132 Unit 1: Polynomial, Rational, and Radical Relationships

LESSON

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UNDERSTAND In a radical equation, there is a variable in the radicand . The radicand is the expression inside the radical symbol ( √

__ ), and the index is the root being taken . In the

equation y 5 √__

x , the radicand is x . The equation takes the square root of x, so the index is 2 . Cube roots are also radicals (with an index of 3) .

UNDERSTAND A radical expression is not a polynomial, but polynomials can be used to solve radical equations . You know that linear equations are solved using inverse operations . Some quadratic equations can be solved by taking the square root of both sides . This is because raising to the nth power and taking the nth root are inverse operations . So, to solve a radical equation, raise both sides to the same power .

Consider the equation √__

x 5 3 . The index is 2, so raise both sides to the second power .

√__

x 5 3

( √__

x )2 5 32

x 5 9

Since √__

9 5 3, the answer checks .

Now consider the equation √__

x 5 23 . Solve by squaring both sides .

√__

x 5 23

( √__

x )2 5 (23)2

x 5 9

Since √__

9 23, the answer does not check . It is extraneous .

It is important to check all answers to radical equations since extraneous solutions can occur . You should also be sure to look at the original equation and use number sense . Without solving the equation √

__ x 5 23, you could have concluded that it has no real solution, since a

square root cannot be negative .

Solving Radical Equations

Radical Equations and Inequalities16

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Lesson 16: Radical Equations and Inequalities 133

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Solve:√__

x 5 x26

Square both sides of the equation .

√__

x 5 x26

( √__

x )2 5 (x26)2

x 5 x2212x 1 36

Check for extraneous solutions .

Substitute each solution into the original equation .

Test: x 5 4

√__

4 0 426

2 22

The solution x 5 4 is extraneous since it does not make the original equation true .

Test: x 5 9

√__

9 5 926

3 5 3

▸The solution to the equation is x 5 9 .

3

1

Solve the resulting equation for x .

x 5 x2212x 1 36

0 5 x2213x 1 36

0 5 (x24)(x29)

x 5 4 and x 5 9

2

Solve: √______

3x 1 1 5 4

TRY

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134 Unit 1: Polynomial, Rational, and Radical Relationships

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UNDERSTAND Solving a radical inequality is much like solving a radical equation, but you will have to take some extra steps . As with other inequalities, the solution will usually be a range, which can be graphed on a number line .

Consider the radical inequality √_____

x 1 3 # 2 . Replace the inequality sign with an equals sign, and solve the equation .

√_____

x 1 3 5 2

( √_____

x 1 3 ) 2 5 22 Square both sides .

x 1 3 5 4 Subtract 3 from both sides .

x 5 1

Now, change the equals sign back to the inequality symbol: x # 1 . Graph the solution .

–8 0 8

Remember that you cannot take the square root of a negative number in the set of real

numbers . So, finding the domain of a radical inequality should be your first step in solving

the inequality . With √_____

x 1 3 # 2, the solution only makes sense when the radicand, x 1 3, is

nonnegative . This means x 1 3 $ 0 . Solve this inequality for x .

x 1 3 $ 0 Subtract 3 from both sides .

x $ 23

Graph this solution on a number line .

–8 0 8

The solution to the original inequality is the range where the two solutions overlap .

–8 0 8

The solution to √_____

x 1 3 # 2 is 23 # x # 1 .

Solving Radical Inequalities

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Lesson 16: Radical Equations and Inequalities 135

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√_____

x 1 6 . x

Determine allowable values for the radicand .

The radicand must be greater than or equal to 0 .

x 1 6 $ 0

x $ 26

Check for extraneous solutions . Use the equation .

Try x 5 3 .

√_____

3 1 6 0 3

√__

9 5 3

Try x 5 22 .

√_______

22 1 6 0 22

√__

4 22

The only solution is x 5 3 .

Since the original inequality has the variable on both sides, test values on either side of x to determine which inequality symbol to use (x . 3 or x , 3) .

The solution is x , 3 .

3

1

Replace the inequality sign with an equals sign, and solve the equation .

