Transcript Ch. 8.6 power point

Chapter 8
Section 6
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
8.6
1
2
3
4
Solving Equations with Radicals
Solve radical equations having square root
radicals.
Identify equations with no solutions.
Solve equations by squaring a binomial.
Solve radical equations having cube root
radicals.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Solving Equations with Radicals.
A radical equation is an equation having a variable
in the radicand, such as
x 1  3
or
3 x  8x  9
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Slide 8.6 - 3
Objective 1
Solve radical equations having
square root radicals.
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Slide 8.6 - 4
Solve radical equations having
square root radicals.
To solve radical equations having square root
radicals, we need a new property, called the squaring
property of equality.
If each side of a given equation is squared, then all
solutions of the original equation are among the
solutions of the squared equation.
Be very careful with the squaring property: Using this property can give a
new equation with more solutions than the original equation has. Because of this
possibility, checking is an essential part of the process. All proposed solutions
from the squared equation must be checked in the original equation.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 8.6 - 5
EXAMPLE 1
Using the Squaring
Property of Equality
Solve.
Solution:

9 x  4
9 x

2
4
2
9  x  16
9  x  9  16  9
x  7
x  7
7
It is important to note that even though the algebraic work may be done
perfectly, the answer produced may not make the original equation true.
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Slide 8.6 - 6
EXAMPLE 2
Using the Squaring Property
with a Radical on Each Side
Solve.
3x  9  2 x
Solution:

3x  9
  2 x 
2
2
3x  9  4 x
3x  9  3x  4 x  3x
x9
9
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Objective 2
Identify equations with no
solutions.
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Slide 8.6 - 8
EXAMPLE 3
Using the Squaring Property
when One Side Is Negative
Solve.
x  4
Solution:
 x
2
  4 
Check:
2
x  4
16  4
4  4
False
x  16

Because x represents the principal or nonnegative square root of x in Example 3,
we might have seen immediately that there is no solution.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 8.6 - 9
Solving a Radical Equation.
Use the following steps when solving an equation with
radicals.
Step 1
Isolate a radical. Arrange the terms so that
a radical is isolated on one side of the
equation.
Step 2
Square both sides.
Step 3
Combine like terms.
Step 4
Repeat Steps 1-3 if there is still a term with a
radical.
Step 5
Solve the equation. Find all proposed solutions.
Step 6
Check all proposed solutions in the original
equation.
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Slide 8.6 - 10
EXAMPLE 4
Using the Squaring Property
with a Quadratic Expression
2
Solve x  x  4x 16.
Solution:
x2 

x 2  4 x  16

2
x2  x2  x2  4 x  16  x2
0  4 x  4 x  16  4 x
4 x 1 6

4
4
x  4
Since x must be a positive number the
solution set is Ø.
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Slide 8.6 - 11
Objective 3
Solve equations by squaring a
binomial.
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Slide 8.6 - 12
EXAMPLE 5
Using the Squaring Property
when One Side Has Two Terms
Solve 2 x 1  10 x  9.
 2 x 1   10 x  9 
4x2  4x 1  10x  9  10x  9  10x  9
2
2
Solution:
4 x  14 x  8  0
 2x 1 2x  8  0
2
2x 1  0
or
2x  8  0
1
x4
x
2
Since x must be positive the solution set is {4}.
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EXAMPLE 6
Rewriting an Equation before
using the Squaring Property
25x  6  x
Solve.
Solution:
25x  6  6  x  6

25x

2
  x  6
2
25x  25x  x2  12 x  36  25x
0  x2  13x  36
0   x  4 x  9
0 x4
0  x9
or
x4
x9
The solution set is {4,9}.
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Slide 8.6 - 14
Solve equations by squaring a binomial.
Errors often occur when both sides of an equation are squared. For
instance, when both sides of
9x  2x  1
are squared, the entire binomial 2x + 1 must be squared to get 4x2 + 4x + 1.
It is incorrect to square the 2x and the 1 separately to get 4x2 + 1.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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EXAMPLE 7
Solve.
Using the Squaring
Property Twice
x 1  x  4  1
Solution:

x 1  1 x  4
 
2
x 1  1 x  4

2
x  1  1  2 x  4   x  4

4  2 x4
2
16  4 x  16

2
32 4x

4
4
x8
The solution set is {8}.
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Slide 8.6 - 16
Objective 4
Solve radical equations having
cube root radicals.
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Slide 8.6 - 17
Solve radical equations having cube
root radicals.
We can extend the concept of raising both sides of
an equation to a power in order to solve radical
equations with cube roots.
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Slide 8.6 - 18
EXAMPLE 8
Solving Equations with
Cube Root Radicals
Solve each equation.
3
Solution:

3
7 x  3 4x  2
 
3
7x

3
4x  2
7x  4x  2
3x 2

3 3
2
x
3
2
 
3
3

3
x2  3 26x  27
 x  
3
2
3
3
26 x  27

3
x 2  26 x  27
0  x 2  26  27
0   x  27 x 1
0  x  27
0  x 1
or
x  27
x 1
27,1
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