Transcript Holt McDougal Algebra 2 5-8

Solving
Radical
Equations
Solving
Radical
Equations
5-8
5-8 and
andInequalities
Inequalities
Warm Up
Lesson Presentation
Lesson Quiz
HoltMcDougal
Algebra 2Algebra 2
Holt
5-8
Solving Radical Equations
and Inequalities
Objective
Solve radical equations and inequalities.
Holt McDougal Algebra 2
5-8
Solving Radical Equations
and Inequalities
Vocabulary
radical equation
radical inequality
Holt McDougal Algebra 2
5-8
Solving Radical Equations
and Inequalities
A radical equation contains a variable within a
radical. Recall that you can solve quadratic
equations by taking the square root of both
sides. Similarly, radical equations can be solved
by raising both sides to a power.
Holt McDougal Algebra 2
5-8
Solving Radical Equations
and Inequalities
Remember!
For a square root, the index of the radical is 2.
Holt McDougal Algebra 2
5-8
Solving Radical Equations
and Inequalities
Example 1A: Solving Equations Containing One
Radical
Solve each equation.
Check
Subtract 5.
Simplify.
Square both sides.
Simplify.
Solve for x.
Holt McDougal Algebra 2
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5-8
Solving Radical Equations
and Inequalities
Example 1B: Solving Equations Containing One
Radical
Solve each equation.
Check
7 3 5x - 7 84
=
7
7
Divide by 7.
7
Simplify.
Cube both sides.
Simplify.
Solve for x.
Holt McDougal Algebra 2
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5-8
Solving Radical Equations
and Inequalities
Check It Out! Example 1a
Solve the equation.
Check
Subtract 4.
Simplify.
Square both sides.
Simplify.
Solve for x.
Holt McDougal Algebra 2
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5-8
Solving Radical Equations
and Inequalities
Check It Out! Example 1b
Solve the equation.
Check
Cube both sides.
Simplify.
Solve for x.
Holt McDougal Algebra 2
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5-8
Solving Radical Equations
and Inequalities
Check It Out! Example 1c
Solve the equation.
Check
Divide by 6.
Square both sides.
Simplify.
Solve for x.
Holt McDougal Algebra 2
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5-8
Solving Radical Equations
and Inequalities
Example 2: Solving Equations Containing Two
Radicals
Solve
Square both sides.
7x + 2 = 9(3x – 2)
Simplify.
7x + 2 = 27x – 18
Distribute.
20 = 20x
1=x
Holt McDougal Algebra 2
Solve for x.
5-8
Solving Radical Equations
and Inequalities
Example 2 Continued
Check
3
Holt McDougal Algebra 2
3 
5-8
Solving Radical Equations
and Inequalities
Check It Out! Example 2a
Solve each equation.
Square both sides.
8x + 6 = 9x
6=x
Simplify.
Solve for x.
Check

Holt McDougal Algebra 2
5-8
Solving Radical Equations
and Inequalities
Check It Out! Example 2b
Solve each equation.
Cube both sides.
x + 6 = 8(x – 1)
Simplify.
x + 6 = 8x – 8
14 = 7x
2 =x
Distribute.
Solve for x.
Check
2
Holt McDougal Algebra 2
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Solving Radical Equations
and Inequalities
Example 3 Continued
Method 2 Use algebra to solve the equation.
Step 1 Solve for x.
Square both sides.
–3x + 33 = 25 – 10x + x2
0 = x2 – 7x – 8
0 = (x – 8)(x + 1)
x – 8 = 0 or x + 1 = 0
x = 8 or x = –1
Holt McDougal Algebra 2
Simplify.
Write in standard form.
Factor.
Solve for x.
5-8
Solving Radical Equations
and Inequalities
Example 3 Continued
Method 2 Use algebra to solve the equation.
Step 2 Use substitution to check for extraneous
solutions.
3
–3 x
6
6 
Because x = 8 is extraneous, the only solution
is x = –1.
Holt McDougal Algebra 2
5-8
Solving Radical Equations
and Inequalities
Check It Out! Example 3a Continued
Method 2 Use algebra to solve the equation.
Step 1 Solve for x.
Square both sides.
