. // a and b have opposite signs and if c is not zero, the graph of 
ax* -f by 2 = c is a hyperbola. 

4. If c j 0, the graph of xy = c is a hyperbola; ifc>Q, one branch 
of the hyperbola lies wholly in quadrant I, and the other in quadrant 
III; if c < 0, the branches are in quadrants II and IV, respectively. 
The coordinate axes are the asymptotes of the hyperbola. 

5. // a quadratic equation in x and y does not involve y 2 or xy, the 
graph of the equation is a parabola whose axis is parallel to the y-axis; 
if the equation does not involve x 2 or xy, the graph is a parabola whose 
axis is parallel to the x-axis. 

ILLUSTRATION 1. The graph of the equation 3# 2 + 7y* = 8 is an ellipse, 
of 5x 2 y 2 = 7 is a hyperbola, and of 4z 2 -f- 4y* = 25 is a circle, whose 
radius is 5/2. 

EXAMPLE 4. Determine the nature of the graph of 

2x* - xy - 3y* = 0. (5) 

SOLUTION. 1. Factor: (2x 3y)(x 4- y) = 0. 

Hence, (5) is satisfied by values (x, y) in case 

(a) 2x - 3y = 0, or (6) x + y = 0. 

Therefore, the set of all points (x, y) satisfying (5) consists of those satisfying 
(a) and those satisfying (6). Or, in other words, the graph of (5) consists 
of the graph of (a) and the graph of (6). 

2. The graphs of (a) and (6) are straight lines through the origin. Hence, 
the graph of (5) consists of these two straight lines. 

Comment. Another case similar to Example 4 was met in finding the 
asymptotes in Example 2. They were the two straight lines which are the 
graph of the equation 9z 2 4y 2 = 0. 

EXAMPLE 5. Determine the nature of the graph of 

5x - 7 = 0. 



SOLUTION. 1. Solve for y: y = |z 2 f x + J. 

2. Thus, y is a quadratic function of x, and therefore the graph of the 
given equation is a parabola whose axis is parallel to the y-axis. To graph 
the equation, we would compute the coordinates of the vertex of the parabola 
and proceed as in Section 132, page 186. 



266 SYSTEMS INVOLVING QUADRATICS 

182. Routine for graphing 

It is important to be able to construct reasonably good graphs 
quickly. Beyond this, it is also essential to have a procedure for im- 
proving on such graphs when the necessity arises. The following 
suggestions are of aid hi constructing graphs quickly for equations of 
the second degree hi x and y. 

1. Refer to the summary of Section 181 and if possible decide on 
the nature of the graph before carrying out details of the work. 

2. When the graph of ax z H- by 2 = c is a circle, find its radius, 17 -> 
and construct the circle with compasses. 

3. When the graph of ax* 4- by* = c is an ellipse, find the x-intercepts 
by placing y ~ and solving for x, and find the y-4ntercepts by placing 
x = in the given equation. Then, sketch the ellipse through the four 
intercept points thus obtained* 

4. When the graph of ax 2 4- by 2 = c is a hyperbola: 

Find its asymptotes by replacing c by and constructing the 
two straight lines which are the graph of ax 2 -f- by* - 0. 

Find the x-intercepts or the y-intercepts. (One set of intercepts 
witt be imaginary because the hyperbola will cut just one of the 
coordinate axes.) 

Sketch the hyperbola through the real intercepts thus found, with 
each branch of the curve approaching the asymptotes smoothly. 

5. When a quadratic equation in (x, y) is linear in one variable, 
solve for it in terms of the other variable and then graph the resulting 
parabola by the method of Section 132. 

ILLUSTRATION 1. To graph 9x 2 4y 2 = 36 quickly, we first note that the 
graph will be a hyperbola. We substitute y = and find that the ^-intercepts 
are real, x = 2; we thus obtain points B and D in Figure 20, page 263. 
We obtain the asymptotes, as in Note 1, page 263. Then we sketch branches 
EDF and GBH through points B and D in Figure 20. 

To improve on a graph as obtained through the preceding sug- 
gestions, or when doubt arises as to the nature of a graph, solve the 
given equation for one variable in terms of the other and compute as 
many points as needed, with Example 2 of Section 181 as a model. 

* Illustrated in Example 3, page 264. 



SYSTEMS INVOLVING QUADRATICS 



267 



1. x* + y* = 9. 
4. 4z 2 + 4y 2 = 9. 
7. 9z 2 - I 2 - 0. 



2. 
5. 
8. 



= 4. 



EXERCISE 97 

Graph each equation on cross-section paper. 

- 36. 3. 4x 2 - j/ 2 - 16. 

6. xy - 6. 

0. 9. 4y 2 9x 2 = 36. 

11. y - a; 2 - 4x + 7. 

\ 

13. 3s 2 + 4xy 4y 2 * 0. 

16. (a; 2y)(3z 2y 6) = 0. 

16. Rewrite statements I and II of page 187 for the graph of 

x = ay 2 + by + c, 

where it is understood that the o>axis is to be horizontal as usual. 



10. (Qx - 3/)(3s + 2y) - 0. 
12. 2y 4s -f 6z 2 - 9. 
14. 36 - * 2 - 9w 2 - 0. 



Graph each equation, with the aid of Problem 16. 
17. x = 4y 2 . 18. x - 2y 2 + Sy - 6. 19. 



9 - 0. 



183. Graphical solution of systems involving quadratics 

EXAMPLE 1. Solve the following system graphically: 

= 1 



(1) 
(2) 

SOLUTION. 1. We graph each equation, on one coordinate system. The 
graph of (1) is the hyperbola and the graph of (2) is the ellipse hi Figure 22. 

2. Any point on the hyperbola has coordinates which satisfy (1), and 
any point on the ellipse has coordinates 
which satisfy (2). Hence, both equations 
are satisfied by the coordinates of A, B, C, 
and Z>, which are the points of intersection 
of the ellipse and the hyperbola: 

A: (x = 9, y = 2). 
B: (x - 3, y - 2). 
C: (x 3, y - - 2). 

D: (x - 3, y - - 2). 

These pairs of values are the solutions of 
the system [(1), (2)] and can be checked by substitution in the given equa- 
tions. 

Only real solutions can be found by the preceding graphical method 
and, usually, solutions can be read only approximately from a graph. 




Fig. 22 



268 SYSTEMS INVOLVING QUADRATICS 

EXERCISE 98 

Solve graphically. 

- (* 2 + 2/ 2 = 16, (2* + y = 3, o 

** -2 = 3. * tf + f -9. * 



9t/ 2 = 36, / 25x 2 + y 2 = 25, 

*i < 

/y.2 Q 

V """ i/ 



f y - 2z 2 - Sx + 9, f 2z 2 - z?/ - 6y 2 = 0, 

= 12. 2 + ? 2 = 4. 



4t/ 2 = 36, 
\25z 2 4- 4y 2 - 100. \z 2 + t/ 2 = 9. 

1 84. Solution of a simple system 

A system consisting of one linear and one quadratic equation in 
two unknowns x and y usually has either (a) two different real solu- 
tions, or (6) two real solutions which are the same, or (c) two im- 
aginary solutions. These possibilities correspond, respectively, to the 
following geometrical possibilities: the straight line, which is the 
graph of the linear equation, (a) may cut the graph of the quadratic 
in two points, or (b) may be tangent to, or (c) may not touch the graph 
of the quadratic. 

EXAMPLE 1 Solve- f 4x 2 - 6^ + 9t/ 2 = 63, (1) 

EXAMPLE 1. bolve. \ 2* - 3y = - 3. (2) 

SOLUTION. 1. Solve (2) for x: x = 3y " 3 - (3) 

Zi 

2. Substitute (3) in (1) : 

' . 63. 



2,2 _ y _ 6 . . (y _ 3 )(y + 2 ) = 0; y = 3 and y = - 2. 
3. In (3), if y = 3, then x = 3; if y = - 2, then x = - 9/2. 



4. The solutions are 



x = 3, y = 3 



and 



x =_i v =_2 



./. Since a solution of a pair of equations in x and y is a pair of 
related values of x and y, it is very essential that each solution should be 
plainly indicated as a pair of values, as in Example 1. 

Note %. A system of the type considered in this section will hereafter be 
called a simple system, 



SYSTEMS INVOLVING QUADRATICS 269 

SUMMARY. To solve a system of one linear and one quadratic equation 
in x and y algebraically: 

1. Solve the linear equation for one unknown in terms of the other, say 
for y in terms of x, and substitute the result in the quadratic equation; 
this eliminates one unknown. 

2. Solve the quadratic equation obtained in Step I and, for each value 
of the unknown obtained, find the corresponding value of the other 
unknown by substitution in the given linear equation. 

EXERCISE 99 

Solve, (a) graphically and (b) algebraically. 



, , <> - -16, 



, , 

\4d + 3c = 50. * \u + 2v - 6. a + 2b - 4. 

Solve algebraically. 

(2x + 7/ + 3 = 0, (z 2 + 92/ 2 = 25, 

\ 2z 2 + y 2 - Qy = 9. \x - 7 + 3y = 0. 



-46 = 12, 
\ a 2 + 2a - - 46 - 12. z 2 + 2z - 4y = 23 - 



y, f 2x2/ + 3?/ + 4x - 1 - 0, 

+ 2y = 5. \2aj + y + 3 = 0. 



/ y - 2x + 1 = 0, t / 2x + z/ = 2o - 1, 

+ 24i/ = 36. \ 4a; 2 + 4x + y = 2a 2 



' - 2xy + 2/ 2 + 8z - 2y = 3. 



+ y z - 2by = 2a 2 -f & 2 . 

185. Solution of a system of two quadratic equations 

When both equations of a system are quadratic, the system usually 
has/owr different solutions, all or two of which may involve imaginary 
numbers. The student should recall his graphical solutions of sys- 
tems of this type where four solutions were obtained. 

Note 1 . The fact stated in the preceding paragraph is a special case of 
the following theorem which is proved in a later course in algebra: 



270 SYSTEMS INVOLVING QUADRATICS 

A system of two integral rational equations in x and y, in which one 
equation is of degree m and the other is of degree n hi x and y, usually has 
mn solutions. 

Thus, a system consisting of an equation of the third degree and a quadratic 
usually has 3 X 2 or 6 solutions. Usually, the algebraic solution of two 
simultaneous quadratics brings hi the solution of a fourth degree equation 
hi one variable. At this stage in algebra, the student is able to solve only 
very simple fourth degree equations. Hence he is not prepared to consider 
the solution of all systems of simultaneous quadratics. Therefore, in this 
chapter we consider only special elementary types of systems. 

1 86. Systems linear in the squares of the variables 

When both equations have the form ax 2 -f 6t/ 2 = c, the system is 
linear in x 2 and y 2 and can be solved for x 2 and y 2 by the methods 
applicable to systems of linear equations. 



EXAMPLE 1. Solve: } * ' '* ~ ""> (1) 



(* 2 
\ x* 



x* + 2y 2 = 34. (2) 

SOLUTION. 1. Multiply by 2 in (1) : 2s a -f 2y 2 50. (3) 

2. Subtract, (3) - (2): z 2 = 16; x = db 4. 

3. Substitute x 2 = 16 in (1): 16 -f 2/ 2 = 25; y 2 = 9; y - 3. 

4. Hence, if x is either 4* 4 or 4, we obtain as corresponding values 
y -f 3 and y 3, and there are four solutions of the system. 



-4,y 



x = 4, y = - 3 



x = 4, y = 



EXERCISE 100 

Solve each system, (a) graphically and (b) algebraically. 



*' ^ - -- . 9. 2 ' is 2 + t/ 2 = 36. " \9z 2 -f W - 16. 



Solve algebraically. 

, / z 2 + 4s/ 2 = 14, / re 2 - t/ 2 = 4, A / 2* 2 - 32/ 2 - 3, 

L i * = S*y2 - Ifi. ' ^ rt " - - 



16. 1 2 2 -f v 2 - 11. ' I 5ar 2 -h 2w 2 - 17. 



f 9s* - 
\ &C 2 - 



9i 2 - %* - 6, ft / ISc 2 - 8 -h w , 

** ^ i <n x r^/v A V 



7. 1 15 - 12d - 20c*. 1 6 - 3^ + 5r*. 



f 7r* -1- Ait 2 = 3ft ( 7r* ft?/ 2 = AQ 

10 <* 11 V ^ ' 12 < 

' \ llr* + 5a* - - 4. ' \ 9x 2 + 2y 2 = 13. \ 4x 2 -h 9y 2 = - 4. 



EXAMPLE 1. Solve: ( * + * ~ 14 J , ft 

\ x 2 - 3xy + 2y 2 - 0. 



SYSTEMS INVOLVING QUADRATICS 271 

1 87. Reduction to simpler systems 

<J> 

(2) 

SOLUTION. 1. Factor (2): (x - 2y)(x - y) = 0. (3) 

2. Therefore, (2) is satisfied if either z 2y = 0, or x y = 0. 

3. Hence, (1) and (2) are satisfied if and only if x and y satisfy one of 
the folio wdng systems: 

( & + yt = 14, T / x* + 2/ 2 - 14, 

l 



4. On solving (I) by the method of Section 184, we obtain two solutions: 
(x = \/7, y = V7) and (x = V7, 2/ = V?). From (II) we obtaui 

(a? = fVTO, y - JVTO); (a; - - fVTO, y - - 



We say that the given system in Example 1 is equivalent to the 
systems I and II because the solutions of the given system consist 
of the solutions of (I) together with those of (II). 

The preceding method applies if, after writing each equation with 
one member zero, we can factor at least one of the other members. 

1 88. Elimination of constants 

A system in which all terms involving the variables are of the second 
degree can sometimes be solved by use of the equation we obtain on 
eliminating the constant terms from the original system. 

EXAMPLE 1. Solve: 

INCOMPLETE SOLUTION. 1. Eliminate the constants. 

