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ADVANCED COURS
SLAUGHT & LENNES
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IN MEMORIAM
FLOR1AN CAJOR1
(T^w'cc^ C^^trv-t:
HIGH SCHOOL ALGEBRA
Hbvancefc Course
BY
H. E. SLAUGHT, Ph.D.
ASSOCIATE PROFESSOR OF MATHEMATICS IN THE UNIVERSITY
OF CHICAGO
N. J. LENNES, Ph.D.
INSTRUCTOR IN MATHEMATICS IN THE MASSACHUSETTS
INSTITUTE OF TECHNOLOGY
JHHc
Boston
ALLYN AND BACON
1908
COPYRIGHT, 1908,
BY H. E. SLAUGHT
AND N. J. LENNES.
m M Sm m.
PREFACE
The Advanced Course of the High School Algebra contains
a review of all topics treated in the Elementary Course, to-
gether with such additional topics as are required to make it
amply sufficient to meet the entrance requirements of any
college or technical school. Its development is based upon
the following important considerations :
1. The pupil has had a one year's course in algebra, involv-
ing constant application of its elementary processes to the
solution of concrete problems. This has invested the pro-
cesses themselves with an interest which now makes them
a proper object of study for their own sake.
2. The pupil has, moreover, developed in intellectual ma-
turity and is, therefore, able to comprehend processes of
reasoning with abstract numbers which were entirely beyond
his reach in the first year's course. This is particularly true
if, in the meantime, he has learned to reason with the more
concrete forms of geometry.
In consequence of these considerations, the treatment
throughout is from a more mature point of view than in the
Elementary Course. The principles of algebra are given in
the form of theorems the proofs of which are based upon a
definite set of axioms.
As in the Elementary Course, the important principles are
used at once in the solution of concrete and interesting prob-
lems, which, however, are here adapted to the pupil's greater
maturity and experience. But relatively greater space and
emphasis are given to the manipulation of standard algebraic
iii
msosisi
IV
PREFACE
forms, such as the student is likely to meet in later work in
mathematics and physics, and especially such as were too com-
plicated for the Elementary Course.
The division of the High School Algebra into two distinct
courses has made it possible to give in the Advanced Course
a more thorough treatment of the elements of algebra than
could be given if the book were designed for first-year classes.
It has thus become possible to lay emphasis upon the pedagogic
importance of viewing each subject a second time in a manner
more profound than is possible on a first view.
Attention is specifically called to the following points :
The scientific treatment of axioms in Chapter I.
The clear and simple treatment of equivalent equations in
Chapter III.
The discussion by formula, as well as by graph, of incon-
sistent and dependent systems of linear equations, pages 40
to I I.
The unusually complete treatment of factoring and the
clear and simple exposition of the general process of finding
the Highest Common Factor, in Chapter V.
The careful discrimination in stating and applying the
theorems on powers and roots in Chapter VI.
The unique treatment of quadratic equations in Chapter VI 1,
giving a lucid exposition in concrete and graphical form of
distinct, coincident, and imaginary roots.
The concise treatment of radical expressions in Chapter X,
and especially — an innovation much needed in this connec-
tion — the rich collection of problems, in the solution of which
radicals are applied.
H. E. SLAUGHT.
N. .1. LENNES.
< ' II I< \iim \M> I'.OSTON,
April, L908.
CONTENTS
CHAPTER I
FUNDAMENTAL LAWS
Axioms of Addition and' Subtraction
Axioms of Multiplication and Division
Theorems on Addition and Subtraction
Theorems on Multiplication and Division .
PAGE
1
CHAPTER II
FUNDAMENTAL OPERATIONS
Definitions
Addition and Subtraction of Monomials .
Addition and Subtraction of Polynomials .
Removal of Parentheses ....
Multiplication and Division of Monomials
Multiplication and Division of Polynomials
11
12
15
17
18
21
CHAPTER III
INTEGRAL EQUATIONS OF THE FIRST DEGREE IN ONE
UNKNOWN
Definitions
Equivalent Equations
Problems in One Unknown
25
27
32
CHAPTER IV
INTEGRAL LINEAR EQUATIONS IN TWO OR MORE VARIABLES
Indeterminate Equations ... ..... .36
Simultaneous Equations in Two Variables 38
VI
CONTENTS
Inconsistent and Dependent Equations
Systems in .More than Two Variables
Problems in Two or More Unknowns
CHAPTER V
FACTORING
Expression of Two, Three, or Four Terms
Factors found by Grouping
Factors found by the Factor Theorem
Solution of Equations by Factoring .
Common Factors and Multiples
CHAPTER VI
POWERS AND ROOTS
Definitions .......
Theorems on Powers and Roots
Roots of Polynomials .....
Roots of Arabic Numbers
PAGE
43
45
47
52
55
■ u
60
Gl
69
73
77
CHAPTER VII
QUADRATIC EQUATIONS
Exposition by means of Graphs
Distinct, Coincident, and Imaginary Roots
Simultaneous Quadratics ....
Special Methods of Solution
Higher Equations involving Quadratics
Relations between the Roots and Coefficients
Formation of Equations with Given Roots
Problems involving Quadratics .
CHAPTER VIII
ALGEBRAIC FRACTIONS
Reduction of Fractions .....
Addition and Subtraction of Fractions
83
87
90
96
106
L08
110
112
11G
120
CONTENTS
Vll
Multiplication and Division of Fractions .
Complex Fractions
Equations involving Algebraic Fractions ....
Problems involving Fractions
CHAPTER IX
RATIO, VARIATION, AND PROPORTION
Ratio and Variation
Proportion ..........
Problems ..........
I'A(iE
123
126
127
133
135
139
141
CHAPTER X
EXPONENTS AND RADICALS
Fractional and Negative Exponents 142
Reduction of Radical Expressions 147
Addition and Subtraction of Radicals 150
Multiplication and Division of Radicals 151
Rationalizing the Divisor . . . . . . . . .155
Square Root of Radical Expressions 156
Equations containing Radicals . . . . . . . .159
Problems involving Radicals 163
CHAPTER XI
LOGARITHMS
Definitions and Principles 167
CHAPTER XII
PROGRESSIONS
Arithmetic Progressions
Geometric Progressions
Harmonic Progressions
CHAPTER XIII
THE BINOMIAL FORMULA
Proof by Induct ion
The General Term
175
181
186
189
191
HIGH SCHOOL ALGEBRA
ADVANCED COURSE
CHAPTER I
FUNDAMENTAL LAWS
1. We have seen in the Elementary Course that algebra, like
arithmetic, deals with numbers and with operations upon num-
bers. We now proceed to study in greater detail the laws that
underlie these operations.
THE AXIOMS OF ADDITION AND SUBTRACTION
In adding numbers we assume at the outset certain axioms.
2. Axiom I. Any two numbers have one and only one
sum.
Since two numbers are equal when and only when they are
the same number, it follows from this axiom that if a = b and
c = d then a + c = 6 -f d.
For if a is the same number as b, and c is the same number as d,
then adding b and d is the same as adding a and c, and by Axiom I
the sums are the same and hence equal.
Therefore from Axiom I follows the axiom usually given :
If equal numbers be added to equal numbers, the sums are equal
numbers.
Since Axiom I asserts that the sum of two numbers is unique,
it is often called the uniqueness axiom of addition.
3. If a = c and b = c then a = b, since the given equations
assert that a is the same number as b. Hence the usual state-
ment: If each of tiro numbers is equal to the same number, they
are equal to each other.
1
2 FUNDAMENTAL LAWS
4. The sum of two numbers, as G and 8, may be found by
ailding 6 to 8 or 8 to 6, in either ease obtaining 14 as the
result. This is a particular case of a general law for all
numbers of algebra, which we enunciate as
Axiom II. The smt) of two numbers is the same in what-
ever order they are added .
This is expressed in symbols by the identity :
a + b = b + a. [See § 37, E. C.*]
Axiom II states what is called the commutative law of addition,
since it asserts that numbers to be added may be commuted or
i . 1 1 1 i rchanged in order.
Definition. Numbers which are to be added are called addends.
5. In adding three numbers such as 5, 6, and 7, we first add
two of them and then add the third to this sum. It is imma-
terial whether we first add 5 and 6 and then add 7 to the sum,
or first add 6 and 7 and then add 5 to the sum. This is a par-
ticular case of a general law for all numbers of algebra, which
we enunciate as
Axiom III. The sum of three numbers is tiie same in
whatever man iter then are grouped.
In symbols we have
a + b + c = a + (b + c).
When no symbols of grouping are used, we understand a + b + c
to mean thai a and b are to be added first and then <■ is to be added
to the sum.
Axiom III states what is called the associative law of addi-
tion, since it asserts that addends may be associated or grouped
in any desired manner.
It is to be noted that an equality may be read in either direction.
Thus a + b + c=a+(b + c) and a + (b + c)=a + b + c
are equivalent statements.
6. If any two numbers, such as 1 ( .> and 25, are given, then
in arithmetic we can always find a number which added to
* E. C. meaus the Elementary Course.
AXIOMS 3
the smaller gives the larger as a sum. That is, we can sub-
tract the smaller number from the larger.
In Algebra, where negative numbers are used, any number
may be subtracted from any other number. That is :
Axiom IV. For any pair- of numbers a and b there is
one and only one number c such that a + c — b.
The process of finding the number c when a and b are given
is called subtraction, b is the minuend, a the subtrahend, and c
the remainder. This operation is also indicated thus, b — a = c.
If a + c = a, then the number c is called zero, and is written
0. That is, a + = a, or a — a = 0.
Adding a to each member of the equality b — a = c, we have
b — a + a.= c + a, which by hypothesis is equal to b. Hence
subtracting a number and then adding the same member gives as
a result the original number operated upon.
Axiom IV is called the uniqueness axiom of subtraction. A
direct consequence is the following : If equal numbers are sub-
tracted from equal numbers, the remainders are equal numbers.
THE AXIOMS OF MULTIPLICATION AND DIVISION
7. Axioms similar to those just given for addition and sub-
traction hold for multiplication and division.
Axiom V. Two numbers have one and only one product.
This is called the uniqueness axiom of multiplication. It is a
direct consequence of this axiom that : If equal numbers are
multiplied by equal numbers, the products are equal members.
8. The product of 5 and 6 may be obtained by taking 5 six
times, or by taking 6 five times. That is, 5-6 = 6-5. This
is a special case of a general law for all numbers of algebra,
which we enunciate as
Axiom VI. The product of two numbers is the same in
whatever order they are multiplied.
In symbols we have a • b = b • a.
4 FUNDAMENTAL LAWS
This axiom states what is called the commutative law of
factors in multiplication.
9. The product of three numbers, such as 5, 6, and 7, may
be obtained by multiplying 5 and 6, and this product by 7, or 6
and 7, and this product by 5. This is a special case of a
general law for all numbers of algebra, which we enunciate as
Axiom VII. The product of three numbers is the same
in whatever manner they are grouped.
In symbols we have abc = a(bc).
The expression ahc without symbols of grouping is understood to
mean that the product of a and b is to be multiplied by c.
This axiom states what is called the associative law of factors
in multiplication.
Principles III and XV of E. C. follow from Axioms VI and VII.
10. Another law for all numbers of algebra is enunciated as
Axiom VIII. The product of the sum or difference of tic<>
numbers and a given number is equal tothe result obtained
by multiplying each number separately by the given num-
ber and then adding or subtracting theproducts.
In symbols we have
a(b -f c) = ab + ac and a(6 — c) = ab — ac.
Axiom VIII states what is called the distributive law of
multiplication.
When these identities are read from left to right, they are equiva-
lent to Principle IV. E. C, and when rend from righl to left (see § 5)
they are equivalent to Principles I and II. E. C. In the form
a(b ± c) = ab ± ac this axiom is directly applicable to the multiplica-
tion of a polynomial by a monomial, and in the form ab ± ac =a(l> ± c),
to the addition and subtraction of monomials having a common factor.
11. Axiom IX. For any two numbers, a and b. provided
a is not equal to zero.' there is one and only one number
c such that a • c = b.
*The symbol for the expression a is not equal to zero is o =f= 0.
AXIOMS 5
Definitions. If ac = b, the process of finding c when a and b
are given is called division, b is the dividend, a the divisor,
c the quotient, and we write b -f- a = c, or - = c. For the case,
a = 0, see §§ 24, 25.
If a • c = a, a =f= 0, then the number c is called unity, and is
written 1. That is, - = 1.
a 7
Multiplying both sides of the equality - = c by a, we have
b a
a • - = ac, which by hypothesis equals b. Hence dividing by a
a
number and, then multiplying by the same number gives as a result
the original number operated upon.
Axiom IX is called the uniqueness axiom of division. As a
direct consequence of this axiom we have : If equal mem-
bers are divided by equal numbers, the quotients are equal
numbers.
12. Axioms I, IV (incase the subtrahend is not greater than
the minuend), V, and IX underlie respectively the processes
of addition, subtraction, multiplication, and division, from the
very beginning in elementary arithmetic. Axioms II, III, VI,
VII, and VIII are also fundamental in arithmetic, where they
are usually assumed without formal statement.
E.g. Axiom VIII is used in long multiplication such as 125 x 235,
where we multiply 125 by 5, by 30, and by 200, and then add the
products.
13. Negative Numbers. Axiom IV, in case the subtrahend is
greater than the minuend, does not hold in arithmetic because
of the absence of the negative number. This axiom therefore
brings the negative number into algebra.
We now proceed to study the laws of operation upon this
enlarged number .system. In the Elementary Course concrete
applications were used to show that certain rules of signs hold
in operations upon positive and negative numbers. We shall
now see that the same rules follow from the axioms just
stated.
6 FUNDAMENTAL LAWS
14. Definitions. If a + b = 0, then b is said to be the negative
of a and a the negative of b. If a is a positive number, that
is an ordinary number of arithmetic, then h is called a negative
number. We denote the negative of a by — a. Hence,
a + ( — a) = 0. </ and — a have the same absolute value.
If a — b is positive, then a is said to be greater than b.
This is written a > ?/. If a — b is negative, then a is said
to be less than b. This is written a < b. If a — 6 = 0, then
a = 6, and if a = 6 then a — 6 = 0. See § G.
THEOREMS ON ADDITION AND SUBTRACTION
Definition. A theorem is a statement to be proved.
A corollary is a theorem which follows directly from some
other theorem.
15. Theorem 1. Adding a negative number is equiva-
lent to subtracting a positive number h(iriit<J the same
absolute value. That is,
a+(-b)=a-b See § 48, E. C.
Proof. Let a + ( - 6) =x. (1)
Such ;i number ./• exists 1 >v Axiom I.
Adding b to each member of (1), a -\- (— 6) -f b = x + b. (2)
By the associative law of addition, § 5, and by §§ 14, 6,
a + (- 6) + 6 = a + [(- 6) + 6] = a + = a. (3)
From (L?) and (•",) by §3, x + b = a. (i)
From (4), by the definition of subtraction, § 6,
a — b = x. (5)
From (1 ) ami (5) by § 3, a + (- 6) = a -b.
It follows from theorem 1 that either of the symbols, +(— b)
or — /-, may replace the other in any algebraic expression.
16. Corollary. ./ -parenthesis preceded by the plus
sign may be*removed without changing I If sign of any
/(■/■in within it. See * 28, E. C-
THEOREMS 7
For, since by the theorem b — a = b + ( — «), each subtraction is
reducible to an addition, so that the associative law, § 5, applies. Thus
a+ (b-c + d) = a + [b + ( - c) + </] = « + b + (- c) + d = a + b - c + d.
Hence an expression may be inclosed in a parenthesis pre-
ceded by the plus sign without changing the sign of any of its
terms.
17. Theorem 2. Subtracting a negative number is
equivalent to adding a positive number having the
same absolute value. That is,
a - (- 6) = a + b. See § GO, E. C.
Proof. Let a-(-b) = x. (1)
From (1) by §2, a -(- &) + (-&) = * + (-&). (2)
From (2) by §§ (3 and 15, a = x + (-b) = x- b. (3)
Hence by the definition of subtraction, a + b—x. (I)
From (1) and (1) by § 3, a- ( - b) = a + b. (5)
It follows from theorem 2 that either of the symbols —(—6)
or + 6 may replace the other in any algebraic expression.
18. Theorem 3. A parenthesis preceded by the minus
sign may be removed by changing the sign of each term
within it. That is,
a _ (6 - c + d) = a - b + c - d. See § 28, E. C.
Proof Let a - (b - c + d) = x. (1)
From (1) by the definition of subtraction,
a = x + (b-c + d). (2)
By §§15, 16, a = x+b + (-c) + d. (3)
Adding ( — b), c, and ( — d) to each member and using § 1,
a + (- b) + c +(- d) = x + b + (- b) + c + (-c) + d + (- <*)• 0±)
From (4), by §§ 11, 15, a-6 + c-rf = a;. (5)
From (1) and (5) by § 0, a - (6 -c + rf) =a-i + c-d. (0)
It follows from equation (6), read from right to left, that an
expression may be inclosed in a parenthesis preceded by a
minus sign, if the sign of each term within is changed.
8 FUNDAMENTAL LAWS
19. Corollary 1. a - b = - (b - a).
For by §§ 15 and 1, a - b = a + (- b) = - b + a. (1)
Hence by § IS, a -b= -(b -a). (2)
20. Corollary 2. -a + (-&) = - (a + 6). See § 48, E. C.
For by §18, - a + (- 6)= - [a - (- ft)]. (1)
Hence by §17, -a + (-6)= - [> + &]• (2)
21. From the identities
a + (— &) = a—b, § 15,
a — ( — b) = a + b, § 17,
a- 6 = - (6 - a), § 19,
-a + (-b)=-(a + b), §20,
it follows that addition and subtraction of positive and nega-
tive numbers arc reducible to these operations as found in urith-
metic, where all numbers added and subtracted are positive,
and where the subtrahend is never greater than the minuend.
E.g. 5 + (- 8) = 5 - 8 = - (8 - 5) = - 3.
5-(-8)=5 + 8 = 13.
- 5-8 = -(5 + 8)= - 13.
THEOREMS ON MULTIPLICATION AND DIVISION
22. Theorem 1. The product of any number and zero is
Z( vit. That is, a ■ = 0.
Proof. By definition oi zero, § 6, a«0 = a(6 — 6).
By the distributive law of multiplication, § 10,
a (J> — b) — ub — ab,
which by definition is zero.
Hence a • = 0.
Notice thai by the commutative law of multiplication, § 8,
a • = • a.
THEOREMS 9
It follows from this theorem and § 9, that a product is zero
if any one of its factors is zero; and conversely, by § 11, if a
product is zero, then at least one of its factors must be zero.
23. Corollary 1. - =0> provided a is not zero.
Since by the theorem = a • 0, the corollary is an immediate con-
sequence of the definition of division (§ 11).
24. Corollary 2. ^ represents any number whatever.
That is, ? = k, for all values of k.
Since = • k, this is an immediate consequence of the definition
of division.
25. Corollary 3. There is no number k such that ^.—k,
provided a is not zero.
This follows at once from k • = for all values of k.
From §§ 24, 25,. it follows that division by zero is to be ruled
out in all cases unless special interpretation is given to the
results thus obtained.
26. Theorem 2. a(-b) = -ab. - See § 63, E. C.
Proof. Let a(-b)=x. (1)
By § 2, a(-b)+ ah = x + ab. (2)
By § 10, a[( - b) + V\ = x + ab. (3)
By §§ 14, 22, a .0 = = x+ ab. (4)
By §14, x = -ab. (5)
Hence, from (1) and (5) a(— &)= — ab. ((5)
27. Theorem 3. (-«)(- 6) =06. See § 63, E. C.
Proof. Let (- a) (- 6) = x. (1)
By §§ 2 and 26, (-a) (-&) + (- a)b = x- ab. (2)
By §§ 10, 14, 22, (- a)(- b + &)=0 = x - ab. (3)
Hence, § 14, x = ab. (4)
From (1) and (4) (- a)(-b)= ab. (5)
10 FUNDAMENTAL LAWS
28. Theorem 4. If the signs of the dividend and divisor
are alike, the quotient is positive; and if unlike, the quo-
tient is negative. See § G7, E. C.
Proof. This theorem is au immediate consequence of the definition
of division and the identities,
(+ a) (+ 6) = ah. «( - b) = - ah and ( - a) ( - b) = ah.
29. Theorem 5. 6 . a =— • See § 191, E. C.
c c
Proof. Let x = - • a. (1)
By §§ 7 and 11, ex = c • - • a = ba. (2)
c
Dividing by c, § 11, x = — . (3)
c
Hence from (1) and (3) - • a = — (4)
c c
30. Theorems. *±l = * + b , a nd°^ = °-*- See § 25,
-j7i i-i c c c ecu
Proof. By § 29, °L+* = 1 _0i±i0 =!.(«+ 6). (1)
c c c
By §§ 10 and 29, - . (a + b) = - ■ a + - . b = - + -• (2)
C CCCC
From (1) and (2), ^^ = « + !i. (:',)
c c c
Similar] v we may show that =
c c c
This theorem states what is called the distributive law of division.
31. In the proofs of the above theorems certain axioms have been
assumed to hold in a more general form than the one in which they
are stated. For example, the commutative law of addition was stated
for two numbers only and has been assumed for more than two.
These extensions can be shown to follow from the axioms as given.
It has Likewise been assumed thai zero and unity, which were defined
respectively as a — a = 0, and - = 1, are the same for all values of a.
CHAPTER II
FUNDAMENTAL OPERATIONS
32. The operations of addition, subtraction, multiplication,
division, and rinding powers and roots are called algebraic
operations.
33. An algebraic expression is any combination of number
symbols (Arabic figures or letters or both) by means of indi-
cated algebraic operations.
E.g. 21, 3 -f 7, 9(b + c), — — — , x 1 + Vy, are algebraic expressions.
34. Any number symbol upon which an algebraic operation
is to be performed is called an operand.
All the algebraic operations have been used in the Elementary
Course. They are now to be considered in connection with the fun-
damental laws developed in the preceding chapter, and then applied
to more complicated expressions. The finding of powers and roots
will be extended to higher cases.
35. One of the two equal factors of an expression is called
the square root of the expression ; one of the three equal fac-
tors is called its cube root; one of the four equal factors, its
fourth root, etc. A root is indicated by the radical sign and a
number, called the index of the root, which is written within
the sign. In the case of the square root, the index is omitted.
E.g. Vi is read the square root ofi ; V8 is read the cube root of 8;
VGi is read the fourth root o/64, etc.
36. A root which can be expressed in the form of an integer,
or as the quotient of two integers, is said to be rational, while
one which cannot be so expressed is irrational.
E. g. v8 = 2, Va 2 + 2 ah + IP- = a + b, and V| = § are rational roots,
while VT and Va 2 + ab + V 1 are irrational roots.
11
12 FUNDAMENTAL OPERATIONS
An algebraic expression which involves a letter in an ir-
rational root is said to be irrational with respect to that letter;
otherwise the expression is rational with respect to the letter.
E.g. a + bVc is rational with respect to a and b, and irrational
with respect to c.
37. An expression is fractional with respect to a given letter
if after reducing its fractions to their lowest terms the letter
is still contained in a denominator.
E.g. — ■ h b is fractional with respect to c and d, but not with
c + d
respect to a and b.
38. Order of Algebraic Operations. In a series of indicated
operations where no parentheses or other symbols of aggrega-
tion occur, it is an established usage that the operations of
finding powers and roots are to be performed first, then the
operations of multiplication and division, and finalty the opera-
tions of addition and subtraction.
E.g. 2 + 3 • 4 + 5 • y/8 - i 2 -=- 8 = 2 + 3 • 4 + 5 • 2 - 16 - 8
= 2 + 12 + 10-2 = 22.
In cases where it is necessary to distinguish whether multipli-
cation or division is to be performed first, parentheses are used.
E.g. In 6 -=- 3 x 2, if the division comes first, it is written (G -f- 3) x
2 = 4, and if the multiplication come first, it is written 6 -=- (3 x 2) = 1.
ADDITION AND SUBTRACTION OF MONOMIALS
39. In accordance with § 10, the sum (or difference) of terms
which are similar with respect to a common factor (§ 78, E. C.)
is equal to the product of this common factor and the sum (or
difference) of its coefficients.
Ex. 1. S ax 2 + 9 ax 2 - 3 ax 2 = (8 + 9 - 3) ax 2 = 14 ax 2 .
Ex. 2. a\'x 2 + y 2 + bVx 2 + y 2 = (ry + b) Vx 2 + y 2 .
Ex
3 x(x-\)(x-2) r(*-l) /x-2 -A *(*-!)
1-2-3 1-2 \ 3 / 1 • 2
_ x+ 1 ■'•(•'■ - 1) _ (x + l).r(x - 1)
3 1-2 1.2-3
ADDITION AND SUBTRACTION 13
EXERCISES
Perform the following indicated operations :
1. 5.x 4 6 2 — 3x*b 2 — 4x- 4 6 2 + 7x*b 2 .
2. 3V*r - 4 - 2Var - 4 + 2 VaT^I - 4 Var - 4.
3. a& 5 c 4 - cZ6 5 c 4 + e&V + /6 5 c 4 .
4. a fi x 4 + 5a 5 x 4 — 5a 6 ar i — Safo 4 .
5. 7a?Y + 5aY - 9x 4 )f + 5a? 5 ?/ 4 .
6. 2a n + a"- 1 + «" +1 = a n '\2a + 1 + a 2 ) = a n - J (l + a) 2 .
7. n(n - l)(n - 2)(n - 3)(n - 4) + «(n - l)(n - 2)(n - 3).
n(n — 1)(« — 2)(rc — 3) is the common factor and n — 4 and 1 are
the coefficients to be added.
8. n(n - l)(n - 2)(n - 3) (n - 4)(w - 5)(« - 6)
+ w(n - 1)(« - 2)(m - 3)(n - 4)(» - 5).
9. n(n - l)(n - 2)(n - 3) + (n - 1)(« - 2)(n - 3).
10. n(n - l)(n - 2) (a - 3)(n - 4) + (n - l)(n - 2)(n - 3).
11. ( a _ 4)(6 + 3) + (a - 1)(6 - 2) + (a + 3) (6 + 3).
First add (a - 4) (ft + 3) and (a + 3) (ft + 3).
12. (x + 2y)(a> - 2y) + (a? - Sy)(x - 2y) - (2x -y)(x- y).
13. (5a - 36) (a - 6) (a + 6) + (26 - 4a) (a - 6)(a + 6)
+ (a - 6) 2 (2a - 6).
14. (7a? 2 + 3y*)(5a> - y)(as + y) + (7a? 2 + 3/)(* + y)(2 y - 4a?)
+ (7x 2 -3y 2 )(x + yy.
15. 2 3 -3 2 -o + 2 4 .3.5.
The common factor is 2 3 • 3 • 5. Hence the sum is
2« . 3 • 5(3 + 2) = 2 8 • 3 • 5 2 .
16. 2 • 3 4 • 7 + 2 2 • 3 8 • 7 2 - 2 4 • 3 3 • 7.
17. 3 4 • 5 7 • 13 + 3 5 • 5 7 • 13 2 .
14 FUNDAMENTAL OPERATIONS
18. o 4 • 7 3 • 1 1 + 5 3 • 7- • 11 - 2 s ■ 3 • 5 s • 7 2 • 11.
19. 3- • 7 18 • 13 15 + 3 21 • 7 17 • 13 15 + 3 24 . 7 17 • 13 15 .
20. 1-2- 3- -n +1 -2-3 ••• n(n +1).
The dots mean that the factors are to run on in the manner indi-
cated up to the number n. The common factor in this case is
1 • 2 • 3 ••• n, and the coefficients to be added are 1 and n + 1. Hence
the sum is 1 • 2 • 3 ••• n(n + 2).
21. 1 -2 -3 ... w +1-2-3 ...«( n + l)
+ 1.2.3...n(n + l)(n + 2).
22. 1 • 2 • 3 ... n + 3 • 4 • 5 • •■ n + 5 • 6 • 7 ••• ».
23. n(n - 1) .•• (w - 6) + ><(« - 1) ... (n - 6)(w - 7).
24. »(w — 1) ... (n — r) + n(% — 1) ••• (n — r)(n — r — 1).
25. na n b + a n b. 26. w ^' ~ ^ a B_1 6 2 + na n_1 6 2 .
_ »(n — l)(w — 2) _o, s . n(w — 1) „_.,,,
27. — ^ — — >-a n -lr-\ — i — L a" -b 3 .
1-2-3 1-2
The common factor is "^ w ~ — '- a n --lfi and the coefficients to be
added are and 1.
1 -2
v(n- l)(n - 2)(n - 3) 3 4 «(n - l)(n - 2) 3& 4
2-3-4 2-3
29 »fo-l)("-2)(*-3)(n-4)
2-3.4-5
w n - 1)Q, - 2)(n - 3)
2-3-4
30 n ( n -1) - (n-r + l)(n - r) , r+1
2.3-r(r + l)
n, » _ 1) ... („ _ r .+ 1) a „_ r&r+1<
ADDITION AND SUBTRACTION 15
ADDITION AND SUBTRACTION OF POLYNOMIALS
40. The addition of polynomials is illustrated by the follow-
ing example.
Add 2a + 3& — 4c and 3a — 26 + 5 c.
The sum may be written thus :
(2 a + 3 b - 4 c) + (3 a - 2 b + 5 c).
By the associative law, § 5, and by § 16, we have,
2 a + 3 6 — 4 c -f 3 a — 2 6 -f- 5 c.
By the commutative law, § 4, and by § 15, this becomes,
2 a + 3 a + 3 6 — 2 6 — 4 c + 5 c.
Again by the associative law, combining similar terms, we have,
5 a + 6 + c.
From this example it is evident that several polynomials may
be added by combining similar terms and then indicating the
sum of these results.
For this purpose the polynomials are conveniently arranged so that
similar terms shall be in the same column. Thus, in the above
example,
1 2a+3 6-4c
3a-26+5c
5 a + b +
41. For subtraction the terms of the polynomials are arranged
as for addition. The subtraction itself is then performed as in
the case of monomials. See §§ 17-19.
Example. Subtract Ax — 2y + 6z from 3x-\-6y — 3z.
3x + 6 y — 3z
4 x - 2 // + G z
The steps are :
'3x -4x = -x; 6y-(-2y) = 8y; - dz -(-f-6z)=-9«.
16 FUNDAMENTAL OPERATIONS
EXERCISES
1. Add 8 x" - 11 x - 7 x 2 , 2 x - (3 x 2 + 10, -5 + 4a; 3 + 9z,
and 13 x- 2 - 5 - 12 ar 3 .
2. Add 5 a 3 -2a -12 -10a 2 , 14- 7a + a 2 - 9a 3 , 3a 2
— 13 a 3 + 4 — 11 a, and 3 — 7 a + 10 a 2 + 4 a 3 .
3. From the sum of 9 m 3 — 3 m 2 + 4 m — 7 and 3 m 2 — 4 m 3
+ 2 m + 8 subtract 4 m 3 — 2 m 2 — 4 + 8 m.
4. From the sum of x 4 — ar 3 — a 2 x~ — a s x + 2 a 4 and 3 ax 3
+ 7 a 2 ^ 2 — 5 a 3 x + 2 a 4 subtract 3 a.- 4 + ax s — 3 a 2 ^ 2 + a?x — a 4 .
5. Add 37a- 4& — 17c + 15d-6/-8ft and 3c -31 a
+ 9&-5d-7i-4 ./".
6. Add 11 g — 10 p — 8 w + 3m, 24m — 17g + 15p — 13 w,
9 n — G m — 4 q — 7/> — 5 n, and 8 a — 4^ — 12 wi + 18 n.
7. From the sum of 13 a — 156 — 7c — 11 d and 7 a — 66
+ 8 c + 3 d subtract the sum of 6 d — 56 — 7c + 2a and 5 c
- 10 d - 28 6 + 17 a.
8. Add 2 3 • 3 4 x 3 - 2 5 • 3 2 a- 2 + 2 2 . 3 3 . 7 x + 2 2 . 3 2 • 5,
2- • :; :i X s - 2 4 • 3 2 • 7 x + 2 4 ■ 3 3 x 2 - 2- • 3 2 • 5 2 , and 2 3 • 3 3 cc 3
-2 ; -3 3 .5 + 2 3 .3 3 x-2 4 -3 4 x 2 .
9. Add (a + 6 — c) wt + (a — 6 + c) n, + (a — 6 — c) A-,
(2 a - 3 6 + c) m + (6 - 3 a + c) n + (4 c + 2 6 + a)ft,
and (6 - 2 c) m +(2 a- 2c + 6)w +(2 6 -2a+ c)k.
10. From the sum of aaj 8 — bx 2 + cos — d and bx 3 + a.'c 2 — dx
+ c subtract (a — 6) or 3 + (c — a) or — (6 + d) x — d + c.
11. From (m — n)(m — w)x 3 + (n — mfx 2 — (n + in) x + 8 sub-
tract the sum of n(m— w)^* 3 — 4(w — m) 2 oj 2 + (ri + m)» — 31
and 2 (h. — m) 2 ^ — m (m — ri)x* — 2 (« + 7?i)x + 25.
12. Add a" + 2 o n+1 + a" +2 and 2 a n - 4 a" +1 + 5 a n+2 and
from this sum subtract 7 a" +1 — 8 a n + a" 4 " 2 .
ADDITION AND SUBTRACTION 17
REMOVAL OF PARENTHESES
42. By the theorems of §§ 15-18, a parenthesis inclosing a
polynomial may be removed with or without the change of
sign of each term included, according as the sign — or + pre-
cedes the parenthesis.
In case an expression contains signs of aggregation, one
within another, these may be removed one at a time, beginning
with the innermost, as in the following example :
a _ { h + c _[<* _ e +f-(g - £)]}
= a - {b + c - Id - e + /- g + K]}
= a-{b + c-d + e-f+g — 7i]
= a — b — c + d — e+f—g+h.
Such involved signs of aggregation may also be removed all
at once, beginning with the outermost, by observing the number
of minus signs which affect each term, and calling the sign of
any term + if this number is even, — if this number is odd.
Thus, in the above example, b and c are each affected by one
minus sign, namely, the one preceding the brace. Hence we write,
a — b — c.
d and/ are each affected by two minus signs, namely the one before
the brace and the one before the bracket, while e is affected by these
two, and also by the one preceding it. Hence we write, d — e + /•
g is affected by the minus signs before the bracket, the brace,
and the parenthesis, an odd number, while h is affected by these
and also by the one preceding it, an even number. Hence we write
-g + h.
By counting in this manner as we proceed from left to right, we
give the final form at once, a — b — c + d — e + / — g + h.
EXERCISES
In removing the signs of aggregation in the following, either
process just explained may be used. The second method is
shorter and should be easily followed after a little practice.
1. 7 -{-4-(4-[-7])-(5-[4-5] + 2)|.
18 FUNDAMENTAL OPERATIONS
2 . _[_(7-{-4 + 9}-13)-(12-3+[-7 + 2])].
3. 6-(-3-[-5.+ 4] + {7-3-(7--19)} + 8)-
4 . 5 + [-(-f-5-3+ll|-15)-3]+8.
5. 4 a- — [3 x — y — \ 3 x — y — (x — y — x) + x\ — 3 ?/].
The vinculum above y — x has the same effect as a parenthesis, i.e.
- 1 1 - x = - (y - x).
6 . 3x i -2tf-(4:a?-\3a?-(y*-2x l ) -3^-^ + 4 a 2 ).
7. 7 a - [3 a - [- 2 a - a + 3 + a] - 2a-5J.
8. / - (- 2 m -n-\l-ml)-(5l-2n-[-3 m + »]).
9. 2d- [3 d + |2cl-(e- 5d)| - (d + 3 e)].
10. 4y-(-2y-[-3y-{-y-y^l}+2y]).
11. 3a?-[8a;-(aj-3)- {-2a; + 6-8a:-l}].
12. a; - (x — \ -4a; - [5 x - 2 .»■ - 5] - [- .r - a- -3] \).
13. 3 » — {y - [3y + 2 2] - (4 .r - [2 y - 3 2] - 3 y - 2 z) +4.r } .
14. a;-(-&-{-3a:-[>-2a; + 5]-4}-[2a;-a;-3]).
MULTIPLICATION OF MONOMIALS
43. Theorem. The product of two powers of the same
base is a power of that base whose exjwnent is the sum
of the exponents of the common base. See § 127, E. C.
Proof. Let b be any number and k and n any positive integers. It
is to be proved that &* . &» = '&*+n
By the definition of a positive integral exponent,
b k = b • 6 -6 ••• to k factors,
and b n — h • h • l> ■■■ to n factors.
Hence, 6* • b n = (J> ■ b ••• to k factors) (b ■ b ■■• to n factors)
= 6 -6 • &'••• to k + n factors,
since the factors may be associated in a single group, § 0.
Hence, by the definition of a positive integral exponent, we have,
6* • h" = // • ".
MULTIPLICATION AND DIVISION 19
44. In finding the product of two monomials, the factors may
be arranged and associated in any manner, according to §§ 8, 9.
E.g. (3 ab 2 ) x (5 a 2 b 3 ) = 3 ab' 2 ■ 5 a 2 b 3 § 9
= 3 • 5 • a ■ a 2 ■ b 2 . b 3 §8
= (3.5)(a.a 2 )(&*.&8) §9
= 15 o 3 6 5 - by the theorem, § 43
The factors in the product are arranged so as to associate those
consisting of Arabic figures and also those which are powers of the
same base. This arrangement and association of the factors is equiv-
alent to multiplying either monomial by the factors of the other in
succession. See § 129, E. C.
45. It is readily seen that a product is negative when it con-
tains an odd number of negative factors ; otherwise it is positive.
For by the commutative and associative laws of factors the negative
factors may be grouped in pairs, each pair giving a positive product.
If the number of negative factors is odd, there will be just one remain-
ing, which makes the final product negative.
EXERCISES
Find the products of the following :
1. 2 s • 3 4 • 4 7 , 2 7 • 3- • 4 2 . 8. a 1 , a 3 *-», a?-**.
2. 3 • 2 4 • 5 2 , 5 • 2 2 • 5, 7 • 2 s ■ 5 3 . 9. a n b m , cW, a l - 3 "b°- 4m .
3. 2 x 2 y s , 5 xY; 2 x*y. 10. 4 ab m , 2 a 3 b n , 3 a G b°-- m - 1 .
4. 5 xy, 2 arty, 4 xy 5 , x?y 2 . 11. 2x m y m+n , 3 x m ~ y»- m + 2 .
5. 3a 5 bc, ab 2 c, a?bc 4 , 4ab 5 c. 12. a d -' 2c+2 b m ~ 3 ' 1 , a*-* -1 ^" 4 * 1 .
6. x n , x n -\ x n+ \ 2 x". 13.3 a; a+3i , 2 x a ~ 2h y c ~ 3 , 2 x 4 - 2a - b y 2c+3 .
7. x m+n ~\ x m -"+\ x 2m . 14. a?*- s b" +1 , a x+s b v -\ 3 a 8 & 2 .
15 3 4a - 2 - 26 . 2" +3_m 3 5 - 4 «+ 26 . o m+2 ~ n
16. a^ +1 ^, &-2*-iy** } y"-* m 17. 7 • 2 3a - 4 , 3 • 2 5 ~ 2a , 5 • 2 3 " .
ig 0~x-l-4y Q _ Ql-5x-4y Q2 , 92-2z
19 3 2 -5'"+3" . O-iaSb Q2-3n+6m _ 95+3>>+5ri
20. (1 + a) 7 - 864 "" • (1 - a) 2+a " 6 , (1 - a)"-"- 1 • (1 + a) 36 " " 6 .
20 FUNDAMENTAL OPEIIATIONS
DIVISION OF MONOMIALS
46. Theorem 1. The quotient of two powers of tJie same
base is a power of that base whose exponent is the exponent
of the dividend minus that of the divisor. See § 154, E. C.
Proof. Let a be any number and let m and khe positive integers of
which m is the greater. We are to prove,
a m ^_ a k _ a m-*.
Since k and m — k are both positive integers, we have, bv § 43,
a*a m -* = a t+m- *=a m . That is, a m_ * is the number -which multiplied
by a k gives a product a m , and hence by the definition of division,
Under the proper interpretation of negative numbers used
as exponents this theorem also holds when m < k. This is
considered in detail in § 177. We remark here that in case
m = k, the dividend and the divisor are equal and the quotient
is unity. Hence a m -=- a'" = a m ~ m — a" — 1. See § 11.
47. Theorem 2. In dividing one algebraic expression bj/
another, all factors common to dividend, and divisor may
be removed or canceled. See §§ 23, 156, 157, E. C.
Proof. We are to show that — = -•
bk b
By definition of division, § 11. — . bk — a {-, (1)
Also - -b = a. (2)
Multiplying (2) by k, - • bk = ak. (3)
b
Hence from (1) and (3), " l - ■ bk = " ■ bk. (4)
bk b
Dividing by bk, «£ = ?. (5)
MULTIPLICATION AND DIVISION 21
-r.. ., EXERCISES
Divide :
1. 4 • 2 4 • 3 7 • 5 2 by 3 • 2 3 . 3 4 • 5. 4. 5 a 5 b 7 c s by 5 a 4 b 7 c 4 d 2 .
2. 5 • 3 7 • 7 4 • 13 5 by 2 • 3 5 • V • 13 2 . 5. x 2n y m z" m by x n y m z m .
3. 3x 7 yh by 2 xhjz. 6. a 8 »-y»+s by a n+6 y 2n+ \
7. a c+3d+2 & d -° c+6 by a c+2< *- 4 .
8. 3 a + 26 - 7 . 53^-2«+4 by 3&+«-8 . 526-2^+3^
9. ft 3+ 2 ™-3»&5 C 7-» by a 2+m^4„ 6 4 c 7-„_
10 /v'4a-26+l ? .e-<i+6„3a+2?i+c by -,26— c+3aya-c+&,w3a+&-2c
11 2 3a_4+76 • 3 3f, - 4c + 6 by 2 2a ~ 5 ~ 7h • 3 2f> - 6r + 7
12. (x - 2) 3m+1 - 3 " • (cc 4- 2)- m+2 - s " by (a; + 2) 1+2 "- 2 " • (x - 2) 1 - 3 »+ 2 "\
13. (x — 2 /)»- 3 ^ 1 . (a; + y y<-M+2 by (a, _ ^-a-sc+ra . ^ + ^-3-26+7^
14. (a 2 - 6 2 )3+«+"* . (a 2 - &2)i-s*-» by (a 2 - 6 2 ) 4+ * ■ (a 2 - 6 2 )- 2 + 26 .
