holt mcdougal algebra 2 5-8 solving radical equations and inequalities 5-8 solving radical equations...
TRANSCRIPT
Holt McDougal Algebra 2
5-8 Solving Radical Equationsand Inequalities 5-8 Solving Radical Equationsand Inequalities
Holt Algebra 2
Warm Up
Lesson Presentation
Lesson Quiz
Holt McDougal Algebra 2
Holt McDougal Algebra 2
5-8 Solving Radical Equationsand Inequalities
Solve radical equations and inequalities.
Objective
Holt McDougal Algebra 2
5-8 Solving Radical Equationsand Inequalities
radical equationradical inequality
Vocabulary
Holt McDougal Algebra 2
5-8 Solving Radical Equationsand 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 Equationsand Inequalities
For a square root, the index of the radical is 2.
Remember!
Holt McDougal Algebra 2
5-8 Solving Radical Equationsand Inequalities
Solve each equation.
Example 1A: Solving Equations Containing One Radical
Subtract 5.
Simplify.
Square both sides.
Solve for x.
Simplify.
Check
Holt McDougal Algebra 2
5-8 Solving Radical Equationsand Inequalities
7
Example 1B: Solving Equations Containing One Radical
Divide by 7.
Simplify.
Cube both sides.
Solve for x.
Simplify.
Check
Solve each equation.
3
77 5x - 7 84
7=
Holt McDougal Algebra 2
5-8 Solving Radical Equationsand Inequalities
Solve the equation.
Subtract 4.
Simplify.
Square both sides.
Solve for x.
Simplify.
Check
Check It Out! Example 1a
Holt McDougal Algebra 2
5-8 Solving Radical Equationsand Inequalities
Cube both sides.
Solve for x.
Simplify.
Check
Check It Out! Example 1b
Solve the equation.
Holt McDougal Algebra 2
5-8 Solving Radical Equationsand Inequalities
Divide by 6.
Square both sides.
Solve for x.
Simplify.
Check
Check It Out! Example 1c
42 42
Solve the equation.
Holt McDougal Algebra 2
5-8 Solving Radical Equationsand Inequalities
Example 2: Solving Equations Containing Two Radicals
Square both sides.
Solve
Simplify.
Distribute.
Solve for x.
7x + 2 = 9(3x – 2)
7x + 2 = 27x – 18
20 = 20x
1 = x
Holt McDougal Algebra 2
5-8 Solving Radical Equationsand Inequalities
Example 2 Continued
Check
3 3
Holt McDougal Algebra 2
5-8 Solving Radical Equationsand Inequalities
Square both sides.
Solve each equation.
Simplify.
Solve for x.
8x + 6 = 9x
6 = x
Check It Out! Example 2a
Check
Holt McDougal Algebra 2
5-8 Solving Radical Equationsand Inequalities
Cube both sides.
Solve each equation.
Simplify.
Distribute.
x + 6 = 8(x – 1)
Check It Out! Example 2b
x + 6 = 8x – 814 = 7x
2 = x Check
2 2
Solve for x.
Holt McDougal Algebra 2
5-8 Solving Radical Equationsand Inequalities
Example 3 Continued
Method 2 Use algebra to solve the equation.
Step 1 Solve for x.
Square both sides.
Solve for x.
Factor.
Write in standard form.
Simplify.–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
5-8 Solving Radical Equationsand Inequalities
Example 3 Continued
Method 2 Use algebra to solve the equation.
Step 2 Use substitution to check for extraneous solutions.
3 –3 6 6 x
Because x = 8 is extraneous, the only solution is x = –1.
Holt McDougal Algebra 2
5-8 Solving Radical Equationsand Inequalities
Method 2 Use algebra to solve the equation.
Step 1 Solve for x.
Square both sides.
Solve for x.
Factor.
Write in standard form.
Simplify.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
Check It Out! Example 3a Continued
Holt McDougal Algebra 2
5-8 Solving Radical Equationsand Inequalities
Method 1 Use algebra to solve the equation.
Step 2 Use substitution to check for extraneous solutions.
4 4 x
Because x = –5 is extraneous, the only solution is x = 1.
Check It Out! Example 3a Continued
2 14 35 5
2 –2
Holt McDougal Algebra 2
5-8 Solving Radical Equationsand Inequalities
Method 2 Use algebra to solve the equation.
Step 1 Solve for x.
Square both sides.
Solve for x.
Factor.
Write in standard form.
Simplify.–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
Check It Out! Example 3b Continued
Holt McDougal Algebra 2
5-8 Solving Radical Equationsand Inequalities
Method 1 Use algebra to solve the equation.
Step 2 Use substitution to check for extraneous solutions.
Check It Out! Example 3b Continued
Holt McDougal Algebra 2
5-8 Solving Radical Equationsand Inequalities
To find a power, multiply the exponents.
Remember!
Holt McDougal Algebra 2
5-8 Solving Radical Equationsand Inequalities
Example 4A: Solving Equations with Rational Exponents
Solve each equation.
(5x + 7) = 3
Cube both sides.
Solve for x.
Factor.
Write in radical form.
Simplify.5x + 7 = 27
5x = 20
x = 4
13
Holt McDougal Algebra 2
5-8 Solving Radical Equationsand Inequalities
Example 4B: Solving Equations with Rational Exponents
Factor.
Solve for x.
Factor out the GCF, 4.
Raise both sides to the reciprocal power.
Simplify.
2x = (4x + 8)12
(2x)2 = [(4x + 8) ]212
4x2 = 4x + 8
Write in standard form.4x2 – 4x – 8 = 0
4(x2 – x – 2) = 0
4(x – 2)(x + 1) = 0
4 ≠ 0, x – 2 = 0 or x + 1 = 0
x = 2 or x = –1
Step 1 Solve for x.
Holt McDougal Algebra 2
5-8 Solving Radical Equationsand Inequalities
Example 4B Continued
Step 2 Use substitution to check for extraneous solutions.
2x = (4x + 8)12
2(2) (4(2) + 8)12
4 1612
4 4
2x = (4x + 8)12
2(–1) (4(–1) + 8)12
–2 412
–2 2 x
The only solution is x = 2.
Holt McDougal Algebra 2
5-8 Solving Radical Equationsand Inequalities
Solve each equation.
(x + 5) = 3
Cube both sides.
Solve for x.
Write in radical form.
Simplify.x + 5 = 27
x = 22
13
Check It Out! Example 4a
Holt McDougal Algebra 2
5-8 Solving Radical Equationsand Inequalities
Factor.
Solve for x.
Raise both sides to the reciprocal power.Simplify.
(2x + 15) = x12
[(2x + 15) ]2 = (x)2 12
2x + 15 = x2
Write in standard form.x2 – 2x – 15 = 0
(x – 5)(x + 3) = 0
x – 5 = 0 or x + 3 = 0
x = 5 or x = –3
Step 1 Solve for x.
Check It Out! Example 4b
Holt McDougal Algebra 2
5-8 Solving Radical Equationsand Inequalities
The only solution is x = 5.
Check It Out! Example 4b Continued
(2(–3) + 15) –312
(–6 + 15) –312
x
(2x + 15) = x12
3 –3
Step 2 Use substitution to check for extraneous solutions.
(2(5) + 15) 512
(2x + 15) = x12
(10 + 15) 512
5 5