Multiply (1) by 2: 2z* + fay - 56. (3) 

Multiply (2) by 7: 7xy + 28y 2 = 56. (4) 

Subtract, (3) - (4): 2z 2 - xy - 28y 2 = 0; or, 

(2x + 7y)(x - 4y) - 0. (5) 

2. To solve [(1), (2)] we may now solve [(2), (5)]. This system is equiva- 
lent to the following simpler systems: 

4t/ 2 = 8, / xy + 4y 2 = 8, 

- 0. (6) 



Instead of using (2) in (6) we could equally well have used (1). Each 
system in (6) has two solutions and thus [(1), (2)] has four solutions. 



272 SYSTEMS INVOLVING QUADRATICS 

EXERCISE 101 

Solve algebraically and graphically. 

1i I y t {% j **' \ y i 

/ vj / 

' H" y)( x "" 2y) = 0. \ (x 2/)(x 4- 3y) = 0. 



. Hereafter in this chapter, leave all surd values in radical form. 
Moreover, unless otherwise stated, to solve a system will mean to solve al- 
gebraicatty. 

f 

Solve by reducing to simpler systems. 

3 ( 2z 2 + 5*s/ - 3y 2 = 0, (3x* + 5xy r 

\2s 2 -f Zxy = 2. \ x 2 + sy = 4. 



, = 0, 

* (a; 2 -f 3y 2 = 7. \ 2z 2 - xy - 2y* = 8. 

Solve by eliminating the constant terms. 

I x* + 3xy = 28, 8 f x 2 - 5o;y + 6t/ 2 = 10, 

\xy + 4y* = 8. ' \x*-xy = 4. 



' 2c 2 - 2cd = 15. * 2a: 2 + 2/ 2 = 5. 



= 11, fx 2 

\t - 2 2 + 3 0. \ x 2 - a^ + 4y z = 40. 



6 = 0, f 2w 2 -f 3mn = 1, 

1 y + 2/ 2 - 35. \9w 2 -f 8n 2 = 9. 



, t/ 2 + 7 = 0, 

\ 2mn -f n 2 = 64. 1D * \x 2 - Zxy - If + 4 - 0. 



*189. Additional devices for reducing to simpler systems 

( * + * = , 27 ' 

I x + y = 3. (2) 



EXAMPLE 1. Solve: * + * = 27 ' 



SOLUTION. 1. Factor (1): 

(* + y)(x* - xy + y*) = 27. (3) 

2. Divide, (3) by (2) : x 2 - xy + y* = 9. (4) 

3. Hence, (z, #) satisfies [(1), (2)] if and only if (x, y} satisfies 

f x + y - 3, (5) 

\ x* - xy + y* = 9. (6) 

The student should complete the solution by solving [(5), (6)] by the 
method of Section 184. 



SYSTEMS INVOLVING QUADRATICS 273 



EXAMPLE 2. Solve: ( * + ^ + * - 20, (7) 

I xy = 5. (8) 

INCOMPLETE SOLUTION. 1. Add, (7) + (8): x* + 2xy + y* 25. (9) 

2. From (9), (x + !/) 2 = 25; hence z + y = 5, or a; + y = 5. 

3. To solve [(7), (8)], we would solve each of the following systems: 

(x + y = 5, ! X +.V = - 5, 

1 xy - 5. \ xy = 5. 

*190. Determination of tangents to curves 

EXAMPLE 1. Find the value of the constant k so that the graphs of the 
equations in the following system will be tangent: 

f x 2 + ?/ 2 = k 2 , (1) 

\x + y = l. (2) 

SOLUTION. 1. If the graph of (2) is tangent to the graph of (1), then 
the two solutions of the system [(1), (2)] must be identical. 

2. Substitute y = 1 - x in (1) : 2z 2 - 2x + (1 - fc 2 ) = 0. (3) 

3. From Step 1, we notice that (3) must have equal roofa. Hence, Us 
discriminant must be zero, or 

(- 2) 2 - 4(2)(1 - fc 2 ) = 0; 8fc 2 - 4 = 0; k = =fc i>/2. 
Thus, the straight line is tangent to the circle if its radius is .707. 

*191. Equations symmetrical in x and y. 

An equation in x and y is said to be symmetrical in x and y in case 
the equation is unaltered when x and y are interchanged. A quadratic 
equation in x and y is symmetrical in x and y if the coefficients of x 2 and 
y 2 are equal and those of x and y are equal. The method of the next 
example applies where each equation is symmetrical in the unknowns. 



- ! <a i 2y = 8, (1) 

EXAMPLE 1. Solve: < . , * ) rt \ 

I 2xy -f x + y = - 4. (2) 

INCOMPLETE SOLUTION. 1. Substitute x u + v; y = u v. (3) 

From (1) : 2w 2 -f 2t^ + 4u = 8, (4) 

From (2) : 2w 2 - W + 2u - - 4. (5) 

* 

2. Solve the system [(4), (5)3 for u and v: 

Eliminate y 2 , [(4) + (5)]: 4w 2 + 6u - 4 = 0. (6) 

3. Solve (6) for u; then obtain v from (4). Each pair of values (u, v) 
when placed in (3) gives a solution of [(1), (2)]. 



274 



SYSTEMS INVOLVING QUADRATICS 



4. 
6. 
8. 



*EXERCISE 102 

Solve by any convenient method. 

\8a 8 + 6 s - 98. 2 * ( a - 6 = 3. 3 * ( y(x 

( a; 2 H- xy H- y 2 = 7, 



+ y) 
+ y) 



40, 
20. 



f 
* \ 



}- 



4, 



- 24, 

- 4 - 0. 

4t^ = 9, 
uv + v = 3. 



f z 2 + 3,2 . 13, 
I OPfl/ =r ft 

^ **y ^^ " 



f4c 2 -h3^-f2/ 2 = 8, 
/ \ xy = 1. 



= 2. 



11. 



= 5. 



12. 



14. 



\ 3j/ icy y* = 4. 
' x + 2y -f 2 = 3, 



13 f * + xy + 2/ 2 - 61, 
" \ icy = 29 - x - y. 



y = 6, 
+ ^ -f = 14. 



16. 



- 2?/ 2 - 1, 



8. 



Find the values of k for which the graphs are tangent. Then, if k is real, 
graph the equations of the resulting system. 

- y = 5. ' 17 ' \Sy + 3x - *. ' 18 * 1 4 + 2y + kx = 0. 



16 



an expression for c in terms of the other constants in case the graphs 
of the two equations in the variables (x t y) are tangent. 

1ft /!/ - mx + c, ^ (y = mx + c, M ( x - my -f c, 

i. < . + 4y2 ^ 36 -w. | fl2a . 2 _ ^2 = (W> 



22. 



the method applying to symmetrical equations. 

4x + y 2 4y = 17, 22 / ^ "~ 3a ^ + 2/ 2 == 1, 

6 = 0. . 1 2x 2 - xy -h 2t/ 2 - 17. 



25. 



24. 

without first clearing of fractions. 
11, 



^ 

=: H- 28 - 0. 



26. 



27. 



- - 7. 



SYSTEMS INVOLVING QUADRATICS 275 

* 

MISCELLANEOUS EXERCISE 103 

Solve graphically. 

f4* 2 + y 2 = 25, 2 fa* + 42,*= 16, f* + y'20, 

L '\2x + y = T. A \* 2 -s/ 2 = 9. * \ z* - 4y 2 = 4. 

^ 

f (a - y _ 2)(* - y - 1) - 0, fi y = 1 - s*, 



6-8. Solve Problems 1, 2, and 4 algebraically and compare the results 
with the solutions previously obtained. 

Graph each equation. 
fc 4z 2 + V - 0. 10. 4z - ty* - 0. 11. 4x - 9y* - 36. 



algebraically. 

-2*+ 1=0,, 



Sx - 4y - 3y*. 4x* - xy - y* - 4. 



17 110 - t> 2 - 9, 

'* 



, 
' 3. V * r' - rA = 75. 



or a; 



, 
- 4 2 = a 2 - 2o6 - 6 2 . 



Solve each problem by introducing two unknowns. 

22. The sum of two numbers is 28 and the sum of their squares is 634. 
Find the numbers. 

23. Find the dimensions of a rectangle whose area is 60 square feet and 
whose diagonal is 13 feet long. 

24. The area of a rectangle is 55 square feet and its perimeter is 49 feet. 
Find the lengths of the sides of the rectangle. 

25. The sum of the reciprocals of two numbers is 3 and the product of the 
numbers is . Find the numbers. 



26. A man divides $500 between two investments at simple interest, a 
first part at twice the interest rate obtained on the second part. The first 
investment grows to the amount $345 in 3 years, and the second to the 
amount $220 hi 4 years. Find the interest rates and the sums invested. 



276 SYSTEMS INVOLVING QUADRATICS 

27. The sum of the squares of the two digits of a positive integral number 
is 65 and the number is 9 times the sum of its digits. Find the original 
number. 

28. Some men row 15 miles downstream on a river to a mountain and then 
climb 12 miles to its summit. They take 9 hours for the journey and, the 
next day, 9 hours to return. Find the rate at which they row in still water 
and their speed in ascending the mountain, if they descended 1 mile per 
hour faster than in ascending, and if the rate of the current of the river is 

1 mile per hour. 

29. 'A weight on one side of a lever balances a weight of 6 pounds placed 
4 feet from the fulcrum on the other side. If the unknown weight is moved 

2 feet nearer the fulcrum, the weight balances 2 pounds placed 9 feet from 
the fulcrum on the other side. Find the unknown weight. 

30. A farmer has 3 fields of equal size and pays each of his workmen $4 
per day. He paid a total of $42 to 2 Workmen after each, working alone, 
plowed one of the fields. These men took 2f days to plow the third field 
when working together. How many days did it take each man to plow a 
field alone? 

31. Two towns on opposite sides of a lake are 33 miles apart by water. 
At 6 A.M., from each town a boat starts for the other town, traveling at uni- 
form speed. The boats pass each other at 9 A.M. One boat arrives at its 
destination 1 hour and 6 minutes earlier than the other. Find the time it 
takes each boat to make the trip across. 

32. A wheel of one automobile makes 96 more revolutions per mile than 
a wheel of a second automobile. If 20 inches were added to the length of the 
radius of a wheel of the first automobile, the result would be the diameter of 
a wheel of the second automobile. Find the diameter of a wheel of each 
automobile, using 22/7 as the approximate value of TT. 



APPENDIX 

NOTE 1. THE IRRATIONALITY OF V2 

If there exists a rational number which is a square root of 2, then there 
exist two positive integers m and n, such that 

(l) 



n 

m 



where is a fraction in lowest terms. In other words, if V2 is rational 

7T 

there exist two integers m and n, without a common factor, such that (1) 
is true. Let us show that this assumption leads to a contradiction. 

1. Square both sides of (1): 



= - ; or > 



2n 2 = m 2 . (2) 

We see that 2 is a factor of the left member of 2n 2 = m 2 ; hence 2 is a factor of 
the right member. Therefore 2 is a factor of m because otherwise 2 could 
not be a factor of m 2 . That is, m = 2k, where k is some positive integer. 

2. Place m = 2k in (2) : 

2n 2 = (2fc) 2 = 4& 2 ; 
w 2 = 2k 2 . (3) 

Consider n 2 = 2& 2 ; since 2 is a factor of the right member, hence 2 is a 
factor of n. 

3. We have shown in Steps 1 and 2 that m and n have 2 as a factor. This 
contradicts our original assumption that m and n had no common factor. 
Hence, the assumed equation 1 has led us to a contradiction, and it follows 
that (1) itself must be false. Therefore no rational number exists which is a 
square root of 2, or, \/2 is an irrational number. 

Comment. We easily verify that (1.4) 2 = 1.96; (1.41)* - 1.9881; 
(1.414) 2 = 1.999396; etc. On considering the sequence of numbers 

1.4, 1.41, 1.414, 1.4142, 1.41421, -, (4) 

we see that the square of each number in (4) is less than 2 but that, on pro- 
ceeding to the right in (4), the squares of the numbers approach 2 as a 
limit. Each number in (4) is a rational number; we refer to these numbers 
in (4) as the successive decimal approximations to A/2. 



278 APPENDIX 

NOTE 2. EXTENSION OF THE INDEX LAWS TO 
RATIONAL EXPONENTS 

A complete proof that the index laws hold for any rational exponents 
could be constructed by showing, hi succession, that the laws hold if the 
exponents are (1) any positive rational numbers and (2) zero, or positive or 
negative rational numbers. Without giving a complete discussion, we shall 
indicate the nature of the methods involved by proving some of the necessary 
theorems. For convenience in details, we shall assume tha$ the base is 
positive. In our proofs, we use the index laws for positive integral exponents 
and the definitions of Sections 113, 114, and 115. 

(m\p mp 
a n ) = a n . 

Proof, (a*)* = [(o) W ] P = (a) mP ; [(3), page 151; (II), page 142] 

(!\p tap 

a n ) = a n . [(3), page 151] 

THEOREM II. // m, n, p, and q are positive integer s t then 

m j> m j mq+np 



I * 2\ n fl / m \*tf/ 2\*m 

Proof. [a* an) = (a*) Uv [(IV), page 143] 

_ a m a pn < (Theorem I) 

(m J>\nq 
a oJ a m *+*. [(I), page 142] 

Therefore, by the definition of an ngth root, 

/ . \A. 

ao = ^a mfl+pn ; n = a 

(m\ 
o> 

r/ !\ 
Suggestion for proof. Compute |_ V */ fl 



In the remainder of this note we shall assume that the index laws have 
been completely established for all positive rational exponents. 

THEOREM IV. Law I of Section 105 holds if the exponents are any positive 
or negative rational numbers. 

Comment. We are assuming that Law I has been established if both 
exponents are positive. Hence, it remains to show that, if h and k are any 
positive rational numbers, then a~ h a~ k = o~*~*, and o*a~* a*~*. 

Incompkte proof. By the definition of a negative power, 

-* a -* =*.--= -; or, 
a* a* a*" 1 "* 



APPENDIX 



279 



NOTE 3. ABRIDGED MULTIPLICATION 

^The following example illustrates a method for abbreviating multiplication 
of numbers with many significant digits when the result is desired with 
accuracy only to a specified number of places. 



EXAMPLE 1. 
places. 