MULTIPLICATION OF POLYNOMIALS
48. Theorem. The product of two polynomials is equal
to the sum of the products obtained by multiplying each
term of one polynomial by every term of the other. See
§ 86, E. C.
Proof. By the distributive law, § 10, we have,
(m + n + h)(a + 6 -f c)= m(a + b + c)
+ n(a + b + c)
+ k(a 4- b + c).
Applying the same law to each part, we have the product,
ma + nib + mc + na + nh + nc + ha + lb -f fee.
This is Principle XIII of the Elementary Course.
22 FUNDAMENTAL OPERATIONS
EXERCISES
Find the following indicated products :
1. (a + &) (a + &), i.e. (a + 6) 2 ; also (a — b) 2 .
2. (a + 6) (a + 6) (a + 6), i. e. (a + &) 3 ; also (a - bf.
3. (a + 6) (a + &) (a + &) (a + 6), i. e. (a + &) 4 ; also (a - 6) 4 .
4. (a 2 + 2 ab + & 2 ) (a 2 + 2 a& + & 2 ) (a + b).
5. (a 2 - 2 a& + ft 2 ) (a 2 - 2 a6 + ft 2 ) (a - 6).
6. (a 2 4- 2 a& + & 2 ) 3 ; also (a + ft) 6 .
7. (a 3 + 3 a 2 6 + 3 aft 2 + ft 3 ) 2 . 9. (a - ft)(« 2 + aft + b 2 ).
8. (a 3 - 3 a 2 & + 3 aft 2 - ft') 2 . 10. (a + b)(a 2 - ab + ft 2 ).
1 1 . (a 2 + 2 aft + ft 2 ) (a 4 + 4 a 8 & + 6 a 2 b 2 + 4 a& 3 + ft 4 ) .
12. (a 2 - 2 a& + ft 2 ) (a 4 - 4 a s b + 6 a 2 b 2 - 4 a& 3 + ft 4 ) .
13. (a — &)(a 8 + d 2 b + aft 2 + ft 3 ).
14. (a + &)(a 3 — a 2 b + a& 2 — 6 s ).
15. (a - &)(a 4 + a 3 6 + a 2 & 2 + a& 8 + ft 4 ).
16. (a + &)(a 4 - a 3 & + a 2 & 2 - a& s + ft 4 )-
17. (a - &)(a s + a 4 ft + a"ft 2 + a 2 6 8 + ab 4 + &*).
18. (a + 6 )( "'' - a 4 ft + a"ft 2 - a 2 lr + ,/ft 4 - ft 5 ).
19. (1 - r)(a + ar + ar 2 + ar 3 ).
20. (1 — r)(a + ca- + or + ar 8 + m A + a?- 5 ).
21. ( a + & + c) 2 . 22. (a + & - c) 2 . 23. (a - 6 - c) 2 .
24. From Exs. 21-23 deduce a rule for squaring a trinomial.
25. (a; + >/ +z + v) 2 . 26. (aj —y + z— r) 2 .
27. From Exs. 25, 26 deduce a rule for squaring a polynomial.
28. (a + & + c)(a 4- 6 - c)(a - ft + c)(& - a + c).
29. (ab + ac + bc)(ab + ac — &c)(a& — ac + 6c)i ac + be — ab ) .
30. (a- b + c + d)(a 4- 6 + c — ri)(a 4- 6 — c 4- d)
(-a + 6 + c + d).
MULTIPLICATION AND DIVISION 23
31. (4 x 3 - 6 xy + 9 y 2 )(2 x + 3 ?/)(4 « 2 + 6 a;# -f 9 y 2 )(2 x-oy).
32. Collect in a table the following products :
(a+b) 2 , (a -hf, (a + b) 3 , (a - bf, (a + b)\
(a - b)\ (a + b) 5 , (a - by, (a + b) G , (a - b) 6 .
33. From the above table answer the following questions:
(«) How many terms in each product, compared with the
exponent of the binomial ?
(l>) Tell how the signs occur in the various cases.
(c) How do the exponents of a proceed ? of b ?
(d) Make a table of the coefficients alone and memorize this.
E.g. For (a + by, they are 1, 5, 10, 10, 5, 1.
34. Make use of the rules in Exs. 24, 27, 33 to write the fol-
lowing products : (a) (2 x — 3 y + 4 z) 2 , (b) (}m 2 -|n J -3r) 2 ,
(c) (4 a.» — 2 ay + 3 m — nf.
^ (3«-6 2 ) 4 - V ° ^ (*>(9*-2y)«.
(/) (i «» - 1 W- (0 ( o 3 _ yV (0 (1-2 x)\
(g) (2 m - n) 5 . V L V (m) ( j L+3 x y m
DIVISION OF POLYNOMIALS
49. According to the distributive law of division, § 30, a
polynomial is divided by a monomial by dividing each term sepa-
rately by the monomial. See § 25, E. C.
r, ab + ac — ad ab , ac ad , , 7
E.g. - = 1 - = b + c — d.
a a a a
A polynomial is divided by a polynomial by separating the
dividend into polynomials, each of which is the product of the
divisor and a monomial. Each of these monomial factors is a
part of the quotient, their sum constituting the whole quotient.
The parts of the dividend are found one by one as the work pro-
ceeds. See §§ 161-163, E.C. This is best shown by an example.
24 FUNDAMENTAL OPERATIONS
Dividend, o 4 + « 3 — 4a 2 + 5 a — 3 la 2 + 2 a — 3, Divisor.
- a 3 -
-a 2
+ 5a -
-3
- a 3 -
-2 a 2
+ 3 a
a 2
+ 2a-
- 3
a 2
+ 2a-
-3
1 st part of dividend : « 4 + 2 a 3 — 3 a 2 ]a 2 — a + 1, Quotient.
2d part of dividend :
3d part of dividend :
The three parts of the dividend are the products of the
divisor and the three terms of the quotient. If after the suc-
cessive subtraction of these parts of the dividend the remain-
der is zero, the division is exact. In case the division is not
exact, there is a final remainder such that
Dividend = Quotient x divisor -f- Remainder.
In symbols we have D = Q • d -f R.
t^. .j EXERCISES
Divide :
1. x' + 5 x A y -f 10 arty 2 + 1 xhf + 5 xy* + f by .r 2 +2 xy+y 2 .
2. .r*-j-.,-y + / by a 4 — x 2 y 2 -\-y\ 6. as 8 — ^ by .r — y\
3. X s — y 5 by x — y. 7. a 8 -f b 6 by a 2 + 6 2 .
4. a 8 — j/ 8 by 37*+ tfy+xy^+y*. 8. ,v'"-)/ 9 byf°-f.
5. .r' + .v 11 by x*-xy-\-y 2 . 9. a 10 — a fl & 5 +& 10 by a 2 - a&+& 2 .
10. 2 a 4 - 3 .rV; + 6 .r/r - a* 8 + 6 6 4 by a- 2 - 2 a?6 + 3 b 2 .
11. 2 .r''' - 5 .i- 5 + ( ; .e 4 - 6 x 8 + ( i .*'- - 4 x + 1 by a 4 - .i- 3 + a- 2 - x + 1 .
12. 26 a 8 6 8 + a 6 + 6 b 6 - 5 a'6 - 17 ab 5 - 2 cW - a?b*
by a 2 -3& 2 -2a&.
13. a 4 + 2.r 3 -7.r-8a + 12 by a- 2 -3 a + 2.
14. 4 //-■ + 4 a 6 4. «-• _ 12 6c - 6 ac + 9 c 2 by 2b + a-Sc.
15. .i' 4 -)- 4 x> f — 4 :cyz + 3 y* + 2 ?/ 2 z — z 2 by a- 2 — 2 a-// + 3 //-'— z.
16. « 2 6 2 c + 3 crb 3 - 3 «&<-"■ - aV + //' - 4 &V + 3 a6 3 c
+ 3 6c 4 — 3 a?b& by b' 2 — c 2 .
CHAPTER III
INTEGRAL EQUATIONS OF THE FIRST DEGREE IN ONE
UNKNOWN
50. When in an algebraic expression a letter is replaced by
another number symbol, this is called a substitution on that letter.
E.g. In the expression, 2 a + 5, if a is replaced by 3, giving 2 • 3 + 5,
this is a substitution on the letter a.
51. An equality containing a single letter is said to be
satisfied by any substitution on that letter which reduces both
members of the equality to the same number.
E.g. 4 x + 8 = 24 is satisfied by x = 4, since 4 • 4 + 8 = 24.
We notice, however, that the substitution must not reduce the
denominator of any fraction to zero.
x 2 — 4
Thus x = 2 does not satisfy — — = 8 although it reduces the left
member of the equation to -, which by § 21 equals 8 or any other
number whatever.
On the other hand, x = 6 satisfies this equation, since
6 2 -4 32
o~T = 4 = 8 '
52. An equality in two or more letters is satisfied by any
simultaneous substitutions on these letters which reduce both
members to the same number.
E.g. 6 a + 3 b = 15 is satisfied by a = 2, 6 = 1 ; a = -, 6 = 2 ; a = 1,
6 = 3, etc.
,2 1.
- is satisfied by x = 3, y — 1, but is not satisfied by
x* — y A
x- + 2 xy + y-
any values of x and y such that x = — y, since these reduce the
denominator (and also the numerator) to zero. See § 24.
25
26 INTEGRAL EQUATIONS OF THE FIRST DEGREE
53. An equality is said to be an identity in all its letters, or
simply an identity, if it is satisfied by every possible substitur
tion on these letters, not counting those which make any
denominator zero.
If an equality is an identity, both members will be reduced
oO the same expression when all indicated operations are per-
formed as far as possible.
The members of an identity are called identical expressions.
Thus in the identity (a + 6) 2 =a 2 +2 ab -f //-, performing the indi-
cated operation in the first member reduces it to the same form as the
second.
54. An equality which is not an identity is called an equa-
tion of condition or simply an equation.
The members of an equation cannot be reduced to the same
expression by performing the indicated operations.
E.g. (x — 2)(x — 3) = cannot be so reduced. This is an equation
winch is satisfied by x = 2 and x — 3. See § 22.
55. In an equation containing several letters any one or more
of them may be regarded as unknown, the remaining ones being
considered known. Such an equation is said to be satisfied by
any substitution on the unknown letters which reduces it to an
identity in the remaining letters.
E.g. x 2 — t' 2 = sx + st is an equation in s, x, or t, or in any pair of
these letters, or in all three of them.
As an equation in x it is satisfied by x = s + /, since this substitu-
tion reduces it to the identity in s and t,
s 2 + 2 st = s 2 + 2 st.
As an equation in s it is satisfied by s = x — t, since this substitu-
tion reduces it to the identity in x and /.
X 2_ t * = X*- ,2.
Any number expression which satisfies an equation in one
unknown is called a root of the equation.
E.g. s + t is a root of the equal ion ./•- — t- = sx + st, when x is the
unknown, and x — t is a root when .s- is the unknown.
EQUIVALENT EQUATIONS 27
56. An equation is rational in a given letter if every term in
the equation is rational with respect to that letter.
An equation is integral in a given letter if every term is
rational and. integral in that letter.
57. The degree of a rational, integral equation in a given
letter is the highest exponent of that letter in the equation.
In determining the degree of an equation according to this defini-
tion it is necessary that all indicated multiplications be performed as
far as possible.
E.g. (x — 2) (x — 3) = is of the 2d degree in x, since it reduces to
x' 2 — 5 x + 6 = 0.
EQUIVALENT EQUATIONS
58. Two equations are said to be equivalent if every root of
either is also a root of the other.
59. Theorem 1. If one rational, integral equation is
derived from another by performing the indicated opera-
tions, then the two equations are equivalent. See § 36,
E. C.
Proof. In performing the indicated operations, each expression is
replaced by another identically equal to it. Hence any expression
which satisfies the given equation must satisfy the other and
conversely.
E.g. 10 x = 50 is equivalent to 3 x + 7 x = 50, since 3 x+ 7 x = 10 x ;
and 8(2 x — 3 y) = 2 y — 1 is equivalent to 16 x — 2±y = 2 y — 1, since
8(2 x - 3t/)ee 16 x -24y.
60. Theorem 2. If any equation is derived from another
by adding the same expression to each member, or by sub-
tracting the same expression from each member, then the
equations are equivalent. See § 36, E. C.
Proof. For simplicity of statement we prove the theorem for the
case where the original equation contains only one unknown, the
proof in the other cases being similar.
Let M=N (1)
be an equation involving one unknown, x, and let A be an expression
which may or may not involve x.
28 INTEGRAL EQUATIONS OF THE FIRST DEGREE
We are to show that equation (1) is equivalent to
M + A=N+\. (2)
and also to M - A = N — A. (3)
(a) Let rj be a root of (1). Then substituting x y for x in (1)
makes M and N identical. Since A = A for any value of x it follows,
§ "_'. that the substitution of x 1 reduces M + A and A* + A to identical
expressions. That is, x = x x satisfies equation (2). Hence any root
of (1) is also a root of (2).
(h) Again if x 1 is a root of (2), its substitution reduces M + A and
N + A to identical expressions, and hence by § 6, it also reduces
M + A — A and N + A — A to identical expressions. That is, x = x x
satisfies equation (1). Hence any root of (2) is also a root of (1).
From (a) and (b) it follows that equations (1) and (2) are equivalent.
In like manner (1) and (3) are shown to be equivalent.
61. Corollary. Any equation can be reduced to an
equivalent equation of the form R — 0.
For if an equation is in the form M — N, then by theorem 1 it
is equivalent to M - i\ r = N — N = 0, which is in the form R = 0.
62. Theorem 3. i/ 1 one equation is derived from <tu<>llicr
by multiplying or dividing each member by the same ex-
pression, then the equations are equivalent, provided the
original equation is not multiplied or divided by zero or
by an expression containing the unknown of theequation.
See § 36, E. C, and the note following it.
Proof. Again consider the case where the original equation con-
tains only one unknown.
Let A be an expression not containing x, and different from zero.
AYe are to show that M = N (1)
is equivalent to M ■ A = N ■ A, (2)
and also to S± -±L. (3)
A A W
If r, is a root of (1), its substitution makes .1/" and X identical,
and hence also M . A and .V • .1 by § 7. That is. r, is a root of (2).
Similarly, if .;-, is a root of (2). then, by § 11, it is a root of (1).
Hence (1) and (2) are equivalent. In like manner, we may show
(1) and (3) equivalent.
EQUIVALENT EQUATIONS 29
63. The ordinary processes of solving equations depend
upon theorems 1, 2, and 3, as is illustrated by the following
examples :
Ex. 1. (x + 4) (./; + 5) = (x + 2) (x + 6). (1)
x 1 + 9 x + 20 = x 2 + 8 x + 12. (2)
x = - 8. (3)
By theorem 1, (1) and (2) are equivalent, and by theorem 2, (2)
and (3) are equivalent. Hence (1) and (3) are equivalent. That is,
— 8 is the root of (1).
Ex.2. | X + | = 4. (1)
2 x + 4 = 12. (2)
2 x = 8. (3)
x = 4. (4)
By theorem 3, (1) and (2) are equivalent. By theorem 2, (2) and
(3) are equivalent. By theorem 3, (3) and (4) are equivalent.
Hence (1) and (4) are equivalent and 4 is the solution of (1).
These theorems are stated for equations, but they apply
equally well to identities, inasmuch as the identities are
changed into other identities by these operations.
64. If an identity is reduced to the form R — 0, § 61, and all
the indicated operations are performed, then it becomes = 0.
See § 53. Conversely, if an equality may be reduced to the
form = 0, it is an identity. This, therefore, is a test as to
whether an equality is an identity.
E.g. (x + 4)' 2 = x 2 + 8x + 16 is an identity, since in x 2 + 8x + 16
— x 2 — 8x — 16 = all terms cancel, leaving = 0.
EXERCISES
In the following, determine which numbers or sets of num-
bers, if any, of those written to the right, satisfy the corre-
sponding equation.
Remember that no substitution is legitimate which reduces
any denominator to zero.
30 INTEGRAL EQUATIONS OF THE FIRST DEGREE
1. 4(a»-l)(a>-2)(a;-3)=3(a;-2)(a>-3). 1,2,3,4.
2. t^> = (x-4:)(x + 6). 2,4,6.
x + 5
X + 3 3 r> o !
3. — ^ = a; — - • 2, d, i.
Var' + 7 A
4. (^-3)^-2) = ar o_ 5a; + 6 , 2,3,0,-2.
x 2 —7x + 10
5. « 2 + 9a + 20 (a+4)(a _ 4)(a+5)j 4> _ 4)5) _ 5 .
a- + 8 a + 16
fa=0, (a = 4, |« = 2,
'•»• + « = * | 6 «S. U = 0. | & =2.
7 369(a-6) _ fl | b J a = ° J J" = 1 > f« = 5 >
a a + 6 2 16 = 0. [6=1. [6 = 4.
s=l, f.r=l, lx = 2,
3 ftt = l, ftt = l, ftt = — 1,
8. ^^^ = (x-2)(i/-l).
K-2/ 1^ = 0. U = l- U
?r — v
9. - = (u* + uv + v*)(u — v).
v? — v 2 I '"
-1. [v = D. \v = 0.
1Q (r-s)(r + s)(r 2 + r) = (/ . 2 _ ^ f _ 3 r = lf , = t .
?" 3 + rs 2 — rs — s 3
7- = 1, s = — 1 ; r= 2, s = — 2 ; r = a, s = — a.
11. a + 6 + c = 6. a = l,6=2,c = 3;
a = 3, 6 = 3, c = ; a = 10, 6 = 0, c = - 4.
a — b + c 3a — 2c + 56 + 2 ,,, n ^ a
12. . T = = ~ T7T~ —' a = 8, 6 = 0, c = 6;
a = l, 6 = 4, c=2; a = 0, 6 =0, c=-4.
13 (a-6)(6-c)(c-a) = (6 _ 6(c _ a) a==2>6 = 1>c = 1 ,
ac — 6c — a- + 6a
a = 3, 6 = 2, c = 3 ; a = 6, 6 = 6, c = 0.
EQUIVALENT EQUATIONS 31
14. (as - 2!) (a; - y) (y - z) = 8 xyz(x 2 - if) (y 2 - z 2 ) (z 2 - x 2 ).
x= 1, y = 1, z — 1 ; ar = l, y = 0, z = l; a; = 1, ?/ = 2, z = 3.
15. x 3 4-3x 2 ?/4-3^ 2 4-?/ 3 = (.c4-?/) 3 . y •
[y = l. [y = 2. I ?/ = 4.
16. Show by reducing the equality in Ex. 15 to the form
R= that it is satisfied by any pair of values whatsoever for
x and y, e.g., for x = 348764, y = 594021. What kind of an
equality is this ?
Which of the following four equalities are identities?
17. 12{x + y) 2 + 17(x + y)-7 = (3x + 3y-l)(4,x + 4,y + 7).
1 8 . ^IzlA 5 _ tt 4 + a s 6 + a 2 6 2 + ab s + &4<
a — b
19. ?£±&! = a 4 _ a 3 b + a 2 & 2 _ ab 3 + & 4
20. 2 (a - 6) 2 + 5(a + &) + 8 ab = (2 a 4- 2 & 4- l)(a 4- b + 1).
Solve the following equations, and verify the results :
21. (2 a + 3)(3 a - 2) = a 2 4- a (5 a + 3).
22. 6 (6 - 4) 2 = - 5 - (3 - 2 b) 2 - 5 (2 + &) (7-2 6).
23. (y - 3) 2 + (.V - 4) 2 - 0/ - 2) 2 - (i, - 3) 2 = 0.
24. (as - 3) (3 x + 4) - (as - 4) (x - 2) = (2a: + 1) (x- 6).
25. 2 (3 r - 2) (4 r 4- 1 ) + (r - 4) 3 = (r + 4) 3 - 2.
26. a 3 — c + b s c + abc = b. (Solve for c.)
27. (&_2) 2 (&-t/)-3%4-(2^4-1)(&-1) = 3-2&. (Findy.)
28. 2 (12 - a;) 4- 3(5 x - 4) + 2 (16 - a:) = 12(3 4- x).
29. (b-a)x-(a + b)x+4:a 2 = 0. (Find x.)
30. (a;— a)(&— c)4-(&— a)(a>-c) — (a— c)(a?— &)=0. (Find*.)
31. r 3 v + s 3 v — 3 r — 3 s 4- 3 v (rs + rs 2 ) = 0. (Find i>.)
32 INTEGRAL EQUATIONS OF THE FIRST DEGREE
32. (aj - 3) (x - 7) - (a? - 5) (a; - 2) + 12 = 2 (a? - 1).
33 . (a + hf + (a- - b) (x -a)- (x + a) (x + 6) = 0. ( Find x.)
34. ny (2/ + ») - (y + m ) (// + ») ( w + ») + ™?/ (2/ + m ) = 0-
(Find //.)
35. (n + i) (J - i + k) - (n - i) (i -j + k) = 0. (Find ».)
36. ^(5a:-l)+fV(2-3.r) + i(4 + a-) = |(l+2.r)- T V
37. (J — m)(»— n) + 22(i» + n) = (J + m)(*+n). (Findz.)
38. a(x — b) — (a + &)(aj + b — a)=b(x—a)+a 2 —b 2 . (Find x.)
39. (m + n)(n + & — y) + (n — m)(&— y)=n(m+b). (Find y.)
.. 3(2a-3 6) 2(3a-5&) , 5(a-b) b ._. , .
40. -^- --^-g- 2 + -L_J = _. (Fmda.)
Solve each of the following equations for each letter in terms of
the others.
41. l(W+w') = l'W'. 43. m,.%(t. 2 -t) = (m + m l )(t — t 1 ).
42. (v — n)d=(v—n ] )d 1 . 44. (m + m^Qi — t) — lm 2 +m.,t.
PROBLEMS
1. What number must be added to each of the numbers 2,
26, 10 in order that the product of the first two sums may
equal the square of the last sum ?
2. What number must be subtracted from each of the num-
bers 9, 12, 18 in order that the product of the first two re-
mainders may equal the square of the last remainder ?
3. What number must be added to each of the numbers
a, b, c in order that the product of the first two sums may
equal the square of the last ?
Note that problem 1 is ;t special case of 3. Explain how 2 may
also be made a special case oE •').
4. What number musl be added to each of the numbers
a, b, c, d in order that the product of the first two sums may
equal the product of the last two '.'
PROBLEMS IN ONE UNKNOWN 33
5. State and solve a problem which is a special case of
problem 4.
6. What number must be added to each of the numbers
a, b, c, d in order that the sum of the squares of the first two
sums may equal the sum of the squares of the last two ?
7. State and solve a problem which is a special case of
problem 6.
8. What number must be added to each of the numbers
a, b, c, d in order that the sum of the squares of the first two
sums may be k more than twice the product of the last two ?
9. State and solve a problem which is a special case of
problem 8.
10. The radius of a circle is increased by 3 feet, thereby in-
creasing the area of the circle by 50 square feet. Find the
radius of the original circle.
The area of a circle is irr 2 . Use 3} for it.
11. The radius of a circle is decreased by 2 feet, thereby
decreasing the area by 36 square feet. Find the radius of the
original circle.
12. State and solve a general problem of which 10 is a
special case.
13. State and solve a general problem of which 11 is a
special case.
How may the problem stated under 12 be interpreted so as to
include the one given under 13?
14. Each side of a square is increased by a feet, thereby
increasing its area by b square feet. Find the side of the
original square.
Interpret this problem if a and b are both negative numbers.
15. State and solve a problem which is a special case of 14,
(1) when a and b are both positive, (2) when a and b are both
negative.
34 INTEGRAL EQUATIONS OF THE FIRST DEGREE
16. Two opposite sides of a square are each increased by a
feet and the other two by b feet, thereby producing a rectangle
whose area is c square feet greater than that of the square.
Find the side of the square.
Interpret this problem when a, b, and c are all negative numbers.
17. State and solve a problem which is a special case of 16,
(1) when a, b, and c are all positive, (2) when a, b, and c are
all negative.
18. A messenger starts for a distant point at 4 a.m., going
5 miles per hour. Four hours later another starts from the
same place, going in the same direction at the rate of 9 miles
per hour. When will they be together ? When will they be
8 miles apart ? How far apart will they be at 2 p.m. ?
For a general explanation of problems on motion, see p. 115, E. C.
19. One object moves with a velocity of i\ feet per second
and another along the same path in the same direction with a
velocity of v 2 feet. If they start together, how long will it re-
quire the latter to gain n feet on the former ?
From formula (2), p. 117, E. C, we have t = — - —
Discussion. Ifw 2 >i'j and n > 0, the value of t is positive, i.e. the
objects will be in the required position some time after the time of
star! ing.
If Vz < '"l and n > 0, the value of ( is negative, which may be taken
to mean that if the objects had been moving before the instant taken
in tlic problem as the time of starting, then they would have been in
the required position some time earlier.
Jiv2 = vi and n=£0, the solution is impossible. See § 25. This
means that the objects will never be in the required position. If
v t ■= i> 2 and n = 0, the solution is indeterminate. See §24. This may
be interpreted to mean that the objects are always in the required
position.
20. State and solve a problem which is a special case of 19
under each of the conditions mentioned in the discussion.
21. At what time after 5 o'clock are the hands of a clock
first in a straight line ?
PROBLEMS IN ONE UNKNOWN 35
22. Saturn completes its journey about the sun in 29 years
and Uranus in 84 years. How many years elapse from con-
junction to conjunction? See figure, p. 119, E. C.
23. An object moves in a fixed path at the rate of v r feet
per second, and another which starts a seconds later moves in
the same path at the rate of v 3 feet per second. In how many
seconds will the latter overtake the former ?
24. In problem 23 how long before they will be d feet
apart ?
If in problem 24 d is zero, this problem is the same as 23. If d is
not zero and a is zero, it is the same as problem 19.
25. A beam carries 3 weights, one at each end weighing
100 and 120 pounds respectively, and the third weighing 150
pounds 2 feet from its center, where the fulcrum is. What is
the length of the beam if this arrangement makes it balance ?
For a general explanation of problems involving the lever, see pp.
120-122, E. C.
26. A beam whose fulcrum is at its center is made to bal-
ance when weights of GO and 80 pounds are placed at one end
and 2 feet from that end respectively, and weights of 50 and
100 pounds are placed at the other end and 3 feet from it
respectively. Find the length of the beam. .
27. How many cubic centimeters of matter, density 4.20,
must be added to 150 ccm. of density 8.10 so that the density
of the compound shall be 5.4? See § 99, E. C.
28. How many cubic centimeters of nitrogen, density
0.001255, must be mixed with 210 ccm. of oxygen, density
0.00143, to form air whose density is 0.001292?
29. A man can do a piece of work in 16 days, another in 18
days, and a third in 15 days. How many days will it require
all to do it when working together?
30. A can do a piece of work in a days, B can do it in b
days, C in c days, and I) in d days. How long will it require
all to do it when working together?
CHAPTER IV
INTEGRAL LINEAR EQUATIONS IN TWO OR MORE
VARIABLES
INDETERMINATE EQUATIONS
65. If a single equation contains two unknowns, an unlimited
number of pairs of numbers may be found which satisfy the
equation.
E.g. In the equation, y = 2 x + 1, by assigning any value to x, a
corresponding value of y may be found such that the two together
satisfy the equation.
Thus, x = — 3, y = — 5 ; x = 0, y = 1 ; x = 2, y = 5, are pairs of
numbers which satisfy this equation.
Fortius reason a single equation in two unknowns is called
an indeterminate equation, and the unknowns are called varia-
bles. A solution of such an equation is any pair of numbers
which satisfy it.
A picture or map of the real (see §§ 135, 195) solutions of
an indeterminate equation in two variables may be made by
means of the graph as explained in §§ 107, 108, E. C.
66. The degree of an equation in two or more letters is the
sum of the exponents of those letters in that one of its terms
in which this sum is greatest. See § 110, E. C.
E.g. y = 2 x + 1 is of the first degree in x and y. y- = 2 x + y and
y = 2 xy + 3 are each of the second degree in x and y.
An equation of the first degree in two variables is called a
linear equation, since it can be shown that the graph of every
such equation is a straight line.
36
INDETERMINATE EQUATIONS 37
67. It is often important to determine those solutions of an
indeterminate equation which are positive integers, and for this
purpose the graph is especially useful.
Ex. 1. Find the positive integral solutions of the equation
3 x + 7 y = 42.
Solution. Graph the equation carefully on cross-ruled paper, find-
ing it to cut the x-axis at x = 14 and the y-axis at y = 6.
Look now for the corner points of the unit squares through which
this straight line passes. The coordinates of these points, if there are
such points, are the solutions required. In this case the line passes
through only one such point, namely the point (7, 3). Hence the
solution sought is x = 7, y = 3.
Ex. 2. Find the least positive integers which satisfy
7x-3y = 17.
Solution. This line cuts the x-axis at x = 2f and the y-axis at
y = — 5|. On locating these points as accurately as possible, the line
through them seems to cut the corner points (5, 6) and (8, 13). The
coordinates of both these points satisfy the equation. Hence the
solution sought is x = 5, y = 6.
EXERCISES
Solve in positive integers by means of graphs, and check:
1.
x + y = 7.
5.
90 -5 x = 9y.
2.
x + y = 3.
6.
5x=29-3y.
3.
x-27=-9y.
7.
140 - 7. r- 10,y = 0.
4.
7 y — 112 = — 4 x.
8.
S-2x- y = 0.
Solve in least positive integers, and check :
9. 7x = 3y + 21. n. 4 x = 9 y — 36.
10. 5x -ky = 20. 12. 5 x - 2 y + 10 = 0.
68. In the case of two indeterminate equations, each of the
first degree in two variables, the coordinates of the point of
intersection of their graphs form a solution of both equations.
38 INTEGRAL LINEAR EQUATIONS
Since these graphs are straight lines, they have only one
point in common, and hence there is only one solution of the
given pair of equations.
E.g. On graphing x + y = 4 and y — x — 2, the lines are found
to intersect in the point (1, 3). Hence the solution of this pair of
equations is . Q
x — 1, y — 6.
EXERCISES
Graph the following and thus find the solution of each pair
of equations. Check by substituting in the equations.
3x — 2y = —2, 7 J8aj = 7y,
x + 7 y = 30. 1 x + 3 = 5 y + 3.
aj + y = 2, [>/ = !>
3» + 2?/ = 3. ' l3// + 4.r = y.
[;
x — 4 y = 1, j 2 .r — 4 ?/ = 4,
a; — w = G // — 3.
10. '"
J-
l2s
x — oy
i. "• i,+.= ;
' l.f-o = i/-l. l3a; + 2y=3.
6. \* = y- 6 ' 12. |-"' = 7 2 '
1 5 y = x + 9. { y = 5.
SOLUTION BY ELIMINATION
69. The solution of a pair of equations such as the foregoing
may be obtained without the use of the graph by the process
called elimination. See pages 154-159, E. C.
70. Elimination by substitution consists in expressing one
variable in terms of the other in one equation and .substituting
this result in the other equation, thus obtaining an equation in
which only one variable appears. See § 116, E. C.
SOLUTION BY ELIMINATION 39
71. Elimination by addition or subtraction consists in making
the coefficients of one variable the same in the two equations
(§ 62), so that when the members are added or subtracted this
variable will not appear in the resulting equation. See § 117,
E. C.
72. Elimination by comparison is a third method, which con-
sists in expressing the same variable in terms of the other in each
equation and equating these two expressions to each other.
As an example of elimination by comparison, solve
^y + x = U, (1)
(2)
(3)
(4)
2y- 5a; =-19
From (1),
x = 14 — 3 y.
From (2),
5
From (3) and
W,
14
o 19 + 27/
3y = ».
o
y = 3.
Solving (5),
Substituting in (1),
x = 5.
(5)
Check by substituting x = 5, y = 3 in both (1) and (2).
In applying any method of elimination it is desirable first
to reduce each equation to the standard form : ax +- by = c.
See § 119, E. C.
EXERCISES
Solve the following pairs of equations by one of the pro-
cesses of elimination.
3 z +■ 2 ?/ = 118, \5x-8% = 7y-U,
x + 5y = 191. \2x = y+%.
3 (6x-3y = 7, \3x + 7y-341 = 7},y + mx,
\2x—2y = 3. ' l2U-+-J?/ = l.
5x-lly-2=±x, [3?/ + 40 = 2.r+- 14,
7.
5 x - 2 y = 63. 19 y - 347 = 5 x - 420
5y-3x + S = 4y + 2x + 7, f 6 y — 5x = 5 x + 14,
4x-2y=3y + 2. ' \3y-2x-6 = 5 + x.
40
INTEGRAL LINEAR EQUATIONS
f(»+5)(y+7)=(a?+l)(y-9)+112, ±Q \73-7y = ox,
11.
12.
' ax = 6?/,
. x + 2/ = c.
13
la;— ?/ = b.
14.
f a# -\-by = c,
Ia- + gy = h.
15. :
16/
.2^-305 = 12.
f3 _ 5 = ( .
x y
2 3
(SB y
- + - = c,
x y
V+'i=h.
I ■'• y
SOLUTION BY FORMULA
73. We now proceed to a more general study of a pair of
linear equations in two variables.
<2x + 3y = 4,
\5x + 6y = 7.
Ex. 1. Solve
Multiplying (1) by 5 and (2) by 2,
5 • 2 x + 5 • 3 y — 5 • 4,
2 • 5 x + 2 • 6# = 2 • 7.
Subtracting (3) from (4), (2 • 6 - 5 • ■'>)// = 2.7 — 5-4.
Solving for #,
y
2-6-5.3
5 - 4 _ - I)
3
(1)
(2)
(3)
(I)
(•'»)
(6)
In like manner, solving for x by eliminating y,
4 • 6 - 7 • 3 _ J5_ _
2-6-5-3 - 3
we have
-1.
(7)
Ex. 2. In this manner, solving,
\7 x + 9 y = 71,
12 a? + 3 y = 48,
we find
71 . 3 _ 48 . 9 , 7 - 48 -2-71
- and y
7 • 3 - 2 • !>
7-3-2-9
* Let - and be the unknowns.
x y
SOLUTION BY FORMULA 41
In Ex. 2, the various coefficients are found to occupy the
same relative positions in the expressions for x and y as the
corresponding coefficients do in Ex. 1.
Show that this is also true in the following:
Ex.3, pa- + 7// = 10, Ex 4 $5x-3y = 8,
t2x — 5 y = 7. 12 a; + 7 y = 19.
A convenient rule for reading directly the values of the un-
knowns in such a pair of equations may be made from the
solution of the following:
Ex. 5. Solve -i x ' w "
{ a 2 x + b 2 y = c 2 .
Eliminating first y and then x as in Ex. 1, we find:
- j,
- 2 - 1 and y =
To remember these results, notice that the coefficients of x and y in
the given equations stand in the form of a square, thus 1 A and that
the denominator in the expressions for both x and y is the cross
product fljAo minus the cross product a 2 b v The numerator in the
expression for x is read by replacing the a's in this square by the c's,
c b
i.e., l , , and then reading the cross products as before. The numera-
c, b 2
tor for y is read by replacing the b's by the c's, i.e., l l , and then
reading the cross products. 2 2
74. To indicate that the coefficients in a pair of equations are
= a x b 2 — a. 2 b x and
a determinant. These are much used in higher algebra.
to be treated as just described, Ave write J 1
\tto Z>9
call ai H
a 2 b 2 \
Since any pair of linear equations in two unknowns may be
reduced to the standard form as given in Ex. 5, it follows that
the values of x and y there obtained constitute a form ula for
tin- solution of any pair of such equations.
42 INTEGRAL LINEAR EQUATIONS
EXERCISES
Keduce the following pairs of equations to the standard form
and write out the solutions by the formula:
i t3x + ±y = 10, 6 jax-by = 0,
\4:X + y = 9. \x — y=c.
1 4 x — 5 y = — 26, f mas + ny = p,
2 * \2x — 3y= -14. ' lraj + sj/ = i.
f (J j/ - 17 = - 5 a, fa(a; + y) - b(x - y)=2a,
l6a»-16=-5y. ' \a(x-y)-b(x + y)=2b.
fK*-S)=— Ky-2)+i^ o J(fc + l> + (fc-2)y = 3a,
H(y-l)=aj-K*-2). ' l(*+3)aj+(* -4)</ =
2 a; - y = 53, / 2 a.f + 2% = 4cr+ fc 2 ,
19 g + 17 >/ = 0. 1 x - 2 y = 2 a - 6.
J (a + b)x — (a — b)y = 4 a&,
1 (a - &)a? + (a + 6)2/ = 2 a 2 - 2 & 2 .
m (a - 6) - i(a - 3 &) = & - 3,
If (a - 6) + |(a + 6) = 18.
3.
( a.
11.
12.
13.
a(« + y) + 6(a?-y) = 2,
ta{x
\a%x
+ y)- &"(» - .'/) = « - &•
,7(a;-5) =3— 2 — as,
14. I v 2
15.
( mx + ra.it/ = ni : '' + 2 7/r'/; + ji 8 ,
t nx + my = m 3 + 2 inn 2 + n 8 .
f (w + a \x — (m — w)t/ = 2 ?m,
1 (m + Z)as — (in — V)y = 2 win.
17.
19.
| : '., 5 m - 7 n + 2) - fc(3 m - 4 n + 7) = w + 3f,
I i' ( 7 m - 3 n + 4) - ±(G m - 5 n + 7) = w - 2.
r (ft + fc)a, + (ft _ jfe) y = 2(//- + ft 2 ),
1 (ft _ fc)as + (ft + Jc)y = 2(ft 2 - F).
i( a + & _ c ). r + Ka - 6 + n.v = a 2 + (b - c) 2 ,
i(a — 6 + c)x + \ (a + 6 - c)?/ = a 2 — (6 — c)\
INCONSISTENT AND DEPENDENT EQUATIONS 43
INCONSISTENT AND DEPENDENT EQUATIONS
75. A pair of linear equations in two variables may be such
that they either have no solution or have an unlimited number
of solutions.
x-2y = -2, (1)
iSx-6y = -12, (2)
Ex. 1. Solve
On graphing these equations they are found to represent two par-
allel lines. Since the lines have no point in common, it follows that
the equations have no solution. See Fig. 1.
Attempting to solve them by means of the formula, § 73, we find :
ar= (-2)(-6)-(-12)(-2) = -12
l(-6)-3(-2) '
l( -12)-3(-2) _ -6
l(-6)-3(-2) *
and
y =
12 — 6
— - and are not numbers. Hence, from this it
But by § 25,
follows that the given equations have no solution. In this case no
solution is possible, and the equations are said to be contradictory.
>-i
r
•^
Ti
\ v
"X^
v>
+i
■---4
<U>' +1
S
V ,],-;.
^
i
X
+- *"'
Ai
1
_^0 ,
?
5
?!T
: V
).0 : 1 : -
Ss
- y l '-ls r -+-
■ L
ii^ -
Fig. 1.
Fig. 2.
Ex. 2. Solve
r 3 x — 6 y = — 6,
\ x-2y = — 2,
(1)
(2)
On graphing these equations, they are found to represent the same
line. Hence every pair of numbers satisfying one equation must
satisfy the other also. See Fig. 2.
44 INTEGRAL LINEAR EQUATIONS
Solving these equations by the formula, we find :
x = (-6)(-2)-(-2)(-6) = and = 3(-2)-l(-6) =
3(-2)-l(-6) 3 (_2)-l(_6)
But by § 24, - may represent any number whatever. Hence we may
select for one of the unknowns any value we please and find from (1)
or (2) a corresponding value for the other, but we may not select
arbitrary values for both x and y.
In this case the solution is indeterminate and the equations
are dependent; that is, one may be derived from the other.
Thus, (2) is derived from (1) by dividing both members by 3.
76. Two linear equations in two variables which have one
and only one solution are called independent and consistent.
The cases in which such pairs of equations are dependent or contra-
dictory are those in which the denominators of the expressions for x
and y become zero. Hence, in order that such a pair of equations
may have a unique solution, the denominator (1^., — aJ^ of the
formula, § 7-3, must not reduce to zero. This maybe used as a test to
determine whether a given pair of equations is independent and con-
sistent.
EXERCISES
In the following, show both by the formula and by the graph
which pairs of equations are independent and consistent, which
dependent, and which contradictory.
| 5 x — 3 y = 5, \ 3x— 6 y+5 = 2 x — 5 y +7,
I 5 a — 3?/ = 9. ' iox+3!/ — l = 3x + 5y+3.
tx-7 + 5y = y-x-2, 2y + 7x = 2 + 6x,
5x + 3y — 4: = 3x — y+3. ' U»-3y = 4 + 3x-5y.
3 j7x-3y-4 = 2x-2, g (5as-3 = 7y + 8,
\x + y — 3=2x — 7. ' L2a5 + 7=4y — 9.
4 ( X -3y = G, (5x + 2y = 6 + 3x + 5y,
1 5 x — 15^ = 18. ' l3x + y = 18 — 3x + 10y.
g 3y-4x-l = 2x-5y+$, 1Q \3x + 4y = 7 + 5y,
i 2 y — 5 x + 8 = 3 x + //. x —y = 6 — 2 x.
EQUATIONS IN MORE THAN TWO VARIABLES 45
SYSTEMS OF EQUATIONS IN MORE THAN TWO VARIABLES
77. If a single linear equation in three or more variables is
given, there is no limit to the number of sets of values which
satisfy it.
E.g. 3 x + 2 y + 1 z = 21 is satisfied by x = 1, y = 3, z = 3f ; x = 2,
y = 2, z = 3£ ; x = 0, y = 0, 2 = 6; etc.
If two linear equations in three or more variables are given,
they have in general an unlimited number of solutions.
E.g. 3 x + 2 y + I z = 21 and a; + y + z = 6 are both satisfied by
x = 2, y = — 1, 2 = 5 ; a; = 3, y = — l£, 2 = 4^ ; etc.
But if a system of linear equations contains as many equa-
tions as variables, it has in general one and only one set of
values which satisfy all the equations.
fx + y + 2 = 6,
E.g. The system -j 3 x — y -\- 2 z = 7,
[2 x + 3 y - z = 5,
is satisfied by a; = 1, y = 2, 2 = 3, and by no other set of values.