2x + 14 = x2 + 6x + 9
0 = x2 + 4x – 5
0 = (x + 5)(x – 1)
x + 5 = 0 or x – 1 = 0
x = –5 or x = 1
Holt McDougal Algebra 2
Simplify.
Write in standard form.
Factor.
Solve for x.
5-8
Solving Radical Equations
and Inequalities
Check It Out! Example 3a Continued
Method 1 Use algebra to solve the equation.
Step 2 Use substitution to check for extraneous
solutions.
2
–2 x
4
4 
Because x = –5 is extraneous, the only solution
is x = 1.
Holt McDougal Algebra 2
5-8
Solving Radical Equations
and Inequalities
Check It Out! Example 3b Continued
Method 2 Use algebra to solve the equation.
Step 1 Solve for x.
Square both sides.
–9x + 28 = x2 – 8x + 16
0 = x2 + x – 12
0 = (x + 4)(x – 3)
x + 4 = 0 or x – 3 = 0
x = –4 or x = 3
Holt McDougal Algebra 2
Simplify.
Write in standard form.
Factor.
Solve for x.
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Solving Radical Equations
and Inequalities
Check It Out! Example 3b Continued
Method 1 Use algebra to solve the equation.
Step 2 Use substitution to check for extraneous
solutions.
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Holt McDougal Algebra 2
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Solving Radical Equations
and Inequalities
Remember!
To find a power, multiply the exponents.
Holt McDougal Algebra 2
5-8
Solving Radical Equations
and Inequalities
Example 4A: Solving Equations with Rational
Exponents
Solve each equation.
(5x + 7)
1
3
=3
Write in radical form.
Cube both sides.
5x + 7 = 27
5x = 20
x= 4
Holt McDougal Algebra 2
Simplify.
Factor.
Solve for x.
5-8
Solving Radical Equations
and Inequalities
Example 4B: Solving Equations with Rational
Exponents
1
2
2x = (4x + 8)
Step 1 Solve for x.
1
2
(2x)2 = [(4x + 8) ]2
Raise both sides to the
reciprocal power.
4x2 = 4x + 8
Simplify.
4x2 – 4x – 8 = 0
Write in standard form.
2
4(x – x – 2) = 0
4(x – 2)(x + 1) = 0
Factor out the GCF, 4.
Factor.
4 ≠ 0, x – 2 = 0 or x + 1 = 0 Solve for x.
x = 2 or x = –1
Holt McDougal Algebra 2
5-8
Solving Radical Equations
and Inequalities
Example 4B Continued
Step 2 Use substitution to check for extraneous
solutions.
2x = (4x + 8)
1
2
2(2) (4(2) + 8)
4
4
2x = (4x + 8)
1
2
1
2
16
4 
The only solution is x = 2.
Holt McDougal Algebra 2
1
2
2(–1) (4(–1) + 8)
–2
–2
1
2
4
2 x
1
2
5-8
Solving Radical Equations
and Inequalities
Check It Out! Example 4a
Solve each equation.
(x + 5)
1
3
=3
Write in radical form.
Cube both sides.
x + 5 = 27
x = 22
Holt McDougal Algebra 2
Simplify.
Solve for x.
5-8
Solving Radical Equations
and Inequalities
Check It Out! Example 4b
(2x + 15)
1
2
=x
Step 1 Solve for x.
1
2
[(2x + 15) ]2 = (x)2
2x + 15 = x2
x2 – 2x – 15 = 0
(x – 5)(x + 3) = 0
x – 5 = 0 or x + 3 = 0
x = 5 or x = –3
Holt McDougal Algebra 2
Raise both sides to the
reciprocal power.
Simplify.
Write in standard form.
Factor.
Solve for x.
5-8
Solving Radical Equations
and Inequalities
Check It Out! Example 4b Continued
Step 2 Use substitution to check for extraneous
solutions.
(2x + 15)
1
2
(2(5) + 15)
(10 + 15)
5
1
2
1
2
=x
5
5
5 
The only solution is x = 5.
Holt McDougal Algebra 2
(2x + 15)
1
2
(2(–3) + 15)
(–6 + 15)
3
=x
1
2
1
2
–3
–3
–3 x