Compute 11.132157(893.214), accurate to two decimal 



SOLUTION. 1. To multiply by 893.214, multiply in succession by 800, 
90, 3, .2, .01, and .004 and add the results (in ordinary multiplication these 
operations are in reverse^ order). Since we desire accuracy in the second 
decimal place, we carry two extra places, or four decimal places, in all items. 

2, In the abridged method, to multiply by 800 we multiply by 8 and 
move the decimal point; all digits of 11.132157 are used in order to obtain 
four significant decimal places. This first operation accurately locates the 
decimal point for the rest of the items. 



ORDINARY METHOD 


ABRIDGED METHOD 


11.132157 
893.214 


vWvV 

11.132157 
893.214 


Multiply 
by 


44528628 
11132157 
2 2264314 
33 396471 
1001 89413 
8905 7256 


8905.7256 
1001.8935 
33.3963 
2.2264 
.1113 
.0444 


800 
90 
3 
.2 
.01 
.004 


9943.398482598 


9943.3975 


Add 


^Result = 9943.40 


Result = 9943.40 



3. To obtain four decimal places when multiplying by 90 we do not need 
the last digit of 11.132157; to indicate this place 'V" over *'7" and 
multiply 11.13215 by 90. Next, place ' V" over "5" and multiply 11.1321 
by 3; then " >/" over "1" at the right and multiply 11.132 by .2; then 
" >/" over "2" and multiply 11.13 by .01; then " V" over "3" and multiply 
II. 1 by .004. Then add and round off to two decimal places, obtaining 
9943.40. 

Note 1. The advantages of the preceding abridged method are obvious. 
As compared with the ordinary method, less labor is involved, the decimal 
point is accurately located, and fewer mistakes will occur in the final addi- 
tion. 

Note g. An abridged method of division can be developed similar in prin- 
ciole to the method of abridged multinlicatioiv abova. . . 



280 APPENDIX 

NOTE 4. A FALLACIOUS PROOF THAT 2 - 1 

The following absurd result that 2=1 illustrates the contradictions that 
arise if division by zero occurs. 

1. Suppose that y 6. 

2. Multiply by y: y 2 = by. 

3. Subtract 6 s : y* - 6> - by - 6*. 

4. Factor: (y - 6)(y + 6) = b(y - 6). 

5. Divide by (y 6) : y 4- 6 = 6. 

6. Since y = 6 (Step 1), 6 + 6 = 6, or 26 =* 6. 

7. On dividing both sides by 6, we obtain 2=1. 

Discussion. In Step 5 we divided by zero, because y 6 = if y 6. 
Hence, Steps 5, 6, and 7 are not valid, because division by zero is not allowed. 

NOTE 5. THE SQUARE ROOT PROCESS OF ARITHMETIC 

Consider finding V7569. Since the radicand has four digits to the left 
of the decimal point, the square root must have two digits to the left of the 
decimal point, because the square of any number of units is less than 100, 
and the square of any number of hundreds is greater than 10,000. We observe 
that 80 is the largest whole number of tens whose square is kss than 7569. 
Hence, we consider finding x, a number of units, so that 

<80 + z) 2 = 7569. (1) 

By the formula for (a -f 6) 2 , from (1) we obtain 

6400 + 2-80- x + a* = 7569; ,(2) 

160* + z 2 = 7569 - 6400; 
160z + z 2 = 1169. (3) 

From (3), z(160 + x) = 1169, 

1169 

x = ieo+T 

An approximation to (4) is obtained if we use the trial divisor 160 in place 
of (160 -f x) ; this gives 

x - 115? _ 7+ / 

x ~ 160 ~ 7 ' (5) 

Then, we take x * 7 and verify that the complete divisor, (160 -f- x) or 

167, gives 

1169 - ,, 
7, exactly. 



Hence, V7669 = 80 -h 7 - 87. We verify by squaring that 87* - 7669. 



APPENDIX 



28? 



In the following examples, the student will observe, in brief form, steps 
corresponding to those just explained in detail by reference to the formula 
for (a + &) 2 . Hereafter, the details will be carried out without making the 
possible contacts with the square of a binomial. Essentially, at each stage 
of the following arithmetical process, we have knowledge of a in a binomial 
(a -f 6) and we obtain an approximation to b so that (a + 6) 2 will be as 
nearly equal as convenient^ at that stage, to the number whose square root is 
being obtained. 

EXAMPLE 1. Find V7569. 

SOLUTION. Arrange 7569 into groups of two figures each, starting at the 
decimal point. After each of the following steps, read the corresponding 
explanation below. 

Step 3 Step 4 

8 87 



Step 1 
8 


Step 2 
8 


75 69 
64 


75 69 
64 




11 69 



160 



75 69 
64 



11 69 



Ififi 
A v/vr 

167 



75 69 
64 



11 69 



^ 

t//\ 
XvJU 

167 


StepS 
8 7 


7569 
64 


11 69 
11 69 



Explanation. 1. The largest perfect square less than 75 is 64. Write 
64 below 75. Write V64, or 8, above 5 of 75. 

2. Subtract 64 from 75. Bring down the next group, 69. 

3. Form the trial divisor: 2X8= 16; annex (160). 

4. Obtain the complete divisor: 1169 -f- 160 = 7 + . Add 7 to 160, forming 
167 as the complete divisor. Write 7 over 9 of 1169. 

5. Find 7 X 167, or 1169. Subtract. Since the remainder is zero, 
V7569 = 87. The complete solution appears as Step 5. It alone would be 
written in an actual solution. 



EXAMPLE 2. Find V1866.24. 

SOLUTION. 1. First trial divisor: 2X4 8. Annex 0, giving 80. 

2. First complete divisor: 266 -* 80 - 3+; 80 + 3 83, 

the complete divisor; write 3 over the right-hand 6 of 66. 43. 2 

3. Place a decimal point in the square root, above that of 
1866.24. 

80- 
83 



4. Second trial divisor. 2 X 43 = 86. Annex (860). 

5. Second complete divisor. 1724 -f- 860 - 2 + ; 

860 + 2 - 862. Place 2 above 4 of 24. 2 X 862 - 1724. 



18 66.24 
16 



266 
249 



OftA 

ouu 
862 



1724 
1724 



Check. We find that (43.2)* - 1866.24, or V1866.24 - 43.2. 



282 APPENDIX 



EXAMPLE 3. Find V645.16. 

SOLUTION. 1. The largest perfect square less than 6 is 4 2 5. 4 



or 2*. Hence, 2 is the first digit of the square root. 

2. After forming the trial divisor, 40, it appears that the next 
figure may be 6 since 245 -*- 40 = 6 + . But 6 X 46 = 276, 
and this is more than 245. Therefore, we must use 5 as the 
second figure of the square root. 

3. To form 500, we take 2 X 25 and annex 0. 2016 -f- 500 
is 4+. Then 500 + 4 = 504. The result is 25.4. 



6 45.16 
4 



245 
225 



504 



2016 
2016 



SUMMABY. To find the square root of a number, written in decimal notation: 

1. Separate the number into groups (or periods) of two figures each, both 
ways from the decimal point. 

2. Below the first group, write the largest perfect square less than that group. 
Above the group, write the square root of this perfect square. 

3. Subtract the perfect square from the first group; bring down the next 
group, thus forming the first remainder. 

4. Form the trial divisor by doubling the part of the root now found, and 
annexing zero. . Divide the first remainder by this trial divisor, taking as the 
quotient only the whole number obtained. (Possibly reduce the number by I.) 
Write this quotient above the next group. 

5. Form the compkte divisor by adding to the trial divisor the new figure 
of the square root found in Step 4. 

6. Multiply the compkte divisor by the new figure of the square root. Write 
the product under the remainder. Subtract. 

7. Continue in this way, following Steps 4 to 6, until the remainder is zero, 
or until you have as many places in the square root as are requested. 

Note 1. As a rule, for a random number N, VN will not be a terminating 
decimal. Then, in finding VAT, we annex zeros at the right in N and carry 
out the square root process to as many decimal places as desired. 

EXERCISE 104 

Find the square root of each number. Obtain the result correct to hundredths 
by carrying out the process to thousandths. 

1. 3969. 2. 134.56. 3. 273,529. 4. 8299.21. 

5. 105,625. 6. 936.36. 7. 40.8321. 8. 2.1904. 

9. 78.354. 10. 15,765. 11. 1643.8. 12. 7.809, 



ANSWERS TO EXERCISES 

Note. Answers to odd-numbered problems are given here. Answers to 
even-numbered problems are furnished free in a separate pamphlet when 
requested by the instructor. 

' Exercise 2. Page 7 

1. 56. 3. - 12. 5. 36. 7. 0. 9. - 8. 11. 56. 

13. 5. 16. - 8. 17. 4. 19. 3. 21. - 9. 23. - 168. 

26. 120. 27. 120. 29. - 360. 31. - 2. 33. - 4. 36. 2. 

37. - 13. 39. 5. 41. 52. 43. f. 46. - 60. 

47. 14.4. 49. 4. 61. 31. 63. - 96. 66. 360. 

Exercise 3. Page 12 

1. 27. 3. - 18. 6. 13. 7. - 16. 9. - 32. 

11. - 35. 13. 0. 16. - 13. 17. - 18.2. 19. - 7. 

21. 13. 23. 61; 29. 26. - 30; 4. 27. - 36; - 70. 

29. 17; - 17. 31. 3.3; - 11.9. 33. - 5. 36. - 7. 

37. 10. 39. - 10. 41. 22. 43. 3. 46. 13. 

47. 42. \ 49. 12. 61. 28; 4. 63. - 40; - 26. 

66. 44; - 32. 67. 100. 69. - 53. 61. - 36.3. 

63. - 14; 14; 0; 0. 66. 3; 9; - 18; - 2. 67. 23; - 23; 0; 0. 

Exercise 4. Page 15 

11. <. 13. <. 16. >. 17. >. 19. <. 21. >. 

29. - 5 < - 3; | - 5 | > | - 3 |. 31. > - 3; | - 3 | > | |. 

33. | - 2 | < 7; - 2 < 7. 36. 2 > - 6; | 2 | < | - 6 |. 

Exercise 5. Page 17 

1. 10. 3. - 24. 6. 5. 7. 8. 9. 44. 

11. 36. 13. - 2. 16. 0. 17. - 24. 19. - 13. 

21. - 2o -f 56 - c. 23. 31 - 5a + y. 26. - 8a + 36 + c. 27. 15o. 

29. 15a. 31. 8a - 12. 33. - 15 + 5a + 30c. 

35. 18 - 12a 4- 156. 37. - (5 - 7a + 46). 39. - (- 6 + 3x + 4y). 

41. 16 - (4a + 6 - 3c). 43. 2ac - (- 3 + 5a - 4c). 

Exercise 6. Page 20 

1. lla. 3. l&c. 6. 6cd. 7. 5x 6a. 

9- 3a - lie. 11. 19c - IScd. 13. - 2a - 26 -f 4; - 60 + 166 - 10. 

16. - x + db - 4c; 7x - lldb + 2c. 17. - 3m - k - 6h; 9m - 9fc + 4A. 

19. - 14x - 3y. 21. - 9ac - 7xy + 46. 23. 9o - 206. 



290 ANSWERS 

^ 

25. 3fl + 14A - 23*. 27. 3a -f 10y - 3. 29. 21a - Sly + 9. 

31. 2t - 3. 33. 2a. 35. - r - s. 37. 2. 39. 13 - 4x. 

41. - 56 + 10. 43. - h + 12* - 36. 

Exercise 7. Page 24 



1-1. 


3. 


5. f 


7. i 


9. f 


11. - i 


13. f. 


15. 3. 


17. 3/4y. 


19. 9/26. 


21. - |. 


23. - 5a/3. 


25. tV 


27. 46/d. 


29. & 


31. f . 


33. 6. N 


35. 3c/5. 


37. ^. 


**. 


12 


c 


5bc 


10 




vJL* "ZTT" * 


43. ~ 


45. ^- 47. f. 


49. 


51. A. 


76 


7 


36 5 


9d 




h 


9W 


vtffc 


55. 4. 57. 


Af. 59. 1 61. 


63. yfy. 


65. - |. 


' 2k' 




c 










Exercise 8. Page 27. 






1. 16. 


3. 100. 


6. 10,000. 


7. 1. 


9. 1. 


11. - 27. 


13. - 125. 


15. - 243. 


17. - 1000. 


19. H. 


21. - i 


23. - 16. 


25. -216. 


27. -75. 


29. 320. 


31. n odd, 


neg. ; n even, pos. 


33. - 120. 




37. J. 


39. y w . 


41. x\ 


43. 6 18 


45 /i*+* 

W Cv 


47. a 8 . 


49. *y. 


W(W2 
</!/ 


53. h*. 


55. 816 4 . 


67. 


59. g- 


61. ft. 


63. -y^S. 




.!5i 4 . 


._ a 4 c 4 

Of _- - * 


69. 27c. 


71. 1 


j JL C4/ */ 


fcHo 4 


166 4 








73. Six 8 . 


7K /r^^-w 

I V t C/ 


1 f*^ * 


79. 


r^t/^ 



1 / ^,4 



125a 

Exercise 9. Page 28 

3. 3fl 2 6. 6. 152 8 . 7. 

9. - 4s 2 *. 11. - 2x 2 j/ 2 . 13. 2as 6 . 15. - 8o 8 6>. 

17. 24rW. 19. 12a; - 3y. 21. 20a - 15. 23. 15z + 12y. 
25. 6s 2 lOa? 4 . 27. 6x 2 2x*. 29. 3^ 2 fc 3hk. 

31. 15t^ 20to* 10w>. 33. 6a 4 6. 35. 24m 6 n 4 . 
37. Qx h+n y l+k . 39. 3^ 44r A; 34 '*. 
41. 14x- 10x 4 + &c - 12x 2 . 43. 3o 4 6 4 -f 15o6 + 18o 2 6 2 - 9o6. 

46. 3 - 2y + 4y 2 . 47. 4a 2 - lOa - 14. 49. 6x - 24x* - 12. 



Exercise 10. Page 30 

1. 3* + x - 12. 3. 2z 2 - 3x - 35. 5. 20a 8 - 43o + 21 

7. 4# - 9* 2 . 9. 2o + ab - 156*. 11. 9r - 25a*. 

