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Holt Algebra 2

8-8 Solving Radical Equations

and Inequalities 8-8

Solving Radical Equations

and Inequalities

Holt Algebra 2

Warm Up

Lesson Presentation

Lesson Quiz

Holt Algebra 2


and Inequalities

Warm Up Simplify each expression. Assume all variables are positive. Write each expression in radical form.

1. 2.

3. 4.

5.

6.

Holt Algebra 2


and Inequalities

Solve radical equations and inequalities.

Objective

Holt Algebra 2


and Inequalities

radical equation

radical inequality

Vocabulary

Holt Algebra 2


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 Algebra 2


and Inequalities

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

Remember!

Holt Algebra 2


and 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


and Inequalities

7

Example 1B: Solving Equations Containing One

Radical

Divide by 7.

Simplify.

Cube both sides.

Solve for x.

Simplify.

Check


3

7

7 5x - 7 84

7 =

Holt Algebra 2


and Inequalities

Solve the equation.

Subtract 4.

Simplify.

Square both sides.

Solve for x.

Simplify.

Check

Check It Out! Example 1a

Holt Algebra 2


and Inequalities

Cube both sides.

Solve for x.

Simplify.

Check

Check It Out! Example 1b

Solve the equation.

Holt Algebra 2


and Inequalities

Divide by 6.

Square both sides.

Solve for x.

Simplify.

Check

Check It Out! Example 1c

42 42Solve the equation.

Holt Algebra 2


and 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


and Inequalities

Example 2 Continued

Check

3 3

Holt Algebra 2


and Inequalities

Square both sides.


Simplify.

Solve for x.

8x + 6 = 9x

6 = x


Check

Holt Algebra 2


and Inequalities

Cube both sides.


Simplify.

Distribute.

x + 6 = 8(x – 1)


x + 6 = 8x – 8

14 = 7x

2 = x Check

2 2

Solve for x.

Holt Algebra 2


and 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


and Inequalities

Solve .

Example 3: Solving Equations with Extraneous

Solutions

Method 1 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


and 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 Algebra 2


and Inequalities

Example 3 Continued


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


and Inequalities


The solution is x = 1.

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



Let Y1 = and Y2 = x +3.

Holt Algebra 2


and Inequalities


Step 1 Solve for x.

Square both sides.

Solve for x.

Factor.


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


and Inequalities



4 4 x

Because x = –5 is extraneous, the only solution is x = 1.


2 14 35 5- - 2 –2

Holt Algebra 2


and Inequalities



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.


Holt Algebra 2


and Inequalities


Step 1 Solve for x.

Square both sides.

Solve for x.

Factor.


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


and Inequalities




Holt Algebra 2


and Inequalities

To find a power, multiply the exponents.

Remember!

Holt Algebra 2


and Inequalities

Example 4A: Solving Equations with Rational

Exponents


(5x + 7) = 3

Cube both sides.

Solve for x.

Factor.

Write in radical form.

Simplify. 5x + 7 = 27

5x = 20

x = 4

1

3

Holt Algebra 2


and 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) 1

2

(2x)2 = [(4x + 8) ]

2

1

2

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


and Inequalities

Example 4B Continued

Step 2 Use substitution to check for extraneous

solutions.

2x = (4x + 8) 1

2

2(2) (4(2) + 8) 1

2

4 16 1

2

4 4

2x = (4x + 8) 1

2

2(–1) (4(–1) + 8) 1

2

–2 4 1

2

–2 2 x

The only solution is x = 2.

Holt Algebra 2


and Inequalities


(x + 5) = 3

Cube both sides.

Solve for x.

Write in radical form.

Simplify. x + 5 = 27

x = 22

1

3


Holt Algebra 2


and Inequalities

Factor.

Solve for x.


reciprocal power.

Simplify.

(2x + 15) = x 1

2

[(2x + 15) ]2 =

(x)

2

1

2

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.


Holt Algebra 2


and Inequalities

The only solution is x = 5.


(2(–3) + 15) –3 1

2

(–6 + 15) –3 1

2

x

(2x + 15) = x 1

2

3 –3

Step 2 Use substitution to check for extraneous

solutions.

