9.1 – Students will be able to evaluate square roots.Students will be able to solve a quadratic equation by finding the square root. 1. 3x +(– 6x) Warm-Up

9.1 – Students will be able to evaluate square roots. Students will be able to solve a quadratic equation by finding the square root.

1. 3x +(– 6x)

Warm-UpSimplify the expression.

2. 5 + 4x + 2

–3 xANSWER

4x + 7ANSWER

3. 4(2x – 1) + x 9x – 4ANSWER

4. – (x + 4) – 6 x – 7x – 4ANSWER

9.1 – Students will be able to evaluate square roots. Students will be able to solve a quadratic equation by finding the square root. Warm-Up

Simplify the expression.

13. (x5)3

14. (– x)3

11. (3xy)3 27x3y3ANSWER

12. xy2 xy3 x2y5ANSWER

x15ANSWER

–x3ANSWER



Find the two square roots of each number.

7 is a square root, since 7 • 7 = 49.

–7 is also a square root, since –7 • –7 = 49.



49 = –7–

49 = 7

100 = 10

100 = –10–

A. 49

B. 100

C. 225

15 is a square root, since 15 • 15 = 225.225 = 15

225 = –15– –15 is also a square root, since –15 • –15 = 225.


A. 25

Check It Out: Example 1

5 is a square root, since 5 • 5 = 25.–5 is also a square root, since –5 • –5 = 25.



25 = –5–25 = 5

144 = 12

144 = –12–

Find the two square roots of each number.

B. 144

C. 289289 = 17

289 = –17–




Additional Example 3A: Evaluating Expressions Involving Square RootsSimplify the expression.

Evaluate the square root.

Add.= 25

Multiply.= 18 + 7

3 36 + 7

3 36 + 7 = 3(6) + 7


Additional Example 3B: Evaluating Expressions Involving Square RootsSimplify the expression.

+ 25 16

3 4

25 16

3 4

+3 4 = +1.5625

Evaluate the square roots.

= 1.25 + 3 4

25 16

= 1.5625.

= 2 Add.


Check It Out: Example 3ASimplify the expression.

Evaluate the square root.

Add.= 14

Multiply.= 10 + 4

2 25 + 4

2 25 + 4 = 2(5) + 4


Check It Out: Example 3BSimplify the expression.

+ 18 t2

1 4

18 t2

1 4

+ 1 4

= + 9

Evaluate the square roots.= 3 + 1 4

18 t2

= 9.

= 3 Add.1 4


3. Evaluate the expression.

A. 17

B. 17

C. 19

D. 72

Lesson Quiz for Student Response Systems


4. Evaluate the expression.

A. 4

B. 8

C. 16

D. 40

Lesson Quiz for Student Response Systems


A Quadratic equation is an equation that can be written in the following standard form:

ax2 + bx + c = 0 where a does not equal 0

If b = 0ax2 + c = 0

These are the type we will work with today.

If b = 0 and c = 0ax2 = 0


Solving Quadratic Equations

a.x2 = 4

b.x2 = 5

c.x2 = 0

d.x2 = -1



a.x2 = 25

b.x2 = 7

c.x2 = 81

d.x2 = -12


Solve the equation.a. 2x2 = 8

SOLUTION

a. 2x2 = 8Write original equation.

x2 = 4 Divide each side by 2.

x = ± 4 = ± 2 Take square roots of each side. Simplify.

ANSWER The solutions are – 2 and 2.


EXAMPLE 1

b. m2 – 18 = – 18 Write original equation.

m2 = 0 Add 18 to each side..

The square root of 0 is 0.m = 0

ANSWER

The solution is 0.


EXAMPLE 1c. b2 + 12 = 5

Write original equation.

b2 = – 7 Subtract 12 from each side.

ANSWER

Negative real numbers do not have real square roots. So, there is no solution.


Solve quadratic equationsSolve the equation.1. c2 – 25 = 0

SOLUTION

c2 – 25 = 0 Write original equation.

c = ± 25 = ± 5 Take square roots of each side. Simplify.

GUIDED PRACTICE

ANSWER

The solutions are – 5 and 5.


EXAMPLE 2

Take square roots of a fractionSolve 4z2 = 9.