13. 2a 2 6 - ab - 15. 15. c 2 ^ - x 2 . 17. a 2 + 60 + 9. 

19. W - 8Wfc -H 16A J . 21. 9a* - 12o6 + 46. 23. oV - 2a6x -f V 



ANSWERS 297 

25. y + 3y* - 10. 27. 3a + a'6 2 - 46*. 29. y - 8. 

31. x - x 2 - llx + 15. 33. 6 - 5x - 6x 2 - x 8 . 35. 2x - 5x 2 - 8x + 5. 

37. 20 - 146 - 36* + 6. 39. 6s 4 - 7x + 12x 2 - 19x + 7. 

41. 2^ - 20y - 6j/ 2 f 25y - 25. 43. 15s 4 - 17x + 12x 2 + 17* - 15. 

45. 25x 2 -f 4y* + 9 + 30* - 12y - 2Qzy. 47. a* -f 6. 

49. 2x - x 2 - 16x +"l5. 51. 6a - llo 8 - 17o + 30. 53. x* - 27y*. 

Exrc!$ 11. Pag* 33 



1. y 2 . 


3. l/x. 


6. x. 


7. 9. 


9. 1/x. 










z 4 


11. 2y*. 


13. 4x. 


15. 7a. 


17. l/5r. 


- ^^ ^ 


- 


. 


Jfc 3 




66 


21 


23. -- 


25. r 


27. tot. 


29 





y> 


/i 




c 2 


Li- 

4c 


33. 7x. 


35. - J. 


07 __ , . 
87 ' S6 2 


89. 6xy. 



41. fa + 56. 43. - a - 46. 45. 3a - 2a 2 . 47. - 2 + 5a*. 

49. r - ! 2 ~ 2x + 3. 53. y 2 - y + 5. 



55. A-^i + r.- 57. 66-3a. 59. - ^ -h 2x + ? - - 

15 5x* 3x 2 2 a; 



61. - 4y + -'- 63. # - a - 



Exercise 12. Ps 36 

1. x + 4. 3. c - 3. 6. 8 - 3. 7. y - 4. 



15. o-6-^--r- 17. x + 2. 19. 3x+ll + - 

2o 4- 6 x 

21. x - 2. 23. 2x - x - 6 - r-^-r- 25. 2y - 3 - 



2x - 3 * 4y 2 - Zy + 2 

27. x - x 2 - 4 H ?_^ 29. x 2 + 3xy + 4y 2 . 81. x 2 - 3x + 9. 

x 3 



33. x 2 + xy + y*. 35. 4w^ + 6u> + 9. 37. a 4 + a^ + 6*. 39. 2x 2 - 3. 

Exercise 13. Page 39 

^ 



15. 

o 

21. 36c. 23. 





2 + 6- 


- a 


6. 5 * 
a 


S< 3 

Oil ^ /Y* 

JU U ^^^ w(f 


-7 


' 8 

I 


17. 6. 

25. 28fec*. 


U * 6 
19. 34. 

O7 80 

Ale $f. 


2o 2 6 2 . 





292 ANSWERS 



29. - 8L. jt. 35.. 37.48. 

35 Q 



39. 300. 41. 18,900. 43. 24a 3 6. 45. 36az 4 . 47. 80/W. 

Exercise 14. Page 41 

1. V- 3. i 6. &. 7. ft. ' ^' 

11.^3*. 13. H. 16. -|. 17. J|. 

21. 2 -3* + 6 */. 23. ^J>. 26 . 



. . 

31. 



13 5r-3fe 106* - 3a 



206 - 15ay + 3a .< 19o -f 94 6 - 



' 20 

._ 12 Art 5x - 9 ._ 2y 2 2 - 4yz - 6y + 9 

45. - 7; - 47. - -- 49. 

4* Ox . 



3 - 13z _. 27y - 10^ + 20 __ 3 - Sab 3 - 4a 2 6* 



Exercise 15. Page 44 



9. . 11. H. 13. 

2 - 3a 5 - 3bc 

5 + 4a' 46c + 3' ' 



5a 2 fe - 3 3 

* ' 



- 2 4x 2 - Zxy* 20x 2 - 6 

+ 156 



2o + 3 2 - 3z 36x - 24 

33. ^. 35. f . 37. - I 39. &. 41. - &. 

43 ! 45 2a-3fe 

36 - 2a* ' 3o + 5/i* 

Exercise 16. Page 46 

1. - 60. 3. 0. 6. f . 7. - 4. 9. 1. 11. 6. 

13. 17. 15. - 32; - 14. 17. 42; - 8. 19. <. 21. >. 

23. 26 -}- c - 3a. 25. a'6 2 - 3a'6. 27. foy - 12zV. 

29. 9Jk - 11 A. 31. 15 - 2o. 33. - f 35. &. 

37. ^. 39. ^f. 41. 24 W. 43. 



ANSWERS 



293 



a 2 *) 2 



45. 


625c 8 d 12 t/ 4 . 


A7 A 7 _ 

Tile J2 5 


AQ . 

TBV* 


53L 63. 








4 


4x* 


55. 


4z 2 - Sx - 21. 




67. 62;* - l; 


3x|/ + 15y. 


69. 


2s 3 - 5s 2 - 8z 4 


- 13s + 15. 


61. a*. 


68. - 






3 


4 -1 


- 8w 


65. 


2z 2 + 4x 3 + 


\j 


A7 


^^y 4^^^ K 
cm ff 


2z -5 


VI. ^, 


to 4 


71. 


31 - 12a 


73 Qy '~ { 


toy* - 20x + 12: 


rt/ 
. 7fg 10 


12 


Id. 


12x 2 y 3 


ID. %%. 


77 


9y 2 - 30z 2 i/ 


70 2 **>- 


3x 2 i/ 


01 5. ! 



4c 



3y - 5x 
Exercise 17. Pase 50 



1. 3.25. 



3. 100,000. 



5. .0001. 



11. + - 2 + , 



9. 3(10 3 ) + 10 2 + 4(10) -f 9. 

15. 5735.35. 17. 14.1192. 

23. - 103.7698; 16.0762. 
29. 326,530. 31. .000317. 



13. 536.437. 
21. 6.64; 3. 
27. 5.32. 



1. 4.914. 

9. .000054322. 



1. 15.326; 15.3. 

7. .034564; .0346. 

13. 31.54; .586. 

19. 2,056,000. 



Exercise 18. Pase 52 

3. 5.993. 6. .51312. 

11. 2.1435402. 



a-5 



7. .0000001. 



19. .0681. 
25. 1.178. 
33. 5.738. 



7. 13.62528. 



Exercise 19. Pase 55 

3. .31486; .315. 
9. 566.5 and 567.5. 
16. 11.4034; .054. 
21. 10 2 (6.7538). 



5. 195.64; 196. 
11. 567.35 and 567.45. 
17. 2738. 
23. 4.5726(10 4 ). 



25. 4.5312(10 6 ); 4.53(10). 27. 7.2200(100; 7.22(10'). 29. 2.6(10) cu. ft 



Exercise 20. Pase 57 



1. 1.37. 
13. 



3. 57.2. 
6- 



6. .263. 
17. 



7. 150. 
19. .625. 



Exercise 21. Pase 62 



9. .02981. 
21. .15. 



11. .286. 
23. .4375. 



1. 
11. - 
21. J. 
31. 8. 
41. 3. 
61. f 



63. - 



3. - 3. 
13. 1. 
23. - 3. 
33. 6. 
43. .36. 
7. 55. 2. 



5. i 


7. 0. 


9. $ 


16. f. 


17. f . 


19. - f . 


26. V- 


27. 15. 


29. . 


35. 2. 


87. 4. 


39. 17. 


46. f. 


47. 4. 


49. f. 


57. - f . 


59. 283.46. 


61. 515.02. 



294 ANSWERS 

Excreta 22. Page 66 



1 3 + 


C 


^ , , , . 


K .-, 


ft 




1 


c 
11. ^ 


5 '3~a 


**6- 

17 -. 


2a 
3 
9a6 15 


IL 2 


1S * 2 
9L ^ 


1T *3fc 


L * 12 
27 9 

41 ^ 

r^ n ^M ir 


2L 36 - a 

9ft . - . 


6a-5c 
_-. 


29> 35 

10. M 9.ft7 r/f 



23. 



* 2a-36 
2a6 
3c ' 

o6c) 
-6d 



n . . 33. a - 

9 m 

k / <\j I a -\- d . I a 

35. a - I (n l)d: n - 3 - : d = - - 

d n 1 



5 o tt >. M + N + P ^^,0,0 

37. a - - - 39. A - - r - -- 4t C - .12n + 6. 

r* 1 3 

Extrcta 23. Page 69 

1. 32.5* and 35.5'. 3. 22.5' and 5'. 

5. 3f. 7. 15; 16; 17. 9. 8'. 11. 40'; 120'. 

13. 13 nickels; 39 dimes; 36 quarters. 16. 80 bu. 

17. 8f hr. 19. 3H da. 21. 2$ hr. 



Excrcist 24. Page 72 

L .05. 3. .0375. 6. 1.263. 7. 7%. 9. 2%. 11. 135%. 
13. 8.32. 16. 37*% of 200. 17. 175% of 200. 19. 452.9, approximately. 
21. 560 dimes. 23. $22,000. 26. 75 Ib. at 70 ff; 25 Ib. at 50*f. 

27. 20 gal. 29. Approximately 88.9 bu. at $1.25 and 111.1 bu. at $.80. 

31. 3 gal. 33. 58&%. 

Exercise 25. Page 74 

1. 21^r ft. from fulcrum on other side. 3. 63& Ib. 

6. 8 ft. from fulcrum on side of 40 Ib. weight. 7. 69& Ib. 



Exercise 26. Page 77 

1. 50 m.p.h. 3. At end 7| hr. 6. 16$ sec. 

til 
7. ~ sec. 9. 310 m.p.h. 11. At end 10 yr. 

x > 

13. 1792 mi.; 7 hr. and 28 min. 16. 1306$ mi.; 7 hr. and 28 min. 

17. Approximately 8.13 hr. 19. At lOft min. after 2 P.M. 



Exercise 27. Page 80 

L $180.00; $5180.00. 3. $48.00; $3048.00. 

ft. $159.00. 7. $2914.98. 9. $42,857.14. 



ANSWERS 295 

It $1000.00. 18. $5000.00. 18. At 5%; gains $17.86. 
17. $4000 at 5%; $3000 at 4%. 

Excreltt 28. Past 83 

1. rfc 5. 3. 11. 5. db $. 7. 3. 9. 9. 

11. 14. 18. |. 16. i. 17. |. 19. x*. 

21. a. 28. a. 25. 2a*. 27. 2^. 29. 7*. 

81. 8*c*. 83. 7*A 36.?. 37. |- 39. ~- 41. 



Extreist 29. Past 86 

1. 15a - 20t>. 3. 4a6x - a*6x. 6. c* ^ 

7. a* + 2ay + y*. 9. 16 - y. 11. 9 - 

13. a 4 - 96*. 16. a* - 60 + 8. 17. x* + lOx + 25. 

19. 4a* - 20a + 25. 2L 4z - 4u* + M* 23. 4a* -f 4a6 + #. 

25. 6 + 5x + x*. 27. x* + 4z - 45. 29. o + 5a6 + 66. 
81. 6x* + 17 + 12. 33. 8y - IQxy + 3a?. 36. 6y -f y - 15. 

37. 21u* + 29u> - 10. 39. 8x + 6xy - 9j/. 41. - 12 + Ifo - 5x*. 

43. - 3aj* + 19x - 20. 46. x* + 4a* + 4. t 47. 4*V - 12xy -f 9y*> 

49. 9 + 246x + 16Wx*. 61. a;* - z + i 63. i - ^ + 4. 

66. <AP - 9z*. 67. a? + .3x - .1. 69. 6 + 1.1* - .1*. 
61. io - #*. 63. .08x* - .2&r - .15. 66. 16 - 8a* + x. 

67. - 14oz + 21x + 7x*. 69. 6 - 7x - 20x*. 
71. 4s* -f 12xy + 90*. 73. 4s* - Sxy + 4y*.' 
76. lOOc* - 300cd + 225(P. 77. 12x* - x* - 6. 

79. 2a* - a*6* - 15M. 8L 21a + a6 - 10M. 

83. 12x - sV - 6. 86. 9u - 15w*e - 14. 

Extrclst 30. Past 88 

L x* + y* H- 2xy + 4 + 4x + 4y. 3. 9 - 12* -f 6y + 4x* - 4xy + &. 

6. 9x + !/* + 25 + 6xy -f lOy + 30x. 7. 16o* -f 6* + c* - 806 - 8oc + 26c. 
9. 4z* - 12ox + 12M* + 9o* + 96* - 18o6*. 

11. x* + 2xy -f y* - 9. 13. 16 - 4a* - 4o6 - 6*. 16. a + 2ab + 6* - x*. 
17. 9a* - Gay + i/* - 16. 19. x* - y -f 2y* - *. 
21. 4x* + 4xy + y* + a* - 60 + 9 + 4ax - 12x + 2ay - 6y. 
23. 4x* + z* 4x* + y* 4y + 4 + 4xy 8x 2yi + 4z. 

26. 4a + 4xy + j/* - * -f 6 - 9. 27. c* - 4cd + 4d* - a* - 2ax - x*. 
29. 4a* + 96* + 16c* + 12o6 + 16ac + 246c. 

31. to* + 25x* + 9a* - lOua + taw - 30ax. 

Exttcist 31. Past 90 

t x(3 + 6). 3. 2x(3y + a). 6. y(2c + <P + 1) 

7. x(36 - a + c). 9. (<-<#- 4a). 



296 ANSWERS 



11. ay*(3ay - 2 + !/*). 13. vto*(Wx - 6 + 5wx 2 ). 

15. (w - z)(w + 2). 17. (8 - xy)(8 + xy). 19. (2x - y)(2x + y). 

21. (6d + ll)(ed - 11). 23. (2a + 36)(2a - 3b). 25. (16a + l)(16a - 1). 

27. ($ + u>)(i - u>). 29. (5w + cd)(5w> - cd). 

