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Chapter 1
Equations and Inequalities
Copyright © 2004 Pearson Education, Inc.
1.1
Linear Equations
Copyright © 2004 Pearson Education, Inc.
Equations

An equation is a statement that two expressions
are equal.

x + 3 =11
9x = 3x + 6x
x2 + 4x – 8 = 0
To solve an equation means to find all numbers that
make the equation true. The numbers are called
solutions or roots of the equation. A number that is a
solution of an equation is said to satisfy the equation,
and the solutions of an equation make up its solution
set. Equations with the same solution set are
equivalent equations.
Copyright © 2004 Pearson Education, Inc.
Slide 1-4
Addition and Multiplication Properties of
Equality
For real numbers a, b, and c:
a = b and a + c = b + c
are equivalent
That is, the same number
may be added to both sides
of an equation without
changing the solution set.
Copyright © 2004 Pearson Education, Inc.
For real numbers a, b, and c:
If c  0,
then a = b and ac = bc
are equivalent.
That is, both sides of an
equation may be multiplied
by the same nonzero
number without changing
the solution set.
Slide 1-5
Solving Linear Equations

Linear Equations in One Variable

A linear equation in one variable is an equation that
can be written in the form
ax + b = 0
where a and b are real numbers with a  0

A linear equation is also called a first-degree equation
since the greatest degree of the variable is one.
Copyright © 2004 Pearson Education, Inc.
Slide 1-6
Solving a Linear Equation
Example: Solve 6p + 2  10(p + 2) = 12p + 14.
Solution: 6p + 2  10(p + 2) = 12p + 14
6p + 2 10p –20 = 12p + 14
4p  18 = 12p + 14
4p + 4p  18 = 12p + 14 + 4p
– 18 = 16p + 14
32 = 16p
1
1
 32  16 p
16
16
2  p
Copyright © 2004 Pearson Education, Inc.
Slide 1-7
Solving a Linear Equation continued

Check:
6p + 2  10(p + 2) = 12p + 14
6(2) + 2  10(2 + 2) = 12(2) + 14
12 + 2 = 24 + 14
10 =  10
Since replacing p with 2 results in a true
statement, 2 is the solution of the given
equation. The solution set is{2}.
Copyright © 2004 Pearson Education, Inc.
Slide 1-8
Clearing Fractions

If an equation has fractions as coefficients, begin by
multiplying both sides by the least common denominator
to clear the fractions.
Example:
Check:
2t  5 1 t  2
 
10
5
6
3  2t  5   6  5t  10
6t  21  5t  10
t  11
2  11  5
1 11  2

10
5
6
17 1
9
  
10 5
6
45
45


30
30

The solution set is {11}
Copyright © 2004 Pearson Education, Inc.
Slide 1-9
Identities, Conditional Equations and
Contradictions
1.
2.
3.
If solving a linear equation leads to a true
statement such as 0 = 0, the equation is an
identity. Its solution set is {all real numbers}.
If solving a linear equation leads to a single
solution such as x = 3, the equation is
conditional. The solution set consists of a
single element.
If solving a linear equation leads to a false
statement such as 10 = 8, the equation is a
contradiction. Its solution set is .
Copyright © 2004 Pearson Education, Inc.
Slide 1-10
Examples of Identifying Types of Equations
a)
b)
c)
4(x  2) + 6x = 2(x + 4)
4x + 8 + 6x = 2x + 8
2x + 8 = 2x + 8
0=0
identity
6x + 5 = 23
6x = 18
x=3
conditional equation
5(3x + 2) = 15x 8
15x + 10 = 15x 8
10 =  8
Copyright © 2004 Pearson Education, Inc.
contradiction
Slide 1-11
Solving for a Specified Variable



A formula is an example of a literal
equation. The methods used to
solve a linear equation can be used
to solve some literal equations for a
specified variable.
Example: Solve p = 6x +2z for z.
Solution: p  6 x  2 z
p  6x  2z
p  6x
z
2
Copyright © 2004 Pearson Education, Inc.
Slide 1-12
Simple Interest Formula

The formula for simple interest is I = Prt.


Example: Stanley has $8000 to invest in a mutual fund. The
funds with higher rates of returns also have greater risk. Stanley
wants a return of at least $300 every six months. What is the
lowest rate of return that will meet his goal?
Solution:
I
r
r
Pt
300
8000 
1
2
r  7.5%

Stanley must chose a rate with a return of at least 7.5%
Copyright © 2004 Pearson Education, Inc.
Slide 1-13
1.2
Applications and Modeling with
Linear Equations
Copyright © 2004 Pearson Education, Inc.
Solving an Applied Problem

Step 1

Step 2

Step 3
Read the problem carefully until you
understand what is given and what is to be
found.
Assign a variable to represent the
unknown value, using diagrams or tables
as needed. Write down what the variable
represents. If necessary, express any other
unknown values in terms of the variable.
Write an equation using the variable
expression(2).
Copyright © 2004 Pearson Education, Inc.
Slide 1-15
Solving an Applied Problem continued

Step 4
Solve the equation.