It may happen, however, as in the case of two variables,
that such a system is not independent and consistent.
Such cases frequently occur in higher work, and a general rule is
there found for determining the nature of a system of linear equa-
tions without solving them : namely, by means of determinants (§ 7:5). In
this book the only test used is the result of the solution itself as
explained in the next paragraph.
78. An independent and consistent system of linear equa-
tions in three variables may be solved as follows :
From two of the equations, say the 1st and 2d, eliminate one of
the variables, obtaining one equation in the remaining two variables.
From the 1st and 3d equations eliminate the same variable, ob-
taining a second equation in the remaining two variables.
Solve as usual the two equations thus found. Substitute the
values of these two variables in one of the given equations, and thus
find the value of the third variable.
The process of elimination by addition or subtraction is usually
most convenient. See § 120, E. C.
46
INTEGRAL LINEAR EQUATIONS
EXERCISES
Solve each of the following systems and check by substitut-
ing in each equation :
x + 5 y — i
2 = 9,
5 x — y + 3 z = 16,
[T x -\- 6 y A- z = 34.
r 8 z - 3 y + x = - 2,
3 x — 5 y — 6 z = — 46,
[y + o x — 4 z = — 18.
[a + 6 + c = 9,
8a + 46 + 2c = 36,
.27 a + 96+3c = 98.
* Use , 7, and - as the unknowns.
a b c
f3 = 2
a b
6.* J
- + r-- = 17,
a b c
3 6
6 c
(18 I — 7 wi — 5 7i , = 161,
1 1 in. — | Z + n = 18,
31 h + 2 m + f Z = 33.
7.
(x + 2y-3z =
> 2 x — 3 ?/ + z =
> .r — 4 y — 7 z
:i + (i + 9
4 i« + & + _ c _ = _4
< 6 ^ 9 ^ 12
9 12 15
fa; + »/ = 16,
9. z + .r = 22,
U + 2 = 28.
f.r + 2y = 26,
10. j3aj + 4z = 56,
[5 y + 6 2 = 65.
[2.T + 3>/-7z = 19,
8. 5ar + 82/ + llz=24,
U x + lly + 4z= 43.
Show that this system is not
independent.
r a + 6 + c = 5,
11. Is a — 5 b + 7 c = 79,
9 a - 11 b = 91.
f I + «i + n = 29£,
12. / + ///- ra = 18i
[l- m + n = L3f.
EQUATIONS IN MOBE THAN TWO VARIABLES 47
13.
14.
I + m + n = a,
, / + m — n = b,
[l — m + n = c.
r 7/ + clz = q,
[ex +fz = r.
15.
a o
= 4,
a c
= 3,
7 + l -
b c
= 2.
16. I
17.
u + 2v + 3x + 4y = 30,
2 u + 3 v + 4 x + 5 y = 40,
3 « + 4 i> + 5 .« + 6 y = 50,
4 u + 5 v + 6 x + 7 ^/ = 60.
= a,
= &,
2/ 2
= c.
18. Make a rule for solving a system of four or more linear
equations in as many variables as equations.
PROBLEMS INVOLVING TWO OR MORE UNKNOWNS
1. A man invests a certain amount of money at 4% inter-
est and another amount at 5%, thereby obtaining an annual
income of $3100. If the first amount had been invested at
5% and the second at 4%, the income would have been
$3200. Find each investment.
2. The relation between the readings of the Centigrade
and the Fahrenheit thermometers is given by the equation
F = 32 + f- C. The Fahrenheit reading at the melting tem-
perature of osmium is 2432 degrees higher than the Centigrade.
Find the melting temperature in each scale.
In the Reaumur thermometer the freezing and boiling points are
marked 0° and 80° respectively. Hence if C is the Centigrade reading
and R the Reaumur reading, then R = ^ C. See § 101, E. C.
3. "What is the temperature Fahrenheit (a) if the Fahren-
heit reading equals ^ of the sum of the other two, (b) if the
Centigrade reading equals | of the Fahrenheit minus the Reau-
mur, (c) if the Reaumur is equal to the sum of the Fahrenheit
and Centigrade ?
48 INTEGRAL LINEAR EQUATIONS
4. Going with a current a steamer makes 19 miles per
hour, while going against a current -| as strong the boat makes
5 miles per hour. Find the speed of each current and the boat.
5. There is a number consisting of 3 digits whose sum is 1 1.
If the digits are written in reverse order, the resulting number
is 594 less than the original number. Three times the tens' digit
is one more than the sum of the hundreds' and the units' digit.
6. A certain kind of wine contains 20 % alcohol and another
kind contains 28%. How many gallons of each must be used
to form 50 gallons of a mixture containing 21.6 r / alcohol ?
7. The area of a certain trapezoid of altitude 8 is 68. If 4
is added to the lower base and the upper base is doubled, the
area is 108. Find both bases.
A trapezoid is a four-sided figure whose upper base b v and lower
base, b„, are parallel, but the other two sides are not. If h is the per-
pendicular distance between the bases, then the area is a = - (/> r + b„).
8. If on her second westward journey the Lusitania had
made 1 knot more per hour, she w 7 ould have crossed in 4
hours and 38 minutes less than she did. But if her speed had
been 4 knots per hour less, she would have required 23 hours
ami 10 minutes longer. Find the time of her passage and her
average speed if the length of her course was 2780 knots.
9. Aluminium bronze is an alloy of aluminium and copper.
The densities of aluminium, copper, and aluminium bronze are
2.6, 8.9, and 7.69 respectively. How many com. of each
metal arc used in 100 ccm. of the alloy ? See § 99, E. C.
10. Wood's metal, which is used in fire extinguishers on
account of its low melting temperature, is an alloy of bismuth,
lead, tin, and cadmium. In 120 pounds of Wood's metal, -| of
the tin plus '. of the lead minus ., 1 „ of the bismuth equals
7 pounds. If \ of the lead and | of the tin be subtracted
from the bismuth, the remainder is ll' pounds. Find the
amount of each metal if 15 pounds of cadmium is used.
PROBLEMS IN TWO OR MORE VARIABLES 49
11. The upper base of a trapezoid is G and its area is 168.
If i the lower base is added to the upper, the area is 210.
Find the altitude and the lower base.
12. A and B can do a piece of work in 18 days, B and C in
24 days, and C and A in 36 days. How long will it require
each man, working alone, to do it, and how long will it require
all working together ?
13. A and B can do a piece of work in m days, B and C in n
days, and C and A in p days. How long will it require each to
do it working alone?
14. A beam resting on a fulcrum balances when it carries
weights of 100 and 130 pounds at its respective ends. The
beam will also balance if it carries weights of 80 and 110
pounds respectively 2 feet from the ends. Find the distance
from the fulcrum to the ends of the beam.
15. A beam carries three weights, A, B, and C. A balance
is obtained when A is 12 feet from the fulcrum, B 8 feet from
the fulcrum (on the same side as A), and C 20 feet from the
fulcrum (on the side opposite A). It also balances when the
distance of A is 8 feet, B 10 feet, and O 18 feet. Find
the weights B and C if A is 50 lbs.
16. At 0° Centigrade sound travels 1115 feet per second
with the wind on a certain day, and 1065 feet per second
against the wind. Find the velocity of sound in calm weather,
and the velocity of the wind on this occasion.
17. If the velocities of sound in air, brass, and iron at 0°
Centigrade are x, y, z meters per second respectively, then
3x+2y— z = 2505, 5x — 2y + z = 151, and x + y + z = 8777.
Find the velocity in each.
18. If x, y, z are the Centigrade readings at the temperatures
which liquefy hydrogen, nitrogen, and oxygen respectively,
then 3 x-Sy+2z= 440, - 8 x + 2 ?/ + 4 z = 903, and x + 4 y
— 62 = 60. Find each temperature in both Centigrade and
Fahrenheit readings.
50 INTEGRAL LINEAR EQUATIONS
19. Two trapezoids have a common lower base. Their alti-
tudes are 8 and 10 respectively, and the sum of their areas is
148. If the upper base of the first trapezoid is multiplied by 2
and that of the second divided by 2, their combined area is
ir»i' ; while if the upper base of the first is divided by 2 and
that of the second multiplied by 2, the combined area is 17<<.
Find the bases of each trapezoid.
20. If x, y, z are the Centigrade readings at the freezing
temperatures of hydrogen, nitrogen, and oxygen respectively,
then we have x + y — 3 z = 199, 2 x — 5 y + z = 328, and
— 4 x + 2 y + 2 z = 156. Find each temperature.
21. If x, y, z are respectively the melting point of carbon.
the temperature of the hydrogen flame in air, and the tempera-
ture of this flame in pure oxygen, then 10 x -+- 2 y + z = 41,892,
15 x _|_ y + 2 z =60,212, and 7 x ±y + z = 29,368. Find each.
22. If a, b, c are the values in millions of the mineral
products of the United States in 1880, 1900, and 1906 re-
spectively, find each from the following relations :
k b c i — () a . b . c rrn b . c
5 a = 1oi2, - -\ h - = 669, a \- - = 37.
8 10 '345 2 7
23. If x, y, z represent in thousands of tons the steel
products of the United States in 1880, 1890, and 1905, find
each from the following relations :
x + y + z = 25,547, 3x + iy-z = 826, x -3y + z = 8 139.
24. If the number of millions of tons of coal mined in the
United States in 1890, L900, and L906 be represented by x, y, z
respectively, find each from the following relations:
E + 1L + A =1 05, x- v + — = 153, 3 x + 2 y - 2 z = 164.
2 30 25 9 17
25. If the values in millions of the farm products of the
United States in 1870, L90<>, and 190<; arc represented by 1. m.
and n respectively, find each from the following relations :
2 1 + m - n = 1633, 3 1 — 2 m + n= 3440, I + in + n = 13,675.
PROBLEMS IN TWO OR MORE VARIABLES 51
26. The sum of the areas of two trapezoids whose altitudes
are 10 and 12 respectively is 284. If the upper base of the
first is multiplied by 3 while its lower is decreased by 2, and
the upper base of the second is divided by 2 while its lower
base is increased by 3, the sum of the areas is 382 ; if the
upper bases of both are doubled and the lower bases of both
divided by 2, the sum of the areas is 322; and if the upper-
bases are divided by 2 while the first lower is doubled and the
second trebled, the sum of the areas is 388. Find the bases of
each trapezoid.
27. Two boys carry a 120-pound weight by means of a pole,
at a certain point of which the weight is hung. One boy
holds the pole 5 ft. from the weight and the other 3 ft.
from it. What proportion of the weight does each boy lift ?
Solution. Let x and y be the required amounts, then 5x is the
leverage of the first boy and 3 y that of the second, and these must be
equal as in the case of the teeter, p. 122, E. C. Hence we have
5 x = 3 y, and x + y = 120.
Solving, we find x = 45, y = 75.
28. If, in problem 27, the boys lift P x and P 2 pounds respec-
tively at distances d x and d 2 , and w is the weight lifted, then
PA = P.A, (1)
P 1 + P 2 = w. (2)
Solve (1) and (2), (a) when P x andP 2 are unknown, (b) when
Pj and w are unknown, (c) when P x and d 2 are unknown.
29. A weight of 540 pounds is carried on a pole by two men at
distances of 4 and 5 feet respectively. How much does each lift ?
30. A weight of 470 pounds is carried by two men, one at a
distance of 3 feet and the other lifting 200 pounds. At what
distance is the latter ?
31. Two men are carrying a weight on a pole at distances of
4 and 6 feet respectively. The former lifts 240 pounds. How
many pounds are they carrying?
CHAPTER V
FACTORING
79. A rational integral expression is said to be completely
factored when it cannot be further resolved into factors which
are rational and integral. Such factors are called prime factors.
The simpler forms of factoring are given in the following
outline.
A monomial factor of any expression is evident at sight,
and its removal should be the first step in every case.
E.g. 4 ax 1 + 2 a-x = 2 ax(2 x + a).
FACTORS OF BINOMIALS
80. The difference of two squares.
E.g. 4 x 2 - 9 z* = (2 x + 3 z 2 ) (2 x - 3 z 2 ) .
81. The difference of two cubes.
E.g. 8 a? - 27 f = (2 x - 3 y) [ (2 x) 2 + ( 2 x) (3 //) + (3 y) 2 ]
= (2 x - 3 //) ( 1 .r- + 6 xy + 9 //-).
82. The sum of tv:o cubes.
E.g. 27 z 8 + 6 1 f = (3 x + 4 y) [(3 sc) 2 - (:*> x) ( 1 //) + (1 //)'-]
= (3a; + l//)(!i./- : - 12 xy + 10 if).
FACTORS OF TRINOMIALS
83. Trinomial squares.
E.g. a 2 + 2 ab + ft 2 = (o + 6) 2 = (a + &)(a + A),
ami a 2 — 2 afi + 6 2 = (a — /;) 2 = (a — 6)(a — /;).
52
FACTORS OF TRINOMIALS 53
84. Trinomials of the form x 2 -\-px + q.
E.g. x 2 + 3 x - 10 = (x + 5) (x - 2).
A trinomial of this form has two binomial factors, x + a and x + b,
if two numbers a and b can be found whose product is q and whose
algebraic sum is p.
85. Trinomials of the form mx 2 + nx + r.
£.<7. 6 a: 2 + 7 j; - 20 = (3 x - 4)(2 x + 5).
A trinomial of this form has two binomial factors of the type
ax + 6 and ex + ^/, if four numbers, a, b, c, d, can be found such that
ac = m, bd — r, and ad + be = n. See § 142, E. C.
86. Trinomials which reduce to the difference of two squares.
E.g. x 4 + x 2 y 2 + if = x 4 + 2 x 2 y 2 + v/ 4 - x 2 y 2 = (x 2 + y-) 2 - x 2 y' 2
= (x 2 + if - xy)(x 2 + y- + xy).
In this case xhf is both added to and subtracted from the expres-
sion, whereby it becomes the difference of two squares. Evidently
the term added and subtracted must itself be a square, and hence the
degree of the trinomial must be 4 or a multiple of 4, since the degree
of the middle term is half that of the trinomial.
Ex. 4 a 8 - 16 a 4 Z/ 4 + 9 b s = 4 a s - 12 aHA + 9 b s - 4 a 4 b*
= (2 a 4 - 3 ft 4 )' 2 - 4 a*h*
= (2 a 4 - 3 b* + 2 a 2 b 2 ) (2 a 4 - 3 ft 4 - 2 a a 6 a ).
EXERCISES ON BINOMIALS AND TRINOMIALS
Factor the following :
1. a s + b 3 . 5. 7 ace 2 - 56 aV. 9. | ^ — -^ rs 2 .
2. a 3 — 6 3 . 6. a 5 -a6 4 . 10. 8r 4 — 27 r.
3. (a + 6) 8 -c 8 . 7. 121 a 7 -4 as/ 4 . 11- (a + &) 2 -c 2 .
4. ( a + &) 8 + c*. 8. £a 8 + T k 6 '- 12- c 2 -(a-&) 2 .
13. 5 c 2 + 7 erf — 6 d 2 . 15. 4 a; 2 — 12 a$ + 9 ?/ 2 .
14. a? 4 — 3afy* + y*. 16. a? 2 + 11 ax? + 30 z 2 .
54 FACTORING
17. G .?•'- — 5 xy — G y 2 . 24. a 2 4- 10 a — 39.
18. 3 aVy 4 - 69 a 2 a«/ 2 4- 336 a 2 . 25. 8 afy 3 - 48 a 2 y 2 z + 72 a 2 */* 2 .
19. 20a 2 6 2 + 23a&a;-2l!B 2 . 26. 4 m 8 - 60 mV + 81 n*.
20. a 4 + 2 a 2 b 2 + 9 6 4 . 27. 35 a 2 * - 6 aV- 9 ft 2 *.
21. 48 crViy - 75 a/. 28. (a + bf— (c — d f.
22. 16 a 4 .r// + 5 I ay 4 . 29. 72 aW - 19 aay 2 - 40 ,y 4 .
23. aj 4 2/ 2 + 2a; 2 2/«4-2 2 . 30. 4(a-3) 6 -376 2 (a-3) 3 +96 4 .
31. 6(s + y)«+6(a*-y»)-6(a!-y)»
32. 9(as -a) 2 - 24(aj - a) (as + a) + 16(as + «) 2 .
33. 12(c + rf) 2 - 7(c + d)(c-d)— 12(c — d) 1 .
34. (a 2 + 5 a — 3) 2 - 25(a 2 + 5 a — 3) + 150.
FACTORS OF POLYNOMIALS OF FOUR TERMS
A polynomial of four terms may be readily factored in case
it is in any one of the forms given in the next three paragraphs.
87. It may be the cube of a binomial.
Ex.1, a 3 - 3 a?b + 3ab"--b 3 = (a - bf.
Ex. 2. 8 x* 4- 36 x 2 y + 54 xtf + 27 f
= (2x)« + 3(2x)*(3*/) + 8(2x)(8y)» + (3y)«
= (2x + 3y) 8 . See Ex. 34, (rf), p. 23.
88. It may be resolvable into the difference of two squares.
In this case three of the terms must form a trinomial square.
Ex. 1. a 2 - c 2 + 2 a& + & 2 = (a 2 + 2 a& + & 2 ) - c 2
= (a + &) 2 - c 2 = (a + 6 + c)(a + & - c).
Ex. 2. 4 a; 2 + z c ' - 4 a' 4 - 1 = z G - 4 x* + 4 .jr - 1
= z 6 - (4x- 4 - 4x 2 + 1) = z & - (2x 2 - I) 2
= (; , + 2x 2 -l)(c :i -2/-+ 1).
FACTORS FOUND BY GROUPING 55
89. A binomial factor may be shown by grouping the terms.
In this case the terms are grouped by twos as in the following
examples.
Ex. 1. ax 4- ay + bx 2 4- bxy = (ax 4- ay) 4- (bx 2 4- bxy)
= a(x + y) + ix(x + y) = (a + for) (a: + y).
Ex. 2. aa; + 6oj + a 2 — 6 2 = (ax + 6a) + (a 2 - b 2 )
- x(a + ft) + (a - ft)(a + ft) = (x + a - ft)(a + ft).
EXERCISES
Factor the following polynomials:
1. a' 3 + 3 x' : y 4- 3 cc?/ 2 + 2/ 3 . 8. a 2 b 2 — a 2 bc 2 n — abn 4- tm 2 c 2 .
2. 8 a 3 - 36 a 2 b + 54 ab 2 - 27 6 3 . 9. 2y 2 + <Lby + 3cy + 6bc.
3. 4 a 4 — 4 a 2 6 2 + 6 4 — 16 x 2 . 10. bcyz + c 2 z 2 + bdy + dcz.
4. 2 aa" 4- 3 6c 4- 2 ac 4- 3 6a\ 11. 5a 2 c + 12ca" — 6 ad— 10 ac 2 .
5. 27 x 3 — 54 a% 4- 36 xy 2 — 8 ?/ 3 . 12. a 2 — 6 2 .i' 2 + acx 2 — bcx 5 .
6. 36 a 4 -24 a 3 + 24 a -16. 13. b z c 2 - chf - Ihf + y 5 .
7- mx.v 2 — mrx — rn 2 x + /•'-'// . 14. a 3 * — 2 a 2 *6* — 2 a*6 24 4- b 3k .
15. m a+i + m a »" 4- m b n b + //" f6 .
1 6. b 2 [f — b(c — d) y 2 4- d (by - c) 4- d 2 .
FACTORS FOUND BY GROUPING
90. The discovery of factors by the proper grouping of terms
as in § 89 is of wide application. Polynomials of five, six, or
more terms may frequently be thus resolved into factors.
Ex. 1. a 2 4- 2 ab 4- b 2 4- 5 a 4- 5 b = (a 4- 6) 2 4- (5 a 4-56)
= (a + ft) (a + ft) + 5(a + ft) = (« + b + 5) (a + ft).
Ex. 2. x 2 — 7 a; + 6 — ax 4- 6 a = or — 7 x + 6 — (aa; — 6a)
= (.r- l)(x - 6)- a(x- 6) = (x - 6) (a; - 1 -a).
Ex. 3. a 2 - 2 a6 + 6 2 - a 2 - 2 .17/ - ?/ 2
= (a 2 - 2 aft + ft 2 ) - (r- + 2 .»•// + //-)
= (a - ft) 2 - (a - y) 2 = (a - b + x - y) (a - ft - x + y) .
56 FACTORING
Ex. 4. ax 2 4- ax — 6 a + sc 2 4 ~ ■>" + 12
= «(./•- + x - 6) + (./- 4 7x4 12)
= a(z + 3)0 - 2) + (x + 3)(.r + 4)
= (•'• + •"') L" ( ■'• --')+ ->- + 4] = (-r + 3) (a* - 2 a + x + 4) .
In some cases the grouping is effective only after a term has
been separated into two parts.
Ex. 5. 2 a 3 + 3 a 2 + 3 a . + 1 = a? + (a 8 + 3 a 2 + 3 a + 1)
= a 3 4(a + l) 8 = (a + a+l)[a 2 -a(a + l) + (a + 1)-]
= (2a + l)(a 2 + a + l).
As soon as the term 2 o 3 is separated into two terms the expression is
shown to be the sum of two cubes.
Again, the grouping may be effective after a term has been
both added and subtracted :
Ex. G. a 4 + 1/ = (a 4 + 2 a 2 6 2 +b*)-2 a 2 b 2
= (a 2 + 6 2 ) 2 -(a&V2) 2
= (a 2 + b- + ab \/2)(a 2 + //- - a&V2).
Tn tliis ease the factors arc irrational as to one coefficient. Such
factors are often useful in higher mathematical work.
EXERCISES
Factor the following :
1. x 2 — 2 xy + y 2 — ax + ay. 3. a 8 — 6 s — a 2 — a6 — ft 2 .
2. a 2 — a& + 6 2 + a 3 4- 6 8 . 4. a- — 2ab+b 2 — x 2 +2xy— y 2 .
5. a 4 + 2 a s 6 — a 2 c 2 + a 2 & 2 — 2 a&c 2 — ft 2 © 2 .
6. .r 4 - // 4 + aa; 2 + ay 2 — -r — //'-'. 7. a 4 + a 2 b 2 + b 4 + a 3 4- & 3 .
In 7 group the first three and the last two terms.
8. a 3 — 1 4- 3 x — 3ar + 3r 3 - ( I roup the last four terms.
9. x s 4- as 2 4- 3 x + y 3 — y 2 4- 3 y.
Gron;> in pairs, the 1st and 1th. 2d and 5th, 3d and 6th terms.
10. x* 4- x*y — xy 3 — y 4 4- •' ,: ' — y 3 .
1 1 . a 4 + -1 <fb + 6 ((-//- 4- 4 ab" + b 4 - x*.
THE FACTOR THEOREM 57
12. x 4 + 4 x 2 z — 4 y- + 4 ?/zo + 4 z 2 — to 2 .
13. 2 a 2 - 12 b 2 + 3 6c? - 5 aft - 9 6c- 6 ac + 2ad.
Group the terms : 2 a 2 — 5 ai — 12 6 2 .
14. a 2 + a6 - 4 ac — 2 6 2 + 4 be + 3 ad - 3 bd.
15. a 3 4- 2 - 3 a, Group thus : (a 3 - 1) + (3 - 3 a).
16. 4 a 2 + a — 8 ax — a; + 4 a; 2 .
17. 3 a 2 — 8 ab + 4 b 2 + 2 ac — 4 6c.
18. £ 8 + ?/ s , also a; 16 + ?/ 1<; .
19. a G + 2 a 3 6 3 + b 6 - 2 a 4 6 - 2 a,6 4 .
20. a 3 - 3 a 2 + 4. Group thus : (a 3 - 2 a 2 ) + (4 - a 2 ).
21. a 2 c — ac 2 — a 2 6 + ab 2 — 6 2 c 4- 6c 2 .
22. orb — a 2 c + b 2 c — a6 2 4- ac 2 — 6c 2 .
23. 3 X s — x 2 — 4 x- 4- 2. Add and subtract - 2 x 2 .
24. 2 x 3 — 11 x 2 + 18 x - 9. Add and subtract 9 x 2 .
FACTORS FOUND BY THE FACTOR THEOREM
91. It is possible to determine in advance whether a poly-
nomial in x is divisible by a binomial of the form x — a.
E.g. In dividing x 4 — 4 x 3 + 7 x- — 7 x + 2 by x — 2, the quotient is
found to be x 3 — 2 x 2 + 3 x — 1.
Since Quotient x Divisor = Dividend, we have
(x-2)(x 3 -2x 2 + 3x- l) = x 4 -4x 3 + 7x 2 -7x + 2.
As this is an identity, it holds for all values of x. For x = 2 the fac-
tor (x — 2) is zero, and hence the left member is zero, § 22.
Hence for x = 2 the right member must also be zero. This is in-
deed the case, viz. :
04 _ 4 . 03 + 7 . 2 2 - 7 • 2 + 2 = 16 - 32 + 28 - 14 + 2 = 0.
Hence if x - 2 is a factor of x 4 - 4 x s + 7 x 2 - 7 x + 2, the latter
must reduce to zero for x = 2.
58 FACTORING
92. Theorem. If a polynomial in x reduces to zero when
a particular number a is substituted for x, thru x — a is a
factor of the polynomial, and if the substitution of a for
x does not reduce the polynomial to zero, then x — a is not
a factor.
Proof. Let D represent any polynomial in x. Suppose D lias been
divided by x — a until the remainder no longer contains x. Then,
calling the quotient Q and the remainder R, we have the identity
D=(2(x-a)+R, (1)
which holds for all values of x.
The substitution of a for x in (1) does not affect R, reduces
Q(x — o) to zero, and may or may not reduce D to zero.
(1) If x = a reduces D to zero, then = + R. Hence R is zero,
and the division is exact. That is, x — a is a factor of D.
(2) If x = a does not reduce D to zero, then R is not zero, and the
division is not exact. That is, x — a is not a factor of D.
93. In applying the factor theorem the trial divisor must
always be written in the form x — a.
Ex. 1 . Factor x i + 6 x"' + 3 x 2 + x + 3.
If there is a factor of the form x — a. then the onty possible values
of a are the various divisors of '■>. namely + 1, — 1, + ; >, — ; >-
To test the factor x + 1, we write it in the form x — (— 1) where
a — — I. Substituting — 1 for x in the polynomial, we have
1-6 + 3-1 + 3 = 0.
Hence x + 1 is a factor.
On substituting + 1, + 3, — 3 for x successively, no one reduces the
polynomial to zero. Hence x — 1, x — 3, x + 3 are not factors.
Ex. 2. Factor 3 a? -x 2 -A x + 2.
If x — a is a factor, then a must be a factor of + 2. We therefore
substitute, +2, —2. + 1, —1 and find the expression becomes zero
when 4-1 is substituted for a;. Hence x — 1 is a factor. The other
factor is found by division to b<> 3 .'- + 2x — 2. which is prime.
1 [ence 3 ,3 _ X 2 _ 4 x + 2 = (x - 1 ) (3 x- + 2 x - 2).
THE FACTOR THEOREM 59
EXERCISES
Factor by means of the factor theorem :
1. 3x i — 2x 2 + 5x — 6. 6. ra 3 + 5m 2 + 7ra + 3.
2. 2x s + 3x 2 -3x-4. 7. x 4 + 3x 5 -3x 2 -7x + 6.
3. 2ar 3 + ar 9 -12a: + 9. 8. 3r J + 5r 2 -7r-l.
4. a 3 + 9 a- 2 + 10 a: + 2. 9. 2 z 3 + 7 z 2 + 4z + 3.
5. a 3 -3a + 2. 10. a 3 - 6 a 2 + 11 a -6.
11. Show by the factor theorem that x k — a k contains the
factor x — a if A; is cm?/ integer.
12. Show that x k — a k contains the factor x + a if & is any
even integer.
13. Show that x k + a k contains the factor x-\-a if Tc is any
ocW integer.
14. Show that x k -\-a k contains neither x-\-a nor x — a as a
factor if k is an even integer.
MISCELLANEOUS EXERCISES ON FACTORING
1. 20a 3 .i%-45a 3 .^ 3 . 4. 16 x 2 - 72 xy + 81 y 2 .
2. 24 am 5 ri 2 - 375 am¥. 5. 162 a 3 6 + 252 a 2 6 2 + 98 a& 3 .
3. 432 a) A s + 54 ars\ 6. 48 afy - 12 afy- 12 a; 2 ?/ +3?/.
7. 12 a 2 te 2 + 8 a&V + 18 a 2 6.r?/ + 12 ab 2 xy.
8. 18 x 3 #- 39 a; 2 ?/ 2 + 18 ay 3 . 16. a s -f. 17. a} 6 -y™.
9. 4ar-9a,v/+6a;-9?/+4a;+6. 18. a 8 + «Y + ?A
10. 6a; 2 -13a-?/ + 6?/ 2 -3.r+2y. 19. a s + a — 2.
11. 6 a 4 + 15 .1/7/° + 9 ?/ 4 . 20. cf — lSaY + y*.
12. 16.« 4 + 24a-7/ 2 + 8 7/ 4 . 21. a I6 -6«y + f.
13. 15 a? 4 + 24 x 2 y 2 + 9 // 4 . 22. .ir 3 + 4ar + 2a — 1.
14. a fi + ?/ 6 . 15. a l2 + y V2 . 23. 3 ar 3 + 2 s: 2 — 7 a; + 2.
60 FACTORING
24. a 8 -3ay + ?/ 8 . 26. a 3 + 9 a 2 + 16 a + 4.
25. a 8 + a 2 + a + l. 27. 2 z 4 + or 3 ?/ + 2 a,- 2 ?/ 2 + a;?/ 3 .
28. ?« 5 + m^i -{- m 8 a 2 + m 2 a? 4- ??ia 4 + a 5 .
29. (x-2y-(y-z) s .
30. a 6 + b G 4- 2 a6(a 4 — a 2 b 2 + & 4 ).
31. afy 5 + x l y 4 z + O/" 3 ^ 2 + xhfz* -j- a*?/z 4 + 2 s .
32. 8a s + 6ab(2a-3b)-27b 3 .
33. a (jb 3 + f)- ax (x 2 - y 2 ) -y\x + ?/).
34. a 3 - & 3 + 3 b-c -3bc 2 + c 8 .
35. a 4 + 2 a 8 6 — 2 ab 2 c - b 2 c 2 .
36. a 4 + 2 a 3 & + a 2 & 2 - a'b 2 - 2 a 2 & 2 c - &V.
SOLUTION OF EQUATIONS BY FACTORING
94. Many equations of higher degree than the first may be
solved by factoring. (See §§ 144-146, E. C.)
Ex. 1. Solve 2 ar 3 - x 2 - 5 x - 2 = 0. (1)
Factoring the left member of the equation, we have
(ar-2)(ar+l)(2x+l) = 0. (2)
A value of x which makes one factor zero makes the whole left
member zero and so satisfies the equation. Hence x = 2, x = — 1,
x = — \ are roots of the equation.
To solve an equation by this method first reduce it to the form
-4=0, and then factor the left member. Put each factor equal to
zero and solve for x. The results thus obtained are roots of the
original equation.
Ex. 2. Solve ar 3 - 12 x 2 = 12 - 35 x. (1)
Transposing and factoring, (x — 4)(x 2 — 8 x + 3) = 0. (2)
I Icnce the roots of (1) are the roots of x — 4 = and x 2 — Sx + 3 = 0.
From x — 4 = 0, x = 4. The quadratic expression x' 2 — 8 x + 3 can-
not be resolved into rational factors. See § 155.
COMMON FACTORS AND MULTIPLES 61
EXERCISES
Solve each of the following equations by factoring, obtain-
ing all roots which can be found by means of rational factors.
1. x 3 + 3 x 2 = 28 x. 6. 2x 3 + 3x = 9x 2 -14.
2. 6x s + 8x-\-5 = 19x 2 . 7. 5x 3 + x 2 -14x + 8 = 0.
3. x 4 + 12 x 2 + 3 = 7x 3 + 9x. 8. 2 x 3 -f-af' = 14 a; — 3.
4. 12x 3 = 20x 2 + 5x + 6. 9. 12 x 4 + U x 3 +l = 3 .r 2 + 4 x.
5. x 3 -4r = 4.i- + 5. 10. x 5 -4x 4 -40x 3 +6-x=58x 2 .
COMMON FACTORS AND MULTIPLES
95. If each of two or more expressions is resolved into
prime factors, then their Highest Common Factor (H. C. F.) is at
once evident as in the following example. See § 182, E. C.
Given (1) x 4 - y 4 = (x 2 + y*)(x + y)(x - y),
(2) x 6 -if = (x 3 + >f) (x 3 - ?/ 3 )
= (x + y)(x 2 - xy + i/)(x - ij)(x 2 + xy + if).
Then (x + y)(x — y) = x 2 - >f is the II. C. F. of (1) and (2).
In case only one of the given expressions can be factored by
inspection, it is usually possible to select those of its factors,
if any, which will divide the other expressions and so to deter-
mine the H. C. F.
Ex. Find the H. C. F. of 6 .f 3 + 4 x 2 - 3 x - 2,
and 2 x 4 + 2 x 3 + x 2 — x — 1.
By grouping we rind :
6 x 3 + 4 x 2 - 3 x - 2 = 2 x 2 (3 x + 2) - (3 x + 2)
= (2x 2 - l)(3x + 2).
The other expression cannot readily be factored by any of the
methods thus far studied. However, if there is a common factor, it
must be either 2 x 2 — 1 or 3 x + 2. We see at once that it cannot be
3 x + 2. (Why ?) By actual division 2 x 2 — 1 is found to be a factor
of 2 x 4 + 2 x 3 + x 2 - x - 1. Hence 2 x 2 - 1 is the H. C. F.
62 FACTORING
96. The Lowest Common Multiple (L. C. M.) of two or more
expressions is readily found if these are resolved into prime
factors. See § 185, E. C.
Ex. 1. Given 6 abx - 6 aby = 2-3 ab(x - y), (1)
SaPx + Sa i y = 2 s a 2 (x + IJ ), (2)
3G b%x? - y*)(x + y) = 2-&V(x -y){x + y)\ (3)
The L. CM. is 2 3 ■ S 2 a 2 b s (x - y)(x + y) 2 , since this contains all
the factors of (1), all the factors of (2) not found in (1), and all
the factors of (3) not found in (1) and {'2), with no factors to spare.
In case only one of the given expressions can be factored by
inspection, it may be found by actual division Avhether or not
any of these factors will divide the other expressions.
Ex. 2. Eind the L. C. M. of 6 x* - x 2 + 4 x + 3, (1)
and 6 s + 3 x 2 - 10 x - 5. (2)
(1) is not readily factored. Grouping by twos, the factors of (2)
are 3 x 2 — 5 and 2 x + 1. Now '■'> ./- — 5 is not a factor of (1). (Why ?)
Dividing (1) by 2 x + 1 the quotient is 3 x 2 — 2 x + 3.
Hence 6 x 3 - x 2 + 4 x + 3 = (2 x + 1) (3 x 2 - 2 x- + 3),
6 x 3 + 3 x 2 - 10 x - 5 = (2 x + 1) (3 x 2 - 5).
Hence the L. C. M. is (2 x + 1) (3 x 2 - 2 x + 3) (3 x 2 - 5).
Ex. 3. Find the L. C. M. of a? + 2 a- -a- 2, (1)
and l0a 3 -3« 2 + 4a + l. (2)
By means of the factor theorem, a — 1, a + 1, and a + 2 are found
to be factors of (1), but none of the numbers, 1, — 1, — 2. when sub-
stituted for a in ("J) will reduce it to zero. Hence (1) and (2) have
no factors in common. The L. CM. is therefore the product of the
two expressions ; viz. (a + \)(<t - l)(a + 2)(10« 3 - 3a 2 + 4 a +1).
97. The H.C. F. of three expressions may be obtained by
finding the H. C. F. of two, and then the H. C. F. of this result
and the third expression. Similarly the L.C.M. of three expres-
sions may be obtained by finding the L. C. M. of two of them,
and then the L. C. M. of this result and the third expression.
This may be extended to any number of expressions.
COMMON FACTORS AND MULTIPLES 63
EXERCISES
Find the H. C. F. and also the L. C. M. in each of the
following :
1. x' 2 4- y 2 , x 6 + ?/ 6 . 2. x- + xy 4- y 2 , x 3 - y 3 .
3. x 2 — 5x — 6, a; 2 -2.x- -3, a; 2 4- 19 .» 4- 18.
4. x 4 — 6.r 2 4-l, ar 3 4-ar — 3.c + l, £ 3 4-3ar4-# — 1.
5. 162a% + 252o 2 6 2 + 9a6 3 , 54a 3 + 42a 2 6.
6. 2.r 3 + x 2 -8a- + 3, a 2 + 2a;-l.
7. 3r 3 + 5?- 2 -7r-l, 3r 2 +8r + l.
8. a 3 — 3 a 2 + 4, ax — ab — 2x + 2b.
9. a 6 + 2 a 3 6 3 + 6 6 - 2 a 4 6 - 2 a& 4 , a 3 -2ab + b 3 .
10. 8 a 3 - 36 a 2 6 + 54 c^ 2 - 27 6 3 , 4a 2 -9 6 2 .
11. 36 o 4 - 9 a 2 -24 a -16, 12 « 3 - 6 a 2 -8 a.
12. 2 if + 4% + 3cy + 6 be, y 2 - 3 by -10 b 2
13. cc 16 — 2/ 1G , a- 8 — ?/ 8 , x 4 — y\
14. m 3 4- 8 m 2 + 7 m, m 3 4- 3 m 2 — m — 3, m 3 — 7m— 6.
98. An important principle relating to common factors is
illustrated by the following example:
Given x 2 + 7x + 10 = (x + 5)(.x + 2), (1)
and a; 2 — x — 6 = (ar — 3) (a; + 2). (2)
Add (1) and (2), 2 x 2 + 6 x + 4 = 2(x + l)(.r + 2). (3)
Subtract (2) from (1), 8x + 1G = S(x + 2). (4)
We observe that x + 2, which is a common factor of (1) and (2),
is also a factor of their sum (3), and of their difference (4). This
example is a special case of the following :
64 FACTORING
99. Theorem 1. A common factor of two expressions is
also a factor of the sum or difference of any multiples of
those expressions.
Proof. Let A and B be any two expressions having the common
factor f. Then if k and I are the remaining factors of A and B
respectively,
A =fk and B= ft.
Also let mA and nB be any multiples of A and B.
Then mA = mfk and nB — nfl, from which we have :
mA ± ?i B = mfk ± nfl = f(mh ± id).
Hence f is a factor of mA ± nB.
100. Theorem 2. // f is a factor of mA ± nB and also of
A. then f is a factor of B, provided n has no factor in com-
mon with A.
Proof. Let / be a factor of n>A ± nB and also of A, where mA
and nB are integral multiples of the expressions A and B.
Then mA ± nB , — , and — may each be reduced to an integral
expression by cancellation.
Now mA ± nB = ?lA ± ilB. Si nC e 2^ is integral, it follows
nB f f f f
that — is also integral. That is, / is a factor of nB. But /
is not a factor of n since it is a factor of A, and by hypothesis
n and A have no factor in common. Hence / is a factor of B.
101. By successive applications of the above theorems it is
possible to find the H. C. F. of any two integral expressions.
Ex. 1. Find the H. C. F. of 9 x* - x- + 2 x - 1, (1)
and 27cc 5 + 8<B 2 -3a; + l.' (2)
Multiplying (1) by Zx and subtracting from (2) we have
•_'7 x B + 8 x 2 - 3 x + 1
27 x 5 - 3 x 3 + 6 x 2 - 3 x
3x 3 + '2x 2 + l (3)
By theorem 1, any common factor of (1) and (2) is a factor of (3).
COMMON FACTORS AND MULTIPLES 65
Calling expressions (1) and (2) B and A respectively of theorem 2,
then (3) is A — 3 x • B ; and since the multiplier, 3 x, has no factor in
common with (2), it follows from the theorem that any common
factor of (3) and (2) is a factor of (1), and also that any common
factor of (3) and (1) is a factor of (2). Hence (1) and (3) have the
same common factors, that is, the same II. C. F. as (1) and (2). There-
fore we proceed to obtain the H. C. F. of
9/ 4 -r+2z-l, (1)
and 3x 3 + 2a»+l. (3)
Multiplying (3) by 3 a; and subtracting from (1) we have
— 6 x 3 — x 2 — x — 1. (4)
By argument similar to that above, (3) and (4) have the same
H. C. F. as (1) and (3) and hence the same as (1) and (2). Multi-
plying (3) by 2 and adding to (4) we have,
3x 2 -x + l. (5)
Then the H. C. F. of (5) and (3) is the same as that of (1) and (2).
Multiplying (5) by x and subtracting from (3), we have
3 x 2 - x + 1 . (6)
Then the H. C.F. of (5) and (0) is the same as that of (1) and (2).
But (5) and (6) are identical, that is, their H. C. F. is ox 2 — x + 1.
Hence this is the H. C. F. of (1) and (2).
The work may be conveniently arranged thus :
(1) 9 x 4 - x 2 + 2 x - 1
27 x 5
+ 8 x 2 - 3 x + 1
(?)
J) x 4 + 6 x 3 + 3 x
27 x 5 -
- 3 x 3 + 6 x 2 — 3 x
(4) - 6 x 3 - x 2 - x - 1
3 x 3 + 2 x 2 +1
(3)
6i 3 +4x 2 +2
3 x 3 — x 2 + x
(5) 3 x 2 -x + 1 3 x- - x + 1 (G )
The object at each step is to obtain a new expression of as low a
degree as possible. For this purpose the highest powers are elimi-
nated step by step by the method of addition or subtraction.
E.g. In Ex. 1, x 5 was eliminated first, then x 4 , and then x 3 .
By theorems 1 and 2, each new expression contains all the factors
common to the given expressions. Hence, whenever an expression is
reached which is identical with the preceding one, this is the H. C.F.
6G FACT OBI Mi
102. The process is further illustrated as follows:
Ex. 2. Find the H. C. F. of 2 ar* -2a?-3x-2,
and 3x 3 — x 2 — 2x—16.
