Rancho Cucamonga High School - Ontario, CA
Algebra
and
Trigonometry
Mathematics
Mathematics
Resource Guide
Guide
Resource
2016−2017
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Table of Contents
SECTION 1:
Basic Properties of
Real Numbers..................................4
SECTION 2:
Linear and Quadratic
Equations..............................................7
2.1 Linear Equations.........................................7
Section 2.1: Exercises................................ 10
2.2 Quadratic Equations.................................. 11
2.2.1 Equations of the Form x2 – p = 0.............. 11
2.2.2 Equations of the
Form k(x + r)2 – p = 0, Where k ≠ 0.......... 12
2.2.3 Equations of the
Form ax2 + bx + c = 0, Where a ≠ 0......... 12
2.2.4 The Discriminant.......................................... 15
Section 2.2: Exercises................................. 16
SECTION 3:
Polynomial Equations..... 17
3.1 When Are Two Polynomials Equal?............ 18
3.2 H
ow Do We Add and
Subtract Polynomials?................................ 18
3.3 How Do We Multiply Polynomials?............ 19
3.4 How Do We Divide Polynomials?..............20
3.5 H
ow Does the Division of Polynomials
Help Us with Polynomial Equations?.......... 24
3.6 Proof of the Rational Root Theorem.............29
3.7 Proof of the Factor Theorem.........................30
SEcTionS 3.1–3.7: Exercises......................... 31
3.8 Complex Numbers...................................... 32
2
3.8.1 How Do We Add
Complex Numbers?......................................33
3.8.2 How Do We Multiply
Complex Numbers?.....................................34
3.8.3 How Do We Divide
Complex Numbers?.....................................35
SEcTion 3.8: Exercises.................................38
SECTION 4:
Functions........................................... 39
4.1 Preliminaries............................................. 39
4.2 Definition of a Function.............................40
4.3 Many-to-One Functions
versus One-to-One Functions..................... 43
4.4 Inverse Functions.......................................44
Section 4: Exercises....................................48
SECTION 5:
Graphing............................................. 49
5.1 What Does the Graph of a
Linear Function y = ax + b Look Like?........49
5.2 What Does the Graph of a Quadratic
Function y = ax2 + bx + c Look Like?........... 51
5.2.1 The Case y = x2. .............................................51
5.2.2 The General Case y = ax2 + bx + c...........53
5.3 What Do the Graphs of
Some Polynomials Look Like?.................... 56
5.4 What Does the Graph of the
Exponential Function y = ax Look Like?...... 57
5.5 What Does the Graph of the Logarithmic
Function y = logax Look Like?.................... 59
Sections 5.1–5.5: Exercises......................... 61
5.6 Transformations of Graphs..........................62
5.6.1 Graphing y = f(x + c)
from the Graph of y = f(x)..........................62
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part I Algebra...............................4
5.6.2 Graphing y = f(x) + C
from the Graph of y = f(x).........................63
8.3 Lines....................................................... 112
5.6.3 Graphing y = f(ax)
from the Graph of y = f(x).........................65
8.3.2 Point-Point Form ....................................... 116
8.3.1 Slope Form.................................................... 115
5.6.4 Graphing y = Af(x)
from the Graph of y = f(x)..........................67
8.3.3 Slope-Point Form....................................... 118
Section 5.6: Exercises.................................68
8.4 Circles......................................................120
8.5 Solving Geometry Problems
Using Coordinate Geometry..................... 121
SEcTion 8: Exercises..................................124
part II Trigonometry. ..126
Non-Polynomial
Equations........................................... 69
6.1 Rational Equations.................................... 69
6.1.1 How Do We Solve
Rational Equations?......................................70
SECTION 6.1.1 Exercises...........................................75
6.1.2 What Do the Graphs of Some
Rational Functions Look Like?..................76
SECTION 6.1.2 Exercises...........................................81
6.2 Exponential Equations.............................. 81
6.2.1 Basic Properties.............................................81
6.2.2 How Do We Solve
Exponential Equations?.............................82
Section 6.2: Exercises.................................84
6.3 Logarithmic Equations...............................84
6.3.1 Basic Properties.............................................84
6.3.2 How Do We Solve
Logarithmic Equations?.............................85
Section 6.3: Exercises.................................89
6.4 Radical Equations......................................90
6.4.1 Method 1.........................................................90
6.4.2 Method 2.........................................................92
Section 6.4: Exercises.................................92
SECTION 7:
Inequalities................................... 94
7.1 Linear Inequalities.....................................94
SECTION 7.1: Exercises...............................100
7.2 Quadratic Inequalities..............................100
7.2.1 Inequalities of the
Form ax2 + bx + c > 0 and a > 0.............101
7.2.2 Inequalities of the
Form ax2 + bx + c < 0 and a > 0............ 104
Section 7.2: Exercises............................... 106
SECTION 8:
Coordinate Geometry... 108
8.1 The Pythagorean Theorem......................... 108
8.2 Points...................................................... 110
SECTION 9:
Trigonometric
Functions. ......................................126
9.1 The Sine Function for Acute Angles............127
9.2 The Tangent Function for Acute Angles . ...129
9.3 The Cosine and Cotangent
Functions for Acute Angles.......................130
9.4 Relations Among
Trigonometric Functions........................... 131
9.5 Trigonometric Functions
of Special Angles...................................... 133
9.6 Trigonometric Functions
of Angles of Any Measure.........................134
9.6.1 Definitions and Properties .....................134
9.6.2 Negative Angles..........................................136
9.7 Trigonometric Identities............................ 137
9.7.1 Sum and Difference Identities................137
9.7.2 Double-Angle Identities...........................140
9.7.3 Half-Angle Identities..................................142
9.7.4 Sum-to-Product Identities.......................143
9.7.5 Product-to-Sum Identities.......................146
SEcTionS 9.1–9.7: Exercises....................... 147
9.8 Graphs of Trigonometric Functions........... 148
9.9 Inverse Trigonometric Functions............... 155
9.10 Trigonometric Equations........................ 163
SEcTionS 9.8–9.10 Exercises..................... 168
9.11 The Law of Sines and Cosines.................. 169
SEcTion 9.11: Exercises............................. 174
9.12 Radians................................................. 174
SEcTion 9.12: Exercises............................. 176
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SECTION 6:
8.3.4 Mutual Positions of Lines......................... 119
part
TI
R
A
P
1
algebra
i:
Part
Algeb
rA
aLGeBRa
Section
1
Basic Properties of
Real
Numbers
SsecTion
ecTIon 1:
Basic
Properties
of numbers
real numbers
1: Basic
Properties
of Real

We will begin this resource guide with a discussion of the real numbers—the numbers with which you
addition, + , and multiplication, ⋅ . When two real numbers are added, the result is a real number. We
often express this fact by saying that  is closed under addition. Likewise, when two real numbers are
multiplied, the result is a real number. Thus,  is also closed under multiplication.
Let us now turn to the basic properties of addition and multiplication. These properties are very important
for understanding the structure of the real numbers. In high school mathematics, however, they are mainly
used to simplify computations, especially computations that involve algebraic expressions.
The numbers 0 and 1 are special—they have the following properties:
1. For any real number x , x + 0 = x
2. For any real number x , x ⋅ 1 = x
No real numbers other than 0 and 1 have these properties. The rest of the properties are as follows:
3. Additive Inverse. For any real number x there is a unique number, denoted by −x , such that
x + (−x) = 0 .
4. Multiplicative Inverse. For any real number x different from 0 , there is a unique number, denoted
by x −1 , such that x ⋅ x −1 = 1 .
5. Associative Law of Addition. For all real numbers x , y , and z , x + ( y + z ) = ( x + y ) + z
6. Commutative Law of Addition. For all real numbers x , y , x + y = y + x
7. Associative Law of Multiplication. For all real numbers x , y , and z , x ⋅ ( y ⋅ z) = (x ⋅ y) ⋅ z
8. Commutative Law of Multiplication. For all real numbers x , y , x ⋅ y = y ⋅ x
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are most familiar. The set of the real numbers is denoted by . This set is equipped with two operations:
9. Distributive Law. For all real numbers x , y , and z , x ⋅ ( y + z ) = x ⋅ y + x ⋅ z
10. Zero Factor Property. For all real numbers x and y , x ⋅ y = 0 if and only if x = 0 or y = 0
or both x and y are equal to 0.
Note that the last property consists of two properties. The first property is: If x ⋅ y = 0 , then x = 0 or
yy ==00 or
bothsecond
x and property
y are equal
second
= 0 or
y = 0property
y =If0x. = 0 and/or y = 0 , then x . y = 0.
. The
is: to
If 0.x The
, then x ⋅ is:
You might be wondering, what about the operations subtraction and division? These two operations are
defined in terms of addition and multiplication, respectively, as follows:
definition
And if y ≠ 0 ,
x
is the real number x ⋅ y −1 .
y
 , the set of all real numbers, includes the integers: 0,±1,± 2,± 3, . The set of all integers is denoted
by  . The set of all positive integers is denoted by  + , and the set of all negative integers is denoted by
 − . The real numbers also include the rational numbers. A rational number is a ratio, or division, of two
2 −4 −23 3
integers. For example, , ,
,
,2.13,− 0.98123 are all rational numbers. Any integer n is rational
3 17 42 −18
n
because it can be written as , which is a ratio of two integers. The set of all rational numbers is denoted
1
by  .
You can see from this description that:
a.  + and  − are subsets of  ,
b.  is a subset of  , and
c.
 is a subset of  .
Figure 1–1 illustrates these relations:
Figure 1–1
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If x and y are real numbers, their difference x − y is the real number x + ( − y ) .
Another important subset of the real numbers is the set of irrational numbers. A real number that is not
rational is called irrational. For example, 2 is irrational. Do you know why
2 is irrational?
theorem
2 is an irrational number.
Proof
1. Assume that
2=
2 is rational. Then by definition
n ≠ 0.
m
, where m and n are integers and
n
m
, then we should be able to reduce this fraction to the lowest term.
n
m
is a reduced fraction.
3. Let m and n be so that
n
m
(Why?)
4. Since 2 = , 2n 2 = m 2 . This implies that m 2 is even, and, hence, m is also even .(why?)
n
5. Since m is even, we can write m = 2k for some integer k .
2=
( )
6. Substituting m = 2k in 2n 2 = m 2 , we get 2n2 = 2k , and so n 2 = 2k 2 . This implies that n 2 is
2
even, and, hence, n is also even.
m
is not a reduced fraction—a contradiction. This shows that
7. Since m and n are both even,
n
our assumption that 2 is rational is not true. Hence 2 is irrational.
We will conclude this section with another set of properties that concern inequalities between real
numbers:
11. For any two real numbers x and y , one and only one of the following is true: x < y , x = y , or xy >< y.x .
12. For any three real numbers x , y , z , if x < y , y < z , then x < z .
13. For any three real numbers x , y , z , if x < y then x + z < y + z .
14. For any three real numbers x , y , z , if x < y and z > 0 , then zx < zy .
15. For any three real numbers x , y , z , if x < y and z < 0 , then zx > zy .

SsecTion
ecTIon 2:
Linear
Quadratic
equations
2: Linear
andand
Quadratic
equations
2 .
Linear
Equations
2 .1 1
linear
equations
The simplest equations are those in the form ax + b = 0 . These equations are called linear equations,
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2. If indeed
We will conclude this section with another set of properties that concern inequalities between real
numbers:
11. For any two real numbers x and y , one and only one of the following is true: x < y , x = y , or y < x .
12. For any three real numbers x , y , z , if x < y , y < z , then x < z .
13. For any three real numbers x , y , z , if x < y then x + z < y + z .
14. For any three real numbers x , y , z , if x < y and z > 0 , then zx < zy .
Section
2
15. For any three real numbers x , y , z , if x < y and z < 0 , then zx > zy .

Linear and Quadratic
SsecTion
ecTIon 2:
Linear
Quadratic
equations
2: Linear
andand
Quadratic
equations
Equations
The simplest equations are those in the form ax + b = 0 . These equations are called linear equations,
because, as we will see in Section 8, they are associated with straight lines. The following equations are
all linear:
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(1) 2 x + 6 = 0
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(2) −3x + 4 = 0
2
(3) − x + 3.1 = 0
5
Equations (1), (2), and (3) have the form ax + b = 0 . In equation (1), a = 2 and b = 6 ; in equation (2),
2
a = −3 and b = 4 ; and in equation (3) a = − and b = 3.1 . Hence, equations (1), (2), and (3) are linear.
5
Now, let’s consider the following equations:
(4) 4 x = 17
(5) 2 x − 9 = 34
(6) −6 + 2x = 4x + 9
These equations do not have the form ax + b = 0 , but they are equivalent to equations of this form. They,
too, are considered linear equations. When are two equations equivalent? Two equations are equivalent
if they have the same solution set, that is to say, if each solution of one equation is also a solution of the
second equation, and vice versa.
ExamplE 2.1a: The equations 2 x − 9 = 34 and 2 x − 43 = 0 are equivalent.
REason: The solution set of each equation consists of the same number: x = 21.5 .
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Now, let’s solve a few linear equations.
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2 .
11
Linear
Equations
2 .
linear
equations
2.1
Linear
Equations
second equation,
equation, and
and vice
vice versa.
versa.
second
second equation, and vice versa.
ExamplE
2 xx −
− 99 =
= 34
34 and
2 xx −
− 43
43 =
= 00 are
ExamplE 2.1a:
2.1a: The
The equations
equations 2
and 2
are equivalent.
equivalent.
ExamplE 2.1a: The equations 2 x − 9 = 34 and 2 x − 43 = 0 are equivalent.
21.5 ..
REason:
REason: The
The solution
solution set
set of
of each
each equation
equation consists
consists of
of the
the same
same number:
number: xx == 21.5
REason: The solution set of each equation consists of the same number: x = 21.5 .
Now,
Now, let’s
let’s solve
solve aa few
few linear
linear equations.
equations.
Now, let’s solve a few linear equations.
solution: We
We can
can add
add 34
34 to
to both
both sides
sides of
of the
the equation
equation without
without changing
changing the
the solution
solution set
set of
of
solution:
solution: We can add 34 to both sides of the equation without changing the solution set of
the
the equation.
equation. And,
And, in
in doing
doing so,
so, we
we get
get aa simpler
simpler equation,
equation, an
an equation
equation that
that is
is easier
easier to
to solve.
solve. Thus:
Thus:
the equation. And, in doing so, we get a simpler equation, an equation that is easier to solve. Thus:
88 xx −
− 34
34 +
+ 34
34 =
= 15
15 +
+ 34
34
8 x − 34 + 34 = 15 + 34
88 xx +
+ 00 =
= 49
49
The
is equivalent to the original equation and is easier to solve. To get x, we simply
8 x +last
0 = equation
49
88 xx =
49
=
49
divide
8 x = 49both sides of the equation by 8.
The last equation is equivalent to the original equation and is easier to solve. To get x, we simply
8 x 49
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byisisis8.Prohibited
= both sides
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x
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x = =6.125
8
8
x = 6.125
The
solution set of the given equation 8 x − 34 = 15 consists of one number: x = 6.125 .
Let’s check that x = 6.125 is indeed a solution to this equation:
The solution set of the given equation 8 x − 34 = 15 consists of one number: x = 6.125 .
8(6.125) − 34 = 49 − 34 = 15
Let’s check that x = 6.125 is indeed a solution to this equation:
8(6.125) − 34 = 49 − 34 = 15
4
1
= 3x − .
7
7
4
1
ExamplE 2.1c: Solve the equation −5 x + = 3x − .
7
7
solution:
ExamplE 2.1c: Solve the equation −5 x +
8
4 4
1 4
solution:
−5x + − = 3x − −
7 7
7 7
4 4 5 1 4
−5x ++ 0 =−3x =− 3x − −
−5x
7 7 7 7 7
5− 5
−5x
+
0
=
3x
−5x = 3x −
7 7
5
−5x −= 3x
3x =
− 3x − 5 − 3x
−5x
7 7
5
−5x =− −
3x5= 3x − − 3x
−8x
7
7
4
from both sides of the equation.
7
4
Subtract from both sides of the equation.
7
Subtract
Subtract 3x from both sides of the equation.
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Subtract 3x from both sides of the equation.
7
7
7
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Rancho Cucamonga High School - Ontario, CA
ExamplE
8 xx −
− 34
34 =
= 15
15 ..
ExamplE 2.1b:
2.1b: Solve
Solve the
the equation
equation 8
ExamplE 2.1b: Solve the equation 8 x − 34 = 15 .
−5x + 0 = 3x −
−5x = 3x −
5
7
5
7
−5x − 3x = 3x −
−8x = −
8x =
5
− 3x
7
5
7
5
7
Multiply both sides of the equation by −1 .
5
x
8x 7
=
8 8
Divide both sides of the equation by 8 .
5
56
In the next series of examples, we will present the solution process without a detailed explanation as was
given in the previous examples.
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ExamplE 2.1d: Solve the equation 2( x − 4) + 3 = 2 x − 5 .
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solution:
2(x − 4) + 3 = 2x − 5
2x − 8 + 3 = 2x − 5
2x − 5 = 2x − 5
Any number x is a solution to the last equation because with any number that we substitute for
x , we get a true statement. Hence, our original equation, 2( x − 4) + 3 = 2 x − 5 , has infinitely many
solutions.
ExamplE 2.1e: Solve the equation x − 7 = 2( x + 3) − x .
solution:
x + 7 = 2( x + 3) − x
x + 7 = 2x + 6 − x
x+7= x+6
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x=
Subtract 3x from both sides of the equation.
ExamplE 2.1e: Solve the equation x − 7 = 2( x + 3) − x .
ExamplE
.
ExamplE 2.1e:
2.1e: Solve
Solve the
the equation
equation xx −− 77 == 2(
2(xx ++ 3)
3) −− xx .
solution:
x + 7 = 2( x + 3) − x
solution:
solution:
x + 7 = 2x + 6 − x
xx ++ 77 == 2(
2(xx ++ 3)
3) −− xx
xx ++ 77 == x2 x+ +6 6 − x
x + 7 = 2x + 6 − x
xx ++ 77 ==isxxno
++ 66number x that satisfies the last equation because if there were such a number, then
There
we would get 7 = 6 , which is absurdity. Hence, the original equation x − 7 = 2( x + 3) − x has no
There
There is
is no
no number
number xx that
that satisfies
satisfies the
the last
last equation
equation because
because ifif there
there were
were such
such aa number,
number, then
then
solution.
has no
we
which is
is absurdity.
absurdity. Hence,
Hence, the
the original
original equation
equation xx −− 77 == 2(
we would
would get
get 77 == 66 ,, which
2(xx ++ 3)
3) −− xx has no
solution.
solution.
Section 2 .2.1
1:
section 2 .1: exercises
Section
Exercises
:: k k k k k k k k k k k k k k k k
kk
k k k k k k k 2 .
kk
k
k
kk
k kSection
k k k k 2 .
k 11k
Exercises
Section
2 .
section
1
:
exercises
section 2 .1: exercises
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
Exercises
Exercises
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
1. Solve the following equations:
1
4
1.
Solve
equations:
(x
8) =
1. a.
Solve the
the+following
following
equations:
3
57
44
 = 11 2   1 
a.
(x
+
8)
3
a. 257
b.
− + 8)+=33x +  + 4  x −  = 9x +17
 x(x
57
 4  3 3   5 
3
2
1
b.
22xx −− 3++ 33xx ++ 2++ 44xx −− 1 == 9x
+17
b. 8x
 − 3 4 5x +2 3   5  9x +17
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−
=
1
c. 
4 
3  5
2
11
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x
+11
x + 3 x −Unauthorized
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e. x + 2x + 3x ++ 33x = 51
2. Sam took a number, multiplied it by 12, added 60 to the result, and got 372. What was Sam’s
initial number?
4
2
from it. Then, he divided 3 by the result and got
3. Tom took a number and subtracted 2
15
3
5
3 . What was Tom’s initial number?
13
4. John left town A at the same time that Mary left town B. They walked toward each other and
met 40 minutes after their departure. John arrived at town B 32 minutes after he met Mary. How
long did it take Mary to arrive at town A?
5. Cities A and B are 70 miles apart. A biker leaves town A at the same time that a bus leaves town
B. They travel toward each other and meet 84 minutes after their departure at a point between A
10
and B. The bus arrives at City
B, Mathematics
stays there forResource
20 minutes,
and then
heads back to City A. The
USAD
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bus meets the biker again 2 hours and 41 minutes after their first meeting. What are the speeds
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kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
5. Cities A and B are 70 miles apart. A biker leaves town A at the same time that a bus leaves town
5. Cities A and B are 70 miles apart. A biker leaves town A at the same time that a bus leaves town
B. They travel toward each other and meet 84 minutes after their departure at a point between A
B. They travel toward each other and meet 84 minutes after their departure at a point between A
and B. The bus arrives at City B, stays there for 20 minutes, and then heads back to City A. The
and B. The bus arrives at City A,
B, stays there for 20 minutes, and then heads back to City B.
A. The
bus meets the biker again 2 hours and 41 minutes after their first meeting. What are the speeds
bus meets the biker again 2 hours and 41 minutes after their first meeting. What are the speeds
of the bus and the biker?
of the bus and the biker?
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We have learned how to solve linear equations. In this section, we will learn how to solve a new kind of
We have learned how to solve linear equations. In this section, we will
learn how to solve a new kind of
equation, quadratic equations. These are equations of the form ax 2 + bx + c = 0 , where a ≠ 0 . We add
equation, quadratic equations. These are equations of the form ax 2 + bx + c = 0 , where a ≠ 0 . We add
the condition a ≠ 0 because if a = 0 , then we would just have a linear equation, bx + c = 0 , which we
the condition a ≠ 0 because if a = 0 , then we would just have a linear equation, bx + c = 0 , which we
have already discussed. When we say “quadratic equation,” we always assume that the coefficient of x 2
have already discussed. When we say “quadratic equation,” we always assume that the coefficient of x 2
is nonzero. We will begin with two special cases of quadratic equations, and then we will deal with the
is nonzero. We will begin with two special cases of quadratic equations, and then we will deal with the
general case.
general case.
2 .2 .1
eQuATIonS
The
form
2 .2 .
eQuations
the
2.2.11Equations
ofof
theof
Form
x2form
–xp22p0
= 0x 22p0
2 .2 .1
of of
Thethe
formform
x p0
2 .2 .1 eQuATIonS
eQuations
x p0
In this case, a = 1 , b = 0 , c = − p . Equations of this form are easy to solve:
In2 this case, a = 1 , b = 0 , c = − p . Equations of this form are easy to solve:
x2 − p = 0
x2 − p = 0
x2 = p
x =p
x = p or x = − pUnauthorized Duplication is Prohibited Outside the Terms of Your License Agreement
10
10
Or,
we just write
10
10
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x = ± AcAdemic
p . Note, however,
that p must be non-negative.
IfGuide
is no real
solution
pResouRce
< 0 , there
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to the given equation.
ExamplE 2.2a: Solve the equation x 2 − 16 = 0.
solution:
x 2 − 16 = 0
x 2 = 16
x = ± 16
x = ±4
2
2 .2 .2eQuATIonS
eQuations
k0
2 .2 .2
of of
The the
formform
k(xr)k(xr)
p0,2p0,
WhereWhere
k0
This type of equationUSAD
can beMathematics
reduced to the
format we
just •discussed.
To do this, we need to divide both
11
Resource
Guide
2016–2017
p
p
p
sides of the equation by k . Then, we get ( x + r ) 2 = (remember that k ≠ 0 ). If ≥ 0 , then x + r = ±
,
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2 .2
Quadratic
equations
2 .2
Quadratic
Equations
2.2
Quadratic
Equations
2 .2
Quadratic
equations
2 .2 Quadratic Equations
2p = 0, 2Where
2 .2 .2
eQuations
the
k0
2.2.2 Equations
ofof
theof
form
k(xform
+
r)2 – k(xr)
k≠
0
2 .2 .2
eQuATIonS
The
form
k(xr)
p0,p0,
WhereWhere
k0
This type of equation can be reduced to the format we just discussed. To do this, we need to divide both
p
p
p
sides of the equation by k . Then, we get ( x + r ) 2 = (remember that k ≠ 0 ). If ≥ 0 , then x + r = ±
,
k
k
k
p
p
and so x = − r ±
. If
< 0 , the equation has no real solution.
k
k
ExamplE 2.2b: Solve the equation 8( x + 3) 2 − 72 = 0 .
solution:
8(x + 3)2 = 72
72
72
2
((xx ++ 1)
3)2 ==
88
22
((xx ++ 1)
3) == 99
−−
xx +1
+ 3==±±√99
3
xx +
+ 13== ±±3
−1 ±± 33
xx =
= −3
xx =
= 02 or
−4
or xx==−6
2.2.3 Equations
ofof
theThe
form
axform
+ 2bx
+axc2bxc0,
= 0, Where
aa0
≠ 0 a0
2 .2 .3
eQuations
of
the
2 .2 .3
eQuATIonS
form
ax
bxc0,
WhereWhere
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To solve this equation, we first must change it into the form we just discussed, k ( x + r ) − p = 0 , which we
2
already know how to solve. Let’s begin with a specific example:
ExamplE 2.2c: Solve 2 x 2 − 3x + 1 = 0 .
solution: We solve this equation by making it an equivalent equation of the form
k ( x + r )2 − p = 0 .
When opening the parentheses in the last equation, we get kx 2 + 2krx + kr 2 − p = 0 . Comparing
the coefficients of this equation with the coefficients of the given equation (see the figure below),
12
we get:  k = 2

2kr = −3
 2
 kr − p = 1
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2
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8(x + 3)2 − 72 = 0
When opening the parentheses in the last equation, we get kx 2 + 2krx + kr 2 − p = 0 . Comparing
the coefficients of this equation with the coefficients of the given equation (see the figure below),
we get:  k = 2

2kr = −3
 2
 kr − p = 1
2
−3
By substituting k = 2 in 2kr = −3 , we get r =
.
4
2
 −3 
1
−3
2
in kr − p = 1 , we get p = 2 ⋅   −1 = .
And, by substituting k = 2 and r =
4
8
4
2
  −3  1
Thus, our equation can be written as: 2  x +   − = 0 .
  4  8
2
Now, we have arrived at an equation of the form k(x + r)2 − p = 0 , which we know how to solve.
  −3  1
2  x +   =
  4  8
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2
  −3 
1
 x +   =
  4  16
2
x+
−3
1
=±
4
4
x=
3 1
±
4 4
x = 1 or x =
1
2
Thus, the solution set of our equation consists of two numbers: x = 1 and x =
1
.
2
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12
ExamplE 2.2d:
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solution: This process is applicable to the general case ax 2 + bx + c = 0 , where a ≠ 0 . As
before, our aim is to bring this equation into the form k(x + r)2 − p = 0 . This form, as has just
been shown, is kx 2 + 2krx + kr 2 − p = 0 .
k = a

Comparing the coefficients of the two equations, we get  2kr = b
kr 2 − p = c

b
Substituting k = a in 2kr = b , we get r =
.
2a
2
 b
b
2
in kr − p = c , we get p = a ⋅   − c .
Substituting k = a and r =
2a
 2a 
Thus, our equation can be written as:
USAD Mathematics Resource Guide • 2016–2017
2
  b    b  2 
a  x +   −  a   − c  = 0 .
13
Comparing the coefficients of the two equations, we get ° NU " E
²NU S " F
±
E
Substituting N " D in NU " E , we get U "
.
D
2
 b
E
in NU S " F , we get p = a ⋅   − c .
Substituting N " D and U "
D
 2a 
Thus, our equation can be written as:
2
  b    b  2 
a  x +   −  a   − c  = 0 .

  2a    2a 

Now we can solve for [ :
2
  b 
 b
a  x +   = a   − c
 2a 
  2a 
2
 b
a ⋅  − c
2
  b 
 2a 
 x +   =
a
  2a 
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2
  b   b  2 c
 x +   =   −
  2a   2a  a
2
  b 
b2 c
 x +   = 2 −
a
  2a  4a
2
  b  b2 − 4ac
 x +   =
4a 2
  2a 
2
© E¹
E DF
[ ª º"t
D
D « »
E
E DF
[" t
D
D E
E DF
[" t
D
D
E t E DF
["
D Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement
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We have just developed a formula for solving quadratic equationsD[ E[ F " . This formula is called
the quadratic formula: [ "
E t E DF
.
D
2.2.4
2.2.4THE
THEDISCRIMINANT
DISCRIMINANT
The expression E DF tells us when a quadratic equation has a real solution, and if it does, whether it
has
expression
is called
the discriminant.
It is denoted by the Greek letter
14 one or two real solutions. This
USAD
Mathematics
Resource
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) (delta).
2a
2 .2 .4
2 .2 .4The
theDIScrImInAnT
discriminant
2.2.4 The Discriminant
The expression b2 − 4ac tells us when a quadratic equation has a real solution, and if it does, whether it
has one or two real solutions. This expression is called the discriminant. It is denoted by the Greek letter
∆ (delta).
From the quadratic formula we developed previously, we get:
−b + b2 − 4ac
−b − b2 − 4ac
or x =
.
2a
2a
−b
2. When ∆ = b2 − 4ac = 0 , the equation has only one real solution: x =
.
2a
1. When ∆ = b2 − 4ac > 0, the equation has two real solutions: x =
It is not always necessary to use the quadratic formula to solve a quadratic equation. Quadratic equations
of the form ax 2 + bx = 0 are easy to solve by factoring, as is shown below:
ax 2 + bx = 0
x ( ax + b) = 0
x = 0 or ax + b = 0
x = 0 or x = −
b
a
Here we utilized the fact that the product of two numbers is zero if and only if at least one of the numbers
is zero.
When using the quadratic formula, we naturally get the same solution:
−b ± b2 − 4ac −b ± b2 − 4a0 −b ± b2 − 0 −b ± b2 −b ± b
=
=
=
=
2a
2a
2a
2a
2a
−2b
b
Thus, x = 0 and x =
= − is the solution set of the equation ax 2 + bx = 0 .
2a
a
x=
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3. When ∆ = b2 − 4ac < 0 , the equation has no real solution.
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Section
2.2
Section 2 .2:
section 2 .2: exercises
Exercises
Exercises
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1. Solve the equations:
a. x 2 − x − 90 = 0
b. 25x 2 + 90x + 81 = 0
c. (3x − 2)(x − 3) = 20
d. 3(5x + 3)(4x 2 −1) = 8(4x 2 −1)2
e. x 2 − 13x + 4 = 0
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3x 2 −14x +11 x + 9 x 2 + x +1
=
−
2
14
5
2
g. x + 2(1+ 8)x + 8 2 = 0 (Do not round off the coefficients.)
f.
2. One of the digits of a two-digit number is 3 less than the other digit. When the digits are
interchanged, we get a new number. The sum of the squares of the new number and the initial
number is 1877. Find the two numbers.
3. Make up a word problem whose solution can be found by solving the equation x ( x + 36) = 180 .
4. Two planes start flying from the same airport at the same time. One plane flies directly north,
and the other flies directly east. After 2 hours from the time they started, the distance between
the two planes is 2000 miles. Find the speeds of each plane if the speed of one is 75% of the
speed of the other.
5. A company has decided to double its production within two years. By how many percents should
the company increase its production each year if it wants the percentage increase in the first year
to be equal to the percentage increase in the second year?
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
secTion
3: Polynomial
equations
SecTIon 3:
Polynomial
equations
So far we have learned how to solve linear and quadratic equations. Now, you may be wondering, what
about equations of similar form but with higher degrees? For example, how do we solve equations like
2 x 3 + 4 x 2 − x + 9 = 0 and −3x 4 + 5 x 3 − 2 x + 12 = 0 ? Equations of this kind are called polynomial equations.
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and the other flies directly east. After 2 hours from the time they started, the distance between
the two planes is 2000 miles. Find the speeds of each plane if the speed of one is 75% of the
speed of the other.
5. A company has decided to double its production within two years. By how many percents should
the company increase its production each year if it wants the percentage increase in the first year
to be equal to the percentage increase in the second year?
Section
kkk3
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
Equations
secTion
3:Polynomial
Polynomial
equations
SecTIon 3:
Polynomial
equations
about equations of similar form but with higher degrees? For example, how do we solve equations like
A polynomial
is an expression
of the
form an x n + an −1 x n −1 +  + a3 x 3 + a2 x 2 + a1 x1 + a0 , where n is a non3
2
4
3
2 x + 4 x − x + 9 = 0 and −3x + 5 x − 2 x + 12 = 0 ? Equations of this kind are called polynomial equations.
with a 0 . a , a , a , , a are called the coefficients
negative integer, and a0 , a1 , a2 , , an are real numbers
A polynomial is an expression of the form an x n + an −1 x n −1 +n + a3 x03 +1 a2 2x 2 + a1nx1 + a0 , where n is a nonDuplication
is Prohibited
Outside the Terms
of Yourthe
License
Agreement
n , the highest
power
in the polynomial,
is called
degree
of the polynomial. The
of the polynomial.Unauthorized
a
0
a
,
a
,
a
,