Step 5
State the answer to the problem.
Does it seem reasonable?

Step 6
Check the answer in the words of the
original problem.
Copyright © 2004 Pearson Education, Inc.
Slide 1-16
Example--Geometry

The perimeter of a rectangle is 108 cm. The
length is 8 cm longer than the width. Find the
dimensions of the rectangle.

Step 1 Read the problem. We must find the
dimensions of the rectangle.

Step 2 Assign a variable.
x = width
x + 8 = length
Copyright © 2004 Pearson Education, Inc.
Slide 1-17
Example--Geometry continued

Step 3 Write an equation.
Perimeter = x + x + x + 8 + x + 8

Step 4 Solve the equation.
108 = x + x + x + 8 + x + 8
108 = 4x + 16
92 = 4x
23 = x
Copyright © 2004 Pearson Education, Inc.
Slide 1-18
Example--Geometry continued

Step 5 State the answer. The width of the
rectangle is 23 cm, the length of the
rectangle is 23 + 8 = 31cm.

Step 6 Check. The perimeter is 108 cm.
31 + 31 + 23 + 23 = 108
Copyright © 2004 Pearson Education, Inc.
Slide 1-19
Motion Problems

In a motion problem, the components distance,
rate, and time, are denoted by the letters d, r,
and t, respectively. (The rate is also called the
speed or velocity. Here, rate is understood to be
constant.) These variables are related by the
equation
d
d
d  rt and its related forms r  and t  .
t
r
Copyright © 2004 Pearson Education, Inc.
Slide 1-20
Example

Nicholas and Catherine are traveling to a math
conference. The trip takes 1 ½ hours for
Catherine and 2 ½ hours for Nicholas, since he
lives 75 miles farther away. Nicholas travels 6
mph faster than Catherine. Find their average
rates.

Step 1 Read the problem. We are looking for
Catherine’s and Nicholas’ average rate.
Copyright © 2004 Pearson Education, Inc.
Slide 1-21
Example continued


Step 2 Assign a variable. Let x = Catherine’s
rate. Then x + 6 = Nicholas’ rate.
Summarize the information in a table.
r
t
d
Catherine
x
1.5
1.5x
Nicholas
x+6
2.5
2.5(x + 6)
Step 3 Write an equation.
2.5(x + 6) = 1.5x + 75
Copyright © 2004 Pearson Education, Inc.
Slide 1-22
Example continued

Step 4 Solve.
2.5( x  6)  1.5 x  75
2.5 x  15  1.5 x  75
x  60

Step 5 State the answer. Catherine’s rate is 60
mph and Nicholas’ rate is 60 + 6 = 66 mph.

Step 6 Check
Distance traveled by Catherine 60(1.5) = 90
Distance traveled by Nicholas 66(2.5) = 165
The difference in the distances 165  90 = 75
Copyright © 2004 Pearson Education, Inc.
Slide 1-23
Work Rate Problems

If a job can be done in t units of time then the rate of
work is 1/t of the job per time unit. Therefore,
rate  time = portion of job completed.

If the letters r, t, and A represent the rate at which work
is done, the time, and the amount of work accomplished,
respectively, then A = rt.

Amounts of work are often measured in terms of the
number of jobs accomplished. For instance, if one job is
accomplished in t time units, then A = 1 and r = 1/t.
Copyright © 2004 Pearson Education, Inc.
Slide 1-24
Mixture Problems

In mixture problems, the rate (percent) of
concentration is multiplied by the quantity to get
the amount of pure substance present. Also, the
concentration in the final mixture must be
between the concentrations of the two solutions
making up the mixture.

Example: How many gallons of a 10% solution
acid solution must be mixed with 5 gallons of a
20% acid solution to obtain a 15% acid solution?
Copyright © 2004 Pearson Education, Inc.
Slide 1-25
Mixture Problems continued



Read the problem. We must find the required
number of gallons of 10% solution.
Assign a variable. Let x = number of gallons of
10% solution.
Strength
Liters of Solution
Liters of pure acid
10%
x
.10x
20%
5
.20(5)
15%
x+5
.15(x + 5)
Write an equation. .10x + .20(5) = .15(x + 5)
Copyright © 2004 Pearson Education, Inc.
Slide 1-26
Mixture Problems continued



Solve. .10 x  .20(5)  .15( x  5)
.10 x  1  .15 x  0.75
.25  .05 x
5 x
State the answer. 5 gallons of 10% solution should
be mixed with 5 gallons of 20% solution, given 5 + 5
= 10 gallons of 15 percent solution.
Check. .10(5) + .20(5) = .15(5 + 5)
0.5 + 1.0 = 1.5
Copyright © 2004 Pearson Education, Inc.
Slide 1-27
Investment Problem

Antonia sold a house for $210,000. He invests
some of the money in an account that pays 4%
interest. He invests the remaining amount in an
investment that pays 7%. His total annual
interest earned is $12,540. How much was
invested in each account?