31. (6a6 + 8x)(6a6 - 8x). 33. a(x - y)(x + y)(x 2 + y 2 ). 

41. (x + 6) 2 . 43. (a - I) 2 . 45. 12>; (w - 6) a . 

47. (x - 9) 2 . 49. (7x + a)'. 51. (8 - a6) 2 . 

53. 12x2; (2x + &)'. 65. 20acd; (2cd - So) 2 . 

57. (3x - 5y)*. 59. (2x 2 - 7) 2 . 61. (2a 2 - St 2 ) 2 . 

63. (72 - 26)(72 + 26). 65. 4w(3t; - w)(3 + w). 

.67. (x - 5j/ 2 ) 2 . 69. x(2o - I) 2 71. 2(7u 2 - 5)(7w 2 + 5) 

73. 25(x - 26 s ) (x + 26 s ). 75. (4x 2 + 25^) (2x - 5t>)(2z + 5t>). 

77. 2(3w - 5). 79. 3(7x - 5y^)(7x + Sw 2 ^). 

81. 4(100). 83. 1600. 85. 280. 

Exercise 32. Page 93 

L (x + 5)(x + 3). 3. (a - 6)(o - 2). 5. (x - 5)(* - 3). 

7. + 7)(* - 3). 9. (x - 6)(x + 3). 11. (w - 6)(w + 8). 

13. (5 + u>)(8 - W). 15. (6 - w) (4 + w). 17. (8 + y)(4 - y). 

19. (9 + Jb)(6 - Jb). 21. (x 4 - 12) (a; + 6). 23. (5a + 7) (a + 1). 

25. (50 - 3)(2x - 1). 27. X 2 (4x - 3)(2x - 1). 29. y(Zy + 5)(y - 1). 

31. (3x - 5)(x + 2). 33. (4w^ + 3)(2w>' - 3). 35. (5a 2 - 7)(3a 2 -f 4). 

37. (7 + 2s) (1 - 3x). 39. (1 - 3x)(9x + 2). 41. (3x + 2)(x + y). 

43. \2w + 5z)(4w - 3). 46. (6u/ + u)(2w - 5u). 47. (3a - 56)(2<z - 6). 

49. (lOa + x 2 )(10a - x 2 ). 61. (x - 2y)(x + 2y)(x 2 + 4y 2 ). 

63. (8a - 3c) 2 . 66. (4 - 3x)(2x + 5). 67. 2x(x - y)(x + y). 

59. (3x 2 + 2y) 2 . 6L (5x + 106)(5a: - 106 2 ). 63. r(2 - 5ft) (1 - 3ft). 

65. (| - 2y)(* + 2y)(^ + 4j/ 2 ). 67. Prime. 

69. (3x 2 + 5)(x 2 - 4). 71. 2 4 (5w - 2)(5to + 2)(25to + 4). 

73. (2y + 2)(2y* - *). 75. - (3a - 66). 

77. (2x 2 - 5)(x* + 3). 79. (3a 2 - 5s/ 2 )(a 2 + 3y 2 ). 



Exercise 33. Page 95 

1. 2(x + 2y). J 3. (c + d)(x + y). 5. (2ft - 3&)(m - 2). 

7. (2d - 5c)(r + ). 9. (3ft - l)(w - 2). 11. (3a + 26) (w - 2k). 

13. (a H- 6)(3c + d). 16. (c + 3d)(r - ). 17. (2x + y)(c - d). 

19. 4(x - 6)(ft - 2c). 21. (x - 2)(x - l)(x + 1). 23. (x + 2)(x 2 + 1). 
25. (x - 3)(x 2 + 1). 27. (a - 3)(a 2 + 1). 29. (3x - 2)(x 2 -f 2). 

31. (2 + x)(r - a). 33. (x - s - 3)(x + 8 + 3). 

35. (22 + w - y)(22 + w + y). 37. (c - 3d - 2x - y)(c - 3d + 2x + y). 
39. (2 + 1 - 3x)(2 + 1 + 3x). 41. (y + z + 2x)(y + z - 2x). 

43. (2o - 3* - l)(2a + 32 + 1). 46. (3x - y + 2)(3x + y - 2). 

47. (4o - 1 + 3x)(4a + 1 - 3x). 49. (6 + c)(x - y)(x + y)(x 2 + y 2 ). 



ANSWERS 297 

51. (z 2 - w)(s 2 4 w - 1). 63. (r 4 3t - a - 6)(r 4 3< 4 a + 6). 

65. (c + 2 - 3d - h)(c 4 2 4 3d + A). 

67. (3x - y - 5o 4 5) (3* - y 4 5a - 6). 59. (a + 6 + 3x)(a + 6 - 3x). 

61. (2o - 36 4 2x + y)(2a - 36 - 2x - y). 

63. (2x - 3y)(2x + 3y)(4x 2 + 9y' + 1). 

Exercise 34. Page 98 

1. x 9 - xy 4 y 2 . 3. a 2 - 3a6 + 96 s . 5. c 4 w>. 

7. 27a 3 - c 8 . 9. 1 - 27x 8 . 11. 6" - 8x. 

13. (d -y)(d? + dy + y 2 ). 15. (y - 3)fo + 3y + 9). 

17. (1 - tO(l 4^4^). 19. ( 4 10)( 2 - 10 + 100). 

21. (1 - 3)(1 + Zx + 9a; 2 ). 23. ( 

26. (6x - i/z)(36x 2 + 6x2/2 + 2/ 2 2 ). 27. (7a - 

29. A 8 - 3h?k 4- 3/iA; 2 - A; 8 . 31. u* -f 9u + 27u -f 27. 

33. &c + 12m 2 + 6w*c + w*. 36. 64z + 48z 2 y + 12xj/ 2 -f 

37. a 6 - 6a 4 x + 12a 2 x 2 - Sx*. 39. c - G^c 2 + 126c - 86. 

41. 8c - 36c 4 z 4- 54c 2 z 2 - 27z 8 . 

43. (x + 2)(x - l)(x 2 + * 4- l)(x 2 - 2x + 4). 

46. (2x - 37/)(4x 2 + Qxy + 9y 2 )(x + y)(x 2 - xy + y 2 ). 47. (a - 

49. (w - 3x) 3 . 61. (c - d - a)(c 2 - 2cd + ffi + <w - a^ + a 2 ). 



Exercise 35. Page 99 

1. (a 2 + a + l)(a 2 - a + 1). 3. (3a 2 + 2a + l)(3a 2 - 2a + 1). 

6. (z 2 + hz + /i 2 )(z 2 - ^z -f /i 2 ). 7. (2i0 2 42aw>43a 2 )(2t0 2 -2au>43a 2 ). 

9. (5a 2 4 5ab 4 26 2 )<5 > a 2 - 5a6 4 26 2 ). 11. (x 2 - 2x 4 2)(x 2 4 2x 4 2). 

13. (z 2 4 4Az 4 8fc 2 )(z 2 - 4^z 4 8A 2 ). 15. (9z 2 4 12xz48x 2 )(9z 2 - 12xz48x 2 ). 

17. (3a 2 4 2oc - 2c 2 )(3a 2 - 2ac - 2C 2 ). 19. (5a 4 3y)(a - y)(5a - 3y)(a 4 y). 

21. (3x 2 4 3xy - 5y 2 )(3x 2 - 3xy - 5y 2 ). 

Exercise 36. Page 101 

1. (2ob) 3 . 3. (2a6) 4 . 5. (5x 2 y) 8 . 7. (4 

9. (a - x)(a 4 x)(a 2 4 * 2 ). 11. (2 - w)(2 4 w)(4 4 t^ 2 ). 

13. (x< 4 y 4 )(^ 2 4 y*}(x -)(* + y). 15. (3 - 2x)(3 4 Jto)(9 + 4x). 

17. (u - l)(w 4 l)(u 2 4 w 4 l)(w 2 - u + 1). 

19. (x - 2y)(x 4 2y)(x 2 4 2xy 4 4y 2 )(x 2 - 2xy 4 4y). 

21. (x 2 4 l)(z* - x 2 4 1). 23. (x 2 + 9)(x -W + 81). 

26. (4 4 o 2 )(2 - a)(2 4 a)(16 4 a<). 

27. (a 4 b)(a 2 - ab 4 6 2 )(a - a 8 6 8 4 6). 

29. (3x 2 - y)(3x 2 4 3/)(9x< 4 J/ 2 ). 31. (5 - 2x)(5 4- 2x)(25 -f 4x*). 

33. (a - 26) (a 4 26) (a 2 4 2a6 4 46 s ) (o 2 - 2a6 4- 46 s ). 

36. (2a - 3x 2 )(4a 2 4 6ax 2 4 9x). 

Exercise 37. Page 103 

L x 6 4 x*y 4 sV + V 4 xy 4 4 y 6 . 3. x* - 2xy 4 4xV - 8xy 4 
6. a 4 o^ 4 aV ay 8 4 y 4 . 



298 . ANSWERS 



7. xi 

9. 2* + w + w^ + w*z + to 4 . It x 4 + z + x* + x + 1. 

13. * - xty + *V - aty 4- *V ~ *Y + *V ~ V + *y* - V*. 

15. a* + 2a + 4. 17. x 4 + y 4 . 19. a - a*b + a6 - 6. 

21. x 4- xy 4- tf. 23. z - xV 4- *V - y f - 

25. 4s> + 6x -f 9. 27. z 4 + 2V + 4x*z* + tote 4- 16x. 

29. (a - t0)(a + w)(o + c). 8t (t* 

88. (1 - y)(l -f y)d + y + i/)d - y 
85. (w - )(ti* + v + )(ti + t*V + ). 
87. (2o - l)(16a* + 8a + 4a + 2a + 1). 

89. (2 + *)(64 - 32x + 16x - 8x + 4x* - 2x 

41. (a - 3a?)(a' + 3aa; + 9x<). 48. Prime. 

45. (ti* + t>)(w tt -!* + *). 47. Prime. 

49. (2 - *)(4 + 2x + x)(64 + &r + x<). 5L (u* 



Extreist 38. Past 105 
1. 3. f. 5. 7. ^4^- 9. 



11. ~- a 13. ^-4-r- 15. ^-T^:- 17. 

23. 



2 

2s - 



' x - 2y 2 3x 4- 2y 

!*-*a. 27.--?- 29. ^. 

a 4- ^ * 4- y c 

3 x ^ x4-3 5 2x 

31. r-r-r-* 83. - . , 35. , . n -rr- 



w * 26 -x 
Excrcis* 39. Pag 107 



( + 3)(* - 2) 4z - 9 2o - 4 



7 . . ,. 

* 1 ^r 



4o6 15(o - 6) 15(x - y) 



17. 



- 4y 6rf - 2c 3(4x - 



13s - 2x + 10 - 19x + 4 6a - 6 

-6x * (2x- l)(3x + 3)' 2o - 3 



- 3xy + 8 
* ' ' *"* 



2x-2y 6(a-n) * x + x - 12 



14 + 2n 6c-5c-H30 Iftc 8 + 36z -f 45 

%FflV^ f^ f 4 \ r * * \ Vv^r ^^ / A ^^ \ X A ^V V ^FUi 



3(1 - n)(n + 4) 2(3c - 2)(c* - 9) (3 - 2x)(8z - 27) 



37. 
43. 



ANSWERS , 299 

12x + llx - 25s - 9 . 13* + 18x - 3 



- 4)(2z - 3) 3(x + l)(2z - 3) 2(9 - 4x)(x - 3) 

9or - t* - 60* -h 81o 4 - 9or* -j-'3or* - 27or 



(r - 3a)(r + 3a) 

Extrcist 40. Past 110 

+ 4) 



Li 3.-- 6. 



9. 

(2a -f 36) (a* - 06 + 6*) 

v 'b-a 

17 . Z 19 . . 2 L 



a + 36 12 2y + 3x 

2 



23. - 

lOsy 1 w 

3x + 4 2y(y -r ~ y i ~ / _ 

29. 3L S3. 



2x(x + 4) 

36. , ' ,- 37. * v * ""^ 39. ^ 

c(n v) y + 5 

4L^ 



ac 36c a 8 6* 2 4a 

-6) 



jiT v***' i^ yy / y^v i **y / MQ 

* 7 - - _ y) 2 -x 3a-26 

(3a - 2)(2a - 



2a(5a - 1) (5o - 3)(3 - 3a) 

ExrcUt 41. Past 115 

1. 14. 3. - 5. 6. i 7. - 11. 9. 3. 

11. 2. 13. - 2. 16. 5. 17. 4. 19. - f . 

21. &. 23. - 5. 26. 1. 27. 3. 29. tf 

31. 4 hr. 33. 380 m.p.h. 36. 15 m.p.h. 

Extrcist 42. Paft 117 

2M-Ja; 3. 3a -f 6. 6. ^^. 7. 2n. 

<5 * 

t _1_ M K C rf 

13. 2a. a 



. 

2o6 6 a 



17. - 19. i 21. 26. 23. r 

Extreii* 43. Pg 118 

1. 0s* - 250 1 . 3. 4x* 4- 12x H- 9. 6. y 4 - 6tcy* 

7. o - 64. 9. (y + 5*)(y - 5). 11. (i - 4y). 



300 ANSWERS 

13. (o - 36) (a' + 3o6 + 96*). 15. (3y + 2 2 ) 2 . 

17. (2 + 7)(f - 3). 19. (4x + 1)(2 - to). 21. 5(i 

23. 2(a + 2)(a + l)(a - 1). 25. (x - a - 36) (x + a + 36). 

36x-f 6 2a - 6 - 5a6 - 56 a 

Ol. 



. . - - 

2y 3x x* a 2 6 2 



83. * " * 36. -*. 37. -2. 39.-^-- 

o 3 c 1 a + 6 

Exercise 44. Page 120 

13. (5, 1); area = 40 sq. units. 16. 9 sq. units. 17. 10 sq. units. 

23. All abscissas are 2. 25. 4 units. 

Exercise 45. Pase 124 

1. (a) 8 and 4; (6) - i, - f, and 0. 
15. (6) Equals if x = 4.4 or 1.6; equals 10 if x = 6.5 or .5. 

