Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring9-6 Solving Quadratic Equations

by Factoring

Holt Algebra 1

Warm Up

Lesson Presentation

Lesson Quiz

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Warm UpFind each product.

1. (x + 2)(x + 7) 2. (x – 11)(x + 5)

3. (x – 10)2

Factor each polynomial.

4. x2 + 12x + 35 5. x2 + 2x – 63

6. x2 – 10x + 16 7. 2x2 – 16x + 32

x2 + 9x + 14 x2 – 6x – 55 x2 – 20x + 100

(x + 5)(x + 7) (x – 7)(x + 9)

(x – 2)(x – 8) 2(x – 4)2

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Solve quadratic equations by factoring.

Objective

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

You have solved quadratic equations by graphing. Another method used to solve quadratic equations is to factor and use the Zero Product Property.

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Example 1A: Use the Zero Product Property

Use the Zero Product Property to solve the equation. Check your answer.

(x – 7)(x + 2) = 0

x – 7 = 0 or x + 2 = 0

x = 7 or x = –2

The solutions are 7 and –2.

Use the Zero Product Property.

Solve each equation.

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Example 1A Continued

Use the Zero Product Property to solve the equation. Check your answer.

Substitute each solution for x into the original equation.

Check (x – 7)(x + 2) = 0

(7 – 7)(7 + 2) 0(0)(9) 0

0 0Check (x – 7)(x + 2) = 0

(–2 – 7)(–2 + 2) 0(–9)(0) 0

0 0

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Example 1B: Use the Zero Product Property

Use the Zero Product Property to solve each equation. Check your answer.

(x – 2)(x) = 0

x = 0 or x – 2 = 0x = 2

The solutions are 0 and 2.

Use the Zero Product Property.

Solve the second equation.

Substitute each solution for x into

the original equation.

Check (x – 2)(x) = 0

(0 – 2)(0) 0

(–2)(0) 00 0

(x – 2)(x) = 0

(2 – 2)(2) 0 (0)(2) 0

0 0

(x)(x – 2) = 0

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Use the Zero Product Property to solve each equation. Check your answer.

Check It Out! Example 1a

(x)(x + 4) = 0

x = 0 or x + 4 = 0x = –4

The solutions are 0 and –4.

Use the Zero Product Property.

Solve the second equation.

Substitute each solution for x into

the original equation.

Check (x)(x + 4) = 0

(0)(0 + 4) 0

(0)(4) 0 0 0

(x)(x +4) = 0

(–4)(–4 + 4) 0(–4)(0) 0

0 0

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Check It Out! Example 1b

Use the Zero Product Property to solve the equation. Check your answer.

(x + 4)(x – 3) = 0

x + 4 = 0 or x – 3 = 0

x = –4 or x = 3

The solutions are –4 and 3.

Use the Zero Product Property.

Solve each equation.

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Check It Out! Example 1b Continued

Use the Zero Product Property to solve the equation. Check your answer.

(x + 4)(x – 3) = 0

Substitute each solution for x into the original equation.

Check (x + 4)(x – 3 ) = 0

(–4 + 4)(–4 –3) 0

(0)(–7) 00 0

Check (x + 4)(x – 3 ) = 0

(3 + 4)(3 –3) 0

(7)(0) 00 0

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

If a quadratic equation is written in standard form, ax2 + bx + c = 0, then to solve the equation, you may need to factor before using the Zero Product Property.

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

To review factoring techniques, see lessons 8-3 through 8-5.

Helpful Hint

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Example 2A: Solving Quadratic Equations by Factoring

Solve the quadratic equation by factoring. Check your answer.

x2 – 6x + 8 = 0

(x – 4)(x – 2) = 0

x – 4 = 0 or x – 2 = 0

x = 4 or x = 2The solutions are 4 and 2.

Factor the trinomial.

Use the Zero Product Property.

