lesson 4.5, for use with pages 266-271 find the exact value. answer 7 –12 1.√49 2.–√144

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Lesson 4.5, For use with pages 266-271

Find the exact value.

ANSWER 7

ANSWER –12

1. √49

2. –√144

Lesson 4.5, For use with pages 266-271

ANSWER 2.3

ANSWER 11 ft

4. The area of half of a square mural is 60 square feet. What is the length of a side of the mural?

12

163. Use calculator to approximate the value of

to the nearest tenth.

82

4.5 Properties of Square Roots

EXAMPLE 1 Use properties of square roots

Simplify the expression.

5= 4a.

80 516=

= 3 14b.

6 21 126= 9 14=

c.

4

81=

4

81=

2 9

d.

7

16=

7

16=

4

7

GUIDED PRACTICEGUIDED PRACTICE for Example 1

271.

3= 327 39=

SOLUTION

982.

2= 798 249=

SOLUTION

GUIDED PRACTICEGUIDED PRACTICE for Example 1

3.

6= 5

SOLUTION

SOLUTION

10 15

10 15 625150 ==

4. 8 28

14= 41416=224= 8 28

GUIDED PRACTICEGUIDED PRACTICE for Example 1

5.

SOLUTION

SOLUTION

9

64=

3 8

9

64

9

64=

6. 15

4

15

4

15

4= =

15

2

GUIDED PRACTICEGUIDED PRACTICE for Example 1

7.

SOLUTION

SOLUTION

11

25

11

25

11

25=

8. 36

49

36

4936

49=

= 5

11

76

=

EXAMPLE 2 Rationalize denominators of fractions.

Simplify (a) 5

2and 3

7 + 2

SOLUTION

(a) 5

2

2

10=

=5

2

=5

2

2

2

(b)

EXAMPLE 2 Rationalize denominators of fractions.

SOLUTION

(b)3

7 + 2=

3

7 + 2 7 – 2

7 – 2

=21 – 3 2

49 – 7 + 7 – 2 2 2

=21 – 3 2

47

EXAMPLE 3 Solve a quadratic equation

Solve 3x2 + 5 = 41.

3x2 + 5 = 41 Write original equation.

3x2 = 36 Subtract 5 from each side.

x2 = 12 Divide each side by 3.

x = + 12 Take square roots of each side.

x = + 4 3

x = + 2 3

Product property

Simplify.

EXAMPLE 3 Solve a quadratic equation

ANSWER

The solutions are and 2 3 2 3–

Check the solutions by substituting them into the original equation.

3x2 + 5 = 41

3( )2 + 5 = 41 ?2 3

41 = 41

3(12) + 5 = 41 ?

3x2 + 5 = 41

3( )2 + 5 = 41 ? – 2 3

41 = 41

3(12) + 5 = 41 ?

EXAMPLE 4 Standardized Test Practice

SOLUTION

15 (z + 3)2 = 7 Write original equation.

(z + 3)2 = 35 Multiply each side by 5.

z + 3 = + 35 Take square roots of each side.

z = – 3 + 35 Subtract 3 from each side.

The solutions are – 3 + and – 3 – 35 35

EXAMPLE 4 Standardized Test Practice

ANSWER

The correct answer is C.

GUIDED PRACTICE for Examples 2, 3, and 4GUIDED PRACTICE

Simplify the expression.

9.

SOLUTION

6

5

5

30=

=6

5

=6

5

5

5

6

5

GUIDED PRACTICE for Examples 2, 3, and 4GUIDED PRACTICE

Simplify the expression.

10.

SOLUTION

9

8=

9

8

=9

8

8

8

9

8

2

4=

3

GUIDED PRACTICE for Examples 2, 3, and 4GUIDED PRACTICE

Simplify the expression.

11.

SOLUTION

17

12=

17

12

=17

12

12

12

17

12

51

12=

2 51

6=

GUIDED PRACTICE for Examples 2, 3, and 4GUIDED PRACTICE

Simplify the expression.

12.