Exercise 46. Page 1ST 

L 7. 3. - 1. 6. f . 7. - 33. 9. |. 

11. 4c - 12C 9 . 13. 9; 6* - b + 3; c< - c 2 + 3; s - 5x + 9. 

15. 4; 4; H*; (x + 2y)/(x - y). 17. - 5; 27; c 2 + 66c. 

Exercise 47. Page 130 

19. x 5; x = 4. 21. Cuts x-axis at (5, 0); y-axis at (0, 3). 

23. Cuts x-axis at ( f, 0); #-axis at (0, f). 25. y = fx - Jgk 

Exercise 48. Page 132 

ATote. In this answer book, in any solution of a system of equations, the values 
of the unknowns will be arranged in their alphabetical order. 

1. (- , - If). 3. (2, 5). 5. (- 2, 3). 7. (- 2J, - f). 

9. (1, f). if. No solution; parallel lines. 

13. No solution; parallel lines. 15. Infinitely many solutions. 

Exercise 49. Page 134 

1. (3, 2). 3. (- 1, - 3). 5. (0, - 4). 7. (2,2). 

9. (- f, |). 11. (i |). 13. (2, f). 15. (5, 2). 

Exercise 50. Page 135 

1. (5, 2). 3. (7, $). 6. (0, 0). 7. <0, 0). 

9. (3, - 2). 17. (- ff, - V). 19. (.42, .19). 21. (- .35, .27). 

23. (2, 3). 25. (5, - 3). 27. (3, 2). 29. (5, - 3). 

Exercise 51. Page 136 

!(-!,- i). 3. (2, 5). 5. (i, - |). 

_ /2 6\ /a 6\ /3M-* M \ 

7 ' la' 2 j ' *' W a) ' U ' UM^' " 9AT2fc/ ' 

13. (26, -3o). 15. (a + 6, 6 -a). 17. (m - n, 2m "" 2n V 



ANSWERS 301 

Exercise 52. Page 138 

1. (1, 2, - 2). 3. (- i - f, |). 6. (f, - f, f). 

7. (ft, - A, A)-' <- 2, 3, 3). 11. (i i - i). 13. (- 1, - 1, 3, 2) 

Exercise 53. Page 140 

1. 30; 120. 3. 41i; 48*. 

5. 5 gal. of 20%; 2 gal. of 50%. 7. 11' by 3'. 

9. 1st, 3 lb.; 2d, 6 Ib. 11. 13, or 26, or 39. 

13. 40 lb. silver; 80 lb. lead. 16. $3000 at 3%; $2500 at 4%; $4500 at 6%. 

17. 465. 19. y = - 4x - 11. 21. y - - Js + 2. 

23. ^ = 2. 26. Land, 90 mi; water, 48 mi. 

Exercise 54. Page 143 

1. 32. 3. - 243. 6. &. 7. Minus. 9. x*+*. 

11. z 16 . 13. 32a". 16. 625zV- 17. - 8z. 19. 16a. 

21. a 2 *. 23. d 2 **. 25. c*cP*. 27. .09cd. 29. - 

Zr 



1 <w <M w 

31. 33. SO. 37. 

d 3 4 6 s a* 

-IS- - 



_ __ > 
a 2 " 64s 2 w*c* 200x 



66. (a) 16; - 16: (6) n odd. 

Exercise 55. Page 147 

1. 8. 3. 9. 6. =fc J. 7. =fc .1. 9. 12. 

11. i 13. f 16. - 3. 17. 5. 19. - 6. 

21. 3. 23. 5. 26. =fc . 27. d. 29. 3. 

31. 3. 33. 57. 36. 4zy 3 . 37. 6. 39. - 2. 

41. 2. 43. 4. 46. 2. 47. - 1. 49. 6. 61. 20. 

63. 20. 66. i 67. - i 69. .1. 61. .1. 63. .2. 

Exercise 56. Page 150 

1. 6. 3. a. 6. x*. 7. *. 9. y 9 . 

11. x 9 . 13. z 8 . 16. 2y. 17. 2y. 19. J. 

21. . 23. &. 26. f. 27. 3s*. 29. - 2z. 

31. x*y*. 33. 2a 2 . 36. - .1. 37. 2xy*. 39. - 2. 

2x 



41. xz*. 43. .2rc. 46. .5x. 47. ^ w . ^-^ 

2y 



Exercise 57. Page 153 

1. 3. 3. 2. 6. J. 7. &. 9. 8. 

11. i. 13. sV. 16. i 17. i 19. 1. 



302 ANSWERS 



21. i 


23. 3. 


26. .6. 27. i 


29. &. 


31. -^ 


r . 33. - 1. 


35. - i 37. - 


5. 39. 10. 


41. 125. 


43. 216. 


45. 625. 47. i 


49. 16. 


1 


y 


KR * K* 3 

65. 57. TT 

d 1 A 4 


ML*. 


6L 


a 

^J^f9 * ^ 


65. 6. 67. ac. 69. J-- 


y* 


4x* 




5a 


"-TO 


78 C 


n. ^ vv 125 8 ^ M 18 J 


I / f 


o * * 


* I e? * 


a*c*d< 


(k* 


oa' 


9 ay 


81. r*. . 


83. 5y-. 85. 5y~. 


87. 4(3- 1 a*x-'y-'). 


89. 8x*y 


-V. 91. 3(1. 


04)-". 93. c(x - 5y)-. 


95. </z. 


97. <^6*. 


99. 5^c. 


101. b#x*. 103. 6*. 


105. ^6c. 


107. V5x\ 


i/. 109. ^49a 4 . 


111. ai 113. 5*ci 


116. (a* - 36)*. 


117. (c - 


3d)^. 


119. (a - 6)i 


121. (4 - x)i. 






Exercise 58. Page 155 




1. X*. 


3. x 4 . 


6. o. 7. 


16. 9. 8x. 


11.^' 


13. 25. 


15. a*. 17. 


- 19 - 


a 9 






x* * 6* 


8 


o 


9y 


xV 


* x* 


* X*' 


* ~x*' 


2? * "36" 


29. 8. 


31. 625. 


S3. 


35. r- 






16 





37. xl 


39. i- 

a* 


4L xi* 


43. ~ 


a* 


aiV 




, 


45. ^r- 


47. 


49. ao*. 


61. - 


fei 


2y 




y 




>- 66 ol6> 


67 3 . 


59. 9m. 


27 


" Six 




61. 25x*. 


63 a 


AfC , r . 


7 6 


v , 

ab 


fi h 


r~j~~~r 


i 


ab 


Oifw^ 


, 


fid 


71 


, iy 


TK 


* 1 


_ I JL _ 

c 1 o 


* i i i . ^ 
a 6* -f- oo -f- a* 


o t* 

3a6 


77 ^ 


- 79. x- - 


- y" 1 . 8t 16x* - yf . 


88. x* - y* 


"' ., _L 1 


a + c 


>* 






85. 15aH 


- 14a* - 8. 


87. a" 4 + 2a-*& 


+ &. 


89. at + 


2ai6* -H 6*. 


91. o" 1 + 2a~V 


4-y 4 . 


93. a- + 


3a"*6 + Sa'V + 


6. 96. 27 - 27y l - 


f 9y* IT*. 


97. 6~a 
irut ^^^ 


;-* - 15x- 4 . 
11 


99. 3"x5. 
nK at fciv mir 


<A< >lm^.^fcMk 

^L^^^Ltt *f w * w tMt w*^jWwfp 

/.*. . ^ ^^-1\/^ i . ._i\ 



109. (3ar> - 6")(3ar + 6-). 111. (2x* - yi)(2xi -f yi). 



ANSWBRS 303 



113. (3x* - 5y*)(3a* + 5y). 115. (2a* - 36*)(2a* + 36*). 

117. (z - 3ar)*. 119- (Sa- 1 - 6-). 121. (6x* + yty. 

123. (Sar 1 - 2y)(or l + y). 125. (2a* + 36*)(4a* - 6a*6* + 96t). 

127. (6 - x*)(36 + 60;* + x*). 120. Sar 1 + jr. 

Extrclst 59. Pagt 158 

1. 3^2; 4.242. 3. 2VS; 4.472. 5. 10V; 14.14. 

7. 3V3; 5.196. 0. 6V2; 8.484. IL .3V6; .6708. 

13. 2^2; 2.520. ^ 15. 3^4; 4.761. 17. - ^5; - 1.710. 

10. - 3^2; - 3.780. 21. tf^'x. 23. y<^P. 

25. x*V&- 27. 3a. 20. 2a^. 31. 

33. 2ayV^. 35. ayi/lV. 37. Sy^^. 30. 

41. - o^. 43. xy*V3ij*. 45. 2d^cd. 47. - 



40. .5xV;. 51. V^. 53. - >^. 55. 

y* 2y* 

57, - 4; ^2T 50. 3Vl + y . 81. aVl + 56. 63. 

atr 

66. x*. 67. 2x^2T 60. ( + ^ ** 71. 8^/2? 73. 

ab 



75. 3V. ' 77. (a - 5fe)V2. 70. (3 - x)^3x. 81. (2x - 

Exercist 60. Pigt 160 

1. Vl5; 3.873. 3. 5V2; 7.070. 5. 6V^ ; 8.484. 

7. 40. 0. 18\/2; 25.45. 11. 5>/42; 32.40. 

13. 54. 15. 9^60; 35.24. 17. - 2^; - 2.884. 

10. V7; 2.646. 21. ^9; 2.080. 23. VH. 25. ? 

i 

^O _ _ 

27. - 20. 3xV5. 31. 3*V2x. 33. 3a6< / 2a6. 



a 

35. 375a. 37. 54s. 30. 6 s ** + 6. 41. - 9 + 7<s/5. 

43. 1. 45. 18 + 13V6. 47. 5. 40. 2>/6 + VlO + 8>/3 + 

51. 27 + 10>/2. 53. 14 - 4>/6. 55. a - 9fy. 57. 6x - 15y + 
50. a + tor + 26Vax. 61. - xyzV*. 63. Vl8a. 

65. V^te. 67. ^27V. 60. ^486. 



Exercist 61. Pgt 162 

1. ^V2; .707. 3. i\/10; .632. 5. iVlO; .7905. 7. J^; .630. 

0. i<^; .6785. 11. - ^^7; - .2646. 13. rb"^; -05477. 

16. - A^30; - .3107. 17. ri^iKJ; .1095. 



304 



ANSWERS 



19. 



' 26 



35. 
2c 

43. - - 



53. 



1. $; .577. 
7. ^v^; 1.291. 

9 ~ 5 3 



13. 



o 



; .057. 



25. 8 ~ 5 ; -.465. 
2 

31. i^5; .342. 



87.- 



2o& 



A 
21. 



29. 



a 



23. 



26 



V 
3z ' 

5to5 



25. 



- 3a 



45. 



a 



39. 
47. 



ax 



49. 



_ , 
5ao 



3) 



61. 4V5. 63. 0. 

Exercise 62. Page 163 

3. fV5; 2.683. 
9. |Vl5; 1.549. 



65. 



a 



a. 



5. 
11. 



; 1.155. 
; 1.890. 



15. 3 - 2V2; .172. 17. 7 ~ e 3 6 ; - .0694. 



; .689. 



17 



27. - i^l8; - .437. 29. - ^100; - .4642. 



33. 



89. 



26c 



35. - 



41. 






2V5 + 4V-3 V6 
14 



45. 



4.900. 



ax* 



9. 
17. 
25. 
33. 
41. 
49. 
67. 



Exercise 63. Page 166 

5. 

13. Vs. 
21. 
29. 

37. 9. 
45. 

63. ^3. 
61. 



3. 
It 
19. 
27. 
35. 

43. 9a. 
51. 
59. 



7. 

15. V^. 
23. 
31. 
39. 
47. 
56. 
63. 



ANSWERS 



305 



65. v^27. 

73. 

81. 



67. o< 
75. 1. 



9L 



97. 

105. v'a. 
113. 2vl2. 

119. 2(V + ^2). 121. 



26 
107. #Zy. 



123. 



Vtf 



a 



69. 



77. 



* 3* 
93. 6^ 

101. 



26 



109. 



4- 



6 -2a 

I /-| 



y* 

95. 3^. 



103. a 
111. 



Exercise 64. Page 168 

3. 1. 5. 3>/3; 5.196. 

9. . 11. fVH; 1.342. 13. 239. 

17. fcVg; .306. 19. v^2; 1.260. 

23. &VI6; .1265. 26. - 



29. 

35. 3zi 

43. - 

a; 8 



61. 



a 



67. 
63. *4 



79. yVx*y. 
87. - : 



i. (a- 



31. 



37. 
45. 



81. 



2z J 
89. 



39. 

47. 



59. - 
65. 

71. Z</x. 73. 6v y 6. 



83. 



-f 



a 6 



7. 125. 
15. 1^12; .572. 

21. 3V2; 4.242. 



33. 



41. 36. 



65. 4; - 3. 



61. 
67. 

75. v*6. 77. 

. 8R (31 - 180)^3 



91. 



97. 



06 



306 ANSWERS 

Exercise 65. Pa 9 172 

L 3i. 3, 6e. 5. St. 7. 5*V2. 9. 

11. #. 13. IT'- 15. it. 17. .3*. 19. ft. 

at .6i. 23. itV5. 25. &iVll. 27. 2W. 29. obi. 

31. 2isV2. 33. 5tfc\/3. 35. SisVVsty. 37. 



39. frdi. 41. ~-- 43. d=fi. 46. 



47. t. 49. - 1. 51. - 1. 53. 10. 55. 13. 

57. - 29 4- 11*. 59. 21 - 20t. 61. 9 + 40*. 63. Si - 40. 

65. 4 + 19i. 67. - 15. 69. 3*V + 10V2. 71. - 20*V - 17. 



73. (4* - 19); (- 34 - 6t); (- 70t - 66). 75. .- 77. 1 + 2*. 

41 

Exercise 66. Page 174 

Note. In simplifying radicals in the solution of an equation, it will be assumed 
that any literal factor of a radicand is positive if this adds to our convenience 
in the reduction. 



1. =fc 5. 3. 3t. 5. =fc ft*. 7. $36; d= 1.183. 