SOLUTION

4z2 = 9. Write original equation.

z2 = 94 Divide each side by 4.

Take square roots of each side.z = ± 94

z = ± 32

Simplify.

ANSWER

The solutions are – and 32

32


Approximate solutions of a quadratic equation Solve 3x2 – 11 = 7. Round the solutions to the nearest

hundredth.

SOLUTION

3x2 – 11 = 7 Write original equation.

3x2 = 18 Add 11 to each side.

x2 = 6 Divide each side by 3.

x = ± 6 Take square roots of each side.

x ± 2.45 Use a calculator. Round to the nearesthundredth.

The solutions are about – 2.45 and about 2.45.


EXAMPLE 1

Solve quadratic equationsSolve the equation.2. 5w2 + 12 = – 8

SOLUTION

5w2 + 12 = – 8 Write original equation.

w = –4 Take square roots of each side. Simplify.

GUIDED PRACTICE

ANSWER

Negative real numbers do not have a real square root.So there is no solution.

5w2 = – 8 –12 Subtract 12 from each side.


Solve quadratic equationsSolve the equation.3. 2x2 + 11 = 11

SOLUTION

2x2 + 11 = 11 Write original equation.

x = 0 The root of 0 is 0.

GUIDED PRACTICE

ANSWER

The solution is 0 .

2x2 = 0 Subtract 11 from each side.

9.1 – Students will be able to evaluate square roots. Students will be able to solve a quadratic equation by finding the square root.EXAMPLE 1

Solve quadratic equationsSolve the equation.4. 25x2 = 16

SOLUTION

25x2 = 16 Write original equation.

Take square roots of each side.

GUIDED PRACTICE

Divided each to by 25.x =1625

x = ± 1625

x = ± 4 5

Simplify.

ANSWER

The solution is – and . 4 5

4 5


Solve quadratic equationsSolve the equation.5. 9m2 = 100

SOLUTION

9m2 = 100 Write original equation.


GUIDED PRACTICE

Divided each to by 9.m =100 9

m = ± 100 9

m = ± 10 3

Simplify.

ANSWER

The solution is – and . 10 3

10 3

9.1 – Students will be able to evaluate square roots. Students will be able to solve a quadratic equation by finding the square root.EXAMPLE 1

Solve quadratic equationsSolve the equation.6. 49b2 + 64 = 0

SOLUTION

49b2 + 64 = 0 Write original equation.


GUIDED PRACTICE

Divided each to by 9.

49b2 = – 64

b2 = –64 49

Subtract 64 from each side.

ANSWERNegative real numbers do have real square root. So there is no solution.

b = – 6449


Solve quadratic equationsSolve the equation. Round the solution to the nearest hundredth.

7. x2 + 4 = 14

SOLUTION

x2 + 4 = 14 Write original equation.


GUIDED PRACTICE

Use a calculation. Round to the nearest hundredth.

x2 = 10 Subtract 4 from each side.

ANSWER The solutions are about – 3.16 and 3.16.

x = 10+–

x = +– 3.16


Solve quadratic equationsSolve the equation. Round the solution to the nearest hundredth.

8. 3k2 – 1 = 0

SOLUTION

3k2 – 1 = 0 Write original equation.


GUIDED PRACTICE


3k2 = 1 Add 1 to each side.

k = +– 0.58

k2 = 1 3

k = +– 13


The solutions are about – 0.58 and 0.58.


Solve the equation. Round the solution to the nearest hundredth.

9. 2p2 – 7 = 2

SOLUTION

2p2 – 7 = 2 Write original equation.



2p2 = 2 + 7 Add 7 to each side.

p = +– 2.12

p2 = 9 2

p = +– 92


GUIDED PRACTICE

The solutions are about – 2.12 and 2.12.



a.x2 + 5 = 21

b.x2 – 2 = 7

c.2x2 = 18

d.3x2 = 75



a.2x2 -8 = 0

b.x2 +25 = 0

c.x2 - 1.44 = 0

d.5x2 = -15



a.3x2 -48 = 0

b.120 - 6x2 = -30

c.12x2 - 60 = 0

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9.1 – Students will be able to evaluate square roots.Students will be able to solve a quadratic equation by finding the square root. 1. 3x +(– 6x) Warm-Up