9-5 Solving Quadratic Equations by Factoring

9-5 Solving Quadratic Equations by 9-5 Solving Quadratic Equations by FactoringFactoring

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

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

Additional 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

Use the Zero Product Property.

Solve each equation.

Additional Example 1A Continued


Substitute each solution for x in 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

Additional 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


Solve the second equation.

Substitute each solution

for x in 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 in 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



Check It Out! Example 1b Continued


(x + 4)(x – 3) = 0

Substitute each solution for x in 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

You may need to factor before using the Zero Product Property. You can check your answers by substituting into the original equation or by graphing. If the factored form of the equation has two different factors, the graph of the related function will cross the x-axis in two places. If the factored form has two identical factors, the graph will cross the x-axis in one place.

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

Helpful Hint

Additional 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

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

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




Additional Example 2B Continued

Check Graph the related quadratic function. Because there are two solutions found by factoring, the graph should cross the x-axis in two places.

The graph of y = x2 + 4x – 21 intersects the x-axis at x = –7 and x = 3, the same as the solutions from factoring.

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Additional 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.




Additional Example 2C Continued

Check Graph the related quadratic function.

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

●

Additional 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.

Additional 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.

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 = 3The only solution is 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.

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

Check Graph the related quadratic function. Because there are two solutions found by factoring, the graph should cross the x-axis in two places.

The graph of y = x2 + 4x – 5 intersects the x-axis at x = 1 and x = –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.


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

Solve the remaining equation.

Factor out the GCF, –1.

Check It Out! Example 2c Continued

Check Graph the related quadratic function. Because there is one solution found by factoring, the graph should cross the x-axis in one place.

The graph of y = –9x2 – 30x – 25 intersects the x-axis at x = , the same as the solution from

factoring.

●

Check It Out! Example 2d


3x2 – 4x + 1 = 0

(3x – 1)(x – 1) = 0 Factor the trinomial.



3x – 1 = 0 or x – 1 = 0

or x = 1

Check It Out! Example 2d Continued

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

Additional 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.



Additional 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.

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.



–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 1.5 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

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.

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9-5 Solving Quadratic Equations by Factoring