Holt Algebra 2

8-8 Solving Radical Equationsand Inequalities

Solve radical equations

Objective

Holt Algebra 2

8-8 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 Algebra 2

8-8 Solving Radical Equationsand Inequalities

The nth root of a real number a can be written as the radical expression , where n is the index (plural: indices) of the radical and a is the radicand. When a number has more than one root, the radical sign indicates only the principal, or positive, root.

n a

Holt Algebra 2

8-8 Solving Radical Equationsand Inequalities

For a square root, the index of the radical is 2.

Remember!

Holt Algebra 2

8-8 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 Algebra 2

8-8 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 Algebra 2

8-8 Solving Radical Equationsand Inequalities

Solve the equation.

Subtract 4.

Simplify.

Square both sides.

Solve for x.

Simplify.

Check

Check It Out! Example 1a

Holt Algebra 2

8-8 Solving Radical Equationsand Inequalities

Cube both sides.

Solve for x.

Simplify.

Check

Check It Out! Example 1b

Solve the equation.

Holt Algebra 2

8-8 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 Algebra 2

8-8 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 Algebra 2

8-8 Solving Radical Equationsand Inequalities

Example 2 Continued

Check

3 3

Holt Algebra 2

8-8 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 Algebra 2

8-8 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 Algebra 2

8-8 Solving Radical Equationsand Inequalities

Raising each side of an equation to an even power may introduce extraneous solutions.

You can use the intersect feature on a graphing calculator to find the point where the two curves intersect.

Helpful Hint

Holt Algebra 2

8-8 Solving Radical Equationsand Inequalities

Example 3 Continued

Method 1 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 Algebra 2

8-8 Solving Radical Equationsand Inequalities

Example 3 Continued

Method 1 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 Algebra 2

8-8 Solving Radical Equationsand Inequalities

Solve .

Example 3: Solving Equations with Extraneous Solutions

Method 2 Use a graphing calculator.

The solution is x = – 1

The graphs intersect in only one point, so there is exactly one solution.

Let Y1 = and Y2 = 5 – x.

Holt Algebra 2

8-8 Solving Radical Equationsand Inequalities

Method 1 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 Algebra 2

8-8 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 Algebra 2

8-8 Solving Radical Equationsand Inequalities

Method 2 Use a graphing calculator.

The solution is x = 1.

The graphs intersect in only one point, so there is exactly one solution.

Check It Out! Example 3a

Solve each equation.

Let Y1 = and Y2 = x +3.

Holt Algebra 2

8-8 Solving Radical Equationsand Inequalities

Method 1 Use a graphing calculator.

Check It Out! Example 3b

Let Y1 = and Y2 = –x +4.

The solutions are x = –4 and x = 3.

The graphs intersect in two points, so there are two solutions.

Solve each equation.

Holt Algebra 2

8-8 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 Algebra 2

8-8 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 Algebra 2

8-8 Solving Radical Equationsand Inequalities

To find a power, multiply the exponents.

Remember!

Holt Algebra 2

8-8 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 Algebra 2

8-8 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 Algebra 2

8-8 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 Algebra 2

8-8 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 Algebra 2

8-8 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 Algebra 2

8-8 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

Holt Algebra 2

8-8 Solving Radical Equationsand Inequalities

Solve for x.

Raise both sides to the reciprocal power.

Simplify.

9(x + 6) = 81

[3(x + 6) ]2 = (9)212

Distribute 9.9x + 54 = 81

Check It Out! Example 4c

9x = 27

x = 3

3(x + 6) = 912

Simplify.

Holt Algebra 2

8-8 Solving Radical Equationsand Inequalities

A car skids to a stop on a street with a speed limit of 30 mi/h. The skid marks measure 35 ft, and the coefficient of friction was 0.7. Was the car speeding? Explain.

Check It Out! Example 6

Use the formula to determine the greatest possible length of the driver’s skid marks if he was traveling 30 mi/h.

The speed s in miles per hour that a car is traveling when it goes into a skid can be estimated by using the formula s = , where f is the coefficient of friction and d is the length of the skid marks in feet.

30fd

Holt Algebra 2

8-8 Solving Radical Equationsand Inequalities

Check It Out! Example 6 Continued

900 = 21d

43 ≈ d

If the car were traveling 30 mi/h, its skid marks would have measured about 43 ft. Because the actual skid marks measure less than 43 ft, the car was not speeding.

Substitute 30 for s and 0.7 for f.

Simplify.

Square both sides.

Simplify.

Solve for d.