Read. We must find the amount invested at
each rate.
Copyright © 2004 Pearson Education, Inc.
Slide 1-28
Investment Problem continued



Assign a variable.
Let x = the dollar amount invested at 4%
Let 210,000  x = amount invested at 7%
Write an equation.
.04x + .07(210,000  x) = 12,540
Solve. .04 x  .07(210,000  x)  12540
.04 x  14,700  .07 x  12,540
.03 x  2160
x  72,000
Copyright © 2004 Pearson Education, Inc.
Slide 1-29
Investment Problem continued

State the answer. Antonio invested $72,000 at
4% and $210,000 - $72,000 = $138,000 at 7%.

Check.
72,000(.04) + 138,000(.07) = 12,540
2880 + 9660 = 12,540
Copyright © 2004 Pearson Education, Inc.
Slide 1-30
1.3
Complex Numbers
Copyright © 2004 Pearson Education, Inc.
Basic Concepts



i2 = 1
i is the imaginary unit
a + bi are called complex numbers


a is the real part
b is the imaginary part
If a  0, then
Copyright © 2004 Pearson Education, Inc.
a  i a
Slide 1-32
Examples

a)
25  i 25  5i

c)
125  i 125  i 25  5  5i 5

d)
3   3  i 3  i 3
i
2
 3
 1  3
2
b)
e)
30  i 30
98 i 98

49
49
i
98
i 2
49
 3
Copyright © 2004 Pearson Education, Inc.
Slide 1-33
Addition and Subtraction of Complex
Numbers

For complex numbers a + bi and c + di,
(a  bi )  (c  di )  (a  c)  (b  d )i
(a  bi )  (c  di )  (a  c)  (b  d )i

Examples
(4  6i) + (3 + 7i)
= [4 + (3)] + [6 + 7]i
=1+i
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(10  4i)  (5  2i)
= (10  5) + [4  (2)]i
= 5  2i
Slide 1-34
Multiplication of Complex Numbers

For complex numbers a + bi and c + di,
(a  bi)(c  di)  (ac  bd )  (ad  bc)i.

The product of two complex numbers is found by
multiplying as if the numbers were binomials and
using the fact that i2 = 1.
Copyright © 2004 Pearson Education, Inc.
Slide 1-35
Examples

(2  4i)(3 + 5i)

(7 + 3i)2
 2(3)  2(5i )  4i (3)  4i (5i )
 7 2  2(7)(3i )  (3i) 2
 6  10i  12i  20i
 49  42i  9i 2
 6  2i  20( 1)
 26  2i
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2
 49  42i  9(1)
 40  42i
Slide 1-36
Powers of i

i1 = i
i5 = i
i9 = i

i 2 = 1
i 6 = 1
i10 = 1

i 3 = i
i 7 = i
i11 = i

i4 = 1
i8 = 1
i12 = 1
and so on.
Copyright © 2004 Pearson Education, Inc.
Slide 1-37
Simplifying Examples

i17
i4
=1

i4
1 1
 1
4
i
1
i17 = (i4)4 • i
= 1•i
=i
Copyright © 2004 Pearson Education, Inc.
Slide 1-38
Property of Complex Conjugates
For real numbers a and b,
(a + bi)(a  bi) = a2 + b2.
The product of a complex
number and its conjugate
is always a real number.
Copyright © 2004 Pearson Education, Inc.

Example
5  3i
2i
(5  3i )(2  i )

(2  i )(2  i )
10  5i  6i  3i 2

4  i2
7  11i

5
7 11i
 
5 5
Slide 1-39
1.4
Quadratic Equations
Copyright © 2004 Pearson Education, Inc.
Quadratic Equation in One Variable

An equation that can be written in the form
ax2 + bx + c = 0,
where a, b, and c are real numbers with a  0, is
a quadratic equation.
Examples:
x2 = 36
Copyright © 2004 Pearson Education, Inc.
5x2 + 5x  7 = 0
Slide 1-41
Zero-Factor Property

If a and b are complex
numbers with ab = 0,
then a = 0 or b = 0 or
both.