,
a
are
real
numbers
with
.
a
,
a
,
a
,

,
an are called the coefficients
negative
integer,
and
®
0
1
2
n
0
1
2
n
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expressions, a0 , a1 x1 , a2 x 2 , a3 x 3 ,  , an −1 x n −1 , an x n , are called the terms of the polynomial. Expressions of
of the polynomial. n , the highest power in the polynomial, is called the degree of the polynomial. The
integer, are called monomials.
the form x k , where1 k is 2a non-negative
expressions, a0 , a1 x , a2 x , a3 x 3 ,  , an −1 x n −1 , an x n , are called the terms of the polynomial. Expressions of
the form x k , where k is a non-negative integer, are called monomials.
ExamplE 3a:
ExamplE 3a:
2
The coefficients of the polynomial 4 x 5 − 3x 4 + x 3 + 3x 2 − 6 x + 1 are:
5
2
2
5
4
3
a5 = coefficients
4 , a4 = −3 ,ofa3the
, a2 = 3 , a41x= −
−63,x a+
this polynomial is n = 5 .
= polynomial
The
0 = 1xand
+ the
3x 2degree
− 6 x + of
1 are:
5
5
2
2 the
5 polynomial is n = 5 .
a5 = terms
4 , a4 of
= −this
3 , apolynomial
, a2 = are:
−6 ,, a03x
= 21,and
of 4x
this
3 , 1a1, =−6x
3 =
.
The
3x 4 , and
x 3 , −degree
5
5
2
The terms of this polynomial are: 1 , −6x , 3x 2 , x 3 , −3x 4 , and 4x 5.
5
ExamplE 3b:
ExamplE 3b:
3
The coefficients of the polynomial 17 x 4 + 0 ⋅ x 3 − 12 x 2 + 0 ⋅ x − are:
4
3 3 3
3
4
2
a4 = coefficients
17 , a3 = 0 , of
a2 the
= −12
, a1 = 0 , a17
The
polynomial
x —+ 04 ⋅ x − 12 x + 0 ⋅ x − are:
0 =
4
4
3
This
polynomial
is
of
degree
.
4
a4 = 17 , a3 = 0 , a2 = −12 , a1 = 0 , a0 =
4
3
This terms
polynomial
of degree 4are:
. − , 0 ⋅ x , −12x 2 , 0 ⋅ x 3, and 17x 4.
The
of thisispolynomial
4
3
The terms of this polynomial are: − , 0 ⋅ x , −12x 2 , 0 ⋅ x 3, and 17x 4.
4
17
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When a coefficient is zero, there is no need to write the corresponding term. For example, instead of
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So far we have learned how to solve linear and quadratic equations. Now, you may be wondering, what
a4 = 17 , a3 = 0 , a2 = −12 , a1 = 0 , a0 =
3
4
This polynomial is of degree 4 .
3
The terms of this polynomial are: − , 0 ⋅ x , −12x 2 , 0 ⋅ x 3, and 17x 4.
4
When a coefficient is zero, there is no need to write the corresponding term. For example, instead of
writing −4x 5 + 0x 4 + 5x 3 + 0x 2 − 4x + 7 , we just write −4x 5 + 5x 3 − 4x + 7 . When all the coefficients
of a polynomial are zeros, the polynomial is called the zero polynomial. For convenience, we denote
polynomials by Q(x), R(x), etc.
3 .11When
Areare
TWo PoLynomIALS
eQuAL?
3 .
When
two Polynomials
Polynomials
equal?
3.1
When
Are
Two
Equal?
ExamplE 3.1a:
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The polynomials P ( x ) = x 3 + x 2 + x and Q ( x ) = 0 ⋅ x 4 + x 3 + x 2 + x + 0 are equal. The polynomials R ( x ) = x 3 + x 2 + x and S ( x ) = x 4 + x 3 + x 2 + x , on the other hand, are not equal because the
S(x).
term x 4 is in R(x)
S(x) but not in R(x).
3 .2 hoW
Do We
SuBTrAcT
PoLynomIALS?
3 .2
how
doADD
WeAnD
add
and
Polynomials?
3.2
How
Do
We
Add
and subtract
Subtract
Polynomials?
Naturally, the sum and difference of two polynomials are defined. We simply add or subtract the
coefficients of the like terms.
ExamplE 3.2a:
Q(x) = −4x 5 + 3x 3 + 2x 2 − 6x + 9
P(x) = 2x 5 + 7x 3 − 2x 2 + 6x +19
Q(x) + P(x) = (−4 + 2)x 5 + (3+ 7)x 3 + (2 + (−2))x 2 + (−6 + 6)x + (9 +19) =
= −2x 5 + 10x 3 + 0x 2 + 0x + 28 = −2x 5 + 10x 3 + 28
Q(x) − P(x) = (−4 − 2)x 5 + (3 − 7)x 3 + (2 − (−2))x 2 + (−6 − 6)x + (9 −19) =
= −6x 5 − 4x 3 + 4x 2 − 12x − 10 = −6x 5 − 4x 3 + 4x 2 − 12x − 10
18
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Two polynomials Q(x) and R(x) are equal if they have the same terms.
Q(x) + P(x) = (−4 + 2)x + (3+ 7)x + (2 + (−2))x + (−6 + 6)x + (9 +19) =
5
2
Q(x) + P(x)
= (−4
+ (3+ 7)x 3 + (25 + (−2))x
+ (−6 + 6)x + (9 +19) =
5
3 + 2)x
2
3
= −2x + 10x + 0x + 0x + 28 = −2x + 10x + 28
= −2x 5 + 10x 3 + 0x 2 + 0x + 28 = −2x 5 + 10x 3 + 28
Q(x) − P(x) = (−4 − 2)x 5 + (3 − 7)x 3 + (2 − (−2))x 2 + (−6 − 6)x + (9 −19) =
5
2
Q(x) − P(x)
= (−4
− 2)x
+ (3 − 7)x 3 + (25− (−2))x
+2(−6 − 6)x + (9 −19) =
5
3
2
3
= −6x − 4x + 4x − 12x − 10 = −6x − 4x + 4x − 12x − 10
= −6x 5 − 4x 3 + 4x 2 − 12x − 10 = −6x 5 − 4x 3 + 4x 2 − 12x − 10
ExamplE 3.2b:
ExamplE 3.2b:
ExamplE 3.2c:
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4
17
17
17
17
P ( x) = 32 x 5 − 2 x3 − 18 x + 19
Q(x) − P(x) = (0 − 32)x 5 + (6 − 0)x 4 + (23 − (−2))x 3 + ( 2 − 0)x 2 + (−3 − (−18))x + (−8 −19) =
= −32x 5 + 6x 4 + 25x 3 + 2x 2 + 15x − 27 .
3.3
How
We
Polynomials?
3 .3How
how
do
We Multiply
multiply
Polynomials?
3 .3
DoDo
We Multiply
Polynomials?
To multiply two polynomials, we multiply each term of one polynomial by each term of the second
polynomial and add like terms.
ExamplE 3.3a:
Q(x) = 5x 3 − 3x 2 + 2x
P(x) = −4x 4 −12x 3 +15
(
)(
)
= (5x ) ⋅ (−4x − 12x + 15) + (−3x ) ⋅ (−4x − 12x + 15) + ( 2x ) ⋅ (−4x − 12x + 15) =
= (5x ) ⋅ (−4x ) + (5x ) ⋅ (−12x ) + (5x ) ⋅ (15) + (−3x ) ⋅ (−4x ) + (−3x ) ⋅ (−12x ) + (−3x ) ⋅ (15)+
+ ( 2x ) ⋅ (−4x ) + ( 2x ) ⋅ (−12x ) + ( 2x ) ⋅ (15) =
Q(x) ⋅ P(x) = 5x 3 − 3x 2 + 2x ⋅ −4x 4 − 12x 3 + 15 =
3
4
3
3
4
3
2
4
3
4
3
3
4
2
4
3
2
3
6
3
5
4
= −20x 7 − 60xUSAD
+ 75xMathematics
+ 12x 6 + 36x 5Resource
− 45x 2 − 8x
− 24x
+ 30x =
Guide
• 2016–2017
= −20x − 48x + 28x − 24x + 75x − 45x + 30x
7
2
3
6
5
4
3
2
19
Rancho Cucamonga High School - Ontario, CA
Q(x) = 65x 7 +13x 3 + 2x 2 − 8
2
Q(x) = 65x 75 +13x4 3 + 2x
−8
3
P(x) = 32x + 4x − 2x + 6x +19
P(x) = 32x 5 + 4x 4 − 2x 3 + 6x +19
Q(x) + P(x) = (65 + 0)x 7 + (0 + 32)x 5 + (0 + 4)x 4 + (13 − 2)x 3 + ( 2 + 0)x 2 + (0 + 6)x + (−8 +19) =
Q(x) + P(x)
= (65 + 0)x 7 + (0 + 32)x 5 + (0 + 4)x 4 + (13 − 2)x 3 + ( 2 + 0)x 2 + (0 + 6)x + (−8 +19) =
= 65x 7 + 32x 5 + 4x 4 +11x 3 + 2x 2 + 6x +11
= 65x 7 + 32x 5 + 4x 4 +11x 3 + 2x 2 + 6x +11
(
)(
)
= (5x ) ⋅ (−4x − 12x + 15) + (−3x ) ⋅ (−4x − 12x + 15) + ( 2x ) ⋅ (−4x − 12x + 15) =
= (5x ) ⋅ (−4x ) + (5x ) ⋅ (−12x ) + (5x ) ⋅ (15) + (−3x ) ⋅ (−4x ) + (−3x ) ⋅ (−12x ) + (−3x ) ⋅ (15)+
+ ( 2x ) ⋅ (−4x ) + ( 2x ) ⋅ (−12x ) + ( 2x ) ⋅ (15) =
Q(x) ⋅ P(x) = 5x 3 − 3x 2 + 2x ⋅ −4x 4 − 12x 3 + 15 =
3
4
3
3
4
3
2
3
4
4
3
3
4
2
4
2
3
3
2
3
= −20x 7 − 60x 6 + 75x 3 + 12x 6 + 36x 5 − 45x 2 − 8x 5 − 24x 4 + 30x =
= −20x 7 − 48x 6 + 28x 5 − 24x 4 + 75x 3 − 45x 2 + 30x
(
)(
)
Q(x) ⋅ P(x) = 5x 3 − 3x 2 + 2x ⋅ −4x 4 − 12x 3 + 15 = −20x 7 − 48x 6 + 28x 5 − 24x 4 + 75x 3 − 45x 2 + 30x
3 .4
how
do
We Divide
divide
Polynomials?
3 .
4 How
DoDo
We Divide
Polynomials?
3.4
How
We
Polynomials?
Division of polynomials is similar to the long division algorithm you learned in elementary school. Before
we show how to divide one polynomial by another, it is worth discussing a few properties of polynomials
Any integer can be factored into the product of other, smaller integers, unless the integer is a prime
number. For example, 12 = 2 2 ⋅ 3 , 60 = 2 2 ⋅ 3 ⋅ 5 , 525 = 3 ⋅ 52 ⋅ 7 , etc. 3 divides 12 because 12 ÷ 3 = 4 , or
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12
18
18
= 3 ⋅ 4 . On the other hand,
5 does
not divide® 12,
because when
we divide
12 by 5, Guide
we get a remainder
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of 2. We can say the same thing using different words: 5 does not divide 12 because there is no integer
whose product with 5 equals 12. In general, we say an integer a ≠ 0 divides an integer b , if there exists
an integer c such that b = ac .
A similar situation exists with polynomials. Polynomials can also be factored into the product of
other polynomials with smaller degrees. For example, 3x 2 + 5 x − 2 = (3x − 1)( x + 2) . Therefore, we say
that the polynomial 3x − 1 divides the polynomial 3x 2 + 5 x − 2 . Also, the polynomial x + 2 divides
the polynomial 3x 2 + 5 x − 2 . We note that 2(x + 2) also divides the polynomial 3x 2 + 5 x − 2 because
3
1
x
2x + 4 . The coefficients of the polynomials can be real numbers and even
2
2
complex numbers in general. But, for computational simplicity, most of our polynomials will have integer
3x 2 + 5x 2 =
(
)
coefficients.
Similarly, each of the polynomials
x − 2 and x + 2 divides the polynomial x 2 − 4 because
x 2 − 4 = ( x − 2)( x + 2) . On the other hand, the polynomial x does not divide x − 2 because there is no
polynomial whose product with x results in x − 2 .
definition
20
A polynomial K ( x) divides a polynomial P ( x) if there exists
USAD Mathematics Resource Guide • 2016–2017
a polynomial Q( x) such that P ( x) = K ( x)Q( x) .
Rancho Cucamonga High School - Ontario, CA
that are analogous to those of whole numbers.
polynomial whose product with x results in x − 2 .
definition
A polynomial K ( x) divides a polynomial P ( x) if there exists
a polynomial Q( x) such that P ( x) = K ( x)Q( x) .
We will now learn how to carry out the division of polynomials.
ExamplE 3.4a: Let P ( x) = 2 x 4 + 5 x 3 + 3 x 2 − 2 x − 8 and K ( x) = x − 1 . Divide P ( x) by K ( x) .
solution: As we said before, the division of polynomials is similar to the division of multithe right.
2x 3 + 7x 2 +10x + 8
3
1. Divide 2x by x . You get 2x .
x −1 2x 4 + 5xUnauthorized
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3
2x − 2x
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2. Multiply
divisor x − 1Guide
and subtract the result from the 19
2x 3 by theResouRce
3
2
polynomial. You get 7x + 3x − 2x − 8.
7x 3 + 3x 2 − 2x − 8
3. Divide
7x 3 − 7x 2
4. Multiply 7x by the divisor x − 1 and subtract the result from the
2
polynomial. You get 10x − 2x − 8.
7x 3 by x . You get 7x 2.
2
10x 2 − 2x − 8
10x 2 −10x
8x − 8
8x − 8
0
5. Divide
10x 2 by x . You get 10x .
6. Multiply 10x by the divisor
polynomial. You get 8x − 8.
7. Divide
8x
by
x − 1 and subtract the result from the
x . You get 8 .
8. Multiply 8 by the divisor
polynomial. You get 0 .
x −1
and subtract the result from the
Thus, P(x) divided by K ( x) is the polynomial Q(x) = 2x 3 + 7x 2 +10x + 8.
We write:
2x 4 + 5x 3 + 3x 2 − 2x − 8
= 2x 3 + 7x 2 +10x + 8
x −1
or
(
)(
)
2x 4 + 5x 3 + 3x 2 − 2x − 8 = 2x 3 + 7x 2 + 10x + 8 x − 1
ExamplE 3.4b: Let P ( x) = 2 x 3 + 7 x 2 + 10 x + 8 and K ( x) = x + 2 . Divide P ( x) by K ( x) .
solution:
x+2
2
Resource Guide • 2016–2017
2xUSAD
+ 3xMathematics
+4
2x 3 + 7x 2 +10x + 8
1. Divide 2x by x . You get 2x .
3
2
21
Rancho Cucamonga High School - Ontario, CA
digit numbers. Each step in the following division process is accompanied by an explanation on
ExamplE 3.4b: Let P ( x) = 2 x 3 + 7 x 2 + 10 x + 8 and K ( x) = x + 2 . Divide P ( x) by K ( x) .
solution:
1. Divide
−(2x 3 + 4x 2 )
2x 3 by x . You get 2x 2 .
2. Multiply 2x by the divisor x + 2 and subtract the result from the
2
polynomial. You get 3x + 10x + 8.
2
3x 2 +10x + 8
3. Divide
− (3x 2 + 6x)
3x 2 by x . You get 3x .
4. Multiply 3x by the divisor x + 2 and subtract the result from the
polynomial. You get 4x + 8.
4x + 8
− (4x + 8)
5. Divide
4x
by
x . You get 4.
6. Multiply 4 by the divisor
polynomial. You get 0.
x+2
and subtract the result from the
0
Therefore, we get: 2x 3 + 7x 2 +10x + 8 = (x + 2)(2x 2 + 3x + 4) .
In both of these examples,
theDuplication
remainder
dividing
is zero.
This
is not always the case, as
P ( xthe
) by
K (ofx)Your
Unauthorized
is of
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Outside
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Agreement
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is shown in the following example.
–2014
 2013
2010–2011
ExamplE 3.4c: Let P(x) = −4x 4 + 9x 3 −12x + 5 and K(x) = x 2 + 2 . Divide P ( x) by K ( x) .
solution: Note, the coefficient of x 2 in polynomial K ( x) is zero.
−4x 2 + 9x + 8
x 2 + 2 − 4x 4 + 9x 3 + 0 ⋅ x 2 −12x + 5
−(−4x
9x + 8x
3
− (9x
−4x 4 by x 2 . You get 4x 2.
2
2. Multiply 4x by the divisor x + 2 and subtract the
3
2
result from the polynomial. You get 9x + 8x .
2
3. Divide
+18x)
3
8x 2 − 30x
− (8x 2
2
2
− 8x )
4
1. Divide
+16)
9x 3 by x 2 . You get 9x .
4. Multiply 9x by the divisor x + 2 and subtract the
2
result from the polynomial. You get 8x − 30x .
2
5. Divide
8x 2 by x 2 . You get 8.
6. Multiply 8 by the divisor x + 2 and subtract the
result from the polynomial. You get − 30x − 11.
2
− 30x −11
Since −30x cannot be divided by x 2 , we have the remainder −30x −11 . Therefore, we get:
22 −4x 4 + 9x 3 + 0 ⋅ x 2 −12x + 5 USAD2 Mathematics
Guide • 2016–2017
−30xResource
−11
x2 + 2
= −4x + 9x + 8 +
x2 + 2
Rancho Cucamonga High School - Ontario, CA
x+2
2x 2 + 3x + 4
2x 3 + 7x 2 +10x + 8
8x − 30x
− (8x 2
6. Multiply 8 by the divisor x + 2 and subtract the
result from the polynomial. You get − 30x − 11.
2
+16)
− 30x −11
Since −30x cannot be divided by x 2 , we have the remainder −30x −11 . Therefore, we get:
−30x −11
−4x 4 + 9x 3 + 0 ⋅ x 2 −12x + 5
= −4x 2 + 9x + 8 +
2
x2 + 2
x +2
or
3
−4x 4 + 9x 2 −12x + 5 = (x 2 + 2)(−4x 2 + 9x + 8) + (−30x −11)
The division of polynomials just shown is based on the following theorem, which we accept without
division algorithm
Let P ( x) and K ( x) be two polynomials, where K ( x) is not the zero polynomial. Then, there
exist two unique polynomials Q( x) and R ( x ) such that P(x) = K (x)Q(x) + R(x) ;
the polynomial Q( x) is the quotient, and the remainder R ( x ) is either the zero
polynomial or of a degree that is less than the degree of K ( x) .
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ExamplE 3.4d:
®
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21
21
In the last example, we saw that −4 x 4 + 9 x 23 − 12 x + 5 = ( x 2 + 2)(−4 x 2 + 9 x + 8) + (−30 x − 11). Here
3
P( x) = −4 x 4 + 9 x3 − 12 x + 5 , K ( x) = x 2 + 2 , Q( x) = −4 x 2 + 9 x + 8 , and R( x) = −30 x − 11. The
degree of the remainder R ( x) is less than the degree of the divisor K ( x) .
ExamplE 3.4e:
We have seen in previous examples that 2 x 3 + 7 x 2 + 10 x + 8 = ( x + 2)(2 x 2 + 3 x + 4)
Here Q( x) = 2 x 2 + 3 x + 4 , and the remainder R ( x ) is the zero-polynomial.
The following theorem relates evaluating a polynomial to the division of polynomials.
USAD Mathematics Resource Guide • 2016–2017
23
Rancho Cucamonga High School - Ontario, CA
proof.
The following theorem relates evaluating a polynomial to the division of polynomials.
remainder theorem
If R is the remainder when a polynomial P ( x) is divided by x − a , then R = P (a ) .
Proof
We use the Division Algorithm. Let Q( x) be the quotient and R be the
remainder when P ( x) is divided by x − a . Thus, P(x) = Q(x)(x − a) + R .
ExamplE 3.4f:
When dividing the polynomial P(x) = 3x 3 − 4x 2 + 5 by x + 2 using the Division Algorithm, we get
the remainder −35 . Using the Remainder Theorem, we obtain the same result:
P(−2) = 3⋅ (−2)3 − 4 ⋅ (−2)2 + 5 = −35
3.5
H
ow
Does
the
of
3 .5
how
does
the Division
division
of Polynomials
Polynomials Help
3 .5 How
Does
the Division
of Polynomials
help
with
Polynomial
equations?
Help Us
Usus
with
Polynomial
Equations?
with
Polynomial
Equations?
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Division of polynomials can be very helpful in solving polynomial equations. We have no general method
of solving polynomial equations of any degree. However, sometimes we can solve a polynomial equation
of a high degree by factoring the polynomial into the product of lower-degree polynomials.
ExamplE 3.5a: Solve the equation x 3 + 3x 2 − 18x − 40 = 0 .
solution: We first factor the polynomial x 3 + 3x 2 − 18x − 40 (we will learn about factoring
polynomials later): x 3 + 3x 2 − 18 x − 40 = ( x + 2)( x 2 + x − 20) . With this factorization, we reduce
the task of solving an unfamiliar equation x 3 + 3x 2 − 18 x − 40 = 0 into solving two familiar
equations: x + 2 = 0 or x 2 + x − 20 = 0 .
The solution of x + 2 = 0 is x = −2 .
24 The solutions of
4.
x = −5 , x = Resource
x 2 + x − 20 USAD
= 0 are
Mathematics
Guide • 2016–2017
Hence, the given equation has three solutions: x = −2 , x = −5, and x = 4 .
Rancho Cucamonga High School - Ontario, CA
Setting x = a in the above equation, we get P(a) = Q(a)(a − a) + R . That is, R = P (a ) .
the task of solving an unfamiliar equation x 3 + 3x 2 − 18 x − 40 = 0 into solving two familiar
equations: x + 2 = 0 or x 2 + x − 20 = 0 .
The solution of x + 2 = 0 is x = −2 .
The solutions of x 2 + x − 20 = 0 are x = −5 , x = 4 .
Hence, the given equation has three solutions: x = −2 , x = −5, and x = 4 .
The question now is: How would one know how to factor the polynomial x 3 + 3x 2 − 18 x − 40 or any
other high-degree polynomial? The following two theorems, together with the Division Algorithm, can
be very helpful in factoring certain polynomials.
Let P(x) = an x n + an −1 x n −1 ++ a1 x + a0 be a polynomial with integer coefficients. If a reduced

m
is a solution to the equation P ( x) = 0 , then m divides a0 and k divides an .
fraction
k
factor theorem
The polynomial x − a is a factor of P ( x) if and only if P (a ) = 0 . The Factor Theorem
follows from the Remainder Theorem, and a direct proof of it by using the Division
Algorithm is presented in Section 3.7. But, before we present the proofs of these
theorems, we will first consider how to apply them to solve polynomial equations.
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ResouRce
x 3 − 2 xGuide
−1 = 0.
ExamplEMATheMATics
3.5b: Solve
23
solution: Let P ( x) = x 3 − 2 x − 1 . In this polynomial, a0 = −1 and an = 1 . By the Rational
m
is a solution to P ( x) = 0 , then m divides −1 and k
Root Theorem, if a reduced fraction
k
divides 1 .
The integers that divide −1 are m = ± 1 .
And, the integers that divide 1 are k = ±1 .
m
, with these values of m and k , is a candidate to solve P ( x) = 0 .
k
m m +1 −1 + 1 − 1
The following are all the possible fractions
:
=
,
,
,
k k +1 −1 −1 +1
Therefore, any fraction
After removing repetitions, we get two candidates for rational solutions to P ( x) = 0 . These are:
+1
+1
= 1 and
= −1
+1
−1
USAD Mathematics Resource Guide • 2016–2017
Now we only need to find out which of these two numbers is a solution to P ( x) = 0 .
25
Rancho Cucamonga High School - Ontario, CA
the rational root theorem
The following are all the possible fractions
m m +1 −1 + 1 − 1
:
=
,
,
,
k k +1 −1 −1 +1
After removing repetitions, we get two candidates for rational solutions to P ( x) = 0 . These are:
+1
+1
= 1 and
= −1
+1
−1
Now we only need to find out which of these two numbers is a solution to P ( x) = 0 .
We substitute x = 1 in P ( x) = 0 , and get 1⋅13 − 2 ⋅1− 1 = −2 ≠ 0 Thus, x = 1 is not a solution.
( )
( )
We substitute x = −1 in the same equation, and get 1⋅ −1 − 2 ⋅ −1 − 1 = 0 Thus, x = −1 is a
3
solution.
By the Factor Theorem, x − (−1) = x + 1 divides x 3 − 2 x − 1 . Let’s do the division:
x 2 − x −1
x +1 x 3 + 0 ⋅ x 2 − 2x − 1
Rancho Cucamonga High School - Ontario, CA
−(x 3 + x 2 )
− x 2 − 2x −1
− (−x 2 − x)
− x −1
− (−x −1)
0
(
)(
)
This shows that x 3 − 2x − 1 = x 2 − x − 1 x + 1
(
)(
)
Solving x 2 − x − 1 x + 1 = 0 , we get
24
x +1 = 0
x2 − x − 1 = 0
x = −1
x=
1± 1+ 4 1± 5
=
2
2
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1+ 5
1− 5
®
x = −1 ResouRce
Thus, our polynomial
equationdecAthlon
has three solutions:
, MATheMATics
x=
,ResouRce
x=
.
mAthemAtics
Guide
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–2014
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2010–2011
Thus, we can see how the Division Algorithm, the Rational Root Theorem, and the Factor Theorem help
us to solve a polynomial equation of degree three.
ExamplE 3.5c: Solve 2 x 3 − 5 x + 3 = 0 .
solution: Let P ( x) = 2 x 3 − 5 x + 3 . In this polynomial, a0 = 3 and a3 = 2 . By the Rational
m
is a solution to P ( x) = 0 , then m divides 3 and k
Root Theorem, if a reduced fraction
k
26 divides 2 .
USAD Mathematics Resource Guide • 2016–2017
The integers that divide 3 are m = ±3, ± 1 .
us
us to
to solve
solve a
a polynomial
polynomial equation
equation of
of degree
degree three.
three.
3
ExamplE
2 xx 3 −
−5
5 xx +
+3
3=
=0
0 ..
ExamplE 3.5c:
3.5c: Solve
Solve 2
3
solution:
solution: Let
Let P
P(( xx)) =
= 22 xx3 −
− 55 xx +
+ 33 .. In
In this
this polynomial,
polynomial,
m
m
is
Root
is aa solution
solution to
to
Root Theorem,
Theorem, if
if aa reduced
reduced fraction
fraction
kk
divides
divides 22 ..
aa00 =
= 33 and
and a
a33 =
=2
2 .. By
By the
the Rational
Rational
P
then m
m divides
divides 3 and
and k
k
P (( xx)) =
=0
0 ,, then
The
are m
The integers
integers that
that divide
divide 3 are
m=
=±
±3,
3, ±
±1
1 ..
And
=±
±2,
2, ±
±1
1 ..
are kk =
And the
the integers
integers that
that divide
divide 22 are
m
m:
The
:
The following
following are
are the
the possible
possible fractions
fractions
kk
m
m
3,
m=+
+3
+3 +3
+3 , −3
−3 , −3
−3
m = +3
=
,
= ,,
,
,
kk +
kk +2
+1
1
+2 −2
−2 +2
+2 −2
−2
m
m
m=+
1, +
1, −
1, −
1
m=+
1,
+1
+1
−1
−1
+1
=
,
,
,
=
,
kk +
kk +
+2
2 −
−2
2 +
+2
2 −
−2
2
+1
1
After
After removing
removing all
all the
the
33
11
P
± ,, ±
± ,,
P (( xx)) =
=0
0 :: ±
±3
3 ,, ±
22
22
3
+3
+
,,
−1
1
−
+
1,
+1
,
−
−1
1
3
−3
−
,,
+1
1
+
−
1,
−1
,
+
+1
1
Rancho Cucamonga High School - Ontario, CA
m
m with these values of m and k is a candidate to solve the equation
Therefore,
with these values of m and is a candidate to solve the equation
Therefore, any
any fraction
fraction
kk
P
P (( xx)) =
=0
0 ..
3
−3
−
−1
1
−
1
−
−1
−
−1
1
repetitions,
repetitions, we
we get
get the
the following
following eight
eight candidates
candidates for
for solutions
solutions to
to
±
±1
1
Now
Now we
we only
only need
need to
to examine
examine which
which of
of these
these 88 numbers
numbers are
are solutions
solutions to
to P(x)
We start
start with
with
P(x) =
=0
0 .. We
the
and ±3 ..
the easy
easy ones:
ones: ±
± 11 and
3
22 ⋅1
⋅13 −
− 55 ⋅1+
⋅1+ 3⋅1
3⋅1 =
= 00
3
22 ⋅⋅ (−1)
(−1)3 −
− 55 ⋅⋅ (−1)
(−1) +
+ 33 =
= 66
x = 1 is
is aa solution
solution
x = −1 is
is not
not aa solution
solution
3
2
2 ⋅⋅ 3
33 −
−5
5 ⋅⋅ 3+
3+ 3
3=
= 42
42
x
x=
is not
not aa solution
solution
= 33 is
3
22 ⋅⋅ (−3)
(−3)3 −
− 55 ⋅⋅ (−3)
(−3) +
+ 33 =
= −36
−36
x
x =
is not
not aa solution
solution
=33 is
You can check on your own that none of the other candidates is a solution to P(x) = 0 .
1 Your
mustLicense
divide
, x −of
Now we can apply
the Factor
Theorem.
Since,Outside
P(1) =
P(x) . After dividing
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Duplication
is
the
Agreement
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−1 , we AcAdemic
get
2x 3 − 5x
+ 3 = (x −1)(2x
+ 2x − 3) .ResouRce
Therefore,Guide
the problem
Q(x)
P(x)
of solving
25
25
the equation 2x 3 − 5x + 3 = 0 is reduced to the problem of solving the equations x − 1 = 0 and
2x 2 + 2x − 3 = 0 .
Solving, we get: x = 1 , x =
−1− 7
−1+ 7
, x=
2
2
Hence, these three values are solutions of the equation 2x 3 − 5x + 3 = 0 .
USAD Mathematics Resource Guide • 2016–2017
ExamplE 3.5d: Solve 2x 4 + 5x 3 + 3x 2 − 2x − 8 = 0 .
27
ExamplE 3.5d: Solve 2x 44 + 5x 33 + 3x 22 − 2x − 8 = 0 .
ExamplE 3.5d: Solve 2x + 5x + 3x − 2x − 8 = 0 .
solution: Let P(x) = 2x 44 + 5x 33 + 3x 22 − 2x − 8 . In this polynomial, a0 = −8 and an = 2 . By the
solution: Let P(x) = 2x + 5x + 3x − 2x − 8 . In this polynomial, a0 = −8 and an = 2 . By the
m
Rational Root Theorem, if a reduced fraction m is a solution to P ( x) = 0 , then m divides −8
Rational Root Theorem, if a reduced fraction k is a solution to P ( x) = 0 , then m divides −8
k
and k divides 2 .
and k divides 2 .
m
Therefore, any fraction m , with these values of m and k , is a solution candidate for P ( x) = 0 .
Therefore, any fraction k , with these values of m and k , is a solution candidate for P ( x) = 0 .
m
The following are all the kpossible fractions m :
The following are all the possible fractions k :
k
m +8 +8 −8 −8
m +8 +8 −8 −8
m = +8, +8 , −8, −8
m = +8, +8, −8, −8
k = +2 , −2 , +2 , −2
k = +1 , −1 , +1 , −1
k +2 −2 +2 −2
k +1 −1 +1 −1
m +4 +4 −4 −4
m +4 +4 −4 −4
m = +4 , +4 , −4 , −4
m = +4 , +4 , −4 , −4
k = +2 , −2 , +2 , −2
k = +1 , −1 , +1 , −1
k +2 −2 +2 −2
k +1 −1 +1 −1
m +1 +1 −1 −1
m +1 +1 −1 −1
m = +1, +1, −1, −1
m = +1, +1, −1, −1
k = +2 , −2 , +2 , −2
k = +1 , −1 , +1 , −1
k +2 −2 +2 −2
k +1 −1 +1 −1
After removing all the repetitions, we get the following candidates as possible solutions for
After removing all the repetitions, we get the following candidates as possible solutions for
1
P ( x) = 0 : ±8, ± 4, ± 2, ± 1, ± 1.
P ( x) = 0 : ±8, ± 4, ± 2, ± 1, ± 2 .
2
Now we only need to examine which of these 10 numbers are solutions to P ( x) = 0 .
Now we only need to examine which of these 10 numbers are solutions to P ( x) = 0 .
We start with the easy ones: ± 1 and ±2.
We start with the easy ones: ± 1 and ±2.
2 ⋅144 + 5 ⋅133 + 3⋅122 − 2 ⋅1− 8 = 0
2 ⋅1 + 5 ⋅1 + 3⋅1 − 2 ⋅1− 8 = 0
26
26
26
26
xx ==11 isis aa solution
solution
2 ⋅ (−1)4 + 5 ⋅ (−1)3 + 3⋅ (−1)2 − 2 ⋅ (−1) − 8 = −6
x = −1 is not a solution
2 ⋅ 24 + 5 ⋅ 23 + 3⋅ 22 − 2 ⋅ 2 − 8 = 72
x = 2 is not a solution
2 ⋅ (−2)4
x = −2
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Outside the Terms of Your License Agreement
3
2
+ 5 ⋅ (−2)
+ 3⋅
(−2)Duplication
− 2 decAthlon
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is
a solution
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2010–2011
You can check on your own that none of the other candidates is a solution to P(x) = 0 . After
dividing 2x 4 + 5x 3 + 3x 2 − 2x − 8 by x − 1 , we get 2x 3 + 7x 2 + 10x + 8 .
(
)(
)
Therefore, we have 2x 4 + 5x 3 + 3x 2 − 2x − 8 = 2x 3 + 7x 2 + 10x + 8 x − 1 .
Now we solve two equations:
1.
K (x) = x − 1 = 0
2. Q( x) = 2 x 3 + 7 x 2 + 10 x + 8 = 0
The first will give us x = 1 . To solve the second equation, we need to factor Q( x) . We have:
P( x) = Q( x) K ( x) .
28
Recall x = −2 solves the equation P ( x) = 0 . Hence, P (2) = Q(−2) K (−2) = 0 .
USAD Mathematics Resource Guide • 2016–2017
Since K (−2) ≠ 0 , Q(−2) = 0 .
Rancho Cucamonga High School - Ontario, CA
The integers that divide a0 = −8 are m = ±8, ± 4, ± 2, ± 1 . And, the integers that divide an = 2
The integers that divide a0 = −8 are m = ±8, ± 4, ± 2, ± 1 . And, the integers that divide an = 2
are k = ±2, ± 1.
are k = ±2, ± 1.
2
( x) will
= 2 x3give
+ 7 xus
+x10
2. Q
The
first
= x1 +. 8To= 0solve the second equation, we need to factor Q( x) . We have:
P
( x) =first
Q( xwill
) K ( xgive
) . us x = 1 . To solve the second equation, we need to factor
The
Q( x) . We have:
P ( x) =xQ=( x−)2Ksolves
( x) . the equation P ( x) = 0 . Hence, P (2) = Q(−2) K (−2) = 0 .
Recall
the=equation
RecallK (x−=2)−≠2 0solves
P ( x) = 0 . Hence, P (2) = Q(−2) K (−2) = 0 .
, Q(−2)
Since
0.
Sincewe
K (use
−2) the
≠ 0 ,Factor
Q (−2)Theorem
= 0.
Now,
again and divide Q( x) by x + 2 to get:
2Now,
x 3 + 7we
x 2 use
+ 10the
x + 8Factor
= (2 x 2Theorem
+ 3 x + 4)(again
x + 2)and divide
Q( x) by x + 2 to get:
2 x 3 solution
+ 7 x 2 + 10
+ 8 = (2 x 2 2x
+ 33 x++7x4)(
2 x + 2)
The
ofx equation
+ 10x + 8 = 0 is reduced to the solution of two equations:
1.Thexsolution
+ 2 = 0 of equation 2x 3 + 7x 2 + 10x + 8 = 0 is reduced to the solution of two equations:
2.1. 2 x 2 + 3 x + 4 = 0
x+2=0
2. 2 x 2the
+ 3 xtwo
+ 4equations,
=0
Solving
we get: x = −2 in the first one and no real solutions in the second one.
Thus, the given equation has two real solutions: x = −2 and x = 1 .
3 .6
Proof
theRational
rational
root
theorem
3 .
6 Proof
of the
Root Theorem
3.6
Proof
ofofRational
the
Root
Theorem
3 .6
Proof
the rational
root theorem
3 .
6 Proof
of theof
Rational
Root Theorem
the rational root theorem
2013
–2014 
2010–2011
2013
–2014 
2010–2011
1
x) = an x nrational
+ an −1 x n −1 +  + a1 xroot
+ a0 , wheretheorem
Let P ( the
all the coefficients are
m
m
If allis the
a solution
to the
integers,
n
na
−1 reduced fraction.
P( x) and
= an xlet
+ kan −be
+  + a1 x1 + a0 , where
Let
coefficients
are
1x
k
m k divides an .
m divides
a0 Ifand
equation
P(m
x) =be0 a, then
reduced
fraction.
is a solution to the
integers,
and let
k
k
Proof
m divides a0 and k divides an .
equation P ( x) = 0 , then
m
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Let a reduced fraction
be
a solution to the equation P ( x) = 0 .
®
AcAdemic
mAthemAtics ResouRce Guide
MATheMATics
ResouRcedecAthlon
Guide k
n
n −1
1
Unauthorized Duplication
the Terms
 m is Prohibited
 mOutside
 mof Your License Agreement

a Guide
0.
Then,
  + an −1®mAthemAtics
 ++ a1 ResouRce
 + a0 = Guide
AcAdemic
MATheMATics
ResouRcendecAthlon
k
k
k




27
27
Multiplying both sides of this equation by k n −1 we get
n
m
an
+ an −1 mn −1 + an − 2 mn − 2 k ++ a1 m1 k n − 2 + a0 k n −1 = 0 .
 k
Since a0 , a1 , a2 ,, an −1 , k and m are all integers,

n −1
n−2
1 n−2
n −1
an −1 m + an − 2 m k ++ a1 m k + a0 k is an integer.
mn 
+  Integer  = 0 .
We have: an
k
Hence, an
mn
must be an integer as well.
k
USAD Mathematics Resource Guide • 2016–2017
This implies that k divides an m n .
29
Rancho Cucamonga High School - Ontario, CA
= −one
2 and
x =1
Thus,
thethe
given
haswe
two
x =solutions:
−2 in the xfirst
Solving
twoequation
equations,
get:real
and no.real solutions in the second one.
We have: an
mn 
+  Integer  = 0 .
k
mn
Hence, an
must be an integer as well.
k
This implies that k divides an m n .
m
is a reduced fraction, k and m have no common factor
k
n
(other than 1), therefore k and m have no common factor too.
Since
In a similar manner, we can prove that m divides a0 . To do this, we multiply both sides
n
n −1
1
 m
 m
 m
kn
.
of the equation an   + an −1   ++ a1   + a0 = 0 by
m
k
k
k




kn
= 0.
We get an mn −1 + an −1 mn − 2 k ++ a1 k n −1 + a0
m

As before, an mn −1 + an −1 mn − 2 k ++ a1 k n −1 is an integer.