9. iv IL -. 13. ijl; .829. 

6a 



15. 30 1.826. 17. i2; .707. 19. * 



. . 

2(1 + c) m T 

33. db *V2; db .707. 35. 14. 37. iVl3; 1.803. 

Exercise 67. Page 176 

1. 5; - 2. 3. - 4; 3. 5. 0; f. 7. 0; . 

9. 0; f. 11. f 13. - 3; i 15. *; . 

17. f ; I 19. - i; - i 21. - 2; - f. 23. f ; - 4. 

25. - J; - f. 27. f ; - 3. 29. f ; - 2. 31. - |; 1. 

33. - ; |. 35. 0; - 37. - 3&; - 26. 39.' - 6; 6. 

-5 ; S' ^ 'I 5 -5' 

49. i 51. - 3; - i 53. 0; ; - f . 55. - 3; |. 

Exercise 68. Page 180 

L 16; (x - 4). 3. c; (x - c). 5. |; ( - ). 

7. f|; ( + *)* ^ - 7; 1. 11. 3; - 7. 13. - 2; - 2, 



ANSWERS 307 

2 :t Vl4 

1C. - ~ - : - 2.871; .871. 17. (2 + t); (2 - t). 



: 



o 2 

4-*- \/TQ 
23. -~-^= -: - 2.786; .120. 25. |; - 1. 

o 

29. 6a; - 3a. 3t 6; - $6. 38. 



- 2 d= V4 + oc - tf - 4HP 
86. - 37. 

9 

Extrcisc 69. P*gt 182 
1.}; - f . 3. f; - J. 5. 1 3t. 7. f ; f. 



2 dr V 1 4- \/2 

- 1<U- 1 Rfift -- . -~ x oi v ^ 

v. ^ . 107 1 l.OOO. 

ttll:i 



2 

5V2. 

OO. ^ . ^.loo, 1. 



- 12ac 



2a 



IT - - . 


.CMJU, .100. 

19. t*. 
l2*Vl3 


"' 4 
23. .3; .5. 

20 32i 


o 

31. t; f. 
37. fd; - id. 

JO ,7. 0^ 


25 ' 2 

~26i 


2 


A ^ OA/ zt ^ y/c "* A^&UAC 

^ii , _-. . 


u> 10* 
r. ?; - 


*O. a , ^C. 
4ft 2 - 5 

1+ V ' 



51. y - x -f 2; y - ^(1 - x). 
53. a; = y 2; x * 1 2y. 

Extrcist 70. Pag 184 
L - 3; t 3. |; - f. 5. 9; - 5. 7. - f ; * 

9. *; f . ' 11. i; i. 13. 2 ^ 5 ; .847; - .047. 15. 0; f- 

o 



17. iA'; .905. 19. =fcfc6. 21. |; - 1. 

23. 6 \/41: 12.403; - .403. 25. 4; - f. 27. 6; i 

- h =* 



oo i wi A 01 L 11 oo 

29. i; ^(1 b). 31. fc; JA. 33. 



C 



35. . 37. 13' by 17'. 39. ^; - tf. 41. 14.928'. 

43. 6.48 yd. ' 46. 20 m.p.h. 47. 6 m.p.h. 



308 ANSWERS 



49. (a) t - ; (6) s - 500' at t - 3.46 sec. and 9.04 sec.; 8 - 0' 

<7 

at t = sec. and t = 12.5 sec. 

Exercise 71. Page 188 

L Vertex (0, 0); axis x * 0; min. = 0. 

3. Vertex (0, 0); axis x 0; max. = 0. 

5. Vertex (0, 5); axis a; = 0; min. =5. 

7. Vertex ( 3, 4); axis x 3; min. = 4. 

9. Vertex (1, 5); axis x = 1; max. = 5. 
11. Vertex (2, 5); axis x = 2; min. = 5. 
13. Vertex (f, 9); axis x = f ; min. = 9. 

15. Min. = - 13. 17. Max. 8. 19. At end 2$ sec. 

27. 30; 30. 29. 7*" by 15". 

Exercise 72, Page 190 

1. f. 3. 3; 3. 5. Roots imag. 7. 1.6; 4.1. 9. Roots imag. 
11. 2.6; - .6. 13. - 3; - 1. 15. (0, 2|); (1, 1*); (3*; - 1). 

Exercise 73. Page 192 

1. Disc. = 9; real, unequal, and rational. 
3. Disc. = 12; real, unequal, and irrational. 
5. Disc. = 0; real, equal, and rational. 
7. Disc. = 1705; real, unequal, and irrational. 
9. Disc. = 0; real, equal, and rational. 

11. Disc. a 16; imaginary and unequal. 19. 5.2; 1.2. 21. 1.2; .2. 
23. Disc. 49; graph is a parabola concave upward, with its axis perpendicular 
to the 3-axis, cutting z-axis in two points, and hence the vertex is below 
that axis. 

26. Disc. = 59; graph is a parabola concave downward, with its axis per- 
pendicular to x-axis, which does not meet that axis and hence lies entirely 
below it. 

27. Disc. = 52; etc. 29. - 2 + 5i. 31. - 6*. 

Exercise 74. Page 195 

1. - 5; - 3. 3. f; f 5. - f ; - |. 7. 0; - tf. 

9. - |; - f. 11. f ; - f 13. - ; - |. 15. y - |. 

17. - r-|-; ^4-- 19 - fir; fr^- 21 - te ' + 1* - 3 - 

. 5 + o 5 -h a 1-fcl+c 

23. x* + 2x - 7. 25. x 2 + 5z -f 6 = 0. 27. 3z - 7x + 2 - 0. 

29. 3z - x - 10 - 0. 31. z 8 - 2 = 0. 33. x - 18 - 0. 

35. 9z 2 4- 4 - 0. 37. x 8 -f 4z - 1 = 0. 39. 2a - 2x - 13 = 0. 



ANSWERS 309 

41. x 9 - Ox + 34 0. 43. a;* - 8z + 20 - 0. 45. x* - 4x + 24 0. 

47. 3z + 4a + 2 - 0. 49. (ftc - 8)(3x + 5). 51. (8z - 15)(3z 4- 4). 
53. 27x a + 12x 32 =. 0. 56. No. (Disc, is not a perfect square.) 

57. (x + 3 + i)(x + 3 - i). 59. (x - $ + ft) (a; -*- 



Exercise 75. Page 197 

1. 1; 2. 3. fc2; 2. 5. 2t; =fc . 7. 3i; =t 

9. |; db i. 11. i? 1. 13. d= 2i; =fc iV. 15. 5; 1. 

17. - 1; - f. 19. 3; 2; - 1. 21. - 1; f; 1(1 =t t 

23. - *; - |. 25. 1; - 4; *(- 3 =t V5). 27. 1; - 3; *(3 V5l). 

29. 1; 3; - 2; - 4. 31. 2; - ^2. 33. ; - 1; *(1 tVf). 

35. d= 2Va; =b i^ea. 37. ; J(~ 1 tV5). 39. 3; *(- 3 db 3tV3). 

41. f; fi. 43. - ; ^(3 3iVi). 45. 4; (- 2 VS). 

47. - 1; J(l db iV). 49. i; i(- 1 =b *V). 51. 1; =t . 

53. 3; 3i. 55. 2; 2t. 57. ; d= ft. 

Exercise 76. Page 200 

1. 7. 3. No sol. 5. 12. 7. - 13. 9. 14. 11. f >/2. 

13. No sol. 15. 4. 17. 9. 19. 4; - 2. 21. 0; f V5. 23. 0. 

25. No sol. 27. 3; - 1. 29. i 31. 4; |. 33. 1; 2; - 3. 



35. o. 37. 0; 46. 39. ; - 41. 

7T t 



43. 1; &. 45. 8; - ^-. 47. 16; if. 49. 16. 

61. =t ^V2. 53. 4. 65. No solution (any principal root is positive). 

57. 243. 59. - 243. 61. - 25$. 

63. ^4 (x~% = 8 has no real solution). 

Exercise 77. Page 203 

1. db 1 3. 2. 6. 0; - J. 

7. H (& = 1 is not a solution because it does not give a quadratic equation). 

9. 10; 2. 11. ^; k does not apply. 

13. - .268; - 3.732. 15. - . 17. &. 19. - i 

21. if. 23. - &. 25. =fc 3V5. 27. - f . 

29. - f. . 31. =fc .816. 33. 1; - 2. 

Exercise 78. Page 206 

1. a 6 + 5o6 + 10a6 J + 10a*& + 5a& + 6 s . 

3. x 8 - 8ofy + 28zV - 56x 5 y + 70x 4 y 4 - 56xV + 28a;V - 8xy -f y. 

5. 16 + 32a 4- 24a 8 -f 8a + a. 

7. 7296 - 14586 B y + 12156y - 540&V + 1356V - 18^ + V*. 

9. a -f So^ + 3a6* + 6. 

IL o - 6a w 6 4- 15a6* - 20a6 -f 15a6 - 6a'6 l <> 

13. x - 



310 



ANSWERS 



15. a* - 

17. a* - 4oir H- 

. , 8x* , 24x 32x* 
19. x - + --- ~ 





a* a* a* a 

25. a" -f 20a& -f 190a6. 
81. 2x* - 30 2ox + 435 

/ _ ^ ^ \ 

85. * n - 
89. 720. 



I5xy* - 
- 40IT 6 + IT 8 . 
16 






27. 1 - + 2.31. 

87. a*" - 



Ig0ou lg 
^ 

29. 1 - 12 >/2 -f- 132. 



+ 



/ ^^ -j \ 



41. 39,916,800. 

Exercise 79. P3t 208 
8. 35z- 5. - 



48. 126. 



7. 



9. - 

n(n-l)...(n~5) 



15. 



21. 



6! 



11. .00056. 
^ 



18. 15x*. 



; 2016xy 10 . 28. 



3/ 



25. 



27. 126o>6; 



29. 10,000 - 4000o + 600a - 40a + o. 3t 96,059,601. 83. 132,651. 
36. 1.127. 87. 1.230. 39. .904. 41. 1.243. 48. .002. 45. 14,776,336. 



Extrcls* 80. Page 211 

1. f . 8. f . 5. s/a. , 7. 

11. *. 18. 30. 15. f. 17. 10"; 8'. 

2L 32'; 21$". 28. 5500 sq. in. 25. 5.366'. 27. 28.7'. 

81. d= 1. 88. 25. 85. db 9i. 87. $ 



9. 

19. 20; 70. 

29. db v^6. 

39. d= (y + 3). 



^ oe 

41. 35. 



A 

48. -T- 
3 



Am 

45. 



27 

77 
175 



. 

47. 



9m 4 


5n 






7. 



Exercise 81. Ps 217 

8.Z- 
. V 1 

9. T 



a* 



15. u varies directly as x 9 and y. 

25. ^. 27. 784'. 29. 2700 Ib. 



18. z is proportional to x*. 

12t/ 

21. - -r^- 28. H - - 
ox 

81. 4| ft. 88. (a) 71f Ib.; (6) 7 sq. in. 87. 5* ft. in diameter. 

89. 1824 r.p.m. 4L /i : /, 1:16. 48. 56%. 45. (10, - 6, 4). 

47. (6, - 2, 4) or (- 6, 2, - 4). 49. (150, 250, 550, 300). 



ANSWERS 3?T 

Exercise 82. Page 223 

L 15; 18; 21; 24; 27; 30. 3. - 18; - 16; - 14; - 12; - 10; - 8. 

9. 13. 11. 14. 13. 151. 16. - 76. 17. 10. 19. I - 78; S - 645- 

21. i - - 72; S - - 882. 23. I - .78; S - 46.86. 

25. d - 16; 8 - 5460. 27. n - 92; 5 - 18,308. 

29. a - 138; S = 2025. ^ 31. n - 21; 5 = 525. 

33. a - - *', I" 126i . ' 35. k - - 1. 

37. 21. 39. 136th. 41. |f. 

Exercise 83. Page 226 

1. 5; 8; 11; 14. 3. 8; 10; 12; 13; 15$. 

5. 10.5; 6; 1.5; - 3; - 7.5; - 12. 7. - i; 1; f ; ; ^. 9. 26. 

11. - 19. 13. 36,270. 15. 2223. 17. 376. 

19. 360'. 21. $11,650. 23. $30,500. 25. 2565 

Exercise 84. Page 230 

1. 5; 15; 45; 135. 3. 4; - 8; 16; - 32. 7. &; ^. 9. or 4 ; or*. 
11. 1.01; (1.01) 8 . 13. 4. 15. f. 17. 729. 19. &. 

21. I - 2916; S = 4372. 23. I = - 1215; S = - 910. 

25. I - 192; -S - 129. 27. Z*>= 3846'; 8 - 



29. W. 81. n - 8; 1275. 33. n = 6; 5 - 111.111. 

35. a - 25; n - 5. 37. n = 11; S = ^^i. 39. i = 80. 
41. f. 43. (4; 8; 16; 32; 64) or (- 4; 8; - 16; 32; - 64). 

46. (12; 36; 108) or (- 12; 36; - 108). 

47. 1; 10; 100; 1000; 10,000; 100,000. 49. 2. 61. 10. 

. _ . M 05^ 7 1 

63. VJJ, if x > 0; - v^ if x < 0. 66. ^^- -- 



_ (1.06)> - (1.06) 4 eA . 1 - (1.02)-" (1.02)" - 

67. - :rr - 59. - -- Ol. 



. . 

.06 .02 ( 1>0 2)i 

63. - A; 2. 66. 8190. 67. $102.30. 



Exercise 85. Pase 233 
1. $95. 3. 227.8" approximately. 6. $3125. 

7. *n(n - 1). 9. $7020. U. 



1 x 

13. 55.339" approximately. 16. At end 2 yr. 

19. 599.59' approximately. 21. (a) 300(1.06)*; (6) 5060 units. . 

23. Approximately 11.1% per year. 27. 12i%. 

Exercise 86. Page 236 

1. i; t; il A- 3. A; i; f; f. 6. 1; f; f; A; 

7. 2. 9. 16. 11. 2*y/(* + y). 



372 



ANSWERS 



1. 14. 
13. * . 