Solve each equation.

x2 – 6x + 8 = 0

(4)2 – 6(4) + 8 016 – 24 + 8 0

0 0

Checkx2 – 6x + 8 = 0

(2)2 – 6(2) + 8 0 4 – 12 + 8 0

0 0

Check

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Example 2B: Solving Quadratic Equations by Factoring

Solve the quadratic equation by factoring. Check your answer.

x2 + 4x = 21x2 + 4x = 21

–21 –21x2 + 4x – 21 = 0

(x + 7)(x –3) = 0

x + 7 = 0 or x – 3 = 0

x = –7 or x = 3

The solutions are –7 and 3.

The equation must be written in standard form. So subtract 21 from both sides.

Factor the trinomial.

Use the Zero Product Property.Solve each equation.

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Example 2B Continued

Solve the quadratic equation by factoring. Check your answer.

x2 + 4x = 21

Check Graph the related quadratic function. The zeros of the related function should be the same as the solutions from factoring.

The graph of y = x2 + 4x – 21 shows that two zeros appear to be –7 and 3, the same as the solutions from factoring.

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Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Example 2C: Solving Quadratic Equations by Factoring

Solve the quadratic equation by factoring. Check your answer.

x2 – 12x + 36 = 0

(x – 6)(x – 6) = 0

x – 6 = 0 or x – 6 = 0

x = 6 or x = 6

Both factors result in the same solution, so there is one solution, 6.

Factor the trinomial.

Use the Zero Product Property.

Solve each equation.

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Example 2C Continued

Solve the quadratic equation by factoring. Check your answer.

x2 – 12x + 36 = 0

Check Graph the related quadratic function.

The graph of y = x2 – 12x + 36 shows that one zero appears to be 6, the same as the solution from factoring.

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Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Example 2D: Solving Quadratic Equations by Factoring

Solve the quadratic equation by factoring. Check your answer.

–2x2 = 20x + 50

The equation must be written in standard form. So add 2x2 to both sides.

Factor out the GCF 2.

+2x2 +2x2

0 = 2x2 + 20x + 50

–2x2 = 20x + 50

2x2 + 20x + 50 = 0

2(x2 + 10x + 25) = 0

Factor the trinomial.2(x + 5)(x + 5) = 0

2 ≠ 0 or x + 5 = 0

x = –5

Use the Zero Product Property.

Solve the equation.

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Example 2D Continued

Solve the quadratic equation by factoring. Check your answer.

–2x2 = 20x + 50

Check

–2x2 = 20x + 50

–2(–5)2 20(–5) + 50–50 –100 + 50–50 –50

Substitute –5 into the original equation.

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

(x – 3)(x – 3) is a perfect square. Since both factors are the same, you solve only one of them.

Helpful Hint

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Check It Out! Example 2a Solve the quadratic equation by factoring. Check your answer.

x2 – 6x + 9 = 0

(x – 3)(x – 3) = 0

x – 3 = 0 or x – 3 = 0

x = 3 or x = 3Both equations result in the same solution, so there is one solution, 3.

Factor the trinomial.

Use the Zero Product Property.

Solve each equation.

x2 – 6x + 9 = 0

(3)2 – 6(3) + 9 09 – 18 + 9 0

0 0

Check

Substitute 3 into the original equation.

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Check It Out! Example 2b

Solve the quadratic equation by factoring. Check your answer.

x2 + 4x = 5

x2 + 4x = 5–5 –5

x2 + 4x – 5 = 0

Write the equation in standard form. Add – 5 to both sides.

Factor the trinomial.

Use the Zero Product Property.

Solve each equation.

(x – 1)(x + 5) = 0

x – 1 = 0 or x + 5 = 0

x = 1 or x = –5

The solutions are 1 and –5.

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Check It Out! Example 2b Continued

Solve the quadratic equation by factoring. Check your answer.

x2 + 4x = 5Check Graph the related quadratic function. The zeros of the related function should be the same as the solutions from factoring.

The graph of y = x2 + 4x – 5 shows that the two zeros appear to be 1 and –5, the same as the solutions from factoring.

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Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Check It Out! Example 2c

Solve the quadratic equation by factoring. Check your answer.

