8-6 Solving Quadratic Equations by Factoring · (2 –2)(2) 0 (0)(2) 0 ... Example 2B: Solving Quadratic Equations by Factoring Solve the quadratic equation by factoring. Check your

Holt McDougal Algebra 1

8-6Solving Quadratic Equations

by Factoring8-6Solving Quadratic Equations by Factoring

Holt Algebra 1

Warm Up

Lesson Presentation

Lesson Quiz




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



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Solve quadratic equations by factoring.

Objective



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



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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.



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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 0✓

Check (x – 7)(x + 2) = 0

(–2 – 7)(–2 + 2) 0

(–9)(0) 0

0 0✓



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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 = 0

x = 2

The solutions are 0 and 2.


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) 0

0 0✓

(x – 2)(x) = 0

(2 – 2)(2) 0

(0)(2) 0

0 0✓

(x)(x – 2) = 0



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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 = 0

x = –4



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✓



by Factoring

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.


Property.Solve each equation.



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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) 0

0 0✓

Check (x + 4)(x – 3 ) = 0

(3 + 4)(3 –3) 0

(7)(0) 0

0 0✓



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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.



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

Helpful Hint



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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 = 2

The solutions are 4 and 2.

Factor the trinomial.


Property. Solve each equation.

x2 – 6x + 8 = 0

(4)2 – 6(4) + 8 0

16 – 24 + 8 0

0 0✓

Check

x2 – 6x + 8 = 0

(2)2 – 6(2) + 8 0

4 – 12 + 8 0

0 0✓

Check



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Example 2B: Solving Quadratic Equations by Factoring


x2 + 4x = 21

x2 + 4x = 21–21 –21

x2 + 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.




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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. ✓

● ●



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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.






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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. ✓

●



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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.



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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.



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

Helpful Hint



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Check It Out! Example 2a


x2 – 6x + 9 = 0

(x – 3)(x – 3) = 0

x – 3 = 0 or x – 3 = 0

x = 3 or x = 3

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




x2 – 6x + 9 = 0

(3)2 – 6(3) + 9 0

9 – 18 + 9 0

0 0✓

Check

Substitute 3 into the

original equation.



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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




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Check It Out! Example 2b Continued


x2 + 4x = 5


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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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.



– 1 cannot equal 0.

Solve the remaining equation.

Factor out the GCF, –1.



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Check It Out! Example 2c Continued


30x = –9x2 – 25


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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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.



by Factoring

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✓

Check

3x2 – 4x + 1 = 0

3 – 4 + 1 0

0 0✓

Check



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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.




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Example 3 Continued

–8 ≠ 0, 2t + 1 = 0 or t – 1= 0Use 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


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.



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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.




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–8 ≠ 0, 2t – 3 = 0 or t + 1= 0Use 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


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



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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



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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 his 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.

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8-6 Solving Quadratic Equations by Factoring · (2 –2)(2) 0 (0)(2) 0 ... Example 2B: Solving Quadratic Equations by Factoring Solve the quadratic equation by factoring. Check your