Arranging the work as in Ex. 1, we have
3x 3 - x 2 - 2 x - 16 (2)
6x 3 - 2x 2 - 4x- 32
6x 3 - 6a: 2 - 9x- 6
(1)
2x 3 -2x 2 - 3x-2
4 x 3 — 4 x* 2 — 6 x — 4
4 a;3 + 5 x 2 _ oe x
(1)
- 9 x 2 + 20 x - 4
9x 2 -18a:
(7)
2x-4
(8)
x-2
4 X 2 + 5a ._ 26 (3)
36 a- 2 + 45 x - 234
-36x 2 + 80x- 16
125 x -250 (5)
x - 2 (6)
Explanation. To eliminate x 3 , we multiply (1) by 3 and (2) by 2
and subtract, obtaining ('■)).
To eliminate x 2 from (3), we need another expression of the second
degree. To obtain this we multiply (1) by 2 and ( ;, >) by x and sub-
tract, obtaining (1).
Using (4) and (3), we eliminate x'-\ obtaining (5). Since (.1) con-
tains all factors common to (1) and (2), and since 125 is not such a
factor, this is discai'ded without affecting the H. C. F.. giving (6).
Multiplying (6) by 9 and adding to (4) we have (7). Discarding
the factor 2 gives (8) which is identical with (6). Hence x — 2 is the
H. C. F. sought.
103. Any monomial factors should be removed from each
expression at the outset. If there are such factors common to
the given expressions, these form a part of the H. C. F.
When this is done, then any monomial factor of any one of
the derived expressions may be at once discarded without
affecting the H. C. F.. as in (5) of the preceding example.
Tti this way a bo the hypothesis of theorem 2 is always fulfilled;
namely, thai at every step the multiplier of one expression shall have
no factor in common with the other expression.
COMMON FACTORS AND MULTIPLES (57
Ex. 3. Find the H. C. F. of 3 X s - 7 x 2 4- 3 x - 2,
and x 4 — x 3 — x 2 — x — 2.
(1) 3x 3 - 7x 2 + 3x- 2 x i - x s - x 2 - x - 2 (2)
12 x 3 - 28 x 2 + 12 x - 8 3 x 4 - 3 x 3 - 3 a; 2 - 3 x - 6
12 x 3 - 18 a: 2 - 3 x - 18 3 a, 4 - 7 x 3 4- 3 .r 2 -2 a:
(1) - 10x 2 + 15 x + 10 4x 3 -0x 2 - x-6 (3)
(5) -5(2x+l)(x-2).
Explanation. To eliminate x 4 , we multiply (1) by x and (2) by 3
and subtract, obtaining (3).
To eliminate x 3 , we multiply (1) by 4 and (3) by 3 and subtract,
obtaining (1).
At this point the work may be shortened by factoring (4) as in
(5). We may now reject, not only the factor — 5, but also 2x + 1,
which is a factor of neither (1) nor (2), since 2x does not divide the
highest power of either expression. But x — 2 is seen to be a factor
of (2), by §§ 91, 92, and hence it is a common factor of (2) and (4)
aud therefore of (1) and (2). Hence x — 2 is the H. C. F. sought.
EXERCISES
Find the H.C.F. of the following pairs of expressions:
1. a? + 6 a 2 + 6 a + 5, a 3 + 4 a 2 - 4 a + 5.
2. x i -2x*-2x i + 5x-2, x i -4x 3 + 6x 2 -5x + 2.
3. 2 a 3 -9. ^-13 z-4, x 3 - 12 x 2 + 31 x + 28.
4. x * _ 5 rf _|_ 3 x - 2, x* - 3 x' + 3 x 2 - 3 x + 2.
5. 2 X s - 9 x 2 + 8 a- - 2, 2 x 3 + 5 or — 5 x + 1.
6. 3 a 4 - 2 a 3 + 10 a 2 - 6 a + 3, 2 a 4 4- 3 a 3 + 5 a 2 + 9 a - 3.
7. 15 .r 4 + 19 a 3 -44 a; 2 -15 A- + 9,
15 x* - 6 a; 3 + 51.» 2 + 11 x - 15.
8. r s + 2^-2r 8 -8r 2 -7r-2, ^-2r 4 - 2r»+ 4»- 2 + r-2.
104. The following theorem enables us to find the L. C. M. of
two expressions by means of the method which has just been
used for finding the H. C. F.
68 FACTORING
Theorem 3. The L.C..M. of two expressions is equal to
the product of either expression and the quotient obtained
by dividing the other by the II. C. F. of the two expressions.
Proof. Let A and B be two expressions whose H.C. F. is F so that
A =mF and B= nF. Hence the L. CM. of A and B is mnF. But
mnF = mnF = mB. Also mnF = nmF = nA. Therefore the L. CM.
is cither mB or nA, where m = A h- F and n — B -=- F.
Ex. Find the L. C. M. of 9 * 4 - a- 2 + 2 x - 1, (1)
and 27 x 5 4- 8 x- — 3 » + 1. (2)
The II. CF. was found in § 101 to be 3 a; 2 -x + 1.
Dividing (1) by 3 x 2 — x + 1 we have 3 x' 2 + x — 1.
Hence the L. C M. of (1) and (2) is
(27 x 5 + 8 x 2 - 3 x + 1) (3 x 2 + x - 1) .
EXERCISES
Find the L. C. M. of each of the following sets.
1. a 4 4- a? + 2 a 2 — a + 3, a 4 + 2 a 3 + 2 a 2 — a + 4.
2. a 3 - 6a 2 4- 11 a - 6, a 3 - 9 a 2 + 20 a - 24.
3. 2 a 3 +3 a 2 6 - 2 a& 2 - 3 & 3 3 2 a 4 - a 8 6 - 2 a 2 & 2 4- 4 a& 3 - 3 6 4 .
4. 2a 8 -a 2 &-13a& 2 -66 8 ,
2 a 4 - 5 a 3 & - 11 aVr 4- 20 oi> 3 + 12 b\
5. 4 a s_ 15 a 2_ 5 a _ 3> 8 a 4_ 34 a s_j_ 5 a -> _ a + 3,
2a 8 -7a 2 +lla-4.
6. a* + a 2 + 1, a 3 + 2 a 2 - 2 a + 3.
7. 2& 8 -& 2 Z-13fcZ 2 4-5Z s , 3fc 3 -l(>fr"-7 + 24A7 2 -7 7 3 .
8. 12 v 4 - 20 r\s - 15 rs-+ 35 rs 8 - 12 s 4 ,
6r 8 -7r 2 s-llrs 2 + 12s 8 .
9. 2 tr-7 a 2 +6a-2, a 3 +2 a 2 - 13 a+10, a 3 + a 2 4-6 a+5.
10. x" 3 — xi f+ yx- — if, 2 x 8 4- .r// 4- •>'!/-+ - ."''•
2. r ! +; i.r-y +3 a*?/ 2 +2^.
CHAPTER VI
POWERS AND ROOTS
105. Each of the operations thus far studied leads to a single
result.
E.g. Two numbers have one and only one sum, § 2, and one and
only one product, § 7.
When a number is subtracted from a given number, there is one
and only one remainder, § 6.
When a number is divided by a given number, there is one and
only one quotient, § 11-
We are now to study an operation which leads to more than
one result ; namely, the operation of finding roots.
Thus both 3 and — 3 are square roots of 9, since 3-3 = 9, and also
(_ 3)(_ 3) = 9 ; this is often indicated by V9 = ± 3. See § 114.
106. The operations of addition, subtraction, multiplication,
and division are possible in all cases except dividing by zero,
which is explicitly ruled out, §§ 24, 25.
Division is possible in general because fractions are admitted to
the number system, and subtraction is possible in general because
negative numbers are admitted. Thus 7 h- 3 = 2J, 5 — 7 ~ — 2.
107. The operation of finding roots is not possible in all
cases, unless other numbers besides positive and negative in-
tegers and fractions are admitted to the number system.
E.g. The number V2 is not an integer since l 2 = 1 and 2 2 = 4.
Suppose V2 = - a fraction reduced to its lowest terms, so that a
and b have no common factor. Then — = 2. But this is impossible,
b~
for if b' 1 exactly divides a 2 , then a and b must have factors in com-
mon. Hence V2 is not & fraction.
69
70 POWERS AND ROOTS
108. If a positive number is not the square of an integer or
a fraction, a number may be found in terms of integers and
fractions whose square differs from the given number by as
little as we please. See p. 228, E. C.
E.g. 1.41, 1.414, 1.4141 are successive numbers whose squares differ
by less and less from 2. In fact (1.4141)- differs from 2 by less than
.0004, and by continuing the process by which these numbers are
found, § 170, E. C, a number may be reached whose square differs
from 2 by as little as we please.
1.41, 1.414, 1.4141, etc., are successive approximations to the
number which we call the square root of 2, and which we represent by
the symbol, V2.
109. Definition. If a number is not the A'th power of an
integer or a fraction, but if its feth root can be approximated
by means of integers and fractions to any specified degree of
accuracy, then such a kth root is called an irrational number.
See § 36.
E.g. V2, V2, a/5, etc., are irrational numbers, whereas V4, V8,
are rational numbers.
It is shown in higher algebra that irrational numbers corre-
spond to definite points on the line of the number scale, § 40,
E. C, just as integers and fractions do.
We, therefore, now enlarge the number system to include
irrational numbers as well as integers and fractions.
It will be found also in higher work that there are other kinds of
irrational numbers besides those here defined.
The set of numbers consisting of all rational and irrational
numbers is called the real number system.
110. Even with the number system as thus enlarged, it is
still no1 possible to find roots in all cases. The exception is
the even root of a negative number.
THE COMPLEX NUMBER 71
E.g. V — 4 is neither + 2 nor - 2, since (+ 2)' 2 = + 4 and (- 2)' 2
= + 4, and no approximation to this root can be found as in the case
of V2.
111. Definition. The indicated even root of a negative num-
ber, or any expression containing such a root, is called an
imaginary number, or more properly, a complex number. All
other numbers are called real numbers.
E.g. V— 4, V— 2, 1 + V — 2, are complex numbers, while 5, a/2,
1 + V2 are real numbers.
Complex numbers cannot be pictured on the line which represents
real numbers, but another kind of graphic representation of complex
numbers is made in higher algebraic work, and such numbers form
the basis of some of the most important investigations in advanced
mathematics.
112. With the number system thus enlarged, by the admis-
sion of irrational and complex numbers, we have the following
fundamental definition.
(-s/7iy = n.
That is, a £th root of any number n is such a number that,
if it be raised to the kt\\ power, the result is n itself.
E.g. (aV2) 3 = 2, (VI) 2 = 4, (V^ = _2.
The imaginary or complex unit is V— 1. By the above
definition we have
(V-l) 2 = -l.
In operating upon complex numbers, they should first be ex-
pressed in terms of the imaginary unit.
E.g. V^2 = V2 ■ V^l, V- 1(3 = VT6 • V^l = 4V^T.
V^I. v/3q= (VI- V^T)(\/U V~l)=2.3(V^T)2=-<>.
\/^4 + \/^9= VI- V^l+ VU • \/3I = (2 + 3)V^T=5\/^l.
V^i6 = Vie- V3i ^vi6 = 4
V^ " V9 • V^l V9 3
72 POWERS AND ROOTS
113. By means of irrational and complex numbers it can be
shown that every number has two square roots, three cube
roots, four fourth roots, etc. See § 195, Ex. 17-20.
E.g. The square roots of 9 are + 3 and — 3. The square roots
of - 9 are ± V^9 = ± 3 V~l. The cube roots of 8 are 2, - 1 + V^3
and — 1 — V— 3. The fourth roots of 16 are +2, — 2, + 2V— 1
and -2V3T.
Any positive real number has two real roots of even degree,
one positive and one negative.
E.g. Vlti = ± 2. The square roots of 3 are ± V3.
Any real number, positive or negative, has one real root of
odd degree, whose sign is the same as that of the number itself.
E.g. ^27 = 3 and v^-32 = - 2.
114. The positive even root of a positive real number, or the
real odd root of any real number, is called the principal root.
The positive square root of a negative real number is also
sometimes called the principal imaginary root.
E.g. 2 is the principal square root of 4, 3 is the principal 4th root
of 81 ; — 4 is the principal cube root of — 01 ; and + V — 3 is the
principal square root of — 3.
Unless otherwise stated the radical sign is understood to in-
dicate the "principal root.
The only exception in this book is in such cases as, v'4 = ± 2.
where it represents either square rout, But in such expressions as
1 + V2, 3 ± V6, etc., the principal root only is understood.
In all cases it is easily seen from the context in what sense the sign
is used.
When it is desired to designate in particular the principal
root, the symbol V is used.
E.g. VlG' = 2, while VlG might stand indifferently for 2. —2.
2/- 1", or _2V^1.
\ 8 — 2. while \ s might represent 2, — 1 + V— 3, or — 1 — V— 3.
THEOREMS ON POWERS ANU ROOTS 73
THEOREMS ON POWERS AND ROOTS
115. Theorem 1. The nth power of the kth power of any
base is the nktli power of that base.
Proof. Let n and k be any positive integers and let b be any base.
Then (&*)» = b k ■ b k . b k ... to n factors. § 124, E. C.
— k+k+k... to n terms _ Q nk^ g 4.3
Hence (b") a = b'" c .
Corollary (b k ) n = (b n ) k = b" k .
E.g. (2 3 ) 2 = (2-) 3 = 26 = 64.
116. Theorem 2. The nth power of the product of several
factors is the product of the nth powers of those factors.
Proof. Let k, r, and n be any positive integers. Then
'a k b') n = (a k b'-) • (a h b r ) • • • to n factors, § 124, E.C.
= (a* • a* . . . to n factors) (6* • b k • ■ ■ to n factors) §§ 8, 9
= (a*)» -(b r Y, §124, E.C.
Hence, (a k b'J'= a nk b nr . §115
E.g.(;2 3 -3- 2 )* = 2 r >-3 i .
117. Theorem 3. The nth power of the quotient of two
numbers equals the quotisnt of the nth powers of those
mi i nbers.
Proof. We have (£- ) = £- • ~ • ? to n factors § 124, E. C.
\b r I If It' b'
a k • a k ■ a k • • • to n factors c , no ^, r ,
= — ; — — . § lyo, rj.L.
b r ■ b 1 ' ■ b 1 ' • • • to ti factors
( a k\n a n»
Hence ' [v) = f^- § 115
„ /2 8 \ 2 2 6 64
E - g - y = 3i = 8i-
74 POWERS AXD ROOTS
118. It follows from theorems 1, 2, and 3 that:
Any positive integral power of a monomial is found by
multiplying the exponents of the factors by the exponent of the
power.
119. Theorem 4. The principal rtli root of the krth
power of any positive real number is a power of that
number ivhose exponent is kr^-r = k.
Proof. Let k and r be positive integers and let b be any positive
real number.
We are to prove that Vb 1 ^ = h k .
From theorem 1, (''*)'' = &**"•
Hence by definition b k is an rtli root of ?/''. and since b k is real and
positive, it is the principal rth root of t,'- r (§ 114).
That is, \/W = b kr ^ r = &*.
E.g. VF = 3^- 2 = 3 2 = 9 ; y/2^ = 2 12 * 4 = 2 3 = 8.
But it does not follow that
^/(3oyn? = ( _ 2)U+ 4 = ( _ o )3 = _ 8)
since (- 2) 12 = (2) 12 and hence ^/(-2) 12 ' = ^2^ = + 8.
The corresponding theorem holds when 6 is negative if r is
odd and also when b is negative if fc is even.
E.#. ^(-2)«' = (-2) 6 - 3 =(-2) 2 =4; v^(- 2) ls ' =(- 2) s = -32.
120. Theorem 5. The principal rtli root of the product
of two positive real numbers equals the product of the
principal rth roots of the number.
Proof. Let a and b be any positive real numbers and lei r be any
positive integer.
We are to prove v«& ■= Va • Vb •
We have ({/? • ^) r = ( Va ) r • V5" )'" § 1 16
= a-&. §112
Hence, a& = (-v^u" • v^ ) r . § 3
THEOREMS ON POWERS AND ROOTS
75
Taking the principal rth root of both members,
we have VctF — Va ' • VV .
When r is even the corresponding theorem does not hold
if a and b are both negative.
9'.
For example, it is not true that \/(— ■!)(— !>)' = V — -i
For V(-4)(- 9) = VW = 6 ; while V- 4' • V^ 1
= 2V^T'.3 V^l'= 6(V3T)2=_6. See § 112
121. Theorem 6. The principal rth root of the quotient
of two positive real numbers equals the quotient of the
principal rth roots of the numbers.
Proof. Let a and b be any positive real numbers and let r be any
positive integer.
w J- ''/" ^«
We are to prove \h= — •
™ vs
Va
We have
(gV = &2Za §§ 117, 112
Hence, taking the principal ?-th root of both members,
we have
r\a _ Va
E.g. J^=^K = ±; <B , =^8: = ^2 = _2.
V 25 v^ 5 ' > 27 </27 3 3
The corresponding theorem does not hold when r is even
if a is positive and b is negative. Thus it is not true that
x
4
- 9
VI 1
3V^T 3(V"^T) 2
2v^T 2
: ~ r=-3^-i'-
But we have
v;
i=%(-»=i
If r is odd, the theorem holds for all real values of a and b.
76 POWERS AND ROOTS
122. From theorems 4, 5, 6, it follows that :
If a monomial is a perfect power of the kth degree, its kth
root may be found by dividing the exponent of each factor by the
index of the root.
In applying the above theorems to the reduction of algebraic
expressions containing letters, it is assumed that the values of the
letters are such that the theorems apply.
EXERCISES
Find the following indicated powers and roots, and reduce
each expression to its simplest form :
1. (crW) 7 . 4. (a x - y Y +xs+ ^. 7. (3"' • 4' • 2y-\
2. (2 a+b • 3° • 5 b ) a - h . 5. (.<ii/V +y )- r -- v . 8. "v 3 1 '" • 2 a • 5 30 '
a'h'r r ' Y „ f?rb 2 >rin\- Q 3 — 2<
V 3 T bc A J * 04rV/ i; "
10. ( a »4»-i 6TO -»c-)» + 13 -^3.^ . 4 ._» . #?-*•
11. (3« + * • 4 B " 7 • 6- 1 )*. 14 . V64T25.256-625 1 .
12. 2 V3 fia • 4- a • 5 8a • 7 4a '. 15. v 27'- 125-64.3".
16. (a - b) m ~ n (b - c) m ~ n (a + 6)™-".
17 / (a - 6) 8 (a g + 2 a& + b 3 )
* (a - &) 4 (a + 6) 2
/ (4x- 2 + 4a ; + l)(4x 2 -4.r + l) '
'v 36 x* - 12 a 2 + 1
19. v(- 343)(- 27)./ (a + 6) 3 « .
20 3/ (- 8)(- 27)(- 125)aW
\(_ 1)(_ 512)(1000)a; 15a i/ 21
ROOTS OF POLYNOMIALS 77
ROOTS OF POLYNOMIALS
123. In the Elementary Course, pp. 221-224, it was shown that
the process for finding the square root of a polynomial is obtained
by studying the relation of the square, a? + 2 ab + b 2 , to its
square root, a + b.
In like manner the process for finding the cube root of a
polynomial is obtained by studying the relation of the cube,
a? + 3 d 2 b + 3 ab' 2 + b :i or a 8 -f- 6(3 a 2 + 3 ab + ft 2 ), to its cube
root, a + b.
An example will illustrate the process.
Ex. 1. Eind the cube root of
27 m 3 + 108 m?n + 144 mn 2 + 64 n 3 .
Given cube, 27 m 3 +108 »r 2 /i+144 mn 2 +64 n 3 13 m+4 n, cube root
a 3 = 27 »i 3 1st partial product
3 a* = 27 m 2
3a6 = 36 inn
62 = 16 n 2
3 a 2 +3 a6+6 2 =27 m 2 +36 mn+16 ?i 2
L08 /c-« M44 //m 2 +(i4 n 3 , 1st remainder
108 m 2 n-|-144 mn 2 +64 ?? 3 = 6(3 « 2 +3 ab+b*)
Explanation. The cube root of the first term, namely om, is the
first term of the root and corresponds to a of the formula. Cubing
3 m gives 27 in 3 which is the a 3 of the formula.
Subtracting 27 m 3 leaves 108 m' 2 n + 111 mri 2 + 61 n 3 , which is the
6(3 a 2 + 3 rtft + 6 2 ) of the formula.
Since b is not yet known, we cannot find completely either factor
of 6(3 a 2 + 3 ab + 6' 2 ), but since a has been found, we can get the first
term of the factor 3 a' 2 + 3 ab-\-b 2 ; viz. 3 rt" 2 or 3(3 m) 2 = 27 ?»' 2 , which
is the partial divisor. Dividing 108 m 2 n by 27 m 2 we have 1 n, which
is the 6 of the formula.
Then 3 a 2 + 3 ab + 6 2 = 3(3 m) 2 + 3(3 m)(i n) + (1 n) 2 = 27 ??i 2
+ 36 ????? + 16 n' 2 is the complete divisor. This expression is then
multiplied by h — i n, giving 108 m 2 n + 111 inn 1 + (if » 3 , which corre-
sponds to 6(3 a' 2 + 3«6 + 6 2 ) of the formula. On subtracting, the
remainder is zero and the process ends. Hence, dm + in is the
required root.
78 POWERS AND BOOTS
Ex. 2. Find the cube root of
33 x 4 - 9 x' + x G - 63 X s + 66 a? - 36 x + 8.
We first arrange the terms with respect to the exponents of x.
x- — 3 x + 2, cube root
(4iven cube, x G — 9 x 5 4- ; >3 x 4 — 63 s 3 + 66 •'"- — 36 x + 8
a 3 = re 6
3 c- = 3 x 4
3 a 2 + 3 a& + b 2 = 3x 4 - 9 x 3 + 9 x 2 _
3 a' 2 = 3(« 2 - 3 x) 2 = 3 x 4 — 18 x 3 27 .■-
3 a' 2 + 3 a'6' + 6' 2 = 3 x 4 - 18 x 3 + 33 x 2 - 18 x + 4
— 9 x5 + 33 x 4 — 63 X s + m x 2 — 36 x + 8
— 9z s + 27 .-■■» — 27 x 3
6 x 4 -- 36. t-3 + 6i ;./•--: 36 x 8
6 x* — 36 x 3 + 66 .r 2 — 36 x 8
The cube root of x 6 , or x 2 , is the first term of the root. The first par-
tial divisor, which corresponds to 3 a 2 of the formula, is •*>(./-)'- = 3 x i .
Dividing— 9x 5 by 3x 4 we have — 3 x, which is the second term of the
quotient, corresponding to b of the formula.
After these two terms of the root have been found, we consider
x 2 —Bx as the a of the formula and call it a'. The new partial divisor
is 3 a' 2 = 3(x 2 — 3 x) 2 = 3 x 4 — 18 x 3 + 27 x 2 , and the new b, which we
call &', is then found to be 2.
Substituting x 2 — ■"'> x for a' and 2 for V in 3 a' 2 + ;1 > a'V + b'-, we
have 3 x 4 — 18 x 3 4- 33 a; 2 — 18 x + 4, which is the complete divisor.
On multiplying this expression by 2 and subtracting, the remainder
is zero. Hence the root is x 2 — 3 x + 2.
In case there are four terms in the root, the sum of the first
three, when found as above, is regarded as the new a, called
a". The remaining term is the new b and is called b". The
process is then precisely the same as in the preceding step.
EXERCISES
Find the square roots of the following:
1. in- + 4 inn + 6 ml+ 4 //-' + 1 2 In + 9 P.
2. 4 x 4 + 8 ax 3 + 4 <r.c + 1 6 tfx 2 + 16 alrx + 1 6 b\
3. 9 a 2 - 6 ab + 30 ac + 6 ad + b- - 10 be - 2 bd 4- 25 c 2
+ 10cd + d 2 .
HOOTS OF POLYNOMIALS 79
4. 9 a 2 - 30 ab - 3 ab 2 + 25 & 2 + 5& 3 + -.
4
5. | era 4 — | o6r 3 2! + 1 arbxrz 2 + & 2 <cV — 4 ab 2 xz" + 4 a 2 b 2 z 4 .
6. a 2 - 6 aft + 10 ac - 14 ad + 9 6 2 - 30 be + 42 &d + 25 c 2
7. | + 6 a- - 17 x 2 - 28 .t- 3 + 49 x\ [ - 70 cd + 49 d 2 .
8. 9 a 6 - 24 a 3 & 4 - 18 aV + 6 a\l 2 + 16 & 8 + 24 &V - 8 b\l 2
9. 25 o 4 "7/* - 70 a" m b 5k + 49 a' 5 '"& 4 *. [ + 9 c 10 - 6 c 5 d 2 + cf*.
10. x w - 8 a 8 ™* + 16 w w - 4 x?y 3 + 16 ^ + 4 f + 6 aV
- 24 z 4 w" - 12 y'Y + 9 z 8 .
Find the cube root of each of the following :
11. cc 3 — 3 x 2 y 4- 3 xy 2 — if + 3 a ,2 z — 6 a:?/ 2 + 3 y 2 z + 3 xz 2
12. 1728 a 6 + 1 728 x 4 f + 576 a- 2 / + 64 f. [ - 3 yz 2 + z 3 .
13. «"' + 3 a 2 b + 3 a 2 c + 3 a& 2 + 6 a&c + 3 ac 2 + & 3 + 3 & 2 c
14. 8« 3 -12(r& + 6a6 2 -& 3 . [ + 3 6c 2 + e 3 .
15. 8 x 6 - 36 a 8 + 114 a- 4 - 207 a- 3 + 285 a; 2 - 225 a- + 125.
16. 27 z'> — 54 az 5 + 63 o 2 z 4 — 44 a¥ + 21 aV — 6 cCz 4- a 6 .
17. 1-9 >/ 2 + 39 f - 99 f + 156 / - 144 y w + 64 y 2 .
18. 125 a 6 - 525 x*y + 60 .r 4 // 2 4- 1547 aty» - 108 a- 2 ?/ 4 - 1701 a// 5
-729/.
19. 64 Z 12 - 576 P + 2160 I 8 - 4320 l G + 4860 Z 4 - 2916 I 2 + 729.
20. a 6 + 6 a s b + 15 a A b 2 + 20 a 3 & 3 + 15 a 2 & 4 + 6 a& 5 + & 6 .
21. a 9 - 9 a% + 36 a 7 b 2 - 84 a 6 & 3 + 126 cc'b 4 - 126 a 4 6 s + 84 a s ¥
-36a 2 & 7 + 9a& 8 -& 9 .
22. a 8 4- 6 a 2 6 - 3 « 2 c + 12 a& 2 - 12 abc + 3 ac 2 + 8 & 3 - 12 6 2 c
+ 6 &c 2 - c 3 .
23. 343 a« - 441 a 5 b + 777 a 4 & 2 - 531 a 8 & 8 + 444 a 2 & 4 - 144 ab 5
+ 64 b\
24. a ls + 12 a 15 + 60 a 12 + 160 a 9 + 240 o fi + 1 92 a 5 + 64.
25. 27 F~ + 189 l n + 198 P - 791 P - 594 I s + 1701 Z 7 - 729 E
80 POWERS AND ROOTS
ROOTS OF NUMBERS EXPRESSED IN ARABIC FIGURES
124. The cube root of a number expressed in Arabic figures, as
in the case of square root, pp. 225-229, E. C, may be found by
the process used for polynomials. An example will illustrate.
Ex. 1. Find the cube root of 389,017.
In order to decide how many digits there are in the root, we
observe that 10 3 = 1000, 100 3 = 1,000,000. Hence the root lies between
10 and 100, that is, it contains two digits. Since 70 3 = 313,000 and
80 3 = 512,000, it follows that 7 is the largest number possible in tens'
place. The work is arranged as follows:
The given cube. 389 017 [70 + 3 , cube root.
« 3 = 70 3 = 3 13 000 1st partial product.
:;„-= 3-70 2 = 14700
3 ab = 3 • 70 • 3 = 630
b 2 = 3 2 = 9
10 017 1st remainder.
3 a 2 + 3 a h + jfl - 15339 j 46Q17 = b (3 a 3 + 3 ab + b-).
Having decided as above that the a of the formula is 7 tens, we cube
this and subtract, obtaining 46,017 as the remaining part of the
power.
The first partial divisor, 3 a 2 = 14,700, is divided into 46,017, giving
a quotient 3, which i> the b of the formula. Hence the first complete
divisor, '■'< u- + Zab + 3 & 2 , is l~>.:j:'>!t and the product, &(3 a- + 3 ab + //-),
is 46,017. since the remainder is zero, the process ends and 7:> is the
cube root sought.
125. The cube of any number from 1 to 9 contains one, two.
or three digits; the cube of any number between 10 and W
contains four, live, or six digits ; the cube of any number
between 100 and 999 contains seven, eight, or nine digits, etc.
Hence it is evident that if the digits o\' a number are separated
into groups of three figures each, counting from units' place
toward the left, the number of groups thus formed is the same
as the number of digits in the root.
ROOTS OF ARABIC NUMBERS
81
Ex. 2. Find the cube root of 13,997,521.
The given cube, 13 907 521 |200 + 40 + 1 = 241, cube root.
u s = 200 8 = 8 000 000
3 a 2 = 120000
3 ab = 24000
b 2 = 1600
145600
3 a'- = 172800
3 a'V = 720
b' 2 = 1
173521
5 997 521
5 824 00 = b (3 a 2 + 3 aft + ft 2 )
17:3 521
173 521 = V (3 a' 2 + 3 a'6' + b' 2 ).
Since the root contains three digits, the first one is the cube root
of 8, the largest integral cube in 13.
The first partial divisor, 3 • 200- = 120,000, is completed by adding
3 ab = 3 <■ 200 ■ 40 = 24,000, and b 2 = 1000.
The second partial divisor, 3 a' -2 , which stands for 3(200 4- 40) 2
= 172,800, is completed by adding 3 a'V which stands for 3 • 240 • 1 =
720, and b 12 which stands for 1, where a' represents the part of the
root already found and b' the next digit to be found. At this step
the remainder is zero and the root sought is 241.
EXERCISES
Find the square root of each of the following:
1. 58,081. 2. 795,564. 3. 11,641,744.
Find the cube root of each of the following :
4. 110,592. 7. 205,379. 10. 2,146,689.
5. 571,787. 8. 31,855,013. 11. 19,902,511.
6. 7,301,384. 9. 5,929,741. 12. 817,400,375.
126. Since the cube of a decimal fraction has three times as
many places as the given decimal, it is evident that the cube
root of a decimal fraction contains one decimal place for every
three in the cube. Hence for the purpose of determining the
places in the root, the decimal part of a cube should be divided
into groups of three digits each, counting from the decimal
point toward the right.
82
POWERS AXD ROOTS
Ex. Approximate the cube root of 34.507 to two places of
decimals.
a* = 3 3 =
3a 2 = 3-3 2 = 27.
3a& = 3-3(.2) = 1.8
&2 = (.2)- = .04
L'S.SJ
3«' 2 = 3(3.2) 2 = 30.72
3a'b' = 3(3.2) (.06)= .48
6' 2 = (.05) 2 = .0025
31.2025
3a"2=3(3.25) 2 =31.6875
3a"b" =3(3.25)(.007) = .06825
b»* = (.Q01)*= .00004
34.567 | 3+ .2 + -05 + .007 = 3.267
27.000
7.507
5.768 ft(3 a' 2 + 3 aft + ft 2 )
1.7 KOI IU(I
1.560125
ft'(3a'* + Sa'6' + 6' 2 )
.238875000
31.755709 1 .222290503 = 6"(3a" 2 + 3 a"5"+ ft" 2 )
.016584407
The decimal points are handled exactly as in arithmetic work.
127. Evidently the above process can be carried on indefi-
nitely. 3.257 is an approximation to the cube root of 34.567.
In fact the cube of 3.257 differs from 34.567 by less than the
small fraction .017. The nearest approximation using two
decimal places is 3.20. If the third decimal place were any
digit less than 5, then 3.25 would be the nearest approximation
using two decimal places. Hence three places must be found
in order to be sure of the nearest approximation to two places.
EXERCISES
Approximate the cube root of each of the following to two
places of decimals.
1. 21.4736.
6.
.003.
11.
.004178.
2. 6.5428.
7.
.3017.
12.
200.002.
3. 58.
8.
.5.
13.
572.271.
4. 2.
9.
.05.
14.
31.7246.
5. 3.
10.
6410.37
15.
54913.416.
16. Approximate
the
square
root in
Exs.
1, 2, 10, 11, and
15 of the above
list.
CHAPTER VII
QUADRATIC EQUATIONS
EXPOSITION BY MEANS OF GRAPHS
128. We saw, § 65, that a single equation in two variables
is satisfied by indefinitely many pairs of numbers. If such an
equation is of the first degree in the two variables, the graph is
in every case a straight line.
We are now to consider graphs of equations of the second
degree in two variables. See § 66.
Ex. 1. Graph the equation y = x 2 .
By giving various values to x and computing the corresponding
values of y, we find pairs of numbers as follows which satisfy this
equation :
(x = 0, f x = 1, f x - - I, I" x = 2, f x = - 2, ( x = 3, J x = - 3, etc>
U = 0. iy = l. ly = l. Ly = 4. \y = 4. ly = 9. iy = 9.
These pairs of numbers correspond to points which lie on a curve
as shown in Figure 3.
By referring to the graph the
curve is seen to be symmetrical
with respect to the y-axis. This
can be seen directly from the
equation itself since x is involved
only as a square and hence, if y = x 2
is satisfied by x = a, y = b, it must
also be satisfied by x = — a, y = b.
It may easily be verified that
no three points of this curve
lie on a straight line. The
curve is called a parabola.
83
V(-3
9)
+ '■>
G
,9)
+ s
CO
+7
+6
%
+8
K-2
l\
-t l
/'
1)
■c
+2
(-1
1)
■tl
1)
:
_
!
-
N^
^?l"
^
2
-)
3
X
-(/.
is
c
,0)
-1
Fig. 3.
84
Q UADRA TIC EQ UA TIONS
■i ■
ta
..ii
+i
(2,5y
+3
v
.»
+2
;
id
i
_
2
S>
+2
i-
1,0)
X-
IX
s
-2
(u
1
-3>
(ft,
:'•)
(-1,
-4
|
Fig. 4.
we have the graph of the equation, as in Figure 4.
Ex. 2. Graph the equation
y = x- + 2x — 3.
Each of the following pairs of
mi ml mt.- satisfies the equation :
f* = o, fx = i, r» = - i,
J a; = 2, fx = -2, far = -3,
[y=5. [y = - 3. ly = 0.
X = — 4,
y = 5.
Plotting these points and draw-
ing a smooth curve through them,
EXERCISES
In this manner graph each of the following:
1. y = x 2 — 1. 7. y = 5x — xr — 4.
2. y — .r + 4 x. 8. ?/ = 4 x — ar' + 5.
3. y = ar + 3 a: — 4. 9. ?/ = a; 2 -f- 5 x — 6.
4. v = .r + 5 a; + 4. 10. y — — x- + x.
5. // = .r — 7x + 6. 11. y = 4ic 2 — 3 a; — 1.
6. y = 3x* — 7x + 2. 12. y = — 4x- + 3x + l.
129. We now seek to find the points at which each of the
above curves cuts the aj-axis. The value of y for all points on
the .r-axis is zero. Hence we put y = 0, and try to solve the
resulting equation.
Thus in Ex. 2 above, if y = 0, a: 2 + 2x - 3 = (x + 3)(a: - 1) = 0,
which is satisfied by x = l and x= —3. Hence this curve cuts the ar-axis
in the two points sc=l, .y = and x= —.'5, // = 0, as shown in Figure 4.
Similarly in Ex. 1, it' y = 0, x- = 0, and hence x = 0. Hence the
curve meets the x-axis in the point x = 0, y = 0, as shown in Figure
3. On this point sec § 131, Ex. 2.
ALGEBRAIC SOLUTION 85
EXERCISES
Find the points in which each of the twelve curves in the
preceding list cuts the a>axis.
Notice that in every case the expression to the right of the equality
sign can be factored, so that when y = the resulting equation in x
may be solved as in § 94.
Ex. 3. Plot the curve y = <B z + 4sc-f-2 and find its intersec-
tion points with the #-axis.
We are not able to factor a; 2 +4a;+2 by inspection. Hence we solve
the equation x' 2 + 4 x + 2 = by completing the square as in § 175,
E. C, obtaining x = — 2 + a/2 and x = — 2 — a/2. Hence the curve
cuts the x-axis in points whose abscissas are given by these values of x.
In making this graph, we first plot points corresponding to integral
values of x, as before ; then, in drawing the smooth curve through
these, the intersections made with the x-axis are approximately the
points on the number scale corresponding to the incommensurable
numbers, - 2 + a/2 and - 2. - a/2. See § 109.
EXERCISES
In this manner, find the points at which each of the follow-
ing curves cuts the x--axis, and plot the curves. For reduction
of the results to simplest forms, see §§ 159, 160, E. C.
1. y = x* + 5x + 3. 5. y = 2x — 5z 2 + 8.
2. i/ = 3x 2 + Sx-2. 6. y = 5 + Sx — 3x 2 .
3. y = 6.x- -4 x 2 + 5. 7. // = 3-9.r-ll.t\
4. y = — 4-2a + 5ar. 8. y = — 2 — 2x + x>.
130. Each of the foregoing exercises involves the solution of
an equation of the general form ax 2 + bx + c = 0. Obviously,
by solving this equation, we shall obtain a formula by means
of which every equation of this type may be solved. See
§ 179, E. C.
The two values of x are :
Xl 2a ' X " 2a
86 QUADRATIC EQUATIONS
EXERCISES
By means of this formula, find the solutions of each of the
following equations :
1. 2 a 2 - 3a -4 = 0. 11. 3 a; -9o* 4- 1 = 0.
2. 3x z + 2aj — 1 = 0. 12. 7a? — 3x-2 = 0.
3. 3 sc 2 — 2 x* - 1 = 0. 13. G x 2 + 7 a- + 1 = 0.
4. 4 3^4- 6 a; 4-1 = 0. 14. 4 .r 4- 5a- 8 = 0.
5. .r - 7 a + 12 = 0. 15. 4 sc 2 - 5 a; - 3 = 0.
6. 5jc 2 + 8aj4-3 = 0. 16. S.r + 3a-5 = 0.
7. 5 a 2 - 8 a; + 3 = 0. 17. 7 as* + as — 3 ="0.
8. 5a 2 4-8a-3 = 0. 18. 7k 2 — a; — 4 = 0.
9. 5 a 2 -8a;- 3 = 0. 19. cc 2 -2aaj = 3&-a 2 .
10. 2a; — 3^4-7 = 0. 20. a 2 — 6aa = 49c 2 - 9« 2 .
2 , o(o + 6) , (a + 6) a;
21. x- -\ *— ! — l = ax 4- i — o •
3 3
22. — 2ar — x — 2 c z a; = — * *•
2 2
23. ar + 2 win = 4 na.
24. a 2 — 2 «a 4- 4 ab = b 2 + 3 a 2 .
25. a* 2 — abx 4- a 2 6 — ax = a& 2 — 6a.
26. a 2 4-9 — C = Ga.
27. /(.< - ' 2 + urn = //?» 2 a 4- /xa.
28.2 (a + 1) a 2 - (a + l) 2 a + 2 (a 4- 1) = 4 x.
29. a 2 + Cd 4- 3 c = (3 c 4- 3 d + 1 )a.
30. x 2 + 2 a 2 4- 3 a — 2 = (3 a + 1) a
131. We now consider the intersections of other straight
lines besides the avaxis with curves like those plotted above.
DISTINCT, COINCIDENT, AND IMAGINARY ROOTS 87
Ex. 1. Graph on the same axes the straight line, y = — 2
and the curve, y = x 2 + 2 x — 3.
This line is parallel to the .r-axis and two units below it. Tt cuts
the curve in the two points whose abscissas are x 1 = — 1 + V2 and
x 2 = - 1 — V'2, as found by substituting — 2 for y in y = x 2 + 2x — 3
and solving the resulting equation in x by the formula, § 130.
Ex. 2. Graph on the same axes y = — 4 and y = x 2 + 2 x — 3.
This line seems not to cut the curve but to touch it at the point whose
abscissa is x = — 1.
Substituting and solving as before, we find,
2 + = _ x
-2+ Vi-
1
2
_ 2 - V4 -
- 4
and xo = — — := = — 1.
2 2
In this case the two values of x are equal, and there is only one point
common to the line and the curve. This is understood by thinking of
the line y — — 2, in the preceding example, as moved down to the position
y= — 4, whereupon the two values of x which were distinct now coincide.
132. From the formula, x = — , it is clear that
2 a
the general equation, ax 2 + bx + c = has two distinct solutions
unless the expression b 2 — 4 ac reduces to zero, in which case the
two values of x coincide, giving x x = ^— = : - and
^ _ _fr_Q _ j>_ 2a 2a
X -~ 2 a ~ 2 a
Ex.1. In 2 x 2 — 9 x + 8 = 0, determine without solving
whether the two values of x are distinct or coincident.
In this case, a = 2, b = — 9, c = 8.
Hence Ifi — 4. ac = 81 — 64 = 17.
Hence the values of x are distinct.
Ex. 2. In 4 x- 2 — 12 x + 9 = 0, determine whether the values
of x are distinct or coincident.
In this case, b 2 — 4 ac = 144 — 4-4-9 = 0. Hence the values of x
are coincident.
88 QUADRATIC EQUATIONS
EXERCISES
In each of the following, determine without solving whether
the two solutions are distinct or coincident :
1. .r-7.<- + 4 = 0. 6. 6 x 2 - 3 x — 1 = 0.
2. 4 <b* + 28 x + 49 = 0. 7. 4 a 2 — 16 x + 16 = 0.
3. 9ar J + 12a; + 4 = 0. 8. 8 x 2 - 13 = 4 x.
4. ;r + 6.r + 9 = 0. 9. 12 a 2 - 18 = 24 a.
. 5 . _ic2 + 9aj + 25 = 0. 10. 16 a?- 56 a? =—49.
133. Definition. A line which cuts a curve in two coincident
points is said to be tangent to the curve.
134. Problem. What is the value of a in y = a, if this line
is tangent to the curve y = x 1 + 5 x + 8 ?
Substituting « for ^/ and solving by means of the formula, we have
-5± V25-4(8-a) j
o
If the line is to be tangent to the curve, then the expression under the
radical sign must be zero so that the two values of x may coincide.