25. 



1. 3. 
15. &. 
29. 3. 
43. 10. 



Exercise 87. Page 239 

3. 22i 6. I 7. flft. 9. i iL 

1K _7_ r 1T JL 1O 5 Q1 19 QQ 

* w 33 * 33 ** TT * IJL * 90 *^ 

27. 3^. 29. 1500". 31 200 sq. in. 33. 12. 

Exercise 88. Page 241 

3. - 1. 6. 64. 7. 81. 9. 10. 1L 1. 13. A 

17. 2. 19. 2. 21. 1000. 23. 6. 26. 2. 27. 2. 

31. 4. 33. i. 35. $. 37. - 1. 39. - 3. 41. 5. 

45. 1. 47. 2. 49. 8. 61. 100. 63. 64. 



1. .7781. 
9. 1.6232. 
17. - .5229. 



Exercise 89. Page 244 

3. 1.5314. 6. 1.4771. 

11. .3680. 13. .7533. 

19. - 2.1549. 21. - 1.7696. 



7. 3.2304. 
16. - .3853. 
23. - 1.3768. 



Exercise 90. Page 247 

1. Ch. = 2; man. = .9356. 

3. Ch. = -2; man. = .700. 

7. Ch. = 6; man. = .325. 

11. 4.4932 - 10. 13. 5. 

19. 1.6355. 21. 7.8949 - 10. 

27. 8.9345 - 10. 29. 5.0043. 

35. 4660. 37. 1.43. 39. 74.0. 

45. .0960. 47. .000900. 



6. Ch. = 3; man. = .5473. 
9. 9.2562 - 10. 
15. - 4. 
23. 0.9759. 
31. 5.1959. 
41. 302. 
49. .264. 



17. -6. 
26. 4.2504. 
33. 243. 
43. .00589. 
51. .00500. 



Exercise 91. Page 251 



1. 3.2615. 

9. 9.7503 - 10. 
17. 6.0910 - 10. 
25. 1379. 
33. 7.695(10 8 ). 



3. 2.7261. 
11. 8.1939 - 10. 
19. 3.4950. 
27. 39.95. 
36. 1.030. 



6. 1.5556. 
13. 4.9546. 
21. 1725. 
29. .0002162. 
37. .00009738. 



7. 9.4790 - 10. 
16. 7.1581 - 10. 
23. 1.459(10). 
31. .4693. 
39. .4236. 



Exercise 92. Page 253 



Note. In some classes, the teacher may desire to teach the use of 5-place log- 
arithms. For the advantage of such classes, in the case of each computation 
problem in the remainder of this chapter, the result obtained by use of 5-place 
logarithms is given in black face type beside the result found with 4-place log- 
arithms. 

1. 24.91; 24.909. 3. .2009; .20086. 5. .006380; .0063797. 

7. - .007667; - .0076660. 9. 51.10; 61.098. 11. .1406; .14061. 

13. 24.56; 24.668. 16. .07808; .078096. 17. 5542; 6644.4. 

19. 27.61; 27.609. 21. .003467; .0034669. 



ANSWERS 313 

23. - 2.627(10-'); - 2.6266(10-*). 26. 1.580(10-'); L6802(10" 6 ). 

27. 38.96; 38.966. 29. (a) 4.792(10 6 ); 4.7922(10 8 ): (6) 8.065; 8.0662. 

Exercise 93. Page .256 

L 5358; 6369.6. 3. .4107; .41082. 6. 1.044; 1.0440. 

7. .9500; .94986. 9. 1.315; L3158. 11. .6030; .60296. 

13. 28.93; 28.936. 16. .1585; .16849. 17. - 1.010; - 1.0099. 

19. 50.32; 60.324. 21. 41.47; 41.470. 23. .1266; .12668. 

26. 2.111; 2.1111. 27. 1.041; 1.0412. 29. .8630; .86268. 
31. 50.12; 60.466. By preliminary use of 7-place table, the results are .5050; 

.60604. 

33. 141.9; 141.82. 36. 215.1; 216.08. 37. .4971; .49714. 

39. .001352; .0013626. 41. .9388; .93896. 43. .3986; .39882. 

46. - 1.916; - 1.9166. 47. 21.76; 21.768. 49. 134.9; 134.84. 

61. - .136;* - .1366.* 63. 1.118; 1.1177. 66. 4.908; 4.9086. 
67. .1730; .17294. 

69. By 4-place table: (a) 2.219(10 4 ); (6) 3.222(10- B ). 

61. .02323; .023229. 63. .0007867; .0007869.* ' 66. 236.1; 236.13. 

Exercise 94. Page 259 

1. 1.341; 1.3410. 3. 1.319; 1.3194. 6. - 5.195; - 6.1923. 

7. 18.1;* 18.02.* 9. 5.63;* 6.634.* IL 4.317; 4.3176. 

13. 2.303; 2.3026. 16. - 14.2;* - 14.20.* 

Exercise 96. Page 261 
1. $4682. 3. $4502. 6. $1203. 7. 5.8%. 9. 18.8 yr. 11. 16 yr. 

Exercise 98. Page 268 

1. (4, .5); (- 2.8, - 2.9). 3. (5, - 3). 

6. (2.1, 1.5); (- 2.1, 1.5). 7. (3.2, 3.7). 9. No real solutions. 

Exercise 99. Page 269 

1. (3, - 4); (- 4, 3). _ 3. (5, - 3); (5, - 3). 

/6 2*W 6 + *V6\ /6 + 2t'v / 6 t v 
5. ^ _ , _ Y ^ > ; 

7. (- 1, '- 1); (- 3, 3). 9. (4, 2); (4, 2). 11. (1, *); (i 2). 

15. (il); (-i3).' 

Exercise 100. Page 270 

1. (1.837, .790); (- 1.837, db .790). 3. (|V2, ); (- |V^, ffl. 
5. (V5, 1); (- V5, 1). 7. (db V2, V); ( \/2, - V5). 

9. (^v^, iv^S); ( |V, iv^S). 11. (V, i^/7)', ( v^, 
* The result is not reliable beyond the last digit given in the answer. 




314 ANSWERS 

Exercise 101. Page 279 



1. (V2, - V5); (- 

3. (i, 1); (- ft, - D;_ (- V2, ftV5); (V2, - 

6. (- V3, V3); (V3, - V); (- 2, 1); (2, - 1). 

7. (14, - 4); (- 4, - 1); (- 14, 4); (4, 1). 
. (I, - ft);- (ft, - *), (- f, i); (- ft, ft). 

(- ft, ft); (i - ft); (_- 2, i); (2, - i). 

13. (ftV2, JV5); (- ftV2, - JV5); (2, 5); (- 2, - 5). 
15. (6, 4); (- 6, - 4); (- 



Exercise 102. Page 274 

X 

1. (- ft, 5); (ft, - 3). 3. <|Vl5, f Vl6); (- |Vl5, - Vl5). 

5. (- 2, 3); (6, - 1). 7. (f, - f); (- 1, - 2); (f, V); (- 1, 1). 

9. (i, 2); (- i - 2); (1, 1); (- 1, - 1). 
11. (2, 1); (1, 2); [ft(- 4 + ,V6), ft(- 4 - iVe)]; 

Cft(- 4 - Vg), J(- 4 + t-s/6)]. 

18. (- 5 + *Vl4, - 5 - iVII); (- 5 - Vl4, - 5 + tVl4); (5, 4); (4, 5). 

15. (ft, 2, - 1); (ft, 2, 1); (- ft, 2, - 1); (- ft, 2, 1); 

(ft, - 2, - 1); (ft, - 2, 1); (- ft, - 2, - 1); (- ft, - 2, 1). 

17. 25. 19. c - V9 + 4m 3 . 21. c = d= V o s + 6m 8 . 

23. (3, 1); (- 3, - 1); (1, 3); (- 1, - 3). 

25. (i, - |); (i - i). 27. (i, 1); (- |, 1). 

Exercise 103. Page 275 

1. (2, 3); (ft, 4). 3. (4.1, 1.8); (- 4.1, =fc 1.8). 

5. (- 1.8, -2.1); (2.5, -5.2). 

7. (|V65, *V35); (- f V65, ftV5). 13. (f, - 2); (4, - 7). 

16. (- 1, 0); (1, 0); (V5, - 3*V2); (- *V2, 3tV^). 

17. (3, 6); (- 3, - 6); (- 4^3, 5^3); (4V3, - 5V). 

19. (10, - 5); (- 10, 5); (ftV, J^V2); (- ftV2, - J^V2). 
21. [ft(a + 1), ft(a - 1)]; [ft (a - 1), ft (a + 1)]. 

23. 12' by 5'. 26. ft; f 27. 81. 29. 3 Ib. 31. 6f hr.; 5ft hr. 

Exercise 104. Page 282 

1. 63 (exact). 3. 523 (exact). 5. 325 (exact). , 7. 6.39 (exact). 

9. 8.85. 11. 40.54. 



INDEX 



Numbers refer to pages. 



Abscissa, 119. 
Absolute value, 4. 
Addition, 8. 
Antilogarithm, 247. 
Approximate values, 52. 
Arithmetic means, 224. 
Arithmetic progression, 221. 
Asymptote, 263. 

Base, of a logarithm, 240. 

for a power, 25. 
Binomial, 28. 
Binomial formula, 207. 
Briggs, 251. 

Characteristic, 244. 
Coefficient, 18. 
Cologarithm, 254. 
Common logarithm, 242. 
Completing a square, 177. 
Complex fraction, 43. 
Complex number, 171. 
Compound interest, 260. 
Conditional equation, 58. 
Conjugate imaginaries, 192. 
Constant, 64. 
Coordinates, 119. 

Decimals, 48. 

terminating, 49. 
Degree, of a polynomial, 61. 

of a term, 61. 
Denominator, 6. 
Dependent equations, 132. 
Dependent variable, 121. 
Difference of numbers, 10. 
Difference of squares, 85. 
Discriminant, 191. 
Dividend, 6. 
Division, 6. 
Divisor, 6. 

Ellipse, 264. 
Equation, 68. 

of a curve, t29. 

of a line, 129. 



Equivalent equations, 59. 
Exponential equation, 258. 
Exponential function, 260. ' 
Exponents, general, 150. 

laws of, 25, 142. 

positive integral, 25. 
Extraneous roots, 114, 198. 

Factor, definition of a, 3. 

Factorial symbol, 206. 

Factoring, 88. 

Fractions, 22, 103. 

Function, definition of a, 121. 

Functional notati6n, 126. 

Fundamental operations of algebra, 3. 

Geometric means, 230. 
Geometric progression, 227. 

infinite, 236. 
Graph, of an equation, 128. 

of a function, 122. 

of a quadratic equation in two 
variables, 262. 

of a quadratic function, 186. 

Harmonic means, 236. 
Harmonic progression, 236. 
Highest common factor, 106. 
Hyperbola, 263. 

Identical equation, 58. 
Imaginary number, 144, 170. 

pure, 171. 

Inconsistent equations, 132. 
Independent variable, 121. 
Index laws, 25, 31, 142, 154, 278. 
Index of a radical, 146. 
Inequalities, 14, 15. 
Infinite series, 238. 
Integral rational polynomial, 28. 
Integral rational term, 28. 
Intercepts of a graph, 128. 
Interpolation, for logarithms, 248. 
Irrational equation, 199. 
Irrational function, 148. 
Irrational number, 147. 



376 



INDEX 



Linear equation, 61. 
Linear function, 122. 
Logarithm, base of a, 240. 

characteristic of a, 244. 

definition of a, 240. 

mantissa of a, 244. 
Logarithmic equation, 258. 
Logarithmic function, 260. 
Logarithms, properties of, 243, 255. 
Lowest common denominator, 40, 106. 
Lowest common multiple, 38, 106. 
Lowest terms, for a fraction, 22, 104. 

Mantissa, 244. 
Maximum value, 187. 
Mean proportional, 212. 
Minimum value, 186. 
Monomial, 28. 

Naperian logarithms, 259. 

Napier, 251. 

Natural logarithms, 251, 259. 

Negative of a number, 9. 

Negative numbers, 4. 

Numerator, 6. 

Numerical value, 4. 

Ordinate, 119. 

Origin of coordinates, 119. 

Parabola, 186. 

Pascal's triangle, 205. 

Percentage, 70. 

Perfect nth power, 100, 148. 

Perfect square, 82, 90. 

Polynomial, 28. 

integral rational, 28. 
Power of a base, 25. 
Prime factor, 88. 
Prime integer, 38. 
Principal root, 145. 
Progressions, arithmetic, 221. 

geometric, 227. 

harmonic, 236. 
Proportion, 210. 

Proportional parts, principle of, 248. 
Pure quadratic equation, 173. 

Quadratic form, 196. 
Quadratic formula, 180. 
Quadratic function, 186. 
Quadratic in one unknown, 



complete, 173. 

discriminant of a, 191. 

graphical solution of a, 189. 

pure, 173. 

Quadratic in two variables, 262. 
Quotient, 6. 

Radicals, 82, 146. 

properties of, 149. 

simplification of, 157, 166. 
Radicand, 82, 146. 
Radius of action, 76. 
Ratio, 6, 210. 
Rational function, 148. 
Rational number, 147. 
Rationalizing a denominator, 161. 
Real number, 1. 
Reciprocal, 44. 
Remainder, in division, 34. 
Repeating decimal, 238. 
Root, of an equation, 59. 

of a number, 145. 
Rounding off numbers, 53. 

Scientific notation for a number, 249. 

Significant digits, 52. 

Signs, laws of, 6. 

Similar terms, 18. 

Simple interest, 79. 

Solution, defined for an equation, 

in one variable, 59. 

in two variables, 127. 
Solution of a system of two equations, 

131, 267. 

Square root, 82, 280. 
Subtraction, 9. 
Surd, 148. 

Systems of equations involving quad- 
ratics, 262. 
Systems of linear equations, 

in three unknowns, 137. 

in two unknowns, 131. 

Terminating decimal, 49. 
Transposing terms, 60. 
Trinomial, 28. 

Uniform motion, 75. 
Unknowns, 58. 

Variables, 64, 121. 
Variation, 213. 
constant of, 213.