kn
kn
= 0 , a0
must be an integer as well.
Again, since  Integer  + a0
m
m
Since m and k have no common factor, m and kn have no common factor either.
Hence m divides a0 . This completes the proof.
3 .
Proof
of
the
factor
theorem
3 .
77Proof
of Unauthorized
the
Factor
Theorem
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is Prohibited Outside
the Terms of Your License Agreement
3.7
Proof
of
the
Factor
Theorem
28
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factor theorem
The polynomial x − a divides a polynomial P ( x) if and only if P (a ) = 0 .
Proof
We have two assertions in this theorem:
1 .
2 .
If x − a divides P ( x) , then P (a ) = 0 .
If P (a ) = 0 , then x − a divides P ( x) .
Proof of 1
Since x − a divides P ( x) , there exists a polynomial Q( x) such that P ( x) = Q( x)( x − a ) .
Substituting a for x , we get: P(a) = Q(a )(a − a) = Q(a) ⋅ 0 = 0 .
Proof of 2
30
By the Division Algorithm
for polynomials,
there exist
two
unique polynomials
USAD Mathematics
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Q ( x) and R ( x ) such that: P(x) = Q(x)(x − a) + R(x) , where either R ( x ) is the
Rancho Cucamonga High School - Ontario, CA
Hence, k divides an .
2 .
If P (Proof
− a divides
a ) = 0 , then xof
1 P( x) .
Proof
of Q1( x) such that P( x) = Q( x)( x − a) .
Since x − a divides P ( x) , there exists
a polynomial
that P
divides P ( x) ,athere
Since x − a Substituting
a polynomial
( x)−such
for xexists
, we get:
P(a) = Q(aQ)(a
a) = Q(a)
⋅ 0( x=) 0=.Q( x)( x − a ) .
, we get: P(a)of
Substituting a for x Proof
= Q(a2
)(a − a) = Q(a) ⋅ 0 = 0 .
Proof
of
By the Division Algorithm for
polynomials,
there2exist two unique polynomials
By
Division
polynomials,
there
exist
two unique
such that: for
, where
either polynomials
Q (the
x) and
R ( x )Algorithm
P(x)
= Q(x)(x − a)
+ R(x)
R ( x ) is the
that:degree
, where
either
is athe
Qzero
( x) and
R ( x ) such
P(x) of
= Q(x)(x
− a)
R(x) the
R ( x )x −
less+ than
degree
of the
.
polynomial
or the
R ( x ) is
zero polynomial or the degree of R ( x ) is less than the degree of the x − a .
We will show that if P (a ) = 0 , then R ( x ) is the zero polynomial, and so x − a divides P ( x) .
and so x −
We will
show that
if P (xa−) =a 0is, then
) is the
x) .r .
1, R ( R
be zero
eitherpolynomial,
the zero polynomial
oraa divides
number,P (say
Since
the degree
of the
x )( xmust
Since the degree of the x − a is 1, R ( x ) must be either the zero polynomial or a number, say r .
Let’s examine these two cases:
Let’s examine
these= two
cases:
Then P(x)
1. Let R ( x ) be the zero polynomial.
Q(x)(x
− a) , and so x − a divides P ( x) .
2. Let R ( x ) = r . Substituting x = a in P(x) = Q(x)(x − a) + r , we get P(a) = (a − a)Q(a) + r .
Since P (a ) = 0 , we get 0 = (a − a)Q(a) + r . This implies that r = 0
that r = 0
Since Pand
(a ) so
= 0x, we
0 = (aP−
r . This implies
− a get
divides
the proof.
( xa)Q(a)
) . This+completes
and so x − a divides P ( x) . This completes the proof.
Before finishing our discussion of polynomial equations, we should note that polynomials of degree three
Before
finishing
discussion
polynomial
equations,
we in
should
that polynomials of degree three
or higher
usuallyour
have
rational of
roots
when they
are studied
high note
school.
or higher usually have rational roots when they are studied in high school.
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
3 .1–3 .7:
sections 3 .1–3 .
7Section
: exercises
Sections
3.1–3.7
Exercises
Section
3 .
1–3 .7:
1k–3 .
k sections
k k k k k k k k k 3 .
kk
k k7
k:kexercises
kkkkkkkkkkkkkkkkkkkkkk
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
Exercises
Exercises
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
1. Solve the following equations
1. Solve the following equations
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a. x 3+ xMATheMATics
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2 = 0 decAthlon
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c. x 3 + 1991x + 1992 = 0
2. Find the quotient and the remainder of the division of P ( x) by K ( x) for
P(x) = 30x 4 − 31x 3 −180x 2 + 7x + 6 , K ( x) = x + 1 .
3. The remainder of the division of 2x 5 − 3x 3 +11x 2 − x + a by x + 2 is 3. Find the value of a .
4. Solve the equation ax 3 − 2 x 2 − 5 x + 6 = 0 , knowing 2 is one of its solutions.
5. Find all rational solutions to the following equations:
USAD
Mathematics Resource Guide • 2016–2017
2
a. 6x 4 + 7x 3 − 22x
− 28x − 8 = 0
31
Rancho Cucamonga High School - Ontario, CA
divides
ber the
zero polynomial.
andget
soP(a)
x−a
1.
= Q(x)(x
P( x+
) .r .
. Substituting
x = a inThen
2. Let R ( x ) =
P(x)P(x)
= Q(x)(x
− a)−+ a)
r ,,we
= (a
− a)Q(a)
3
2
4.
, knowing 2 is one of its solutions.
4. Solve
Solve the
the equation
equation ax
ax 3 −− 22xx 2 −− 55xx ++ 66 == 00 , knowing 2 is one of its solutions.
5.
5. Find
Find all
all rational
rational solutions
solutions to
to the
the following
following equations:
equations:
4
3
2
a.
a. 6x
6x 4 ++ 7x
7x 3 −− 22x
22x 2 −− 28x
28x −− 88 == 00
3
2
b.
b. xx 3 ++ 3x
3x 2 −−18x
18x −− 40
40 == 00
6
4
2
c.
c. xx 6 −− xx 4 −− xx 2 ++11 == 00
6
5
4
x = −3 and xx == −−44 solutions
6.
?
solutions to
to the
the equation
equation xx 6 ++ bx
6. For
For what
what values
values of
of bb and
and cc are
are x = −3 and
bx 5 +
+ cx
cx 4 =
= 00 ?
7.
7. Solve
Solve the
the following
following equations:
equations:
2
2
a.
a. (x
(x 2 ++ xx +1)(x
+1)(x 2 ++ xx ++ 2)
2) == 12
12
8.
does not divide ( x) = x − 2 (Hint:
Assume that there exists a polynomial
8. Prove
Prove that
that K
K ((xx)) == xx does not divide P
P ( x) = x − 2 (Hint: Assume that there exists a polynomial
such that
, and conclude that ( x) must
be a number. Proceed from
Q
Q((xx)) such that P(x)
P(x) =
= Q(x)K
Q(x)K(x)
(x) , and conclude that Q
Q ( x) must be a number. Proceed from
there
does not exist.)
there to
to conclude
conclude that
that such
such Q
Q((xx)) does not exist.)
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
3 .8
complex
numbers
3 .
Numbers
3.8
Complex
Numbers
3 .8
complex
numbers
3 .88 Complex
Complex
Numbers
Earlier
Earlier we
we concluded
concluded that
that there
there are
are no
no real
real solutions
solutions to
to aa quadratic
quadratic equation
equation when
when its
its discriminant
discriminant is
is less
less
than
than zero.
zero. What
What is
is meant
meant by
by this
this is
is that
that there
there are
are no
no real
real solutions,
solutions, numbers
numbers of
of the
the set
set of
of real
real numbers
numbers 
 ..
For
For example,
example, if
if the
the discriminant
discriminant is
is −−11,, there
there are
are no
no real
real solutions
solutions because
because there
there is
is no
no real
real number
number xx for
for
2
x
=
−
1
2
which
which x = −1.. That
That is,
is, the
the expression
expression −−11 does
does not
not represent
represent aa real
real number.
number.
Mathematicians
Mathematicians encountered
encountered certain
certain problems
problems in
in science
science and
and mathematics
mathematics that
that necessitated
necessitated extending
extending
the
the real
real numbers
numbers into
into aa new
new set
set of
of numbers
numbers called
called complex
complex numbers.
numbers. This
This was
was done
done by
by defining
defining the
the
expression
Duplication
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Outside
of Your
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a number that
solvesis the
equation
denoted
this number by the letter i.
−1 asUnauthorized
x 2 the
= −Terms
1 . They
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30 new number, in combination
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This
the real® numbers,
generates
the complex
numbers.
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Let’s consider the equation x 2 − 2 x + 5 = 0 .
Solving, we get: x =
2 ± 4 − 4 ⋅ 5 2 ± −16 2 ± 4 −1 2 ± 4i
=
=
=
= 1± 2i
2
2
2
2
The expressions 1 + 2i and 1 − 2i are complex numbers; they are solutions to the equation x 2 − 2 x + 5 = 0.
A complex number is an expression of the form a + bi , where a and b are real numbers. The set of all
complex numbers is denoted by  . Clearly, any real number r is a complex number since it can be written
as r + 0i . Hence,  is a subset of  . With this relation, we can extend Figure 1–1 from Section 1 of the
32
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guide, as shown in Figure 3–1:
Rancho Cucamonga High School - Ontario, CA
b.
(x +1)(x
+1)(x ++ 3)(x
3)(x ++ 5)(x
5)(x ++ 7)
7) +15
+15 == 00
b. (x
A complex number is an expression of the form a + bi , where a and b are real numbers. The set of all
complex numbers is denoted by  . Clearly, any real number r is a complex number since it can be written
as r + 0i . Hence,  is a subset of  . With this relation, we can extend Figure 1–1 from Section 1 of the
guide, as shown in Figure 3–1:
Figure 3–1
Every complex number z = a + ib corresponds to the point (a, b) in the coordinate plane, and, conversely,
every point (a, b) in the coordinate plane corresponds to the complex number z = a + ib . (See Figure
3–2 below.)
y
z
b
a
0
x
Figure 3–2
3 .8 .
hoW
comPlex
numBers?
3 .8 .1
hoW
Do
We
ADD
comPLex
numBerS?
3.8.11How
Dodo
WeWe
Addadd
Complex
Numbers?
Addition and subtraction of complex numbers are defined as follows:
Let z1 = a1 + b1i and z2 = a2 + b2i be two complex numbers; then z1 ± z2 = (a1 ± a2 ) + (b1 ± b2 )i .
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31
ExamplE 3.8a:
1. (5 + 3i) + (4 − 5i) = (5 + 4) + (3 − 5)i = 9 − 2i
2. (3+ 15i) − (4 + 5i) = (3 − 4) + (15 − 5)i = −1+ 10i
1 3  2
 1 1 2 3
1
1
3.  + i  −  − 2i  + i =  −  +  − −2 +  i = − + 3i
4
6
2 4  3
 4  2 3  4
( )
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Properties 1–8 from Section 1 hold for the complex numbers as well.
1. (5 + 3i) + (4 − 5i) = (5 + 4) + (3 − 5)i = 9 − 2i
2. (3+ 15i) − (4 + 5i) = (3 − 4) + (15 − 5)i = −1+ 10i
1 3  2
 1 1 2 3
1
1
3.  + i  −  − 2i  + i =  −  +  − −2 +  i = − + 3i
4
6
2 4  3
 4  2 3  4
( )
3 .8 .2
hoW
multiPly
comPlex
numBers?
3 .8 .2
hoW
Do
We
muLTIPLy
comPLex
numBerS?
3.8.2 How
Dodo
WeWe
Multiply
Complex
Numbers?
Multiplication of complex numbers is defined as follows:
Let z1 = a1 + b1i and z2 = a2 + b2i be two complex numbers. Then
z1 ⋅ z2 = a1 ⋅ (a2 + b2 i) + b1i ⋅ (a2 + b2 i) = a1 a2 + a1b2 i + a2 b1i + b1b2 i 2 =
( )
Notice that i = ( −1) =
2
−1 + (a1b2 + a2 b1 )i = a1 a2 − b1b2 + (a1b2 + a2 b1 )i .
2
2
−1 ⋅ −1 = −1 .
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= a1 a2 + b1b2
ExamplE 3.8b:
1. (4 +15i) ⋅ (17 − 2i) = 4 ⋅17 + 4 ⋅ (−2)i +15i ⋅17 +15 ⋅ (−2)i 2 = 68 − 8i + 255i − 30 ⋅ (−1) =
= 68 + 30 + (255 − 8)i = 98 + 247i
 1   1 
 1  1
1  1
1 4
1
2. 1+ i  ⋅  4 − i  = 1⋅ 4 +1⋅ − i  + i ⋅ 4 + −  i 2 = 4 − i + i − ⋅ (−1) =
3 5
5 3 15
 3   5 
 5  3
1  4 1
2
1
= 4 + +  −  i = 4 +1 i
15  3 5 
15 15
3. (3 − 5i)2 = 32 − 2 ⋅ 3⋅ 5i + (5i)2 = 9 − 30i + 25 ⋅ (−1) = 9 − 25 − 30i = −16 − 30i
4. (12 + 3i) ⋅ (12 − 3i) = 122 − (3i)2 = 144 + 9 = 153
In the last example, we multiplied two complex numbers that differ by the sign in front of 3i . One is
referred to as the conjugate of the other. In general, we say that a − bi is the conjugate of a + bi , and vice
versa: a + bi is the conjugate of a − bi .
The product of a complex
number
and isitsProhibited
conjugate
is a the
realTerms
number:
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(a + bi) ⋅ (a − bi) =
2
2 2
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The conjugate of a complex number z is denoted by z . Therefore, we have: z ⋅ z = a 2 + b 2 .
z
zz
a 2 + b2
=
=
=1
And, z ⋅ 2
a + b2 a 2 + b2 a 2 + b2
34 the last equality, we see that
USADzMathematics
Resource
• 2016–2017
From
is the reciprocal
of zGuide
because
their product is 1. As with real
2
2
a +b
−1
The conjugate of a complex number ] is denoted by ] . Therefore, we have: ] š ] " D E .
zz
a 2 + b2
z
=
=
=1
And, z ⋅ 2
a + b2 a 2 + b2 a 2 + b2
]
is the reciprocal of ] because their product is 1. As with real
D E
numbers, the reciprocal of ] is denoted by ] .
From the last equality, we see that
D EL
D
E
]
If] " D EL , then ] " " " L.
D E
D E D E
D E
As with real numbers, the absolute value of a complex number ] , also denoted by ] , is defined as the
distance of ] from the origin. Therefore: ] " D E .
Let ] " D EL and ] " D EL be two complex numbers, and ] | . And, let ] be a complex number
]
such that: ] " .
]
We rewrite the last expression as ] " ] š ] .
Since ] "
]
D E
"
D
D E
E
D E
L , we have:
]
D E L š D E L
D E L š D E L
] š]
] " ] š ] " D EL š " " " .
D E L š D E L
] š ]
D E
D E
or
]
] š]
]" " ] ] š ]
Substituting in the last equation ] " D EL and ] " D EL and multiplying, we get:
D E L š D E L
D D EE D E D E
]" " L .
D E L š D E L
D E
D E
This shows that the result of dividing two complex numbers is a complex number.
EXAMPLE 3.8c:
1.
L š L L L L L L
L
"
"
"
"
" L
L š L
L
L
2.
"
L š
L š
L
L
š š L L L ®
LMATHEMATICS
2013
–2014 †
ACADEMIC
MATHEMATICS
RESOURCE GUIDE
2010–2011
RESOURCEDECATHLON
GUIDE
L
"
L
L
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" is Prohibited Outside the Terms of Your
" License Agreement
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"
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3.8.3
HOW
DIVIDE
COMPLEX
NUMBERS?
3.8.3
HOW
DO
WE
DIVIDE
COMPLEX
NUMBERS?
3.8.3 How
doDO
WeWE
Divide
Complex
Numbers
2.
2 −i
3 − 4i
=
=
( )(
( 3 − 4i) ⋅ (
)=
3 + 4i)
2 ⋅ 3 + 2 ⋅ 4i − i 3 − 4i
( 3) − (4i)
2
6+4 4 2 − 3
+
i
7
7
2
=
(
)
3+ 4
=
ExamplE 3.8d: Solve the equation 5x 2 − 6x + 5 = 0 .
solution: 5 x 2 − 6 x + 5 = 0
∆ = 36 − 5 ⋅ 5 ⋅ 4 = 36 − 100 = −64
The equation does not have real solutions.
6 ± −64 6 ± 64 ⋅ −1 6 ± 8 ⋅ i 3 4
=
=
= ± i
10
5 5
10
10
3 4
Thus, x = ± i .
5 5
ExamplE 3.8e: Solve the equation x 3 + 3x 2 + 7x + 10 = 0 .
3
2
solution: x + 3x + 7x + 10 = 0
By the Rational Root Theorem, a rational solution to this equation must be one of the following:
m +10 +10 −10 −10
,
=
,
,
,
+1 −1 +1 −1
k
m +2 +2 −2 −2
,
= ,
,
,
k +1 −1 +1 −1
m +5 +5 −5 −5
,
= , , ,
k +1 −1 +1 −1
m +1 +1 −1 −1
,
= , , ,
k +1 −1 +1 −1
After removing all the repetitions, we get the following candidates for rational solutions:
±1, ± 2, ± 5, ± 10 .
Of all these, only x = −2 is a rational solution of the equation.
By the Factor Theorem, the polynomial x 3 + 3x 2 + 7x + 10 is divisible by x + 2 . We get:
x 3 + 3x 2 + 7x +10 = (x + 2)(x 2 + x + 5)
(x + 2)(x 2 + x + 5) = 0
x+2=0
or
34
x2 + x + 5 = 0 .
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Solving the first equation, we get x = −2 .
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−1± −19 −1± −1 ⋅ 19 −1± i 19
.
Solving the second equation, we get x =
=
=
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x=
x2 + x + 5 = 0 .
Solving the first equation, we get x = −2 .
Solving the second equation, we get x =
−1± −19 −1± −1 ⋅ 19 −1± i 19
=
=
.
2
2
2
Thus, the solutions to the given equation are: x = −2 , x =
−1 ± i 19
.
2
ExamplE 3.8f: Solve the equation ( 2 + i)x 2 + 2x − ( 2 − i) = 0 .
solution: ( 2 + i)x 2 + 2x − ( 2 − i) = 0
x=
x=
−2 ± 4
2( 2 + i)
1
=
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∆ = 4 + 4( 2 − i)( 2 + i) = 4 + 4 ⋅ (2 +1) = 16
−1± 2
2+i
2 +i
or
x=
−3 .
2 +i
Each of these solutions can be expressed in the form of a + bi . For this, we utilize the division of
complex numbers as we learned earlier:
1
=
1⋅
2 −i
)
=
2 −i
2 1
=
− i
2 +1
3 3
) ( 2 − i)
−3⋅ ( 2 − i)
−3 2 + 3i
−3
=
=
=−
2 +1
2 + i ( 2 + i) ⋅ ( 2 − i)
2+i
(
(
2+i ⋅
2+i
Thus, the given equation has the solutions: x =
2 1
− i and x = − 2 + i .
3 3
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37
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
Section
Section 3 .3.8
8:
Section 3.8: exercises
Exercises
Exercises
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
1. Find the conjugate z and the absolute value z of the given complex numbers z :
a. 4 − 13i
2
b. (1− i)
c.
(3+ i)2
2 − 5i
a. z1 + z2
b. z1 ⋅ z2
c.
z1 + z2 i
z2
3. Present the following complex number in the standard form a + bi :
(2 + i)3 − (2 − i)3
(2 + i)2 − (2 − i)2
4. Solve the following equation in the set � : x 5 + 2x 4 + 2x 3 + 2x 2 + x = 0
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

section
4: Functions
SecTIon 4:
functions
4.1
Preliminaries
4 .1 Preliminaries
The term function is often used in everyday language to indicate dependencies between quantities. For
example, we say “the commuting time from work to home is a function of traffic density”; “the price
of gasoline is a function of supply from oil-producing countries”; “population growth is a function of
time”; etc. In order to understand the properties of such dependencies, it is necessary to express them
mathematically. For example, to study the growth rate of a population, it is necessary to model the size
of the population at a given time. The equation P (t ) = 67.38(1.026)t is an example of such a model; it
models the population growth in a particular country at any given time. The equation says that the size of
the population (in millions) at a given time t is the number P(t) = 67.38(1.026)t . For example, the initial
population (when the measurement started) is 67.38 million ( P(0) = 67.38(1.026)0 = 67.38 ), and after 5
38
36
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2. Given z1 = −i + 3 , z2 = 3i −1 , find:
1
2
b. z1 ⋅ z2
c.
z1 + z2 i
z2
3. Present the following complex number in the standard form a + bi :
(2 + i)3 − (2 − i)3
(2 + i)2 − (2 − i)2
4. Solve the following equation in the set � : x 5 + 2x 4 + 2x 3 + 2x 2 + x = 0
Section
kkk4
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

FUNCTIONS
section
4: Functions
SecTIon 4:
functions
The term function
function is often used in everyday language to indicate dependencies between quantities. For
example, we say “the commuting time from work to home is a function of traffic density”; “the price
of gasoline is a function of supply from oil-producing countries”; “population growth is a function of
time”; etc. In order to understand the properties of such dependencies, it is necessary to express them
mathematically. For example, to study the growth rate of a population, it is necessary to model the size
of the population at a given time. The equation P (t ) = 67.38(1.026)t is an example of such a model; it
models the population growth in a particular country at any given time. The equation says that the size of
the population (in millions) at a given time t is the number P(t) = 67.38(1.026)t . For example, the initial
population (when the measurement started) is 67.38 million ( P(0) = 67.38(1.026)0 = 67.38 ), and after 5
years, it is 76.607 ( P(5) = 67.38(1.026)5 = 76.607 ).
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linear equations—equations
equations—equations of
of the
the form
form ax
ax++bb =
=0 —
—and
and quadratic equations—
Earlier we dealt with linear
equations of the form ax 2 + bx + c = 0 . The expressions ax + b and ax 2 + bx + c are functions. We can
think of the expression 2 x − 7 in terms of a dependency between quantities: the value of 2 x − 7 depends
on the value of x , or we can say 2 x − 7 is a function of x . For x = 3 , 2 x − 7 = −1 ; for x = 0 , 2 x − 7 = −7 ;
for x = 2 , 2x − 7 = 2 2 − 7  −4.1716 . Functions of the form y = ax + b are called linear functions, and
functions of the form y = ax 2 + bx + c , where a ≠ 0 , are called quadratic functions.
When thinking of an expression as a function, we use symbols such as f ( x ) , g ( x ) , k ( x ) , etc. For example,
the function f ( x ) = 2 x − 7 indicates that for any value of x , there corresponds a value f ( x ) (read f of
x ). For example, x = 3 corresponds to f (3) = −1 ; x = 0 corresponds to f (0) = −7 ; x = 2 corresponds
to f ( 2) = 2 2 − 7 , etc. Often, instead of f ( x ) = 2 x − 7 we write y = 2 x − 7 , meaning the value of x
corresponds to the value y = 2 x − 7 .
4.2
Definition
of a Function
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4 .2 Definition
of aMathematics
Function
39
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4.1
Preliminaries
4 .1 Preliminaries
4.1
Preliminaries
corresponds to the value y = 2 x − 7 .
4.2
Definition
of a Function
4 .2 Definition
of a Function
4.2 Definition of a Function
The function f ( x ) = 2 x − 7 admits any value of x . That is to say, for any real number x , there is a
corresponding value f ( x ) = 2 x − 7 . Likewise, the function u( x ) = 3x 2 − 2 x + 4 admits any value of x .
1
Not all functions admit all real values. For example, the function g ( x ) = does not admit x = 0 , because
x
1
f (0) = is not a number. Likewise, the function h( x ) = x admits only non-negative numbers.
0
A number that can be admitted into a function is called an input, and its corresponding number under
the function is called an output. Thus, for example, for the function h ( x ) = x 2, any real number can be an
input. Can any real number be an output for this function? The answer is obviously no. For example, –1
cannot be an output, because there is no real number x for which x 2 = −1 .
outputs of a function is called the range of the function. Thus, the domain of h ( x ) = x 2 is the set of all
real numbers. The range of this function is the set of all non-negative numbers.
ExamplE 4.2a: What are the domain and range of the function h( x ) = x ?
answEr: Both the domain and range of this function are the set of all non-negative numbers.
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the domain and range of a function requires a little more work.
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ExamplE 4.2b: What are the domain and range of the function k(x) =
2
?
x−3
answEr: Clearly, any real number different from 3 is in the domain of this function. So, the
domain consists of real numbers x ≠ 3 .
What about the range of this function?
By definition, a real number a is in the range if there exists x such that
2
=a.
x−3
Multiplying both sides of this equation by x − 3 , we get 2 = a ( x − 3) .
Clearly a = 0 is not in the range of the function because otherwise we get 2 = 0( x − 3) = 0 , which
is not possible.
40 On the other hand, any number
USAD
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a ≠Mathematics
0 is in the range.
To show
this,• 2016–2017
we need to show that for any
a ≠ 0 , the equation
2
= a is solvable for x . Indeed:
37
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The set of all possible inputs of a function is called the domain of the function; and the set of all the
With this background, we can now introduce the formal definition of a function.
With this background, we can now introduce
the formal definition of a function.
DeFinition
DeFinition
A function f is a correspondence between two sets, A and B , that assigns
DeFinition
A function f is a correspondence
between two sets, A and B , that assigns
to each element x in A one and only one element f ( x ) in B .
DeFinition
A function
a correspondence
that
A andf ( B
f iselement
x in A onebetween
B assigns
and onlytwo
onesets,
element
.
to each
x ), in
A function
a correspondence
that
A andf ( B
f iselement
x in A onebetween
B assigns
and onlytwo
onesets,
element
.
to each
x ), in
in A one notation:
and only one
each element
This state of affairs istoindicated
by thex following
f : Aelement
→ B f ( x ) in B .
This
affairs
is indicated
by the following
: A →ofB the function. You might think of a
calledofthe
domain
of the function,
and B isnotation:
called thef range
A is state
Unauthorized
Duplication
is Prohibitednotation:
Outside the Terms
of Your
License Agreement
This
statefofasaffairs
is indicated
the following
f : A→
a machine
with anbyinput
and output.
The domain
ofBf is all the numbers that can be fed
function
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This
state of affairs is indicated
by decAthlon
the followingmAthemAtics
notation: f : A
→ B License
®
igure
4–1.)
The
into the machine, and
the
range
is
all
the
numbers
that
come
out
of
the
machine.
(See
F
38
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38
machine
38
38
38
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decAthlon
never produces
two outputs
foris the
same
input.the Terms
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x
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ff
ff(x)
(x)
Figure 4–1
Thinking of a function as a machine helps us see how functions are composed of more elementary
41
USAD Mathematics Resource Guide • 2016–2017
2
functions. For example, the function f (x) = (x + 2) is composed of two functions: g(x) = x + 2 and
Rancho Cucamonga High School - Ontario, CA
is not
in the
range
of the function
Clearly a = 0both
we get 2 otherwise
Multiplying
sides
of this
equation
by x − 3 , because
= a ( x − 3) .we get 2 = 0( x − 3) = 0 , which
is not
in the
range
of the function
Clearly a = 0both
we get 2 otherwise
Multiplying
sides
of this
equation
by x − 3 , because
= a ( x − 3) .we get 2 = 0( x − 3) = 0 , which
is not possible.
a = 0 is not in the range of the function because otherwise we get 2 = 0( x − 3) = 0 , which
Clearly
is
not possible.
a = 0 hand,
is notany
in the
range aof≠the
because
which
Clearly
2 =to0(show
x − 3)that
= 0 ,for
0 isfunction
in the range.
To otherwise
show this, we
we get
need
any
On
thepossible.
other
number
is
not
On
the other hand, any2 number a ≠ 0 is in the range. To show this, we need to show that for any
isa not
≠ 0 ,possible.
the equation
= a is solvable for x . Indeed:
2− number
x
3
a ≠ 0 is in
On
the
other
hand,
any
x . range.
a ≠ 0 , the equation
= a is solvable
forthe
Indeed:To show this, we need to show that for any
2
x
−
3
a ≠ 0 is in
On
the= other
hand, any2 number
a
x . range.
ax ≠
= a is solvable
forthe
Indeed:To show this, we need to show that for any
2− 03 , the equation
2
=a
a ≠ 03 , the equation x − 3 = a is solvable for x . Indeed:
2x =2− a(x
−
3)
x−3
=a
2x =2− a(x
3 − 3)
2
=a
x =−=a(x
3x − −3 3)
22
a = x−3
2 − 3)
2a
= a(x
x2 == x −+ 33
2
x2a = a + 3
= a2x − 3
2
xa =summarize,
+3
To
the domain of the function k ( x ) =
consists of all the real numbers different
2
2
a
x =summarize,
+3
To
the domain of the function k ( x ) = x − 3 consists of all the real numbers different
from 0 .
froma3 , and its range consists of all the real numbers
x 2−different
3
To
summarize,
domain
of of
theall
function
( x) =
consists
of all
3 , and its the
0 . the real numbers different
range
consists
the realknumbers
different
from
from
x 2− 3
To
summarize,
domain
of of
theall
function
( x) =
consists
of all
3 , and its the
0 . the real numbers different
range
consists
the realknumbers
from
from
x −different
3
from 3 , and its range consists of all the real numbers different from 0 .
With this background, we can now introduce the formal definition of a function.
With this background, we can now introduce the formal definition of a function.
Figure 4–1
Thinking of a function as a machine helps us see how functions are composed of more elementary
functions. For example, the function f (x) = (x + 2)2 is composed of two functions: g(x) = x + 2 and
h ( x ) = x 2 . Any given input x corresponds to the output g ( x ) = x + 2 . This output, in turn, is an input
for the function h . We get: h(g(x)) = h(x + 2) = (x + 2)2 = f (x) . Thus, f is a composition of h and g :
f (x) = h(g(x)) . Figure 4–2 below illustrates this composition.
xx
g(x)
h
ff(x)=h(g(x))
(x)=h (g(x))
Figure 4–2
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ExamplE
4.2c: Consider
the
functions® mAthemAtics
. These
h(x) = xGuide
g ( x) = 2 x − 3 and
functions can be
composed in two ways: h( g ( x)) and g (h( x)) .
1. For a given input x, we get the output g ( x) = 2 x − 3 , which, in turn, is an input for the
function h . Thus, we have h(g(x)) = g(x) = 2x − 3 .
2. For a given input x, we get the output h( x) = x , which, in turn, is an input for the function
(
)
g . We have g h(x) = 2h(x) − 3 = 2 x − 3 .
ExamplE 4.2d: The function f (x) =
(
)
2
x 2 − 9 + 3 can be looked at as a composition of three
functions: g(x) = x − 9 , h(x) = x and k(x) = (x + 3)2 .
2
42
1. Every non-negative output of the function g is an input of the function h :
Mathematics Resource Guide • 2016–2017
h(g(x)) = g(x) = x 2 − USAD
9;
39
Rancho Cucamonga High School - Ontario, CA
g
( x − 9 + 3) can be looked at as a composition of three
= x The
− 9 ,function
and k(x) = (x + 3) .can be looked at as a composition of three
functions: g(x)
h(x) = xf (x)
ExamplE
4.2d:
f (x) ==(( xx −−99++33)) can
be looked at as a composition of three
ExamplE
4.2d:
The function
functions: g(x) = x − 9 , h(x) = x and k(x) = (x + 3) .
ExamplE 4.2d: The function f (x) =
2
2
2
2
2
2
2
22
2
functions: g(x) = x 2 − 9 , h(x) = x and k(x) = (x + 3)2 .
1. Every non-negative output of the function g is an input of the function h :
2.2.
3.
3.3.
h(g(x))
= g(x) = output
x 2 − 9 ; of the function g is an input of the function h :
Every
non-negative
Every non-negative output of the function g is an input of the function h :
2
h is an input of the function k :
Every
output
of the
function
h(g(x))
g(x)
h(g(x))
== g(x)
== xx2 −−99;; 2
2
2
k h =output
h + 3 of= thexfunction
− 9 + 3 h; is an input of the function k :
Every
is an input of the function k :
Every
output of
the function h
2
2
2
2
2
Finally, every output of the function k determines the function f :
kk hh == hh++33 == xx2 −−99++33 ;; 2
2
f (x) = kevery
h g(x)
= ofxthe
− 9function
+ 3 . k determines the function f :
Finally,
output
k determines the function f :
Finally, every output of the function
2
2
2
(x)==kk hh g(x)
g(x) == xx2 −−99++33 ..
ff(x)
() ( ) (
(( )) (( )) ((
( ( ))
(( (( ))))
(
((
)
))
)
))
4.3
Functions
4.3
M
Many-to-one
any-to-One
Functions
versus Functions
One-to-One
4 .3
Many-to-One
Functions
versus One-to-One
versus
one-to-one
Functions
4.3 Many-to-one
Many-to-one Functions
Functions
4.3
Functions
4 .3
Many-to-One
Functions
versus One-to-One
One-to-One Functions
Functions
4 .3
Many-to-One
FunctionsFunctions
versus
versus
one-to-one
Functions
versus
one-to-one
Is it possible that different inputs of a function correspond to one output of the same function? The
= −same
2same
=
2 andof
answer
is yes. that
For different
example, inputs
for theoffunction
correspond
toThe
the
h ( x )correspond
= x 2 , the inputs
possible
function
correspond
to one
onexoutput
output
ofxthe
the
function?
IsIs itit possible
that different
inputs of
aa function
to
function?
The
2
same output
. Such
functions
arefunction
often called many-to-one
functions,
meaning
more than
one
4For
answer
yes. For
example,
for the
the
the inputs
inputs
and
correspond
to the
the
xx==−−22that
xx==22 and
answer
isis yes.
example,
for
function hh((xx))==xx2 ,, the
correspond
to
input corresponds
to one
output. are often called many-to-one functions, meaning that more than one
same
output 44.. Such
Such
functions
same
output
functions
are often called many-to-one functions, meaning that more than one
inputcorresponds
corresponds toone
one output.
input
We have already to
learnedoutput.
that any quadratic equation ax 2 + bx + c = 0 for which the discriminant
−b − b2 − 4ac
−b + b2 − 4ac
2
2
has
two distinct
solutions:
and
.discriminant
This shows
∆
=
b
−
4
ac
>
0
x
=
x
=
2
for
which the
the discriminant
We
have
already
learned
that
any
quadratic
equation
ax
+
bx
+
c
=
0
We have already learned that any quadratic 1 equation2aax + bx + c = 02 for which
2a2
2
−b−− bb2 −−4ac
−b++ bb2 −−4ac
4ac
4ac
2
−b
−b
that
quadratic
functions
(for
which
the discriminant
is positive)
are
many-to-one
functions.
2 − 4ac that
has
two
distinct
solutions:
and
This
shows
∆
=
b
>
0
x
=
x
=
Is
it
possible
functions
are
one-to-many:
that
is,
one
input
of
a
function
corresponds
to more
than
one
1
2
and
.. This
shows
∆ = b − 4ac > 0 has two distinct solutions: x1 =
x
=
2
2a
2a
2a
2a
Is
it possible
functions
arewhich
one-to-many:
thatThis
is, one
input ofisaare
function
corresponds
to more
than one
output
of thethat
same
function?
The
answer
no.
possibility
excluded
in
order
to avoid
confusion.
If
that
quadratic
functions
(for
thediscriminant
discriminant
positive)
many-to-one
functions.
Unauthorized
Duplication
is is
Prohibited
Outside
the Termsare
of
Your
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Agreement
that
quadratic
functions
(for
which
the
isispositive)
many-to-one
functions.
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output
of the
function?
is ano.
This
possibility
isMatheMatics
excluded
inResouRce
order tosuch
avoidasconfusion.
we
forsame
example,
function,
upon evaluating
an expression
f ( x )The
= ± answer
x to be
( x2010–2011
) +–2014
x +If1
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®
we
example,
to outcome
be a function,
expression
such
f ( xwhether
) = ± decAthlon
xthe
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x +1
40
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mAthemAtics
Guide
AcAdemic
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= 4 , we for
4 + 1evaluating
= ResouRce
7ResouRce
+ResouRce
1ResouRce
= 3 . Guide
at xallowed,
wouldn’t
know
is 2 +upon
or −2 +an4Guide
®
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at x = 4 , we wouldn’t know whether the outcome is 2 + 4 + 1 = 7 or −2 + 4 + 1 = 3 .
Finally, there are functions that are one-to-one, meaning that two distinct inputs must correspond to two
Finally,
there are functions that are one-to-one, meaning that two distinct inputs must correspond to two
distinct outputs.
distinct outputs.
ExamplE 4.3a: Any linear function y = ax + b , for which a ≠ 0 , is one-to-one.
ExamplE 4.3a: Any linear function y = ax + b , for which a ≠ 0 , is one-to-one.
proof: Take any two distinct inputs x1 and x2 ; their corresponding outputs y1 = ax1 + b and
proof:
any
distinct
inputsthis
their
corresponding
must
be two
different.
Assume
is not
namely
that y1 = y2outputs
. Then: y1 = ax1 + b and
x1 and
x2 ;so,
y2 = ax2 + bTake
y2 1=+ax
ax
b=
axb2 must
+ b be different. Assume this is not so, namely that y1 = y2 . Then:
2 +
ax1 +
= bax=2 ax2 + b USAD Mathematics Resource Guide • 2016–2017
= ax
ax22 = 0
ax1 −
43
Rancho Cucamonga High School - Ontario, CA
1.1.
2.
proof: Take any two distinct inputs x1 and x2 ; their corresponding outputs y1 = ax1 + b and
proof:
any
distinct
inputsthis
their
corresponding
x1 and
x2 ;so,
must
be two
different.
Assume
is not
namely
that y1 = y2outputs
. Then: y1 = ax1 + b and
y2 = ax2 + bTake
y2 =+ax
ax
b=
axb2 must
+ b be different. Assume this is not so, namely that y1 = y2 . Then:
2 +
1
ax1 +
b = ax2 + b
ax
1 = ax2
ax1 −
= ax
ax2 = 0
ax
1
2
ax
a (1x1−−axx22 )==00
a ( x1 −ax≠2 )0=, x0 = x .
Since
1
2
Since a ≠ 0 , x1 = x2 .
This, of course, is not possible since we started with two distinct inputs x1 and x2 . Hence, y1 ≠ y2 .
This, of course, is not possible since we started with two distinct inputs x1 and x2 . Hence, y1 ≠ y2 .
ExamplE 4.3b: y = x is one-to-one
proof: As before, take any two distinct inputs x1 and x2 . We are to show that their correspondproof:
inputs Assume
. Weis are
thatthat
their
x1 and x2this
and any
are different.
notto
so,show
namely
ing
outputsAsybefore,
= x take
y =twoxdistinct
y correspond= y . Then:
1
1
ingx outputs
x2 y1 =
1 =
( xx =) =x( x )
(x =x x) = ( x )
2
1
1
1
1
2
2
2
2
2
2
x1 and y2 =
1
2
x2 are different. Assume this is not so, namely that y1 = y2 . Then:
2
2
2
x1 = x2
This of course is not possible since we started with two distinct inputs x1 and x2 . Hence, it is not
This that
of course
true
y1 = y2is. not possible since we started with two distinct inputs x1 and x2 . Hence, it is not
true that y1 = y2 .
4.4
inverse
Functions
4.4
Inverse
Functions
4 .4 Inverse
Functions
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Let f be a function.
If there exists a function g such that y = f ( x ) if and only if x = g ( y ) , then we say
that f is invertible and g is the inverse of f . The function g is unique (see the next page for a proof)
and is denoted by f −1 .
ExamplE 4.4a: The inverse of the function f ( x ) = x 3 is g ( x ) =
proof: If y = x 3 then 3 y = x , and conversely, if
3
3
x.
y = x , then y = x 3 . Thus, g(x) = f −1 (x) .
4
4
y=x
3
3
3
1
44
–6
–4
–2
y=3:
x
x
2
2
1
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2
4
–66
–4
–2
2
4
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ExamplE 4.3b: y = x is one-to-one
proof: If y = x 3 then 3 y = x , and conversely, if
proof: If y = x 3 then 3 y = x , and conversely, if
3
3
y = x , then y = x 3 . Thus, g(x) = f −1 (x) .
y = x , then y = x 3 . Thus, g(x) = f −1 (x) .
4
4
3
2
y=x3
y=x3
4
3
2
1
2
1
1
1
–6
–4
–2
–6
–4
–2
–1
2
4
–66
–4
–2
2
4
–66
–4
–2
–1
–2
–2
–3
–2
–3
–3
–4
–3
–4
–4
–4
x −1
ExamplE 4.4b: The inverse of f ( x ) = 2 x + 1 is g ( x ) =
.
x 2− 1
ExamplE 4.4b: The inverse of f ( x ) = 2 x + 1 is g ( x ) =
.
2
y −1
proof: y = 2 x + 1 if and only if x =
. Thus g ( x ) = f −1 ( x )
y 2− 1
proof: y = 2 x + 1 if and only if x =
. Thus g ( x ) = f −1 ( x )
4
2
3
2
–6
–4
–2
–1
–2
4
6
2
4
6
4
3
2
2
1
1
–4
2
4
3
yy=2x
x+1
yy=2x
x+1
2
1
–6
–1
–1
–2
4
3
y=3:
x
x
3
y= :
x
x
3
2
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4
3
1
2
4
–6 6
2
4
–6 6
–4
–4
–2
–2
–1
–1
–2
–1
–2
–2
–3
–2
–3
–3
–4
–3
–4
–4
–4
y= x – 1
2
y= x – 1
2
2
4
6
2
4
6
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42
42
®
x + 5 the Terms
2Outside
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Duplication
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ExamplE 4.4c:
Find
the
inverse
of
.
f ( x) =mAthemAtics
®
−3 x + 7
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7y −5
2x + 5
7x − 5
if and only if x =
(check). Hence the function g ( x) =
is
3
y
2
+
3
x
+
2
−3 x + 7
2x + 5
.
the inverse of the function f ( x) =
−3 x + 7
7
The domain of f is the set all real numbers different from , and the domain of g is the set of all
3
2
real numbers different from − .
3
solution: y =
USAD Mathematics Resource
Guide • 2016–2017
theoreM
45
real numbers different from −
3
.3
theoreM
theoreM
An invertible function f can have only one inverse function.
An invertible function f can have only one inverse function.
ProoF
ProoF
Let f1 and f 2 be two different inverse functions of f .
Let f1 and
f 2 be two different inverse functions of f .
Then y = f ( x ) if and only if x = f1 ( y )
Then y = f ( x ) if and only if x = f1 ( y )
and
and
y = f ( x ) if and only if x = f ( y )
y = f ( x ) if and only if x = f 2 ( y2 )
Rancho Cucamonga High School - Ontario, CA
This implies that
f ( y ) if and only if x = f ( y ) . We see then that if
This implies that x =x =
f1 ( y1 ) if and only if x = f 2 ( y2 ) . We see then that if
f1 and f 2 are both inverse functions of f , then f1 = f 2 .
f1 and
f 2 are both inverse functions of f , then f1 = f 2 .
theoreM
theoreM
The graph of −1 is the reflection of f in the line y = x .
The graph of f −1f is the reflection of f in the line y = x .
ProoF
ProoF
Implied from the definition of an inverse function is that ( x0 , y0 ) is on the
Implied from the definition of an inverse function is that A( A
x0 , y0 ) is on the
graph of f if and only if A '( y0 , x0 ) is on the graph of−1f −1. Notice that A and
graph of f if and only if A '( y0 , x0 ) is on the graph of f . Notice that A and
A ' are symmetric with respect to the line y = x . This is so, because the midpoint
A ' are symmetric with respect to the line y = x . This is so, because the midpoint
x +y y +x 
between A and A ' is M
 x +0 y 0y, +0 x 0 , and M is on the line y = x .
between A and A ' is M  0 20 , 0 20  , and M is on the line y = x .
2 
 2
ExamplE 4.4d:
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43
3
The property we have just proven can clearly be seen in the graphs of f ( x ) = x and f −1 ( x ) = 3 x
which we discussed earlier. This property is useful because if we know the graph of an invertible
function f , then we can graph its inverse f −1 . The two graphs are symmetric with respect to
y= x.
y=x3
4
y=x
y=x
3
y=3:
x
2
1
46
USAD Mathematics Resource Guide • 2016–2017
–6
–4
–2
2
4
6
43
which we discussed earlier. This property is useful because if we know the graph of an invertible
function f , then we can graph its inverse f −1 . The two graphs are symmetric with respect to
y= x.
y=x3
4
y=x
y=x
3
y=3:
x
2
1
–6
–4
–2
2
4
6
–1
–2
–3
Not all functions are invertible of course. For example, y = x 2 is not invertible. To see why, we graph
y = x 2 , and then graph its symmetric image with respect to the line y = x . (See Figure 4–3 below.) We
see that the resulting graph is not a function because for one input, x , there are two outputs, y1 and y2 .
5
y=x
y=x
4
3
•
y2 •
2
1
–6
–4
•&
–2
2
4
•
xx
6
8
–1
-2
y1 •
•
–3
–4
–5
Figure 4–3
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–4
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
Section
Section 4:4
Section 4: exercises
Exercises
Exercises
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
1. Find the domain and the range of the following functions:
3
2− x
b. f ( x) = x − 5
a. f ( x) =
2
c. f ( x) = 3 + ( x − 2)
a. f ( x) = x − 3
5
b. f ( x) = x − 3
2
3. a. h( x) = x , g ( x) = 2 − x . Find h( g ( x)) and g (h( x)) .
b. h( x) = 6 x − 1 , g ( x) = x + 1 , k ( x) =
1
. Find h( g ( x)) , h( g (k ( x))) and k ( g (h( x))) .
x
2
− 5.7 x
.
4. Find the range and domain of f ( x) = 7
1
23 + x
6
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

