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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 oddnumbered or
the evennumbered 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 oddnumbered problems are provided
in the text, and answers for evennumbered 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 FOURPLACE LOGARITHMS OF NUMBERS 284
in COMPOUND AMOUNT: (1K*)* 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 178 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. Thirtyfive 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. 63 = 6X3 = 6(3) = 18. We read 6X3, 63,
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,
(41) 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
=45 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 (5417)   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= 128 + 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 + 7w9; 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; NQ = 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 = Qx, or 5= 0,
which is contradictory.
EXERCISE 3
Compute each sum mentally.
1. 20 + 7. 2.  28  9. 3.  13  5.
4. 364. 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 54321 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 <8and  7 < 8. 23.  4 <  9 and 4 < 9.
24.   3 1 > 1 1 but  3 < 0. 26.   8  > 3 but  8 < 3.
26.  5 <"7but 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
53o& + 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 righthand 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 lefthand 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 5a6cand 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,
84512 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  03 ( l)(o  3) 3Q
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) (SXSX8
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 = 55 = 25. 5 3 = 5* 5' 5 = 125.
If m is any positive integer,* we define a m by the equation
9
a m = aaa 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 = 3333 = 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 = (aao)(ao) = o0'Oaa = 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 = (aaa)(aaa) = aaaaaa = 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 = (a0'o)(666)
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  x2
(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
aaa
aaa
aa
T"
1
a
aaaaa aa
= a 2 ; (Divide out aaa)
= = (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
. 7T7  = quotient H 377  (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
2x3 "^ X 2x3
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 83 + 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) = 65 + s + 32s = 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 = 22255 = 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 = 257a 4 6.
2. Hence, LCM  2 2 57a 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+ir 2 5y + 5*
2,6 __a 4 c _d _ 7
3 *3+3 3* 888'
.3 5,76 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. ,55,; 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
35'8 or 120. Hence, to add the fractions, we. change them to new fractions
whose denominator is 120:
1 4_ ? 4 1 = 2Q . 324 715 = 20 + 72 + 105 197
6 + 5 + 8 620 + 524" 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,
220 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. y5 + 3. 17 ' To30 + 5' 18 ' " 2 + 6 + 15
 3 3 3o6 7
a3 . 642 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 
y2
Ki  2  4* 4a  5 3a + 2
51. g  1  2J 52. ? 2 +
2x 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 3020 10 5
, , 2 15o6
3a + yr
T Q 56 56
ILLUSTRATION 3. = =
6o67
36
15o6 + 2 36 45a6
56 606  7 30a6  35
where we divided out 6 from numerator and denominator.
265 265
36 36 26  5 1 265
ILLUSTRATION 4.
46 46 36 46 126  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. fr^ 13. g^4 14.
* i + 5 "* 1  f "'5
5c~f
J2c'
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).
. (7X3) ( 2) (5) (
* 14 " (3X6) ~ (2X4)
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 2h3 4A7
_ 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 HinduArabic 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
tenthousandths"
* 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 fcplace 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 sevenplace 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 lefthand side, we verify that the product is the same as the
righthand 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).
Lefthand side: 3  3( 5) = 3 + 15 = 18.
Righthand 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(z4); 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
Lefthand side: ^  j; = ___ == 1_2= 1.
o A o
347 10
Righthand 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
Righthand 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 Fahrenheitcentigrade 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 kilowatthours, 5^ per kilowatthour for the next 20 kilo
watthours, and 2%i per kilowatthour for all over 30 kilowatthours. Write
an expression for the total cost, C, of (a) 60 kilowatthours; (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 (580 h 4250) or 1400, and for those
at the right is
(4 100 + 5200), 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 40pound weight is placed 6 feet from the fulcrum on a lever, and
a 60pound weight 8 feet from the fulcrum on the other side. Where should
a 30pound 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 8foot
lever to lift a 300pound 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 400meter
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 800meter 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 300yard 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 400meter 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 righthand member.
SPECIAL PRODUCTS AND FACTORING 85
I. a(x + y) = ax f
II. (*4y)(*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 + yz + 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).
iLLUSniiATiON 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 ~ ah ~ 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 (x3)(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 m42 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 2z2y 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 43(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* + x6 ^
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) :
169 16 + 46 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(3c2d) 6c  4d 3(3c  2d) 2(3c 
(52)  (73) _ 11
32(3c2c*) 6(3c2d)'
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 2x41
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 4x 2 6 2^43^2 x 6  4
ADVANCED TOPICS IN FRACTIONS J09
39 3x2 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 = fj ^ 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 9u25
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 2c4da63a
a&' 63 *6c26c
4 tf
hxhy cwbw, 
abac 3xZy 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.
