Standard 8 Solve a quadratic equation

Solve 6(x – 4)2 = 42. Round the solutions to the nearesthundredth.

6(x – 4)2 = 42 Write original equation.

(x – 4)2 = 7 Divide each side by 6.

x – 4 = ± 7 Take square roots of each side.

7 x = 4 ± Add 4 to each side.

ANSWER

The solutions are 4 + 6.65 and 4 – 1.35.7 7

Solve a quadratic equation

Standard 8

CHECKTo check the solutions, first write the equation so that 0 is on one side as follows: 6(x – 4)2 – 42 = 0. Then graph the related function y = 6(x – 4)2 – 42. The x-intercepts appear to be about 6.6 and about 1.3. So, each solution checks.

EXAMPLE 1 Solve quadratic equations

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

1. 2(x – 2)2 = 18

GUIDED PRACTICE Solve a Quadratic Equation

ANSWER –1, 5

2. 4(q – 3)2 = 28 ANSWER 0.35, 5.65

3. 3(t + 5)2 = 24 ANSWER –7.83, –2.17

EXAMPLE 1 Solve quadratic equations

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.

Solve quadratic equationsEXAMPLE 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.

Solve quadratic equationsEXAMPLE 1

c. b2 + 12 = 5Write original equation.

b2 = – 7 Subtract 12 from each side.

ANSWER

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

EXAMPLE 2 Take square roots of a fraction

Solve 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.

Take square roots of a fractionEXAMPLE 2

ANSWER

The solutions are – and 32

32

Approximate solutions of a quadratic equation

EXAMPLE 3

Solve 3x2 – 11 = 7. Round the solutions to the nearesthundredth.

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.

Approximate solutions of a quadratic equation

EXAMPLE 3

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

ANSWER

The solutions are about – 2.45 and about 2.45.

EXAMPLE 1 Solve quadratic equations

Solve the equation.

1. c2 – 25 = 0

GUIDED PRACTICE

ANSWER –5, 5.

2. 5w2 + 12 = – 8 ANSWER no solution

3. 2x2 + 11 = 11 ANSWER 0

EXAMPLE 1 Solve quadratic equations

Solve the equation.

4. 25x2 = 16

GUIDED PRACTICE

ANSWER 4 5

4 5– ,

5. 9m2 = 100 ANSWER 103– ,

103

6. 49b2 + 64 = 0 ANSWER no solution

EXAMPLE 1 Solve quadratic equations

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

7. x2 + 4 = 14

GUIDED PRACTICE

ANSWER – 3.16, 3.16

8. 3k2 – 1 = 0 ANSWER – 0.58, 0.58

9. 2p2 – 7 = 2 ANSWER – 2.12, 2.12

Standard 8 Complete the square

Find the value of c that makes the expression x2 + 5x + c a perfect square trinomial. Then write the expression as the square of a binomial.

STEP 1Find the value of c. For the expression to be a perfect square trinomial, c needs to be the square of half the coefficient of bx.

Find the square of half the coefficient of bx.

22 = 25

4c = 5

Standard 8 Complete the square

STEP 2Write the expression as a perfect square trinomial. Then write the expression as the square of a binomial.

Substitute 25 for c.4

x2 + 5x + c = x2 + 5x +254

Square of a binomial52

2+x =

GUIDED PRACTICE

Find the value of c that makes the expression a perfect square trinomial. Then write the expression as the square of a binomial.

1. x2 + 8x + c ANSWER 16; (x + 4)2

2. x2 12x + c

3. x2 + 3x + c

ANSWER 36; (x 6)2

ANSWER ; (x )294

32

EXAMPLE 2 Solve a quadratic equation

Solve x2 – 16x = –15 by completing the square.

SOLUTION

Write original equation.x2 – 16x = –15

Add , or (– 8)2, to each side.

– 16 2

2x2 – 16x + (– 8)2 = –15 + (– 8)2

Write left side as the square of a binomial.

(x – 8)2 = –15 + (– 8)2

Simplify the right side.(x – 8)2 = 49

EXAMPLE 2 Standardized Test Practice

Take square roots of each side.x – 8 = ±7

Add 8 to each side.x = 8 ± 7

ANSWER

The solutions of the equation are 8 + 7 = 15 and 8 – 7 = 1.

EXAMPLE 2 Standardized Test Practice

CHECK

You can check the solutions in the original equation.

If x = 15:

(15)2 – 16(15) –15 ?=

–15 = –15

If x = 1:

(1)2 – 16(1) –15 ?=

–15 = –15

EXAMPLE 3 Solve a quadratic equation in standard form

Solve 2x2 + 20x – 8 = 0 by completing the square.

SOLUTION

Write original equation.2x2 + 20x – 8 = 0

Add 8 to each side.2x2 + 20x = 8

Divide each side by 2.x2 + 10x = 4

Add 10 2

2, or 52, to each side.x2 + 10x + 52 = 4 + 52

Write left side as the square of a binomial.

(x + 5)2 = 29

EXAMPLE 3 Solve a quadratic equation in standard form

Take square roots of each side.x + 5 =

± 29

Subtract 5 from each side.x = –5 ± 29

ANSWER

The solutions are – 5 + 29 0.39 and – 5 - 29 –10.39.

GUIDED PRACTICE

4. x2 – 2x = 3

ANSWER 1, 3

5. m2 + 10m = –8

ANSWER 9.12, 0.88

6. 3g2 – 24g + 27 = 0

ANSWER 1.35, 6.65