holt mcdougal algebra 1 8-6 solving quadratic equations by factoring 8-6 solving quadratic equations...

Holt McDougal Algebra 1

8-6 Solving Quadratic Equations by Factoring8-6 Solving Quadratic Equations

by Factoring

Holt Algebra 1

Warm UpWarm Up

Lesson PresentationLesson Presentation

Lesson QuizLesson Quiz



8-6 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



Solve quadratic equations by factoring.

Objective



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



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.



Example 1A Continued


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



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.


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



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



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



Check It Out! Example 1b


(x + 4)(x – 3) = 0

x + 4 = 0 or x – 3 = 0

x = –4 or x = 3

The solutions are –4 and 3.





Check It Out! Example 1b Continued


(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



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.



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

Helpful Hint



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.



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



Example 2B: Solving Quadratic Equations by Factoring


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.


Use the Zero Product Property.Solve each equation.



Example 2B Continued


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.

● ●



Example 2C: Solving Quadratic Equations by Factoring


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.






Example 2C Continued


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.

●



Example 2D: Solving Quadratic Equations by Factoring


–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


Solve the equation.



Example 2D Continued


–2x2 = 20x + 50

Check

–2x2 = 20x + 50

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

Substitute –5 into the original equation.



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

Helpful Hint



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.




x2 – 6x + 9 = 0

(3)2 – 6(3) + 9 09 – 18 + 9 0

0 0

Check

Substitute 3 into the original equation.



Check It Out! Example 2b


x2 + 4x = 5

x2 + 4x = 5–5 –5

x2 + 4x – 5 = 0

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




(x – 1)(x + 5) = 0

x – 1 = 0 or x + 5 = 0

x = 1 or x = –5




Check It Out! Example 2b Continued


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.

●●



Check It Out! Example 2c


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.


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

Solve the remaining equation.

Factor out the GCF, –1.



Check It Out! Example 2c Continued


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.

●



Check It Out! Example 2d


3x2 – 4x + 1 = 0

(3x – 1)(x – 1) = 0

or x = 1




3x – 1 = 0 or x – 1 = 0

The solutions are and x = 1.



Check It Out! Example 2d Continued


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



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.




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


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.



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.




–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


0 –16(1.5)2 + 8(1.5) + 24

0 –36 + 12 + 24 0 0

Check It Out! Example 3 Continued

t = 1.5



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



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.

holt mcdougal algebra 1 8-6 solving quadratic equations by factoring 8-6 solving quadratic equations...

Documents