3xl
ADVANCED TOPICS IN FRACTIONS
_ oxf 6x
17.
20.
23.
26.
a
IU
29 .
9x 2  1
o 2 a 2
XO* ." ' "
4x + 5
ill.
43
3x
18. LI.
?_
*
z ^
2o
a 2
._!
19. ^ j
6 a
2
li
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
.. A2 2_V_? L_\
V a + 3/U + 2 3 a/
64. ? 66.
A 2a
x r7 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 rx
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* + t2
* + 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 _ 54to ^ _3 2 2sl
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
3y2 2y5 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
654321
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
zcoordinate and to the ordinate as the ycoordinate.
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 crosssection 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 singlevalued function of a;; if just two values of y correspond to each
value of x, then y is a twovalued 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 zaxis) 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 brokenline 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 yaxis 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 hardsurfaced 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)  33 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 xintercept of the graph. The
ordinate of any. point where the graph meets the 2/axis is called a
yintercept of the graph. To find the xintercept (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 yintercept.
2. Place y = and compute x, to find the xintercept.
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 xintercept 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 xaxis
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 yaxis 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. s7.
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,
5x2y = 21.
6.
9.
f 3x + 4 = 0,
x + 2y = 0.
y = 14.
3z + 5y = 0.
1015. Solve Problems 16 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 2f 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.
MH,
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 twodigit 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
yaxis 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 twodigit 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 righthand
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 twodigit 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 threedigit 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 625mile
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 = (ao a)(a*aa 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 afil'ji $' (n factors)
1 1
_____
n m) factors a] a*a a a
nOT
EXPONENTS AND RADICALS 143
IV. Law for finding a power of a product: (ab) n = a n b".
Proof. (ab) n = abab 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) _ aa  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 righthand 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. 32.
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
rf2*
' 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\* /827afa\* = /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. (3xy) 2 . 26. (5arV 3 ) 2 . 27. (Gar 1 !/ 3 ) 2 . 28.
29. 32i 80. 271 31. 125*. 32. 216i
33. (4V)t. 34. (a: 4 !/ 2 )*. 35. (27a 8 a*)i 36. (25ar 2 )*.
25 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 ^fs 62. i~ 63.
67. PTr  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 42
67 ' a l + ' ** /,2 J.A* 69 ' x^J_,r2* 70 
rt l J_ Al 41 _ a 
71. " + L 72. T^ fl _.. 73. " " 74.
a 2 _ 52 42 _ a 2 a  8 _ 6  32 + ^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/493 = 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
tf1
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
V34
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: >. = =.<
V3V2 + V5 V3V6 + 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 \164c 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. jj
\/ 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. V27F.
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.
3i4
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 tfab 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 12 222 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 . 43.162 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  43 ( 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 ~* 32*

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 crosssection area of a chimney, its socalled 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 zaxis, 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 xaxis; 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 zaxis 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 zunit in order to be able to plot all points from
the table of values on the available part of the crosssection 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 xaxis, 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 xaxis; 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 xaxis 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 xaxis.
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 xaxis in two points when and only when equation 1 has
unequal real roots;
II. touches the xaxis in just one pointy or is tangent to the xaxis, when
and only when the roots are equal;
III. does not meet the xaxis 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 xaxis in two points
is tangent to xaxis
does not touch xaxis
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 _ 445 =  71
imaginary numbers
4z 2  4x + 1 =
( 4) 2 44 =
real; equal; rational
% * 4
4z 2  3x  5
(3) 2 + 445  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 zaxis. 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  : rfs=  (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 = 22, 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 69 = 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.
= 32*. 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*43 = 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. V3f 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 xaxis.
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 tle 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 (62) 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 (102) 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 (76) 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! = 12345 = 120. 31 = 123 = 6.
1L = 1234567 1 J_
10! 12345678910 8910 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 iy
(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 ' ^. ixnr+i^ri. (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 2t/ ' 22
*// 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 yaxis 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 zaxis. 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  2 163 = 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 crosssection 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 footpounds 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 castiron 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 = npJ)> (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 + 3n3; 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 n1 .
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 n1 , (1)
and Sr ar f ar 2 + ar 8 f f ar n1 + 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  23 2  1458 79R
S== 13 = 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 n1 , (4), and (5).
EXAMPLE 3. If S = 750, r = 2, and Z = 400, find n and a.
a rl
SOLUTION. 1. From S
l
trrr\ < K/ %
750 = = =~; hence, a = 50.
J. ~~ &
2. From I = ar n ~\ 400 = 50(2*'); 2r~ l = = 8;
2ni _ 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)* + (102) + (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. )m100)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 299f 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 socalled 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 ni 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] 7377 = 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 midpoints of the sides of the 1st square; a 3d
square is drawn by connecting the midpoints 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 . Fourplace 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 fourplace 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 fourplace 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 fourplace
or a fiveplace 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
righthand 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
lightyear, 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 Pf 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 (15501617). 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 (15561631), who was aided by Napier.