Example:
15 x 2  x  2
15 x 2  x  2  0
(5 x  2)(3 x  1)  0
5x  2  0
2
x
5
Copyright © 2004 Pearson Education, Inc.
or
or
3x  1  0
1
x
3
Slide 1-42
Square Root Property

If x2 = k, then x  k

Examples
or x   k .
x  19
x 2  36
x   19
x  6i
2
( x  5) 2  16
x  5  4
x 54
x 54 9
x  5 4 1
Copyright © 2004 Pearson Education, Inc.
Slide 1-43
Completing the Square
To solve ax2 + bx + c = 0, a  0, by completing the square:

Step 1

Step 2

Step 3
If a  1, multiply both sides of the
equation by 1/a.
Rewrite the equation so that the constant
term is alone on one side of the equal
sign.
Square half the coefficient of x, and add
this square to both sides of the equation.
Copyright © 2004 Pearson Education, Inc.
Slide 1-44
Completing the Square continued

Step 4

Step 5
Factor the resulting trinomial as a
perfect square and combine terms on the
other side.
Use the square root property to
complete the solution.
Copyright © 2004 Pearson Education, Inc.
Slide 1-45
Example a = 1






x2  6x  12 = 0
Step 1 Not necessary since a = 1.
Step 2
x2  6x = 12
2
1
Step 3 x2  6x + 9 = 12 + 9
 2 (6)  9
Step 4
(x  3)2 = 21
Step 5
x  3   21
x  3  21
Copyright © 2004 Pearson Education, Inc.
Slide 1-46
Example a  1

Complete the square.
6 x 2  14 x  1  0
7
1
x  x 0
3
6
7
1
2
x  x
3
6
7
49 1 49
2
x  x
 
3
36 6 36
2
Copyright © 2004 Pearson Education, Inc.
Slide 1-47
Example a  1 continued
2
7
55

x  
6  36

7
55
x 
6
36
7
55
x 
6
6
 7  55 
The solution set is 
.
 6 
Copyright © 2004 Pearson Education, Inc.
Slide 1-48
Quadratic Formula

The solutions of the quadratic equation
ax2 + bx + c = 0, where a  0 are
b  b 2  4ac
x
.
2a
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Slide 1-49
Example

Solve: 3x2 = x  5.
3x 2  x  5  0
a  3, b  1, c  5
(1)  (1) 2  4(3)(5)
x
2(3)
1  1  60
x
6
1  59
x
6
1  i 59
x
6
Copyright © 2004 Pearson Education, Inc.
Slide 1-50
Solving for a Variable that is Squared

Solve the given formula for r, V =  r2h
V
 r2
h
V

 r2
h
 V
h

r
h h
V h
r
h
Disregard the negative solution since r can not be negative.
Copyright © 2004 Pearson Education, Inc.
Slide 1-51
Discriminant

The quantity under the radical in the quadratic
formula b2  4ac, is called the discriminant.
Discriminant
Number of Solutions
Kind of Solutions
Positive, perfect square
Two
Rational
Positive, but not a
perfect square
Two
Irrational
Zero
One (a double solution)
Rational
Negative
Two
Nonreal complex
Copyright © 2004 Pearson Education, Inc.
Slide 1-52
Examples

Determine the number of distinct solutions and
tell whether they are rational, irrational, or
nonreal complex numbers.
3x2 + x  5 = 0
b2  4ac =
12(4)(3)(5)= 61
Two irrational
solutions
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x2  12x = 36
x2  12x + 36 = 0
b2  4ac =
(12)2(4)(36)= 0
One rational double
solution
3x2  x + 2 = 0
b2  4ac =
(1)2(4)(2)(3)= 23
Two nonreal
complex solutions
Slide 1-53
1.5
Applications and Modeling with
Quadratic Equations
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Problem Solving

When solving problems that lead to quadratic
equations, we may get a solution that does not
satisfy the physical constraints of the problem.
For example, if x represents a width and two
solutions of the quadratic equation are 9 and 1,
the value 9 must be rejected since a width must be
a positive number.
Copyright © 2004 Pearson Education, Inc.
Slide 1-55
Example

A piece of machinery is capable of producing
rectangular sheets of metal such that the length
is three times the width. Furthermore, equalsized squares measuring 5 in. on a side can be
cut from the corners so that the resulting piece
of metal can be shaped into an open box by
folding up the flaps. If specifications call for the
volume of the box to be 1435 in.3, what should
the dimensions of the original piece of metal be?
Copyright © 2004 Pearson Education, Inc.
Slide 1-56
Solution


Step 1
Read the problem. We must find the
dimensions of the original piece of metal.
Step 2
Assign a variable. The length is 3 times the
width, let x = width and 3x = length
The box is formed by cutting 5 + 5 = 10 inches from both
the length and width. The length and width of the box are
shown on the figures.
Copyright © 2004 Pearson Education, Inc.
Slide 1-57
Solution continued