SecTIon 5:
Graphing
section
5: Graphing
The graph of a function f is the set of all ordered pairs ( x, f ( x )) , where x belongs to the domain of f .
In this section, we will learn what the graphs of different functions look like.
5.1
What
theofGraph
a
5 .1 What
DoesDoes
the Graph
a Linear of
Function
yaxb Look Like?
Linear Function yaxb Look Like?
In Section 8 we will prove that the graph of a linear function y = ax + b is a line. The coefficients a
and b determine the form of the graph. For x = 0 , y = b . Therefore, (o, b) is the point where the line
intersects the y-axis. This number b is called the y-intercept. The graph is shown in Figure 5–1.
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2. Determine which of the functions have an inverse. If the function is invertible, find its inverse.
5
b. f ( x) = x − 3
2
3. a. h( x) = x , g ( x) = 2 − x . Find h( g ( x)) and g (h( x)) .
b. h( x) = 6 x − 1 , g ( x) = x + 1 , k ( x) =
1
. Find h( g ( x)) , h( g (k ( x))) and k ( g (h( x))) .
x
2
− 5.7 x
.
4. Find the range and domain of f ( x) = 7
1
23 + x
Section
6
k k k5
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

Graphing
SecTIon 5:
Graphing
section
5: Graphing
The graph of a function f is the set of all ordered pairs ( x, f ( x )) , where x belongs to the domain of f .
5.1
W
hat Does
the
Graph
of aaLinear Function
5.1
What
theof
Graph
5 .1 What
DoesDoes
the Graph
a Linear of
Function yaxb Look Like?
Linear
Function
yaxb
Look
Like?
y = ax + b Look Like?
In Section 8 we will prove that the graph of a linear function y = ax + b is a line. The coefficients a
and b determine the form of the graph. For x = 0 , y = b . Therefore, (0,
(o, bb)) is the point where the line
intersects the y-axis. This number b is called the y-intercept. The graph is shown in Figure 5–1.
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45
(0, b)
b)
y=ax+b
y = ax + b
x
○
Figure 5–1
When a = 0 , we have y = 0 ⋅ x + b = b . We see that the output value for any x is b . Hence, the graph is
simply a line parallel to the x-axis. Its y-intercept is b . (See Figure 5–2.)
yy
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49
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In this section, we will learn what the graphs of different functions look like.
When a = 0 , we have y = 0 ⋅ x + b = b . We see that the output value for any x is b . Hence, the graph is
simply a line parallel to the x-axis. Its y-intercept is b . (See Figure 5–2.)
yy
yy=b
=b
(0, b)
x
○
Figure 5–2
Figure 5–3 below.)
b
(why?). This point is called the x-intercept. (See
a
yy
y =y=ax+b
ax + b
– –b , 0
a
xx
○
Figure 5–3
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When a > 0 , the function
This means
theoffollowing:
any numbers x1 and x2 ,
y = ax
+ b is increasing.
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if x1 ≥ x2 , then ax1 + b ≥ ax2 + b (See Figure 5–4.).
y
y =y=ax+b
ax + b
axax
++b
b
2 2
ax1 1++b
b
ax
○ xx1
1
xx22
x
Figure 5–4
50
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When a ≠ 0 , the graph intersects the x-axis at x = −
Figure 5–4
ProoF
Since a > 0 and x1 ≥ x2 , a ( x1 − x2 ) ≥ 0 . Therefore, ax1 − ax2 ≥ 0 . Therefore, ax1 ≥ ax2 .
Adding b to both sides of this inequality, we get: ax1 + b ≥ ax2 + b .
When a < 0 , the function y = ax + b is decreasing. This means the following: For any numbers x1 and x2 ,
if x1 ≥ x2 , then ax1 + b ≤ ax2 + b .
ProoF
Since a < 0 and x1 ≥ x2 , a ( x1 − x2 ) ≤ 0 . Therefore, ax1 − ax2 ≤ 0 . Therefore, ax1 ≤ ax2 .
y
y=ax+b
y = ax + b
ax1 +
+b
ax
b
1
ax
b
ax2 2++b
○
xx11
xx22
x
Figure 5–5
5.2 W
hat Does the Graph of a Quadratic
5.2
What
the of
of a
5 .2 What
DoesDoes
the Graph
a Quadratic
2Graph
2 y = ax + bx 2+ c Look Like?
Function
Function yaxFunction
bxc Look
Like?
Quadratic
yax
bxc Look Like?
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47
22 2
5.2.1
Case
y yx
= xyx
5 .2 .1 The
The
cASe
5.2.1
the
caSe
We can observe several properties of the graph of the function y = x 2 .
ProPerty 1: The graph has a minimum at x = 0 .
It easy to see that (0,0) is the lowest point in the graph of this function. It is so because for any real
number x , x 2 ≥ 0 .
ProPerty 2: The graph is symmetric.
The graph of y = x 2 is symmetric with respect to the y-axis. This means that any line parallel to the
51
USAD Mathematics Resource Guide • 2016–2017
x-axis that intersects the graph will intersect it in two points that are equidistant from the y-axis.
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Adding b to both sides of the inequality, we get: ax1 + b ≤ ax2 + b .
ProPerty 2: The graph is symmetric.
The graph of y = x 2 is symmetric with respect to the y-axis. This means that any line parallel to the
x-axis that intersects the graph will intersect it in two points that are equidistant from the y-axis.
Let y = d be such a line, and let P and Q be the intersection points of the line with the graph. Also, let
Y be the intersection of the line with the y-axis. We are to show that PY = QY . Clearly the coordinates
of Y are (0, d ) . To find the coordinates of P and Q , we solve the system
 y = x2

y = d
Solving the system, we get
x2 = d
x=± d
Thus, P = ( d , d ) and Q = (− d , d ) . From this, we get that PY = QY = d . Hence, the graph y = x 2 is
symmetric.
ProPerty 3: The function y = x 2 is increasing on the positive side of the x-axis and decreasing on the
negative side of the x-axis. This means:
A. If x1 > x2 ≥ 0 , then x12 > x2 2 (the function is increasing)
And
B. If 0 ≥ x1 > x2 , then x12 < x2 2 (the function is decreasing)
Proof of A: x12 > x2 2 if and only if x12 − x2 2 > 0 if and only if ( x1 − x2 )( x1 + x2 ) > 0 .
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(1)
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Since x1 > x2 ≥ 0 , (2) x1 − x2 > 0
From (1) and (2), ( x1 − x2 )( x1 + x2 ) > 0 .
Therefore, x12 − x2 2 > 0 . Hence, x12 > x2 2 .
Proof of B: The proof for B is similar to the Proof of A, and you can complete the proof for B on
your own as an exercise.
The three properties just discussed determine the graph of the function y = x 2 . The graph is shown in
Figure 5–6:
yy
yy=x
= x22
52
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y=d
y=d
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y=d
The three properties just discussed determine the graph of the function y = x 2 . The graph is shown in
Figure 5–6:
yy
yy=x
= x22
yy=d
=d
(
d, 0)
○ (0, 0)
x
(:
d, 0)
Figure 5–6
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2
5 .2 .2 The
The
GenerAL
cASe
bxc
5.2.2
the
GeneraL
yax
5.2.2
General
Case
ycaSe
=yax
ax2 +
bx 2+bxc
c
2
b  b 2 − 4ac

In Section 2, we showed that y = ax 2 + bx + c = a  x +
 −
2a 
4a
2

2
b − 4ac
b 

Let k = −
. Then, y = a  x +
 +k .
4a
2a 
2

b 

As before, we can observe several properties of the graph of y = a  x +
 +k .
2a 

ProPerty 1: When a > 0 , the graph has a minimum at x = −
b
.
2a
 b 
Proof: We are to show that for any x , y  −  ≤ y ( x ) .
 2a 
2
b 
 b 
 b
(1) y  −  = a  −
+  + k = a⋅0 + k = k
 2a 
 2a 2a 
2
b 

(2) y ( x) = a  x +
 +k
2a 

2
2
b 
b 


Since a > 0 and  x +
 ≥ 0 , a x +
 ≥ 0.
2a 
2a 


Therefore:
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b 

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(3)
k
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≤
+
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2a 

 b 
From (1), (2), and (3), we get y  −  ≤ y ( x ) .
 2a 
ProPerty 2: When a < 0 , the graph has a maximum at x = −
49
b
.
2a
Proof: The proof is very similar to the proof of Property 1, and you can complete it on your own as
an exercise.
ProPerty 3: The graph is symmetric with respect to the line x = −
b
.
2a
Proof: We are to show
that
any line parallel
to the
x-axis
that intersects the graph will intersect it53
at
USAD
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• 2016–2017
two points that are equidistant from the line x = −
b
.
an exercise.
ProPerty 3: The graph is symmetric with respect to the line x = −
b
.
2a
Proof: We are to show that any line parallel to the x-axis that intersects the graph will intersect it at
two points that are equidistant from the line x = −
b
.
2a
Let y = d be such a line, and let P and Q be the intersection points of the line with the graph. Also, let
Y be the intersection of the line y = d with the line x = −
b
. We will show that PY = QY .
2a
 b

,d .
 2a 
Clearly, the coordinates of Y are  −
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To find the coordinates of P and Q , we solve the system
2

b 

=
+
y
a
x


 +k
2a 


y = d

Solving the system, we get
2
b 

a x +
 +k =d
2a 


b
d −k
±
x = −
2a
a

y = d

 b
 b
d −k 
d −k 
Thus, P =  −
and Q =  −
+
,
d
−
, d  .

 2a

 2a
a
a




Now we calculate PY and QY , using the distance formula (which we will prove in Section 8):
2
 b
d −k
PY =   − +
 2a
a

 b 
d −k
2
 +  + (d − d ) =
a
 2a 
 b
d −k
QY =   − −

a
  2a
 b 
d −k
2
 +  − (d − d ) =
a
 2a 
2
y Agreement
= ax 2 + bx + c is symmetric
Hence: PY = QY . This
completes
the proof
that the
graph
theoffunction
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Note that Property 3 implies that the quadratic function has no inverse since it is not a one-to-one
function.
ProPerty 4: When a > 0 , the function y = ax 2 + bx + c is increasing on the right of x = −
decreasing on the left of x = −
yy=ax
= ax2+bx+c
+ bx + c
54
b
.
2a
y
b
xx=
=–
2a Resource Guide • 2016–2017
USAD Mathematics
b
and
2a
ProPerty 4: When a > 0 , the function y = ax + bx + c is increasing on the right of x = −
b
.
decreasing on the left of x = −
2a
b
xx=
=–
2a
yy=ax
= ax2+bx+c
+ bx + c
2a
and
y
– b ,0
2a
2a
x
x
○ (0, 0)
22
4ac
0, – bb –– 4ac
4a
Figure 5–7
Proof: We are to prove:
b
, ax12 + bx1 + c ≥ ax2 2 + bx2 + c
2a
b
B. For any x1 ≥ x2 ≥ −
, ax12 + bx1 + c ≤ ax2 2 + bx2 + c
2a
We will prove A here. The proof for B is similar, and you can complete it on your own as an exercise. As
before, we write:
2
b 

(1) ax + bx1 + c = a  x1 +
 +k
2a 

2
b 

(2) ax2 2 + bx2 + c = a  x2 +
 +k
2a 

b
Since x1 > x2 > −
,
2a
2
2
b  
b 

(3)  x1 +
 >  x2 +

2a  
2a 

2
1
Since a > 0
2
b 
b 


(4) a  x1 +
 > a  x2 +

2a 
2a 


2
Adding k to both sides of this inequality, we get
22
22
bb  Unauthorized
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is Prohibited Outside the Terms of Your License Agreement


(5)
(5) aaxx11++  ++kk >>aaxx22 ++  ++kk
®
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51
51
From
From(1),
(1),(2),
(2),and
and(5),
(5),we
weget
getax
asdesired.
desired.
ax1122 ++bx
bx11++cc≥≥ax
ax2222 ++bx
bx22 ++cc,,as
The
Thegraph
graphof
of aaquadratic
quadraticfunction
functionisiscalled
calledaaparabola,
parabola,and
andthe
theminimum
minimumor
ormaximum
maximumpoint
pointisiscalled
calledthe
the
vertex
vertexof
of the
theparabola.
parabola.
ProPerty
theset
setof
ofall
allreal
realnumbers.
numbers.IfIfaa>>00,,
ProPerty5:
5:The
Thedomain
domainof
ofthe
thequadratic
quadraticfunction
function yy==ax
ax22 ++bx
bx++cc isisthe
 bb 
 bb  55
USAD Mathematics Resource Guide • 2016–2017
the
the range
range of
of the
the quadratic
quadratic function
function isis the
the set
set of
of all
all numbers
numbers yy for
for which
whichyy≥≥aa−−  ++bb−− ++cc.. IfIf
 22aa
 22aa
22
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A. For any x1 ≥ x2 ≥ −
vertex of the parabola.
ProPerty 5: The domain of the quadratic function y = ax 2 + bx + c is the set of all real numbers. If a > 0 ,
2
 b 
 b 
the range of the quadratic function is the set of all numbers y for which y ≥ a  −
 + b  −  + c . If
 2a 
 2a 
2
 b 
 b 
a < 0 , the range of the quadratic function is the set of all numbers y for which y ≤ a  −  + b  −  + c .
 2a 
 2a 
You can complete the proof of this property on your own as an exercise.
5.3
hat Do
5.3 W
What
Do the
the Graphs
Graphsof
ofSome
Some Polynomials
5 .3 What Do the Graphs
Some Polynomials Look Like?
Polynomials
Look of
Like?
Look Like?
solution: We first find where the polynomial intersects the x-axis. To do this, we need to solve
the equation x 3 − 2 x 2 − 5 x + 6 = 0 .
We use the Rational Root Theorem to find that this equation has three solutions: x = 1 , x = −2 ,
and x = 3 .
Also, substituting x = 0 in the polynomial, we find that the polynomial intersects the y-axis at
y = 6.
We also note that as x gets larger moving toward infinity, y gets larger moving toward infinity;
and as x gets smaller moving toward negative infinity, y gets smaller moving toward negative
infinity.
We can use this information to sketch the graph of the polynomial:
y = x3 – 2x2 – 5x + 6
lled the
fa > 0,
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
 + c . If

b 
+c.
a
56
3
ExamplE 5.3b:USAD
Graph Mathematics
the polynomial y = xResource
+1.
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ExamplE 5.3a: Graph the polynomial y = x 3 − 2 x 2 − 5 x + 6 .
ExamplE 5.3b: Graph the polynomial y = x 3 + 1 .
solution: We first find where the polynomial intersects the x-axis. To do this, we need to solve
the equation x 3 + 1 = 0 . This equation has one solution: x = −1 . The polynomial intersects the
y-axis at y = 1 . We also note that as x gets larger toward infinity, y gets larger toward infinity;
and as x gets smaller toward negative infinity, y gets smaller toward negative infinity. We can use
this information to sketch the graph of the polynomial:
5.4
W
hat
Does
the
of
5.4
What
Does
the Graph
Graph
ofthe
theExponential
5 .4 What
Does
the Graph
of
the Exponential
x
exponential
Function
ya
Look Like?
Function
ya x Look
Function
y =Like?
ax Look
Like?
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The function y = a x is called the exponential function. The exponential function is only defined for a > 0
since if a > 0 , y = a x is positive for any real number x , and so its graph is above the x-axis.
Another feature of this function is that when a > 1 , the function is increasing, and when 0 < a < 1 , the
function is decreasing.
ExamplE 5.4a:
x
1
Consider the functions y = 2 and y =   :
2
x
In y = 2 x , as x gets larger, 2 x gets larger.
7
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y = 2x
6 Guide • 2016–2017
x
2
2
57
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y = x3 + 1
1
2
x
Consider the functions y = 2 x and y =   :
In y = 2 x , as x gets larger, 2 x gets larger.
7
x
2
2
y = 2x
6
5
4
3
x
2
1
2
–8
–6
–4
x1
–2
2
x2
4
6
8
–1
–2
That is: if x1 > x2 , then 2 x1 > 2 x2 . For example, 2 4 > 23 .
 11 
1
1
In yy==( )x =
= x , on the other hand, as x gets larger,   gets smaller.
 22  2
2
x1
x2
3
2
1 1
1
1
That is: if x1 > x2 , then   <   . For example,   <   .
2 2
2
2
x
x
x
y=
x
1
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x
2
x1
x2
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The domain of the exponential function is the set of all real numbers, whereas its range is the set of
10
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1
The domain of the exponential function is the set of all real numbers, whereas its range is the set of
all positive real numbers. Since the exponential function is increasing, it is one-to-one, and thus has an
inverse. The inverse of the exponential function is called the logarithmic function, which we will study
next.
14
16
The properties of the logarithmic function can be derived from the properties of the exponential function
because the logarithmic function is the inverse of the exponential function, as the following definition
indicates.
DeFinition
c
c = log a b if and only if a = b .
The function y = log a x is called the logarithmic function. a is called the base of the logarithm.
ProPerty 1: The expression y = log a x is equivalent to a y = x . Since a y is defined only for a > 0 , and
Unauthorized
Duplication
is Prohibited
Outside
Terms of Your
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Agreement of the logarithmic
x must
(as we showed
earlier),
bethepositive.
Thus,
the domain
a y is always positive
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ProPerty 2: Another important condition to observe is that a , the base of the logarithm, must be
different from 1. This is so because for a = 1 , log a x is not a function. Recall, a function cannot be one-tomany. If a = 1 , then the input x = 1 would have infinitely many outputs. For example, log1 1 = 4 because
14 = 1 ; log1 1 = −3 because 1−3 = 1 ; log1 1 = 6.9 because 16.9 = 1 ; log1 1 = 5 because 1 5 = 1 , etc. Thus, the
logarithmic function y = log a x is meaningful only for x > 0 and 1 ≠ a > 0 .
ProPerty 3: Since the function admits only positive inputs, its graph is located in the half plane to the
right of the y-axis.
ProPerty 4: Next, we can observe that the x-intercept of the logarithmic function is x = 1 , because
log a 1 = 0 . Recall that y = log a x is defined only for x > 0 . Hence, there is no y-intercept—the function
does not intersect the y-axis.
ProPerty 5: Finally, y = log a x is increasing when a > 1 and decreasing when 0 < a < 1 .
Putting all these properties together, we can see that the graph of y = log a x when a > 1 is of the following
form:
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12
5.5
W
hat
Does
the
Graph
of
5.5
What
Does
the of
Graph
ofthe
theLogarithmic
5 .5 What
Does
the Graph
the Logarithmic
Logarithmic
ylog
x Look Like?
Function
ylog aFunction
xy Look
Like?
a Like?
Function
= log
x
Look
a
ProPerty 5: Finally, y = log a x is increasing when a > 1 and decreasing when 0 < a < 1 .
Putting all these properties together, we can see that the graph of y = log a x when a > 1 is of the following
form:
y = log2x
log2x2
log2x1
x2
Figure 5–8
Duplication
Outside
the Termsform:
of Your License Agreement
0 < ais <Prohibited
1 is of the
And, the graph of Unauthorized
following
y = log a x when
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y = log 1 x
2
x1
x2
log1x1
2
log1x2
2
Figure 5–9
x
1
The graph of y = log 1 x is symmetric to the graph of y =   as is shown in Figure 5–10. This is
2
2
because the two functions are the inverse of one another.
60
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Rancho Cucamonga High School - Ontario, CA
x1
Figure 5–9
x
1
The graph of y = log 1 x is symmetric to the graph of y =   as is shown in Figure 5–10. This is
2
2
because the two functions are the inverse of one another.
x
y
y = log 1 x
Figure 5–10
The domain of the logarithmic function is the set of all positive real numbers, whereas its range is the set
of all real numbers.
In science, the most frequently used base is the number e = 2.718281828459045 23536 …. . Logarithms
with base e are called natural logarithms. The number e is irrational since it cannot be expressed as a
repeating decimal. The natural logarithm is written ln x and is defined to be the inverse function of e x .
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Sections
5.1-5.5
Section 5 .1–5 .5:
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SectionS 5.1–5.5:
exercises
Exercises
Exercises
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1. Determine the maximum or minimum and axis of symmetry of the following quadratic
functions. Graph each parabola including the vertex and the axis of symmetry.
a. f ( x) = x 2 − 6 x + 8
b. f ( x) = − x 2 + 6 x − 9
c. f ( x) = 2 x 2 − 3 x − 3
2. Graph the following functions:
x
1
a. f ( x) =  
3
x
3
b. f ( x) =  
2
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3. Sketch the graphs
of the
following functions:
a. f ( x) = 4 x 3 − 8 x 2 − 15 x + 9
61
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2
2. Graph the following functions:
x
1
2. Graph
functions:
a. f ( xthe
) = following
x
 13 
a. f ( x) =   x
3
b. f ( x) =   x
3
b. f ( x) =  2 
2
3. Sketch thegraphs
of the following functions:
3. Sketch
2 the following functions:
a. f ( x)the
= 4graphs
x3 − 8 xof
− 15 x + 9
4.
4.
5.
5.
3
2
a.
9 x − 45
b. ff (( xx)) =
= 4x x4 −−68xx3 −−415
x 2x++54
4
3
2
b. f ( x) = x − 6 x − 4 x + 54 x − 45
Let f ( x) = x 2 + px + q . Find the values of p and q if the minimum of this function is (1, −2) .
Let f ( x) = x 2 + px + q . Find the values of p and q if the minimum of this function is (1, −2) .
Find the inverse of the following functions.:
Find
x the following functions.
a. f (the
x) =inverse
3 − 22−of
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5.6
transformations
ofGraphs
Graphs
5 .6 Transformations
of Graphs
5.6
Transformations
of
5.6
transformations
of
5 .6 Transformations of Graphs Graphs
It is recommended that you review the section on the composition of functions (Section 4) before
It is recommended
studying
this section.that you review the section on the composition of functions (Section 4) before
studying this section.
If we know the graph of a function y = f ( x) , we are able to construct the graphs of the functions
If we know ,the graph of ,a function y, and
, we are able to construct the graphs of the functions
y = f ( x + c) y = f ( x) + C y = f (ax) = f (yx)= Af ( x) , where a , c , A , and C are constants.
y = f ( x + c) , y = f ( x) + C , y = f (ax) , and y = Af ( x) , where a , c , A , and C are constants.
5 .6 .1 GrAPhInG
yfyf
(xc)
from
The the
GrAPh
of yfoF
(x)yf (x)
5.6.1
GraPhinG
(xc)
FroM
GraPh
5 .6 .1 Graphing
GrAPhInG
from
TheGraph
GrAPh
(x)yf (x)
5.6.1
GraPhinG
FroM
the
GraPh
oF
5.6.1
y yf
= fyf
(x(xc)
+(xc)
c) from
the
ofof
y =yf
f (x)
The function y = f ( x + c) is the composition of two functions: y = x + c and y = f ( x) . For any input x ,
the composition
of two
functions:
The
function
and
any input x ,
y = xfor
+ cthe
= f ( x +x c+)cis. This
y = f ( xy) =. For
we first
get theyoutput
output, in turn,
becomes
an input
function
f ( x) , giving the
we first get the output x + c . This output, in turn, becomes an input for the function y = f ( x) , giving the
output f ( x + c).
58
58
exAmPles:
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(1) y = ( x + 2) 2 is the composition y = x + 2 and y = x 2 : to any given x , we first add the 2, and then we
square the resulting sum.
(2) y = 2 x − 4 is the composition of y = x − 4 and y = 2 x : to any given x , we first subtract 4, and then we
raise 2 to the power of x − 4 .
(3) y = log 2 ( x − 5) is the composition of y = x − 5 and y = log x : to any given x , we first subtract 5, and
then we take the log of x − 5 .
Notice that the two functions y = f ( x) and y = f ( x + c) have the same output for any two inputs that
differ by c . For the inputs x and x − c , we have f ( x) = f (( x − c) + c) . On the basis of this fact, we can
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now construct the graph of y = f ( x + c) from the graph of y = f ( x) .
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2− x
 3+ x 
a. ff (( xx)) =
= 32 −
b.
log 3 
−2
 3 + x 
b. f ( x) = 2 − log 3  x 
 x 
then we take the log of x − 5 .
Notice that the two functions y = f ( x) and y = f ( x + c) have the same output for any two inputs that
differ by c . For the inputs x and x − c , we have f ( x) = f (( x − c) + c) . On the basis of this fact, we can
now construct the graph of y = f ( x + c) from the graph of y = f ( x) .
For any point ( x0 , y0 ) on the graph of y = f ( x) , we have y0 = f ( x0 ) . What input x of the function
(
)
y = f ( x + c) gives the output y0 ? The answer is x = x0 − c because f (x0 + c) = f (x0 − c) + c = f (x0 ) = y0 .
Conversely, if ( x0 − c, y0 ) is on the graph of y = f ( x + c) , then ( x0 , y0 ) is on the graph of y = f ( x) .
The conclusion that we can draw from this is that to construct the graph of the function y = f ( x + c)
from the graph of y = f ( x) , we simply shift each point of the graph y = f ( x) by c units along the x-axis.
We move the graph by c units to the left if c > 0 and by c units to the right if c < 0 .
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ExamplE 5.6a: Construct y = 2 x − 4 .
The function y = 2 x − 4 is of the form y = f ( x + c) where f ( x) = 2 x and c = −4 < 0 . Therefore, the
graph of y = 2 x − 4 is the result of shifting the graph of y = 2 x along the x-axis to the right by 4
units.
x–4
5.6.2
y yf
= fyf
(x)
+(x)C
Cisfrom
the
Graph
of
yof
= yf
f (x)
Unauthorized
Duplication
Prohibited
Outside
the
Terms of
Your
License
Agreement
5.6.2Graphing
GraPhinG
FroM
the
GraPh
oF
5 .6 .2
GrAPhInG
(x)C
from
The
GrAPh
(x)yf (x)
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The function y = f ( x) + C is the composition of two functions: y = f ( x) and y = x + C . For any input
x , we get the output f ( x) . This output, in turn, becomes an input for the function y = x + C , giving the
output f ( x) + C .
exAmPles:
(1) y = x 2 − 3 is the composition of y = x 2 and y = x − 3 : We first square an input x , and then add −3
to the result.
x
x
1
1
1
(2) y =   + 2 is the composition of y =   and y = x + 2 : We first raise to the power x and then
3
3
3
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add 2 to the result.
(1) y = x − 3 is the composition of y = x and y = x − 3 : We first square an input x , and then add −3
to the result.
x
x
1
1
1
(2) y =   + 2 is the composition of y =   and y = x + 2 : We first raise to the power x and then
3
3
3
add 2 to the result.
1
3 1
and then add − to the result.
3
1
3
(3) y = log 2 x − is the composition of y = log 2 x and y = x − : We first compute the log of an input x
Notice that for any input x , the outputs y = f ( x) + C and y = f ( x) differ by C units. This fact helps us
construct the graph of y = f ( x) + C from the graph of y = f ( x) . For any point ( x0 , y0 ) on the graph
of y = f ( x) , we have y0 = f ( x0 ) . Clearly, the point ( x0 , y0 + C ) belongs to the graph of y = f ( x) + C
(because y0 + C = f ( x0 ) + C ).
graph of y = f ( x) .
The conclusion that we can draw from this is that to construct the graph of the function y = f ( x) + C , we
simply shift the graph y = f ( x) by C units along the y-axis. The graph moves up by C units if C > 0 , and
it moves down by C units if C < 0 .
ExamplE 5.6b: Construct the graph of y = log 2 x − 3 .
The function y = log 2 x − 3 is of the form y = f ( x) + C where f ( x) = log 2 x and C = −3 < 0 .
Therefore, moving f ( x) = log 2 x down the y-axis by 3 units results in the graph of y = log 2 x − 3 .
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5.6.3
GraPhinG
(ax)
FroM
the
GraPh
oF
5 .6 .3
GrAPhInG
yfyf
(ax)
TheResource
GrAPh
of •yf
(x)yf (x)
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Conversely, if the point ( x0 , y0 + C ) belongs to the graph of y = f ( x) + C , then the point ( x0 , y0 ) is on the
5.6.3
GraPhinG
yf
FroM
the
5 .6 .3 Graphing
GrAPhInG
(ax)(ax)
from
GrAPh
yfoF
(x)yf (x)
5.6.3
y yf
= f(ax)
from
theThe
Graph
ofGraPh
yof
= f(x)
The function y = f (ax) , a ≠ 0 , is the composition of y = ax and y = f ( x) . For any input x , we first get the
output ax . This output, in turn, becomes an input for the function y = f ( x) , giving the output y = f (ax) .
exAmPles:
2
x
x
(1) y =   is the composition of y = and y = x 2 (How?)
2
2
(2) y = 23 x is the composition of y = 3 x and y = 2 x (How?)
1
1
x is the composition of y = x and y = log 2 x (How?)
7
7
Let ( x0 , y0 ) be a point on the graph of y = f ( x) , then y0 = f ( x0 ) . What input x for the function y = f (ax)
x
 x 
gives the output y0 ? Clearly, the answer is x = 0 . Indeed: f (ax0 ) = f  a 0  = f ( x0 ) = y0 .
a
 a
x

This shows that if the point ( x0 , y0 ) belongs to the graph of y = f ( x) , then point  0 , y0  belongs to the
a

 x0

graph of y = f (ax) . And, conversely, if the point  , y0  belongs to the graph of y = f (ax) , then the
a

point ( x0 , y0 ) belongs to the graph of y = f ( x) .
The relation between the graph of y = f (ax) and the graph of y = f ( x) depends on the value of a :
k
If a > 1 , then we get the graph of y = f (ax) by “shrinking” the graph of y = f ( x) along the x-axis by
a factor of
k
1
.
a
Unauthorized
the Terms of the
Yourgraph
Licenseof
Agreement
If 0 < a < 1 , then
we get theDuplication
graph of isy Prohibited
by “expanding”
= f (ax) Outside
y = f ( x) along the x-axis
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by a factor of
a
.
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61
You can investigate the cases a < −1 and −1 < a < 0 on your own as an exercise. (Hint: note that for any
function y = g ( x ) the graphs of y = g ( x ) and y = g ( − x ) are symmetric with respect to the y-axis.)
ExamplE 5.6c: Construct the graph of y = 23 x .
The function y = 23 x is of the form y = f (ax) , a ≠ 0 , where f ( x) = 2 x and a = 3 > 1 . To get the
1
along the x-axis toward the
graph of y = 23 x , we “shrink” the graph of f ( x) = 2 x by a factor of
3
y-axis.
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(3) y = log 2
ExamplE 5.6d: Construct the graph of y = 2−3 x .
The function y = 2−3 x is of the form y = f (ax) , a ≠ 0 , where f ( x) = 2 x and a = −3 < −1 . We first
1
along the x-axis
obtain the graph of y = 23 x by expanding the graph f ( x) = 2 x by a factor of
3
3x
toward the y-axis. Following this, we reflect the graph of y = 2 with respect to the y-axis to
obtain the graph of y = 2 −3 x .
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5.6.4
GraPhinG
yAf
FroM
the
GraPh
5 .6 .4
GrAPhInG
yAf
(x) (x)
from
TheResource
GrAPh
of• 2016–2017
yfoF
(x)yf (x)
Rancho Cucamonga High School - Ontario, CA
The function y = 23 x is of the form y = f (ax) , a ≠ 0 , where f ( x) = 2 x and a = 3 > 1 . To get the
1
along the x-axis toward the
graph of y = 23 x , we “shrink” the graph of f ( x) = 2 x by a factor of
3
y-axis.
5.6.4
y yAf
= Af(x)
from
theThe
Graph
ofGraPh
y of
= f(x)
5.6.4
GraPhinG
yAf
(x)
FroM
the
5 .6 .4 Graphing
GrAPhInG
(x)
from
GrAPh
yfoF
(x)yf (x)
To obtain the graph of the function y = Af ( x) , A ≠ 0 , from the graph of the function y = f ( x) , we
reason as before:
1. ( x0 , y0 ) is on the graph of y = f ( x) if and only if ( x0 , Ay0 ) is on the graph of y = Af ( x) (Why?)
2. Therefore, to plot the graph of y = Af ( x) , we simply transform each point ( x0 , y0 ) on the graph
(
)
y = f ( x) to the point x0 , Ay0 .
From there:
3. If A > 1 , then we get the graph of y = Af ( x ) by “expanding” the graph of y = f ( x) along the y-axis
4. If 0 < A < 1 , then we get the graph of y = Af ( x ) by “shrinking” the graph of y = f ( x) along the
y-axis by a factor of A .
5. If A = −1 , we get the graph of y = Af ( x ) by simply reflecting the graph of y = f ( x) with respect to the
x-axis.
You can work out the cases −1 < A < 0 and A < −1 as an exercise on your own. (Hint: note that for any
function y = g ( x ) , the graphs of y = g ( x ) and y = − g ( x ) are symmetric with respect to the x-axis.
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ExamplE
Construct
the decAthlon
graph
− log 1 x .
63
63
3
The function y = − log 1 x is of the form y = Af ( x) where f ( x) = log 1 x and A = −1 . Therefore,
3
3
the graph of y = − log 1 x is symmetrical to the graph of f ( x) = log 1 x with respect to the x-axis.
3
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USAD Mathematics Resource Guide • 2016–2017
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by a factor of A .
1
4
ExamplE 5.6f: Construct the graph of y = − x 2 .
1
1
. Therefore, the
4
4
1
graph of y = − x 2 is obtained in two steps: first we shrink the graph of f ( x) = x 2 along the
4
1
The function y = − x 2 is of the form y = Af ( x) where f ( x) = x 2 and A = −
x-axis.
4
, and then we take the reflection of the latter graph with respect to the
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2013
decAthlon mAthemAtics ResouRce
ResouRce
 k
2010–2011
k k k k k k k k k AcAdemic
kkkkk
k k k k k k k k k kMatheMatics
k k k kGuide
k
k k kGuide
kkk
k–2014
kk
Section 5 .5.6
6:
Section
Section 5.6: exercises
Exercises
Exercises
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1. Construct the graphs of the following functions:
a. y = −( x + 3) 2 − 6
1
2
c. y = − log 3 x + 4
b. y = log 2 ( x + )
2. Find the inverse of the following functions and sketch both the function and its inverse in the
same graph:
1
a. f (x) = ln 2x
3
b. f ( x) = e −2 x + 2
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
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section
6: non-polynomial
equations
SecTIon 6:
non-Polynomial
equations
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y-axis by a factor of
b. y = log 2 ( x + )
2
c. y = − log 3 x + 4
2. Find the inverse of the following functions and sketch both the function and its inverse in the
same graph:
1
a. f (x) = ln 2x
3
b. f ( x) = e −2 x + 2
Section
k k k6
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
non-polynomial
equations
section
6: non-polynomial
equations
SecTIon 6:
non-Polynomial
equations

