1. ALGEBRA 1 Lesson 9-5 Warm-Up

2. ALGEBRA 1

3. “Factoring to Solve Quadratic Equations” (9-5) How do you solve a quadratic equation when b  0? Rule: To solve a quadratic equation (ax 2 + bx + c = 0) in which b  0, you can use the Zero Product property which says: If ab = 0, then a = 0 or b = 0 You can apply the Zero Product Rule to find solutions to a quadratic equation by factoring the equation, making one of the factors equal 0, and solving for the other factor. Example: Solve (x + 2)(x - 3) = 0 Method 1: Zero Product Rule: If (x + 2)(x - 3) = 0, then either (x + 2) = 0 or (x - 3) = 0 according to the Zero Product Rule. x + 2 = 0 Let (x + 2) = 0 -2 -2 Subtract 2 from both sides x = -2 x - 3 = 0 Let (x - 3) = 0 +3 +3 Subtract 2 from both sides x = 3 The solutions are x = -2 and 3.

4. ALGEBRA 1 “Factoring to Sove Quadratic Equations” (9-5) Method 2: Identity Property of Multiplication: : If (x + 2)(x - 3) = 0, and we divide both sides by one of the factors, (x + 2) or (x - 3), then the result is the other factor equals 0. So, the solutions are -2 and 3. 1 (x + 2)(x – 3) = 0 Divide both sides by x + 2 x + 2 x + 2 1 x – 3 = 0 Identity Property 1(x – 3) = x – 3 + 3 +3 Add 3 to both sides. x = 3 1 (x + 2)(x – 3) = 0 Divide both sides by x - 3 x - 3 x - 3 1 x + 2 = 0 Identity Property 1(x + 2) = x + 2 - 2 -2 Subtract 2 from both sides. x = -2

5. ALGEBRA 1 “Factoring to Sove Quadratic Equations” (9-5) How do you check solutions to a quadratic equations? Check: To check the solutions of a quadratic equaltion, substitute the variable for each solution separately and see if you get 0 to make the equation a true statement.. (x + 2)(x - 3) = 0Given (-2 + 2)(-2 - 3) = 0 Substitute x = -2 into the factored equation (0)(-5) = 0 Simplify 0 = 0  True Statement (x + 2)(x - 3) = 0Given (3 + 2)(3 - 3) = 0Substitute x =3 into the factored equation (5)(0) = 0 Simplify 0 = 0  True Statement

6. ALGEBRA 1 Solve (2x + 3)(x – 4) = 0 by using the Zero-Product Property. (2x + 3)(x – 4) = 0 2x + 3 = 0 or x – 4 = 0Use the Zero-Product Property. 2x = –3 Solve for x. Substitute 4 for x. (2x + 3)(x – 4) = 0 [2(4) + 3][(4) – 4] 0 (11)(0) = 0 x = – 3232 orx = 4Check: Substitute – for x. 3232 (2x + 3)(x – 4) = 0 [2(– ) + 3][(– ) – 4] 0 3232 3232 (0)(– 5 ) = 0 1212 Factoring to Solve Quadratic Equations LESSON 9-5 Additional Examples

7. ALGEBRA 1 Solve x 2 + x – 42 = 0 by factoring. x 2 + x – 42 = 0 (x + 7)(x – 6) = 0Factor x 2 + x – 42. x + 7 = 0orx – 6 = 0Use the Zero-Product Property. x = –7orx = 6 Solve for x. Factoring to Solve Quadratic Equations LESSON 9-5 Additional Examples

8. ALGEBRA 1 “Factoring to Sove Quadratic Equations” (9-5) How do you solve a quadratic equation not in standard form?? To solve a quadratic equation not in standard form, such as ax 2 + bx = c, make the right side equal to zero by “undoing” everything from the right side. Once the equation is in standard form, factor it, and then solve for each factor as before. ax 2 + bx = c ax 2 + bx = cSubtract c from both sides to undo it -c -cfrom the right side. ax 2 + bx – c = 0Standard form. Example: Solve 2x 2 - 5x =88 2x 2 - 5x =88 Given -88 -88Subtract 88 from both sides 2x 2 - 5x -88 =0 Simplify (2x + 11)(x – 8) =0 Factor 2x 2 - 5x =88 2x + 11 = 0 x - 8 = 0 Let (2x + 11) = 0 and (x - 8) = 0 - 11 -11 +8 +8 Add 8 or subtract 11 from both sides 2x = -11 x = 8 2 2 Divide each side by 2. x = - or -5.5 11 2

9. ALGEBRA 1 Solve 3x 2 – 2x = 21 by factoring. 3x 2 – 2x – 21 = 0Subtract 21 from each side. (3x + 7)(x – 3) = 0Factor 3x 2 – 2x – 21. 3x + 7 = 0orx – 3 = 0Use the Zero-Product Property. 3x = –7Solve for x. x = – or x = 3 7373 Factoring to Solve Quadratic Equations LESSON 9-5 Additional Examples

10. ALGEBRA 1 The diagram shows a pattern for an open-top box. The total area of the sheet of materials used to make the box is 130 in. 2. The height of the box is 1 in. Therefore, 1 in.  1 in. squares are cut from each corner. Find the dimensions of the box. Define: Let x = width of a side of the box. Then the width of the material = x + 1 + 1 = x + 2 The length of the material = x + 3 + 1 + 1 = x + 5 Words: length  width = area of the sheet Factoring to Solve Quadratic Equations LESSON 9-5 Additional Examples

11. ALGEBRA 1 (continued) Equation: (x + 2) (x + 5) = 130 x 2 + 7x + 10 = 130Find the product (x + 2) (x + 5). x 2 + 7x – 120 = 0Subtract 130 from each side. (x – 8) (x + 15) = 0Factor x 2 + 7x – 120. x – 8 = 0orx + 15 = 0Use the Zero-Product Property. x = 8or x = –15Solve for x. The only reasonable solution is 8. So the dimensions of the box are 8 in.  11 in.  1 in. Factoring to Solve Quadratic Equations LESSON 9-5 Additional Examples

12. ALGEBRA 1 1.Solve (2x – 3)(x + 2) = 0. Solve by factoring. 2.6 = a 2 – 5a3.12x + 4 = –9x 2 4.4y 2 = 25 –2, 3232 –1, 6 – 2323 ± 5252 Factoring to Solve Quadratic Equations LESSON 9-5 Lesson Quiz