(2(5) + 15) 5 1

2

(2x + 15) = x 1

2

(10 + 15) 5 1

2

5 5

Holt Algebra 2


and Inequalities

Solve for x.


reciprocal power.

Simplify.

9(x + 6) = 81

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

2

1

2

Distribute 9. 9x + 54 = 81

Check It Out! Example 4c

9x = 27

x = 3

3(x + 6) = 9 1

2

Simplify.

Holt Algebra 2


and Inequalities

A radical inequality is an inequality that contains a variable within a radical. You can solve radical inequalities by graphing or using algebra.

A radical expression with an even index and a negative radicand has no real roots.

Remember!

Holt Algebra 2


and Inequalities

Example 5: Solving Radical Inequalities

Method 1 Use a graph and a table.

Solve .

On a graphing calculator, let Y1 = and Y2 = 9. The graph of Y1 is at or below the graph of Y2 for values of x between 3 and 21. Notice that Y1 is undefined when < 3.

The solution is 3 ≤ x ≤ 21.

Holt Algebra 2


and Inequalities

Example 5 Continued

Method 2 Use algebra to solve the inequality.

Step 1 Solve for x.

Subtract 3.

Solve for x.

Simplify.

Square both sides.

2x ≤ 42

x ≤ 21

Holt Algebra 2


and Inequalities

Example 5 Continued


Step 2 Consider the radicand.

The radicand cannot be negative.

Solve for x.

2x – 6 ≥ 0

2x ≥ 6

x ≥ 3

The solution of is x ≥ 3 and x ≤ 21, or 3 ≤ x ≤ 21.

Holt Algebra 2


and Inequalities



On a graphing calculator, let Y1 = and Y2 = 5. The graph of Y1 is at or below the graph of Y2 for values of x between 3 and 12. Notice that Y1 is undefined when < 3.

The solution is 3 ≤ x ≤ 12.

Solve .

Holt Algebra 2


and Inequalities


Step 1 Solve for x.

Subtract 2.

Solve for x.

Simplify.

Square both sides.

x – 3 ≤ 9

x ≤ 12


Holt Algebra 2


and Inequalities




Solve for x.

x – 3 ≥ 0

x ≥ 3


The solution of is x ≥ 3 and x ≤ 12, or 3 ≤ x ≤ 12.

Holt Algebra 2


and Inequalities



Solve .

The solution is x ≥ –1.

On a graphing calculator, let Y1 = and Y2 = 1. The graph of Y1 is at or above the graph of Y2 for values of x greater than –1. Notice that Y1 is undefined when < –2.

Holt Algebra 2


and Inequalities


Step 1 Solve for x.

Solve for x.

Cube both sides.

x + 2 ≥ 1

x ≥ –1


Holt Algebra 2


and Inequalities




Solve for x.

x + 2 ≥ 1

x ≥ –1


The solution of is x ≥ –1.

Holt Algebra 2


and Inequalities

Example 6: Automobile Application

The time t in seconds that it takes a car to

travel a quarter mile when starting from a

full stop can be estimated by using the

formula , where w is the weight

of the car in pounds and P is the power

delivered by the engine in horsepower. If

the quarter-mile time from a 3590 lb car is

13.4 s, how much power does its engine

deliver? Round to the nearest horsepower.

Holt Algebra 2


and Inequalities

Example 6 Continued Use the formula to determine the amount of horsepower the 3590 lb car has if it finishes the quarter-mile in 13.4s.

Substitute 13.4 for t

and 3590 for w.

Cube both sides.

Solve for P.

Simplify.

2406.104P ≈ 709,548.747

P ≈ 295 The engine delivers a power of about 295 hp.

Holt Algebra 2


and 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.

30 fd

Holt Algebra 2


and 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.

Holt Algebra 2


and Inequalities

Lesson Quiz: Part I

Solve each equation or inequality.

x = 16 1.

2.

3.

4.

x = –5

x = 14

x = 2

x ≥ 59 5.

Holt Algebra 2


and Inequalities

Lesson Quiz: Part II

137,258 ft3

6. The radius r in feet of a spherical water tank

can be determined by using the formula ,

where V is the volume of the tank in cubic feet. To the nearest cubic foot, what is the volume of a spherical tank with a radius of 32 ft?

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