( √_____

x 1 6 ) 2 5 x2

x 1 6 5 x2

0 5 x22x26

0 5 (x23)(x 1 2)

x 5 3 and x 5 22

2

Graph the complete solution .

Graph the areas of overlap for x , 3 and x $ 26 .

▸ –8 0 8

4

Why is 26 included in the solution to √

_____ x 1 6 . x, but 3 is excluded?

DISCUSS

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136 Unit 1: Polynomial, Rational, and Radical Relationships

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Cube both sides of the equation .

( 3 √______

7x 1 6 ) 3 5 x3

7x 1 6 5 x3

0 5 x327x26

Check for extraneous solutions .

Test: x 5 21

( 3 √_________

7(21) 1 6 ) 3 0 213

21 5 21

Test: x 5 22

( 3 √_________

7(22) 1 6 ) 3 0 223

28 5 28

Test: x 5 3

( 3 √________

7(3) 1 6 ) 3 0 33

27 5 27

The square root of a negative is not a real number, but the cube root of a negative number is a real number . There are no extraneous solutions .

▸The solutions are x 5 22, 21, and 3 .

3

1

Solve the resulting cubic polynomial .

Use the rational roots theorem to find possible roots .

p

__ q 5 61, 62, 63, 66

Try 21 .

(21)327(21)26 5 21 1 726

5 0

So, 21 is a root and (x 1 1) is a factor .

Use synthetic division to divide by (x 1 1) .

�1 1

1 �1 �6 0

0 �7 �6

�1 1 6

Factor the resulting quadratic factor .

x22x26 5 (x 1 2)(x23)

Write the polynomial equation in factored form .

0 5 (x 1 1)(x 1 2)(x23)

The solutions are 21, 22, and 3 .

2

EXAMPLE A Solve: 3 √______

7x 1 6 5 x

Graph the solution to 3 √______

7x 1 6 , x .

TRY

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Lesson 16: Radical Equations and Inequalities 137

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Solve: 3 1 √_________

(5x210) 1 8 # 8

TRY

Test values on either side of 8 to determine the solution to the inequality .

Test: x 5 0

0 # 24 False

Test: x 5 18

6 # 14 True

▸The complete solution is x $ 8 .

4

Determine allowable values for the radicand .

The radicand must be greater than or equal to 0 .

2x $ 0

x $ 0

The domain is x $ 0 .

Check for extraneous solutions .

Test: x 5 8

√____

2 ? 8 0 824

√___

16 5 4

Test: x 5 2

√____

2 ? 2 0 224

√__

4 22

The only solution is x 5 8 .

3

1

Solve the related equation .

( √___

2x ) 2 5 (x24)2

2x 5 x228x 1 16

0 5 x2210x 1 16

0 5 (x28)(x22)

The solutions are x 5 8 and x 5 2 .

2

EXAMPLE B Solve: √___

2x # x24

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138 Unit 1: Polynomial, Rational, and Radical Relationships

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Isolate the radical expression .

This will make it much simpler to square both sides .

x24 5 √_____

x 1 2

Check for extraneous solutions .

Use the original equation .

Test: x 5 2

22√_____

2 1 2 0 4

0 4

Test: x 5 7

72√_____

7 1 2 0 4

4 5 4

▸The only solution is x 5 7 .

3

1

Solve .

Square both sides .

(x24)2 5 ( √_____

x 1 2 ) 2

x228x 1 16 5 x 1 2

x229x 1 14 5 0

(x22)(x27) 5 0

The solutions are x 5 2 and x 5 7 .

2

EXAMPLE C Solve: x2√_____

x 1 2 5 4

Solve: 2x 5 √______

5x21 1 1

TRY

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Lesson 16: Radical Equations and Inequalities 139

Problem Solving

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READ

The period of a pendulum is the amount of time required for it to swing from one side to the other and back . A pendulum’s period in seconds, P, is related to the length of the pendulum in meters, L, by the following equation:

P 5 2p √___

L ___ 9 .8

A pendulum in a grandfather clock has a period of 2 .5 seconds . Find the length of the pendulum to the nearest tenth of a meter .