Step 3

Step 4
Write an equation. V = lwh
Volume = length  width  height
1435 = (3x  10)(x  10)(5)
Solve the equation.
1435  15 x 2  200 x  500
0  15 x 2  200 x  935
0  3 x 2  40 x  187
0  (3 x  11)( x  17)
3 x  11  0
x
Copyright © 2004 Pearson Education, Inc.
or
11
or
3
x  17  0
x  17
Slide 1-58
Solution continued

Step 5
State the answers. Only 17 satisfies
the restriction. Thus, the dimensions of
the original piece of metal should be 17
in. by 3(17) = 51 in.

Step 6
Check. The length of the bottom of the
box is 51  2(5) = 41 in. The width is
17  2(5) = 7 in. The height is 5 in.
So, the volume of the box is
41(7)(5) = 1435 in.3
Copyright © 2004 Pearson Education, Inc.
Slide 1-59
The Pythagorean Theorem


In a right triangle, the sum of the squares of the
lengths of the legs is equal to the square of the
length of the hypotenuse.
a 2 + b2 = c2
Leg a
Hypotenuse
c
Leg b
Copyright © 2004 Pearson Education, Inc.
Slide 1-60
Example

A kite is flying on 70 feet of string. Its vertical
distance from the ground is 15 feet more than its
horizontal distance from the person flying it.
Assuming that the string is being held at ground
level, find its horizontal distance from the person
and its vertical distance from the ground.
Copyright © 2004 Pearson Education, Inc.
Slide 1-61
Example continued

Step 1

Step 2

Step 3
Read the problem. We are finding the
horizontal and vertical distance.
Assign a variable.
Let x = the horizontal distance
Let x + 15 = the vertical distance
Write an equation.
x2 + (x + 15)2 = 702
70 ft of string
x + 15
x
Copyright © 2004 Pearson Education, Inc.
Slide 1-62
Example continued

Step 4
Solve:
x 2  ( x  15) 2  70 2
x 2  x 2  30 x  225  4900
2 x 2  30 x  4675  0
a  2, b  30, c  4675
30  (30) 2  4(2)( 4675)
x
2(2)
30  900  37400
x
4
30  10 383
x
4
x  56.43
x  41.43
Copyright © 2004 Pearson Education, Inc.
Slide 1-63
Example continued

Step 5
State the answer. Since x represents a
length, the horizontal distance could
only be 41.43 feet. The vertical
distance would be 41.43 + 15 = 56.43
feet.

Step 6
Check. The lengths 41.43, 56.43 and
70 satisfy the Pythagorean theorem.
Copyright © 2004 Pearson Education, Inc.
Slide 1-64
Modeling with Quadratic Equations


The bar graph shows the sales of SUV’s in the
United States, in millions. The quadratic
equation S = .016x2 + .124x + .787 models sales
of SUV’s from 1990 to 2001, where S represents
sales in millions and x = 0 represents 1990, x =
1 represents 1991, and so on.
a) Use this model to predict sales in 2000 and
2001. Compare the results to the actual figures
of 3.4 million and 3.8 million from the graph.
Copyright © 2004 Pearson Education, Inc.
Slide 1-65
Modeling with Quadratic Equations
continued

b) According to the
model, in what year
do sales reach 3
million? (Round down
to the nearest year.)
Is the result accurate?
Copyright © 2004 Pearson Education, Inc.
Slide 1-66
Solution

a) S  .016 x 2  .124 x  .787
S  .016(10) 2  .124(10)  .787
S  3.6 million
For 2000, x = 10
S  .016(11) 2  .124(11)  .787
S  4.1 million

For 2001, x = 11
The predictions are greater than the actual
figures of 3.4 and 3.8 million.
Copyright © 2004 Pearson Education, Inc.
Slide 1-67
Solution continued

b)
3  .016 x 2  .124 x  .787
0  .016 x 2  .124 x  2.213
.124  (.124) 2  4(.016)( 2.213)
x
2(.016)
x  16.3
or x  8.5

Reject the negative solution and round 8.5 down to 8
The year 1998 corresponds to x = 8. The model is a bit
misleading, since SUV sales did not reach 3 million until
2000.
Copyright © 2004 Pearson Education, Inc.
Slide 1-68
1.6
Other Types of Equations
Copyright © 2004 Pearson Education, Inc.
Rational Equations

A rational equation is an equation that has a
rational expression for one or more terms. Since
a rational expression is not defined when its
denominator is 0, values of the variable for
which any denominator equals 0 cannot be
solutions of the equation. To solve a rational
equation, begin by multiplying both sides by the
least common denominator of the terms of the
equation.
Copyright © 2004 Pearson Education, Inc.
Slide 1-70
Example
Solve:
3x  1
6
3x