That is, 25 - 4(8 - a) = 0, or a = J.
On plotting the curve, the line y = | is found to be tangent to it.
EXERCISES
In the first 18 exercises on p. 8G obtain equations of curves
by letting the left members equal y. Then find the equations
of straight lines, y = <t, which are tangent to these curves.
135. Problem. Find the intersection points of the curve
?/ = .r 2 + 3 x + 5 and the line y = 21.
Substituting for y and solving for x we have
_ - 6 4 V36 -40 -6 + 2V-1
Xl 4 4
-3+ v
o
- 3 - V
/ - 1
^ _ _ 6 - V36 _40_-6-2V-l_
- 1
DISTINCT, COINCIDENT, AND IMAGINARY ROOTS 89
These results involve the imaginary unit already noticed in
§ 112. Numbers of the type a + &V— 1 are discussed further
in § 195. For the present we will regard such results as merely
indicating that the conditions stated by the equations cannot
be fulfilled by real numbers. This means that the curve and
the line have no jjoint in common, as is evident on constructing
the graphs.
By proceeding as in § 134 we find that the line y = -^ is tangent
to the curve y = x' 2 + 3 x + 5. Clearly all lines y = a, in which a >-j l ,
are above this line and hence cut this curve in two points.
All such lines for which a < y are below the line y = y and hence
do not meet the curve at all.
Solving y = a and y — x- -f- 3 x + 5 for x by first substituting a for y
we have
- 3± V4 a - 11
x = — .
2
If a >-V" th e number under the radical sign is positive, and there are
two real and distinct values of x. Hence the line and the curve meet
in two points.
If a<-V-) the number under the radical sign is negative. Con-
sequently the values of x are imaginary and the line and the curve do
not meet.
Hence we see that the conclusions obtained from the solution of
the equations agree with those obtained from the graphs.
136. From the two preceding problems it appears that it is
possible to determine the relative positions of the line and the
curve viitliout completely solving the equations. Namely, as
soon as y is eliminated and the equation in x is reduced to
the form ax 2 + bx + c = 0, we examine lr — 4 ac as follows :
(1) If lr — 4 ac > 0, i.e. positive, then the line cuts the curve
in two distinct points.
(2) If b 2 — 4 ac = 0, then the line is tangent to the curve.
See § 133.
(3) If b 2 — 4 ac<0, i.e. negative, then the line does not cut
the curve.
90 QUADRATIC EQUATIONS
137. Problem. Find the points of intersection of
y = a?+3x + 13 (1), and y+3x=7 (2).
Eliminating y and reducing the resulting equation in x to the form
ax- + bx + c = 0, we have x 2 + 6 x + (5 = 0.
Solving, a?j = — 3 + VS, x 2 = — 3 — VVi.
Substituting these values of z in (li) and solving for y, we have
.r. = - 3 + V3 f ./•„ = - 3 - VS
,_ and { , r _
yi = 16 - 3 V3 [ y 2 = lb" + 3 V3
which are the points in which the line meets the curve.
Here b' 2 — 4 ac = 12, which shows in advance that there are tivo
points of intersection.
EXERCISES
In each of the following determine without graphing whether
or not the line meets the curve, and in case it does, find the
intersection points :
(y = 2x*-3x-4:, 6 (y = 5x* + 8x + 3,
' [y-x = 3. ' [2y-5x-2 = 0.
\ y = 2x* + 2x-l, \y = 5a?-Sx + 3,
\2y = x-l. ' [3-x = 3y.
y = 3 .<•-' - 2 x - 1, 8 [ y = - 5 x- +Sx- 3,
4.
2x-y = 4. I 2-4//-.i- = 0.
?/ = 4.r + 6.r + l, 9 \y = — 5x 2 — Sx — 3,
x = y + 5. \ 5 y — 3 a? = 8.
5 (y=* 2 -7a- + 12, 1Q fy = 3a>-3a» + 7,
[5a; — 2/ = — 1. " \— 5 — 3x + 2y=0.
138. Problem. G-raph the equation jb 2 + y 2 = 25.
Writing the equation in the form // = ± V^o — x' 2 , and assigning
values to x, we compute the corresponding values of // as follow 3 :
J x = 0, J a? = ± 5, | x = 3, f x = - 3, f x = 4, f z = - 4,
ljf=±5. U = 0. ly = ±4. ly = ±4. U = ±3. |y = ±3.
DISTINCT, COINCIDENT, AND IMAGINARY ROOTS 91
1
HI
5)
(-3,41
JM>
■2
NJ(U)
(-*,3)
\ *
r^\
ft
>£*
,>
-"I
—a
y
>v
»J
(-n.o)
(0,0)
x i
(5,0)
X-
axis
(ri
,jk
/r''
1)
(-3
-4)|V
(M)
(0,^5)
FlG. 5.
Evidently, for x greater than 5 in absolute value, the corresponding
y's are imaginary, and for each x between — 5 and + 5 there are two
y's equal in absolute value, but
with opposite signs.
It seems apparent that these
points lie on the circumference
of a circle whose radius is 5, as
shown in Figure 5. Indeed, if we
consider any point x v y l on this
circumference, it is evident that
x \~ + Hi 2 = -5, since the sum of
the squares on the sides of a right
triangle is equal to the square
on the hypotenuse. (See figure,
p. 207, E. C)
The equation x- + y 2 = 25 is,
therefore, the equation of a
circle with radius 5. Similarly, x' 2 + y 2 = r 2 is the equation of a
circle with center at the point (0, 0) and radius r.
139. Problem. Find the points of intersection of the circle
x 2 + y 2 = 25 and the line x -f y = 7.
Eliminating y from these equations, and reducing the equation in
x to the form ax 2 + bx + c = 0, we have
x 2 - 7 x + 12 = 0.
From which x x = 4, x„ = 3.
Substituting these values of x in x + y = 7, we have y 1 = 3, y 2 = 4.
Hence x x = 4, y { = 3 and x 9 = 3, y 2 = 4 are the required points.
Verify this by graphing the two equations on the same axes.
140. Problem. Find the points of intersection of the circle
x 2 + y 2 = 25 and the line 3 x + 4 y = 25.
Eliminating y and solving for x, we find x = — ^— = 3.
Hence x x = x 2 = 3, from which ?/, = y 2 = 4.
Since the two values of x coincide, and likewise the two values of
y, the circumference and the line have but one point in common.
Verify by graphing the line and the circle on the same axes.
92 QUADRATIC EQUATIONS
141. Problem. Find the points of intersection of
x 2 + y 2 = 25
and x + y = 10.
Substituting for y and solving for x we have
_ 20 ± v400 - 600 _ 20 ± V- 200
4 4
20 ± 10 V^2 10 ± 5 v/^2 "
4 2
The imaginary values of a: indicate that there is no intersection
point. Verify by plotting.
EXERCISES
In each of the following determine by solving whether
the line and the circumference meet, and in case they do,
find the points of intersection. Verify each by constructing
the graph.
.r 2 + ,v 2 =16, 5 \x> + y 2 =l, p 2 -f ?/ 2 =12,
x + ii = 4. \ x + y = 8. [ x — y = 6.
x 2 + y 2 = 36, 6 (.r 2 + y- = 8, 1Q {.r°- + y- = 4,
4.
4 x 4- y = 6. [ x — ?/ = 4. { 2 x — 3 y = 4.
r.r- + r = 2o, 1^ + ^ = 41, 1-^+^=40,
\2oj + y=-5. * \x-Sy = 7. ' \.r + 2,/ = 10.
ra 2 + ? /- o = 20, a {x* + y*=:29, 12 (.r 2 + /
[2x + y = 0. \3x-7y=-29. ■ \x + y = 9.
142. Problem. Graph on the same axes the circle, x 2 +y 2 =b 2 ,
and the lines, 3 x + 4y = 20, 3 a + 4 // = 25, and 3 sc + 4ty = 30.
The first lino cuts the circumference in two distinct points, the
second seems fco be tangent to it. and the third does not meet it. Ob-
ser\e that the three lines are parallel. See Figure 0.
DISTINCT, COINCIDENT, AND IMAGINARY ROOTS 93
In order to discuss the
relative positions of such
straight lines and the circum-
ference of a circle, we solve
the following equations simul-
taneously :
x 2 + y 2 — 1~ (1)
3x + ±y = c (2)
Eliminating y by substitution,
and solving for x, we find
s "In
V
IflJUS.
v rv^
<&$*
»
1 X
•2
L^^n!^
8
It
(-
-..111
*
(0,0)
(5,0)
X-
axis
CO,-
5)
3 c ± 4 V-25 r 2 - c 2
(3)
Fig. 6.
The two values of x from (3) are the abscissas of the points of
intersection of the circumference (1) and the line (2).
These values of x are real and distinct if 25 r 2 — c 2 is positive, real
and coincident if 25 r 2 — e 2 = 0, and imaginary if 25 r 2 — c' 2 is negative.
Now 25 r 2 — c 2 is positive if r = 5, c = 20; zero if r = 5, c = 25 ; and
negative if r = 5, c = 30.
Hence these results obtained from the solution of the equations
agree with the facts observed in the graphs above.
143. Definition. Letters such as c and r in the above solution
to which any arbitrary constant values may be assigned are
called parameters, while x and y are the unknowns of the
equations.
EXERCISES
Solve each of the following pairs of equations.
Give such values to the parameters involved that the line (a) may
cut the curve in two distinct points, (b) may be tangent to the curve,
(c) shall fail to meet the curve.
1.
| X 2 +tf = 4:,
\ ax + 3 y = 16.
x 2 + y 2 =16,
2 x + by = 12.
(x 2 + y 2 = 25,
\2x + Zy = c.
y- = 8 X,
3x + 4y = c.
94
Q UADRA TIC EQ UA TIOXS
(5tf = 2px,
J x + y = c.
f y = xr -f maj + /»,
j aj + ?/ = 4.
f ?/ = ma; 2 — nx — 4,
{ x — 3 y = 8.
( i y = 2ar ! -3a: + l,
\ 2 a; - by - 1 = 0.
10.
11.
12.
f y = 3 a;- + ??ix — m,
U',' + 2y+l=0.
// = ma? + 2 »a:,
2 // - 6a; -5 = 0.
( ?/ = .r + //.>; + 1,
{ ax + -J i, = 10.
| iB 2 + r = r 2 ,
( ax -f- by = c.
144. Problem. Graph the equation — + - 1
= 1.
25 16
Writing the equation in the form y = ± f V25 — x-, and assigning
values to a:, we compute the corresponding values of y as follows :
x — 0, I" x = ± 5, fx = l, f .<• = — 1,
.'/ = ± 4,
2
y = 0, { // = ± 3.9, = ± 3.9.
x = 3, f x = 4,
2/ = ±3.2, U = ±2.4.
Evidently if x is greater than 5
in absolute value, the correspond-
ing values of y are imaginary.
Plotting these points, they are
found to lie on the curve shown in
Figure 7. This curve is called
an ellipse.
EXERCISES
Solve the following pairs
of equations.
In this way determine whether
the straight line and the curve
intersect, and in case they do,
determine the coordinates of the intersection points. Verify each by
constructing the graphs.
'£ + £ = !, o te + jd-1,
(0,1)
'":
^.J.*)
■4.2
*) f
H
,(!
- 1)
1
(-5I0)
JM)
(5,0^
i x
axis
(
1.--'
d»
',|
■2.4)
(
&
3,7
(0,-4)
Fig. 7.
L6 9
3a; + 4y = 12.
2.
49 16
2 a? — 7 y = !
DISTINCT, COINCIDENT, AND IMAGINARY BOOTS 95
3.
5.
x 2 + 4 y- = 25,
2 x -y = L
8a»+2jf=ll,
x — 3y = 7.
6.
7. J
1,
\ st+yl
25 9
2*-y=14
y=2a£— 3as+4,
y— 4a— 8=0. 9.
1-^ + ^ = 16,
I x + y = 7 -
64 12 10.
4y-2a-=4.
36 45
-5 x+6y = 10.
[- + ^- = 1,
49 25
la + y-12.
When arbitrary constants are introduced, in the equations of a
straight line and an ellipse, we may determine values for
these constants so as to make the line cut the ellipse, touch it,
or not cut it, as in the case of the circle, § 142.
EXERCISES
Solve each of the following pairs simultaneously.
Give such values to the constants that the line shall (a) cut the
curve in two distinct points, (b) be a tangent to the curve, (c) have
no point in common with the curve.
In case (b) is found very difficult, this may be omitted.
1.
k 2 +^=i,
a 2 16
{8x + 5y=4Q.
5.
16 + 25 '
[ ax + Ay =20.
9.
' a? 8 + ?/ 2 = r 2 ,
ax — 3 y = 4.
2.
•j 25 b-
[4x4-15?/ =60.
6.
f - r ~ _l y 2 1
i36 + 16 = 1 '
10.
'5x 2 +Sy 2 =W,
hx — ky == 8.
[ ax + 6 y — 60 =
0.
3.
1 *- + -2L = l,
25 16
[ 4 x — 5 y = c.
7.
\ v 2 i if -.
.36 + 25" 1 '
[5x + by = 60.
11.
x 2 + 7?/ 2 = 144,
aa5 + 6y = 12.
4.
[ or i ^_ 1
16 + 25 '
8.
j a lr
12. ■
ar + 4 y* = r-,
[5x-by = 20.
[bx-2y = 5.
_ ax + by = c.
96
QUADRATIC EQUATIONS
SPECIAL METHODS OF SOLUTION
145. We have thus far solved simultaneously one equation
of the second degree with one of the first degree. After sub-
stitution each has reduced to the solution of an ordinary quad-
ratic, namely, of the form, ax 2 + bx + c = 0.
While this is an effective general method, yet some im-
portant special forms of solution are shown in the following
examples :
x 2 + f = a, (1)
x - y = b. (2)
Ex. 1. Solve
Square both members of (2) and subtract from (1).
2 xy = a — b' 2 .
Add (1 ) and (3) . x 2 + 2 xy + >/ 2 = 2 a- ft 2 .
Hence x + y = ± V'2 a — b-.
From (2) and (5), adding and subtracting
V2 a - b- + b
and •
V2 a - V 2 - b
F =
-V 2q-6 a -6
2
Ex. 2. Solve \ ,
[x-y = b.
From (1) ( ./■ - ?/)(.r -f //) = a.
Substituting b for x — y in ('•]), x + y = - •
ft
Then (2) and (1) may be solved as above.
Ex. 3. Solve
x + y = a,
xy = b.
Multiply (2) by 1. subtract from the square of (1). and get
x 2 — 2 xy + >/- = a' 2 - 1 6,
w hence, a; — y = ± \ a 2 — 46.
Then (1) and (1) may be solved as in Ex. 1.
(3)
(1)
(5)
(1)
(2)
(3)
(4)
(1)
(2)
( ; 5)
(1)
SPECIAL METHODS OF SOLUTION
97
The equations i ,
1 I xy = b,
may be solved in a similar manner.
146. We are now to study the solution of a pair of equations
each of the second degree. See § 66.
Consider x 2 + y = a, (1)
x + tf = b. (2)
Solving (1) for y and substituting in (2) we have,
x + a 2 — 2 ax 2 + x* = b,
which is of the fourth degree and cannot be solved by any
methods thus far studied. There are, however, special cases in
which two equations each of the second degree can be solved
by a proper combination of methods already known.
147. Case I. When only the squares of the unknowns enter the
equations.
Example. Solve f ^ + b /\ = ° u
[a 2 x- + b 2 y = c 2 .
These equations are linear if x 2 and y 2 are regarded as the
unknowns.
Solving for a; 2 and y 2 as in § 73, we obtain,
''A ,.i __ a \ c -y — 2 Ci
Hence, taking square roots,
y
a i^2 — °2^1
2 - C A ^
a x b. 2 — a 2 bj
- J a i C 2 ~ a ?Pl }
* ajb 2 — a 2 b^
- _ J C A -~<iA
^a l 6i-a 8 & 1 ,
.. -J c A - C A
a A - «2 6 i '
\
' ajb 2 — aj) x
= Al
^A T2
r/.c, — a„c.
a x b. 2 - a 2 b^
c 2 b t
a x b 2 — a.pi
y*
VS
a x b., — a 2 /)j
In this case there are four pairs of numbers which satisfy the two equa-
tions. This is in general true of two equations each of the second degree.
98
QUADRATIC EQUATIONS
y\s"
1-,
.1,
■W/
(k
. a\
/
"N
//
Iff
\\
\
/
| / .r
-axis
(0,0)
\
\ V—
■2
\\
A V V
e
i/l/l /
/
\
s>
/ / y
/
Lfl
:.' ^Ov.
^s { y
v;
>
^x >
^>J__
j i
^j!
Example. Solve simul-
taneously, obtaining re-
sults to one decimal
place :
(1)
: 25. (2)
36 16
Fig. 8.
Clear (1) of fractions and
proceed as above. Verify
the solution by reference
to the graph given in
Figure 8.
EXERCISES
Solve simultaneously each of the following pairs of equa-
tions and interpret all the solutions in each case from the
graph in Figure 8 :
36
16
= 1,
.<•-'
+ if =
36.
r 9
.1-
36
16
= 1,
.r
+ f =
16.
3.
X~
36
16
= 1,
X 2
+ if =
49
' X 2
36
9
16
= 1,
ar
+ f =
9.
2.
148. Problem. Graph the equation —
25 16
1.
Writing the equation in the form y — ±^Vx- — 25, and assigning
values to z, we compute the corresponding values of y exactly or
approximately as follows:
\x=±5, \x = 6\, (x=-6\, ix = 7, fx=-7, fx=8, fa;=-8,
b = 0, U=±3,ly=±3, \y=±S.9, \y=±3.9, { y= ±5, \y=±5.
SPECIAL METHODS OF SOLUTION
99
Evidently when x is less than 5 in absolute value, y is imaginary,
and as x increases beyond 8 in absolute value, ij continually increases.
Plotting these points, they are found to lie on the curve as shown in
Figure 9. This curve is called a hyperbola.
(
-S,5)j\
'h
yjcs.o
(-7
,3.9)
\
1
(7,3
9)
(-6)
i,*r
r^,\::,)
( 5,(
)
(0,0
(5,0
.r-(
xis
.
(
-6*.
3 V
V(CK,-3)
(-7,-
3.9)
,(V
S.9)
(-
3 .- 5 >J/
(8,-5)
^-
Fig. 9.
EXERCISES
Solve each, of the following pairs of equations.
Construct a graph similar to the one in Figure 8 which shall contain
the hyperbola ^ — ^— = 1 and the circles given in Exs. 1, 2, and 3.
Construct another graph containing the same hyperbola and the
ellipses given in Exs. 4, 5, and 6. From these graphs interpret the
solutions of each pair of equations.
25 16
x 2 + y- = 16.
25 16
a* +>/== 25.
x T —-i
L6
\a? + y* = 36.
100
QUADRA TIC EQUA TIONS
*. J!-1
25 16" '
-^ + ^ = 1.
36 16
a;'
25
L6
r = l,
25
^+- ?/ "=l.
25 16
95 i« '
7 + '
16 9
16
7. Graph the equation xy = 9.
( > raph a;;/ = 8 on the same axes with each of the following :
8. x 2 + y- = 16. 9. x 2 +y 2 = 25.
11.
25 16 '
12.
25 10.24
14.
x 2 y 2 _ 1
25 16
15.
x 2 y 2 1
25 9
10. x 2 + / = 4.
13. ^ + ^ = 1.
16 4
16.
16
£ = 1.
4
17. From those graphs in Exs. 8 to 16, in which the curves-
meet, determine as accurately as possible by measurement the
coordinates of the points of intersection or tangency.
18. Solve simultaneously the pairs of equations given in
Exs. 8 to 10, after studying the method explained in § 150,
Ex. 1. Compare the results with those obtained from the
graphs.
19. Solve Exs. 11 to 16 by the method explained in § 149,
and compare the results with those obtained from the graphs.
149. Case II. When all firms containing the unknowns are of
the second degree in the unknovms.
Example. Solve
2x 2 -3xy + ±y 2 = 3,
3 x 2 — 4 xy + 3 y 2 = 2.
Put y = vx in (1) and (2), obtaining
"U 2 (3-4i< + 3r 2 )=2.
Hence from (3) and (4),
3
2 - 3 v + 4 v 2
, and also x' 2
3 — 4 v + 3 v' 2
(1)
(2)
(3)
(4)
(5)
SPECIAL METHODS OF SOLUTION
101
From (5)
3 2
(6)
2 -3u + 4i> 2 3-4v + 3 v 2 '
» 2 — 6 w + 5 = 0.
(7)
Hence
v = 1, and f = 5.
(8)
From y =
vx,
y = x, and y = 5 x.
(9)
If # = x, then from (1) and (2),
y = i,
If ?/ = 5 a:, then from (1) and (2),
1
1 , f x = — 1.
' and ■{ '
12/
y
V29
V29
and <
X =
i
V29
y =
5
V29
Verify each of these four solutions by substituting in equations
(1) and (2).
150. There are many other special forms of simultaneous
equations which can be solved by proper combination of the
methods thus far used. Also, many pairs of equations of a
degree higher than the second in the two unknowns may be
solved by means of quadratic equations.
The suggestions given in the following examples illustrate
the devices in most common use.
The solution should in each case be completed by the student.
3^ + ^ = 58,
Ex. 1. Solve
Adding twice (2) to (1) and taking square roots, we have
./■ + y = 10, and x + y = — 10.
(1)
(2)
(3)
Each of the equations (3) may now be solved simultaneously with
(2), as in Ex. 3, p. 96.
102
QUADRATIC EQUATIONS
Ex. 2. Solve
M = 5,
x y
4 + ^ = 13.
Let - = a and - = b. Then these equations reduce to
x y
f a + b - 5,
1 a 2 + b 2 = 13.
(3) and (4) may then be solved as in Ex. 1, p. 96.
Ex.3. Solve (x* + f + x + y = 8,
la 7/ = 2.
Add twice (2) to (1), obtaining
x 2 + 2 xy + y' 2 + x + y = 12.
Let x + y = a. Then (3) reduces to
a 2 + a = 12,
or, o = 3, a = — 4.
Hence x + ?/ = 3, and x + y — — 4.
Now solve each equation in (5) simultaneously with (2).
Ex. 4. Solve
xY + xY = 272,
X* + y*=10.
(1)
(2)
(3)
(4)
(1)
(2)
(3)
(4)
(«)
(1)
(2)
In (1) substitute a for z 2 # 2 . Then
a 2 + a = 272, whence o = 16, and — 17.
Hence xy = ± VT6 = ± 4, and ± V- 17.
Each of these equations may now be solved simultaneously with
(2), as in Ex. 1, p. 101.
Ex. 5. Solve
• jgS _ ?/ 3 _ 117j
. a; - y = 3.
(1)
(2)
SPECIAL METHODS OF SOLUTION 103
By factoring, (1) becomes
(x-yX* + xy + y*)=117. (3)
Substituting 3 for x — y, we have
x 2 + xy + y 2 = 39. (4)
(2) and (4) may now be solved by substitution as in §§ 140-144.
Ex.6. Solve K + / = 513, (1)
x + y=9. (2)
Factor (1) and substitute 9 for x + y. Then proceed as in Ex. 5.
Ex.7. Solve (.^ + ^ = 126, (1)
\x + !, = 9. (2)
Factoring (1) and substituting 9 for x + y, we have
ay = 14. (3)
(2) and (3) may then be solved as in Ex. 3, p. 96.
Ex.8. Solve (^ + 2^ = 54^, (1)
U + y = 6. (2)
Factor (1) and substitute 6 for x + y, obtaining
x 2 — xy + y 2 = 9 xy. (3)
(2) and (3) may now be solved by substitution, as in §§ 140-144.
Ex.9. Solve (**-»* = 63, (1)
I a? + xy + y 2 = 21. (2)
Factor (1) and substitute 21 for x 2 + xy + y 2 , then proceed as in
Ex. 8.
Ex.10. Solve ( 3 s + 9 s = 243. (1)
UV + .wr = 162. (2)
Multiply (2) by 3 and add to (1), obtaining a perfect cube.
Taking cube roots, we have
x + y = 9. (3)
(1) and (3) are now solved as in the preceding example.
104 QUADRATIC EQUATIONS
Ex. 11. Solve K + .7< = 641, (1)
[x + y = 7. (2)
Raise (2) to the fourth power and subtract (1), obtaining
4 x 3 y + 6xV + 4 xy 3 = 1760. (3)
Factoring, 2 xy (2 x 2 + 3 xy + 2 y-) = 1760. (4)
Squaring (2) we have
o x 2 + 4 x>/ + 2 y 2 = 98, (5)
or 2 x 2 + 3 ./vy + 2 y- = 98 - x#. (6)
Substituting (6) in (4), we have
2 xy (98 - xy) = 1760, (7)
or x°-y 2 - 98 x,y + 880 = 0. (8)
In (8) put xy = a, obtaining
tt 2 _ gg a + 880 = 0. (9)
The solution of (9) gives two values for xy, each of which may now
be combined with (2) as in Ex. 3. p. UG.
EXERCISES
Solve each of the following pairs of equations :
r + rs + sr
r — s = 3.
63,
5.
| x 2 + y- = a,
{ xy = b.
3 a? + 2 y 2 = 35,
2x*-3y 2 = 6.
(3 x* + 2 xy = 16,
1 4 .r - 3 xy = 10.
7.
4.
13.
14.
8.
C a 2 -f- db + & 2 = ~,
( a 2 _ a 6 + 6 2 = 19.
3 a -2?/ = 6,
;i.r_ 2. -7/ + 4/ = 12
a + 6 + aft = 11,
(a + 6) 2 + a 2 6 2 = Gl.
9.
10.
11.
12.
,-< + y 5 = 91,
x + .?/ = 7.
x- + ?/ 2 = a,
x 2 -y- = b.
.r-— '.) xy = 0,
:,,■- + .sy 2 =9.
i + i = W,
1 + 1=1.
15.
s+A + i = 49,
CT ah b~
1+1=8.
a 6
SPECIAL METHODS OF SOLUTION 105
^4a 2 -2o& = & 2 -16, 27> f (x- 4) 2 + G/ + 4) 2 = 100,
5a 2 = 7a6 — 36. }a+y = 14.
17.
19.
21.
22.
3a 2 -9 ? / 2 = 12, fay + 2/ -ha = 17,
2x-3y= 14. { a?y + f + a 2 = 129.
18. l-'- + ^/ + r = «, 29 f5 + a 2 = o («-&),
x 2 + y = 6. | a + 52 _ o ( a _ 6 ^
rf + jf + a + y-18, r(13.,) 2 + 2 // 2 = 177,
^ = 6 - 1(2^-13^ = 3.
j x 2 + y 2 + a; — ?/ _ 36, r/Q\ 2 /or;\2
31.
9 25
fx- 2 -5a7/ + ?/ 2 = -2,
{■^ + 7.^ + r = 22. i*" y
« 2 + 6a^ + & 2 = 124, 32. j a ' 2 + . ? / 2 = 20 >
a + 6 = 8. [5.r-3y/ 2 = 28.
a 2_ 3r(6 + 2 & 2 = 0, 33. f^=-6-3ajy,
23 - J2« 2 + a&-& 2 = 9. ' l-2^ = /-24.
24 (x- + f + 2x + 2y = 27, 34 ( a + y + Va~+2/ = 12,
xy = -12. " 1^ + ^ = 189.
a 2 + r - 5 x — 5 y = - 4, 35 f -« 4 + x 2 y 2 + ;'/ 4 = 133,
xy = 5. 1 a 2 - xy + y 2 = 7.
26. (( 7 +*)( 6 + ») = 80 » 36. (* + *# + // = 29,
\x + y = 5. { x 2 + ay + y 2 = 61.
37 (2x*-5xy + 3x-2y = 22, 3g (x + y = U,
\5xy + 7x-8y-2 x 2 = 8. { x 2 + // 2 = 3026.
39 f7 / / 2 -r,.r' + 20.r + 13//-29,
{ 5 (x - 2) 2 - 7 y 2 - 17 // = - 17.
106 QUADRATIC EQUATIONS
40 ((3x + 4y)(7x-2y)+3x + 4y = U,
\(3x + 4 : y)(7x-2y)-7'x + 2y = 30.
41. p + ." = 4 >
1 x> + x*y + afy 2 + aft/ 3 + a*/ 4 4- /' = 364.
42. f aj5 -2/ ! = 31 J 44. (v l +y*—xy=80,
1 •'• - ?/ = 1 • I « — y - xy = - 8.
43 fa* + 2/* = 82, 45 r8a + 8&-a&-a 2 = 18,
{ x 2 + t/ 2 + 2 x 2 y 2 = 28. [5a+5& — ft 2 - a& = 24.
46.
47.
48.
(a 3 + xh, + xtf + f) (x + y) = 325,
(or* - x*y + xy z — f) (x — y) = 13.
2(^ + 4) 2 -50/-7) 2 = 7o,
7 (x + 4) 2 + 15 (?/ - 7) 2 = 1075.
( a s + tf = (a + b)(x-y),
\ x- — xy + //" = a — b.
HIGHER EQUATIONS INVOLVING QUADRATICS
151. An equation of a degree above the second may often be
reduced to the solution of a quadratic after applying the factor
theorem. See § 92.
Example. Solve 2 x^+x 2 - 10 .r + 7 = 0. (1)
By the factor theorem, x — 1 is found to be a factor.
giving (x - 1) (2 x 2 + 8 x - 7) = 0. (2)
Hence by § 22, x - 1 = and 2 x 1 + 3 a; - 7 = 0. 05)
From x - 1 = 0, x = 1. (4)
From 2 ./■- f 3 x - 7 = 0, x = ~ 3 ± vfi5 . (5)
4
Hence (4) and (5) give the three roots of (1).
HIGHER EQUATIONS INVOLVING QUADRATICS 107
EXERCISES
Solve each of the following equations :
1. 7 a- 3 - liar +4 a- = 0. 5. 28 a; 3 - 10 x 2 -U x = 6.
2. 3 a- 4 + a- 3 4- 2 ar 4- 24 a; = 0. 6. a- 4 -3.x- 3 + 3 x 2 - x = 0.
3. 3a- 3 -16ar+23a,--6=0. 7. 4 a,- 8 + 12a 2 - 3 x - 9=0.
4. 5 a- 3 4- 2 ar 4- 4 a- =-7. 8. a- 4 - 5 a; 3 4- 2 ar 4- 20 a- =24.
9. 6 a,- 3 4- 29 ar- 19 a; = 16.
10. 15 .e 4 4- 49 a; 3 -92 a,- 2 4- 28 a; = 0.
EQUATIONS IN THE FORM OF QUADRATICS
152. If an equation of higher degree contains a certain
expression and also the square of this expression, and involves
the unknown in no other way, then the equation is a quadratic
in the given expression.
Ex. 1. Solve x A + 7r = 44. (1)
This may be written, (x 2 ) 2 4- 7(.r 2 ) = 44, (2)
which is a quadratic in x 1 . Solving, we find
x 2 = 4 and x 2 = - 11. (3)
Hence, x = ± 2 and x = ± V-ll. (4)
Ex.2. Solve a- + 2 4-3VaT+2 = 18. (1)
Since x + 2 is the square of Vx + 2, this is a quadratic in Vx +2.
Solving we find Vx + 2 = 3 and Vx + 2 = — 6. (2)
Hence x 4-2 = 9 and £+2=36, (3)
Whence x = 7 and x = 34. (1)
Ex.3. Solve (2x 2 -l) 2 -r>(2a- 2 -l)-14 = 0.
First solve as a quadratic in 2x- — 1 and then solve the two result-
ing quadratics in x.
108 QUADRATIC EQUATIONS
Ex. 4. Solve or - 7 x + 40 - 2 Vo 2 -7x + 09 = - 20. (1)
Add 29 to each member, obtaining
x 2 - 7 x + 09 - 2Vx 2 -7x + 69= 3. (2)
Solve (2) as a quadratic in Vx 2 — 7x + 69, obtaining
Vx' 2 - 7a; + 69 = 3 and Vx 2 - 7 x + 69 = - 1, (3)
whence x 2 — 7 x + 69 = 9 or 1 . (4)
The solution of the two quadratics in (i) will give the four values
of x satisfying (1).
EXERCISES
Solve the following equations :
1. x 6 + 2.^ = 80. 2. o.r-4-2V5.r-4 = 03.
3. (2 — a; + x 2 ) 2 + x 2 — x = 18.
4. a 2 -3a+-4-3Va 2 -3a + 4= -2.
5. 3 a 6 - 7 a 8 - 1998 = 0.
6. .r - 8. r + 16 + GVar'- 8 a + 10 = 40.
7 - ( a+ I) !+4 ( a+ «) =2L
8. a 8 -97 a 4 + 1296 = 0.
9. a 2 - 3 « + 4 + Va 2 — 3 a + 15 = 19.
10. (5 x - 7 + 3 .r 2 ) 2 + 3 a 2 + 5 .v - 247 = 0.
11. V7 a - 6 -4 V7 a- 6 + 4 = 0.
RELATIONS BETWEEN THE ROOTS AND THE COEFFICIENTS OF A
QUADRATIC
153. If in the general quadratic, ax 2 + bx + c = 0, we divide
b c
both members by a and put - =p, - = q, we have x i -\-px-\-q=0.
a a
ci- — V + Vp 2 — 4 a , « — » — V/r — 4 q
Solving, x, = — i — ! 1 i , and xr = — ±- — ±- 2 .
RELATIONS OF COEFFICIENTS AND ROOTS 109
2 n
Adding «i and x 2 , x 1 + x 2 = — ~ = —p. (1)
Multiplying x 1 and x 2 , x 1 x. 2 = ^—— s ^—~ — ^ = q. (2)
Hence in a quadratic of the form x 2 -\- p>x -\- q = 0, the sum of
the roots is — p, and the product of the roots is q .
r P1 • o a b 2 4 c b 2 — 4 ac
lne expression p — 4 5 = = .
a 2 a a 2
Hence p 2 — 4 g is positive, negative, or zero, according as
b 2 — 4 ac is positive, negative, or zero.
Hence, as found on pp. 87, 89, the roots of
ax 2 + bx + c = 0, or x 2 + p.x* + q = are :
reaZ and distinct, ifb 2 — 4 ac > 0, or p 2 — • 4 g > 0, (3)
reaZ and equal, if b 2 — 4 ac — 0, or p 2 — 4 g = 0, (4)
imaginary, if b 2 — 4 ac < 0, or j> 2 — 4 g < 0. (5)
By means of (1) to (0), we may determine the character of
the roots of a quadratic without solving it.
Ex. 1. Determine the character of the roots of
8 a: 2 - 3 x - 9 = 0.
Since b 2 — 4 ac = 9 — 4 • 8( — 9) = 297>0, the roots are real and dis-
tinct. Since b' 2 — 4 ac is not a perfect square, the roots are irrational.
Since q = — f = x,z 2 , the roots have opposite signs.
Since p = — | or — p = f = x x + x 2 , the positive i - oot is greater in
absolute value.
Ex. 2. Examine 3 a- 2 + 5 » + 2 = 0.
Since b' 2 — 4 ac = 25 — 4 • 3 • 2 = 1 > 0, the roots are ? - eoZ and distinct.
Since 6 2 — 4 «c is a perfect square, the roots are rational.
Since q — § = x x x 2 , the roots have the same sign.
Since — p = — ^ = x l + x 2 , the roots are both negative.
Ex. 3. Examine x 2 — 14 x + 49 = 0.
Since p 2 — 4 9 = 196 — 4 • 49 = 0, the roots are real and coincident.
Ex. 4. Examine x 2 — 7 x + 15 = 0.
Since p 2 — 4 5 — 49 — 4 • 15 = — 11, the roots are imaginary.
110 QUADRATIC EQUATIONS
EXERCISES
Without solving, determine the character of the roots in each
of the following :
1. 5a 2 — 4a;- 5 = 0. 9. 16m 2 + 4 = 16m.
2. 6 x 2 + 4 x + 2 = 0. 10. 25 a 2 - 10 a = 8.
3. . c -' _ 4 x + 8 = 0. 11. 20 - 13 6 — 15 6 2 = 0.
4. 2 + 2 a 2 = 4 a;. 12. 10 y- + 39 y + 14 = 0.
5. 6 x + 8 ar = 9. 13. 3 a 2 + 5 a + 22.
6. 1 - a 2 = 3 a, 14. 3 a 2 - 22 a + 21 = 0.
7. a - 30 = 3 a 2 . 15. 5 b- +66 = 27.
8. 6a 2 + 6 = 13a. 16. Ctt-17 = lla 2 .
FORMATION OF EQUATIONS WHOSE ROOTS ARE GIVEN
154. Ex. 1. Form the equation whose roots are 7 and — 4.
From (1) and (2), § 153, we have
./■j + x 2 = — p = 7 + ( — 4) = 3. Hence p = — 3.
And XjZ 2 = </ = 7(- 4) = - 28.
Hence a. -2 + px + q = becomes a; 2 — 3 x — 28 = 0.
In case the equation is to have more than two roots, we pro-
ceed as in the following example:
Ex. 2. Form the equation whose roots are 2, 3, and 5.
Recalling the .solution by factoring, we may write the desired equa-
tion in the factored form as follows :
(x-2)(x-3)(x - 5) =0.
Obviously 2, 3, and 5, are the roots and the only roots of this equa-
tion. Hence the desired equation is:
(x - 2)(x - 3)0 - 5) = x 3 - 10 x 1 + 31 x - 30 = 0.
EQUATIONS WITH GIVEN ROOTS 111
EXERCISES
Form the equations whose roots are :
1. 3, -7. 4. 5, -4,-2. 7. -5, -6.
2. b, c. 5. V5, - V5. 8. - 6 + &, — 6 - k.
3. a, — b, — c. 6. a — Vo, a + V3. 9. V — 1, — V— 1.
10. a, - 6. 11. 8 + V3, 8 - V3. 12. 2, 3, 4, 5.
13. 3 + 2 V^I, 3-2 V = T. 14. S-V^l, 5 + V^T.
1C -, , , o ,^ — b + v'lr— 4ac — 6 — V& 2 — 4a c
15. 1, -iy, i, o. lb. — , — — .
2 a 2 a
155. An expression of the second degree in a single letter
may be resolved into factors, each of the first degree in that
letter, by solving a quadratic equation.
Ex. 1. Factor 6 a,- 2 — 17 x + 5.
This trinomial may be written, 6(x 2 — ig- x + |).
Solving the equation, x 2 — l 7 - x + £ = 0, we find x 1 = | and x 2 = §.
Hence by the factor theorem, § 92, x — * and x — f are factors of
x 2 - J /x + f. And finally
6(x 2 — 17 x + 5) = 6(x - i)(x - |) = 3(x - \) ■ 2(x - f)
= (%x- l)(2x — 5).
This process is not needed when the factors are rational, but
it is applicable equally well when the factors are irrational or
imaginary.
Ex. 2. Factor 3<c 2 + 8a>-7 = 3(ar 4- £ a; — £).
Solving the equation x 2 + § x — ^ = 0, we find,
- 4 + VWf , - 4 - V37
x, = ! and x = .
1 3 3
Hence as above :
3 x 2 + 8 x
-4 + V37"ir -4 -V37"
.i[,_^±v»][._
-A^I-fl^bTl
112 QUADRATIC EQUATIONS
EXERCISES
In exercises 1 to 16, p. 110, transpose all terms of each equa-
tion to the first member, and then factor this member.
PROBLEMS INVOLVING QUADRATIC EQUATIONS
In each of the following problems, interpret both solutions
of the quadratic involved :
1. The area of a rectangle is 2400 square feet and its perim-
eter is 200 feet. Find the length of its sides.
2. The area of a rectangle is a square feet and its perimeter
is 2 b feet. Find the length of its sides. Solve 1 by substitu-
tion in the formula thus obtained.
3. A picture measured inside the frame is 18 by 24 inches.
The area of the frame is 288 square inches. Find its width.
4. If in problem 3 the sides of the picture are a and b and
the area of the frame c, find the width of the frame.
5. The sides a and b of a right triangle are increased by the
same amount, thereby increasing the square on the hypotenuse
by 2 Jc. Find by how much each side is increased.
Make a problem which is a special case of this and solve it by sub-
stitution in the formula just obtained.
6. The hypotenuse c and one side a are each increased by
the same amount, thereby increasing the square on the other
side by 2 k. Find how much was added to the hypotenuse.
Make a problem which is a special case of this and solve it by
substituting in the formula just obtained.
7. A rectangular park is SO by 120 rods. Two driveways of
equal width, one parallel to the longer and one to the shorter
side, run through the park. AVhat is the width of the drive-
ways it their combined area is 591 square rods?
8. If in problem 7 the park is a rods wide and b rods long
and the area of the driveways is c square rods, find their
width.
PROBLEMS INVOLVING QUADRATICS 113
9. The diagonal of a rectangle is a and its perimeter 2 b.
Find its sides.
Make a problem which is a special case of this and solve it by sub-
stituting in the formula just obtained.
10. If in problem 9 the difference between the length and
width is b and the diagonal is «, find the sides. Show how
one solution can be made to give the results for both problems
9 and 10.
11. Find two consecutive integers whose product is a.
Make a problem which is a special case of this and solve it by sub-
stituting in the formula just obtained.
What special property must a have in order that this problem may
be possible. Answer this from the formula.
12. A rectangular sheet of tin, 12 by 16 inches, is made into
an open box by cutting out a square from each corner and
turning up the sides. Find the size of the square cut out if
the volume of the box is 180 cubic inches.
The resulting equation is of the third degree. Solve it by factor-
ing. See § 151. Obtain three results and determine which are appli-
cable to the problem.
13. A square piece of tin is made into an open box contain-
ing a cubic inches, by cutting from each corner a square whose
side is b inches and then turning up the sides. Find the
dimensions of the original piece of tin.
14. A rectangular piece of tin is a inches longer than it is wide.
By cutting from each corner a square whose side is b inches and
turning up the sides, an open box containing c cubic inches is
formed. Find the dimensions of the original piece of tin.
15. The hypotenuse of a right triangle is 20 inches longer
than one side and 10 inches longer than the other. Find the
dimensions of the triangle.
16. If in problem 15 the hypotenuse is a inches longer than
one side and b inches longer than the other, find the dimen-
sions of the triangle.
114 QUADRATIC EQUATIONS
17. The area of a circle exceeds that of a square by 10
square inches, while the perimeter of the circle is 4 less than
that of the square. Find the side of the square and the radius
of the circle.