In the previous sections, we learned how to solve linear equations, quadratic equations, and some
new type of equation that is not a polynomial equation.
6.1
rational
equations
6 .1 Rational
Equations
6.1
Rational
Equations
The first non-polynomial equations we will learn about in this section are called rational equations. A
rational equation is an equation that involves only polynomials or rational expressions. By “rational
expression,” we mean a fraction whose numerator and denominator are polynomials.
exAmPles:
1.
2.
3.
1
1
− x = 0 is a rational equation. It involves the polynomials x and 0 and the rational expression .
x
x
x + x 3 − 2 = x is not a rational equation since x is not a rational expression.
x−3
+ 7 x 2 + 10 = 5 x is a rational equation since it involves only polynomials ( 7 x 2 + 10 and 5x)
2
( x − 4)
x−3
and a rational expression
( x − 4) 2
.
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4. 2 x ( x − 4) 2 = 0 is
a polynomial equation, but we can view it as a rational equation since we can write
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2 x( x −MatheMatics
4) 2
it as
1
= 0.
5. ( x − 5)( x − 4) − ( x − 2)(2 x + 9) =
6.
65
65
17
is a rational equation for the same reason given in 1 and 3.
2 x + 1.9
( x + 3)( x + 5)
11x + 13
x−2
is not a rational equation (why?)
−
=4+
3
10
7
A polynomial admits any real number x ; that is, its domain is the set of all real numbers. A rational
expression, on the other hand, may not admit all real numbers. For example, the rational expression
17
+ 1 does not admit the values x = ±2 2 because the denominator x 2 − 8 = 0 is zero for these
2
69
x −8
USAD Mathematics Resource Guide • 2016–2017
values. We say that the rational expression is not defined for x = ±2 2 or the domain of the rational
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techniques for solving higher-degree polynomial equations. In this section, we will learn how to solve a
3
10
7
A polynomial admits any real number x ; that is, its domain is the set of all real numbers. A rational
expression, on the other hand, may not admit all real numbers. For example, the rational expression
17
+ 1 does not admit the values x = ±2 2 because the denominator x 2 − 8 = 0 is zero for these
2
x −8
values. We say that the rational expression is not defined for x = ±2 2 or the domain of the rational
17
function f (x) = 2
+1 is the set of all real numbers x ≠ ±2 2 . On the other hand, the expression
x
−
8
1
− 4 is defined for all real numbers because there is no real number x for which the denominator
2
5x + 1
1
− 4 is the set of all real numbers.
5 x 2 + 1 is zero. So, the domain of the rational function f ( x) = 2
5x + 1
6.1.1
hoW
SoLve
rationaL
eQuationS?
6 .1 .1
hoW
DoDo
WeWe
SoLve
rATIonAL
eQuATIonS?
6.1.1 How
Do
We
Solve
Rational
Equations?
dealt with like fractions. However, since they involve variables, we must pay attention to their domains.
x2 − 1
Consider, for example, the rational expression A( x ) =
.
( x − 1)( x + 3)
x +1
One might hastily simplify this expression into B ( x ) =
by factoring the numerator into ( x − 1)( x + 1)
x+3
and canceling the expression
expressing x − 1 , which appears in both the numerator and denominator.
1
Is A( x ) = B ( x ) for all x ? The answer is no because while for x = 1 , B ( x ) = , A( x ) is not defined for
2
x = 1 . The lesson we learn from this is that before we simplify an expression, we must pay attention to the
domain of the expression. Thus, for example, we can say A( x ) = B ( x ) for all x ≠ 1 .
Likewise,
x 2 − 4 x + 4 ( x − 2) 2
=
= x − 2 for all x ≠ 2 .
x−2
x−2
Let us now turn to solving rational equations.
ExamplE 6.1a: Solve the equation
5
− 4x = 0 .
x−2
solution:
1. We first exclude the numbers that cannot be solutions. Since x − 2 appears in the
denominator,
x= 2 cannot
be a solution.
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4. Since a fraction is zero if and only if its numerator is zero, we get 5 − 4 x ( x − 2) = 0 .
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2. Next, as we do with fractions, we write all expressions in equivalent forms with the same
5
4 x ( x − 2)
−
=0.
denominator. In this case, x − 2 is a common denominator. We write:
x−2
x−2
5 − 4 x ( x − 2)
=0.
3. Now we add the two rational expressions as we do with fractions:
x−2
5. Now we proceed to solve this familiar equation:
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2
5 − 4 x + 8x = 0
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When solving rational equations often we need to simplify rational expressions. Rational expressions are
3. Now we add the two rational expressions as we do with fractions:
5 − 4 x ( x − 2)
=0.
x−2
4. Since a fraction is zero if and only if its numerator is zero, we get 5 − 4 x ( x − 2) = 0 .
5. Now we proceed to solve this familiar equation:
5 − 4 x2 + 8x = 0
4 x2 − 8x − 5 = 0
8 ± 64 + 80
x=
8
8 ± 12
x=
8
6. The solutions to the equation are: x =
6x
= 4.
x−9
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ExamplE 6.1b: Solve the equation
5
1
, x =− .
2
2
solution:
1.
x ≠ 9 since the expression
6x
is not defined for x = 9 .
x−9
6x
=4
x−9
6x
− 4 = 0 Subtract 4 from both sides.
3.
x−9
6x
x−9
−4
= 0 Write all expressions in an equivalent form with the same denominator.
4.
x−9
x−9
2 x + 36
= 0 Add the rational expressions.
5.
x−9
2.
6. 2 x + 36 = 0 A rational expression is zero if and only if the numerator is zero.
7. 2 x = −36 Solve
8.
x = −18
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In the next set of examples, you are expected to explain most of the steps on your own.
x2 + 1 − 2 x
− x = 0.
ExamplE 6.1c: Solve the equation
x −1
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71
In the next set of examples, you are expected to explain most of the steps on your own.
ExamplE 6.1c: Solve the equation
x2 + 1 − 2 x
− x = 0.
x −1
solution: Condition: x ≠ 1 .
1.
2.
x2 + 1 − 2 x
−x=0
x −1
x 2 + 1 − 2 x − x ( x − 1)
=0
x −1
3. x 2 + 1 − 2 x − x ( x − 1) = 0
4.
x2 + 1 − 2 x − x2 + x = 0
6.
x =1.
Since x = 1 was excluded as a solution, there is no solution to this equation. You can see how
important it is to specify the domains of the expressions involved beforehand.
ExamplE 6.1d: Solve the equation
x +1 x − 3
=
.
x −5 x +6
solution: Condition: x ≠ 5 and x ≠ −6 .
1.
2.
x +1 x − 3
=
x −5 x +6
x +1 x − 3
−
=0
x −5 x +6
3. A common denominator here is ( x − 5)( x + 6) . Hence:
4.
68
68
( x + 1)( x + 6) ( x − 3)( x − 5)
−
=0
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5. ( x + 1)( x + 6) − ( x − 3)( x − 5) = 0
( x − 5)( x + 6)
6. ( x 2 + 7 x + 6) − ( x 2 − 8 x + 15) = 0
7. x 2 + 7 x + 6 − x 2 + 8 x − 15 = 0
8. 15 x − 9 = 0
72 9. x =
9
15
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5. 1 − x = 0
7. x 2 + 7 x + 6 − x 2 + 8 x − 15 = 0
8. 15 x − 9 = 0
9.
x=
9
15
10. The solution to the given equation is x =
ExamplE 6.1e: Solve the equation
9
.
15
2x − 9 2x + 1
1
+
=
.
1− x
x + 1 1 − x2
solution:
2x − 9 2x + 1
1
+
−
=0
1− x
x + 1 1 − x2
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1.
2. First, we factor 1 − x 2 .
3. 1 − x 2 = (1 − x )(1 + x )
4. Thus, (1 − x )(1 + x ) is a common denominator for the three rational expressions in the given
equation.
5. Condition: x ≠ 1 and x ≠ −1 .
6. Changing rational expressions into equivalent forms that have the same common denominator,
we get
7.
(2 x − 9)( x + 1) (2 x + 1)(1 − x )
1
+
−
=0
(1 − x )( x + 1)
( x + 1)(1 − x ) (1 − x )( x + 1)
(2 x − 9)( x + 1) + (2 x + 1)(1 − x ) − 1
=0
(1 − x )( x + 1)
8. (2 x 2 + 2 x − 9 x − 9) + (2 x − 2 x 2 + 1 − x) − 1 = 0
9. −6 x − 9 = 0
99
= –1.5
10. xx == – = 1.5
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xx == 1.5
11.
The given equation
has only Guide
one solution:
.
–1.5.
ExamplE 6.1f: Solve the equation
69
69
2x
1
1
− 2
=
.
x − 9 x + 14 x − 3x + 2 x − 1
2
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solution: As before,
we find a common
denominator
the rational fractions in this equation.
For this, as we did in the previous examples, we factor the given denominators into unfactorable
73
ExamplE 6.1f: Solve the equation
2x
1
1
− 2
=
.
x − 9 x + 14 x − 3x + 2 x − 1
2
solution: As before, we find a common denominator to the rational fractions in this equation.
For this, as we did in the previous examples, we factor the given denominators into unfactorable
expressions.
1.
x − 1 is unfactorable.
3.
x 2 − 9 x + 14 = ( x − 2)( x − 7)
4.
x 2 − 3x + 2 = ( x − 1)( x − 2)
5. Thus, our equation can be written as
2x
1
1
−
−
=0
( x − 2)( x − 7) ( x − 1)( x − 2) x − 1
6. ( x − 1)( x − 2)( x − 7) is a common denominator of the three rational expressions in our
equation, because each denominator of the rational expressions divides it.
7. Thus, we have the condition: x ≠ 1 , x ≠ 2 , x ≠ 7 .
8. Writing these rational expressions in equivalent forms that have this common denominator,
2 x ( x − 1)
x−7
( x − 2)( x − 7)
−
−
=0
( x − 1)( x − 2)( x − 7) ( x − 1)( x − 2)( x − 7) ( x − 1)( x − 2)( x − 7)
− x– 7
x2 x+2 6x
9. Adding, we get
=0
2)(xx −
– 7)(
7)(xx –− 1)
1)
((x
x −– 2)(
we get
10. Solving, we get (x
x 2 +−7)(x
x = 0– 1) = 0
0 or
11. xx == –7
or xx==11
= –7
0.
12. Since x = 1 cannot be a solution, the only solution to the equation is xx =
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2. Next we factor the other two denominators:
x−3
=0.
x − 5x + 6
xx −− 33
ExamplE
ExamplE 6.1g:
6.1g: Solve
Solve the
the equation
equation x 22 − 5 x + 6 == 00 ..
+ 6which the denominator x 2 − 5 x + 6 is 0 , and we
solution 1: First we find the values xof −x5 xfor
ExamplE 6.1g: Solve the equation
2
exclude these values.
2
solution
solution 1:
1: First
First we
we find
find the
the values
values of
of xx for
for which
which the
the denominator
denominator xx 2 −− 55xx ++ 66 is
is 00,, and
and we
we
2
exclude
these
1.
Solving
x values.
− 5 x + 6 = 0 , we get x = 2 , x = 3 .
exclude
these
values.
2
1.
Solving
== 00 ,,xwe
≠ 3get
. xx == 22 ,, xx == 33 ..
2.
we
get
1. Condition:
Solving xx 2 −−x55x≠
x ++266and
xx ≠≠ 22to
xx ≠≠the
33 .. equation:
and
2.
Condition:
3.
we proceed
solve
and
2. Now
Condition:
x−3
3.
we
= 0 to
4.
3. Now
Now
we proceed
proceed
to solve
solve the
the equation:
equation:
2
x − 5x + 6
xx −− 33
4.
3 = 0 == 00
5.
4. xx 2−
x 2 −− 55xx ++ 66
xx == 33 x = 3 was excluded as a solution, our equation has no solution.
6.
7.
6. Since
x−3
33 was
7.
excluded
aa solution,
our
=0
solution
Factoring
outas
denominator,
we get 2has
7. Since
Since xx ==2:
was
excluded
asthe
solution,
our equation
equation
has no
no solution.
solution.
x − 5x + 6
xx −− 33
x−3
solution
out
=0
≠ 3denominator,
1.
solution 2:
2: Factoring
Factoring
outxthe
the
denominator, we
we get
get x 22 − 5 x + 6 == 00
( x − 3)( x − 2)
x − 5x + 6
1x−3
x −=30 = 0
1.
2.
=0
1.
3)(
( xx −− 2)
3)(xx −− 2)
2)
xx ≠≠ 33
1
1 the= denominator
0
2.
3.
is different from zero for all x , the equation has no solution.
2. Since
( x − 2) = 0
( x − 2)
3.
3. Since
Since the
the denominator
denominator is
is different
different from
from zero
zero for
for all
all xx ,, the
the equation
equation has
has no
no solution.
solution.
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Section 6 .1 .1:
Section 6.1.1: exercises
Section
6.1.1
k
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
k
Exercises
Section
6 .1 .1:
Exercises
k Section
k
k k k k k k k6.1.1:
k k k k kexercises
k
kkkkkkkkkkkkkkkkkkkkkkkk
Section
6.1.1:
exercises
Exercises
k
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
k
1. Solve the following equations:
1.
3 x the
−
2 following
x + 2 equations:
1. Solve
Solve
the
=following equations:
a.
x −3 x +3
33xx −− 22 xx ++222
== 8 x + 3 x + 2
a.
a. xx+−0.5
b. x − 33 + xx ++233 =
9 x + 3 9 x − 1 3x − 1
xx ++ 0.5
88xx22 ++ 33 xx ++ 22
0.5
=
b.
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xx22 x−−+
111==33x4xDuplication
−−11
c. 9 x + 3 9 +
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x+3
3− x
2
−
=
e.
d. 
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5.
6.
− 33 == 00
5. xx −=
x2 − 2x + 1
x +1
c. x 2 − 2 x + 1 + x + 1 = 4
c.
x −3 + 3− x = 4
d.
d.
e.
e.
f.
f.
x −3
3− x
1− 2x
2x +1
8
1 −2 2 x + 2 x2 + 1 =
82
6 x + 3 x + 14 x − 7 x = 12 x − 3
6 x 2 + 3 x 14 x 2 − 7 x 12 x 2 − 3
x+3
3− x
2
= 2
x 2+ 3 − 2 3 − x
4 x − 9 − 4 x + 12 x + 9 = 2 x − 3
4 x 2 − 9 4 x 2 + 12 x + 9 2 x − 3
2x + 7
3
1
+ 2 3
= 1
22 x + 7
x + 5 x − 6 + x + 9 x + 18 = x + 3
x 2 + 5 x − 6 x 2 + 9 x + 18 x + 3
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
6.1.2
What
the
GraPhS
oF SoMe
6 .1 .2 WhAT
DoDo
The
GrAPhS
of Some
6.1.2
What
Do
the
GraPhS
oF
SoMe
6 .1 .2
WhAT
Do
The
GrAPhS
of
Some
rationaL
FunctionS
Look
Like?
rATIonAL
funcTIonS
Look
LIke?
6.1.2
What
Do
the
Graphs
of
Some
Rational
Functions Look Like?
rationaL
FunctionS
Look
rATIonAL funcTIonS
Look
LIke?Like?
y
y
x
x
Figure 6–1
Figure 6–1
1
The graph of the function f ( x) = 1 is shown in Figure 6–1.
The graph of the function f ( x) = x is shown in Figure 6–1.
x
Note that the function is not defined for x = 0 . As x gets closer to zero along the positive x-axis, y gets
Note that the function is not defined for x = 0 . As x gets closer to zero along the positive x-axis, y gets
larger; and as x gets closer to zero along the negative x-axis, y gets smaller, and in both cases, the graph
larger; and as x gets closer to zero along the negative x-axis, y gets smaller, and in both cases, the graph
approaches the y-axis.
approaches the y-axis.
72
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2. Find the domain of the following functions:
2. Find the domain of the following functions:
2
a. y = 22
a. y = x2 + 2
x + 2 x −3
b. y =
x −3
b. y = 2 x33 + x22 − 13 x + 6
2 x + x − 13 x + 6
Also, as x gets farther from zero along the x-axis, y gets closer to zero, and the graph approaches the
x-axis. The graph never intersects the x-axis or the y-axis. In general, rational functions can approach,
without intersecting, certain lines. These lines are called asymptotes. The lines x = 0 and y = 0 are
1
asymptotes of the function f ( x) = .
x
ExamplE 6.1h: Graph the function f ( x) =
x
.
x −3
solution:
3
.
x −3
2. You can get this by the Division Algorithm or by noting that
x
( x − 3) + 3
3
f ( x) =
=
= 1+
x −3
x −3
x −3
3. From this, we see that f ( x) is of the form f ( x) = Ag ( x) + C where g ( x) =
C = 1.
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1. We first note that f ( x) = 1 +
1
, A = 3 , and
x −3
4. We can graph this function in three steps:
Step 1: Graph the function g ( x) =
along the x-axis by 3 units.
1
1
. We can graph this function by shifting the graph of
x −3
x
y
x
3
1
. We can graph this function by expanding the graph of
x −3
x −3
by a factor of 3 along the y-axis to the right.
Step 2: Graph the function
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Step 3: Graph the function 1 +
by 1 unit along the y-axis.
x
3
3
. We graph this function by moving the graph of
up
x −3
x −3
y
ExamplE 6.1i: Graph the function f ( x) =
solution: We first note that f ( x) = 2 −
x
2x + 5
.
x+4
3
.
x+4
1. You can get this by the Division Algorithm or by noting that
2 x + 5 2( x + 4) − 3
3
f ( x) =
=
= 2−
x+4
x+4
x+4
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y
.
2. From here we see that f ( x) is of the form f ( x) = Ag ( x) + C where g ( x) =
C = 2.
1
, A = −3 , and
x+4
We can graph this function in four steps:
1
1
. We graph this function by shifting the graph of
x+4
x
along the x-axis by 4 units to the left.
Step 1: Graph the function g ( x) =
6
yy =
1
x+4
4
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2
-4
-5
-2
-4
Step 2: Graph the function
a factor of 3 along the y-axis.
3
1
. We graph this function by expanding the graph of
by
x+4
x+4
y
x
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with respect to the y-axis.
3
3
. We graph this function by reflecting the graph of
x+4
x+4
y
Step 4: Graph the function 2 −
up by 2 units along the y-axis.
x
3
3
. We graph this function by moving the graph of −
x+4
x+4
y
x
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Section 6 .1 .2:
SectionS 6.1.2: exercises
Exercises
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1. Graph the following functions:
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Step 3: Graph the function −
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Section 6 .6.1.2
1 .2:
SectionS 6.1.2: Section
exercises
Exercises
Exercises
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1
)=−
a. f ( xthe
1. Graph
following
x − 2 functions:
1 13
b. f ( x) = − −
a.
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1x + 1 3
c. f ( x) = −
b.
2x − 3x + 1
x +1
( x)k= k k k k k k k k k k k k k k k k k k k k k k k k k k k k k k k k k
c. kf k
kkk
x −3
6.2
exponential
equations
6 .2 Exponential
Equations
6.2 Exponential Equations
6.2
exponential
equations
6.2.1
BaSicProPerTIeS
ProPertieS
6 .2
Exponential
Equations
6 .2 .1
BASIc
6.2.1
Basic
Properties
6.2.1
BaSic
ProPertieS
Recall
meanings
of
expressions of the form a n . For example:
6 .2 .1theBASIc
ProPerTIeS
25 = 2 ⋅the
2 ⋅ 2meanings
⋅2⋅2
Recall
of expressions of the form a n . For example:
1
9252 = 2 ⋅92 ⋅ 2 ⋅ 2 ⋅ 2
1 1
9272 3== 93 27
1
−23
1
627 == 32 27
6
1
The
a n is called a power. a is called the base of the power, and n is called the exponent of the
6−2 =expression
2
6
power. Recall the following important properties of powers: For any real numbers a > 0, b > 0 a ≠ 0 , and
The expression a n is called a power. a is called the base of the power, and n is called the exponent of the
real numbers m, and n, the following properties, 1 through 6, hold true:
power. Recall the following important properties of powers: For any real numbers a > 0, b > 0 a ≠ 0 , and
0
1. anumbers
= 1 ; m, and n, the following properties, 1 through 6, hold true:
real
2.
⋅ a1n; = a m + n
1. a 0m =
m
n
m−n
3. a m ÷
2.
⋅: aan == aam + n
m n n
n m −nn
4.
3. (aab÷:) a= =a a⋅ b
n
n
 a  n an n
5. (ab) = a n ⋅ b
4.
b
b
n
n
a a
6. ( a n ) m == ann⋅m
5.
b
b
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When
6.
( a nm) mand
= ann⋅mare integers, the above properties, 1 through 6, hold true for all real numbers a ≠ 0,b ≠ 0.
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kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
n
n
a a
5.   = n
b
b
6. ( a n ) m = a n⋅m
When m and n are integers, the above properties, 1 through 6, hold true for all real numbers a ≠ 0,b ≠ 0.
And, for a > 0 and n a natural number,
1
n
7. a = n a .
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77
x
We note that (−2) = −8 has the solution x = 3.
6 .2 .2
hoW
DoDo
We Solve
SoLve
exPonenTIAL
eQuATIonS?
6.2.2
hoW
We
SoLve
exPonentiaL
eQuationS?
6.2.2
How
Do
We
Exponential
Equations?
The basic idea for solving exponential equations of the form a x = b is to write b as the power with the
becomes: 3x−4 = 34 . Since a b = a c if and only if b = c , we get x − 4 = 4 . Hence, the solution to the equation
is x = 8 .
Let’s now turn to solve more complicated exponential equations.
.
ExamplE 6.2a: Solve 32 x +1 − 2 ⋅ 9 x = 5 − 32 x −1
solution:
All the powers involving the variable x can be written as powers with base 3:
32 x +1 − 2 ⋅ (32 ) x = 5 − 32 x −1
32 x +1 − 2 ⋅ 32 x = 5 − 32 x −1
Notice that each exponent in the last equation involves 2x:
32 x +1 = 32 x ⋅ 31 = 3 ⋅ 32 x
32 x
3 = 3 ⋅3 =
3 2x
3
2x
2x
3⋅ 3 − 2 ⋅ 3 = 5 −
3
2 x −1
2x
−1
Writing the given equation in terms of 32 x , we get
32 x
3⋅ 3 − 2 ⋅ 3 = 5 −
3
2x
2x
Let 32 x = t . Now, our equation turns into a linear equation:
t
3⋅ t − 2 ⋅ t = 5 − .
3
82 Solving for t, we get
t=
15
= 3.75
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base a. Consider, for example, the equation 3x−4 = 81.. We write 81 = 34 , and accordingly our equation
š [ š [ " Let [ " W . Now, our equation turns into a linear equation:
W
š W š W " .
Solving for t, we get
W " " Thus, [ " .
78
Recall that D E " F if and only if ORJ D F " E . Therefore: [ " if and only if [ " ORJ ,
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if and only if [ " ORJ
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.
Thus: [ " ORJ is the only solution to the given equation.
EXAMPLE 6.2b: Solve the equation [ [
"
[
[ SOLUTION: Notice [ " [
Hence, our equation can be written as [ [
"
[
[ [
We separate between the powers of 4 and the powers of 5: š š " š š [
Factoring out , we get: [
–
" [
[
[
[
[ "
[
We will compute the part on the right of the equation separately:
1
1  1 3  2
2
5
+
3
1
1
1
5
5
52 ⋅ 52 + 1 4
3
−
2+
2
2
2
2
2
2
2
2
2


5 +5
(25 + 1)4
26 ⋅ 4
2⋅4
4 ⋅4
4
4
4 2

52 = 
=
=
=
=
=
=
=
=


3
3
3
3
3
3
5
1+
1
4 + 4−2
5
2
2
2
2
2
2
2
2
4+ 2
5 4 ⋅ 4 +1
(64 + 1)5
65 ⋅ 5
5⋅5
5⋅5
5
5
4
(
[
©¹ ©¹
We have ª º " ª º
«» «»
Hence: [ " .
)
The solution to the given equation is [ "
.
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k kMathematics
k k k k kResource
k k k kGuide
kkk
k k k k k k k k k k k k k k k83
k
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SECTION 6.2: Exercises
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Using a calculator, we can find that solution [ is approximately 0.60155.
2
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Section 6 .2:
Section
6.2
Section 6.2: exercises
Exercises
Exercises
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x following
−5
1. Solve the
equations:
3 7
a.   =  
 73  x  73Unauthorized
 −5
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a.  x2 +x −0.5
= 
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b.
2
=
4
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3
79
2
b. 2 x 1++ xx−2 0.5 = 4
22
c. 10
− 101− x = 99
2
2
c. 10x211+− xx − 101x1− x 3= 99
d. 5
⋅ 0.2 = 25
1
1
x
d. 5 xx − x ⋅x0.2
= 3 25
e. 4 − 3 −0.5 = 3x + 0.5 − 22 x −1
e. 4 x x− 3x −0.5 =x 3x + 0.5 −x+212 x −1
f. 27 − 13 ⋅ 9 + 13 ⋅ 3 − 27 = 0
f. 27 x −x 13 ⋅ 9xx + 13 ⋅ 3x+x 1 − 27 = 0
g. 3 ⋅16 + 36 = 2 ⋅ 81
g. 3 ⋅16 x + 36 x = 2 ⋅ 81x
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6.3
Logarithmic
equations
6 .3 Logarithmic
Equations
6.3 Logarithmic
Equations
6.3
Logarithmic
equations
6 .3
Logarithmic
Equations
6.3.1
BaSic
ProPertieS
6 .3 .1
BASIc
ProPerTIeS
6.3.1 Basic
Properties
6.3.1
BaSic
ProPertieS
6 .3 .1
BASIc
ProPerTIeS
The simplest
logarithmic
equation is of the form log a x = b . Its solution is x = a b . For example, the solution
to the equation log 2 x = 4 is x = 2 4 = 16 . Logarithmic equations can, however,b be more complicated. For
The simplest logarithmic
equation is of the form log a x = b . Its solution is x = a . For example, the solution
example, to solve the equation log4 2 x + log 4 x 2 = 17 we need to apply some properties of logarithms.
to the equation log 2 x = 4 is x = 2 = 16 . Logarithmic equations can, however, be more complicated. For
2
example,
the equation
we need to only
applyfor
some
2 x + loglog
4 xa x= 17
a >properties
0 and a ≠ of
1 , logarithms.
is meaningful
and its domain
Before
wetodosolve
so, recall
that thelog
function
is the positive real numbers. Also, before proceeding, you should make sure that you understand the
Before we do so, recall that the function log a x is meaningful only for a > 0 and a ≠ 1 , and its domain
meaning of log c = b . Here are a few identities that will help you review this meaning:
is the positive areal numbers. Also, before proceeding, you should make sure that you understand the
meaning
= bsame
. Here
a few
that will help you review this meaning:
log 2 32of=log
5 isa cthe
as are
saying
25identities
= 32 .
1 = 5 is the same as saying 25 =−332 . 1
log 24 32
= −3 is the same as saying 4 =
.
64
64
1
1
1
−3
log84 2 2==−3 isis the
the same
same as
as saying
saying 4 8 == 2 .2 .
64
64
2
1
2 =is not
is the
same as saying
8 = 2 is2no
. real number b such that 6b = −36 .
log 68 (2−36)
meaningful
because there
2
b such that
−36)now
is not
because there
is no real In
number
6b = −36
We log
will6 (turn
tomeaningful
the basic properties
of logarithms.
the following,
1) through
8),.let a be a positive
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real number different from 1. Then,
We will turn now to the basic properties of logarithms. In the following, 1) through 8), let a be a positive
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2
log 6 ( −36) is not meaningful because there is no real number b such that 6b = −36 .
We will turn now to the basic properties of logarithms. In the following, 1) through 8), let a be a positive
real number different from 1. Then,
1) blogbea 1a=positive
0
Let
real number. Then
log
Let
beaa baa==positive
real number. Then
log
2) ba
3)
b1
log
Let
bea b a=positive
real number. Then
3) b
b
ba be
Let
a positive
real number
and pis be
any real
number.
Then,
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log a b
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a be
b
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a=ppositive
real
number and
p be any mAthemAtics
real number. Then,
4) blog
a b = p log a b
3)
80
80 Let
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Let
be ab ppositive
and p be any real number. Then,
4) b
= p log areal
b number
blog
Let
bea a positive
real
number and q be any real number different from zero. Then,
4) blogbea bap positive
= 1p log bb number and q be any real number different from zero. Then,
Let
5) log a q b = log aareal
q
1
Let
ab positive
and q be any real number different from zero. Then,
b number
= log a real
5) b
blogbe
Let
and
real numbers. Then,
a q d be positive
1q
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b = log
5) blogand
ab
Let
positive
real numbers. Then,
6) log aa qbdd =be
qlog a b + log a d
b d =belog
Let
positive
6) blog
logand
bd
logreal
d numbers. Then,
7)
= log a abb−+log
aa
a ad
db
6) log
log a bd
log abb−+log
log add
7)
==log
Let b bea a positive
different from 1 , and let c be any positive real number. Then,
a real number
a
db
7)
log
=log
logbacbreal
− lognumber
a d
Let
bea ac positive
different from 1 , and let c be any positive real number. Then,
8) blog
a d=
log
a
logbb c real number different from 1 , and let c be any positive real number. Then,
Let
be ac positive
8) blog
Property
(8)
a is=called the Change of Base formula.
log
logbb ac
8) log
Property
(8)
a cis=called the Change of Base formula.
log byou
a will be guided as to how to prove these properties.
In the next section,
Property (8) is called the Change of Base formula.
In the next section, you will be guided as to how to prove these properties.
6.3.2
hoW
We
SoLve
eQuationS?
6 .3 .2
hoW
DoDo
Wewill
SoLve
LoGArIThmIc
eQuATIonS?
In
the next
section,
you
be guided
as to LoGarithMic
how to prove these
properties.
6.3.2
hoW
SoLve
LoGarithMic
eQuationS?
6 .3 .2 hoW
DoDo
We We
SoLve
LoGArIThmIc
eQuATIonS?
6.3.2
How
Do We
WeWe
Solve
Logarithmic
Equations?
6.3.2
hoW
SoLve
LoGarithMic
eQuationS?
6 .3 .2 hoW
DoDo
SoLve
LoGArIThmIc
eQuATIonS?
ExamplE 6.3a: Solve the equation log5 x = 3 .
ExamplE 6.3a: Solve the equation log5 x = 3 .
ExamplE
Solve
the equation
log5 x = 3 log
. 5 x = 3 if and only 53 = x . Hence, x = 53 = 125
solution:6.3a:
By the
definition
of a logarithm,
is
the solution By
to this
equation. of a logarithm, log5 x = 3 if and only 53 = x . Hence, x = 53 = 125
solution:
the definition
is
the solution By
to this
equation. of a logarithm, log5 x = 3 if and only 53 = x . Hence, x = 53 = 125
solution:
the definition
is the solution to this equation.
.
ExamplE 6.3b: Solve the equation log10 ( x 2 − 9 x + 10) = log10 ( x − 6) .
ExamplE 6.3b: Solve the equation log10 ( x 2 − 9 x + 10) = log10 ( x − 6) .
ExamplE 6.3b:
Solve
. set of the positive real
( x 2logarithmic
− 9 x + 10) =function
log10 ( x −is6)the
solution:
Recall
thatthe
theequation
domainlog
of 10the
numbers. Hence,
we must
the following
conditions:
solution:
Recall
that state
the domain
of thetwo
logarithmic
function is the set of the positive real
1. x 2 − 9Hence,
x + 10Recall
>we
0 must
numbers.
the following
conditions:
solution:
that state
the domain
of thetwo
logarithmic
function is the set of the positive real
2.
1. x 2−−69>Hence,
x 0+ 10 >we
0 must state the following two conditions:
numbers.
Since
function is one-to-one, log10 ( x 2 − 9 x + 10) = log10 ( x − 6) if and only
0+ 10 > 0
2.
1. x 2−the
−69>xlogarithmic
2
Mathematics
Resource Guide
x 2 −x9−xthe
6.
Since
function
is one-to-one,
log10 (•x2016–2017
− 9 x + 10) = log10 ( x − 6) if and only
6+ >10logarithmic
0= x − USAD
2.
2
2
85
numbers. Hence, we must state the following two conditions:
1.
x 2 − 9 x + 10 > 0
2.
x−6>0
Since the logarithmic function is one-to-one, log10 ( x 2 − 9 x + 10) = log10 ( x − 6) if and only
Solving
the last equation, we get two solutions: x = 8 or x = 2 .
x 2 − 9 x + 10 = x − 6 .
x = 8 or1xand
= 2 .2 above.
Solving
last whether
equation,
we get
twosatisfy
solutions:
Now, wethe
check
these
values
inequalities
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Now,
whether
these values
1 and
2 above.
Substituting x = 8 in these inequalities, we get:
81
81
82 − 9 ⋅ 8 + 10 = 2 > 0
Substituting x = 8 in these inequalities, we get:
82− 6 = 2 > 0
8 − 9 ⋅ 8 + 10 = 2 > 0
x = 8 satisfies conditions 1 and 2, and hence it is a solution.
8−6 = 2 > 0
x = 8 satisfiesx conditions
1 and 2,
hence
= 2 in inequality
1, and
we get
Substituting
22 −it9is⋅ 2a+solution.
10 = −4 < 0
Since x = 2 does not satisfy at least one of conditions 1 and 2 , it is not a solution.
ExamplE 6.3c: Solve the equation log 3 x + log 3 ( x + 8) = 2 .
ExamplE 6.3c: Solve the equation log 3 x + log 3 ( x + 8) = 2 .
solution: log 3 x + log 3 ( x + 8) = 2
Condition:
solution: log 3 x + log 3 ( x + 8) = 2
1. x > 0
Condition:
2. x + 8 > 0
1. x > 0
log 3xx+( x8+>8)
By Property 6
0=2
2.
log
x ( x3 +x (8)x += 8)
32 = 2
By Property 6
2
xx (2 x++88)
x ==93
2
+ 88xx −= 99 = 0
xx 2 +
xx 2=+−88x ±− 9 64
= 0+ 36
2
−
8
±
64 + 36
= 8 ± 10
xx =
2 2
8 ± 10
xx =
= 9 , 2x = −1
x = 9 ,xx==−−11violates condition 1, the only solution our equation has is x = 9 .
Since
Since x = −1 violates condition 1, the only solution our equation has is x = 9 .
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x = 2not
in satisfy
inequality
1, we
getof22conditions
Substituting
− 9 ⋅ 2 + 10 1= and
−4 <2 0, it is not a solution.
at least
one
Since x = 2 does
ExamplE 6.3d: Solve the equation log 3 x + 2 log 3 ( x + 8) = 4
solution:
Condition:
1.
x>0
2.
x +8> 0
log 3 x + log 3 ( x + 8) 2 = 4
By Property 4
log 3 x ( x + 8) 2 = 4
By Property 6
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log 3 x + 2 log 3 ( x + 8) = 4
x ( x + 8) 2 = 43
x 3 + 16 x 2 + 64 x = 81
x 3 + 16 x 2 + 64 x − 81 = 0
By the Rational Root Theorem, since x = 1 is a solution to the last equation, the polynomial
x 3 + 16 x 2 + 64 x − 81 is divisible by x − 1 .
When dividing x 3 + 16 x 2 + 64 x − 81 by x − 1 , we get x 2 + 17 x + 81 .
Hence, the last equation can be written as ( x − 1)( x 2 + 17 x + 81) = 0 .
Equation x 2 + 17 x + 81 = 0 has no real solution (verify).
Therefore x = 1 is the only solution to the last equation.
Since x = 1 satisfies conditions 1 and 2, it is the (only) solution to the given equation.
.
ExamplE 6.3e: Solve the equation log 32 ( x − 1) − log 3 ( x − 1) − 2 = 0
solution:
Condition: x − 1 > 0 .
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87
83
83
(
)
22
log3232(x
(x 1)
1) == log
log33(x
(x 1)
1) .
We note that log
Let log 3 ( x − 1) = t .
We have the quadratic equation t 2 − t − 2 = 0 .
Solving for t we get t = −1 or t = 2 .
Hence: log 3 ( x − 1) = −1 or log 3 ( x − 1) = 2 .
Solving each of these two equations, we get: 3−1 = x − 1 or 32 = x − 1
x=
4
or x = 10 .
3
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Since these two values satisfy the condition x − 1 > 0 , they are both solutions to the given equation.
ExamplE 6.3f: Solve the equation ln x + ln 2 x = eln1
solution:
Condition: x > 0 .
First we note that eln1 = 1 (why?)
ln x + ln 2 x = eln1
ln x ⋅ 2 x = 1 (why?)
2x 2 = e
x2 =
e
2
x=±
e
2
Since x > 0 , the equation has only one solution: x =
84
88
e
.
2
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Section 6.3
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Section 6 .3:
Section 6.3: exercises
Exercises
Exercises
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1. Simplify the following expressions:
a. log 3
b. log
c.
3 3
3
3
25 − log 3 7
58
81
log10 8 + log10 18
2 log10 2 + log10 3
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d. log 3 2 ⋅ log 4 3 ⋅ log 5 4 ⋅ log 6 5 ⋅ log 7 6 ⋅ log8 7 (Hint: Use the Change of Base Formula.)
2. Use the Change of Base Formula and a calculator to evaluate the following logarithms, correct
to four decimal places:
a. log 3 19
b. log 2 3
3. Solve the following equations:
a. log 0.5 x = −1
b. log 22 ( x + 1) − log 1 ( x + 1) = 5
4
c. log 2 x + log 4 x + log8 x = 11
d. log 2 x +
4
=5
log x 2
e. 2 log10 x − log10 4 = − log10 (5 − x 2 )
f. ln x + log x e = 2
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6.4
radical
equations
6 .4 Radical
Equations
The following are examples of radical equations:
k
x+5 = 2
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89
85
f. ln x + log x e = 2
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6.4
radical
equations
6 .4 Radical
Equations
6.4
Radical
Equations
The following are examples of radical equations:
k
x+5 = 2
k
x −4 + 2− x =1
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k
5 x − 1− xMatheMatics
=1
k
85
2
3
x + 3 x =7
In all these equations, the variable x appears under the radical sign, and none of them is of the types we
The fact that an equation includes a radical does not necessarily make it a radical equation. For example,
the following equations are not radical equations:
k
x− 2 = 5
k
x2 + 3 3 = 5
k
x2 − 2 x = 3
The first two are not radical because the variable x does not appear under the radical sign. The third
equation is not radical because it is equivalent to x − 2 x = 3 , which is not radical.
6.4.1Method
MethoD
6 .4 .1
meThoD
6.4.1
11
1
One useful method for solving a radical equation of the form
n
u = n v is to raise both sides of the
equation to the power n . In this way, we get a simpler equation of the form u = v . We must be careful,
however, for the two equations may not be equivalent. This is the case because not every solution to
the second equation is necessarily a solution to the first equation. For example, consider the following
equation:
x+2 = x.
1. Squaring both sides, we get x + 2 = x 2
2. Solving this quadratic equation, we get two solutions: x = −1 and x = 2 .
3. Note that while x = 2 is a solution to our equation, x = −1 is not.
Let’s demonstrate this method further by working through several examples:
90 ExamplE 6.4a: Solve theUSAD
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have learned about so far. In this section of the resource guide, we will learn how solve such equations.
ExamplE
ExamplE 6.4a:
6.4a: Solve
Solve the
the equation
equation 55xx −−11 −− 33xx −− 22 == 22xx ++11
solution:
solution:
(
)
55xx −−11 −− 33xx −− 22 == 22xx ++11 ..
2
5x −1 − 2 5x −1 ⋅ 3x − 2 +
(
3x − 2
) =(
2
2x +1
)
2
Square both sides
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8 x − 3 − 2 (5 x − 1)(3x − 2) = 2 x + 1
−2 (5 x − 1)(3x − 2) = −6 x + 4
2 (5 x − 1)(3x − 2) = 6 x − 4
15 x 2 − 13x + 2 = 3x − 2
15 x 2 − 13x + 2 = 9 x 2 − 12 x + 4
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86
86
86
86
Square both sides
6x2 − x − 2 = 0
2
1
, x=−
3
2
1
x = − cannot be a solution to our original equation because some of the expressions in the
2
1
original equation are not defined for this value. (For example, when x = − is substituted in
2
the expression 5 x − 1 , we get a square root of a negative number.) On the other hand, all the
2
expressions in our original equation are defined for x = .
3
2
Hence x = is the only solution to our original equation.
3
x=
ExamplE 6.4b: Solve the equation x − 5 + 3 − x = 1 .
solution:
(
)
2
x − 5 + 2 x − 5 ⋅ 3− x +
(
3− x
) =1
2
Square both sides
x − 5 + 2 x − 5 ⋅ 3− x + 3− x = 1.
2 − x 2 + 8 x − 15 = 3
− x 2 + 8 x − 15 =
3
2
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2 − x 2 + 8 x − 15 = 3
− x 2 + 8 x − 15 =
3
2
99
2
−
− xx 2 +
+ 88 xx −
− 15
15 =
=
44
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This
This equation
equation has
has no
no solution.
solution. Hence,
Hence, our
our original
original equation
equation too
too has
has no
no solution.
solution.
6.4.2
MethoD
6 .4 .2
meThoD
6.4.2
MethoD
2
6 .4 .2
meThoD
6.4.2 Method
222 2
ExamplE
ExamplE 6.4c:
6.4c: Solve
Solve the
the equation
equation
solution:
solution: Note
Note that
that if
if we
we let
let
33
33
2
3
(( xx −
− 1)
1) 2 −
− 3 xx −
− 11 =
= 12
12 ..
2
xx −
− 11 =
= tt ,, we
we will
will get
get the
the quadratic
quadratic equation:
equation: tt 2 −
− tt =
= 12
12 ..
Solving
= 44 or
=−
−33 ..
Solving for
for tt ,, we
we get
get that
that tt =
or tt =
Therefore,
Therefore,
33
xx −
− 11 =
= 44 or
or
33
xx −
− 11 =
=−
−33 ..
Raising
− 11 =
= 64
64 or
− 11 =
=−
−27
27
or xx −
Raising both
both sides
sides of
of each
each equation
equation to
to the
the third
third power,
power, we
we get
get xx −
xx =
= 65
65 or
=−
−26
26 ..
or xx =
By
By checking
checking each
each of
of these
these values
values in
in the
the original
original equation,
equation, we
we can
can conclude
conclude that
that both
both x
x=
= 65
65 and
and
xx =
=−
−26
26 are
are solutions
solutions to
to our
our equation.
equation.
k
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
k
Section
Section 6 .6.4
4:
Section 6.4: exercises
Exercises
Exercises
k
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
k
1.
1. Solve
Solve the
the following
following equations:
equations:
92
a.
a.
2
33 −
− 22 xx 2 =
= 11
b.
b.
4
55 +
+ 22 xx =
= 10
10 −
− 33 4 55 +
+ 22 xx
2
c.
+ 1)
1) xx 2 −
− xx −
− 66 =
= 66 xx +
+ 66
c. (( xx +
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Now
Now we
we will
will learn
learn aa second
second method
method for
for solving
solving radical
radical equations.
equations. It
It is
is called
called solution
solution by
by substitution.
substitution.
a.
3 − 2x = 1
b.
5 + 2 x = 10 − 3 4 5 + 2 x
c. ( x + 1) x 2 − x − 6 = 6 x + 6
d.
x − 3 + Unauthorized
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e.
x + x + 11 + x − x + 11 = 4
f.
3 x − 15 − 2 x = 3 − 5 − x
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
7 .1 Linear
Inequalities
7.1
Linear
Inequalities
In Section 1, we listed the important set of properties of the real numbers concerning inequalities. These
are:
1. For any two real numbers x and y , one and only one of the following is true: x < y , x = y , or y < x .
2. For any three real numbers x , y , and z , if x < y , y < z , then x < z .
3. For any three real numbers x , y , and z , if x < y then x + z < y + z .
4. For any three real numbers x , y , and z , if x < y and z > 0 , then zx < zy .
5. For any three real numbers x , y , and z , if x < y and z < 0 , then zx > zy .
These properties are needed to solve inequalities. Let’s consider a few examples:
ExamplE 7.1a: Solve the inequality 3x − 4 > 5 .
We are asked to find the set of all numbers x that satisfy this inequality.
Solution:
3x − 4 > 5
Add 4 to both sides of this inequality to get the following inequality:
3x > 9
Divide both sides of the inequality by 3 to get the following inequality:
x>3
The solution set consists of all the numbers that are greater than 3. For example, x = 4 satisfies the
inequality because 3 ⋅ 4 − 4 = 8 > 5 . On the other hand, 2 does not satisfy the inequality because
3⋅ 2 − 4 = 2 < 5 .
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section
7: inequalities
SecTIon 7:
Inequalities
d.
x −3 + 6− x = 3
e.
x + x + 11 + x − x + 11 = 4
f.
3 x − 15 − 2 x = 3 − 5 − x
Section
k k k7
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

section
7: inequalities
SecTIon 7:
Inequalities
Inequalities
In Section 1, we listed the important set of properties of the real numbers concerning inequalities. These
are:
1. For any two real numbers x and y , one and only one of the following is true: x < y , x = y , or xy >< y.x .
2. For any three real numbers x , y , and z , if x < y , y < z , then x < z .
3. For any three real numbers x , y , and z , if x < y then x + z < y + z .
4. For any three real numbers x , y , and z , if x < y and z > 0 , then zx < zy .
5. For any three real numbers x , y , and z , if x < y and z < 0 , then zx > zy .
These properties are needed to solve inequalities. Let’s consider a few examples:
ExamplE 7.1a: Solve the inequality 3x − 4 > 5 .
We are asked to find the set of all numbers x that satisfy this inequality.
Solution:
3x − 4 > 5
Add 4 to both sides of this inequality to get the following inequality:
3x > 9
Divide both sides of the inequality by 3 to get the following inequality:
x>3
The solution set consists of all the numbers that are greater than 3. For example, x = 4 satisfies the
inequality because 3 ⋅ 4 − 4 = 8 > 5 . On the other hand, 2 does not satisfy the inequality because
3⋅ 2 − 4 = 2 < 5 .
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7 .1 Linear
Inequalities
7.1
Linear
Inequalities
7.1
Linear
Inequalities
ExamplE 7.1b: Solve the inequality 8 x − 2 > 2 x − 3 .
Solution:
8x − 2 > 2 x − 3
Add 2 to both sides of this inequality to get the following inequality:
8x > 2 x − 1
Add −2x to both sides of this inequality to get the following inequality:
6 x > −1
Divide both sides of this inequality by 6 to get the following inequality:
1
6
Thus, the solution set of the given inequality consists of all the numbers that are greater than −
1
.
6
ExamplE 7.1c: Solve the inequality −3x + 12 > 45 .
Solution:
−3x + 12 > 45
Add −12 to both sides of this inequality to get the following inequality:
−3x > 33
Divide both sides of the inequality by −3 to get the following inequality:
x < −11
If you feel you understand the process of solving an inequality, there is no need to explain each
step. Nor is there a need to say explicitly the meaning of the last expression. The answer x < −11
is understood to mean “the solution set of the given inequality consists of all the numbers that
are smaller than −11 .”
1
8
ExamplE 7.1d: Solve the inequality 4( x − 12.4) ≤ 8 x − .
Solution: You should justify each step in the following process on your own:
4( x − 12.5) ≤ 8 x −
1
8
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x>−
4 x − 50 ≤ 8 x −
1
8
32 x − 400 ≤ 64 x − 1
−32 x − 400 ≤ −1
−32 x ≤ 399
399
15
= −12
32
32
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x≥−
ExamplE 7.1e: Solve the inequality 2( − x + 3) > −2 x + 19 .
Solution:
2( − x + 3) > −2 x + 19
−2 x + 6 > −2 x + 19
Look carefully at this inequality. Clearly, there is no number that satisfies it because if there were
such a number, then we would get 6 > 19 , which is not true. Thus, the given inequality has no
solution.
ExamplE 7.1f: Solve the inequality −22(3x − 3) > −6(11x + 1) − 35 .
Solution:
−22(3x − 3) > −6(11x + 1) − 35
−66 x + 66 > −66 x − 6 − 35
−66 x + 66 > −66 x − 41
Since 66 > −41 , the given inequality has infinitely many solutions: any number x satisfies it.
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5( x − 3) > 3x − 7
ExamplE 7.1g: Solve the system of inequalities 
2 x + 3 > 36 − x
Solution: This time, we are looking for the set of all the numbers that satisfy two inequalities.
We maintain the two inequalities together while solving each separately:
5 x − 15 > 3x − 7