 .
4
x3 4
Solution: The least common denominator is 4(x + 3),which is
equal to 0 if x = 3. Therefore, 3 cannot possibly be a
solution of this equation.
6  3x
 3x  1
4  x  3 

 4  x  3

x 3 4
 4
(3 x  1)  x  3  24  3 x  x  3
3 x 2  9 x  x  3  24  3 x 2  9 x
 x  21  0
x  21
Check the result we find that the solution set is {21}.
Copyright © 2004 Pearson Education, Inc.
Slide 1-71
Another Example
Solve:
8 x
8
16

 2 .
x 1 x 1 x 1
Solution:
8 x
8
16

 2
x 1 x 1 x 1


16
 8 x 
 8 
 x  1 x  1 

   x  1 x  1 
   x  1 x  1 
 x 1 
 x 1
  x  1 x  1 
8 x  x  1  8  x  1  16
Copyright © 2004 Pearson Education, Inc.
Slide 1-72
Another Example continued
8 x 2  8 x  8 x  8  16
8 x 2  8  0
8( x 2  1)  0
( x  1)( x  1)  0
x  1
Neither proposed solution is valid, so the solution set is 0.
Copyright © 2004 Pearson Education, Inc.
Slide 1-73
Equations with Radicals

To solve an equation in which the variable appears in the
radicand, we use the following power property to
eliminate the radical.
Power Property

If P and Q are algebraic expressions, then every solution of the
equation P = Q is also a solution of the equation Pn = Qn, for any
positive integer n.


When using the power property to solve equations,
the new equation may have more solutions than the
original equation.
When an equation contains radicals (or rational
exponents), it is essential to check all proposed
solutions in the original equations.
Copyright © 2004 Pearson Education, Inc.
Slide 1-74
Radicals
Solving an Equation Involving Radicals

Isolate the radical on one side of the equation.

Raise each side of the equation to a power that is the
same as the index of the radical so that the radical is
eliminated.
If the equation still contains a radical, repeat Steps 1 and 2.

Solve the resulting equation.

Check each proposed solution in the original equation.
Copyright © 2004 Pearson Education, Inc.
Slide 1-75
Example
Solve x  x  7  5
x5  x7
 x  5
2


x7
Isolate the radical.

2
Square both sides of the equation.
x 2  10 x  25  x  7 Simplify. The result is a quadratic.
x 2  11x  18  0
 x  9  x  2   0
Put the quadratic in standard form.
Factor.
Solve for x.
x  9 or x  2
Copyright © 2004 Pearson Education, Inc.
Slide 1-76
Example continued

Check:
If x = 9, then
9  97 5
9  45
99



Check
If x = 2, then
?
2  27 5 ?
?
?
2  35
28
As the check shows, only
9 is a solution, giving the
solution set {9}
Copyright © 2004 Pearson Education, Inc.
Slide 1-77
Solving an Equation Containing Two
Radicals

Example: Solve

Solution:
x  2  3x  4  2.
x  2  3x  4  2
x  2   3x  4  2

    3x  4  2 
x  2    3x  4   2  2   
x2
2
2
2

3x  4  22
x  2   3x  4   4 3x  4  4
2 x  6  4 3 x  4
Copyright © 2004 Pearson Education, Inc.
Slide 1-78
Solving an Equation Containing Two
Radicals continued
x  3  2 3x  4
 x  3
2

 2 3x  4

2
x 2  6 x  9  4  3x  4 
x 2  6 x  9  12 x  16
x2  6x  7  0
 x  7  x  1  0
x  7 or x  1
Copyright © 2004 Pearson Education, Inc.
Slide 1-79
Solving an Equation Containing Two
Radicals continued

