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Holt McDougal Algebra 1 8-6 Solving Quadratic Equations by Factoring Solve quadratic equations by factoring. Objective

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Holt McDougal Algebra 1

8-6 Solving Quadratic Equations by Factoring

Objective

Holt McDougal Algebra 1

8-6 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 McDougal Algebra 1

8-6 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 McDougal Algebra 1

8-6 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 McDougal Algebra 1

8-6 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 McDougal Algebra 1

8-6 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 McDougal Algebra 1

8-6 Solving Quadratic Equations by Factoring

Example 2A: Solving Quadratic Equations by Factoring

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 McDougal Algebra 1

8-6 Solving Quadratic Equations by Factoring

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.

Factor the trinomial.

Use the Zero Product Property.Solve each equation.

Holt McDougal Algebra 1

8-6 Solving Quadratic Equations by 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.

Factor the trinomial.

Use the Zero Product Property.

Solve each equation.

Holt McDougal Algebra 1

8-6 Solving Quadratic Equations by Factoring

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.

Holt McDougal Algebra 1

8-6 Solving Quadratic Equations by Factoring

Example 2D: Solving Quadratic Equations by Factoring

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 McDougal Algebra 1

8-6 Solving Quadratic Equations by Factoring

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.

Holt McDougal Algebra 1

8-6 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 McDougal Algebra 1

8-6 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.