2 x + 3 > 36 − x
5 x − 3x > −7 + 15

2 x + x > 36 − 3
 x > 4.5

 x > 11
The solution set of the given system consists of all the numbers that are greater than 4.5 and
greater than 11. Clearly, any number that is greater than 11 is also greater than 4.5. Thus, the
solution set is the set all the numbers x that are greater than 11. Or we simply write: x > 11 .
Some linear inequalities involve absolute values. The absolute value of a number is the distance of the
number from zero on the number line. For example, the absolute value of 5 , is 5 , and the absolute value
of −5 is also 5. The absolute value of zero is obviously zero.
The absolute value of a number x is denoted by x . The absolute value of a number x depends on where
x is located on the number line: if x is located to right of 0 , then x = x ; if x is located to left of 0 ,
then x = − x ; and if x is zero, then x = 0 .
The above paragraph can be written in more compact form:
if x > 0
x

x = 0
if x = 0
 − x if x < 0

The outputs of the absolute-value function
are always non-negative. Its graph consists of two
linear functions: y = x for x ≥ 0 , and y = − x for x < 0 . (See Figure 7–1.)
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2 x > 9

3x > 33
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y = | x|
Figure 7–1
Let’s now solve a few linear inequalities involving absolute values.
ExamplE 7.1h: Solve x < 3 .
Solution: We are looking for all the numbers x whose distance from zero is smaller than 3.
The only numbers x that satisfy this inequality are −3 < x < 3 .
–3
3
0
ExamplE 7.1i: Solve x ≥ 6 .
Solution:
We are looking for all the numbers x whose distance from zero is greater than or equal to 6. The
only numbers x that satisfy this inequality are: x ≥ 6 or x ≤ −6 .
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ExamplE 7.1j: Solve x + x <
1
.
2
Solution: This inequality is a bit harder. The algebraic definition of absolute value can be
very helpful in solving this kind of inequality. We will deal with two cases:
CaSE 1: x ≥ 0 .
In this case x = x , and so the given inequality is: x + x <
1
2
Solving:
2x <
1
4
So, when x ≥ 0 the solution of the given inequality is x <
Hence, any x where 0 ≤ x <
Rancho Cucamonga High School - Ontario, CA
x<
1
2
1
.
4
1
is a solution to the inequality.
4
CaSE 2: x < 0 .
In this case, x = − x , and so the given inequality is: x − x <
1
.
2
Solving:
0⋅ x <
1
2
1
We are looking for all the numbers x for which 0 ⋅ x < . Obviously, any multiple of zero is zero,
2
1
1
which is less than . Therefore, any number x satisfies the inequality 0 ⋅ x < .
2
2
But, remember that we are dealing with the case x < 0 . Hence the solution in this case is x < 0 .
Summarizing:
The solution to the given inequality consists of two sets of numbers:
1
The set of all the numbers x for which 0 ≤ x <
4
The set of all the numbers x for which x < 0 .
Combining these two sets, we get:
The solution to the given inequality is x <
1
.
4
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kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
Section 7 .1:
SectIon 7.1: exercises
Section
7 .7.1
1:
Exercises
Section
SectIon 7.1: exercises
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
Exercises
Exercises
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
1. Solve the inequalities:
1. Solve the inequalities:
a. −3 x + 21 > 0
a. −3 x + 21 > 0
b. 2( x − 2) − 5(1 − 3 x) < 2
c. 6(2 − x) ≥ 5 − 2 x
2x
− 1 < 3 − 2(1 − 2 x)
d.
23x
− 1 < 3 − 2(1 − 2 x)
d.
3 − 3 − 7 x + x + 1 < 14 − 7 − 8 x
e. 13
107 x x 2+ 1
3−
7 −28 x
+
< 14 −
e. 13 −
10
2
2
 x 7 5x 7
 2 − 4 > 2 − 8
x 7 5x 7
f. 
 − >
 22 x +41 2 −1 −
2x
< 5− 8
f. 
 2 x4+ 1
1 −32 x

< 5−
 4
3
g. x − 2 ≤ 5 − x
g. x − 2 ≤ 5 − x
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
7.2
Quadratic
Inequalities
7 .2
Inequalities
7.2Quadratic
Quadratic
Inequalities
7.2Quadratic
Quadratic
Inequalities
7 .2
Inequalities
The following are examples of quadratic inequalities:
The
following are examples of quadratic inequalities:
k x 2 − 3x + 6 < 0
k
k
x 2 − 3x + 6 < 0
5 x 2 < 125
k
k
5 x 2 < 125
( x − 3)( x + 5) > 0
k
k
( x −2 3)( x + 5) > 0
− x − 3x ≤ 0
k
k
− x 2 − 3x ≤ 0
(− x + 5)(3 − x) ≤ 6
(− x + 5)(3 − x) ≤ 6
Any quadratic inequality can be brought by algebraic operations into one of two forms:
k
Any quadratic
inequality can be brought by algebraic operations into one of two forms:
2
1. ax + bx + c > 0 and a > 0
ax 22 + bx + c > 0 and a > 0
1. ax
+ bx + c < 0 and a > 0
2.
2
2. ax + bx + c < 0 and a > 0
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b. 2( x − 2) − 5(1 − 3 x) < 2
c. 6(2 − x) ≥ 5 − 2 x
For example, the inequality −2 x 2 + 3 x − 9 > 0 can be changed into the second form by multiplying both
sides of the inequality by −1 . It becomes: 2 x 2 − 3 x − 9 < 0 . (Note that a , the coefficient of x 2 , is positive).
7.2.1
IneQuaLItIeS
the
a0
7 .2 .1 Inequalities
IneQuALITIeS
ofof
The
form
a0
7.2.1
of the
Form
ax2form
+ ax
bx2bxc0
+ax
c >2bxc0
0 andAnD
a > and
0
We first observe that when a > 0 , the function ax 2 + bx + c has a minimum. Next we ask: Does the
parabola intersect the x-axis? And, if it does, at how many points does it intersect the x-axis?
Recall that the answer to these questions depends on the value of the discriminant ∆ = b2 − 4ac .
Case 1: When ∆ = b2 − 4ac < 0 and a > 0 , the parabola does not intersect the x-axis, and it looks like
Figure 7–2
This implies that for any value of x , ax 2 + bx + c > 0 . Hence: When ∆ = b2 − 4ac < 0 and a > 0 , the
solution to the inequality ax 2 + bx + c > 0 is −∞ < x < ∞ .
Case 2: When ∆ = b2 − 4ac = 0 , the parabola touches the x-axis at one point: x = −
b
2a
bb
2a
2a
Figure 7–3
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this:
b
, ax 2 + bx + c > 0 . Hence: When
2a
b
.
∆ = b2 − 4ac = 0 and a > 0 , the solution to the inequality ax 2 + bx + c > 0 is any number x ≠ −
2a
This implies that for any value of x that is different from −
Case 3: When ∆ = b2 − 4ac > 0 and a > 0 , the parabola intersects the x-axis at two points:
−b − b2 − 4ac
−b + b2 − 4ac
and x2 =
.
2a
2a
x < x1
x > x2
x1
x2
Figure 7–4
This implies that for any value of x for which x > x2 or x < x1 , ax 2 + bx + c > 0 .
Hence: When ∆ = b2 − 4ac > 0 and a > 0 , the solution to the inequality ax 2 + bx + c > 0 is x < x1 or
x > x2 .
ExamplE 7.2a: Solve the inequality x 2 + 6 x > −10 .
Solution: First we bring this inequality into standard form ax 2 + bx + c > 0 by adding 10 to
each side: x 2 + 6 x + 10 > 0
We find the value ∆ for x 2 + 6 x + 10 :
∆ = 62 − 10 ⋅ 4 = 36 − 40 = −4 < 0 .
Since a = 1 , the quadratic function has a minimum, and its graph is above the x-axis. Therefore,
any real number x satisfies the given inequality.
To show this, we write: −∞ < x < ∞ .
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x1 =
ExamplE 7.2b: Solve x 2 + 6 x + 9 > 0 .
Solution: We find the value of ∆ for x 2 + 6 x + 9
∆ = 62 − 9 ⋅ 4 = 36 − 36 = 0 .
Since a = 1 , the graph of the quadratic function touches the x-axis from above at one point.
To find this point, we solve: x 2 + 6 x + 9 = 0 .
Solving, we get x = −3.
ExamplE 7.2c: Solve x 2 + 6 x + 5 > 0 .
Solution: The value of ∆ for x 2 + 6 x + 5 is ∆ = 62 − 5 ⋅ 4 = 36 − 20 = 16 > 0
Since a = 1 , the function has a minimum, and it intersects the x-axis at two points. These points
are the solutions to the equation x 2 + 6 x + 5 = 0 .
Solving, we get:
x=
−6 ± 4
2
x = −5
or
x = −1 .
–5
–1
For x < −5 or x > −1 , y = x 2 + 6 x + 5 is positive.
Thus, the solution to the given inequality is the set: x < −5 or x > −1 .
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This implies that any x ≠ –3 satisfies the given inequality.
2
2
7.2.2
IneQuaLItIeS
a0
7 .2 .2
IneQuALITIeS
ofof
Thethe
form
a0
7.2.2 Inequalities
of the
Form
ax2form
+ax
bxbxc0
+ax
c <bxc0
0 andAnD
a > and
0
Recall that when a > 0 the function ax 2 + bx + c has a minimum. As before we ask: Does the parabola
intersect the x-axis, and if it does, at how many points does it intersect the x-axis?
The answer to this question depends on the value of the discriminant ∆ = b2 − 4ac . As before, we have
three cases:
Figure 7–5
This implies that there is no x for which ax 2 + bx + c < 0 . Hence: When ∆ = b2 − 4ac < 0 and a > 0 , the
inequality ax 2 + bx + c < 0 has no solution.
Case 2: When ∆ = b2 − 4ac = 0 , the parabola touches the x-axis at one point: x = −
–
b
2a
b
2a
Figure 7–6
This implies that there is no x for which ax 2 + bx + c < 0 . Hence: When ∆ = b2 − 4ac = 0 and a > 0 , the
inequality ax 2 + bx + c < 0 has no solution.
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Case 1: When ∆ = b2 − 4ac < 0 , the parabola does not intersect the x-axis, as is shown in Figure 7–5.
Case 3: When ∆ = b2 − 4ac > 0 , the parabola intersects the x-axis at two points:
x1 =
−b − b2 − 4ac
−b + b2 − 4ac
and x2 =
.
2a
2a
x1 < x < x2
x1
x2
This implies that for any value of x between x1 and x2 , ax 2 + bx + c < 0 . Hence: When ∆ = b2 − 4ac > 0
and a > 0 , the solution to the inequality ax 2 + bx + c < 0 is x1 < x < x2 .
ExamplE 7.2d: Solve the inequality x 2 + 8 x + 18 < 0 .
Solution: ∆ = 82 − 18 ⋅ 4 = 64 − 72 = −8 < 0
Therefore, the graph of the function y = x 2 + 8 x + 18 does not intersect the x-axis. Since a = 1 , the
graph is above the x-axis.
Therefore, the given inequality has no solution.
ExamplE 7.2e: Solve the inequality x 2 + 8 x + 16 < 0 .
Solution: ∆ = 82 − 16 ⋅ 4 = 64 − 64 = 0
Therefore, the graph of the function y = x 2 + 8 x + 16 touches the x-axis at one point.
Since a = 1 , the graph is above the x-axis.
Hence, the given inequality has no solution.
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Figure 7–7
ExamplE 7.2f: Solve the inequality x 2 + 8 x + 7 < 0 .
Solution: ∆ = 82 − 7 ⋅ 4 = 64 − 28 = 36 > 0
Therefore, the graph of the function y = x 2 + 8 x + 16 intersects the x-axis at two points. These
points are the solutions to the equation x 2 + 8 x + 7 = 0 .
Solving, we get
x=
−8 ± 6
2
x = −7
x = −1
Since a = 1 , the function y = x 2 + 8 x + 16 has a minimum.
Hence, for −7 < x < −1 the inequality holds.
Section 7.2
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
Section 7 .2:
SectIon 7.2: exercises
Exercises
Exercises
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
1. Solve the inequalities:
a. x 2 − 5 x + 6 < 0
b. 4 x 2 − 4 x + 1 ≤ 0
c. 81 − (3 + 2 x) 2 < 0
d. 3 + 2x ≤ x 2
e. 5 − x 2 ≥ −4 x
f.
3
>0
2x −1
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or
g.
2x − 9
<1
x−6
2. Find the domain of the following functions:
a. y = (3 x − 2)( x − 5)
b. y =
1
112 x + 64 + 49 x 2
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

section
8: coordinate
Geometry
SecTIon 8:
coordinate
Geometry
through equations (and sometimes through inequalities). In this way, an investigation of an equation
representing a geometric object can reveal important properties of the geometric object. The most basic
geometric object is the point. A basic idea of coordinate geometry is the creation of a correspondence
between the set of all ordered pairs of real numbers and the set of all points in the plane: For any point
in the plane, there corresponds exactly one ordered pair of real numbers ( x, y ) , and, conversely, for any
ordered pair of real numbers ( x, y ) , there corresponds exactly one point in the plane.
8.1
The
Pythagorean
8 .1 The
Pythagorean
Theorem Theorem
Before we start with coordinate geometry, we will first prove an ancient theorem, called the Pythagorean
Theorem. We will need this theorem to prove the distance formula.
Pythagorean Theorem: In a right triangle, the square of the hypotenuse is the sum of the squares of the
other two sides of the triangle.
Proof: Consider a right triangle with sides a and b and hypotenuse c as shown in Figure 8–1 below.
We are to prove that a 2 + b 2 = c 2 .
Figure 8–1
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The goal of coordinate geometry is to describe geometric objects, such as lines and circles, algebraically
g.
2x − 9
<1
x−6
2. Find the domain of the following functions:
a. y = (3 x − 2)( x − 5)
b. y =
1
Section 112 x + 64 + 49 x
k k k8
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
2

section
8: coordinate
Geometry
Geometry
SecTIonCoordinate
8:
coordinate
Geometry
The goal of coordinate geometry is to describe geometric objects, such as lines and circles, algebraically
representing a geometric object can reveal important properties of the geometric object. The most basic
geometric object is the point. A basic idea of coordinate geometry is the creation of a correspondence
between the set of all ordered pairs of real numbers and the set of all points in the plane: For any point
in the plane, there corresponds exactly one ordered pair of real numbers ( x, y ) , and, conversely, for any
ordered pair of real numbers ( x, y ) , there corresponds exactly one point in the plane.
8.1
The
Pythagorean
Theorem
8 .1 The
Pythagorean
TheoremTheorem
8.1
The
Pythagorean
Before we start with coordinate geometry, we will first prove an ancient theorem, called the Pythagorean
Theorem. We will need this theorem to prove the distance formula.
Pythagorean Theorem: In a right triangle, the square of the hypotenuse is the sum of the squares of the
other two sides of the triangle.
Proof: Consider a right triangle with sides a and b and hypotenuse c as shown in Figure 8–1 below.
We are to prove that a 2 + b 2 = c 2 .
c
a
b
Figure 8–1
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through equations (and sometimes through inequalities). In this way, an investigation of an equation
Four copies of the right triangle are assembled to form Figure 8–2 below. In this figure we have two
squares: the outer square with side a + b , and the inner square with side c .
b
a
b
c
a
c
c
a
c
b
b
a
Let A be the area of the outer square: On the one hand, A = (a + b) 2 . On the other hand, A is the sum
1 
of the areas of the four triangles and the area of the inner square. That is: A = 4  ab  + c 2
2 
1 
Therefore, (a + b) 2 = 4  ab  + c 2 .
2 
This leads to: a 2 + 2ab + b 2 = 2ab + c 2
Hence: a 2 + b 2 = c 2 , as was required.
The Converse of the Pythagorean Theorem: If the square of one side of a triangle is equal to the sum of
the squares of the other two sides, then the triangle is a right triangle.
Proof: Let ABC be a triangle such that
(1) AC 2 + BC 2 = AB 2 .
We are to prove that C = 90° .
Construct a triangle AB′C such that B′C is perpendicular to AC and B′C = BC , as is shown in Figure 8–3:
A
B'
C
B
Figure 8–3
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Figure 8–2
Since triangle AB′C is a right triangle, according to the Pythagorean Theorem,
(2) AC 2 + B ' C 2 = AB '2
Since B′C = BC , we get
(3) AC 2 + BC 2 = AB '2
From (1) and (3), we get AB′ = AB .
The triangles ABC and AB′C are congruent by the side-side-side triangle congruence theorem (SSS).
Hence C ' = C = 90 .
8.2
Points
8 .2 Points
8.2
Points
Distance between two points: Given two points P ( x1 , y1 ) and Q( x2 , y2 ) , find the distance d between them.
SoluTion: In Figure 8–4, QL ⊥ PR
2
By the Pythagorean Theorem, PQ = QL + PL
2
QL = x1 − x2
PL = y1 − y2
Therefore d = ( x1 − x2 ) 2 + ( y1 − y2 ) 2
This is called the distance formula.
yy
xx 1
P
1
dd
yy11
xx22
L
Q
yy22
xx
0
R
Figure 8–4
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This completes the proof.
Midpoint: Given two points P ( x1 , y1 ) and Q( x2 , y2 ) , find the midpoint M ( x, y ) between them.
SoluTion: We solve this problem for the case where the segment PQ is not parallel to the y-axis (see
Figure 8–5). You can pursue the case where PQ is parallel to the y-axis on your own as an independent
yy
exercise.
Q
M
P
xx
O
1
M
(x 0)
, 0)
M11 (x,
1
Q
Q1 (x(x2, 0)
, 0)
1
2
Figure 8–5
Since PQ is not parallel to the y-axis, x1 ≠ x2 .
From the points P, Q, and M, drop lines parallel to the y-axis, and let P1 ( x1 , 0) , Q1 ( x2 , 0) , and M 1 ( x, 0)
be the intersections of these lines with the x-axis, respectively.
From geometry, we know that the point M 1 is the midpoint of PQ
1 1 . This implies that PM
1
1 = M 1Q1 .
Therefore, x − x1 = x − x2 .
Solving for x , we get either x − x1 = x − x2 , which is not possible since x1 ≠ x2 , or x − x1 = −( x − x2 ) , which
x +x
implies that x = 1 2 .
2
y + y2
 x + x2 y1 + y2 
. Thus, M  1
In a similar way, we find that y = 1
;
 is the midpoint between P ( x1 , y1 )
2
2 
 2
and Q( x2 , y2 ) .
Dividing a Segment into a Given ratio: Given two points P ( x1 , y1 ) and Q( x2 , y2 ) , find the point C on the
PC m
= , where m and n are two positive integers.
CQ n
PC m
=
(see Figure 8–6).
SoluTion: Let C ( x, y ) be the point on the segment PQ such that
CQ n
segment PQ such that
From the points P, C, and Q, drop lines parallel to the y-axis, and let P1 ( x1 , 0), Q1 ( x2 , 0), C1 ( x, 0) be the
intersections of these lines with the x-axis. From geometry, we know that
PC
PC
1 1
.
=
C1Q1 CQ
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P
0)
P1 (x
(x 1,, 0)
Q
C
P
PP1 (x1, 0)
1
(x 1 , 0)
C1 (x,
(x 0)
, 0)
1
Q
Q1 (x2, 0)
1
(x , 0)
xx
2
Figure 8–6
Therefore,
Rancho Cucamonga High School - Ontario, CA
x − x1 m
=
x2 − x n
n( x − x1 ) = m( x2 − x)
nx − nx1 = mx2 − mx
x(n + m) = nx1 + mx2
x=
nx1 + mx2
n+m
In a similar way, we find that y =
ny1 + my2
.
n+m
 nx + mx2 ny1 + my2 
Thus, C  1
,
.
n+m 
 n+m
8.3
Lines
8 .3 Lines
8.3
Lines
In the previous section, we dealt with points in the plane. In this section, we will deal with straight lines.
We will begin with the following problem:
Perpendicular Bisector: Let P ( x1 , y1 ) and Q( x2 , y2 ) be two points in the plane, and let l be the perpendicular
bisector of the segment PQ .1 Find the equation of l .
1
The perpendicular bisector of the segment PQ is the line that goes through the midpoint of PQ and is perpendicular to PQ.
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yy
l
A (xx,, yy)
P (xx11,, yy11 )
(x2, yy22))
Q (x
O
xx
Figure 8–7
P and Q .
By the distance formula,
AP = ( x − x1 ) 2 + ( y − y1 ) 2
AQ = ( x − x2 ) 2 + ( y − y2 ) 2
Thus, A( x, y ) is on l if and only if
( x − x1 ) 2 + ( y − y1 ) 2 = ( x − x2 ) 2 + ( y − y2 ) 2 .
Squaring both sides and expanding, we get x 2 − 2 x1 x + x12 + y 2 − 2 y1 y + y12 = x 2 − 2 x2 x + x2 2 + y 2 − 2 y2 y + y2 2.
The last equation is equivalent to 2( x2 − x1 ) x + 2( y2 − y1 ) y = x2 2 + y2 2 − x12 − y12 .
Thus, we have obtained a formula for the equation of the perpendicular bisector of a segment given by its
endpoints P ( x1 , y1 ) and Q( x2 , y2 ) . The formula is 2( x2 − x1 ) x + 2( y2 − y1 ) y = x2 2 + y2 2 − x12 − y12 .
1
ExamplE 8.3a: Find the equation of the perpendicular bisector of P (2, − 3) and Q(− ,5) .
2
Solution: The equation of the perpendicular bisector of the given segment PQ is:
1
1
2(− − 2) x + 2(5 − (−3)) y = (− ) 2 + 52 − 22 − (−3) 2 .
2
2
49
.
Simplifying, we get −5 x + 16 y =
4
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SoluTion: Any point A( x, y ) is on the perpendicular bisector l if and only if it is equidistant from
Multiplying both sides by 4 , we get that the equation of the perpendicular bisector of PQ is
−20 x + 64 y = 49 .
The formula we have obtained for a perpendicular bisector will help prove several important theorems.
The first of these theorems concerns the equation of a line in the plane:
TheoreM 1: A line in the plane has the equation ax + by + c = 0 , where a, b, c are real numbers, and a
and b are not both zero.
Proof: Let l be a line in the plane, and regard l as the perpendicular bisector of a certain segment, say,
By the Perpendicular Bisector Formula, l has the equation 2( x2 − x1 ) x + 2( y2 − y1 ) y = x2 2 + y2 2 − x12 − y12 .
Let a = 2( x2 − x1 ) , b = 2( y2 − y1 ) , and c = −( x2 2 + y2 2 − x12 − y12 ) .
We get that l has the equation ax + by + c = 0 .
Now, we must show that a and b are not both zero. This is easy: notice that since P and Q are different
points, x2 − x1 ≠ 0 or y2 − y1 ≠ 0 . Hence, a and b cannot both be zero.
ExamplE 8.3b: Find the equation of the line that goes through the points A (−2,3) and B (0, 4) .
Solution: Let ax + by + c = 0 be the equation of the line through A and B. Recall that a and
b are not both zero.
Since A and B are on the line, their coordinates satisfy this equation. Namely:
(1) − 2 a + 3b + c = 0
(2) 0 + 4b + c = 0
Solving for a and b in terms of c , we get the following equation:
c
c
(3) x − y + c = 0 .
8
4
Multiplying both sides of the equation (3) by 8 and then dividing both sides by c, we get that the
equation of the line that goes through the points A (−2,3) and B (0, 4) is x − 2 y + 8 = 0 .
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PQ . Let ( x1 , y1 ) and ( x2 , y2 ) be the coordinates of P and Q , respectively.
8
4
Multiplying both sides of the equation (3) by 8 and then dividing both sides by c, we get that the
equation of the line that goes through the points A (−2,3) and B (0, 4) is x − 2 y + 8 = 0 .
Since we divided both sides by c , we must show that that c ≠ 0 . Indeed, if c = 0 then from equation
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, and so from
equation
(1) a = 0Outside
. Butthe
and bof Your
are not
both
zero. Hence, c ≠ 0 .
(2) we get b = 0Unauthorized
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8.3.1
Form
8 .3 .1
SLoPe
form
8.3.1Slope
SLoPe
Form
When b ≠ 0 , we can write the equation of a line ax + by + c = 0 as y = −
If we denote −
a
c
x− .
b
b
a
c
= m and − = d , we get the equation y = mx + d .
b
b
A line whose equation is of the form y = mx + d cannot be vertical; that is, it cannot be parallel to the
equation. Take two distinct points with the same x -coordinate: A( x1 , y1 ) and B ( x1 , y2 ) . Since y1 ≠ y2 , it
is not possible that y1 = mx1 + d and y2 = mx1 + d .
The equation y = mx + d is called the slope-intercept form of a line. The reason for this name has to do
with the geometric meanings of the coefficients m and d.
Take any two distinct points A( x1 , y1 ) and B ( x2 , y2 ) on the line whose equation is y = mx + d . Clearly,
(1) y1 = mx1 + d
(2) y2 = mx2 + d
First note that x1 ≠ x2 . It is so because if x1 = x2 , then from equations (1) and (2), y1 = y2 , which implies
that A( x1 , y1 ) and B ( x2 , y2 ) are the same points. But, we are given that they are distinct points.
Now, subtracting equation (1) from equation (2), we get
(3) y2 − y1 = m( x2 − x1 ) .
Since x1 ≠ x2 , we divide both sides of equation (3) by x2 − x1 to get m =
y2 − y1
.
x2 − x1
Thus, m is independent of the choice of the points A( x1, y1) and B ( x2 , y2) on the line. m is called the slope
of the line. It is also called the rise over run. Geometrically, m represents the “steepness” of the line: the
larger | m | is, the steeper the line is.
Now, in the equation y = mx + d what does d represent?
Note that for x = 0 , y = d . Thus, d is the y-intercept—it is the point where the line intersects the y-axis.
The equation y = mx + d has an advantage over the equation ax + by + c = 0 : the values m and d tell us
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y-axis. The reason is simple: there are no two distinct points with the same x -coordinate that satisfy this
Note that for x = 0 , y = d . Thus, d is the y-intercept—it is the point where the line intersects the y-axis.
The equation y = mx + d has an advantage over the equation ax + by + c = 0 : the values m and d tell us
how the line is positioned in the coordinate system. The equation y = mx + d is called the slope form of
a line.
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8.3.2
PoinT-PoinT
Form
8 .3 .2 Point–Point
PoInT-PoInT
form
8.3.2
Form
Determining a line by two points. The equation of a non-vertical line that goes through two distinct points
y −y
P( x1 , y1 ) and Q( x2 , y2 ) is y − y1 = 2 1 ( x − x1 ) .
x2 − x1
Since P ( x1 , y1 ) is on l , its coordinates satisfy equation (1). Namely:
y2 − y1
x1 + d
x2 − x1
y − y1
(3) d = y1 − 2
x1
x2 − x1
(2) y1 =
Substituting for d in equation (1), we get
(4)
y=
y2 − y1
y − y1
x + y1 − 2
x1
x2 − x1
x2 − x1
Moving y1 to the left-hand side and factoring out
(5)
y − y1 =
y2 − y1
( x − x1 ) .
x2 − x1
y2 − y1
, we get
x2 − x1
Theorem 1 asserts that any line in the plane has an equation of the form ax + by + c = 0 . What about the
converse of this theorem? Namely: Does every equation of the form ax + by + c = 0 represent a line?
The following theorem answers this question affirmatively.
TheoreM 2: The graph of the equation ax + by + c = 0 , where a and b are not both zero, is a line.
Proof: Consider first the case where b ≠ 0 .
Recall that in this case the equation ax + by + c = 0 had the form y = mx + d , where m = −
a
c
and d = − .
b
b
A graph of an equation is the set of all points ( x, y ) that satisfy the equation. Let γ be the graph of the
equation y = mx + d .
We will show that there exists a line l that has this equation. This will prove that γ and l are identical.
Hence, γ is a line.
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Proof: Let l be the line through P ( x1 , y1 ) and Q( x2 , y2 ) . We have seen that l has the equation
y −y
y = mx + d , where m = 2 1 . Hence l has the equation:
x2 − x1
y2 − y1
(1) y =
x+d .
x2 − x1
We will show that there exists a line l that has this equation. This will prove that γ and l are identical.
Hence, γ
Take any
a
B 1,
110
110
b
is a line.
c
two distinct points on γ . For simplicity’s sake, we will take the points A 0,
and
b
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. (Check that
these points
are indeed
on γ .)
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b
Let l be the line between A and B. Using the Point-Point Form, the equation of l is
 a c  c
− − −− 
c
b b  b
y=
x− .
1− 0
b
a
c
Simplifying, we get y = − x − .
b
b
So, the graph γ and the line l have exactly the same equation.
Position of a line. How does the position of the line whose equation is ax + by + c = 0 depend on the
coefficients a , b , and c ?
SoluTion: We will distinguish among several cases:
CaSe 1: a = 0 and b ≠ 0 .
c
In this case, the equation of the line is y = − . (See Figure 8–8.) This line is parallel to the x-axis and
b
c
passes through the point (0, − ) (Why?)
b
y
O
xx
B
y = –c/b
(0, –c/b)
–c/b)
Figure 8–8
CaSe 2: a ≠ 0 and b = 0 .
c
In this case, the equation of the line is x = − . (See Figure 8–9.) This line is parallel to the y-axis and
a
c
passes through the point (− , 0) (Why?)
a
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111
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You can work through the case where b = 0 as an exercise on your own.
CaSe 2: a ≠ 0 and b = 0 .
c
In this case, the equation of the line is x = − . (See Figure 8–9.) This line is parallel to the y-axis and
a
c
passes through the point (− , 0) (Why?)
a
yy
x = – c/a
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O
A
111
111
x
(–c/a, 0)
(–c/a,
Figure 8–9
CaSe 3: c = 0 .
In this case, the equation of the line is ax + by = 0 . (See Figure 8–10.) This line goes through the origin
yy
(Why?)
ax + by = 0
O
x
Figure 8–10
8.3.3
Form
8 .3 .3
SLoPe-PoInT
form
8.3.3Slope–Point
SLoPe-PoinT
Form
Determining a line by slope and one point. The equation of a line with slope m that passes through the
point P1 ( x1 , y1 ) is y − y1 = m( x − x1 ) .
Proof: Recall that the slope of a line is independent of the two distinct points we choose on the line
to calculate it.
y − y1
other
than P1 ( x1 , yResource
.
Let
1 ) . Then Guide=• m
118 P( x, y ) be any point on the line
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to calculate it.
Let P ( x, y ) be any point on the line other than P1 ( x1 , y1 ) . Then
y − y1
=m.
x − x1
x1 we get: Duplication
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8.3.4
Positions
of Lines
8 .3 .4
muTuAL
PoSITIonS
of LIneS
8.3.4Mutual
muTuaL
PoSiTionS
oF LineS
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If we are given two lines by their equations, can we determine whether the lines are parallel to each other?
Can we determine whether they are perpendicular to each other? The following two theorems answer
these questions.
TheoreM 4: Let l1 and l2 be two distinct lines, whose equations are y = m1 x + d1 and y = m2 x + d 2 ,
respectively. l1 and l2 are parallel if and only if m1 = m2 .
Since the two lines are different, d1 ≠ d 2 .
Clearly, there is no ordered pair ( x, y ) that satisfies both equations (why?).
Hence, l1 and l2 have no point in common.
This proves that l1 and l2 are parallel.
Now let l1 and l2 be two distinct parallel lines. Therefore, the equations of these lines do not have any
common solution.
Therefore, when subtracting the two equations, we get that the equation 0 = (m2 − m1 ) x + (d 2 − d1 ) has no
solution.
Since this equation has no solution, m1 = m2 .
This is so because if m1 ≠ m2 , then this equation does have a solution. It is x =
d1 − d 2
.
m2 − m1
TheoreM 5: Let l1 and l2 be two lines with non-zero slopes whose equations are y = m1 x + d1 and
1
y = m2 x + d 2 , respectively. l1 and l2 are perpendicular to each other if and only if m2 = − .
m1
Proof: Let P1 ( x1 , y1 ) and P2 ( x2 , y2 ) be two distinct points on l1 . Then, the slope of l1 is
m1 =
y2 − y1
(Note that x2 − x1 ≠ 0 .)
x2 − x1
We have seen that the perpendicular bisector l ' of P1 P2 has the equation 2( x2 − x1 ) x + 2( y2 − y1 ) y =
x22 + y22 − x12 − y12 .
2( x2 − x1 )
x −x
1
From here, the slope of l ' is m ' = −
=− 2 1 =− .
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y1 )
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Proof: Let m1 = m2 = m . Then, the equations of l1 and l2 are, respectively, y = mx + d1 and y = mx + d 2 .
We have seen that the perpendicular bisector l ' of P1 P2 has the equation 2( x2 − x1 ) x + 2( y2 − y1 ) y =
x22 + y22 − x12 − y12 .
2( x2 − x1 )
x −x
1
=− 2 1 =− .
2( y2 − y1 )
y2 − y1
m1
l
l
From geometry, weUnauthorized
can conclude
that
is
perpendicular
to
and
only if l ' is parallel to l2 .
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From here, the slope of l ' is m ' = −
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On the other hand, l ' is parallel to l2 if and only if m2 = m2′ = −
1
.
m1
113
113
8.4
Circles
8 .4 Circles
8.4
Circles
A circle is the set of all points in the plane that are equidistant from a given point in the plane. What is
the equation of the circle? Let’s begin with a simple case where a circle with radius r is centered at the
origin. By the Pythagorean Theorem (see Figure 8–11), for any point A( x, y ) on the circle, x 2 + y 2 = r 2 .
A ((x,
x, yy))
( 0, yy )
rr
O
x
x
((x,
x, 00))
Figure 8–11
Thus, x 2 + y 2 = r 2 is the equation of a circle with radius r centered at the origin.
Now let’s take a circle with a center at any arbitrary point C (m, n) in the plane, and let r be the radius of
the circle. (See Figure 8–12.) By the distance formula, ( x − m) 2 + ( y − n) 2 = r 2
yy
A (x,
( x y)
,y)
rr
C (m,
( m n)
,n)
O
x
Figure 8–12
Thus, ( x − m) 2 + ( y − n) 2 = r 2 is the equation of a circle with radius r centered at C (m, n) .
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yy
Figure 8–12
Thus, ( x − m) 2 + ( y − n) 2 = r 2 is the equation of a circle with radius r centered at C (m, n) .
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8.5 Solving Geometry Problems Using
8.5 Solving Geometry Problems
8 .5
Solving
GeometryGeometry
Problems
Using Coordinate Geometry
Coordinate
using
Coordinate
Geometry
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TheoreM: The diagonals of a parallelogram bisect each other.
Proof: We are given a parallelogram ABCD . To prove this theorem using coordinate geometry, we
must first introduce a coordinate system. We choose the coordinate axes so that AB is on the x-axis and
the point A is at the origin (see Figure 8–13).
D(c, d)
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y
C(x, d)
B(b, 0)
x
Figure 8–13
Therefore, B = (b, 0) .
Let D = (c, d ) .
Since CD is parallel to AB, the y-coordinate of C is d.
Let C = ( x, d ) .
To find x, we use the fact that BC is parallel to AD. We will consider two cases:
CaSe 1. AD and BC are not parallel to the y-axis.
Since they are parallel to each other, their slopes must be equal. Hence,
d
d
=
.
c x −b
From this we get x = b + c .
CaSe 2. AD and BC are parallel to the y-axis.
In this case, c = 0 and x = b = b + c .
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So, in both cases we get
.
121
CaSe 2. AD and BC are parallel to the y-axis.
In this case, c = 0 and x = b = b + c .
So, in both cases we get C = (b + c, d ) .
To show that the diagonals bisect each other, it is sufficient to show that the midpoint of AC is the same
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The midpoint M 1 of AC has the coordinates: x =
115
115
1
1
1
1
⋅ (0 + (b + c)) = (b + c) , y = (0 + d ) = d .
2
2
2
2
1 
1
Therefore, M 1 =  (b + c), d  .
2 
2
The midpoint M 2 of BD has the coordinates: x =
1
1
1
1
⋅ (b + c) = (b + c) , y = (0 + d ) = d .
2
2
2
2
Hence: M 1 = M 2 .
This implies that the diagonals of a parallelogram bisect each other.
TheoreM: The three perpendicular bisectors of the sides of a triangle are concurrent.
Proof: Let ABC be any triangle positioned in a coordinate system as shown in Figure 8–14:
y
C(c, 0)
K(x, y)
A(a, 0)
B(b, 0)
x
Figure 8–14
A = (a, 0) , B = (b, 0) , C = (0, c) .
Let us find the equation of the perpendicular bisector of AB.
Take any point K = ( x, y ) on the perpendicular bisector of AB. Then, since it is equidistant from A and
B, we have: ( x − a ) 2 + ( y − 0) 2 = ( x − b) 2 + ( y − 0) 2
Opening parentheses and simplifying, we get:
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1
(1) x = (a + b) .
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1 
1
Therefore, M 2 =  (b + c), d  .
2 
2
B, we have: ( x − a ) 2 + ( y − 0) 2 = ( x − b) 2 + ( y − 0) 2
Opening parentheses and simplifying, we get:
(1) x =
1
( a + b) .
2
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The equations of the
other two perpendicular bisectors can be found in a similar way. They are
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2ax − 2cy = a − c and 2bx − 2cy = b 2 − c 2 .
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2
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To find the point of intersection of the last two bisectors, we solve their system of equations by subtracting
the second from the first. We get:
2ax − 2bx = a 2 − b 2 ,
From (1) and (2) we get that the point of intersection lies on the first perpendicular bisector. Hence, the
three perpendicular bisectors are concurrent.
TheoreM: The three medians of a triangle are concurrent, and the distance from any vertex to the
intersection point is two-thirds the length of the median drawn from that vertex.
Proof: Let P1 P2 P3 be a triangle in a coordinate system as shown in Figure 8–15:
P1 = ( x1 , y1 ) , P2 = ( x2 , y2 ) , P3 = ( x3 , y3 ) .
y
x1+x2+x3
y1+y2+y3
x
Figure 8–15
 x + x2 y1 + y2
The midpoint of P1 P2 has the coordinates  1
,
2
 2

.