Check: x = 7
7  2  3 7  4  2
9  25  2
82

?
?
Check: x = 1
1  2  3  1  4  2 ?
?
1 1  2
22
The solution set is {1}
Copyright © 2004 Pearson Education, Inc.
Slide 1-80
Quadratics
Equations Quadratic in Form
An equation is said to be quadratic in form if it
can be written as
au2 + bu + c = 0,
where a  0, and u is some algebraic expression.
Copyright © 2004 Pearson Education, Inc.
Slide 1-81
Example
x4  2x2 + 1 = 0
(x2)2  2x2 + 1 = 0
u2  2u + 1 = 0
(u  1)(u  1) = 0
u  1= 0
u=1
To find x, replace u with x2.
x4 = (x2)2
Let u = x2
Solve the quadratic equation.
Zero-factor Property
x2 = 1
x = 1
x = 1
Checking in the original problem, the solution set is {1}.
Copyright © 2004 Pearson Education, Inc.
Slide 1-82
Another Example
Solve: (x + 1)2/3  (x + 1)1/3  2 = 0.
Solution: Since (x + 1)2/3 = [(x + 1)1/3]2, let u = (x + 1)1/3.
u2  u  2 = 0
(u  2)(u + 1) = 0
u  2 = 0 or
u = 2 or
Substitute.
Factor.
u+1=0
u = 1
Now replace u with (x + 1)1/3
(x + 1)1/3 = 2 or (x + 1)1/3 = 1
[(x + 1)1/3]3 = 23 or [(x + 1)1/3]3 = (1)3 Cube each side.
x + 1 = 8 or
x + 1 = 1
x = 7 or
x = 2
Copyright © 2004 Pearson Education, Inc.
Slide 1-83
Another Example continued
Check: (x + 1)2/3  (x + 1)1/3  2 = 0
If x = 7, then
If x =  2, then,
(7 + 1)2/3  (7 + 1)1/3  2 = 0. (2 + 1)2/3  (2 + 1)1/3  2 = 0 ?
82/3  81/3 2 = 0 ?
(1)2/3  (1) 1/3  2 = 0 ?
422=0?
1+12=0?
0=0
0=0
True
True
Both check, so the solution set is {2, 7}
Copyright © 2004 Pearson Education, Inc.
Slide 1-84
1.7
Inequalities
Copyright © 2004 Pearson Education, Inc.
Inequality

An inequality says that one expression is
greater than, greater than or equal to, less than,
or less than or equal to, another.

Properties of Inequality
For real numbers a, b, and c:
1. If a < b, then a + c < b + c
2. If a < b and if c > 0, then ac < bc,
3. If a < b and if c < 0, then ac > bc.
Copyright © 2004 Pearson Education, Inc.
Slide 1-86
Linear Inequalities

A linear inequality in one variable is an
inequality that can be written in the form
ax + b > 0, where a and b are real numbers with
a  0. (Any of the symbols , <, or  may also be
used.)

Always remember to reverse the direction of the
inequality symbol when multiplying or dividing by
a negative number.
Copyright © 2004 Pearson Education, Inc.
Slide 1-87
Example

Solve 5 x  7  13.
5 x  7  13
5 x  7  7  13  7
5 x  20
5 x 20

5
5
x4

)
–5
0
5
The original inequality is satisfied by any real number
less than 4. The solution set can be written {x|x < 4}.
Copyright © 2004 Pearson Education, Inc.
Slide 1-88
Open Interval Notation
Copyright © 2004 Pearson Education, Inc.
Slide 1-89
Half-open interval
Copyright © 2004 Pearson Education, Inc.
Slide 1-90
Closed Interval and All Real Numbers
Copyright © 2004 Pearson Education, Inc.
Slide 1-91
Example

Solve 9x  3x + 8  3(x + 4)
6 x  8  3 x  12
6 x  8  8  3 x  12  8
6 x  3x  4
6 x  3x  3x  3x  4
3x  4
3x 4

3 3
4
x
3
Copyright © 2004 Pearson Education, Inc.
–5
]
0
5
In interval notation the solution
set is (, 4/3].
Slide 1-92
Three-Part Inequalities

Solve 5 < 7 + 2x < 21
5  7  7  7  2 x  21  7
12  2 x  14
–10
12 2 x 14


2
2
2
6  x  7
(
)
–5
0
5
10
The solution set is the interval (6, 7).
Copyright © 2004 Pearson Education, Inc.
Slide 1-93
Quadratic Inequalities

A quadratic inequality is an inequality that can be
written in the form ax2 + bc + c < 0 for real numbers a, b,
and c with a  0. (The symbol, < can be replaced with >,
, or .)

Solving a Quadratic Inequality



Step 1 Solve the corresponding quadratic equation.
Step 2 Identify the intervals determined by the
solutions of the equation.
Step 3 Use a test value from each interval to
determine which intervals form the solution
set.
Copyright © 2004 Pearson Education, Inc.
Slide 1-94
Example



Solve x2  x  20 < 0
Step 1 Find the value that satisfy x2  x  20 = 0.
(x + 4)(x  5) = 0
x+4=0
or
x5=0
x = 4
or
x=5
Step 2 Determine the intervals
Interval A
Interval B
Interval C
(, 4)
(4, 5)
(5, )
–5
Copyright © 2004 Pearson Education, Inc.
0
5
Slide 1-95
Example continued

Step 3 Choose a test value in each interval to see if it satisfies the
original inequality. If the test value makes the interval true,
then the entire interval belongs to the solution set.
Interval