30x = –9x2 – 25

–9x2 – 30x – 25 = 0

–1(3x + 5)(3x + 5) = 0

–1(9x2 + 30x + 25) = 0

–1 ≠ 0 or 3x + 5 = 0

Write the equation in standard form.

Factor the trinomial.

Use the Zero Product Property. – 1 cannot equal 0.

Solve the remaining equation.

Factor out the GCF, –1.

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Check It Out! Example 2c Continued

Solve the quadratic equation by factoring. Check your answer.

30x = –9x2 – 25Check Graph the related quadratic function. The zeros of the related function should be the same as the solutions from factoring.

The graph of y = –9x2 – 30x – 25 shows one zero and it appears to be at , the same as the solutions from factoring.

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Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Check It Out! Example 2d

Solve the quadratic equation by factoring. Check your answer.

3x2 – 4x + 1 = 0

(3x – 1)(x – 1) = 0

or x = 1

Factor the trinomial.

Use the Zero Product Property.

Solve each equation.

3x – 1 = 0 or x – 1 = 0

The solutions are and x = 1.

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Check It Out! Example 2d Continued

Solve the quadratic equation by factoring. Check your answer.

3x2 – 4x + 1 = 0

3x2 – 4x + 1 = 0

3(1)2 – 4(1) + 1 0 3 – 4 + 1 0

0 0

Check3x2 – 4x + 1 = 0

3 – 4 + 1 0

0 0

Check

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Example 3: Application

The height in feet of a diver above the water can be modeled by h(t) = –16t2 + 8t + 8, where t is time in seconds after the diver jumps off a platform. Find the time it takes for the diver to reach the water.

h = –16t2 + 8t + 8

0 = –16t2 + 8t + 8

0 = –8(2t2 – t – 1)

0 = –8(2t + 1)(t – 1)

The diver reaches the water when h = 0.

Factor out the GFC, –8.

Factor the trinomial.

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Example 3 Continued

–8 ≠ 0, 2t + 1 = 0 or t – 1= 0 Use the Zero Product

Property.

2t = –1 or t = 1 Solve each equation.

It takes the diver 1 second to reach the water.

Check 0 = –16t2 + 8t + 8

Substitute 1 into the original equation.

0 –16(1)2 + 8(1) + 8

0 –16 + 8 + 8 0 0

Since time cannot be negative, does not make sense in this situation.

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Check It Out! Example 3

What if…? The equation for the height above the water for another diver can be modeled by h = –16t2 + 8t + 24. Find the time it takes this diver to reach the water.

h = –16t2 + 8t + 24

0 = –16t2 + 8t + 24

0 = –8(2t2 – t – 3)

0 = –8(2t – 3)(t + 1)

The diver reaches the water when h = 0.

Factor out the GFC, –8.

Factor the trinomial.

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

–8 ≠ 0, 2t – 3 = 0 or t + 1= 0 Use the Zero Product Property.

2t = 3 or t = –1 Solve each equation.

Since time cannot be negative, –1 does not make sense in this situation.

It takes the diver 1.5 seconds to reach the water.

Check 0 = –16t2 + 8t + 24

Substitute 1 into the original equation.

0 –16(1.5)2 + 8(1.5) + 24

0 –36 + 12 + 24 0 0

Check It Out! Example 3 Continued

t = 1.5

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Lesson Quiz: Part I

Use the Zero Product Property to solve each equation. Check your answers.

1. (x – 10)(x + 5) = 0

2. (x + 5)(x) = 0

Solve each quadratic equation by factoring. Check your answer.

3. x2 + 16x + 48 = 0

4. x2 – 11x = –24 •

10, –5

–5, 0

–4, –12

3, 8

Holt Algebra 1

9-6 Solving Quadratic Equations by Factoring

Lesson Quiz: Part II

1, –7

–9

–2

5 s

5. 2x2 + 12x – 14 = 0

6. x2 + 18x + 81 = 0

7. –4x2 = 16x + 16

8. The height of a rocket launched upward from a 160 foot cliff is modeled by the function h(t) = –16t2 + 48t + 160, where h is height in feet and t is time in seconds. Find the time it takes the rocket to reach the ground at the bottom of the cliff.