Use 3i as the value of it.
18. If in problem 17 the area of the circle exceeds that of
the square by a square inches, while its perimeter is 2 b inches
less than that of the square, find the dimensions of the square
and the circle.
Determine from this general solution under what conditions the
problem is possible.
19. Find three consecutive integers such that the sum of
their squares is a.
Make a problem which is a special case of this and solve it by
means of the formula just obtained. From the formula discuss the
cases, a = 2, a = 5, a = 14. Find another value of a for which the
problem is possible.
20. The difference of the cubes of two consecutive integers
is 397. Find the integers.
21. The upper base of a trapezoid is 8 and the lower base is
3 times the altitude. Find the altitude and the lower base if
the area is 78.
See problem 7, p. 48.
22. The lower base of a trapezoid is 4 greater than twice
the altitude, and the upper base is \ the lower base. Find
the two bases and the altitude if the area is 52^-.
23. The lower base of a trapezoid is twice the upper, and its
area is 72. If £ the altitude is added to the upper base, and
the lower is increased by \ of itself, the area is then 120.
Find the dimensions of the trapezoid.
24. The upper base of a trapezoid is equal to the altitude,
and the area is 48. If the altitude is decreased by 4, and the
upper base by 2, the area is then 14. Find the dimensions of
the trapezoid.
PROBLEMS INVOLVING QUADRATICS 115
25. The upper base of a trapezoid is 4 more than 4 the
lower base, and the area is 84. If the upper base is decreased
by 5, and the lower is increased by i the altitude, the area is
78. Find the dimensions of the trapezoid.
26. The area of an equilateral triangle multiplied by V3,
plus 3 times its perimeter, equals 81. Find the side of the
triangle.
See problem 15, p. 236, E. C.
27. The area of a regular hexagon multiplied by V3, minus
twice its perimeter, is 504. Find the length of its side.
See problem 20, p. 237, E. C.
28. If a times the perimeter of a regular hexagon, plus V3
times its area, equals b, find its side.
29. The perimeter of a circle divided by -rr, plus V3 times
the area of the inscribed regular hexagon, equals 78f. Find
the radius of the circle.
30. The area of a regular hexagon inscribed in a circle plus
the perimeter of the circle is a. Find the radius of the circle.
31. One edge of a rectangular box is increased 6 inches,
another 3 inches, and the third is decreased 4 inches, making
a cube whose volume is 862 cubic inches greater than that of
the original box. Find its dimensions.
32. Of two trains one runs 12 miles per hour faster than the
other, and covers 144 miles in one hour less time. Find the
speed of each train.
In a township the main roads run along the section lines, one half
of the road on each side of the line.
33. Find the area included by the main roads of a township
if they are 4 rods wide.
34. If the area included by the main roads of a township is
11,190 square rods, find the width of the roads.
35. Find the width of the roads in problem 34 if the area
included by them is a square rods.
CHAPTER VIII
ALGEBRAIC FRACTIONS
156. An algebraic fraction is the indicated quotient of two
algebraic expressions.
ii
Thus — means n divided by d.
d *
From the definition of a fraction and § 11, it follows that
the product of a fraction and its denominator equals its numerator.
That is, d • -= n.
a
REDUCTION OF FRACTIONS
157. The form of a fraction may be modified in various ways
without changing its value. Any such transformation is called
a reduction of the fraction.
The most important reductions are the following :
(A) By manipulation of signs.
— n n — n h — a a — h a — b
E.g.
d d — d — d c — d c — d d — c
(B) To loivest terms.
X* + X 2 + 1 _ (x 2 + X + 1)(3? 2 - X + 1)
' 9 ' X 6 - 1 ~ (x - l)(x 2 + X + l)(x + l)(x 2 - X + 1)
1
(x- l)(x + l)
(C) To integral or mixed expressions.
2 a * + x* + x + 2 ., 1 + -r + l =2z ^x-1,
•' X 2 + 1 X 2 + 1 X 2 + 1
116
E.g.
REDUCTION OF FRACTIONS 117
(D) To equivalent fractions having a common denominator.
.g. — - — and become respectively ^ — "' and
x + 3 z + 2 ( x + 3)(x + 2)
^— - - — *■ — ; a + 1 and become respectively a ~ ~ and -
(a;+3)(ar + 2) a-1 a-1 a-1
158. These reductions are useful in connection with the
various operations upon fractions. They depend upon the
principles indicated below.
Reduction (.4) is simply an application of the law of signs in
division, § 28. It is often needed in connection with reduction (Z>).
See § 159.
Reduction (B) depends upon the theorem, § 47, — = -, by which
bk b
a common factor may be removed from both terms of a fraction. It is
useful in keeping expressions simplified. This reduction is complete
when numerator and denominator have been divided by their H. C. F.
See §§ 95-102.
Reduction (C) is merely the process of performing the indicated
division, the result being integral when the division is exact, otherwise
a mixed expression.
In case there is a remainder after the division has been carried
as far as possible, this part of the quotient can only be indicated.
Thus »_, + *
in which D is dividend, d is divisor, q is quotient, and i? is
remainder.
Reduction (D) depends upon the theorem of § 47, -= — , by which
b kb
a common factor is introduced into the terms of a fraction.
A fraction is thus reduced to another fraction whose denominator
is any required multiple of the given denominator.
If two or more fractions are to be reduced to equivalent fractions
having a common denominator, this denominator must be a common
multiple of the given denominators, and for simplicity the L. C. M. is
used.
118 ALGEBRAIC FRACTIONS
EXERCISES
Keduce the following so that the letters in each factor shall
occur in alphabetical order, and no negative sign shall stand
before a numerator or denominator, or before the first term of
any factor.
n — m — (c — a) (d — c)
b — a (a — b) (6 — c)
(b — a)(c — d) ^ (b — a)(c— b)(c — a)
9. — ■
x(s— r-
- ( x — y)
-o
(b — a) (c — i
-(x-y)(z-
— (6 — o)(c-
-d)
<
10.
11.
(a — b) (b — c) (c — a)
(c — & — a) (6 — a — c)
3 (a — c) (6 — c) (c — a)
(3c-2q)f46-a)d
(a — 6) (c — b)(c — a) ( — a + 6) (a — &)(c —a)
g _ a(c + 6 ) ^ -(-r-.s)( g -Q(f-r)
b (c — a) (n — m) (—k — m — I)
Reduce each of the following to lowest terms :
a 4 -b
a 6 - b
13. a 4 ~& 4 , 18 . <* + 2a? + 2x + l
14 c 2 -(a-&) 2
(a + c) 2 -6 2
19.
- _ 7 a# 2 — 56 a^ar* 9ft
' 28^(1 -64 a 6 * 6 )'
16 m 3 +5ffl' + 7m + 3 21
///-' + 4m + 3
17 a 3 - 7 a + 6 22
a 3 - 7 a 2 + 14a- 8* ' cr + ab + lr + a + b
x* + a
;3_ iC 2_2.T_2
2a?-
_ ar 2 _ 8 a; — 3
2X 3 -
3 or — 7 .i' -f 3
4 jc 8 +
■8 a 2 — 3 a: + 5
6^-
5 .r + 4 ,v — 1
a 2 -
#V + ?/ 2 + * — ?/ + 3
^ + 2/
3 + x 2 -y 2 + 3x + 3y
a 4 + a
2 b 2 + & 4 + a 3 + '< :;
23.
REDUCTION OF FRACTIONS 119
X- 4 + 4 x?y + 6 x 2 y 2 + 4 an/ 3 + y 4 — a 4
oj 2 + 2 xy + y 2 — a 2
x 2 y — x 2 z + y 2 z — xy 2 -f- a-z 2 — ?/z 2
a 2 — (y + ») & + y z
25 2 a: 4 -a? -20 a- 2 + 16 x -3
3 ic 4 + 5 x 3 — 30 x 2 — 41 a; + 5
3 a s _ 8 a 2 b - 5 air + 6 b s
26.
27.
a 3 + C M _ 9 a &2 _ 9 53
2 r 3 + r 2 s + rs 2 + 2 s 3
2 ?- 4 + rs + 3 rW -f- rs 3 + 2 s 4
Reduce each of the following to an integral or mixed
expression :
r 4 4- 1 a -4 ^ 5
28. ^-X±. 30. _^_. 32.
a; + 1 x — 1 c 3 + c 2 — c + 1
29. £±1. 31. *_. 33. ^"» + l
a; + 1 a" + a + 1 ar + a? + 1
„ a 4 + a 2 6 2 + & 4 QO ar'-ar'-a + l
,54. • 00. — — •
a — 0. x 3 -f- ar -f a; — 1
3 a 3 — 3 a 2 + 3 a — 1 4 m 4 — 3 m 3 + 3
a — 2 2 m 2 — 2 m + 1
Reduce each of the following sets of expressions to equiva-
lent fractions haviner the lowest common denominator:
38.
39.
a; 4 — 3 xry 2 + ?/ 4 x 2 — xy — y 2 ' x 2 + xy—y 2
a + b a b
5 a 2 c + 12 cd — 6 ad — 10 «o 2 ' 5ac — 6d' a — 2c
40 a^ 2 + ?/ 2 « + :>/— x 2 + xy+j/ 2
x s + y 3 + x 2 — xy + y 2 x 2 — xy + y 2 x + y + 1
120 ALGEBRAIC FRACTIONS
x y
41.
(a — b)(c — b)(c — a) (a— b)(b — c)(a — c)
42. uJ5_, _^_, -i_, d
6 + c c + a a -f- o
43.
44.
_(6-a)(&-c)(a-c)
6 — c a — b c — a
(a— c)(a — b)' (c — a)(6 — c)' (6 — a)(c — b)
m — n a + 2 a + 3
a 3 — 6 dr + 11 a — 6 ' a 2 — 4 a + 3 ' a 2 — 3 a 4- 2
If o, b, m are positive numbers, arrange each of the follow-
ing sets in decreasing order. Verify the results by substitut-
ing convenient Arabic numbers for a, b, m.
Suggestion. Reduce the fractions in each set to equivalent frac-
tions having a common denominator.
a 2 a 3a ._ m 2 m 3 m
45. -, t:, -• 46.
a + l' a + 2' a + 3 " 2m + l' 3m4-2' 4m + 3
a 4- 3 b a + b a -\- Ab
47.
a + 4 b ' a + 2 b ' a + 56
48. Show that, for a different from zero, neither - - nor
" d + a
n ~ a can equal -, unless n = d. State this result in words,
d — a d
and fix it in mind as an impossible reduction of a fraction.
ADDITION AND SUBTRACTION OF FRACTIONS
159. Fractions which have a common denominator are added
or subtracted in accordance with the distributive law for
division, § 30.
rm a . 6 c a -f b — c
1 hat is, - + - — : = — —z
d d d d
In order to add or subtract fractions not having a common
denominator, they should first be reduced to equivalent frac-
tions having a common denominator.
ADDITION AND SUBTRACTION OF FRACTIONS 121
When several fractions are to be combined, it is sometimes
best to take only part of them at a time. In any case it is
advantageous to keep all expressions in the factored form as
long as possible.
Ex. —J , i + 1
(x - 1)0 - 2) (2 - x)(x -3) (3 - x)(± - x)
Taking the first two together, we have
1 1 2x-4 2
(x - l)(s-2) (*-2)(x-8) (x-l)(x-2)(a!-8) (x-l)(x-3)
Taking this result with the third,
2 1 3 x- 9 3
(x-l)(x-3) (x-8)(x-4) (s-l)(x-8)(x-4) (x-l)(x-4)
If all are taken at once, the work should be carried out as follows :
The numerator of the sum is
(x - 3)(x - 4) + (x- 1)0 - I) + (x - 1)0 - 2).
Adding the first two terms with respect to (x — 4), we have
20-2)(z-4) + (x- l)(x-2>
Adding these with respect to (x — 2), we have 3(x — 3)0 — -)•
tt «. • 3(x- 3)(x - 2)
Hence the sum is * -^ -f-
(x - l)(x- 2)0 - 3)0 - I) O — !)(*— *)
EXERCISES
Perform the following indicated additions and subtractions :
1 2 3 4 3 5 1
x-3 x — 4 x-5 40 + 3) 80 + 5) 80+1)
2 0-1) 0-2 2 0-3)
1 7 13
12 + 1) 3 0-2) 40-3)
122
ALGEBRAIC FRACTIOXS
9
5
, -3,4 „ 5a; + 6 3a- -4
(.'• + l) 2 ' x + 1 ' x — 2 ' x 2 -f a: + 1 ar — a; + 1
6 1
x - 2 Q 1 4 x - 8
3 (x + 1) 3 (x- — a; + 1) .",(./• + 2 ) ' 5 ( .r + 1 )
2 11
9 1 ■
(a; - 2) 2 a; - 2 a; + 1
° 1 r 4- 2
10 1 ^ -
(a;~2) 2 6(a>-2) 5(x» + l)
1\
1 1 1 ' *
(as_l)« (a,_i) («"_!)
12.
1 1 3 1 3 1 1
2 (1 - 3 xf 8 (1 - 3 a;) 2 ' 32 (1 - 3 x) ' 32 (1 + a?)
13
1 12
(1 - a) (2 - a) (2 - a) (a - 3) ' (3 - a)(a - 1)
14
a; y yz xz
(z - ?/) (a? - z) (x- z) (x - y) (y - x){y - z)
15
1 2a -5 5 a 2 - 3 a - 2
a - 1 a 2 - 2 a + 1 (a - l) 8
16.
1 1 2m+2
m 2 + m + 1 nr — m + 1 m 4 + »i 2 + 1
17.
11 2
6 2 - 3 b + 2 ' V 1 - 5 6 + 6 b- - 4 b + 3
18.
r + s s + t r + t
(r-t) (s - (r - s)(t - r) (t - s) (s - r)
19 p- + <f | (f-pr ( 9-°+p 9
(y> - 7 K P + '•) (q - r)(q - p) (r - q)(r+p)
20. ;! - 4 - 1
5 x- - 18 x + 9 4 ar' - 11 a; - 3
MULTIPLICATION AND DIVISION OF FRACTIONS 123
MULTIPLICATION AND DIVISION OF FRACTIONS
160. Theorem. Th e produ ct of two fractions is a fraction
whose numerator is the product of the given numerators
and whose denominator is the product of the given de-
nominators.
Proof. We are to prove that - • - = — •
1 b d bd
Let
Then
X = -
b
d
bdx = hi
'a
Kb
d)
§7
bdx = b ■
a
b '
d
§8
bdx = an.
§H
_ an
X ~bd
a n _ an
b ' d~M
§2
Hence,
Therefore,
Corollary 1 . A fraction is raised to any power by raising
numerator and denominator separately to that power.
t-i i ,i i ,i a a a' 2 a a a a 3
ior by the above theorem, - • - = — , _.-.- = —, etc.
J b b b 2 b b b b 3
Corollary 2. A fraction multiplied by itself inverted
equals + 1.
For n - ■ i = ^.= + l and-^ • f_^=^= + l.
d ?i nd d \ nl nd
161. Definitions. If the product of two numbers is + 1, each
is called the reciprocal of the other. Hence from Cor. 2, the
reciprocal of a fraction is the fraction inverted.
Also, since from ab = l we have « = - and & = -, it follows
b a
that if two numbers are reciprocals of each other, then either
one is the quotient obtained by dividing 1 by the other.
■MM
124 ALGEBRAIC FRACTIONS
162. Theorem. To divide by any number is equivalent to
multiplying by its reciprocal.
n 1
Proof. We are to prove that n ~- d or - = n — This is an im-
mediate consequence of § 29.
Corollary 1. To divide a n umber by a fraction is equiva-
lent to multiplying by the fraction inverted.
For by § 161 the reciprocal of the fraction is the fraction inverted.
Corollary 2. , / fraction, is divided by an integer by m 7/1-
tiph/in'J its denominator or dividing its numerator by
that integer.
For - + a = - ■ - = — , Cor. 1 and § 160
d d a ad
and - -*■ a — , since — = by $ 4 1 .
d d ad d
In multiplying and dividing fractions their terms should at
once be put into factored forms.
When mixed expressions or sums of fractions are to be mul-
tiplied or divided, these operations are indicated by means of
parentheses, and the additions or subtractions within the
parentheses should be performed first, § 38.
Ex. Simplify
1 , 2a 2 \ f 1 1
1 — a +
1+uJ VI + a 1 — a
« 4 -l
Performing the indicated operations within the parentheses, we have
r l + a 2 . 2d I 3 a 8 _ 1 + a" a 2 - 1 3 a 3 _ 3 a 2
l-l+o 'o ! -lJ'fl 4 -l l + o' 2a ' (V- l)(a 2 +l) 2(a + l)'
EXERCISES
Perform the following indicated operations and reduce each
result to its simplest form.
1 ,r 4 -f x-y- + y* m a* 2 — f .
X s — y 3 x 3 + f
MULTIPLICATION AND DIVISION OF FRACTIONS 125
a 2 — b' 2 x' 2 + ucx 2 — bcx 3 _ — ay 2 — bxy 2 — cx 2 y 2
30 ~ a *--~War r + 24 a - 16 "*" 20 a; 2 - 15 ao 2 - 30 aV
20 rV + 23 rsf — 21 1 2 12 wm 3 - 28 mnhj — 24 ran?/ 2
8 //r'/< 3 — 48 m 2 n 2 y + 72 mr'ny* 10 rV + 24 rs£ — 18 t 2
4. f a ftVa' , 6 2 a & , 1 N. a 5 + 6 g
\ 6 aj\b 2 a 2 6 c< y a — 6
f . 3 w+ 2 i
L : 2s + 3,
t
5.
my*- 1
a; 2/y\cc W
s-yyo i 2 -V
*+
iA'
■« -2/
« — 6 a + 67 Ur &-y Va 2 6 2
10.
7. 1 +
1-
a — £
>/i + n m — n
m — n m + n
x 2 + y 2 ar — y'
x 2 -y 2 x 2 + y :
a + &y V a 2 — 6 2
. . 2ir \ fm + n . m—n
mi — »y \m — ra m + h
r+r +
2.H/ 2 +2.v
x+y + z
+
x + 1/ (as + ^/) 2 y V ( •*• + yf—z\
x- - y
(x + yf
x+y x—y
v—y x+y
1 +
* + y
a' 2 + ab + b 2 a +- b f , , b s — a 2 b
11. - — ! • ■ — ■ • a -\
a' 2 — ub + b'- a? — b V a + b
12.
m + mix m? — mir — mf + nr m'-ir + mil 3 + rr
m 2 + >i 2 m 3 n — /i 4
13. |ay» +a fy_2^r
x-y
x- + y-
m 4 -
2 m 3 + mr
■*■
m s n + 2 m
V 2 +
mn
m* -
-u*
1
-I)
or + y 2 I
n _
31
r
f {
v r
*yj
126 ALGEBRAIC FRACTIONS
COMPLEX FRACTIONS
163. A fraction which contains a fraction either in its
numerator or in its denominator or in both is called a complex
fraction.
Since every fraction is an indicated operation in division,
any complex fraction may be simplified by performing the indi-
cated division.
It is usually better, however, to remove all the minor
denominators at once by multiplying both terms of the com-
plex fraction by the least common multiple of all the minor
denominators according to § 47.
* + ± ± + ± .6
I- i 3 2 V\ 21 2x + 3a; ;" /•
r or example, = = ^ = — _ — ,
2i_ 2 _3 I'ljP _ 3\ 4 a: 2 - 9 4 a: 2 - 9
3 2 V 3 2/
A complex fraction may contain another complex fraction
in one of its terms.
E ( , — has the complex fraction
fl 4^ « +
1 a - 1
a +
a — 1
in its denominator. This latter fraction is first reduced by multiply-
ing its numerator and denominator by a — 1, giving
a + 1 a' 2 — 1
a+ 1 «»-« + !
a — 1
Substituting this result in the given fraction, we have
1 _ 1 _ a 2 - a + 1
„ + -JL±1 ~ „ + Z 2 - 1 "a' + a-l 1
1 a 2 - a + 1
a + -
a — 1
COMPLEX FRACTIONS 127
EXERCISES
Simplify each of the following,
m 2 + mn
1
m 2 — ir
m n
m
— n m 4 n
a'-b 4
2.
a 2
-2ab + b 2
a 2 4 ab
•A.
a—b
x 5 — 3 x 4 y + 3 xhf 2 — x 2 y s
x s y — y 4
x> -2 x 4 y 4 x A f
xhf 4- xf 4 1
1 +-^-+4^-,
a 2
a 2
a-2+-
a
a 2 4 1 4 -,
a'
a
2 1
-24--—
a or
1
1
1
1-1
a 4 x a — as a,"
a-\-x a — x a- — a; 2
+ -^ + ^1 l- 1 -
a 4 *' «> — x a - 4 *
iC
a — x" a + ,c a+^
10.
o , q 7/1 3 — n 3
■»i" — mr« 4 "" « 4
■m 4 »• o _i_ 3
7/i"4w-4»i"H ■ — & +
.»■
EQUATIONS INVOLVING ALGEBRAIC FRACTIONS
164. In solving a fractional equation, it is usually conven-
ient to clear it of fractions, that is, to transform it into an equiv-
alent equation containing no fractions.
In case no denominator contains any unknown this may be
done by multiplying both members by the L. C. M. of all the
denominators, § 62.
128 ALGEBRAIC FRACTIONS
When, however, the unknown appears in any denominator,
multiplying by the L. C. M. of all the denominators may or may
not introduce new roots, as shown in the following examples.
It may easily be shown, that multiplying an integral equation
by any expression containing the unknown always introduces new
roots.
Ex.1. Solve JL_ + ^!_ = 2. (1)
x-2 x-'S v '
Clearing of fractions by multiplying by (x — 2)(.r — 3), and simpli-
fying, we have . „ ,_ _ rt , lwo _.
J & ' 2 ./ - - 13 x + 20 = (x - 4) (2 x - 5) = 0. (2)
The roots of (2) are 4 and 2 1 ,, both of which satisfy (1). Hence
no new root was introduced by clearing of fractions.
Ex.2. Solve — 1— = . (1)
jc_1 (a;-l)(a;_2)
Clearing of fractions, we have,
a; -2 = 1. (2)
The only root of (2) is x = 3, which is also the only root of (1).
Hence no new root was introduced.
Ex. 3. Solve ^fll = i. (1)
x 2 - 4
Clearing of fractions and simplifying, we have,
a! 2_ a; _2 = ( a; _2)(s + l) = 0. (2)
The roots of (2) are 2 and — 1. Now x = — 1 is a root of (1). but
x = 2 is not, since we are not permitted to make a substitution which
reduces a denominator to zero, § 50. Hence a new root has been in-
troduce'! and (1) and (2) are not equivalent.
x - 2
If the fraction ~r, r is first reduced to lowest terms, we have the
x 2 — 4
eiiuation -i
— — = 1. (3)
x + 2 K }
Clearing of fractions, x + 2 = 1. (4)
Now (:'>) and ( I) are equivalent, — 1 being the only root of each.
EQUATIONS INVOLVING FRACTIONS 129
Ex.4. Solve _i^__^±l = i. (i)
x- — 1 X — 1
dealing of fractions and simplifying,
x' 2 - x = x(x - 1) = 0. (2)
The roots of (2) are and 1. x = satisfies (1), but x = 1 does
not, since it is not a permissible substitution in either fraction of (1).
Hence a new root has been introduced.
165. Examples 3 and 4 illustrate the only cases in which new
roots can be introduced by multiplying by the L. C. M. of the
denominators.
This can be shown by proving certain important theorems,
the results of which are here used in the following directions
for solving fractional equations :
(1) Reduce all fractions to their lowest terms.
(2) Multiply both members by the least common multiple
of the denominators.
(3) Reject any root of the resulting equation which reduces
any denominator of the given equation to zero. The remain-
ing roots will then satisfy both equations, and hence are the
solutions desired.
If when each fraction is in its lowest terms the given equation con-
tains no two which have a factor common to their denominators, then
no new root can enter the resulting equation and none need to be re-
jected. See Ex. 1 and Ex. 3 after being reduced.
If, however, any two or more denominators have some common
factor x — a, then x = a may or may not be a new root in the resulting
equation, but in any case it is the only possible kind of new root
which can enter, and must be tested. Compare Exs. 2 and 4.
Ex.5. Examine 3x + 7 + — ^±1 = 8.
3^ + 2 as + 11 a? + 3x + 2 x — 1
Since each fraction is in its lowest terms and no two denominators
contain a common factor, then clearing of fractions will give an equa-
tion equivalent to the given one.
130 ALGEBRAIC FRACTIONS
Lx. 6. Examine — ! — — = 4.
ar + ox + G xr + 3 x 4- *
Each fraction is in its lowest terms, but the two denominators have
the factor x + 2 in common. Hence x = — 2 is the only possible new-
root which can enter the resulting integral equation, but on trial it is
fou ml not to be a root. Hence the two equations are equivalent.
EXERCISES
Determine whether each of the following when cleared of
fractions produces an equivalent equation, and solve each.
2. ±±**±± = 2x + S.
3 x 2 — 7 x + 3 x- — 4
3 3 | ° | 1 _Q
2^-a-l f-1 s+1
. — t C . Ju O it -*
4. r- = — 1.
2 a - 1 a- + 1 a - 1
1 . .,, a + x , 9 6 + x
u-b ' — = air —
3(aj-l) a,- 2 -l 4 b
6 . 2a-l + l = 3a . 1Q ^^hi,
a 2 3« — 1 1 — okb
2 3 fi "
.;■ — a .r — 6 .t — c .r — 40 10 — .'•
1 1 ."> + x- (') — .<• .r — 4 c
a — x a 4- .'■ a 2 — or cc — 4 6 — a; d
13 « | 24 = 2(«-4) 1.
2a-l 4a--l 2 a + 1 9
a a — 1 _ a 2 + a — 1
a — \ a a 2 — a
15. -J 2_ = i + _J
a 2 — 4 2 — a 3(a + 2)
EQUATIONS INVOLVING FRACTIONS 131
ax + b . ox + d ax — b . ex — d
16. 1 = 1 •
a + bx e + dx a — bx e — dx
(a — x)(x — b) _ , _ x + m — 2 n. _ n -4- 2 m — 2 a- _
(a — .r) — (a; — 6) x + ?/t -f 2 n n — 2 m 4- 2 .«
(a — xf — (x— b) 2 _ 4 afr
(a — x) (x — b) a 2 — b 2
20 1 + 3x 9 ~ ll a; = ll (2a-~3y
5 4- 7 a; 5 — 7x 25 — 49 x 2
x + 2a ,x — 2a _ 4a&
2 6 — oj 2 6 + a; 4 ft 2 — a; 2
22. JL+ 7x = -J—
.r -2 24 (as + 2) a~ - 4
23. 3? + ft ( a;-« = 2 (a 2 + 1)
a; — a a: + a (1 + «) (1 — a)
24. * = * = »=£. 25. -1 **- = b.
x + hi n + .f a; — a x 1 — a 2
26. _4_ + 4(3*-l) = 3j?L ± l.
3 a; + 1 2 a; + 1 3 x + 1
2s + 3 + T-3s s-7 =Q
2(2 x - 1) 3 x- 4 2(a; + 1)
28 . J_ + «_^ = _J_+£±^.
a — 6 a; a + b x
1 p
29
3 (m + » ■)" w + » 2 (m + n)
p 2 x p
30 - /2 + 2 ?/ ~ 2 + y _. ?/ .
y* + 5y + 6 ?/ + 3 y + 2*
132 ALGEBRAIC FRACTIONS
__ 5 7 _ 8 a: 2 — 13 a; — 64
2 a; + 3 3 a; — 4 6 ar* + sc — 12
«• ^'-,-^-
33.
sc 3 — a 8 a; — a x 2 + ckb + a?
1 _ l> x 5 _ 6 a; 8 1 - 3 a 8
3 - 4 x- 7 — 8 aj 3 21 - 52 cc + 32 a~
34. m ~9 | n-'p = m-q n-p
x — n x — q x — p x — in
35. -* ^- + - -5- - = 0.
aj-3 a; -2 4.^ -20a; + 24
or q i
36. — ; + _i_ = 0.
.c :i + 27 x- — 3 x + 9 as + 3
37.
.c
9 .'■ — 7 s» — 9 &• — 8 x — 7 x —
- 5 x — 2 x — 4 a- — 5 x — 4 a; — 2
38 q — ^ + h - c ) ( x ~ h + c ) .
(6 + c + x) (6 + c — x)
PROBLEMS
1. Find a number such that if it is added to each term of
the fraction f and subtracted from each term of the fraction
i| the results will be equal.
2. Make and solve a general problem of which 1 is a
special case.
3. Three times one of two numbers is 4 times the other.
If the sum of their squares is divided by the sum of the num-
bers, the quotient is 42$ less than that obtained by dividing
the sum of the squares by the difference of the numbers. Find
the numbers.
PROBLEMS 133
4. The sum of two numbers less 2, divided by their differ-
ence, is 4, and the sum of their cubes divided by the difference
of their squares is If times their sum. Find the numbers.
5. The circumference of the rear wheel of a carriage is
4 feet greater than that of the front wheel. In running one
mile the front wheel makes 110 revolutions more than the rear
wheel. Find the circumference of each wheel.
6. State and solve a general problem of which 5 is a special
case, using b feet instead of one mile, letting the other num-
bers remain as they are in problem 5.
7. In going one mile the front wheel of a carriage makes
88 revolutions more than the rear wheel. If one foot is added
to the circumference of the rear wheel, and 3 feet to that of
the front wheel, the latter will make 22 revolutions more than
the former. Find the circumference of each wheel.
8. State and solve a general problem of which 7 is a special
case, using a instead of 88, letting the other numbers remain
as they are.
9. The circumference of the front wheel of a carriage is
a feet, and that of the rear wheel b feet. In going a certain
distance the front wheel makes n revolutions more than the
rear wheel. Find the distance.
10. State and solve a problem which is a special case of
problem 9, using the formula just obtained.
11. There is a number consisting of two digits whose sum,
divided by their difference, is 4. The number divided by the
sum of its digits is equal to twice the digit in units' place
plus -|- of the digit in tens' place. Find the number.
12. There is a fraction such that if 3 is added to each of its
terms, the result is 4, and if 3 is subtracted from each of its
terms, the result is ^. Find the fraction.
13. State and solve a general problem of which 12 is a
special case.
134 ALGEBRAIC FRACTIONS
14. A and B working together can do a piece of work in
6 days. A can do it alone in 5 days less than B. J low long
will it require each when working alone'.'
15. State and solve a general problem cf which 14 is a
special case.
16. On her second westward trip the Mauritania traveled
625 knots in a certain time. If her speed had been 5 knots
less per hour, it would have required 6\ hours longer to cover
the same distance. Find her speed per hour.
17. By increasing the speed a miles per hour, it requires
b hours less to go c miles. Find the original speed. Show
how problem 16 may be solved by means of the formula thus
obtained.
18. A train is to run d miles in a hours. After going c miles
a dispatch is received requiring the train to reach its destina-
tion b hours earlier. What must be the speed of the train for
the remainder of the journey ?
19. A man can row a miles down stream and return in b
hours. If his rate up stream is c miles per hour less than
down stream, find the rate of the current, and the rate of the
boat in still water.
20. State and solve a special case of problem 19.
21. A can do a piece of work in a days, B can do it in b days,
and ( 1 in c days. How long will it require all working to-
gether to do it ?
22. Three partners, A, B, and C, are to divide a profit of p
dollars. A had put in a dollars for m months, B had put in
b dollars for // months, and C c dollars for t months. What
share of the profit does each get?
23. State and solve a problem which is a special case of the
preceding problem.
CHAPTER IX
RATIO, VARIATION, AND PROPORTION
RATIO AND VARIATION
166. In many important applications fractions are called
ratios.
E.g. f is called the ratio of 3 to 5 and is sometimes written 3 : 5.
It is to be understood that a ratio is the quotient of two num-
bers and hence is itself a number. We sometimes speak of the
ratio of two magnitudes of the same kind, meaning thereby
that these magnitudes are expressed in terms of a common unit
and a ratio formed from the resulting numbers.
E.g. If, on measuring, the heights of two trees are found to be 25
feet and 35 feet respectively, we say the ratio of their heights is ff or f.
167. Two magnitudes are said to be incommensurable if there
is no common unit of measure which is contained exactly an
integral number of times in each.
E.g. If a and d are the lengths of the side and the diagonal of a
square, then d' 2 — a' 2 + a' 2 . § 151, E. C. Hence, — = - or - = — -. But
d 2 '2 d -y/2
since V'2 is neither an integer nor a fraction (§ 108), it follows that a
and d have uo common measure, that is, they are incommensurable.
168. In many problems, especially in Physics, magnitudes
are considered which are constantly changing. Number ex-
pressions representing such magnitudes are called variables,
Avhile those which represent fixed magnitudes are constants.
E.g. Suppose a body is moving at a uniform rate of 5 ft. per
second. If t is the number of seconds from the time of starting
and s the number of feet passed over, then s and t are variables.
135
13(3 RATIO, VARIATION^ AND PROPORTION
The variables s and t, in case of uriform motion, have a. fixed ratio;
namely, in this example, s:t = 5 for every pair of corresponding
values of s and t throughout the period of motion.
169. When two variables are so related that for all pairs of
corresponding values, their ratio remains constant, then each
one is said to vary directly as the other.
E.g. If s : t = k (a constant) then s varies directly as t, and t varies
directly as s.
Variation is sometimes indicated by the symbol oc. Thus
sac t means s varies as t, i.e. - = k or s = kt.
t
170. When two variables are so related that for all pairs of
corresponding values their product remains constant, than each
one is said to vary inversely as the other.
E.g. Consider a rectangle whose area is A and whose base and
altitude are b and h respectively. Then, A = h • b.
If now the base is multiplied by 2, 3, 4, etc., while the altitude is
divided by 2, 3, 4, etc., then the area will remain constant. Hence,
b and h may both run/ while .4 remains constant.
The relation b ■ h = A may be written b = A • - or h — A ■ -• It
1 1 h b
mav also be written b : — = A or h: ~ — A, so that the ratio of either
h b
b or /< to the reciprocal of the other is the constant A. For this rea-
son one is said to vnr;/ i/tn rsr/// as tit? other.
171. If y = A.r, k being constant and x and y variables, then
y varies directly as x". If y = — , then y varies inversely as x 2 .
xr
If y = k ■ ivx, then y varies jointly as w and x. If y oc wx, then y
x w if x is constant and y cc x if w is constant. If y = A - • — , then
a;
y varies directly as w and inversely as x.
Example. The resistance offered by a wire to an electric
current varies directly as its length and inversely as the area
of its cross section.
If a wire | in. in diameter has a resistance of r units per mile,
find the resistance of a wire \ in. in diameter and 3 miles long.
RATIO AND VARIATION 137
Solution. Let R represent the resistance of a wire of length I and
cross-section area s = tt • (radius) 2 . Then R = k ■ - where k is some
constant. Since R = r when 1 = 1 and s = ^(y^) 2 , we have
1 , 77T
r — I- . or k =
r ~ k _7r_ 256
256
Hence, when I = 3 and s = 7r(|) 2 , we have,
j, _ TIT 3 _ 3
256 ' jr 4 ? '
64
That is, the resistance of three miles of the second wire is
| the resistance per mile of the first wire.
PROBLEMS
1. If zee id, and if z = 27 when w = 3, find the value of z
when w = 4^.
2. If 2 varies jointly as to and a*, and if z = 24 when ro = 2
and x = 3, find z when w = '3\ and x = 7.
3. If z varies inversely as w, and if 2 = 11 when w = 3,
find 2 when w = G6.
4. If z varies directly as to and inversely as x, and if z = 28
when zc = 14 and x = 2, find z when w = 42 and x = 3.
5. If 2 varies inversely as the square of w, and if z = 3
when ^c = 2, find z when w = 6.
6. If 7 varies directly as m and inversely as the square
of rZ, and q = 30 when m — 1 and d = T ^, find q when m = 3
and d = GOO.
7. If ?/ 2 oc x s , and if ^/ = 16 when x = 4, find ?/ when a* = 9.
8. The weight of a triangle cut from a steel plate of uni-
form thickness varies jointly as its base and altitude. Find
the base when the altitude is 4 and the weight 72, if it is
known that the weight is 00 when the altitude is 5 and base 6.
138 RATIO, VARIATION, AND PROPORTION
9. The weight of a circular piece of steel cut from a sheet
of uniform thickness varies as the square of its radius. Find
the weight of a piece whose radius is 13 ft., if a piece of
radius 7 feet weighs 196 pounds.
10. If a body starts falling from rest, its velocity varies
directly as the number of seconds during which it has fallen.
If the velocity at the end of 3 seconds is 96.6 feet per second,
find its velocity at the end of 7 seconds ; of ten seconds.
11. If a body starts falling from rest, the total distance
fallen varies directly as the square of the time during which it
has fallen. If in 2 seconds it falls 64.4 feet, how far will it
fall in 5 seconds ? In 9 seconds ?
12. The number of vibrations per second of a pendulum
varies inversely as the square root of the length. If a pen-
dulum 39.1 inches long vibrates once in each second, how
long is a pendulum which vibrates 3 times in each second ?
13. Illuminating gas in cities is forced through the pipes by
subjecting it to pressure in the storage tanks. It is found that
the volume of gas varies inversely as the pressure. A certain
body of gas occupies 49,000 cu. ft. when under a pressure of 2
pounds per square inch. What space would it occupy under
a pressure of 2i pounds per square inch ?
14. The amount of heat received from a stove varies in-
versely as the square of the distance from it. A person sitting
15 feet from the stove moves up to 5 feet from it. How much
will this increase the amount of heat received?
15. The weights of bodies of the same shape and of the same
material vary as the cubes of corresponding dimensions. If a
ball 3* inches in diameter weighs 14 oz., how much will a ball
of the same material weigh whose diameter is 3. 1 , inches?
16. On the principle of problem 15, if a man 5 feet 9 inches
tall weighs 165 pounds, what should be the weight of a man of
similar build 6 feet tall ?
PROPORTION 139
PROPORTION
172. Definitions. The four numbers a, b, c, d are said to be
proportional or to form a proportion if the ratio of a to b is equal
to the ratio of c to d. That is, if -=-• This is also sometimes
' b d
written a:b::c:d, and is read a is to b as c is to d.
The four numbers are called the terms of the proportion ;
the first and fourth are the extremes; the second and third the
means of the proportion. The first and third are the antece-
dents of the ratios, the second and fourth the consequents.
If a, b, c, x are proportional, x is called the fourth propor-
tional to a, b, c. If a, x, x, b are proportional, x is called a mean
proportional to a and b, and b a third proportional to a and x.
173. If four numbers are proportional when taken in a given
order, there are other orders in which they are also proportional.
E.g. If ii, b, c, d are proportional in this order, they are also pro-
portional in the following orders: a, c, b, d; b, a, d, c; b, d, a, c;
c, a, d, b ; c, d, a, b ; d, c, b, a : and d, b, c, a.
Ex. 1. Write in the form of an equation the proportion cor-
responding to each set of four numbers given above, and show
how each may be derived from - = — See § 196, E. C.
b d
Show first how to derive - = - (1), and then - = - (2).
c d a c
t\ • i a + b c + d /0 v ,a — 6 c — d . , s
Derive also — ! — = (.3), and = (1).
a c a c
In (1) the original proportion is said to be taken by alterna-
tion, and in (2) by inversion ; in (3) by composition, and in (4)
by division.
Ex. 2. From - = - and (1), (2), (3), (4) obtain the following.
See pp. 279-281, E. C.
a±b _ c ±d a ±b _a a±c _a a ±b _a — b
b d c ±d c ' b±d b' c + d cTd'
When the double sign occurs, the upper signs are to be read together
and the lower si<nis together.
X
■YZ^Z}. MCALS
•:r won! exponent
■ - -
■
-------
/ r jfguarmeum» aAikk gov-
- .-
-
V :" = :
V - V = v"-'
' = C' n " -'-'-'
% 116
-
•■• - -
-rii
■
FBACTIONA VD NEGATIVE EXPONENTS 143
/I'll r
Similarly, from \b' t? • ft*. ..to r factoi . l>%
we show that ■ \b" J = (i'h ) .
Hence, = vV = (Vft ) '. See ;' 119
Thus a posit i- ina] exponent means a roo£ 0/ a power
or a power of a roi e numerator indicating the power and the
denominator indica the root.
E.g. J= Vol = ■ </ )-': v v •; J I, or (v^) 2 = 2 2 = 4.
177. Assuming I I to hold also for negative exponents,
and letting £ be a tive number, integral or fractional, we
determine as follow the meaning of 6 ' (read b exponent nega-
tive t).
By LawT, V ■ b~ l = ft° = 1. .: 16
Therefore, ft-' = -• §11
ft'
Hence a numberwith a negative exponent means the same
as the reciprocal qjthe number with, a positive exponent of the
same absolute i
E.q. a" 2 =-: L"*=i- = ! = 1.
a 8 ' 4 | 23 8
178. It thus appars that fractional and negative exponents
simply provide n ways of indicating operations already well
known. Sometime one notation is more convenient and s<
times the other.
Fractional and cgative exponents are also called powers.
E.g. x? may be 3ad x to the f power, and x~* may be read x to the
— 4 th power.
The limitation- a to principal roots and tl i the base,
imposed in theoi • on powers and roots in Chapter VT,
sarily ap] lv to cot responding theorems in this chapter. S
§§ 114-122."
CHAPTER X
EXPONENTS AND RADICALS
FRACTIONAL AND NEGATIVE EXPONENTS
174. The meaning heretofore attached to the word exponent
cannot apply to a fractional or negative number.
E.g. Such an exponent as § or — 5 cannot indicate the number of
times a base is used as a factor.
It is possible, however, to interpret fractional and negative
exponents in such a way that the laws of operations which gov-
ern positive integral exponents shall apply to these also.
175. The laws for positive integral exponents are :
I. a'" ■ a" = a m+n . § 43
II. a'" h- a" = a"' - ". § 46
III. (a'")" = a"'". § 115
IV. (a'" ■ b")p = a"'Pb"P. § 116
V. (a'" -f- s") p = a'"P -7- s"P. § 117
176. Assuming Law I to hold for positive fractional expo-
nents and letting /• and s be positive integers, we determine as
follows the meaning of 1/ (read l> exponent r divided by s).