Thus, the point that is two-thirds the way down the median from P3 has the coordinates
1
1
1 ⋅ x3 + 2 ⋅ ( x1 + x2 ) x + x + x
1⋅ y3 + 2 ⋅ ( y1 + y2 ) y + y + y
3
2
2
.
x=
= 1 2 3 , y=
= 1 2
3
3
3
3
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• 2016–2017
can simplyGuide
interchange
indices. (Replace 3 by 1, 1 by 2, and
To find the corresponding
for P1 , we Resource
by 3.) We do the same for P .
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2 x ( a − b) = a 2 − b 2 ,
1
(2) x = (a + b) .
2
1 2


2
2


Thus, the point that is two-thirds the way down the median from P3 has the coordinates
1
1
1 ⋅ x3 + 2 ⋅ ( x1 + x2 ) x + x + x
1⋅ y3 + 2 ⋅ ( y1 + y2 ) y y y
+ + 3
2
2
.
x=
= 1 2 3 , y=
= 1 2
3
3
3
3
To find the corresponding point for P1 , we can simply interchange indices. (Replace 3 by 1, 1 by 2, and 2
by 3.) We do the same for P2 .
But the above expressions for x and y are invariant under a permutation of the indices. This proves the
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But the
above
for x anddecAthlon
y are invariant
under a permutation
the indices. This proves 117
the
theorem.
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Section 8:
SeCTion 8: exercises
Section
8
Exercises
Section
8:
k SeCTion
k k k k k k k k8:
k k exercises
kkkkkkkkkkkkkkkkkkkkkkkkkkkk
Exercises
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
1. Which of the given points A(3, − 4) , B (1, 0) , C (0,5) , D(0, 0) , and E (0,1) belong to the following
circles?of the given points A(3, − 4) , B (1, 0) , C (0,5) , D(0, 0) , and E (0,1) belong to the following
1. Which
a. ( x − 1) 2 + ( y + 3) 2 = 9
circles?
2
+1)y 2 += (25
b.
y + 3) 2 = 9
a. (xx −
b. x 2 + y 2 = 25
2. Find which of the following equations represent a circle. If it is a circle, identify the radius and
the center
2. Find
whichofofthe
thecircle.
following equations represent a circle. If it is a circle, identify the radius and
2
+ y 2 of
+ 8the
x − 4circle.
y + 40 = 0
a. xcenter
the
xx22 +
−=1=−10
x 2y+22 −
y+248−x 4−x42−y2+=y40
b.
a. b.
x2 +
x 2y+2 −
y 24−x 4−x2−y2=y−=1 −1
b. b.
3. The vertices of a triangle ABC are A(4, 6) , B (−4, 0) , C (−1, − 4) . Find the equation of the median
CM.vertices of a triangle ABC are A(4, 6) , B (−4, 0) , C (−1, − 4) . Find the equation of the median
3. The
CM.
4. What are the values of a and b in the equation of the line ax + by = 1 going through the points
?
(1, 2) and
4. What
are (2,1)
the values
of a and b in the equation of the line ax + by = 1 going through the points
(1, 2) and (2,1) ?
5. Three vertices of a parallelogram ABCD are A(1, 0) , B (2,3) , C (3, 2) . Find the coordinates of
fourth
vertex,
find the coordinates
of A
the
point
of intersection
thethe
diagonals.
5. the
Three
vertices
of aand
parallelogram
ABCD are
, B (2,3)
, C (3, 2) . of
Find
coordinates of
(1, 0)
the fourth vertex, and find the coordinates of the point of intersection of the diagonals.
6. The distance of the point A from the x-axis is q1 , and its distance from the y-axis is q2 , and its
= q2 the
= q3y-axis
fromof
thethe
point
theiscoordinates
of A if q1from
.
B (3,
6) is qthe
3 . Find
q1 , and its distance
6. distance
The distance
point
A from
x-axis
is q2 , and its
distance from the point B (3, 6) is q3 . Find the coordinates of A if q1 = q2 = q3 .
7. Find the point that is one-fourth of
the distance from the point A(–1, 2) to the point B(6, 7)
alongthe
the point
segment
7. Find
thatAB.
is one-fourth of the distance from the point A(–1, 2) to the point B(6, 7)
3
124 along the segment AB. USAD Mathematics Resource Guide • 2016–2017
8. Find the point that is five-sevenths of the distance from the point A(− , 2) to the point B (2, 2)
2
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Exercises
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
1
2
distance from the point B (3, 6) is q3 . Find the coordinates of A if q1 = q2 = q3 .
7. Find the point that is one-fourth of the distance from the point A(–1, 2) to the point B(6, 7)
along the segment AB.
3
8. Find the point that is five-sevenths of the distance from the point A(− , 2) to the point B (2, 2)
2
along the segment AB.
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125
part
II
T
R
PA
2
TRIGONOMETRY
Part II:
TrigonomeTry
Trigonometry
Section
9
TRIGonometric
Functions
Ssection
ecTIon 9: trigonometric
Trigonometric
functions
Functions

Trigonometry means triangle measurement. Trigonometry is used to compute the lengths of segments
astronomy, geography, and navigation. To illustrate how trigonometry is used in measurement, let’s
consider the following example.
In Figure 9–1, AD is the height of a mountain sitting on a plane. There is no direct way to measure
the height of the mountain. However, with a special instrument, we can measure the angle in which
we see the summit A of the mountain from a point B on the plane. Let’s denote the measure of
this angle by β . Now, along the segment BD we move a meters toward the mountain, and again
measure the angle in which we see the summit A . Let’s denote the measure of this angle by γ . The
numbers β , γ , and a are sufficient to compute the height AD . In this section of the resource guide,
you will learn trigonometric tools for finding measurements of sides and angles of a triangle.
A
β
B
D
Figure 9–1
We will begin with the trigonometric functions of an acute angle.
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and measures of angles, and therefore it is an important tool in many branches of science, such as
9.1
the
Sine
Function
for
acuteAngles
angles
9 .
1 The
SineSine
Function
for Acutefor
Angles
9.1
The
Function
Acute
Take an acute angle α and position it in a coordinate system so that its vertex is at the origin O and one
of its sides is on the positive x-axis. Let P ( x, y ) be a point on the second side of the angle, and let r be
the distance between P and the origin O . (See Figure 9–2.)
yy
xx
rr
PP(x,
(x , y)
y)
yy
xx
Figure 9–2
y
y
gets larger, and
As α varies from 0° to 90° , the ratio
also varies. Moreover, as α gets larger,
r
r
y
y
as α gets smaller,
gets smaller. The ratio
is called the sine of α and is written as sin α .
r
r
Since α is acute, and y is positive, the sin α is therefore positive. And, since y is smaller than r ,
sin α is smaller than 1 .
We summarize this as follows: For 0° < α < 90° , 0 < sin α < 1 .
At 90° , y = r . Hence: sin 90° = 1 .
At 0° , y = 0 . Hence: sin 0° = 0 .
Is sin α a function? As defined, there is no guarantee that sin α is a function, or in other words there
is no guarantee that for any input α , there exists exactly one output sin α . Thus, to insure that sin α
is a function, we must prove the following theorem, which asserts that the value sin α is independent
of the choice of the point P( x, y ) on the second side of the angle.
Theorem: Let angle α be an acute angle such that its vertex is at the origin and one of its sides is
on the positive x-axis, as is shown in Figure 9–3. Let P( x, y ) and P '( x ', y ') be two distinct points on
y y'
the non-horizontal side of the angle. Call OP = r and OP ' = r ' . Then, = .
r r'
Proof: Drop two perpendiculars PC and PC ' as shown in Figure 9–3.
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O
y
x
P(x, y)
x'
P'(x', y')
r'
y'
y
x
Clearly, OPC and OP ' C ' are similar. Hence,
y y'
= .
r r'
One of the important applications of the sine function is expressed in the following theorem:
Theorem: The area of any triangle is one-half the product of two sides of the triangle
and the sine of the angle between them. That is, the area of  ABC is given by the formula:
1
area ( ABC ) = AB ⋅ AC ⋅ sin α .
2
Proof: We will prove this theorem for the case where α is acute. The theorem, however, is also valid
when α is not acute.
The area of any triangle ABC (see Figure 9–4) is half the product of a side of the triangle and the
1
altitude to that side. Hence, area ( ABC ) = AB ⋅ CH .
2
Figure 9–4
Point C is at distance CH from the line AB. This distance can be seen from the angle CAB (let’s call
CH
= sin α . From here, CH = AC ⋅ sin α . Substituting the value of CH into the
it α ). Therefore, ratio
AC
1
1
initial formula, we get area ( ABC ) = AB ⋅ CH = AB ⋅ AC ⋅ sin α .
2
2
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Figure 9–3
Therefore, area ( ABC ) =
1
AB ⋅ AC ⋅ sin α .
2
9 .2
Tangent
Function
for Acute for
Angles
9.2The
the
tangent
Function
forAcute
acuteAngles
angles
9.2
The
Tangent
Function
As with the sine function, take an acute angle α and position it in a coordinate system so that its vertex
is at the origin O and one of its sides is on the positive x-axis. Let P ( x, y ) be a point on the second side
x
P(x, y)
r
y
Figure 9–5
y
also varies but always remains positive. Moreover, as α gets
x
y
larger, y gets larger and x gets smaller, and so
gets larger. As α gets smaller, y gets smaller and
x
y
gets smaller.
x gets larger, and so
xr
As α varies from 0° to 90° , the ratio
y
is called the tangent of α , and is written as tanα . Indeed tanα is the slope of the line
x
going through the origin O and the point P. For 0 < α < 90 , 0 < tanα < ∞ .
The ratio
As with the sine function, here too we must guarantee that tan α is a function. That is, we must show
that the value of tan α is independent of the choice of the point P( x, y ) on the second side of the
angle.
Theorem: Let angle α be an acute angle such that its vertex is at the origin and one of its sides is
on the positive x-axis, as is shown in Figure 9–6. Let P( x, y ) and P '( x ', y ') be two distinct points on
the non-horizontal side of the angle. Then
y y'
= .
x x'
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of the angle. (See Figure 9–5.)
yy
xx
P(x,
y)
P(x ,y)
P'(x',y')
P'(x',
y')
r'r'
x'
x'
O
yy
y'
y'
xx
C'
C
Figure 9–6
are similar. Hence,
y y'
= .
x x'
9.3
T
the
he
Cosine
and
Functions
9.3
Cosine
and Cotangent
Cotangent
9 .3 The
Cosine
and Cotangent
Functions for
Acute Anglesfor
Functions
for acute angles
Acute Angles
In addition to the functions sine and tangent, we will define two additional trigonometric functions.
The four functions are needed to write trigonometric formulas more effectively.
The cosine of an angle 0° < α < 90° is denoted by cos α and is defined as follows: cos α = sin(90° − α ) .
The cotangent of an angle 0° < α < 90° is denoted by cot α and is defined as follows: cot α = tan(90° − α )
From here, it is easy to see that sin α = cos(90° − α ) and that tan α = cot(90° − α )
Here is why:
By the definition of cosine, cos(90° − α ) = sin(90° − (90° − α )) = sin α .
By the definition of cotangent, cot(90° − α ) = tan(90° − (90° − α )) = tan α .
Thus, in a right triangle ABC (see Figure 9–7), we have the following relations:
a
= cos B
c
b
2. cos A = = sin A
c
a
3. tan A = = cot B
b
1. sin A =
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Proof: Drop two perpendiculars PC and PC ' as shown in Figure 9–6. Clearly, OPC and OP ' C '
4. cot A =
b
= tan B
a
a
b
c
Figure 9–7
There are six basic relations among trigonometric functions. They are referred to as basic identities:
1. tan α =
sin α
cos α
2. cot α =
cos α
sin α
3. tan α ⋅ cot α = 1
4. sin 2 α + cos 2 α = 1
5. tan 2 α + 1 =
1
, cos α ≠ 0
cos 2 α
6. 1 + cot 2 α =
1
, sin α ≠ 0
sin 2 α
Often
1
1
is denoted by csc α and
by sec α .
sin α
cos α
Thus identities 5 and 6 can be written as:
7. tan 2 α + 1 = sec 2 α
8.
1 + cot 2 α = csc 2 α .
To prove these formulas for an acute angle α , let  ABC be a right triangle in C with the acute angle
BAC = α . Let AC = b , CB = a , and AB = c , as shown in Figure 9–8.
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9 .4 Relations
among
Trigonometric
Functions
9.4
relations
among
trigonometric
Functions
9.4
Relations
among
Trigonometric
Functions
C
90°
aa
bb
B
cc
A
Figure 9–8
sin α
cos α
Proof: Since sin α =
tan α =
sin α
.
cos α
2. cot α =
sin α a b a
a
b
a
= : = . By definition, = tan α . Therefore,
and cos α = , the ratio
cos α c c b
c
b
c
cos α
sin α
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1. tan α =
cos α b a b
b
= : =
, we have:
sin α c c a
c
b
cos α
.
Since = cot α , we get cot α =
a
sin α
Proof: Since cos α =
3. tan α ⋅ cot α = 1
Proof: By definition, cot α =
b
a
a b
and tan α = . Hence: tan α ⋅ cot α = ⋅ = 1 .
a
b
b a
4. sin 2 α + cos2 α = 1
Proof: By the Pythagorean Theorem, in a right triangle ABC, we have a 2 + b 2 = c 2 .
Dividing both sides of the equality by the non-zero c 2 , we get
Since
a 2 b2 c 2
+
=
=1
c2 c2 c2
a
b
= sin α and = cos α , we get: sin 2 α + cos2 α = 1 .
c
c
5. tan 2 α + 1 =
1
cos2 α
2
sin 2 α
sin 2 α + cos2 α
1
 sin α 
Proof: tan α + 1 = 
+
1
=
+
1
=
=
.

2
2
cos α
cos α
cos2 α
 cos α 
2
6. cot 2 α + 1 =
1
sin 2 α
2
cos2 α
cos2 α + sin 2 α
1
 cos α 
Proof: cot α + 1 = 
+
1
=
+
1
=
= 2 .

2
2
sin α
sin α
sin α
 sin α 
2
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9.5
trigonometric
Functions
of Special
SpecialAngles
angles
9 .5
Functions
of Special Angles
9.5Trigonometric
Trigonometric
Functions
of
We will now show how the identities we just proved can be used to compute the trigonometric functions
for the angles 30°, 45°, 60°, and 90°.
First, let’s determine the values of all trigonometric functions for 30º and 60º. For this, consider a right
triangle ABC ( AC = b , CB = a , and AB = c ) with acute angles A = α = 30° and B = β = 60° . From
geometry, we know that the hypotenuse c is twice the side opposite the 30º. Hence,
1
.
2
Using the identity sin 2 α + cos2 α = 1 , it is easy to get cos 30° =
We have cos 30° = 1 − sin 2 30° = 1 −
3
.
2
1
3
3
.
=
=
4
4
2
1
. Also, sin 60° = cos 30° , cos 30° =
2
sin α
sin 30° 1
= :
, we compute tan 30° =
Using the identity tan α =
cos α
cos 30° 2
Since sin 30° = cos 60° , cos 60° =
3
.
2
3
1
=
.
2
3
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sin 30° =
a 1
= , and so:
c 2
And
sin 60°
3 1
: = 3.
=
cos 60° 2 2
cos 30°
3 1
cos α
: = 3.
=
, we get cot 30° =
Using the identity cot α =
sin α
sin 30°
2 2
tan 60° =
And
cot 60° =
cos 60° 1 3
1
= :
=
.
sin 60° 2 2
3
Now let’s compute the values of all trigonometric functions for the angle 45º. For this we take a right
isosceles (i.e., having two equal sides) triangle ABC ( AC = a , CB = a , and AB = c ) with β = α = 45° .
By the Pythagorean Theorem, a 2 + a 2 = c 2 . Hence: 2a 2 = c 2.
From this, we get
Since
a
1
.
=
c
2
a
1
= sin 45° = cos 45° , we have: sin 45° = cos 45° =
.
c
2
And
tan 45° = cot 45° =
1 1
:
= 1.
2 2
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We will now conclude with the values of the trigonometric functions for angle 90º. We have already seen
that: sin 90° = 1 and sin 0° = 0 .
From these two facts, we can calculate the rest:
cos 90° = 1 − sin 2 90° = 1 − 1 = 0 .
tan 90° =
sin 90° 1
= =∞.
cos 90° 0
cot 90° =
cos 90° 0
= =0.
sin 90° 1
α
30º
45º
60º
90º
sin α
1
2
1
2
3
2
1
cos α
3
2
1
2
1
2
0
tan α
1
3
1
3
−
cot α
3
1
1
3
0
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The following table summarizes the values of the special angles:
9.6
Functions of
ofAngles of Any
9.6
Ttrigonometric
rigonometric
Functions
9 .6 Trigonometric
Functions
of Angles of Any
Measure
angles of any Measure
Measure
9 .6 .1
DefInITIonS
ProPerTIeS
9.6.1
DeFInItIonS
anD
ProPertIeS
9.6.1 Definitions
andAnD
Properties
The definitions of the trigonometric functions for acute angles we learned in the previous sections are
extendable to angles of any measure. As before, we take any angle α and position it in a coordinate
system such that its vertex is at the origin O and one of its sides is on the positive x-axis. Let P( x, y )
be a point on the second side of the angle, and let PO = r . Here we think of the angle α as the angle
generated by moving counterclockwise from the positive x-axis to the segment PO . See Figure 9–9 .
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yy
PP(x,
(x ,y)y)
xx
rr
yy
xx
O
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Figure 9–9
The definitions of the four trigonometric functions for such an angle α are as follows:
y
r
x
2. cos α =
r
y
3. tan α = , x ≠ 0
x
x
4. cot α = , y ≠ 0
y
1. sin α =
From these definitions, it is easy to derive the following identities:
l
l
sin α
cos α
cos α
cot α =
sin α
tan α =
Since r is positive, the sign of the sine function (that is, whether the function is positive or negative) is
determined by the sign of y , whereas the sign of cosine is determined by the sign of x .
l
For 0 < α < 90° , y > 0 and x > 0 . Hence: sin α > 0 and cos α > 0
l
For 90° < α < 180° , y > 0 and x < 0 . Hence: sin α > 0 and cos α < 0
l
For 180° < α < 270° , y < 0 and x < 0 . Hence: sin α < 0 and cos α < 0
l
For 270° < α < 360° , y < 0 and x > 0 . Hence: sin α < 0 and cos α > 0
α == 00,
°, 90,
90°, 180,
180°, 270,
270°,360
In addition, the following table gives the values of sine and cosine for 
360°::
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α
0º
90º
180º
270º
360º
sin α
0
1
0
−1
0
cos α
1
0
−1
0
1
As for tan α and cot α , these functions are not defined for all angles. tan α is not defined for α = 90°
and α = 270° because for these values x = 0 ; cot α is not defined for α = 0° , α = 180° , and α = 360°
because for these values y = 0 .
l
For 0° < α < 90° , tan α > 0 and cot α > 0
l
For 90° < α < 180° , tan α < 0 and cot α < 0
l
For 180° < α < 270° , tan α > 0 and cot α > 0
l
For 270° < α < 360° , tan α < 0 and cot α < 0
In addition, the following table gives the values of tangent and cotangent for
 == 00,
90,
180,
360°::
α
°, 90
°, 180
°, 270,
270°,360
α
0º
90º
180º
270º
360º
tan α
0
–
0
−
0
cot α
−
0
−
0
−
9.6.2Negative
Negative
aNgles
9 .6 .2
neGATIve
AnGLeS
9.6.2
Angles
Using the above identities, we can determine the values of trigonometric functions for negative angles. As
before, we take any angle α and position it in a coordinate system so that its vertex is at the origin O and
one of its sides is on the positive x-axis. And, as before, we let P ( x, y ) be a point on the second side of the
angle, and PO = r . Here, however, we think of the angle α as the angle generated by moving clockwise
from the positive x-axis to the segment PO . In the exercises at the end of this section, you will prove the
following relations using identities we will learn in this section:
1. sin(−α ) = − sin α
2. cos(−α ) = cos α
3. tan(−α ) = − tan α
4. cot(−α ) = − cot α
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From the definitions of tangent and cotangent we get:
9.7
trigonometric
9 .7 Trigonometric
Identitiesidentities
9.7 Trigonometric Identities
9.7.1
sum
aNd
differeNce
ideNtities
9 .7 .1
Sum
AnD
DIfference
IDenTITIeS
9.7.1 Sum
and
Difference
Identities
The following identities are helpful in computing the measure of many angles:
1. sin(α + β ) = sin α cos β + cos α sin β
2. cos(α + β ) = cos α cos β − sin α sin β
3. sin(α − β ) = sin α cos β − cos α sin β
4. cos(α − β ) = cos α cos β + sin α sin β
Proofs of Identities 1 and 2:
These identities hold for α and β of any measure. Here we will prove them only for the case where
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0 < α + β < 90 .
We first construct a figure like that shown in Figure 9–10.
B
α sinβ
F 90°
1
90°
D
cosβ
sin(α+β)
β
A
α
α+β
cos(α+β)
90°
90°
C
E
Figure 9–10
In this figure we have:
BC = sin(α + β )
AC = cos(α + β )
BD = sin β
AD = cos β
sin(α + β ) = BC = BF + DE
cos(α + β ) = AC = AE − FD
BF = BD ⋅ cos α = sin β cos α
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DE = AD ⋅ sin α = cos β sin α
AE = AD ⋅ cos α = cos β cos α
FD = BD ⋅ sin α = sin β sin α
Therefore, sin(α + β ) = sin α cos β + cos α sin β
and
cos(α + β ) = cos α cos β − sin α sin β .
This completes the proofs for Identities 1 and 2.
We will now prove Identities 3 and 4:
Let α = (α − β) + β . Then,
sinα = sin((α − β) + β) = sin(α − β)cos β + cos(α − β)sin β
cosα = cos((α − β) + β) = cos(α − β)cos β − sin(α − β)sin β .
We treat these two equations as the system of equations with two unknowns sin(α − β ) and cos(α − β ).
Multiplying the first equation by cos β , and the second by − sin β and adding, we get:
sinα cos β − cosα sin β = sin(α − β) ⋅ (cos 2 β + sin 2 β) = sin(α − β) ⋅1 = sin(α − β) .
This completes the proof for Identity 3.
To prove Identity 4, solve the system of equations 1 and 2 for cos(α − β ) . To do this, multiply the
first equation by sin β , and the second equation by cos β , and then add them. You will get:
sinα sin β + cosα cos β = cos(α − β) ⋅ (cos 2 β + sin 2 β) = cos(α − β) ⋅1 = cos(α − β) .
The following identities are also useful:
1. tan(α + β ) =
integer.
2. tan(α − β ) =
integer.
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tan α + tan β
, where α + β ≠ (1 + 2n )90°, α ≠ (1 + 2n )90°, β ≠ (1 + 2n)90°, n is an
1 − tan α ⋅ tan β
tan α − tan β
where α − β ≠ (1 + 2n )90°, α ≠ (1 + 2n )90°, β ≠ (1 + 2n )90°, n is an
1 + tan α ⋅ tan β
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Proofs of Identities 3 and 4:
sinα cos β cosα sin β
+
sin(α + β) sinα cos β + cosα sin β cosα cos β cosα cos β
tan(α + β) =
=
=
=
cos(α + β) cosα cos β − sinα sin β cosα cos β sinα sin β
−
cosα cos β cosα cos β
( tanα ) ⋅1+1⋅ ( tan β ) = ( tanα ) + ( tan β ) .
=
1− tanα tan β
1− tanα tan β
Proof:
The proof for the second identity is similar.
ExamplE 9.7a: Find cos135°
Solution:
1
2
−1⋅
1
2
=−
1
2
.
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cos135° = cos(90° + 45°) = cos90°cos 45° − sin 90°sin 45° = 0 ⋅
ExamplE 9.7b: Find tan 210°
1
tan180° + tan 30°
3 = 1 .
tan 210° = tan(180° + 30°) =
=
1 − tan180°⋅ tan 30° 1 − 0 ⋅ 1
3
3
Solution:
0+
ExamplE 9.7c: Find sin 300°
sin 300° = sin(270° + 30°) = sin 270° cos 30° + co
Solution:
sin 300° = sin(270° + 30°) = sin 270° cos 30° + cos 270° sin 30° == (−1) ⋅
= (−1) ⋅
3
1
3
+ 0⋅ = −
.
2
2
2
3
1
3
+ 0⋅ = −
.
2
2
2
ExamplE 9.7d: Find cot 75°
1
1 − tan 45° tan 30°
3 = 3 −1 = 4 − 2 3 = 2 − 3 .
cot 75° = cot(45° + 30°) =
=
tan 45° + tan 30° 1 + 1
2
3 +1
3
Solution:
1−
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Summarizing: we have the following Sum and Difference Identities:
1. sin(α + β ) = sin α cos β + cos α sin β
2. cos(α + β ) = cos α cos β − sin α sin β
3. sin(α − β ) = sin α cos β − cos α sin β
4. cos(α − β ) = cos α cos β + sin α sin β
integer.
6. tan(α − β ) =
integer.
tan α + tan β
, where α + β ≠ (1 + 2n )90°, α ≠ (1 + 2n )90°, β ≠ (1 + 2n)90°, n is an
1 − tan α ⋅ tan β
tan α − tan β
, where α − β ≠ (1 + 2n )90°, α ≠ (1 + 2n )90°, β ≠ (1 + 2n )90°, n is an
1 + tan α ⋅ tan β
9 .7 .2
IDenTITIeS
9.7.2DouBLe-AnGLe
double-aNgle
ideNtities
9.7.2.
Double-Angle
Identities
From the above identities, it is easy to prove the following identities. You can complete the proofs of these
identities as an exercise on your own:
1. sin 2α = 2 sin α cos α
2. cos 2α = cos 2 α − sin 2 α
3. tan 2α =
2 tan α
1 − tan 2 α
ExamplE 9.7e: Given that sin α =
cot 2α .
4
where 0 < α < 90°, find sin 2α , cos 2α , tan 2α and
5
Solution:
2
9
3
4
cos α = ± 1 − sin 2 α = ± 1 −   = ±
=± .
25
5
5
Since 0° < α < 90° , we take cos α positive, and so cos α =
3
.
5
4 3 24
=
5 5 25
From here, we have sin 2α = 2sin α cos α = 2 ⋅ ⋅
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5. tan(α + β ) =
2
2
7
 3  4 
cos 2α = cos α − sin α =   −   = −
25
 5  5
2
2
4
sin α 5 4
tan α =
= = .
cos α 3 3
5
1
1 3
= =
cot α =
tan α 4 4
3
4
8
2⋅
2 tan α
3 = 3 = − 24
=
tan 2α =
2
2
7
1 − tan α
7
4
−
1−  
9
 3
1
7
cot 2α =
=− .
tan 2α
24
Thus, if sin α =
cot 2α = −
7
.
24
4
24
7
24
where 0 < α < 90° , then sin 2α =
, cos 2α = − , tan 2α = − ,
5
25
25
7
ExamplE 9.7f: Given that cos 2α =
4
and 90° < α < 180°, find tan α + cot α .
5
Solution: Recall that tan 2 2α + 1 =
1
. Hence:
cos 2 2α
4
1−
 
1− cos 2 2α
9
3
5
tan 2α = ±
=±
=±
=± .
2
2
16
4
cos 2α
4
 
5
2
Since 90° < α < 180° , so 180° < 2α < 360° , but cos 2α > 0 , therefore 270° < 2α < 360° and
3
tan 2α = − .
4
To find tan α , we use the identity tan 2α =
We have: −
2 tan α
.
1 − tan 2 α
3
2 tan α
=
, 1 − tan 2 α ≠ 0 .
4 1 − tan 2 α
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To find tan 2α and cot 2α , we first find tan α and cot α :
Solving for tan α , we get: 3 tan 2 α − 8 tan α − 3 = 0
1
3
tan α = 3 or tan α = − .
Since 90° < α < 180° , tan α = −
1
and, therefore, cot α = −3 .
3
1
3
Now we are ready to compute tan α + cot α : tan α + cot α = − − 3 = −3
1
3
9 .7 .3
hALf-AnGLe
IDenTITIeS
9.7.3
Half-aNgle
ideNtities
9.7.3 Half-Angle
Identities
We can also obtain the following identities for half angles:
1 − cos α
2
2
α
1 + cos α
2. cos = ±
2
2
α
sin α
3. tan = ±
2
1 + cos α
α
1 + cos α
4. cot = ±
2
sin α
α
=±
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1. sin
Proof:
Proof of (1):
 α
α
α 
α
α
α
cosα = cos  2  = cos 2 − sin 2 = 1− sin 2  − sin 2 = 1− 2sin 2
2
2 
2
2
2
 2
sin
α
2
=±
1 − cos α
2
Proof of (2):
 α
α
α
α 
α
α
cosα = cos  2  = cos 2 − sin 2 = cos 2 − 1− cos 2  = 2cos 2 −1
2
2
2 
2
2
 2
cos
α
2
=±
1 + cos α
2
Proof of (3):
α
sin
2
α
2 = ± 1− cosα = ± 1− cosα ⋅ 1+ cosα = ± 1− cos α = ± sinα
tan =
=
α
2
1+ cosα
1+ cosα 1+ cosα
±(1+ cosα)
(1+ cosα)2
cos
sin α 2
, where α ≠ 180°n and n is an integer.
=
(1 + cos α )
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Proof of (4): You should complete the proof of (4) on your own as an exercise.
ExamplE 9.7g: Find sin15°.
Solution:
3
1− cos30°
2 =± 2− 3 =± 2− 3 .
sin15° = ±
=±
2
2
4
2
2− 3
.
Since 0° < α < 90° , we take the positive value: sin15° =
2
ExamplE 9.7h: Simplify the expression A = 0.125cos 4α + sin 2 α cos 2 α .
Solution: (Justify each step using the identities you have learned thus far.)
A = 0.125cos 4α + (sin α cos α ) 2
(2sin α cos α ) 2
4
(sin 2α ) 2
A = 0.125cos 4α +
4
1
A = 0.125cos 4α + sin 2 2α .
4
A = 0.125cos 4α +
1
1 1− cos 4α
A = 0.125cos 4α + sin 2 2α = 0.125cos 4α + ⋅
= 0.125cos 4α + 0.125(1− cos 4α) =
4
4
2
= 0.125cos 4α + 0.125 − 0.125cos 4α = 0.125.
Thus, A = 0.125 .
9.7.4
Identities
9 .7 .4 Sum–to–Product
Sum-To-ProDucT
IDenTITIeS
9.7.4
Sum-to-Product
IdentItIeS
In the previous sections, we learned identities of trigonometric functions for the sum and difference
of angles. In this section, we will learn identities that will help us deal with the sum and difference of
trigonometric functions such as sin18° + cos 42° .
Recall the following identities:
1. sin( x + y ) = sin x cos y + cos x sin y
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1−
2. sin( x − y ) = sin x cos y − cos x sin y
Adding Equations 1 and 2, we get sin( x + y ) + sin( x − y ) = 2sin x cos y
Subtracting Equations 1 and 2, we get sin( x + y ) − sin( x − y ) = 2 cos x sin y
Let x + y = α and x − y = β
Adding these two equations, we get: x =
α +β
2
Subtracting these two equations, we get: y =
α −β
2
Substituting the values of x and y in sin( x + y ) + sin( x − y ) = 2sin x cos y , we get:
sin α + sin β = 2sin
α +β
2
cos
α −β
2
.
sin α − sin β = 2 cos
α +β
2
sin
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Substituting the values of x and y in sin( x + y ) − sin( x − y ) = 2 cos x sin y , we get:
α −β
2
In a similar manner, using sum and difference identities for cosine, we can prove that:
1. cos α + cos β = 2 cos
α+β
cos
α −β
2
2
α+β
α −β
sin
2. cos α − cos β = −2sin
2
2
You can complete this proof on your own as an exercise.
ExamplE 9.7i: Compute sin18° + cos 42° .
Solution:
sin18° + sin 42° = 2sin
18° + 42°
18° − 42°
1
cos
= 2sin 30°cos(−12°) = 2 ⋅ cos12° = cos12° .
2
2
2
We will now derive similar formulas for tangent and cotangent :
tanα + tan β =
sinα sin β sinα cos β + cosα sin β sin(α + β)
+
=
=
cosα cos β
cosα cos β
cosα cos β
and
sinα sin β sinα cos β − cosα sin β sin(α − β)
, where
−
=
=
cosα cos β
cosα cos β
cosα cos β
α ≠ 90°(1 + 2n ), β ≠ 90°(1 + 2n ) , and n is an integer.
tanα − tan β =
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Similarly, cotα + cot β =
cosα cos β sin β cosα + cos β sinα sin(β + α)
+
=
=
sinα sin β
sinα sin β
sinα sin β
and
cotα − cot β =
integer.
cosα cos β sin β cosα − cos β sinα sin(β − α)
, α ≠ 180° ⋅ n, β ≠ 180° ⋅ n , and n is an
−
=
=
sinα sin β
sinα sin β
sinα sin β
Summarizing: we have the following Sum-to-Product Identities:
1. sin α + sin β = 2sin
3.
4.
5.
6.
cos
α −β
2
2
α+β
α −β
sin α − sin β = 2 cos
sin
2
2
α+β
α −β
cos α + cos β = 2 cos
cos
2
2
α+β
α −β
cos α − cos β = −2sin
sin
2
2
sin(α ± β )
, α ≠ 90°(1 + 2n ), β ≠ 90°(1 + 2n ) , and n is an integer.
tan α ± tan β =
cos α cos β
sin( β ± α )
, α ≠ 180° ⋅ n, β ≠ 180° ⋅ n , and n is an integer.
cot α ± cot β =
sin α sin β
ExamplE 9.7j: Compute (without calculator)
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2.
α+β
2sin 2 49° −1
.
cos53° − cos37°
Solution: (Justify each step in the solution.)
1− cos98°
−1
1− cos98° −1 cos98°
2sin 49° −1
2
=
=
=
cos53° − cos37° −2sin 45°sin8°
− 2 sin8°
2 sin8°
2
2
Notice, cos 98° = cos(90° + 8°) = − sin 8° .
Thus,
− sin 8°
1
.
=−
2 sin 8°
2
ExamplE 9.7k: Simplify the expression
sin 7 β + sin11β
and find its value for β = 30° .
cos10 β − cos8β
Solution:
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sin 7 β + sin11β 2sin 9 β cos(−2 β )
.
=
−2sin 9 β sin β
cos10 β − cos8β
sin 7 β + sin11β 2sin 9 β cos(−2 β )
cos 2 β
.
=
=−
cos10 β − cos8β
sin β
−2sin 9 β sin β
cos 2 β
cos 60°
1 1
For β = 30° , we have −
=−
= − : = −1
sin β
sin 30°
2 2
Recall that cos(−2 β ) = cos 2 β . Hence:
9 .7 .5
ProDucT-To-Sum
IDenTITIeS
9.7.5
Identities
9.7.5Product–to–Sum
Product-to-Sum
IdentItIeS
The following are three more useful identities; they allow us to transfer the product of trigonometric
functions into the sum or difference of trigonometric functions.
1
cos(α − β) − cos(α + β)
2
1
2. cosα cos β = cos(α + β) + cos(α − β)
2
1
3. sinα cos β = sin(α + β) + sin(α − β)
2
(
)
(
(
)
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1. sinα sin β =
)
Use the sum and difference identities to prove these identities.
ExamplE 9.7l: Compute (without a calculator) A = sin15°cos7° − cos11°cos79° − sin 4°sin86° .
Solution:
1
1
1
sin 22° + sin8° − cos90° + cos68° − cos(−82°) − cos90° .
2
2
2
1
Opening the parentheses, we get A = sin 22° + sin8° − cos68° − cos82° .
2
A=
(
)
(
)
(
(
)
)
(Recall that cos(−82°) = cos82° .) Note, cos 68° = cos(90° − 22°) = sin 22° and
cos82° = cos(90° − 8°) = sin 8° .
We have: A =
1
sin 22° + sin8° − sin 22° − sin8° = 0 .
2
(
)
ExamplE 9.7m: Show that tan 30° + tan 40° + tan50° + tan 60° =
Solution: Recall that tan α + tan β =
8cos 20°
3
.
sin(α + β )
.
cos α cos β
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Hence: tan 30° + tan 40° + tan50° + tan 60° = (tan 30° + tan 60°) + (tan 40° + tan50°) =
sin(30° + 60°) sin(40° + 50°) sin 90°
sin 90°
4
1
+
=
+
=
+
=
cos30°cos60° cos 40°cos50°
3 1 cos 40°cos50°
3 cos 40°cos50°
⋅
2 2
4 cos 40° cos 50° + 3
=
.
3 cos 40° cos 50°
(
)
(
)
(
)
Now we can use the identity cosα + cos β = 2cos
α + β cos α − β .
2
2
4(cos10° + cos30°) 4 ⋅ 2cos 20°cos(−10°) 8cos 20° cos10° 8cos 20°
=
=
=
.
We get
3cos10°
3 cos10°
3
3 cos10°
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
k
k
Sections 9.1 – 9.7
Section
9 .1–9 .7:
SectIonS 9.1–9.7:
exercises
Exercises
Exercises
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
kk
k
k
7 270° < α < 360°
1. Find sin 2α and cos 2α if cos α =
,
.
25
3
1
2. Find sin(α + β ) if sin α = , cos β = − , 90° < α < 180° , 180° < β < 270° .
4
5
3. An 18-ft ladder leans against a building so that the angle between the ground and the ladder is
70 degrees. At what height does the ladder reach the building?
4. In a triangle ABC, the length of BC is 1 inch, and angles A and B are 30° and 45° , respectively.
Find the other two sides of the triangle.
5. Simplify the following expressions:
(
)
2222
a. (cos
cos222218°
18–cos
a.
− cos72
72°) • :2cos27
2cos 27°
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1
To proceed, we use cosα cos β = cos(α + β) + cos(α − β) :
2
1
4 ⋅ cos90° + cos(−10°) + 3
4cos 40°cos50° + 3 = 2
= 2(0 + cos10°) + 3 = 4cos10° + 2 3
3
3 cos 40°cos50°
3 cos10°
3 ⋅ 1 cos90° + cos(−10°)
(0 + cos10°)
2
2