EXERCISE 91
^i
Find the fourplace 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 fourplace 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 fiveplace logarithms to obtain
fourplace .accuracy, and with at least fourplace logarithms to obtain
threeylace 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 fourplace 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 fourpl&ce log NUN has more than four
significant digits, round off AT to four significant digits.
EXERCISE 92
Compute by use of fourplace 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= 8.491410
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  8.327910
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 fourplace 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 sevenplace log 1.04 = 0.0170333, compute Problems
31 and 32 and compare with the less accurate former answers.
Compute by use of fourplace 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 * / 891 A0 / 47.5317
' (46.3) 3 " \163 X .62* ^ \.031 X .964*
10*V?78 10i"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 crosssection 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 = 1250.
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
realvalued 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 towvalued function of y, as
shown in (2), or y as a towvalued function
of x. The graph of the equation consists
of the graphs of the two singlevalued 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 yaxis.
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
xintercepts:
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 xaxis. 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 xaxis and yaxis.
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 xaxis and yaxis.
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 yaxis;
if the equation does not involve x 2 or xy, the graph is a parabola whose
axis is parallel to the xaxis.
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 yaxis. 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 xintercepts
by placing y ~ and solving for x, and find the y4ntercepts 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 xintercepts or the yintercepts. (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 crosssection 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*,
68. 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 + 280 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 righthand 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 oddnumbered problems are given here. Answers to
evennumbered 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
11.
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. 663a. 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. o6^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 5r3fe 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 2a3fe
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.
a5
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 .  .
6a5c
_.
29> 35
10. M 9.ft7 r/f
23.
* 2a36
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. ^4r 15. ^T^: 17.
23.
2
2s 
' x  2y 2 3x 4 2y
!**a. 27.? 29. ^.
a 4 ^ * 4 y c
3 x ^ x43 5 2x
31. rrr* 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
* ' ' *"*
2x2y 6(an) * x + x  12
14 + 2n 6c5cH30 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 'ba
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 3a26
(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
2MJa; 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).
36xf 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 xaxis at (5, 0); yaxis at (0, 3).
23. Cuts xaxis 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
4y 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 + 4V3 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 3axis, cutting zaxis 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 xaxis, 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 1fcl+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$.
63. ^4 (x~% = 8 has no real solution).
Exercise 77. Page 203
1. db 1 3. 2. 6. 0;  J.
7. H (& = 1 is not a solution because it does not give a quadratic equation).
9. 10; 2. 11. ^; k does not apply.
13.  .268;  3.732. 15.  . 17. &. 19.  i
21. if. 23.  &. 25. =fc 3V5. 27.  f .
29.  f. . 31. =fc .816. 33. 1;  2.
Exercise 78. Page 206
1. a 6 + 5o6 + 10a6 J + 10a*& + 5a& + 6 s .
3. x 8  8ofy + 28zV  56x 5 y + 70x 4 y 4  56xV + 28a;V  8xy f y.
5. 16 + 32a 4 24a 8 f 8a + a.
7. 7296  14586 B y + 12156y  540&V + 1356V  18^ + V*.
9. a f So^ + 3a6* + 6.
IL o  6a w 6 4 15a6*  20a6 f 15a6  6a'6 l <>
13. x 
310
ANSWERS
15. a* 
17. a*  4oir H
. , 8x* , 24x 32x*
19. x  +  ~
a* a* a* a
25. a" f 20a& f 190a6.
81. 2x*  30 2ox + 435
/ _ ^ ^ \
85. * n 
89. 720.
I5xy* 
 40IT 6 + IT 8 .
16
27. 1  + 2.31.
87. a*" 
Ig0ou lg
^
29. 1  12 >/2 f 132.
+
/ ^^ j \
41. 39,916,800.
Exercise 79. P3t 208
8. 35z 5. 
48. 126.
7.
9. 
n(nl)...(n~5)
15.
21.
6!
11. .00056.
^
18. 15x*.
; 2016xy 10 . 28.
3/
25.
27. 126o>6;
29. 10,000  4000o + 600a  40a + o. 3t 96,059,601. 83. 132,651.
36. 1.127. 87. 1.230. 39. .904. 41. 1.243. 48. .002. 45. 14,776,336.
Extrcls* 80. Page 211
1. f . 8. f . 5. s/a. , 7.
11. *. 18. 30. 15. f. 17. 10"; 8'.
2L 32'; 21$". 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 5place 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 5place
logarithms is given in black face type beside the result found with 4place 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 7place 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 4place 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 + ts/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.