, , . ( -Y - -
By definition, \lf J = Ir ■ Ir ■•• to s factors,
r r r
which by Law I = h* • ■■■ to s terms = 6 s = b r .
Hence. Ir is one of I lie s equal factors of l r .
r l
That is, b* - \ 7/ . and in parfcictHar 6* = VB . See § 114
142
FRACTIONAL AND NEGATIVE EXPONENTS 143
/ IV I 1 =
Similarly, from \b" J = b* ■ b s •••to r factors, = b",
we show that . 6 s = \ff ) = (vV) .
Hence, 6* = </F = (VFY. See § 119
Thus a positive fractional exponent means a root of a power
or a power of a root, the numerator indicating the power and the
denominator indicating the root.
E.g. J = Iffi = (V^) 2 ; 8*= ^6? = 4, or (v^)*= 2 2 = 4.
177. Assuming Law I to hold also for negative exponents,
and letting t be a positive number, integral or fractional, we
determine as follows the meaning of b~* (read b exponent nega-
tive t).
By Law I, V ■ Jr* = b° = 1. §46
Therefore, b~< = -■ § 11
Hence a number with a negative exponent means the same
as the reciprocal of the number with a positive exponent of the
same absolute value.
' 9 ' a a 2 ' 4 f 2 8 8*
178. It thus appears that fractional and negative exponents
simply provide new ways of indicating operations already well
known. Sometimes one notation is more convenient and some-
times the other.
Fractional and negative exponents are also called powers.
2
E.g. x s may be read x to the j power, and x~ 4 may be read x to the
— -ith -power.
The limitations as to principal roots and the sign of the base,
imposed in theorems on powers and roots in Chapter VI, neces-
sarily apply to the corresponding theorems in this chapter. See
§§ 114-122.
144 EXPONENTS AND RADICALS
In any algebraic expression, radical signs may now be re-
placed by fractional exponents, or fractional exponents by
radical signs.
In a fraction, any factor may be changed from numerator to
denominator, or from denominator to numerator, by changing
the sign of its exponent.
Z/—T 5/ I— 7/ 3/ — % 3 ' 4 2
Ex. 1. vr + 3 vx 3 • Vi/ + 5 w v y- = x 3 -f 8z T y* + ox'y*.
r.x. 2. — = ata -, siuce aox z = ab • — = — •
r< o t-a 5 •> 1 '"'"
Lx. •>. alr 6 c- = «c- • — =
¥ 63
- 4 1 1
Ex. 4. 32 5 = -±- = — i =
82 * (^32>
1 _ 1
2 4 ~ 16
EXERCISES
(n) In the expressions containing radicals on p. 151, replace these
by fractional exponents.
(b) Replace all positive fractional exponents on this page by
radicals.
(<■) Change all expressions containing negative exponents to equiv-
alent expressions having only positive exponents.
179. Fractional and negative exponents have been defined
so as to conform to Law I, §§ 176, 177. We now show thai
when so defined they also conform to Laws II, III, IV, and V.
To verify Law II. Since by Law I. n m -" ■ n n = n m , for m and n inte-
gral or fractional, positive or negative, it follows by § 11 that
a m •*- a n — a m ~ n for all rational exponents.
To verify Law III. Let r and .< be positive integers, and let k be
any positive or negative integer or fraction. Then we have :
r r* r
(1) ( f ,*y = </(a*y' = Vci*? = a ' = a' , by §§ 170, 115.
(2) (a*)" =— L_ = _J_ = «"V*. by § 177 and (1).
(a k y a' ' *
Hence (a*)" = a" 1 ' for all rational values of n and k.
FRACTIONAL AND NEGATIVE EXPONENTS 145
To verify Law IV. Let m and n be positive or negative integers or
fractions, and let r and s be positive integers, then we have
(1) («W/')' = </'(('"'/>")'■' = </a mr b nr ', by §§ 170. 115,
= <fa^ . </b^' = a s ' m • b 1 ' ", by §§ 120, 176.
(2) (a m b»)~'= — J— = r l ^ =a"'"-ft"'"", by § 177 and (1).
Hence (a m b n )p = a pm b pn for all rational values of m, n, and p.
(n»*\ p a m P
— ) = — for all rational values of
m, n, ana /), since
(^)" = (a™ ■ b n )P = a m P ■ b-'P = — , § 177 and Law IV.
V (>" I b n P
180. From Laws III, IV, and V, it follows that any mono-
mial is affected with any exponent by multiplying the exponent
of each factor'of the monomial by the given exponent.
Ex. 1. (ah- 2 c*)~$ = a~* ' V* ' ~ 2 c~% " 3 = a~h%~ 2 .
Ex 2 ( 3 "*** ) "" = 3 ~* a ~ lx ~* = h ^ 2
^ l>!/i ' b~h~ 2 $ax*
Ex.3.
/ 8x 9
\ 27 »<
27 //'\ 3 _ 27 3 y 2 _ 3 //' 2
S T 9 / 1 •) r 3
EXERCISES
Perform the operations indicated by the exponents in each
of the following, writing the results without negative expo-
nents and in as simple form as possible :
1.
(M) 4 -
5.
(afV) *.
9.
(¥)~ f -
13.
(.0009)1
2.
(!i) 4 -
6.
25*
10.
GV) f -
14.
(.027)*.
3.
(«)*■
7.
25~l
11.
(0.25)*.
15.
(32 a- 5 6 10 )*.
4.
(27 a- fl )i
8.
25°.
12.
(0.25)"*.
16.
8* • 4 _ i
146 EXPONENTS AND RADICALS
17. («\ . 19. (J 2 ) *(A) '• ( *V ' Ul) •
18. (27aVV^-J. 20. tytf.(ttt)-*. WvJUry
23. \/81a- 4 6 8 (-27a 3 &- c )-*. 26 - Vl6a-*&'-". -\/8a 3 &- 6 .
24 w-^y*. 27 ( _ 2 - 5 „-,,- T5( _o-i (r ^-. ) ,
\m-7J v V" /
/a»&-"\* . Art-\-\ 28 . /81r-^Y9^-^V^
25 - (^J A^J \,626r-A; i,2or^V
29. Prove Law III in detail for the following cases:
(1) (a»)~«, (2) (a~»)«, (3) (a"-)""*-
30. Prove Law IV in detail for the following cases :
(1) (ar*b- f )~; (2) (a~ k b- l f\
Multiply :
31. ar 2 + x-^y- 1 + y~ 2 by ar 1 — jr 1 .
2 II S 1 1
32. a- 3 — x*y* + y 3 by x 3 + y*
4 8 1 22 13 4 1 1
33. as* + **y* + « a y* + •*■■'' //■'■ +^/ 5 by X s — y=.
34. \ (/ ; + a/6 2 by ^e?— VW.
35. as ■' + x%y% + y* by a?» — y f -
36. as — Sajty^+SasV -1 — if* l, y a! * — 2a?*y~* + y~ 1 -
37. a>* + ■'•.'/' + *V + y* b y •*"- - y*
Divide :
n. 8 11 14 i, i i
38. /- — .(•' • n + ./■-// — as% 1 + « :! Z/ 3 — ?/> by x'—x 3 i/ + y.
39. 3 a £ _ a &f + 4a6 2 - 3 ah + &* - 4 6 s by 3 a* - b* + 4 & 2 .
40. .c- - 3 .'■■' + G as* — 7 x + T> as* — .°» x 1 * + 1 by .r' — a^ + 1 .
41. ixh- 2 -17xh 2 + 16x~h%by 2 x l - - b' 2 - i x~h\
REDUCTION OF RADICAL EXPRESSIONS 147
Find the square root of :
42. 4 x 2 — 4 xy% + 4 xz~* + y% — 2 yh~% + z~\
43. «"* — 2 aM + Z>3 + 2 a'^c 2 4- c 4 — 2 bh-.
44. 6" * - 2 &~M + c* + 2 ZT^d* + 2 &~le~* _ 2 <&# + d*
+ 2 dV^ — 2 cV* + e _1 .
Find the cube root of : 45 i a s _ | a * h l + Q ab _$ & f .
1, *
46. a c -3a 5 + 5a 3 -3a-l. 47. a" 1 4-3a _t & 5 + 3 a" T 6 T + 6 5 .
REDUCTION OF RADICAL EXPRESSIONS
181. An expression containing a root indicated by the radical
sign or by a fractional exponent is called a radical expression.
The expression whose root is indicated is the radicand.
3/— 2 .
E.g. v5 and (1 -f x) 3 are radical expressions. In each case the
index of the radical is 3.
The reduction of a radical expression consists in changing
its form 'without changing its value.
Each reduction is based upon one or more of the Laws I to
V, § 175, as extended in § 179.
182. To remove a factor from the radicand. This reduction is
possible only when the radicand contains a factor which is a
perfect power of the degree indicated by the index of the root,
as shown in the following examples :
Ex.1. V72 = V3(TT2 = VM V2 = 6 V2.
Ex. 2. (aWyrf = ("Y • z 2 )* = («V)' ■ (**)* = ay*x%.
This reduction involves Law IV, and may be written in
symbols thus : r
v x h y = v x kr V/ = * V/-
148 EXPOXEXTS AND RADICALS
EXERCISES
In the expressions on p. 154, remove factors from the radi-
cands where possible.
In the case of negative fractional exponents, first reduce to equiva-
lent expressions containing only positive exponents.
183. To introduce a factor into the radicand. This process
simply retraces the steps of the foregoing reduction, and hence
also involves Law IV.
Ex. 1. 6 V2 = VW 2 ■ V2 = VW^2 = V72. See § 112
Ex. 2. ay' 2 x J = V(a//' 2 ) 3 - Vx' 2 = V(ay' 2 ) 3 x- = Va 3 y G x' 2 .
Ex. 3. x Vy = y/x r Vy — Vx*y.
EXERCISES
In the expressions on p. 154, introduce into the radicand any
factor which appears as a coefficient of a radical.
184. To reduce a fractional radicand to the integral form.
This reduction involves Law IV or Law V, and may always be
accomplished.
Ex. 1. v'| = vT| = V Z V.15= |\/15. Law IV
Ex.2. (<l^)K(^^) l =i«^n\=(«*-^ 1 . LawV
\a + hi \(u + l>y 2 J [( a + h yji a+b
Ex. 3. — - = Ai 1 — = ^/y _ o.
</5 * J
T t , , r\ a r ab rl Vab r - 1 1 ,.—
In svmbols, we have \/ = \j— - L — = — — — = 7 -sjab r ~ K
* * o r vA^
EXERCISES
In the expressions on p. 154, reduce each fractional radicand
to the integral form.
In case negative exponents arc involved, firsl reduce to equivalent
expressions containing only positive exponents.
REDUCTION OF RADICALS 149
185. To reduce a radical to an equivalent radical of lower index.
This reduction is effective when the radicand is a perfect power
corresponding to some factor of the index.
Ex.1. ^/8 = 8''=8(^) = (8^ = 2^ = V2.
Ex. 2. \/a 2 + 2 ab + b 2 = yj Va 2 + 2 ab + 6 2 = Va + 6.
This reduction involves Law III as follows :
ii n ^
from which we have
r Vx = </i'x = i / <j x . See § 1 14
By this reduction a root whose index is a composite number
is made to depend upon roots of lower degree.
E.g. A fourth root may be found by taking the square root twice;
a sixth root, by taking a square root and then a cube root, etc. In
the case of literal expressions this can be done only when the radicand
is a perfect power of the degree indicated by the index of the root.
But when the radicand is expressed in Arabic figures, such roots may
in any case be approximated as in § 127.
EXERCISES
In the expressions on p. 154, make the reduction above indi-
cated where possible.
In the case of arithmetic radicands, approximate to two places of deci-
mals such roots as can be made to depend upon square and cube roots.
186. To reduce a radical to an equivalent radical of higher index.
This reduction is possible whenever the required index is a
multiple of the given index. It is based on Law III as follows :
r r t rt
x~ s = (xs)t = x st . See § 179
/- 1 1 I R 6,
Ex.1. Va . =« 2 = (a 2 ) 3 =a 6 = Va 3 .
Ex. 2. Vb = b* = b% = ^6*.
150 EXPONENTS AND RADICALS
Definition. Two radical expressions are said to be of the
same order when their indicated roots have the same index.
By the above reduction two radicals of different orders may
be changed to equivalent radicals of the same order, namely, a
common multiple of the given indices.
E.g. Va and Vb in Exs. 1 and 2 above.
EXERCISES
In Exs. 3, 4, 6, 17, 18, 23, 28, 30, on p. 154, reduce the
corresponding expressions in the first and second columns to
equivalent radicals of the same order.
187. In general, radical expressions should be at once
reduced so that the order is as low as possible and the radicand
is integral and as small as possible. A radical is then said to
be in its simplest form.
ADDITION AND SUBTRACTION OF RADICALS
188. Definition. Two or more radical expressions are said
to be similar when they are of the same order and have the
same radicands.
E.q. 3V7 and 5V7 are similar radicals as are also aVx 4 and
i,A. '
If two radicals can be reduced to similar radicals, they may
be added or subtracted according to § 10.
Ex. 1. Find the sum of \8, \ 50, and V98.
By § 182, V8 = 2V2, V50 = 5V2, and V98 = 7V2.
Hence Vs + V50 + V9S = 2 v/2 + 5 V2 + 7V2 = 14 V2.
Ex. 2. Simplify V| - V20 + V3~I.
By § 1 8 1. \ ! 1 V5, V20 = 2 a o, V3i = >/v = 4 v ' = •• % •"'•
Hence V\ - V20 + \'% = l V5 - 2 Vo + | Vo = - V3.
MULTIPLICATION OF RADICALS 151
If two radicals cannot be reduced to equivalent similar radi-
cals, their sum can only be indicated.
E.g. The sum of V'J and vo is V2 + Vo.
Observe, however, that
VIo + Vij = V2 • v?) + ^ ■ V3 = V2 ( V5 + \/3).
EXERCISES
(a) In Exs. 1, 2, 5, 7, 8, 19, 20, 21, p. 154, reduce each
pair so as to involve similar radicals and add them.
(b) Perform the following indicated operations :
1. V28 + 3V7-2V63. 5. V|+ V63 + 5V7.
2. a/24 - ^81 - -v^. 6. V99-11VS+ V44.
3. Va? + vo" - \/32a. 7. 2 VJ + 3 Vf + Vl75.
4. 2V48-3VI2 + 3VJ. 8. V=b + G V} - Vl2.
9. V9 + V27 + V324.
10. ;« 2 + Va 3 + a 2 b — V(a 2 — b'-)(a — b).
MULTIPLICATION OF RADICALS
189. Radicals of the same order are multiplied according to
§ 120 by multiplying the radicands. If they are not of the same
order, they may be reduced to the same order according to § 186.
E.g. y/a-y/b ■= orfe* = a s b^ = Va 3 • Vb 2 = yVa s b' 2 .
In many cases this reduction is not desirable. Thus, Vx*- Vy 3 \s
written x 3 y^ rather than vx*y 9 .
Radicals are multiplied by adding exponents when they are
reduced to the same base with fractional exponents, § 176.
E.g. a/7 2 • Vx 3 = x% . x% = x% + * = x&.
152 EXPONENTS AND RADICALS
190. The principles just enumerated are used in connection
with § 10 in multiplying polynomials containing radicals.
Ex. 1. 3 V2 + 2 V5 Ex. 2. 3 V2 + 2 V5
2 V2 - 3 V5 3 V2 - 2 Vo
6-2 + 4 VlO 9-2 +6V10
- 9 VlO -6-5 - 6 VlO -4-5
12 - 5 VlO - 30 18 - 20
Hence (2 V2 + 2 V5)(2 V2 - 3 V5) = - 18 - 5 VlO,
and (3 V2 + 2 V5)(3 V2 - 2 V5) = 18 - 20 = -2.
EXERCISES
(a) In Exs. 21 to 38, p. 154, find the products of the
corresponding expressions in the two columns.
(b) Find the following products:
i. (3 + Vii)(3 - Vii).
2. (3 V2 + 4 V5)(4 V2 - 5 V5).
3. (2 + V3 + \/5)(3 + VB - V5).
4. (3 V2 - 2 V18 + 2 V7)(2 V2 - Vl8 - V7).
5. (Va - V6)(Va + V6)(a 2 + ab + b 2 ).
6. (Vvl3+3)(VVl3-3).
7. (V2 + 3 VB)(V2 + 3 V6).
8. (3 a - 2 V. ; i-)(4 a + 3 Vx).
9. (3 V3 + 2 V6 - 4 V8)(3 V3 - 2 V6 + 4 V8).
(Va + V6 - ^Jc)(y/a - V6 + Vc).
10.
DIVISION OF RADICALS 153
11. (a — Vb — -Vc)(a + V6 + Vc).
12. (2Vl + 3V! + 4V!)(2V!-5Vf).
13. (Va 2 + ^(Va 2 + \/a^ + ^ 3 ).
14. (V^ 3 -i/ 3 ) 3 .
DIVISION OF RADICALS
191. Radicals are divided in accordance with Laws II and V.
That is, the exponents are subtracted when the bases are the
same, and the bases are divided when the exponents are the same.
See §§ 179, 121.
-n -. 5/—, IS 2 3 2.3 .II
Ex. 1. yV h- V.s 3 = a 3 -5- a; 2 = a; 5 2 = x x °.
Ex. 2. a?* -s- y* = ( -)*= {xy~ l y= <fx 2 y~ 2 .
Ex.3. Va-V6 = a«-6S = (~f = Va 3 6- 2 .
EXERCISES
(a) In each of the Exs. 1 to 20, on p. 154, divide the
expression in the first column by that in the second.
(6) Perforin the following divisions :
1. ( Vu 3 + 2 Vo* - 3 Va) h- 6 Va.
2. (Va + V& - c) -*- Vc.
3. (2 V9 + 3 Vl2 - 4 Vl5) -h V3.
4. (4 V7 - 8 V2i + 6 V42) - 2 V7.
154 EXPONENTS AND RADICALS
EXERCISES
1. 3V45, 2V125. 21. 3(50)2, 4(72)1
2. a*, ak 22. a*, ai
3. x-5, x ? . 23. <'.<S fcci
4. 3Vx- 2 ?/, 2i/x\ ■ 24. a*&^ aV.
>r 7 3/ o, s 3/— 5^9 „ i 4,7 3 2,1
5. dva-b-, cvao- 25. Dr'tt-h, m*n*ls.
6. 7V(a + Z>) 3 , ll\/(a-6) 6 . 26. 5VaW, 3Va 3 //c l;f .
7. A^a 4 , al 27. -Va 8 </&">, V 6 9 a/o".
/ — s 3
8. nv»l, Wl 5 .
9. Vf, Vf
10 Vi? V^
1U. V 72 , v 54-
11. Vf|, Vff.
12. — , rVF.
6" 2
13. a -2 & 8 ,
a°
28.
8*,
165
29.
25~\
125-3.
30.
9t,
8i
31.
(*)*,
(A)"*-
32.
5 a* 3 ,?/ 5
x 3 y 5 z-
„ .5 8 5,3
3a -/-- i a *& 5
00. ,
1, _i
6 4 4a"*&~* 5a~^6" s
14. A '£!£!. 34 dd - * A
»H -7 ' 4- ' ,-i -i' 1
3 ft -| " dVr- 3
15- T 7 ^' ^~* 35 - 5 ( a + & )" 3 ' 3(a + &)-*.
16. ^ ^J. 36 - (" 64 >* ~ 64i -
IT. tt WL 3? - ( ^ <*>*
- 3 r- 3
is. s/^, - &\ ;
19. IS 1 . \ .'32. 39. -v / 4tt 8 -12a 2 6+12a6 2 -46 8 .
1 V»3
SO. Vl2, 48*. 40. V(3a-2&)(9a 2 -4& 2 ).
BATIONALIZING THE DIVISOR 155
192. Rationalizing the divisor. In case division by a radical
expression cannot be carried out as in the foregoing examples,
it is desirable to rationalize the denominator when possible.
Ex. 1.
Ex.
V2^ V2- Vo = VlO
^5~V5'V5 5
Va -\/a(y/a + Vo) _ a + Vab
Va-V& ( Va - V6)( Va + V&) a-6
Evidently this is always possible when the divisor is a mono-
mial or binomial radical expression of the second order.
The number by which numerator and denominator are multi-
plied is called the rationalizing factor.
For a monomial divisor, Vx, it is Vx itself. For a binomial divisor,
Va;± Vy, it is the same binomial with the opposite sign, y/xT Vy.
EXERCISES
Reduce each of the following to equivalent fractions having
a rational denominator.
1.
2 _
V5'
7
V5
+ V3
V27
V3
+ VH
2 —
V7
2 +
V7
V2
-V3
Va; + Vy _
Va; — Vy
3V3-2V2
3 V3 + 2 V2
3.
4. = ^=- 9.
Va 2 + 1 — Va
2 -l
Va 2 — 1 + V«
Vx + 1 + Va; -
+ 1
-1
V# + 1 — Va; ■
-1
Va — b — Va
+-■&
10.
V2 + V3 Va - & + Va + b
156 EXPONENTS AND RADICALS
193. In finding the value of such an expression as
— —^ — =, the approximation of two square roots and division
by a decimal fraction would be required. But v i "*" v ^ equals
114-2V21 V7-V3
— C— — — which requires only one root and division by the
integer 4.
EXERCISES
Find the approximate values of the following expressions to
three places of decimals.
1 3V5 + 4V3 g 7V5 + 3V8
V5 - V3
2. ^
V7 - V2
4V3
V3 - V2
2Vo -
-3V2
5V19
-3V7
3V7
- \ 19
3V2-
-V5
V5-
6 V2
5V^-
-7 a/13
4 11V5-3V3 g
2 V5 + V3 3 v 13 - 7 V6
194. Square root of a radical expression. A radical expression
of the second order is sometimes a perfect square, and its
square root may be written by inspection.
E.g. The square of Va ± Vb is a +b ± 2 Vab. Hence if a radical
expression can be put into the form x ± 2v/. where x is the sum of
two numbers a and b whose product is y, then va ± Vb is the square
root of x + 2 v^.
Example. Find the square root of 5 -f- v24.
Since o + V24 = 5 + 2V6, in which 5 is the sum of 2 and 3, and 6 is
their product, we have V5 + y/94, = V2 + VS.
IMA GIN A HIES 157
EXERCISES
Find the square root of each of the following :
1. 3 - 2 V2. 3. 8 - V60. 5. 24 - 6 Vf.
2.7 + V40. 4. 7 + 4 V3. 6. 28 + 3 Vl2.
195. Radical expressions involving imaginaries. According to
the definition, § 112, (V^T) 2 = - 1. Hence, (V — l) 3
= (V^T) 2 V^T = -V^T and (V^I) 4 =(V^1) 2 (V^1) 2
= (-l)(-l)= + l.
The following examples illustrate operations with radical
expressions containing imaginaries.
Ex.1. V" zr 4 + V" =r 9 = V4V :r T + V9 V^l
= (2 + 3) V^l = 5 V^l.
Ex. 2. V^4 • V^9 = v'4 • V9 • ( V^) 2 = - 2 • 3 = - 6.
Ex 3 y/ ~~ 1 = ^ ^^ = V ^ = 2
V^~9 V9V^1 V9 3
Ex. 4. V^2 • V^3 • V^6 = V2 • V3 • V6 ■ (V^T) 3
= -V36V^T==-6V^T.
Ex. 5. Simplify ($ + | V^3) 3 .
We are to use %(l+\/~3\/— 1) three times as a factor. Re-
serving (J) 3 = £ as the final coefficient, we have,
1 + V3 V^l - 2 + 2V3 V^T
i + va V3T i + V3V^1
1 + V3 V^T - 2 + 2 V3 i/-T
V3~V^T-3 -2V3V3l_6
1 + 2 V3 V - 1 - 3 - 2 - 6 =
Hence Q + £ V^3) 8 = i(- 8) = - 1.
158
EXPONENTS AND RADICALS
EXERCISES
Perform the following indicated operations.
1. V- 16 + V^ + V-25.
2. V^ 4
■V— x 2 .
3. 3 + 5V :r T-2y
4. (2 + 3V^l)(3 + 2V^l).
5. (2 + 3V^I)(2-3V^T).
6. (4.4-5V :r 3)(4-5V :r 3).
7. (2 V2 - 3 y/r^S) (3 V3 + 2 V=2).
8. (V^3 + V^2)(V ::: 3-V^2).
9. (3 Vo + 2 V^7)(2 V5 - 3 V^T).
10. (-|_+V^3)(-l-iV^3)'-'.
Rationalize the denominators of
11.
12.
13.
1-V-l
3
V3+V^l
1-V-l
l+V-i
14.
15.
16.
V2+V"
-3
V2-V"
^3
5
2 - 3 V -
5
x 4- y V^
^1
xV— 1 — #
17. Solve ic 4 — 1 = by factoring. Find four roots and verify
each.
18. Solve ar' + 1 = by factoring and the quadratic formula.
Find three roots and verify each.
19. Solve a? — 1 = as in the preceding and verify each root.
20. Solve x 6 — 1 = by factoring and the quadratic formula.
EQUATIONS CONTAINING RADICALS 159
SOLUTION OF EQUATIONS CONTAINING RADICALS
196. Many equations containing radicals are reducible to
equivalent rational equations of the first or second degree.
The method of solving such equations is shown in the fol-
owing examples.
Ex. 1. Solve 1 + Va- = V3 + x. (1)
Squaring and transposing, 2vs = 2. (2)
Dividing by 2 and squaring, x = 1. (3)
Substituting in (1), 1 + 1 = V3 + 1 = 2.
Observe that only principal roots are used in this example.
If (1) is written 1 + Var = - V3 + x, (4)
then (2) and (3) follow as before, but x = 1 does not satisfy (4). In-
deed algebra furnishes no means whereby to obtain a number which will
satisfy (4).
Ex. 2. Solve Vx + 5 = * - 1. (1)
Squaring and transposing. x' 2 — 3 x — 4 = 0. (2)
Solving, x — 4 and x — — 1.
x = 4 satisfies (1) if the principal root in Vx + 5 is taken. x= —I
does not satisfy (1) as it stands but would if the negative root were
taken.
tt q « i V4.r + 1- V3a;-2 _1 m
Ex.3. Solve — ===== ==• (.- 1 ;
V4a + l + V3a;-2 5
Clearing of fractions and combining similar radicals.
2V4x-+ 1 = 3V3x -2. (2)
Squaring and solving, we find x = 2.
This value of x satisfies (1) when all the roots are taken positive and
also when all are taken negative, but otherwise not.
EXPOXENTS AXD RADICALS
Zx ~l -1. (1)
The fraction in the second member should be reduced as follows :
3x-l (VWx - l)(VS~x + 1) r—
— = — — = V3x + l.
V3 x - 1 V3 x - 1
Hence, (1) reduces to V2 x + 3 = V3x + 1 - 1 = a/3x. (2)
Solving, x = 3, which satisfies (1).
If we clear (1) of fractions in the ordinary manner, we have
(V3 x - 1) V2 x + 3 = - V3 x + 3 x. (2')
Squaring both sides and transposing all rational terms to the
second member,
2 x y/3 x-6 Vfilc = 3 x 2 - 8 x - 3. (3)
Factoring each member,
2(x - 3) V¥x = (x - 3) (3 x + 1), (4)
which is satisfied by x = 3.
Dividing each member by x — 3, squaring and transposing, we have
9ar 2 -6a;+l= (3 x - 1)- = 0, (5)
which is satisfied by x = 3.
Equation (1) is not satisfied by x = i, since the fraction in the
second member is reduced to {J by this substitution. Sec i ."in.
The root x = ± is introduced by clearing of fractions without
first reducing the fraction to its lowest terms, Vox— 1 being a
factor of both the numerator and the denominator. See § 165.
Ex.5. Solve (] ~ x -V^= *~ 3 • (1)
V0 — x V& — 3
Reducing the fractions by removing common factors, we have
V(fZ x - y/3 = vT^~3. (2)
Squaring, transposing, and squaring again,
x 2 -9x+18 = 0, (3)
whence x = 3, x = 6.
But neither of these is a root of (1). In this ease (1) has no root.
EQUATIONS CONTAINING RADICALS 161
197. In solving an equation containing radicals, we note the
following :
(1) If a fraction of the form ®~ is involved, this
V« — V6
should be reduced by dividing numerator and denominator by
s/a — V& before clearing of fractions.
(2) After clearing of fractions, transpose terms so as to
leave one radical alone in one member.
(3) Square both members, and if the resulting equation still
contains radicals, transpose and square as before.
(4) In every case verify all results by substituting in the
given equation. In case any value does not satisfy the given
equation, determine whether the roots could be so taken that
it would. See Ex. 3.
EXERCISES
Solve the following equations :
1. Va- 2 + 7a;-2-Va; 2 -3a,' + 6 = 2.
2. V3?/-V3//-7
b
V3.V-7
fr/-l = Vfy-l . t 8. V5^ + l = l_ 5 ^~- 1
V&v + l 2 V5a;-fl
4. V5aj-19+V3a; + 4 = 9. „ 4
9- =V3.
Va; 2 + a 2 — x
-Vx- + a 2 + x
x
4 + g + V8a; + ar =1>
V a + Vox + x' 1 = Va — Va\ 4 + x — V8« + x 2
7.
y-* =^y-^~ l +2 Vi. ii. "-* 4- - T+ft -=v^
V?/+V? ° Va — a- Va-+a
12. x ~ a _ = V * + V " + 2V«.,
V^ — Va - 1 ■
162 EXPONENTS AND RADICALS
13. — a — V m — n = •
Vm — y V# — n
i4. 2v;n7i + 3y^ = 7f '± ; K
Va- — a
1 5 . V2 g + 7 + V2.r + 14 = "V4 x + 35 + 2 V4 x 2 + 42 a; - 21.
16. Vo; -3 + Vw + 9"= Va; + 18 + Va; - 6.
17. V.7+7 - Va; — 1 = Vx + 2 + V.c - 2.
18. a V# + b — c V& — y = V6 (a 2 + c 2 ) .
1 9. i/ V// — c — Vy" + c 3 + c V# +• c = 0.
20. v? + y*»^vw | v^
Vm Va; Vw Vm
12
21. \14 + Va?+\6-v*=
\ 6 - \ x
91
22. V 3 g + V3 a? + 13 = — =.
V3 * + 13
23. V6g + 3 + Vg + 3 = 2g+3.
24. v x - a + Vft - x = Vft
- Va; — a+ Va;— ft /a;— a
^5. — - — == "V ■
\ x a \ x—b *x—b
26. V(g - l)(g - 2) + V(g— 3)(x - 4) = V2.
27. V2 x + 2 + V7 + 6 .« = V7a;+72.
2 8 . V 2 ( i bx + Va 2 — 6a; = Va 2 + bx.
a -+-g-f \ a* .'•-' c
29
+ .'• — Va 2 •
30. Va? - 2 .r + 4 + V3 ar + 6 a; + 12 = 2 Va; 2 + g + 10.
PROBLEMS
163
PROBLEMS
1. Find the altitude drawn to the longest side of the tri-
angle whose sides are 6, 7, 8.
Hint. See figure, p. 235, E. C. Calling % and 8 — x the segments
of the base and h the altitude, set up and solve two equations involv-
ing x and it.
2. Find the area of a triangle whose sides are 15, 17, 20.
First find one altitude as in problem 1 .
3. Find the area of a triangle whose base is 16 and whose
sides are 10 and 14.
4. Find the altitude on a
side a of a triangle two of
whose sides are a and a third b.
A three-sided pyramid all of
whose edges are equal is called a
regular tetrahedron. In Figure 10
AB, A C, AD, BC, BD, CD are all
equal.
5. Find the altitude of a
regular tetrahedron whose edges
are each 6. Also the area of the
base.
Hint. First find the altitudes AE and DE and then find the alti-
tude of the triangle AED on the side DE, i.e. find AF.
6. Find the volume of a regular tetrahedron whose edges
are each 10.
The volume of a tetrahedron is \ the product of the base and
the altitude.
7. Find the volume of a regular tetrahedron whose edges
are a.
8. In Figure 10 find EG if the edges are a.
9. If in Figure 10 EG is 12, compute the. volume.
Use problem S to find the edge, then use problem 7 to find the
volume.
Fig. 10.
164 EXPONENTS AND RADICALS
10. Express the volume of the tetrahedron in terms of EG.
That is if EG = b, find a general expression for the volume in
terms of b.
11. If the altitude of a regular tetrahedron is 10, compute
the edge accurately to two places of decimals.
12. Express the edge of a regular tetrahedron in terms of
its altitude.
13. Express the volume of a regular tetrahedron in terms
of its altitude.
14. Express the edge of a regular tetrahedron in terms of
its volume.
15. Express the altitude of a regular tetrahedron in terms
of its volume.
16. Express EG of Figure 10 in terms of the volume of the
tetrahedron.
17. Find the edge of a regular tetrahedron such that its vol-
ume multiplied by V2, plus its entire surface multiplied by V3,
is 114.
The resulting equation is of the third degree. Solve by factoring.
18. An electric light of 32 candle power is 25 feet from a
lamp of G candle power. "Where should a card be placed
1 iet ween them so as to receive the same amount of light from
each '.'
Compare problem b5. p. 141. Compute result accurately to two
places of decimals.
19. "Where must the card be placed in problem 18 if the
lamp is between the card and the electric light?
Notice that the roots of the equations in IS and 19 are the same.
Explain what this means.
20. State and solve a general problem of which 18 and 19
are special cases.
PROBLEMS 165
21. If the distance between the earth and the snn is 93
million miles, and if the mass of the snn is 300,000 times that
of the earth, find two positions in which a particle would be
equally attracted by the earth and the sun.
The gravitational attraction of one body upon another varies in-
versely as the square of the distance and directly as the product of
the masses. Represent the mass of the earth by unity.
22. Find the volume of a pyramid whose altitude is 7 and
whose base is a regular hexagon whose sides are 7.
The volume of a pyramid or a cone is \ the product of its base and
its altitude.
23. If the volume of the pyramid in problem 22 were 100
cubic inches, what would be its altitude, a side of the base
and the altitude being equal ? Approximate the result to two
places of decimals.
24. Express the altitude of the pyramid in problem 22 in
terms of its volume, the altitude and the sides of the base
being equal.
25. If in a right prism the altitude is equal to a side of the
base, find the volume, the base being an equilateral triangle
whose sides are a.
The volume of a right prism or cylinder equals the product of
its base and its altitude.
26. Find the volume of the prism in problem 25 if its base
is a regular hexagon whose side is a.
27. Express the side of the base of the prism in problem 25
in terms of its volume. State and solve a particular problem
by means of the formula thus obtained.
28. Express the side of the base of the prism in problem 26
in terms of its volume. State and solve a particular problem
by means of the formula thus obtained.
106
EXPONENTS AND RADICALS
In Figures 11 and. 12 the altitudes are each supposed to be
three times the side a of the regular hexagonal bases.
Fig. 11.
Fig. 12.
29. Express the difference between the volumes of the pyra-
mid and the circumscribed cone in terms of a.
The volume of a cone equals \ the product of its base and altitude.
30. Express a in terms of the difference between the volumes
of the cone and pyramid. State and solve a particular problem
by means of the formula thus obtained.
31. Express the volume of the pyramid in terms of the dif-
ference between the areas of the bases of the cone and the
pyramid. State a particular case and solve by means of the
formula first obtained.
The lateral area of a right cylinder or prism equals the perimeter
of the base multiplied by the altitude.
32. Express the difference of the lateral areas of the cylin-
der and the prism in terms of a.
The following four problems refer to Figure \2. In each case state
a particular problem and solve by means of the formula obtained.
33. Express a in terms of the difference of the lateral areas.
34. Express the volume of the prism in terms of the differ-
ence of the perimeters of the bases.
35. Express the volume of the cylinder in terms of the dif-
ference of the lateral areas.
36. Express the sum of the volumes of the prism and cylin-
der in terms of the difference of the areas of the bases.
CHAPTER XI
LOGARITHMS
198. The operations of multiplication, division, and finding
powers and roots are greatly shortened by the use of logarithms.
The logarithm of a number, in the system commonly used,
is the index of that power of 10 which equals the given number.
Tims, 2 is the logarithm of 1U0 since 10- = 100.
This is written log 100 = 2.
Similarly log 1000 = 3, since 10 3 = 1000,
and log 10000 = 1, since 10 4 = 10000.
The logarithm of a number which is not an exact rational
power of 10 is an irrational number and is written approxi-
mately as a decimal fraction.
Thus, log 71 = 1.8092 since 10 1SG92 = 74 approximately.
In higher algebra it is shown that the laws for rational ex-
ponents (§ 179) hold also for irrational exponents.
199. The decimal part of a logarithm is called the mantissa,
and the integral part the characteristic.
Since 10" = 1, 10 l = 10, 10- = 100, 10 3 = 1000, etc., it follows
that for all numbers between 1 and 10 the logarithm lies be-
tween and 1, that is, the characteristic is 0. Likewise for
numbers between 10 and 100 the characteristic is 1, for num-
bers between 100 and 1000 it is 2, etc.
200. Tables of logarithms (see p. 170) usually give the man-
tissas only, the characteristics being supplied, in the case of
whole numbers, according to § 199, and in the case of decimal
numbers, as shown in the examples given under § 201.
167
108 LOGARITHMS
201. An important property of logarithms is illustrated by
the following:
From the table of logarithms, p. 170, we have :
log 376 = 2.5752, or 376 = 10 2 - 5752 . (1)
Dividing both members of (1) by 10 we have
37.6 = 10 2 - 5752 - 10 1 = lO 2 - 5752 - 1 = 10 157 * 2 .
Hence, log 37.6 = 1.5752,
Similarly, log 3.76 = 1.5752 — 1 = 0.5752,
log .376 = 0.5752 — 1, or 1.5752,
log .0376 = 0.5752 - 2. or 2.5752,
where 1 and 2 are written for — 1 and — 2 to indicate that the char-
acteristics are negative while the mantissas are positive.
Multiplying (1) by 10 gives
log 3760 = 2.5752 + 1 = 3.5752,
and log 37600 = 2.5752 + 2 = 4.5752.
Hence, if the decimal point of a number is moved a certain
number of places to the right or to the left, the characteristic of
the logarithm is increased or decreased by a corresponding
number of units, the mantissa remaining the same.
From the table on pp. 170, 171, we may find the mantissas
of logarithms for all integral numbers from 1 to 1000. In this
table the logarithms are given to four places of decimals,
which is sufficiently accurate for most practical purposes.
E.fj. for log 4 the mantissa is the same as that for log 40 or for log
400.
To find log .0376 we find the mantissa corresponding to 376, and
prefix ilif characteristic 2. See above.
Ex. 1. Find log 876.
Solution. Look down the column headed N to 87, then along this
line to the column headed 6, where we liud the number 9425, which is
the mantissa. Hence log 876 = 2.9425.
LOGARITHMS 169
Ex. 2. Find log 3747.
Solution. As above we find log 3740 = 3.5729,
and log 3750 = 3.5740.
The difference between these logarithms is 11, which corresponds to
a difference of 10 between the numbers. But 3740 and 3747 differ
by 7. Hence, their logarithms should differ by -fa of 11, i.e. by 8.1.
Adding this to the logarithm of 3740, we have 3.5737, which is the
required logarithm.
The assumption here made, that the logarithm varies directly
as the number, is not quite, but very nearly, accurate, when the
variation of the number is confined to a narrow range, as is
here the case.
Ex. 3. Find the number whose logarithm is 2.3948.
Solution. Looking in the table, we find that the nearest lower loga-
rithm is 2.3945 which corresponds to the number 218. See § 199.
The given mantissa is 3 greater than that of 248, while the man-
tissa of 249 is 17 greater. Hence the number corresponding to
2.3948 must be 248 plus T \ or .176. Hence, 248.18 is the required
number, correct to 2 places of decimals.
Ex. 4. Find log .043.
Solution. Find log 43 and subtract 3 from the characteristic.
Ex. 5. Find the number whose logarithm is 4.3949.
Solution. Find the number whose logarithm is 0.3949, and move
the decimal point 4 places to the left.
EXERCISES
Find the logarithms of the following numbers :
1. 491. 6. .541. 11. .006. 16. 79.31.
2. 73.5. 7. .051. 12. .1902. 17. 4.245.
3. 2485. 8. 8104. 13. .0104. 18. .0006.
4. 539.7. 9. 70349. 14. 2.176. 19. 3.817.
5. 53.27. 10. 439.26. 15. 8.094. 20. .1341.
170
LOGARITHMS
N
1
2
3
4
5 6 7
8
9
10
0000 0U43 0086 0128 0170
0212 025:; H204 0334 0374
11
0414
04.:.:;
0492 0531
0509
0607 0645
0682 0710 0755
12
0702
0828
0864 0899
0934
0969 1004
1038 1072
1100,
13
1139
1173
1206 1239
1271
1303 1335
L367 1300
1150
14
1401
1492
152:; 1553 j 1584
1614 1044
1673 1703
1732
15
1701
1790
1818
1847 1875
1903
1931 1959 L987
2014
16
2H41 2068
2095
2122 2148
2175
2201 2227
2353
2279
17
2304 2330
2355
2380 2405
2430
2455 2480
2504
2520
18
2553 2577
2001
2625 2648
2672
2005 27 is
27 12
2705
19
2788 2810
2833
2856 2878
2900
2923 20 15
2967 2989
20
3010
3032
3054
3075
3096
31 is
3139 3160
3181
520 1
21
3222
324:1
3263
3284
3304
3324
5315
330,5
33s5
3404
22
3424
3444
3464
3483
3502
3522
354 1
3560
3,570
3598
23
3617
3636
3655
3674
3692
3711
3729
3747
370,0
37S4
24
3802
3820
3838
3856
3874
3892
3909
3927
30 15
3962
25
3979
3997
4014 4031
4048
lot; 5
4082
4099
1110
4133
26
1 1 51 1
H66
4183
42(10
4216
12:12 4210 4205
42-1
4298
27
4314
1330
4346
1362
4378
4393
4400
4425
4440
4450,
28
4472
4487
4502
4518
4533
4548
4564
1570
4594
4609
29
4624
4639
4654 4669
4683
L698
4713
172-
4742
4757
.{()
4771
4786
4800 4814
4829
1*43
4857
4871 4886
4900
31
4914
1928
4042
4055
4000
4983
4907
5011 5024
5038
32
5051
5065
5079
5092
5105
5119
5132
5145 5159
5172
33
5185
5 19W
5211
5224 5237
5251)
5203
5270 5289
5302
34
5315 5328
5340
5353 5366
5378
5391
5403
5410
5428
.;.-,
5441
5453
5465 5478 5490
5502
5514
5527
553,0
5551
36
5563
5575
5587 5500 5611
50,23
5635
5047
5658
5670
37
5682
5694 5705
57 17 5729
5740 5752
5763
5775
5786
38
5798
5809
5821
5832 5843
5855 5866
5877
5sss
5S99
39
5911
5922
5933
5944 1 5955
5966 5! 17 7
5988
5000
6010
40
6021
oo:;i
6042
6053
606 1
0075
6085
0090
0107
6117
41
6128
6138
6149
6160
0170
0180
619] 0,201
0212
0,222
42
6232
6243
6253
020.'!