Test Value
True/False x2  x  20 < 0
(, 4)
5
(5)2 (5) 20 < 0
10 < 0 False
(4, 5)
0
0  0  20 < 0
(5, )
10
(10)2 (10) 20 < 0
70 < 0 False
True
Since the values in interval B make the inequality true, the solution set
is (4, 5).
Copyright © 2004 Pearson Education, Inc.
Slide 1-96
Rational Inequalities

Step 1

Step 2

Step 3
Rewrite the inequality, if necessary, so that 0
is on one side and there is a single fraction on
the other side.
Determine the values that will cause either
the numerator or denominator of the rational
expression to equal 0. These values
determine the intervals on the number line to
consider.
Use a test value from each interval to
determine which intervals form the solution
set.
Copyright © 2004 Pearson Education, Inc.
Slide 1-97
Example

Solve

1
x8
6
1  0
x8
6
x8

0
x8 x8
6 x 8
0
x8
x  2
0
x8
Copyright © 2004 Pearson Education, Inc.

Step 2

The quotient possibly
changes sign only where xvalues make the numerator
or denominator 0. This
occurs at
x  2 = 0 or x + 8 = 0
x = 2 or
x = 8
x = 2
These form the intervals
(, 8), (8, 2) and
(2, ).
Slide 1-98
Example continued

Step 3 Choose test values.
Interval
(, 8)
(8, 2)
( 2, )

Test Value
10
4
0
True/False
6
1
10  8
False
6
1
4  8
True
6
1
08
False
6
1
x8
The solution set is (8, 2].
Copyright © 2004 Pearson Education, Inc.
Slide 1-99
1.8
Absolute Value Equations and
Inequalities
Copyright © 2004 Pearson Education, Inc.
Properties of Absolute Value



|a| gives the distance from a to 0 on a number
line
1. For b > 0, |a| = b if and only if a = b or a = b.
2. |a| = |b| if and only if a = b or a = b.
For any positive number b:
 3. |a| < b if and only if b < a < b.
 4. |a| > b if and only if a < b or a > b.
Copyright © 2004 Pearson Education, Inc.
Slide 1-101
Solving Absolute Value Equations
Solve |4  7x| = 10
4  7 x  10
or
4  7 x  10
7 x  6
or
7 x  14
or
x2
6
x
7
Check the solutions by substituting them in the
original absolute value equation. The solution
set is {6/7, 2}.
Copyright © 2004 Pearson Education, Inc.
Slide 1-102
Example

Solve |3x  5| = |x  6|
or
3x  5  x  6
2 x  1
3 x  5  ( x  6)
3x  5   x  6
1
x
2


4 x  11
11
x
4
Check these solutions.
The solutions are 1
11
 and .
2
4
Copyright © 2004 Pearson Education, Inc.
Slide 1-103
Absolute Value Inequalities

Solve: |3x + 4| > 8
3x  4  8
or
3x  4  8
3x  12
3x  4
x  4
4
x
3
The solution set of the final compound inequality
is (, 4)  (4/3, )
Copyright © 2004 Pearson Education, Inc.
Slide 1-104
Example

Solve |3x + 4| < 8
3x  4  8
8  3 x  4  8
12  3 x  4

4
4  x 
3
The final inequality gives the solution set
(4, 4/3).
Copyright © 2004 Pearson Education, Inc.
Slide 1-105
Example

Solve |2x + 1|  3 < 10
2 x  1  3  10
2 x  1  13
13  2 x  1  13
14  2 x  12

7  x  6
The final inequality gives the solution set
(7, 6).
Copyright © 2004 Pearson Education, Inc.
Slide 1-106
Special Cases


Solve |2  5x|  4
Solution: Since the
absolute value of a
number is always
nonnegative, the
inequality is always true.
The solution set includes
all real numbers, written
(, ).
Copyright © 2004 Pearson Education, Inc.


Solve |4x  7| < 3
Solution: There is no
number whose absolute
value is less than 3 (or
less than any negative
number). The solution set
is .
Slide 1-107
Special Cases

Solve |5x + 15| = 0

Solution: The absolute value of a number will
be 0 only if that number is 0. Therefore,
|5x + 15| = 0 is equivalent to 5x + 15 = 0, which
has a solution set {3}.
Copyright © 2004 Pearson Education, Inc.
Slide 1-108
Using Absolute Value to Model Tolerance

Suppose y = 2x + 1 and we want y to be within
.01 unit of 4. For what values of x will this be
true?
y  4  .01

Solution:
2 x  1  4  .01
2 x  3  .01
.01  2 x  3  .01
2.99  2 x  3.01
1.495  x  1.505
Copyright © 2004 Pearson Education, Inc.
Slide 1-109