SOLUTION

19

21

399

21=

=19

21

=19

21

21

21

19

21

SOLUTION

– 6

7 – 5=

– 6

7 – 5 7 + 5

7 + 5

=– 42 – 6 5

49 – 7 + 7 – 5 5 5

=– 21 – 3 5

22

for Examples 2, 3, and 4GUIDED PRACTICE

13. – 6

7 – 5

SOLUTION

2

4 + 11=

2

4 + 11 4 – 11

4 – 11

=8 – 2 11

16 – 4 + 4 – 1111 11

=8 – 2 11

5

for Examples 2, 3, and 4GUIDED PRACTICE

14. 2

4 + 11

SOLUTION

– 1

9 + 7=

– 1

9 + 7 9 – 7

9 – 7

=– 9 + 7

81 – 9 + 9 – 77 7

=– 9 + 7

74

for Examples 2, 3, and 4GUIDED PRACTICE

15. – 1

9 + 7

SOLUTION

4

8 – 3=

4

8 – 3 8 + 3

8 + 3

=32 + 4 3

64 – 4 + 4 – 3 3 3

=32 + 4 3

61

for Examples 2, 3, and 4GUIDED PRACTICE

16. 4

8 – 3

SOLUTION

for Examples 2, 3, and 4GUIDED PRACTICE

17.

Solve the equation.

5x2 = 80

Write original equation.5x2 = 80

x2 = 16 Divide each side by 5.

x = + 16 Take square roots of each side.

x = + 4 4

x = + 4

Product property

Simplify.

for Examples 2, 3, and 4GUIDED PRACTICE

ANSWER

The solutions are and . 4 4–

Check Check the solutions by substituting them into the original equation.

5x2 = 80

5(– 4)2 = 80 ?

80 = 80

5(16) = 80?

5x2 = 80

80 = 80

5(16) = 80?

5(4)2 = 80 ?

SOLUTION

for Examples 2, 3, and 4GUIDED PRACTICE

18.

Solve the equation.

z2 – 7 = 29

Write original equation.

z2 = 36

z = + 36 Take square roots of each side.

z = + 6 6

z = + 6

Product property

Simplify.

z2 – 7 = 29

Add 7 to each side.

for Examples 2, 3, and 4GUIDED PRACTICE

ANSWER

The solutions are and . 6 6–

Check the solutions by substituting them into the original equation.

29 = 29

z2 – 7 = 29

(6)2 – 7 = 29 ?

?36 – 7 = 2929 = 29

z2 – 7 = 29

(– 6)2 – 7 = 29 ?

36 – 7 = 29 ?

SOLUTION

for Examples 2, 3, and 4GUIDED PRACTICE

19.

Write original equation.

Take square roots of each side.

Division property

3(x – 2)2 = 40

3(x – 2)2 = 40

Divide each side by 3.(x – 2)2 =403

+(x – 2) =403

x =40

32 +

x =40

32 + 3

3

=120

32 +

for Examples 2, 3, and 4GUIDED PRACTICE

=4(30)

32 +

= 2 30

32 +

ANSWER

The solutions are . 2 30

32 +

EXAMPLE 5 Model a dropped object with a quadratic function

Science Competition

For a science competition, students must design a container that prevents an egg from breaking when dropped from a height of 50 feet. How long does the container take to hit the ground ?

EXAMPLE 5 Model a dropped object with a quadratic function

SOLUTION

h = – 16t 2 + h0 Write height function. 0 = – 16t 2 + 50 Substitute 0 for h and 50 for h0.

– 50 = – 16t 2 Subtract 50 from each side.5016 = t2 Divide each side by – 16.

50

16+ = t2 Take square roots of each side.

+ 1.8 t Use a calculator.

EXAMPLE 5 Model a dropped object with a quadratic function

ANSWER

Reject the negative solution, – 1.8 , because time must be positive. The container will fall for about 1.8 seconds before it hits the ground.

GUIDED PRACTICE for Example 5GUIDED PRACTICE

What If? In Example 5, suppose the egg container is dropped from a height of 30 feet. How long does the container take to hit the ground?

20.

SOLUTION

h = – 16t 2 + h0 Write height function. 0 = – 16t 2 + 30 Substitute 0 for h and 30 for h0.

– 30 = – 16t 2 Subtract 30 from each side.3016 = t2 Divide each side by – 16.

30

16+ = t2 Take square roots of each side.

+ 1.4 t Use a calculator.

GUIDED PRACTICE for Example 5GUIDED PRACTICE

ANSWER

Reject the negative solution, – 1.4, because time must be positive. The container will fall for about 1.4 seconds before it hits the ground.

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