6274
02 S(
0,20 1 030 1
6314
03,25
43
6335
6345
6355
6365
6375
6385
0305 6405
0,415
0125
44
6435
6444
6454
6464
6474
6484
0103, 0503
0,513,
0,522
45
6532
0.", 4 2
6551
6561
6571
6580
c,5: mi
6599 6609
0,0 is
46
6628
6637
oo4';
6656
6665
6675
6684
6693 0,702
0712
47
0721
6730
6739
0710
6758
070,7
0770
6785 0,701
6803
48
6812
6821 6830
6839 6848
7857
osoo,
6875 6884
0,so5
49
6902 6911 |6920
6928 j 6937
6946
0055
0004 0072
oosi
50
6990 1 6998
7(107
7010 7024
7033 7042
7050 7059
7067
51
7076 7084
7093
7101 7110
71 is 7120,
7135 7143 7152
52
7160 7168
7177
7185 710:;
7202 7210
72 is 7220 723,5
53
72 1:: 7251
7259
7207 7275
7284 7202
73.00 7308 7310,
54
7324 7332 7340 7348 7356
7304 7372 7380 7388 73.90
LOGARITHMS
171
N
1
2
3
4
5
6
7
8
9 |
55
7404 7412
7410
7427
7435
7443
7451 7459
7466
7474
56
7482 74DO
7497
7505
7513
7520
7528
7536
7543
7551
57
7559 75(3(3
7574
7582
7589
7597
7(504
7612
7619
7627
58
7634 | 7(342
7649
7657
7664
7672
7(570
7686
7094
7701
59
7709 7716
7723
7731
7738
77 15
7752
7760
7767
7774
60
7782
7789
7796
7803
7810
78 18
7825
7832
7839
7846
61
7853
7860
7868
7875
7882
7889
7896
7003
7910
7917
62
7924
7931
7938
7945
7952
7959
7966 ; 7973
7980
7087
63
7993
8000
8007
8014
8021
8028
8(135 8041
8048
8055
64
8002
800!)
8075
8082
8080
8096
8102 ; 8100
811(5
8122
65
8129
813(5 8142
8149
HI 50
8162
8169
8176
8182
8189
66
8195
8202
8209
8215
8222
8228
8235
8241
8248
8254
67
8261
8267
8274
8280
8287
8293
8209
8300
8312
8319
68
8325
83:: 1
8338
8344
8351
8357
8363
8370
8376
8382
69
8388
8395 | 8401
8407
8414
8420
842(i
8432
8430 8445
70
8451
8457
84(53
8470
8476
8482
8488
S10-J
8500
850(3
71
8513
8519
8525
8531
8537
8543
8549
8555
8561
8507
72
8573
8579
8585
8591
8597
800:5
86(19
8615
8621
8627
73
8633
8639
8(545
8651
8657
8(563
8669
8675
8(381
8680
74
8692
8(598
8704
8710
8716
8722
8727
8733
8739
8745
75
8751
8756
87(52
87(58
8774
8779
8785 1 8791
8797
8802
76
8808
8814
8820
8825
8831
8837
8842
8848
8854
8859
77
8865
8871
8876
8882
8887
8893
8809
8904
8910 1 8915
78
8921
8927
8032
8938
8943
8949
8954
8060
8965 8971
79
8976
8982
8987
8993
8998
9004
0000
0015
9020 9025
80
9031
903(3
9042
9047
9053
9058
9063
OdOO
9074
9079
81
9085
9000
9096
9101
9106
9112
9117
9122
9128
9133
82
9138
9143
9149
9154
9159
9165
9170
9175 ; 9180
9186
83
9191
919(5
9201
920(3
0212
9217
9222
9227
9232
9238
84
9243
9248
9253
9258
9263
9209
9274
9279
9284
9289
85
9294
9299
9304 9309
9315
0320
9325
9330
9335
0340
86
9345
9350
0355 03(50
9365
0370
9375
9380
9385
9390
87
9395
9400
9405 1 9410
9415
9420
9425
9430
9435
9440
88
9445
9450
9455 1 9460
9465
9469
9474
9479
9484
9489
89
9494
9499
9504 9509 | 9513
9518
9523
9528
0533
9538
90
9542
9547
9552 9557
9562
9566 9571
9576
9581 9586
91
9590
9595
9600 9(505
9609
9614
9010
9624
9628
9(533
92
9(338 9643
9647 9(552
9657
9661
9606
9671
9675
9680
93
9685
9(589 ' 9694 i 9(599
9703
9708
9713
9717
9722
9727
94
9731
973(5
9741
9745
9750
9754
9759
9763 9768
9773
95
9777
9782
9786
9791 i 9795
9800
9805
9809
9814
9818
96
9823
9827
9832
9836 | 9841
9845
9850
9854
9859
9863
97
98(38
9872
9877
9881 1 9886
9890
9894 0800
9903
9008
98
9912
9917
9921
9926 9930
0034
9939 i 9943
9948
0052
99
9956
9961
9005 \ 9969 i 9974
9978
9983 , 9987
9901 9906
172 LOGARITHMS
Find the numbers corresponding to the following logarithms:
21. 1.3179. 26. 2.9900. 31. 1.5972. 36. 0.2468.
22. 3.0146. 27. 0.1731. 32. 1.0011. 37. 0.1357.
23. 0.7145. 28. 0.8974. 33. 2.7947. 38. 2.0246.
24. 1.5983. 29. 0.9171. 34. 2.5432. 39. 1.1358.
25. 2.0013. 30. 3.4015. 35. 0.5987. 40 . 4.0478.
202. Products and powers may be found by means of loga-
rithms, as shown by the following examples.
Ex. 1. Find the product 49 x 134 x .071 x 349.
Solution. From the table,
log49 = l.f>902 or 49 = lO 1 -** 02 .
log 134 = 2.1271 or 134 = 10 2 - 1271 .
log .071 =2.8513 or .071 = 10 2 - 8618 .
log 349 = 2.5428 or 349 = 10 2 - 6428 .
Since powers of the same base are multiplied by adding exponents,
§ 17G, we have 49 x 131 x .071 x 349 = 10*- 2114 .
Hence log (49 x 131 x .071 x 319) = 5.21 11.
The number corresponding to this logarithm, as found by the
method used in Ex. 3 above, is 162704. By actual multiplication the
product is found to be 162698.914 or 162099 which is the nearest
approximation without decimals. Hence the product obtained by
means of logarithms is 5 too large. This is an error of about 32^55 of
the actual result and is therefore so small as to be negligible.
Ex. 2. Find (4.05) 20 .
Solution. Log l.o.l = 0.0212 or 10 - 0212 = 1.05.
Hence (1.05) 20 = (10 - 0212 ) 20 = lO* - 02 * 2 '- 20 = 10 - 424 ,
or log (1.05) 2° = 0.4240.
Hence (1.05) 20 is the number corresponding to the logarithm 0.4240.
LOGARITHMS 173
Since logarithms are exponents of the base 10, it follows from
the laws of exponents (see § 198) that
(a) The logarithm of the product of two or more numbers is
the sum of the logarithms of the numbers.
(b) The logarithm of a power of a number is the logarithm of
the number multiplied by the index of the power.
That is,
log (a b ■ c) — log a -f log b + log c, and log a" — n log a.
EXERCISES
By means of the logarithms obtain the following products
and powers :
1. 243 x 76 x .34. 7. 5.93 x 10.02. 13. (49) a x .19 x 21 2 .
2. 823.68 x 370. 8. 486 x 3.45. 14. .21084 x (.53) 2 .
3. 216.83 x 2.03. 9. (.02) 2 x (0.8). 15. 7.865 x (.013) 2 .
4. 57 2 x (.71) 2 . 10. (65) 2 x (91) 3 . 16. (6.75) 3 x (723) 2 .
5. 510x(9.1) 3 . 11. (84) 2 x(75) 3 . 17. (1.46)* x (61.2)".
6. 43.71 x (21) 2 . 12. (.960) 2 (49) 2 . 18. (3.54) 3 x (29.3) 2 .
19. (4.132) 2 x (5.184) 2 . 20. 1946 x 398 x .08.
203. Quotients and roots may be found by means of logarithms,
as shown by the following examples.
Ex. 1. Divide 379 by 793.
Solution. From the table,
log 379 = 2.5786 or 10 2 - 6 ™ 5 = 379.
log 793 = 2.8993 or 10 2 -^ _ 793.
Hence by the law of exponents for division, § 175,
379 + 793 = 10 2 -67S6-2.8993.
Since in all operations with logarithms the mantissa is positive, write
the first exponent 3.5786 - 1 and then subtract 2.8993.
Hence log (379 - 793) =.6793 - 1 = 1.6793.
Hence the quotient is the number corresponding to this logarithm.
174 LOGARITHMS
Ex. 2. By means of logarithms approximate V 42- x 37 5 .
By the methods used above we find
log(42 2 x 37 5 ) = 11.0874 or 10 U <>8H = 42- x 37 5 .
Hence V42 2 x 37 5 = (lO 11 - 0874 )? = 10 3 = lO 3 ® 58 .
That is, log V42 2 x 37 5 = 3.6958.
Hence the result sought is the number corresponding to this
logarithm.
It follows from the laws of exponents (see § 198) that
(a) TJie logarithm of a quotient equals the logarithm of the
dividend minus the logarithm of the divisor.
(b) TJie logarithm of a root of a number is the logarithm of
the number divided by the index of the root.
That is
log - = log a — log b and log V a = ^—- •
b n
EXERCISES
By means of logarithms approximate the following quo-
tients and roots :
1. 45.2-5-8.9. 4. V196 x 25G. 7 . Vl5 x ^67.
2. 231.18 + 42. 5. 5334 x. 0237 4, ^2Ux^34T.
27.43x3.246
3. .04Q05 + . 327. 6 . </69-s--^21. 9 - (5184)*+ (38124)*.
10. (6.75) 3 + (2.132) 2 . s/ 13* x .31' x 4.31 3 '
lb. \j — — == .
11. #105 + ^76. v V71xv41xV51
12. (91125)* + (576)*. ,! 4 , x >r>7 3 x 4l >a
17. \/-rr=
13. (3.040) 3 - (.0005) 3 . x s!/o.2 x V.83 x V23
14. (29.3)Uv(3,i: 1 . ig / mx<mx</7 \
15. S^39 x </W x \ 87. W4 7 x a/7 1 x (.003)*
CHAPTER XII
PROGRESSIONS
ARITHMETIC PROGRESSIONS
204. An arithmetic progression is a series of numbers, such
that any one after the first is obtained by adding a fixed num-
ber to the preceding. The fixed number is called the common
difference.
The general form of an arithmetic progression is
a, a + d, a + 2d, a + 3d, • ■•,
where a is the first term and d the common difference.
E.g. 2, 5, 8, 11, 14, ••• is an arithmetic progression in which 2 is the
first term and 3 the common difference. Written in the general form,
itwouldbe ^ 2 + 3j 2 + 2 . 3) + 3-3, 2+4-3, ....
205. If there are n terms in the progression, then the last
term is a-\-(n — l)d. Indicating the last term by I, we have
/=:a+(n-l)d. I
An arithmetic progression of n terms would then be written
in general form, thus,
a, a + d, a + 2d, •• , a+(n-2)d, a+(n — l)d.
EXERCISES
1. Solve I for each letter in terms of all the others.
In each of the following find the value of the letter not given, and
write out the progression in each case.
a = 2,
fa = 3,
fo = l,
(a=7,
d = 2,
3.
\d = 5,
4.
11 = 15,
5.
\n = :n,
n = 7.
{ I = 43.
I = 15.
U = 91.
175
176 PROGRESSIONS
a = 4, (a =3, rd = — 5, fo = ll,
d = — 3, 8. |d = — 5, 10. |n = 13, 12. Z = — 39,
n = 18. I — - 32. / = - 63. \d = -5
a =3,
[d = -5,
d = —
5,
10.
a = 13,
Z = -
32.
I / = - 63.
c? = 7,
fo = -3,
« = 9,
U = -27.
w = 8,
11.
2 = 24
a = x,
d = i, 9. |n = 8, 11. J7i = 9, 13. ll = y,
n = 7. [1 = 24:. [z = -27. U = «-
206. The sum of an arithmetic progression of n terms may be
obtained as follows :
Let s n denote the sum of n terms of the progression. Then,
s„ = a + [a + d] + [a + 2 rf] + ... + [a + (« - 2>/] + [a + ( n - l)r/]. (1)
This may also be written, reversing the order of the terms, thus,
s„ = [a + (» - 1>/] + [a +(n - 2)d] + ••■ + [a + 2 d] + [a + d] + a. (2)
Adding (1) and ('2), we have
2 s„ = [2 a + (» - 1)'/] + [2 a + (n - 2)d + d]
+ ... + [2 a + („ - 2)d + d] + [2 a + (n - l)d].
The expression in each bracket is reducible to '2 n + (>i — l)r/,
which may also be written a + [« + (» — IV] = a + I, by § 205.
Since there are n of these expressions, each a + I, we have
2s n = n(a + I).
Hence s„ = - (a + /). II
&
This formula for the sum of n terms involves a, I, and n, that
is, the first term, the last term, and the number of terms.
207. In the two equations,
l = a+(n-l)d, I
.-5(0 + 1), II
there are live letters, namely, a, d, I, n, s. If any three of
these are given, the equations I and II may be solved simul-
taneously to find the other two, considered as the unknowns.
ARITHMETIC PROGRESSIONS 177
The solution of problems in arithmetic progression by means
of equations I and II is illustrated in the following examples:
Ex. 1. Given n = ll, 1 = 23, s = 143. Find a and d.
Substituting the given values in I and II,
23 = a+(ll-l)d. (1)
U3 = V(« + 23). (2)
From (2), a = 3, which in (1) gives d = 2.
Ex. 2. Given d = 4, n = 5, s = 75. Find a and ?.
From I and II, I = a + (5 - 1)4, (1)
75=|(a + Q. (2)
Solving (1) and (2) simultaneously, we have a = 7, / = 23.
Ex.3. Given rf = 4, 1 = 35, s = 1G1. Find a and n.
From I and II, 35 = a + (n - 1)4, (1)
161=|(a + 35). (2)
From (1) a = 39 - in,
which in (2) gives 161 = -(74 - 4 n) = 37 » - 2 n 2 .
Hence n = - 2 2 3 , or 7.
Since an arithmetic progression must have an integral number of
terms, only the second value is applicable to this problem.
Ex. 4. Given d = 2, I = 11, s = 35. Find a and n.
Substituting in T and U, and solving for a and n, we have
a = 3, n — 5, and a = — 1, n = 7.
Hence there are two progressions,
- 1, 1, 3, 5, 7, 0, 11,
and 3, 5, 7, 0, 11,
each of which satisfies the given conditions.
178 PROGRESSIONS
EXERCISES
In each of the following obtain the values of the two letters
not given.
If fractional or negative values of n are obtained, such a result in-
dicates that the problem is impossible. This is also the case if an
imaginary value is obtained for any letter. In each exercise interpret
all the values found.
r s = 96, f.s = 88, \d=-l, [d=6,
1. ? = 19, 4. /=-7, 7. |w=4l, 10. j 2=49,
U=2. U=-3. [l=-3o. [s=232.
r s =34, fn=18, (1 = 30, U=7,
2. < = 14, 5. a=4, 8. s=162, 11. \d=n,
|r/ = 7, \n=U, (a =30, r s =14,
3. I 1 = 27, 6. ja=7, 9. » = 10, 12. r/=:3,
[s = 187. U=14. [s=120. [z = 4.
In each of the following call the two letters specified the unknovms
and solve for their values in terms of the remaining three letters
supposed to be known.
13. a. <i. 15. a, n. 17. d, I. 19. <!, s. 21. /, s.
14. a, I 16. a, s. 18. <l. n. 20. /. n. 22. ?j, s.
208. Arithmetic means. The terms between the first and Hie
last of an arithmetic progression are called arithmetic means.
Thus, in 2, .">, 8, 11, 11, 17, the four arithmetic means between 2
and 17 are ."J, 8, 11, 11.
If the first and the last terms and the number of arith-
metic means between them are given, then these means can be
found.
For we have given a, I, and n. Hence <1 can lie found and the
whole series constructed.
ARITHMETIC PROGRESSIONS 179
Example. Insert 7 arithmetic means between 3 and 19.
In this progression a = 3, / = 19, and n = 9.
Hence, from 1 = a + (n — l)d we find d = 2 and the required means
are 5, 7, 9, 11, 13, 15, 17.
209. The case of one arithmetic mean is important. Let A
be the arithmetic mean between a and I. Since a, A, I are in
arithmetic progression, we have A = a 4- d, and 1 = A + d.
Hence A-l = a-A
A^"^ 1 - HI
EXERCISES AND PROBLEMS
1. Insert 5 arithmetic means between 5 and — 7.
2. Insert 3 arithmetic means between — 2 and 12.
3. Insert 8 arithmetic means between — 3 and — 5.
4. Insert 5 arithmetic means between — 11 and 40.
5. Insert 15 arithmetic means between 1 and 2.
6. Insert 9 arithmetic means between 2| and — 11.
7. Find the arithmetic mean between 3 and 17.
8. Find the arithmetic mean between — 4 and 16.
9. Find the tenth and eighteenth terms of the series
4, 7, 10, -.
10. Find the fifteenth and twentieth terms of the series
-8, -4,0,.-..
11. The fifth term of an arithmetic progression is 13 and
the thirtieth term is 49. Find the common difference.
12. Find the sum of all the integers from 1 to 100.
13. Find the sum of all the odd integers between and 200.
14. Find the sum of all integers divisible by 6 between 1
and 500.
15. Show that 1 4- 3 + 5 + ••• + n = n 2 where n is any odd
integer.
180 PBOGEESSIONS
16. In a potato race 40 potatoes are placed in a straight line
one yard apart, the first potato being two yards from the
basket. How far must a contestant travel in bringing them
to the basket one at a time ?
17. There are three numbers in arithmetic progression whose
sum is 15. The product of the first and last is 31- times the
second. Find the numbers.
18. There are four numbers in arithmetic progression whose
sum is 20 and the sum of whose squares is 120. Find the
numbers.
19. If a body falls from rest 16.08 feet the first second,
48.24 feet the second second, 80.40 the third, etc., how far will
it fall in 10 seconds ? 15 seconds ? t seconds ?
20. According to the law indicated in problem 19 in how
many seconds will a body fall 1000 feet ? s feet ?
If a body is thrown downward with a velocity of v feet per second,
then the distance, s, it will fall in t seconds is v t feet plus the distance
it would fall if starting from rest.
That is, 5 = V Q t + \ gf, where g = 32.10.
21. In what time will a body fall 1000 feet if thrown down-
ward with a velocity of 20 feet per second ?
22. "With what velocity must a body be thrown downward
in order that it shall fall 360 feet in 3 seconds ?
23. A stone is dropped into a well, and the sound of its
striking the bottom is heard in 3 seconds. How deep is the
well if sound travels 1080 feet per second?
A body thrown upward with a certain velocity will rise as far as it
would have to fall to acquire this velocity. The velocity (neglecting
the resistance of the atmosphere) of a body starting from rest is gt
where g = 32.16 and / is the number of seconds.
24. A rifle bullet is shot directly upward with a velocity of
2000 feet per second. How high will it rise, and how long be-
fore it will reach the ground '.'
GEOMETRIC PROGRESSIONS 181
25. From a balloon 5800 feet above the earth, a body is
thrown downward with a velocity of 40 feet per second. In
how many seconds will it reach the ground ?
26. If in Problem 25 the body is thrown upward at the rate
of 40 feet per second, how long before it will reach the ground ?
GEOMETRIC PROGRESSIONS
210. A geometric progression is a series of numbers in which
any term after the first is obtained by multiplying the pre-
ceding term by a fixed number, called the common ratio.
The general form of a geometric progression is
a, an, ar, ar\ •••, ar"\
in which a is the first term, r the constant multiplier, or com-
mon ratio, and n the number of terms.
E.g. 3, 6, 12, 24, 48, is a geometric progression in which 3 is the
first term, 2 is the common ratio, and 5 is the number of terms.
Written in the general form it would be 3, 3 • 2, 3 • 2' 2 , 3 • 2 3 , 3 • 2 4 .
211. If I is the last or nth term of the series, then
/ = ar"~\ I
If any three of the four letters in I are given, the remaining
one may be found by solving this equation.
EXERCISES
In each of the following find the value of the letter not given :
4. r = -2, 7. r = -3 10.
11.
3. r = -3, 6. r = 2, 9. r = -f, 12
a = — 1,
f a = — \,
r = -2,
7.
\ r = h
n = 9.
[n = 0.
1 = 1024,
U = 6.
r = -2,
8.
n = 11.
1 = 1024,
r z=-i6
U- 1.
[n = S.
r = 2,
9.
>i = ll.
182 PROGRESSIONS
212. The sum of n terms of a geometric expression may be
found as follows :
If s„ denotes the sum of n terms, then
s„ = a + ar + ar- + ■■■ + ar' 1 - 2 + ar n ~ l . (1)
Multiplying both members of (1) by r, we have
rs n = ar + ar'- + ar 3 + ••• + ar'- 1 + ar". (2)
Subtracting (1) from (2). and canceling terms, we have
?>•„ — s„ = ar™ — a.
(3)
Solving (3) for s n we have
_ ar"-a _a(r"-l)
r— 1 r — 1
This formula for the sum of w terms of a geometric series
involves only a, r, and. n.
Since ar" = r • ar nl = r • /, s" may also be written :
^ = W-a o-W. m
r — 1 1 — r
This formula involves only r, I, and a.
213. From equations I and II or I and III any two of the
numbers a, I, r, s, and n can be found when the other three are
given, as in the following examples.
Ex. 1. Given n = 7, r = 2, s = 381. Find a and I.
From I and III, Z = a-2 fi =6ia, (1)
381 = y~~ = 21 -a. (2)
Substituting / = 61a in (2), we obtain a — 3, and / = 192.
Ex. 2. Given a = - 3, I = - 243, s = - 183. Find r and ».
From I and III. - 213 = (- Sy»~\ (1)
_ 183 = -243r + 3 , (L))
r — 1
GEOMETRIC PROGRESSIONS
183
From (2) r = - 3. (3)
From (1) 81 = (-3)*" 1 . (4)
Since ( — 3) 4 = 81, we have n — 1 = 4 or n = 5.
EXERCISES
1. Solve II for a in terms of the remaining letters.
2. Solve III for each letter in terms of the remaining letters.
In each of the following find the terms represented by the
interrogation point.
3.
a = 1,
r = 3,
n = 5,
s = ?
a
S = f ¥5
s = 635,
r = 2,
n = 7,
a = ?
n =
a —
13,
f»
?
1-
?i = 5,
£ = 1296, 9.
— ?
8 = 1050 1, 10.
Z=?
a =
9
5,
2
"3*
9.
2 J
= 7,
1 6.
8l>
81?
214. Geometric means. The terms between the first and the
last of a geometric progression are called geometric means.
Thus in 3, 6, 12, 24, 48, three geometric means between 3 and 48
are 6, 12 and 24.
If the first term, the last term, and the number of geometric
means are given, the ratio may be found from I, and then the
means may be inserted.
Example. Insert 4 geometric means between 2 and 64.
We have given a = 2,1 = 64, n = 4 + 2 = 6. to find r.
From [, 64 = 2 • r 6 - 1 or r 5 = 32 and r = 2.
Hence, the series is 2, 4, 8, 16, 32, 64.
215. The case of one geometric mean is important. If G is
the geometric mean between a and b, we have — = —
a G
Hence,
Va6.
184 PROGRESSIONS
216. Problem. In attempting to reduce § to a decimal, Ave
find by division .006 •••, the dots indicating that the process
goes on indefinitely.
Conversely, we see that .006 • ■ • = T % + jfa + y^- + • • ., that
is, a geometric progression in which a = T 6 TF , r = T 1 jJ , and n is
not fixed but goes on increasing indefinitely.
As n grows large, / grows small, and by taking n sufficiently large,
I can be made as small as we please. Hence formula III, § I'll', is to
be interpreted in this case as follows :
a - rl 10 10 6 - I
1 - /• Y _ J_ 9
10
in which / grows small indefinitely as n increases indefinitely, so that
by taking n large enough s n can be made to differ as little as we please
from = - = •
9 9 3
In this case we say s H approaches las a limit as n increases
indefinitely.
Observe that this interpretation can apply only when the
constant multiplier r is a proper fraction.
EXERCISES AND PROBLEMS
1. Insert 5 geometric means between 2 and 128.
2. Insert 7 geometric means between 1 and 7 i^.
3. Find the geometric mean between 8 and 18.
4. Find the geometric mean between T V and \.
5. Find the fraction which is the limit of .333 •••.
6. Find the fraction which is the limit of .1666 ■•-.
7. Find the fraction which is the limit of .08333 •••.
8. Find the 13th term of - - 1 /, 4, -3 •••.
9. Find the sum of 15 terms of the series — 243, 81, — L'7 • • •.
10. Find the limit of the sum | + | + | + £+ ..., as the
number of terms increases indefinitely.
GEOMETRIC PROGRESSIONS 185
Given
Find
G
iven
Find
11. a, r,
n
l,s
15.
a, n, I
s, r
12. a, r,
s
I
16.
r, n, I
s, a
13. r, n,
s
I, a
17.
r, I, s
a
14. a, r,
I
s
18.
a, I, s
r
19. The product of three terms of a geometric progression
is 1000. Find the second term.
20. Four numbers are in geometric progression. The sum
of the second and third is 18, and the sum of the first and
fourth is 27. Find the numbers.
21. Find an arithmetic progression whose first term is 1
and whose first, second, fifth, and fourteenth terms are in geo-
metric progression.
22. Three numbers whose sum is 27 are in arithmetic pro-
gression. If 1 is added to the first, 3 to the second, and 11 to
the third the sums will be in geometric progression. Find
the numbers.
23. To find the compound interest when the principal, the
rate of interest, and the time are given.
Solution. Let p equal the number of dollars invested, r the rate of
per cent of interest, t the number of years, and a the amount at the
end of t years.
Then a = p(l + r) at the end of one year.
a — p ( 1 + r) (1 + r) = p (1 + r) a at the end of two years.
and a — p(l + r)' at the end of t years.
That is, the amount for t years is the last term of a geometric pro-
gression in which p is the first term, 1 + r is the ratio, and t + 1 is the
number of terms.
24. Show how to modify the solution given under problem
23 when the interest is compounded semiannually; quarterly.
25. Solve the equation a=p(\. + r)' forp and for r.
186 PROGRESSIONS
26. Solve a =p(l + '">' f° r t.
Solution, log a = logp(l + /•)' = log/) 4- log (1 + r) 1
= log p + / loo- (1 + r). (See § 202.) Hence i = loga ~ log;j -
log (1 + r)
27. At what rate of interest compounded annually will
$1200 amount to $1800 in 12 years'.'
28. At what rate of interest compounded semiannually will
a sum double itself in 20 years ? in 15 years ? in 10 years ?
29. In what time will $8000 amount to $13,500, the rate
of interest being 3^ % compounded annually ?
30. In what time will a sum double itself at 3 %, 4 %, 5 ' , .
compounded semiannually '.'
The present value of a debt due at some future time is a sum such
that, if in vested at compound interest, the amount at the end of the
time will equal the debt.
31. What is the present value of $2500 due in 4 years,
money being worth 3. 1 , >J interest compounded semiannually?
32. A man bequeathed $50,000 to his daughter, payable on
her twenty-fifth birthday, with the provision that the present
worth of the bequest should be paid in case she married before
that time. If she married at 21, howmuch would she receive,
interest being A'/ (l per annum and compounded quarterly?
33. What is the rate of interest if the present worth of
si' 1.000 due in 7 years is $19,500?
34. In how many years is $5000 due if its present worth
is $3500, the rate of interest being 3| <f compounded annually?
HARMONIC PROGRESSIONS
217. A harmonic progression is a series whose terms arc
the reciprocals of the corresponding terms of an arithmetic
progression.
E.g. 1. . . . ••• is a harmonic progression whose terms are the
:i ."> 7 it
reciprocals of the terms of the arithmetic progression 1. ">. 5, 7. '.) ■•-.
HARMONIC PROGRESSIONS 187
The name harmonic is given to such a series because musical strings
of uniform size and tension, whose lengths are the reciprocals of the
positive integers, i.e. 1, -, -, - •••, vibrate in harmony.
2 3 4
The general form of the harmonic progression is
1 _1 i_ 1 j
a' a + </'a + 2</' '"' a + (n-l)d
It follows that if a, b, c, d, e, ... are in harmonic progression,
then-,-,-,-,-, ••• are in arithmetic progression. Hence, all
a bode
questions pertaining to a harmonic progression are best an-
swered by first converting it into an arithmetic progression.
218. Harmonic means. The terms between the first and the
last of a harmonic progression are called harmonic means be-
tween them.
Example. Insert five harmonic means between 30 and 3.
This is done by inserting five arithmetic means between -^l and i.
By the method of § 20S the arithmetic series is found to be ^, r \, T 2 5 ,
iff) 3 7 o' Jo' h Hence, the harmonic series is 30, 12, -^, f£, - 3 ^, f S, 3.
219. The case of a single harmonic mean is important.
Ill
Let a, H, I be in harmonic progression. Then -, — , - are in
arithmetic progression. a
M
Hence, by § 209, i = ?-_i or H = f^- / -
220. The arithmetic, geometric, and harmonic means be-
tween a and I are related as follows :
We have seen A = a , G = vV, // = — — -.
2 a + I
Hence, A = «±i ^ a l = «±i.
G 1 2 2 al
Therefore, — = -, or A = -.
G* H G II
That is, G is a mean proportional between .1 and //. See § 172.
188 PROGRESSIOXS
EXERCISES AND PROBLEMS
1. Insert three harmonic means between 22 and 11.
2. Insert six harmonic means between \ and - 2 #.
3. The first term of a harmonic progression is i and the
tenth term is ^V Find the intervening terms.
4. Two consecutive terms of a harmonic progression are 5
and 6. Find the next two terms and also the two preceding
terms.
5. If a, b, c are in harmonic progression, show that
a -=- c = (a — b)^-(b — c).
6. Find the arithmetic, geometric, and harmonic means
In 'tween :
(a) 16 and 36 ; (b) m + n and m — n ; (c) and
m + n m — n
7. The harmonic mean between two numbers exceeds their
arithmetic mean by 7, and one number is three times the other.
Find the numbers.
8. If x, y, and z are in arithmetic progression, show that
nix, my, and mz are also in arithmetic progression.
9. x, y, and z being in harmonic progression, show that
x 1/ , z . .
-, and — ate m harmonic progression,
x + y + z x + y + z x + y + z
and also that — — — •—, and are in harmonic progression.
y + z x + z .'• + .7
10. The sum of three numbers in harmonic progression is 3,
and the first is double the third. Find the numbers.
11. The geometric mean between two numbers is \ and the
harmonic mean is !. Find the numbers.
12. Insert n harmonic means between the numbers a ami //.
CHAPTER XIII
THE BINOMIAL FORMULA
221. In Chapter II the following products were obtained :
(« + 6)2 - a 2 + -2 ah + ft 2 .
(a + b) 8 = a 3 + 3 a 2 b + 3 air + b 3 .
(a + hy = a 4 + i a s b + 6 a 2 b* + 1 ab 3 + 6*.
(a + b) 5 = « 5 + 5 a 4 !* + 10 a 8 6 2 + 10 a?b 3 + 5 ^A 4 + b 5 .
By a study of these the following facts may be observed :
1. Each product has one term more than the number of units in
the exponent of the binomial.
2. The exponent of a in the Jirst term is the same as the exponent
of the binomial, and diminishes by unity in each succeeding term.
The exponent of b in the last term is the same as the exponent of
the binomial, and diminishes by unity in each preceding term.
3. The sum of the exponents in each term is equal to the expo-
nent of the binomial.
4. The coefficient of the first term is unity; of the second term,
the same as the exponent of the binomial ; and the coefficient of
any other term may be found by multiplying the coefficient 'of the
next preceding term by the exponent of a in that term and dividing
this product by a number one greater than the exponent of b in
that term.
5. The coefficients of any pair of terms equally distant from the
ends are equal.
Statements 2 and 4 form a rule for writing out any power
of a binomial up to the fifth. Let us find (a + b) 6 .
189
190 THE BINOMIAL FORMULA
Multiplying (a + b) 5 by a + b, we have
(a + b) 5 (a + 6) = a 6 + 5 a 5 i + 10 a 4 * 2 + 10 o 8 A 3 + 5 a 2 i 4 + <//< 5
r/ 5 /;+ 5a 4 fe- + 10 r rV,* 8 + 10 aHA + 5 ab 5 + b 6
Hence (a + b) 6 = a 6 + 6 « 5 6 + 15 rt 4 // 2 + 20 a% 3 + 15 a 2 b* + 6 ab 5 +V 6
From this it is seen that the rule holds also for (a + b) 6 .
PROOF BY MATHEMATICAL INDUCTION
222. A proof that the above rule holds for nil positive
integral powers of a binomial may lie made as follows:
First step. Write out the product as it would be for the nth
power on the supposition that the rule holds.
Then the firs! term would be a n ami the lust term b". The second
terms from the ends would lie na n ~ 1 b and nab n ~ l . The third terms
i- ,i t ,11 n(n — 1) „ ojo i n(n — 1) .>,„ r rl
i roni the ends would be — ^ — — - a n ~ 2 b 2 and — ^ — --' <i -/>»--. I lie
1-2 1-2
fourth terms from the ends would be
u(n — l)(/i - 2) _ .,,„ , h(i) - \)(n — 2) „,„_»
— y - v — — ^ a n ~W and — * — — ^ - a s b" ■:
1-2-3 1-2.3
and so on, giving by the hypothesis,
(„ 1 /,)•> = a" + >«i<>-ib + ^~±)<v i --b~+ ■■■ +l^!l^ln-b»- 2 +tiob" i /.".
1-2 1 • 2
Second step. Multiply this expression by a + b and see if
the result can be so arranged as to conform to the same rule.
Then, (a + b)"(a + b)
, . . , >t(» - }) ,,„ „. , ,
= «" fl + na"b + -±-— -a"- l b- + ■ ■ ■ + na-b"- 1 + ab"
a"h + )in"-Vi- + • • • + — k - — — a 2 6 n_1 + «a6 n + 6 n+1 .
Hence adding,
(a + b) n+1 = a n+1 + (n + l)a"b + [~ " v " ~ ^ + n~| a"" 1 *) 2 + • • •
+ T„ + El " ~ ^ "l a 2 6 n-l + („ + l) a jn + &»+l.
Combining the terms in brackets,we have,
(a + b) n+i = a"+ l + (n + })>,»/> + ( " " } > ) ", /»-'//- + • • ■
PROOF BY INDUCTION 191
The last result shows that the rule holds for (a + b) n + 1 if it
holds for (a + by. That is, if the rule holds for any positive inte-
gral, exponent, it holds for (he next higher integer.
Third step. It was found above by actual multiplication that
the rule does hold lor (a + bf. Hence by the above argument
we know that the rule holds for (a + by.
Moreover, since we now know that the rule holds for (a + l>)~, we
conclude by the same argument that it holds for (a + 6) 8 , and if for
(u + b) s , then for (a + b) 9 , and so on.
Since this process of extending to higher powers can be car-
ried on indefinitely, we conclude that the live statements in
§221 hold for all positive integral powers of a binomial.
The essence of this proof by mathematical induction consists
in applying the supposed rule to the nth power and finding
that the rule, does hold for the (// -fl)th power if it holds for
the nth power.
223. The general term. According to the rule now known to
In ild for any positive integral exponent, we may write as many
terms of the expansion of {a + &)" as may be desired, thus :
(a + b)" = a" + na" '6 + /7 ^=-^a» -6-
1 • 2
„(„_1 )( „_2) /i(fi-l)(/i-2)(ii-3) 4ft4 ■ ... i
^ 1.2- 3 1- 2-3-4
From this result, called the binomial formula, we see :
(1) The exponent of b in any term is one less than the number of
that term, and the exponent of a is n minus the exponent of b. Hence
the exponent of b in the (Jc + l)st term is I; and that of a is n - k.
(2) In the coefficient of any term the last factor in the denominator
is the same as the exponent of b in that term, and the last factor in
the numerator is one more than the exponent of (7.
Hence the (/V + l)st term, which is called the general term is
„(„_l)(„_ 2)( /7-3V--(/7-* + l) a „, / , M n
1-2-3-4-5-*
192 THE BINOMIAL FORMULA
224. The process of writing out the power of a binomial is
called expanding the binomial, and the result is called the ex-
pansion of the binomial.
Ex. 1. Expand (x — y) 4 .
In this case a = x, b = — y, n = 4.
Hence substituting in formula I,
(I - v y = z> + i x\ - ,,) + ±ii=±> x-J(> .,)'-•+ 4 "~, n j i t ~'- ) i-( - y)»
+ 4(4-l)(4-2X4- 3 )
1 • o • 4
4 13 , 4-3 o o 4.3-2 3 , 4- 3-2.1 4 ,.„
Hence (x - y) 4 = .r 4 - 4 x 3 ,-/ + 6 .'-//- - 4 xy* + >/*. (3)
Notice that this is precisely the same as the expansion of (x + y)*
except that every other term beginning with the second is negative.
Ex. 2. Expand (1-2 y) 5 .
Here 0= 1, b = — 2 //, n = .1.
Since the coefficients in the expansion of (a + h) h are 1, 5, 10, 10, •">. 1.
we write at once,
(1-2^)6= l* + 5.1*.(-2y) +10.P.(-2y)2
+ 10 ■ l 2 • (- 2 //) 3 + 5 • 1 • (- 2 //) 4 + (- 2 y)5
= 1 - 10 y + 40 //- - 80 y 8 + 80 1/ - 32 y 6 .
Ex. 3. Expand f-+|Y-
Remembering the coefficients just given, we write at once,
l + !V-(!V + »aYfeUio(!W!V+io'!W«
r/ \x/ \3/ \./7 \3/ \x/ \3
"©O'+GD'e+i
1 :, v 10 y* h}>l *_£ + jL
x 5 Sx* 9 x 3 27 x 3 81 x 243"
EXPANSION OF BINOMIALS 193
In a similar manner any positive integral power of a
binomial may be written.
Ex. 4. Write the sixth term in the expansion of (x— 2 y) w
without computing any other term.
From II, § 223, we know the (k + l)st term for the nth power of
a + b, namely, . . OA , , 1
n (n - 1 ) ( n - 2) ■ • • ( « - k + 1 ) aH _ tfjt
2.3-4... k
In this case a = x, b = — 2 >/, >i = 10, k + 1 = 6. Hence k = 5.
Substituting these particular values, we have
10(10-l)(10-2)-(10-5 + l) xl0 _ 5( _ 2 ?/)5
2 • 3 • 4 • 5
10 • 9 • 8 • 7 • fi ., QO ~
2 • 3 • 4 • 5 ^ J '
= - 32 - 252 j 5 // 5 = - 8064 j- 5 // 5 .
EXERCISES
1. Make a list of the coefficients for each power of a binomial
from the 2d to the 10th.
Expand the following :
2. (x-yf. 9. (x'-^y. n (t_t
3. (2x + 3f. 10 - (ar' + ST 8 ) 5 - ^ *
4. (3*4-2^. "■ <«-»>'■ 18. ^|-jfV5Y
12. r*4-t/V. V.t y
5. (3+2/) 5 -
12. (oj + 2/) 9 .
1
19.
13. (m-n)» /V» , % ftV
6. (.r 3 + 2/) 6 . 14. (>-* + s 2 ) 4 . ' \^V *n
7. (^-^) 6 . 15. (c--d-^. ^ /c^_j^
8. (a?-y 2 ) 7 . 16. (V«-V&) 6 . ' Vvd 4
21. (2<ftr*-3&y-*) 4 . 22. (3 xy-z - x~ 3 yy.
194 THE BINOMIAL FORMULA
In each of the following find the term called for without
finding any other term :
23. The 5th term of (a 4-6) 12 .
24. The 7th term of (3 x — 2y) n .
25. The 6th term of (V.v- \ y) w .
26. The Oth term of (x - y i .
27. The 8th term (if ( \ m - \ n i 18 .
28. The 7th term of (a 2 b — ab .
29. The (ith term of («— or 1 ) 2 *.
30. The 11th term of {xhj— x~ 2 y- x )
31. The 5th term from each end of the expansion of (n — h ,-".
32. The 7th term from each end of (a\ a — b~\ b) 21 .
33. "Which term, counting from the beginning, has the same
coefficient as the 7th term of (a + &) 10 ? Verify by finding both
coefficients. How do the exponents differ in these terms '.'
34. "What other term has the same coefficient as the l'.Mh
term of (a + &) 24 ? How do the exponents differ '.' Find in the
shortest way the 21st term of (a -I-&) 25 .
35. Find the 87th term of (a 4- &) 90 .
3G. Find the 53d term of (a^ — &^) 56 .
37. What other term lias the same coefficient as the 5th
term in flic expansion of U'4-?/) 19 ?
38. Expand [c 4-&) 4- c] 3 by the binomial formula.
39. Expand [1 + (2 x + 3 //)]' by the binomial formula.
40. Expand (2x — 3y+ iz) s by the binomial formula.
41. Write the Qc + 1) si term of (a + b) n . Write the (n+1) si
term of (« + &)". Show that the next and also all succeeding
terms after the (ra + l isl term have zero coefficients, thus prov-
iii- I hat there are exactly /' + 1 terras in the expansion.
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