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Example 1 Use Properties of Square Roots Simplify the expression. a . 18 b . 2 18 c . 9 5 SOLUTION 9 a . 18 = 2 = 2 3 c . 9 5 5 9 = = 5 3 4 b . = = 5 2 10 20 = 5 2

Example 1 Use Properties of Square Roots Simplify the expression. a. 18 b. 2 18 c. 9 5 SOLUTION 9 a. 18 = 2 = 2 3 c. 9 5 5 9 == 5 3 4 b. = = 5 2 1020 =

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Page 1: Example 1 Use Properties of Square Roots Simplify the expression. a. 18 b. 2 18 c. 9 5 SOLUTION 9 a. 18 = 2 = 2 3 c. 9 5 5 9 == 5 3 4 b. = = 5 2 1020 =

Example 1 Use Properties of Square Roots

Simplify the expression.

a. 18 b. 2 • 18 c.9

5

SOLUTION

9 •a. 18 = 2 = 23

c.9

5 5

9= =

5

3

4•b. = = 5210 20 = 5•2

Page 2: Example 1 Use Properties of Square Roots Simplify the expression. a. 18 b. 2 18 c. 9 5 SOLUTION 9 a. 18 = 2 = 2 3 c. 9 5 5 9 == 5 3 4 b. = = 5 2 1020 =

Example 2 Rationalize the Denominator of a Fraction

Simplify .2

5

SOLUTION

2

5 5

2= Quotient property of square roots

5

2= •

2

2

2

2Multiply by .

=10

2Simplify.

Page 3: Example 1 Use Properties of Square Roots Simplify the expression. a. 18 b. 2 18 c. 9 5 SOLUTION 9 a. 18 = 2 = 2 3 c. 9 5 5 9 == 5 3 4 b. = = 5 2 1020 =

Checkpoint

Simplify the expression.

1.

Use Properties of Square Roots

12

2. 3•15

3.3

7

ANSWER 32

ANSWER 53

ANSWER21

3

Page 4: Example 1 Use Properties of Square Roots Simplify the expression. a. 18 b. 2 18 c. 9 5 SOLUTION 9 a. 18 = 2 = 2 3 c. 9 5 5 9 == 5 3 4 b. = = 5 2 1020 =

Example 3 Solve a Quadratic Equation

Solve .=x 2 + 1 13

Subtract 1 from each side. = 12x 2

SOLUTION

=x 2 + 1 13 Write original equation.

=x +– 12 Take the square root of each side.

=x +– 4 3• Product property of square roots

=x +– 32 Simplify.

ANSWER The solutions are and . 32 32–

Page 5: Example 1 Use Properties of Square Roots Simplify the expression. a. 18 b. 2 18 c. 9 5 SOLUTION 9 a. 18 = 2 = 2 3 c. 9 5 5 9 == 5 3 4 b. = = 5 2 1020 =

Example 3 Solve a Quadratic Equation

CHECK 32 32–Substitute and into the original equation.

=x 2 + 1 13 =x 2 + 1 13

+ 1 13=?4 3• + 1 13=?4 3•

=12 + 1 13 =12 + 1 13

+ 1 1332( )2 =? + 1 13=?32( )2–

Page 6: Example 1 Use Properties of Square Roots Simplify the expression. a. 18 b. 2 18 c. 9 5 SOLUTION 9 a. 18 = 2 = 2 3 c. 9 5 5 9 == 5 3 4 b. = = 5 2 1020 =

Checkpoint

Solve the equation.

Solving a Quadratic Equation

4. =x 2 4– 14

5. =x 2 3 13+

ANSWER 2,3 23–

ANSWER 10, 10–

=3y 2 246. ANSWER 2,2 22–

Page 7: Example 1 Use Properties of Square Roots Simplify the expression. a. 18 b. 2 18 c. 9 5 SOLUTION 9 a. 18 = 2 = 2 3 c. 9 5 5 9 == 5 3 4 b. = = 5 2 1020 =

Example 4 Use a Quadratic Equation as a Model

Skydiving A skydiver jumps from an airplane that is 6000 feet above the ground. The skydiver opens her parachute when she is 2500 feet above the ground.

a. Write an equation that gives the height (in feet) of the skydiver above the ground as a function of time

(in seconds).

b. For how many seconds does the skydiver fall before opening her parachute?

SOLUTION

a. The initial height of the skydiver is h0 6000.=

Write falling object model.–= 16t 2h + h0

Page 8: Example 1 Use Properties of Square Roots Simplify the expression. a. 18 b. 2 18 c. 9 5 SOLUTION 9 a. 18 = 2 = 2 3 c. 9 5 5 9 == 5 3 4 b. = = 5 2 1020 =

Example 4 Use a Quadratic Equation as a Model

Substitute 6000 for h0.–= 16t 2h + 6000

b. The height of the skydiver when she opens her parachute is h 2500. Substitute 2500 for h in the model from part (a). Solve for t.

=

Write model from part (a).–= 16t 2h + 6000

Substitute 2500 for h.–= 16t 22500 + 6000

Subtract 6000 from each side.–= 16t 23500–

Divide each side by 16.t 2=16

3500–

––

Page 9: Example 1 Use Properties of Square Roots Simplify the expression. a. 18 b. 2 18 c. 9 5 SOLUTION 9 a. 18 = 2 = 2 3 c. 9 5 5 9 == 5 3 4 b. = = 5 2 1020 =

Example 4 Use a Quadratic Equation as a Model

Take the square root of each side.t =16

3500–

––+

Use a calculator.–+ t ≈15

ANSWER

Reject the solution 15, because time must be positive. The skydiver falls for about 15 seconds before opening her parachute.

Page 10: Example 1 Use Properties of Square Roots Simplify the expression. a. 18 b. 2 18 c. 9 5 SOLUTION 9 a. 18 = 2 = 2 3 c. 9 5 5 9 == 5 3 4 b. = = 5 2 1020 =

Checkpoint

7. Skydiving A skydiver jumps from a plane that is 5000 feet above the ground. The skydiver opens his parachute when he is 2000 feet above the ground.

Use a Quadratic Equation

a. Write an equation that gives the height (in feet) of the skydiver above the ground as a function of time (in seconds).

ANSWER –= 16t 2h + 5000

b. For how many seconds does the skydiver fall before opening his parachute?

ANSWER about 14 sec