3 
4 cos10° +
2  4(cos10° + cos30°)

3
4cos10° + 2 3
= cos 30° . Hence
Recall that
=
=
2
3cos10°
3cos10°
3cos10°
(
) (
b. cos 2 α + cos 2 β − cos α + β cos α − β
c.
1 + cos α + cos 2α + cos 3α
cos α + 2 cos 2 α − 1
d.
sin 58° cos 52° + sin 52° cos 58°
cos 72° cos 37° + sin 72° sin 37°
)
6. Prove:
a. sin(−α ) = − sin α
b. cos(−α ) = cos α
d. cot(−α ) = − cot α
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
9.8
Graphs
ofTrigonometric
trigonometric
Functions
9.8
Graphs
of
9 .8 Graphs
of Trigonometric
Functions Functions
You
to to
verify
thatthat
the the
graphs
of the
functions,
sine, cosine,
You can
canuse
usea acalculator
calculator
verify
graphs
of six
thetrigonometric
six trigonometric
functions,
sine,tangent,
cosine,
cotangent,
secant, andsec,
cosecant
like the
shownshown
here for
each:
tangent, cotangent,
andlook
csc look
likegraphs
the graphs
here
for each:
y = sin x
Figure 9–11
It is important to note in this graph (Figure 9–11) that for any integer k ,
l
sin(α + 360°k ) = sin α
l
sin(180°k − α ) = sin α
l
sin( −α ) = − sin α
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c. tan(−α ) = − tan α
You can use the identities you learned in the previous sections to verify these identities.
yy==cos
x
cosx
1
–540°
–360°
–180°
–α
180°
0° α
540°
360°
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–1
Figure 9–12
It is important to note that in this graph (Figure 9–12), for any integer k ,
l
cos(α + 360°k ) = cos α
l
cos( −α ) = cos α
You can use the identities you learned in the previous sections to verify these identities.
tanxx
yy== tan
–540°
–360°
–180°
0°
α
180°
180°+α
360°
540°
Figure 9–13
It is important to note in this graph (Figure 9–13) that for any integer k ,
l
tan(α + 180°k ) = tan α
l
tan( −α ) = − tan α
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You can use the identities you learned in the previous sections to verify these identities.
y = cot x
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Figure 9–14
It is important to note in this graph (Figure 9–14) that for any integer k ,
l
cot(α + 180°k ) = cot α
l
cot( −α ) = − cot α
You can use the identities you learned in the previous sections to verify these identities.
y = sec x
1
–540°
–360°
180°
–180°
360°
540°
0°
–1
Figure 9–15
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y = csc x
1
–540°
–360°
180°
–180°
360°
540°
0°
Figure 9–16
Before proceeding, it is recommended that you revisit Section 5.6.
ExamplE 9.8a: Graph the function y = 4 − 2 cos x .
Solution: This function is of the form y = Af ( x) + C , where f ( x) = cos x , A = −2 , C = 4 .
The graph of this function can be obtained from the graph of y = cos x in the following steps:
Step 1: Construct the graph of y = 2 cos x by expanding the graph of y = cos x along the y-axis
by a factor of 2.
2cos x
yy == 2cosx
1
–360°
–180°
180°
360°
540°
0°
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STEP 2: Reflect the graph of \ " FRV [ with respect to the x-axis to get the graph of \ " FRV [ .
y = – 2cos x
y = –2cosx
1
–360°
–180°
180°
360°
0°
STEP 3: Move the graph of \ " FRV [ along the y-axis by 4 units to get the graph of
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\ " FRV [ .
2cosx
yy==44––2cos
x
1
–360°
–180°
180°
360°
0°
EXAMPLE 9.8b: Graph the function \ " FRV [ s .
SOLUTION:
First we rewrite this function as \ " FRV [ s . This function is of the
form \ " $I D[ F & , where I [ " FRV [ , $ " , & " , D " , F " s .
The graph of this function can be obtained from the graph of \ " FRV [ in the following steps:
STEP 1: Move the graph of \ " FRV [ along the x-axis by s to the left to get the graph of
\ " FRV [ s .
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yy== cos
(x+30˚)
cos(x+30°)
1
–360°
–180°
180°
360°
0°
y y==cos
2(x+30˚)
cos2(x+30°)
1
–360°
–180°
180°
360°
0°
Step 3: Construct the graph of y = 5cos 2( x + 30°) by expanding the graph of y = cos 2( x + 30°)
along the y-axis by a factor of 5.
y y==55cos2(x+30°)
cos 2(x+30˚)
1
–360°
–180°
180°
360°
0°
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Step 2: Construct the graph of y = cos 2( x + 30°) by shrinking the graph of y = cos( x + 30°)
1
along the x-axis by a factor of .
2
Step 4: Reflect the graph of y = 5cos 2( x + 30°) with respect to the y-axis to get the graph of
y = −5cos 2( x + 30°) .
yy==5–5cos2(x+30
cos 2(x+30˚)
°)
1
–360°
–180°
180°
360°
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0°
Step 5: Move the graph of y = −5cos 2( x + 30°) along the y-axis by 1 unit to get the graph of
yy== 5–5cos2(x+30°)+1
cos 2(x+30˚)+1
y = −5cos 2( x + 30°) + 1 .
1
–360°
–180°
180°
360°
0°
Graphs of the trigonometric functions are periodic. This means that there exists a number T for which
f ( x + T ) = f ( x) for any x in the domain of the function. T is called the period of the function. For the
cosine and sine functions, the period is 360° because cos(α + 360°k ) = cos α and sin(α + 360°k ) = sin α .
For the tangent and cotangent functions, the period is 180° because tan(α + 180°k ) = tan α and
cot(α + 180°k ) = cot α .
The period of a function can change if the function is of the form y = f (ax) a ≠ 0 . Recall from
Section 5.6 that the graph of the function y = f (ax) shrinks or expands along the x-axis by a factor
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1
. From here we conclude that the period of y = f (ax) is determined from the period of y = f ( x)
a
T
by the following formula: T1 = , where T1 is the period of y = f (ax) and T is the period of y = f ( x) .
a
of
ExamplE 9.8c: Find the period of the function y = −5cos(2 x + 60°) + 1.
Solution: By the formula just discussed, the period is: T1 =
360°
= 180°.
2
When a periodic function is of the form y = Af ( x) , its graph expands or shrinks along the y-axis by the
l
The amplitude of the function y = −5cos(2 x + 60°) + 1 is −5 = 5 .
l
The amplitude of the function y = 4 − 2 cos x is −2 = 2 .
9.9
Inverse
Trigonometric
Functions
9 .
9 Inverse
Trigonometric
Functions Functions
9.9
Inverse
Trigonometric
Clearly, the sine function is a many-to-one function because for a value of y between −1 and 1, there are
infinitely many values of x such that sin x = y , as is shown in Figure 9–17:
= sin
sinxx
yy =
1
α1
α2
α3
–90 °
0° α 4
90°
α5
α6
α7
–1
Figure 9–17
Recall from Section 4 that this implies that the function y = sin x is not invertible. However, if we
restrict the domain of the function to −90° ≤ x ≤ 90° , the function y = sin x over this interval is oneto-one, and, hence, it is invertible.
The inverse of the function y = sin x over the interval −90° ≤ x ≤ 90° is called arcsine and is denoted
by y = arcsin x . Recall, that the domain of function f will be the range of function f −1 (see Section
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factor A . The absolute value A is called the amplitude of the function.
4). Since we restrict the domain of y = sin x to be −90° ≤ x ≤ 90° , the range of y = arcsin x is
−90° ≤ y ≤ 90° . Likewise, the range of y = sin x is the domain of y = arcsin x . Hence the domain
4). Since we restrict the domain of y = sin x to be −90° ≤ x ≤ 90° , the range of y = arcsin x is
). Sinceofwe
of y = sin x to be −90° ≤ x ≤ 90° , the range of y = arcsin x is
y =restrict
arcsin xthe
is: domain
−1 ≤ x ≤ 1 . In summary: y = arcsin x if only if sin y = x and −90° ≤ y ≤ 90° . The
−90° ≤ y ≤ 90° . Likewise, the range of y = sin x is the domain of y = arcsin x . Hence the domain
90° ≤ y graph
≤High
90° .school
yoften
=
sin use
x is9–18.
Likewise,
thex range
of in
the
–1 domain of y = arcsin x . Hence the domain
books
sin
instead of arcsin. Consider the following example: sin–1
y =mathematics
arcsin
is shown
Figure
of
if sin y = x and −90° ≤ y ≤ 90° . The
of y = arcsin x is: −1 ≤ x ≤ 1 . In summary: y = arcsin x if only
1
x is:
y = arcsin
x if only
x and
° ≤1y=≤ sin
90°90.
if sin y = 90,
. The
f y = arcsin
−1 ≤can
x ≤use
1 . In
= 90,
so we
thissummary:
easy (though
semi invalid)
step:
and−90
thus
The graph of
sin
y = arcsin x
graph of y = arcsin x is shown in Figure 9–18.
raph of y = arcsin x is shown in Figure 9–18.
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y = arcsin x
y = arcsin x
Figure 9–18
exampleS:
Figure 9–18
Figure 9–18
1
1
xampleS:
2
2
1
1
arcsin 0 ==90
30°°because
30°° == 0
becausesin
l1 arcsin
1sin90
arcsin = 30° because
sin 30° =
2
2
2
2
exampleS:
l
arcsin = 30° because sin 30° =
because
l arcsin 0
90°not
90°there
= 0 is no angle α for which sin α = 2 .
does
existsin
since
2=
arcsin 0 = 90° because sin 90° = 0

3
3
since there
no°)angle
arcsin 2−does not
= −exist
60° because
sin(is−60
= − α for which sin α = 2 .
does
not
exist
since
there
is
no
angle
for
which
α
arcsin 2
2 sin α = 2 .
 2 

3
3
 arcsin
l
3   −
3 °) = −
 = −60° because sin(−60
arcsin  −
sin(−60°) = −
2
 =−602° because

2
 2 
ExamplE 9.9a: For any a ≤ 1 , arcsin(− a ) = − arcsin a .
l
ExamplE 9.9a: For any a ≤ 1 , arcsin(− a ) = − arcsin a .
ExamplE
9.9a: Let:
For any a ≤ 1 , arcsin(− a ) = − arcsin a .
proof:
(1) arcsin(− a ) = α , −90° ≤ α ≤ 90° .
proof: Let:
proof: Let:
Therefore,
(1) arcsin(sin
− a )α==α−,a−. 90° ≤ α ≤ 90° .
(1) arcsin(− a ) = α , −90° ≤ α ≤ 90° .
Dividing both
Therefore,
sin αsides
= − aof. this equation by −1 , we get − sin α = a .
Therefore, sin α = − a .
Dividing both sides of this equation by −1 , we get − sin α = a .
Dividing both sides of this equation by −1 , we get − sin α = a .
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Recall that sin(−α ) = − sin α . So, sin(−α ) = a .
Note that −90° ≤ −α ≤ 90° . Thus, −α = arcsin a .
Therefore:
(2) α = − arcsin a
From (1) and (2), we get arcsin(− a ) = − arcsin a .
The cosine function, too, is not one-to-one because for a value of y between −1 and 1, there are infinitely
many values of x such that cos x = y , as is shown in Figure 9–19:
yy==cosx
cos x
1
α1
α2
α3
α4
0°
α5
180°
α6
α7
α8
–1
Figure 9–19
This implies that the function y = cos x is not invertible. However, if we restrict the domain of the
function to 0° ≤ x ≤ 180° , the function y = cos x over this interval is one-to-one, and, hence, it is
invertible.
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 1
1
ExamplE 9.9b: arcsin −  = −arcsin   = −30° .
 2
2
The inverse of the function y = cos x over the interval 0° ≤ x ≤ 180° is called arccosine and is denoted
by y = arccos x . Again, since we take the interval 0° ≤ x ≤ 180° as the domain of the function y = cos x ,
the range of the function y = arccos x is 0° ≤ y ≤ 180° . Likewise, the range of y = cos x is the domain
of y = arccos x . Hence the domain of y = arccos x is: −1 ≤ x ≤ 1 . In summary: y = arccos x if and
only if cos y = x and 0° ≤ y ≤ 180° . The graph of y = arcsin x is shown in Figure 9–20.
f(x) = cos–1(x)
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y = arccos x
Figure 9–20
exampleS:
2
2
= 45° because cos 45° =
2
2
l
arccos
l
arccos1 = 0° as cos 0° = 1
l
arccos(−3) does not exist because there is no angle α for which cos α = −3 .
l
1
 1
arccos  −  = −120° because cos120° = − .
2
 2
ExamplE 9.9c: For any a ≤ 1 , arccos(− a ) = 180° − arccos a .
proof: Let:
(1) arccos(− a ) = α , 0° ≤ α ≤ 180° .
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Therefore, cos α = − a .
Using the fact that cos α = − cos(180° − α ) , we get −a = cos α = − cos(180° − α )
or just a = cos(180° − α ) .
Note that 0° ≤ 180° − α ≤ 180°. Thus, 180° − α = arccos a .
Therefore, (2) α = 180° − arccos a .

2
 2
 = 180° − arccos   = 180° − 45° = 135° .
ExamplE 9.9d: arccos −

 2 
2


 
As with the sine and cosine functions, the tangent and cotangent functions too are not one-to-one because
for any value of y there are infinitely many values of x such that tan x = y and cot x = y , as is shown in
the following figures:
yy =
tanx
= tan
x
α1
α 2 –90 °
0°
α 3 90°
α4
α5
Figure 9–21
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From (1) and (2), we get arccos(− a ) = 180° − arccos a .
Figure 9–22
This implies that the functions y = tan x and y = cot x are not invertible. However, if we appropriately
restrict their domains, we can make them invertible functions. For the tangent function, we only
consider the interval −90° < x < 90° , and for the cotangent function, we only consider the interval
0° < x < 180° .
The inverse of the function y = tan x over the interval −90° < x < 90° is called the arctangent, and is
denoted by y = arctan x . The domain of this function is −∞ < x < ∞ because the range of y = tan x
is any number. The range of y = arctan x is −90° < y < 90° . Note that since y = tan x is undefined for
all α = ±90°(1 + 2k ), k ∈  , the angles 90° and –90° will be excluded from the range of y = arctan x .
The inverse of the function y = cot x over the interval −90° < x < 90° is called the arccotangent, and is
denoted by y = arc cot x . The domain of this function is also −∞ < x < ∞ and the range is 0° < y < 180° .
Note, angles 0° and 180° will be excluded from the range of y = arc cot x for y = cot x is undefined for
all α = 180°k , k ∈  .
Thus: y = arctan x if only if tan y = x and −90° < y < 90° . y = arccot x if and only if cot y = x and
0° < x < 180° .
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y = cot x
y = arctan x
Figure 9–23
y = arccot x
Figure 9–24
l
arctan1 = 45° because tan 45° = 1
l
 3
3
arccot   = 60° because cot 60° =
3
 3 
l
arctan − 3 = −60° because tan(−60°) = − 3
l
arc cot(−1) = 135° because cot135° = −1
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ExamplEs:
( )
ExamplE 9.9e: Prove that for all −1 ≤ y ≤ 1 cos(arccos y ) = y .
proof: For any y and x , −1 ≤ y ≤ 1 and 0° ≤ x ≤ 180° , arccos y = x if and only if cos x = y .
Substituting x in the last equation, we get cos x = cos(arccos y ) = y .
In the same way, we can prove that for any y , −1 ≤ y ≤ 1 , sin(arcsin y ) = y ; and for any y ∈ 
l
tan(arctan y ) = y
l
cot(arc cot y ) = y (You can complete the proof of this on your own as an exercise.)
ExamplE 9.9f: Prove that for all −90° ≤ x ≤ 90° arcsin(sin x) = x .
proof: For any y , −1 ≤ y ≤ 1 and any x , −90° ≤ x ≤ 90° , arcsin y = x if and only if sin x = y .
Substituting y in the first equation, we get arcsin y = arcsin(sin x) = x .
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In the same way, we can prove that for any x , 0° ≤ x ≤ 180° , arccos(cos x) = x ; And for any x ,
−90° < x < 90° , arctan(tan x) = x ; And for any x , 0° < x < 180° , arc cot(cot x) = x . (You can complete
the proof of this on your own as an exercise.)
ExamplEs:
l
3
3
)=
2
2
arcsin(sin 30°) = 30°
l
arctan(tan(−60°)) = −60°
l
arc cot(cot 270°) is not defined (why?)
l
cos(arccos
Solution: Let arcsin
Then, sin α =
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2
ExamplE 9.9g: Compute cos(arcsin ) .
5
2
= α where −90° ≤ α ≤ 90° .
5
2
, 0° ≤ α ≤ 90° (why?)
5
2
21
2
.
Therefore, cos α = 1 − sin 2 α = 1 −   = ±
5
5
21
.
Since 0° ≤ α ≤ 90° , cos α =
5
21
=α .
From here we have arccos
5
21
2
= arcsin .
Therefore, arccos
5
5
2
21
21
)=
.
Hence, cos(arcsin ) = cos(arccos
5
5
5
4
ExamplE 9.9h: Compute sin(arctan(− )) .
5
4
Solution: Let arctan(− ) = α , where −90° < α < 90° .
5
4
Then tan α = − , −90° < α < 0° (why?)
5
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To find sin α , we first find cot α using the identity cot α ⋅ tan α = 1 and then we find sin α using
1
.
the identity cot 2 α + 1 =
sin 2 α
cot α =
1
5
=− .
tan α
4
1
16
4
.
=±
=±
25
cot α + 1
41
41
+1
16
4
.
Since −90° < α < 0° , sin α = −
41
sin α = ±
1
=±
2
From here, arcsin(−
4
) =α .
41
4
4
4
.
Hence, sin(arctan( − )) = sin(arcsin( −
)) = −
5
41
41
9.10
Trigonometric
Equations
9 .10 Trigonometric
Equations
9.10
Trigonometric
Equations
The simplest trigonometric equations are equations such as sin x = a , cos x = a , tan x = a and cot x = a .
Since the range of the function y = sin x is −1 ≤ y ≤ 1 , the equation sin x = a has a solution if and only
if −1 ≤ a ≤ 1 .
1
2
ExamplE 9.10a: Solve the equation sin 2 x = − .
Solution: sin 2 x = −
1
2
 1
2 x = arcsin  −  + 360°k
 2
or
 1
2 x = 180° − arcsin  −  + 360°k
 2
1
 1
Since arcsin  −  = − arcsin = −30° , we get 2 x = −30° + 360°k
2
 2
or 2 x = (180° − (−30°)) + 360°k = 210° + 360°k .
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4
4
Therefore, arctan(− ) = arcsin( −
).
5
41
Thus: x = −15° + 90°k
or x = 105° + 180°k
ExamplE 9.10b: Solve the equation
Solution:
1
sin 2 x = 3 .
2
1
sin 2 x = 3
2
sin 2 x = 2 3 .
ExamplE 9.10c: Solve the equation cos( x − 45°) =
Solution: cos( x − 45°) =
x − 45° = arccos
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Since 2 3 > 1 , the equation has no solution.
3
.
2
3
2
3
+ 360°k
2
or
x − 45° = − arccos
Since arccos
3
+ 360°k
2
3
= 30° , we get x − 45° = 30° + 360°k
2
or
x − 45° = −30° + 360°k
Thus: x = 75° + 180°k
or
x = 15° + 180°k
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ExamplE 9.10d: Solve the equation cos(−2 x) = 1 .
Solution: cos(−2 x) = 1
−2 x = arccos1 + 360°k
or −2 x = − arccos1 + 360°k
Since arccos1 = 0° , we get −2 x = 0 + 360°k
x = −180°k
2
3
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Since k varies over all the integers, we can write the solution as: x = 180°k .


ExamplE 9.10e: Solve the equation tan  x − 30°  = 3 .
2
3


Solution: tan  x − 30°  = 3
2
x − 30° = arctan 3 + 180°k
3
2
x − 30° = 60° + 180°k
3
x = 135° + 270°k .
ExamplE 9.10f: Solve the equation cot 2 x = −
Solution: cot 2 x = −
1
.
3
1
3
 1 
2 x = arc cot  −
 + 180°k
3

2 x = 120° + 180°k
x = 60° + 90°k .
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ExamplE 9.10g: Solve the equation sin 4
x
x 1
− cos 4 = .
2
2 2
x
x 1
− cos 4 =
2
2 2
1
 2x
2 x 
2 x
2 x
 sin − cos   sin + cos  =
2
2 
2
2 2

1
x
x
x
x
Since sin 2 + cos 2 = 1 and sin 2 − cos 2 = − cos x , we get − cos x =
2
2
2
2
2
1
Multiplying both sides of the equation by −1 , we get cos x = −
2
 1
x = ± arccos  −  + 360°k
 2
1
 1
Since arccos  −  = 180° − arccos = 180° − 60° = 120° , we get x = ±120° + 360°k .
2
 2
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Solution: sin 4
ExamplE 9.10h: Solve the equation cos x + 3 sin x = 2 .
Solution: cos x + 3 sin x = 2
1
3
cos x +
sin x = 1 .
2
2
1
3
.
Notice that sin 30° = and cos 30° =
2
2
Hence, the equation can be written as sin 30° cos x + cos 30° sin x = 1 .
Using one of the sum identities we learned earlier, we get sin(30° + x) = 1 .
30° + x = 90° + 360°k
x = 60° + 360°k .
ExamplE 9.10i: Solve the equation cos 2 x − cos 6 x = 0 .
Solution: −2sin
2x − 6x
2x + 6x
sin
=0
2
2
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−2sin(−2 x) sin 4 x = 0
2 sin 2 x sin 4 x = 0
In the last equation the product is equal to zero. Therefore, sin 2 x = 0
or sin 4 x = 0
Solving, we get 2 x = 180°k
or 4 x = 180°k
That is, x = 90°k
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or x = 45°k
It is not difficult to see that the last set of solutions comprises the first one. Thus: x = 45°k .
ExamplE 9.10j: Solve the equation 6sin 2 x − 5sin x + 1 = 0 .
Solution: Let sin x = t . Then, we get 6t 2 − 5t + 1 = 0 .
5 ±1
12
1
t=
2
1
or t = .
3
t=
Hence, sin x =
1
1
or sin x = .
3
2
From the first equation we get x = 30° + 360°k
or x = 150° + 360°k
1
3
From the second equation we get x = arcsin + 360°k
1
3
or x = 180° − arcsin + 360°k
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x
x cos x = 2 .
ExamplE
+ cos x = 2 .
sin 2 +
ExamplE 9.10k:
9.10k: Solve
Solve the
the equation
equation sin
2
Since
= sin
sin xx and
Since the
the functions
functions yy =
and yy =
= cos
cos xx
xx
xx
sin
+ cos
= 22 if
sin 2 +
cos xx =
sin 2 and
if and
and only
only if
if each
each of
of the
the addends
addends sin
and
2
2
Solution:
Solution:
cannot
cannot exceed
exceed the
the value
value 1,
1,
cos
is 1.
1. Hence,
Hence, we
we have
have the
the
cos xx is
following
following system
system of
of equations:
equations:
 xx
sin
sin =
=1
1
 2
2
cos x = 1
cos x = 1
xx
Solving
= 90
+ 360
90°° +
360°°kk
Solving the
the first
first equation,
equation, we
we get
get 2 =
2
Solving
Solving the
the second
second equation,
equation, we
we get
get xx =
= 360
360°°kk ..
The
180°° +
+ 720
720°°kk11 =
= 360
360kk22 ??
and kk22 such
such that
that 180
The question
question now
now is:
is: Are
Are there
there integers
integers kk11 and
This
180°° =
= 360
360°°((kk22 −
− 22kk11 ))
This is
is equivalent
equivalent to
to 180
1
1 k − 2k .
The
= k22 − 2k11 .
The last
last equation,
equation, in
in turn,
turn, is
is equivalent
equivalent to
to 2 =
2
This
This obviously
obviously is
is not
not possible,
possible, since
since the
the difference
difference between
between two
two integers
integers cannot
cannot be
be aa fraction.
fraction.
Hence,
Hence, our
our equation
equation has
has no
no solution.
solution.
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Section
9 .9.8
8–9 .–10:
Sections
9.10
SEcTionS
Exercises
SEcTionS 9.8–9.10:
9.8–9.10:Exercises
Exercises
Exercises
Exercises
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1.
1. Which
Which of
of the
the following
following expressions
expressions are
are meaningful?
meaningful?
a.
a. arccos(
arccos(−
−1)
1)
b.
b. tan(arctan(
tan(arctan(−
−3))
3))
22
c.
sin(arcsin(−
− 2 ))
))
c. sin(arcsin(
2
2.
2. Compute
Compute the
the following:
following:
24
24
a.
tan(arcsin 25 ))
a. tan(arcsin
25
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xx =
= 180
180°° +
+ 720
720°°kk ..
1
b. cos(arctan( − ))
3
3. Solve the following equations:
(
)
a. sin −4x + 30° = −
b. 1 − 2sin 2 x =
(
1
2
7
sin x
3
)
(
)
c. sin 15° + x + sin 45° − x = 1
d.
3 sin x − cos x = 3
f. 1 + cos x + cos 2 x = 0
x
x
− sin 4
2
2
1
= 25
h. cot 4 2 x +
sin 4 2 x
g. sin 2 x = cos 4
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9.11 The Law of Sines and Cosines
9.11
The
Law
of SineS
and CoSineS
9 .11 The
Law
of Sines
and Cosines
So far, we have learned how the trigonometric functions can help us solve geometric problems involving
right triangles. In this section, you will learn two theorems that will help us solve geometric problems
involving any triangle.
The Law of Sines: Let ABC be any triangle, with sides a , b , c , and angles α , β , and γ , as shown in
a
b
c
Figure 9–25. Let r be the radius of the circle circumscribing the triangle. Then
=
=
= 2r .
sin α sin β sin γ
Proof:
Let ABC be any triangle, with sides a , b , c , and angles α , β , and γ , as shown in Figure 9–25. We
are to prove:
a
b
c
.
=
=
sin α sin β sin γ
a
b
c
2)
=
=
= 2r .
sin α sin β sin γ
1)
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e. cos 2 x − 3sin x cos x = −1
C
γ
aa
bb
α
A
β
cc
B
Figure 9–25
Proof of 1): We know that
l
l
1
ab sin γ
2
1
area ( ∆ABC ) = bc sin α
2
1
area ( ∆ABC ) = ac sin β .
2
area ( ∆ABC ) =
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l
Hence:
1
1
ab sin γ = bc sin α
2
2
1
1
bc sin α = ac sin β
(2)
2
2
(1)
From (1), we get
(3)
a
c
.
=
sin α sin γ
From (2), we get
(4)
a
b
.
=
sin α sin β
From (3) and (4), we get
(5)
a
b
c
.
=
=
sin α sin β sin γ
Proof of 2): It is sufficient to prove that
a
= 2r .
sin α
Circumscribe the triangle ABC.
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CaSe 1: α = 90°
The center of the circle is on the side BC as is shown in Figure 9–26.
C
In this case, BC = a = 2r .
γ
a
2r
2r
=
=
= 2r .
Hence:
sin α sin 90° 1
bb
a
a
O
A α
β
B
Figure 9–26
CaSe 2 α ≠ 90°
Let CA1 be a diameter. Clearly triangle A1BC is a right triangle (why?) with angle CBA1 = 90° .
C
γ
a
a
bb
A
O
α
β
cc
B
A1
Figure 9–27
We have: sin A1 =
Therefore,
a
a
.
=
A1C 2r
a
= 2r .
sin A1
Since angle A1 and angle α are inscribed within the same arc, they are equal, and, hence, sin α = sin A1 .
This shows that
a
= 2r , as was required.
sin α
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cc
Note, however, that the points $ and % can be on the same side of the diameter &$ , as is shown in
FIGURE 9–28.
C
γ
b
b
aa
A α
O
cc
β
A1
B
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FIGURE 9–28
In this case, F $ " , or $ " F .
Still, VLQ $ " VLQ s F " VLQ F , and so the relation
D
" U still holds.
VLQ F
EXAMPLE 9.11a: In the triangle ABC $& " , *$ " s , *& " s . Find side AB.
SOLUTION: To find side $% " F we will use the Law of Sines:
Since *% " s s s " s , we have F "
š VLQ s
"
VLQ s
š
F
"
.
VLQ % VLQ s
" š " .
C
60°
aa
12
75°
A
cc
B
The Law of Cosines: Let $%& be any triangle, with sides D , E , F , and angles �, �, and �, as shown in
FIGURE 9–29. Then:
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a 2 = b2 + c 2 − 2bc cos α
C
b = a + c − 2ac cos β
γ
2
2
2
c = a + b − 2ab cos γ
2
2
2
aa
bb
α
β
cc
A
B
Figure 9–29
Proof: We will prove the first equality a 2 = b2 + c 2 − 2bc cos α . The proof for the other two are analogous.
Situate the triangle in a coordinate system as is shown in Figure 9–30.
C
(b cos
, b αsin
bsin
) )
C (bcos
α, 
β
a
a
bb
γ
α
A
cc
B
(c,
c 0)
xx
Figure 9–30
We have A = (0, 0) , B = (c, 0) and C = (b cos α , b sin α ) (Why? Hint: drop the altitude CH from the vertex
C to the segment AB and consider the right triangle ACH).
2
2
2
2
2
2
2
2
2
By the distance formula: BC = a = (b cos α − c ) + (b sin α − 0) = b cos α + b sin α − 2bc cos α + c =
b2 (sin 2 α + cos2 α ) − 2bc cos α + c 2 = b2 + c 2 − 2bc cos α .
Thus, a 2 = b2 + c 2 − 2bc cos α .
ExamplE 9.11b: In the triangle ABC A = 60° and AB = 6 6 , AC = 12 . Find side BC.
Solution: By the Law of Cosines Theorem, we have:
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yy
( )
22
1
BC 22 = 6 6 +1222 − 2 ⋅ 6 6 ⋅12 ⋅ cos60° = 216 +144 −144 6 ⋅ = 360 − 72 6 .
2
Then, BC = 360 − 72 6 = 36(10 − 2 6) = 6 10 − 2 6 .
Section 9.11
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Section 9 .11:
Exercises
SeCTion 9.11: exercises
Exercises
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1. Find all the sides and angles of the triangle ABC if
C
γ
a
a
bb
b. A = 80°, a = 16, b = 10
c. a = 14, b = 18, c = 20 .
α
2. Determine the angles of a triangle ABC if its sides are
A
cc
β
B
a. 5, 4, and 4
b. 17, 8, and 15
c. 9, 5, and 6.
3. Find the bisectors in the triangle with sides 5, 6, and 7.
4. A group of students was assigned to determine the width of a river. They marked two trees on
one bank of the river, A and B, and measured the distance between them. The distance was 70
yards. On the other bank of the river, adjacent to the water, they marked another tree C. They
also found that CAB = 12°30′ and ABC = 72°42′ . The students determined the width of the
river from this information. What is the river’s width?
5. Use the Law of Cosines to prove the Pythagorean Theorem.
6. State the converse of the Pythagorean Theorem and prove it using the Law of Cosines.
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9.12
Radians
9 .12 Radians
9.12
Radians
So far we have measured angles by degree units. In this section, we will learn how to measure angles by
length units, called radians. To define the radian measure of an angle α , take any circle with radius r
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a. A = 60°, B = 40°, c = 14
centered at the vertex of F , as shown in FIGURE 9–31, and let O be the length of the arc intercepted by F .
centered at the vertex of F , as shown in FIGURE 9–31, and let O be the length of the arc intercepted by F .
O is considered to be positive if F is a positive angle and negative if F is a negative angle.
O is considered to be positive if F is a positive angle and negative if F is a negative angle.
l
l
r
r
FIGURE 9–31
FIGURE 9–31
O
The radian measure of the angle F is defined to be the ratio, O .
The radian measure of the angle F is defined to be the ratio, U .
U
For this definition to be meaningful, we must show that the radian measure of the angle F is independent
For this definition to be meaningful, we must show that the radian measure of the angle F is independent
of the choice of the radius U . In the exercises at the end of this section, you will be guided to justify this
of the choice of the radius U . In the exercises at the end of this section, you will be guided to justify this
fact.
fact.
O
F ininradians
radian is
Since the ratio, O , is independent of the radius U , we may choose U " . So, the measure of �
F ininradians
radian is
Since the ratio, U , is independent of the radius U , we may choose U " . So, the measure of �
the length of theUarc intercepted by F in a circle of radius 1. When we say the radian measure of an angle
the length of the arc intercepted by F in a circle of radius 1. When we say the radian measure of an angle
F is, for example, 1.5, we mean that the length of the arc intercepted by F in a circle of radius 1 is .
F is, for example, 1.5, we mean that the length of the arc intercepted by F in a circle of radius 1 is .
Thus, it easy to see that:
Thus, it easy to see that:
1. If the degree measure of an angle F is s , then the radian measure of F is U .
1. If the degree measure of an angle F is s , then the radian measure of F is U .
2. If the degree measure of an angle F is s , then the radian measure of F is U .
2. If the degree measure of an angle F is s , then the radian measure of F is U .
U
3. If the degree measure of an angle F is s , then the radian measure of F is U .
3. If the degree measure of an angle F is s , then the radian measure of F is .
And in general:
And in general:
U
4. If the degree measure of an angle F is Qs , then the radian measure of F is U š Q .
4. If the degree measure of an angle F is Qs , then the radian measure of F is š Q .
We can also convert radians to degrees:
We can also convert radians to degrees:
© ¹
5. If the radian measure of an angle F is 1, then the degree measure of F is ª© º¹s .
5. If the radian measure of an angle F is 1, then the degree measure of F is ǻ U ȼ s .
« U »
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α
α
And in
in general:
general:
And
π
 π
 
α is
is ss ,, then
then the
the degree
degree measure
measure of
of α
α is
6. If
If the
the radian
radian measure
measure of
of an
an angle
angle α
⋅⋅ ss  °° ..
is 
6.
 180
180 
33π
π , we mean that the radian measure of the angle whose degree measure is 135°
When
135°° =
=
, we mean that the radian measure of the angle whose degree measure is 135°
When we
we write
write 135
44
3π
π
3
..
is
is
44
examPLeS:
examPLeS:
l
l
l
l
l
l
π
π ⋅ 360° = 2π
360
360°° =
=
⋅ 360° = 2π
180
180°°
π
3π
135°° =
135°° =
= π ⋅⋅135
= 3π
135
180
44
180°°
180
180°° ⋅ 2.5 ≈ 143°
2.5
=
⋅ 2.5 ≈ 143°
2.5 =
π
π
π 180
180°° π
π
π
30°° ..
=
= 30
=
⋅⋅ =
66
π
π 66
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Section 9 .9.12
12:
Section
SeCTion 9.12: exercises
Exercises
Exercises
Exercises
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1.
1. Find
Find the
the radian
radian measure
measure for
for each
each angle.
angle.
°° 
15°
15°
30°
30°
45°
45°
60°
60°
90°
90°
120°
120°
150°
150°
180°
180°
225°
225°
270°
270°
315°
315°
360°
360°
radians
radians
2.
2. Determine
Determine to
to which
which quarter
quarter each
each of
of the
the following
following angles
angles belongs:
belongs:
22π
3
π
π
5
π
7
π
2
π
15
2
π , 3π , π , 5π , 7π , 2π , 35π , 146π , 15 π , 1, 8, 2 .
,
, ,
,
,
, 35π , 146π ,
π , 1, 8, π .
33
44 66 18
22
18 12
12 55
π
3.
3. Find
Find the
the radian
radian measures
measures of
of the
the angles
angles whose
whose degree
degree measures
measures are:
are:
a.
a. 45
45°°30
30′′
b.
b. 12
12′′
4.
4. Solve
Solve the
the following
following equations:
equations:
a.
tan(3sin π
π xx)) =
= 33
a. tan(3sin
b.
sin(cos 33 xx)) =
= 11
b. sin(cos
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