PLAN

Substitute the value of the period for P, and solve for L .

SOLVE

Substitute for P into the equation .

5 2p √___

L ___ 9 .8

Solve for L:

( )2 5 ( 2p √___

L ___ 9 .8 ) 2

5 22p

2

L m

CHECK

Substitute the value of L into the equation that relates P and L, and solve .

When L 5 , P .

▸ To the nearest tenth of a meter, the pendulum is m long .

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Practice

140 Unit 1: Polynomial, Rational, and Radical Relationships

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Identify the radicand and index of each equation.

1. y 5 3 √_____

x 1 1

radicand:

index:

3. 3 √_______

2x25 1 1 5 6

radicand:

index:

2. y 5 4 √___

2x

radicand:

index:

4. √_______

1 __ 2 x 1 10 5 50

radicand:

index:

Solve.

5. √______

3x 1 1 5 5 6. √______

9x 1 1 5 10

REMEMBER Raise both sides of the equation to the power equal to the index.

7. 3 √______

2x 1 1 1 1 5 4

9. √_______

2n 1 8 5 n

8. √______

x 1 20 5 x

REMEMBER Check for extraneous solutions.

10. 2n 5 √_______

11n 1 3

Choose the best answer.

11. Solve: √______

x 1 16 5 x24

A. x 5 0 only

B. x 5 9 only

C. x 5 0 and x 5 9

D. There is no solution .

12. Solve: √_______

3n 1 4 1 4 5 3

A. x 5 21 only

B. x 5 1 only

C. x 5 21 and x 5 1

D. There is no solution .

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Lesson 16: Radical Equations and Inequalities 141

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Solve.

13. √_______

2x22 5 x25

15. x2√_____

x21 5 3

14. √_____

x 1 1 1 5 5 x

16. 3 √___

4x 5 x

Solve.

17. A pendulum’s period in seconds, P, is related to the length of the pendulum in meters, L, by the following equation:

P 5 2p √___

L ___ 9 .8

A pendulum in a grandfather clock has a period of 1 .7 seconds . Find its length to the nearest tenth of a meter .

The pendulum’s length is m .

Solve each inequality. Graph the solution on the number line.

18. √_____

x 1 2 . 3

–8 –7 –6 –5 –4 –2 –1–3 0 1 2 3 4 5 6 7 8

19. √______

2x 1 6 # 4

–8 –7 –6 –5 –4 –2 –1–3 0 1 2 3 4 5 6 7 8

20. √______

2x21 # x

–8 –7 –6 –5 –4 –2 –1–3 0 1 2 3 4 5 6 7 8

21. √______

3x 1 1 . x21

–6 –5 –4 –2 –1–3 0 1 2 3 4 5 6

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142 Unit 1: Polynomial, Rational, and Radical Relationships

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Solve.

22. You can estimate s, the speed in miles per hour, at which a car is moving when it goes into a skid . Use the formula s 5 √

____ 21d , where d is the length of the skid marks in feet .

During an accident, a driver claims to have been driving 42 miles per hour . If the driver’s estimate of his speed is accurate, about how long would the skid marks be?

The skid marks would be about ft long .

The actual skid marks are about 115 ft . Is the driver’s estimate of his speed accurate? Explain .

Solve.

23. EXTEND Solve: √_____

x 1 8 2√__

x 5 2

Isolate the first radical .

√_____

x 1 8 5

Square both sides of the equation .

x 1 8 5

Isolate the radical expression in the resulting equation .

5 √__

x

Square both sides of the resulting equation .

5 x

Show that your answer is correct by substituting it into the original equation .

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Lesson 16: Radical Equations and Inequalities 143

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24. EXTEND Solve: 3 √_____

x 1 4 5 √__

x

Raise both sides of the equation to the same power so that all radicals are eliminated .

To do this, raise both sides to the power . Then, gather all terms on one side .

0 5

Find all real zeros of the resulting equation .

The only real zero is x 5 .

Check the answer .

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