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CHAPTER 3:
Quadratic Functions and
Equations; Inequalities
3.1 The Complex Numbers
3.2 Quadratic Equations, Functions, Zeros, and
Models
3.3 Analyzing Graphs of Quadratic Functions
3.4 Solving Rational Equations and Radical
Equations
3.5 Solving Equations and Inequalities with
Absolute Value
Copyright © 2009 Pearson Education, Inc.
3.4
Solving Rational Equations
and Radical Equations


Solve rational equations.
Solve radical equations.
Copyright © 2009 Pearson Education, Inc.
Rational Equations
Equations containing rational expressions are called
rational equations.
Solving such equations requires multiplying both
sides by the least common denominator (LCD) to
clear the equation of fractions.
Copyright © 2009 Pearson Education, Inc.
Slide 3.4 - 4
Example
x8 x3

 0.
Solve:
3
2
Solution: Multiply both sides by the LCD 6.
 x  8 x  3
6

 60

 3
2 
x8
x3
6
 6
0
3
2
2 x  8   3x  3  0
2x 16  3x  9  0
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5x  25  0
5x  25
x5
Slide 3.4 - 5
Example (continued)
The possible solution is 5.
x8 x3

0
Check:
3
2
58 53

? 0
3
2
3 2

3 2
1 1 0
0 0 TRUE
The solution is 5.
Copyright © 2009 Pearson Education, Inc.
Slide 3.4 - 6
Example
x2
9

.
Solve:
x3 x3
Solution: Multiply both sides by the LCD x  3.
x2
9
 x  3
x  3
x3
x3
x2  9
x  3 or x  3
Copyright © 2009 Pearson Education, Inc.
Slide 3.4 - 7
Example (continued)
The possible solutions are –3 and 3.
Check x = –3:
x2
9

x3 x3
3
2
9
?
3  3
3  3
9
9
TRUE
6
6
The number 3 checks,
so it is a solution.
Copyright © 2009 Pearson Education, Inc.
Check x = 3:
x2
9

x3 x3
3
2
9
?
3 3
3 3
9
9
Not Defined
0
0
Division by 0 is not defined,
so 3 is not a solution.
Slide 3.4 - 8
Radical Equations
A radical equation is an equation in which variables
appear in one or more radicands. For example:
2x  5  x  3  1
The Principle of Powers
For any positive integer n:
If a = b is true, then an = bn is true.
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Slide 3.4 - 9
Solving Radical Equations
To solve a radical equation we must first isolate the
radical on one side of the equation.
Then apply the Principle of Powers.
When a radical equation has two radical terms on one
side, we isolate one of them and then use the principle
of powers. If, after doing so, a radical terms remains,
we repeat these steps.
Copyright © 2009 Pearson Education, Inc.
Slide 3.4 - 10
Example
Solve
3x 1  4.
Check x = 5:
Solution

3x  1

2
 42
3x  1  4
3 5  1 ? 4
3x 1  16
15  1
16
4
3x  15
x5
4 TRUE
The solution is 5.
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Slide 3.4 - 11
Example
Solve: 5  x  7  x.
Solution: First, isolate the radical on one side.
x7  x5
 x  7   x  5
2
2
x  7  x 2 10x  25
0  x 2 11x 18
0  x  9 x  2 
x9 0
x9
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or x  2  0
or x  2
Slide 3.4 - 12
Example (continued)
The possible solutions are 9 and 2.
Check x = 2.
Check x = 9.
5 x7  x
5 x7  x
5 97 ? 9
5 27 ? 2
5  16
54
9
9 TRUE
5 9
53
8
2 FALSE
Since 9 checks but 2 does not, the only solution is 9.
Copyright © 2009 Pearson Education, Inc.
Slide 3.4 - 13
Example
x3 x5  4
Solve:
Solution:
x3  4 x5
 x  3   4 
2
x5

2
x  3  16  8 x  5  x  5
x  3  21 8 x  5  x
24  8 x  5
3 x5

3  x5
9 x5
4x
2
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
2
Slide 3.4 - 14
Example (continued)
We check the possible solution, 4, on a graphing calculator.
Since y1= y2 when x = 4, the number 4 checks. It is the
solution.
Copyright © 2009 Pearson Education, Inc.
Slide 3.4 - 15