Algebra unit 10.4

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UNIT 10.4 FACTORING TO UNIT 10.4 FACTORING TO SOLVE QUADRATIC EQUATIONS SOLVE QUADRATIC EQUATIONS

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Unit 10.4

Transcript of Algebra unit 10.4

Page 1: Algebra unit 10.4

UNIT 10.4 FACTORING TO UNIT 10.4 FACTORING TO SOLVE QUADRATIC EQUATIONSSOLVE QUADRATIC EQUATIONS

Page 2: Algebra unit 10.4

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 ContinuedUse 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 0

Check (x – 7)(x + 2) = 0(–2 – 7)(–2 + 2) 0

(–9)(0) 00 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 = 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) 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 = 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

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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.

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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) 0

0 0Check (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 = 0x = 4 or x = 2

The 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 0

16 – 24 + 8 0 0 0

Checkx2 – 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 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.

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Example 2B ContinuedSolve 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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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.

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Example 2C ContinuedSolve the quadratic equation by factoring. Check your answer.

x2 – 12x + 36 = 0Check 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 Solve the quadratic equation by factoring. Check your answer.

–2x2 = 20x + 50The 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.

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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.

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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 Solve the quadratic equation by factoring. Check your answer.

x2 – 6x + 9 = 0(x – 3)(x – 3) = 0

x – 3 = 0 or x – 3 = 0x = 3 or x = 3

Both 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 0

9 – 18 + 9 0 0 0

Check

Substitute 3 into the original equation.

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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.

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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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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.

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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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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.

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

3x2 – 4x + 1 = 0

3x2 – 4x + 1 = 03(1)2 – 4(1) + 1 0

3 – 4 + 1 0 0 0

Check3x2 – 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.

Factor the trinomial.

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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.

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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.

Factor the trinomial.

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–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

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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 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.

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