Lesson 9-5 Warm-Up

“Factoring to Solve Quadratic Equations” (9-5) How do you solve a quadratic equation when b  0? Rule: To solve a quadratic equation (ax2 + 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.

“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. (x + 2)(x – 3) = 0 Division Property of Equality x + 2 x + 2 1 1 x – 3 = 0 Identity Property: 1(x – 3) = x – 3 + 3 +3 Addition Property of Equality x = 3 (x + 2)(x – 3) = 0 Division Property of Equality x - 3 x - 3 1 1 x + 2 = 0 Identity Property: 1(x + 2) = x + 2 - 2 -2 Subtraction Property of Equality x = -2

“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) = 0 Given (-2 + 2)(-2 - 3) = 0 Substitute x = -2 into the factored equation (0)(-5) = 0 Simplify 0 = 0  True Statement (3 + 2)(3 - 3) = 0 Substitute x =3 into the factored equation (5)(0) = 0 Simplify

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

Solve x2 + x – 42 = 0 by factoring. Factoring to Solve Quadratic Equations LESSON 9-5 Additional Examples Solve x2 + x – 42 = 0 by factoring. x2 + x – 42 = 0 (x + 7)(x – 6) = 0 Factor x2 + x – 42. x + 7 = 0 or x – 6 = 0 Use the Zero-Product Property. x = –7 or x = 6 Solve for x.

“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 ax2 + 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. ax2 + bx = c ax2 + bx = c Subtract c from both sides to undo it -c -c from the right side. ax2 + bx – c = 0 Standard form. Example: Solve 2x2 - 5x =88 2x2 - 5x = 88 Given -88 -88 Subtract 88 from both sides 2x2 - 5x -88 = 0 Simplify (2x + 11)(x – 8) = 0 Factor 2x2 - 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

Solve 3x2 – 2x = 21 by factoring. Factoring to Solve Quadratic Equations LESSON 9-5 Additional Examples Solve 3x2 – 2x = 21 by factoring. 3x2 – 2x – 21 = 0 Subtract 21 from each side. (3x + 7)(x – 3) = 0 Factor 3x2 – 2x – 21. 3x + 7 = 0 or x – 3 = 0 Use the Zero-Product Property. 3x = –7 Solve for x. x = – or x = 3 7 3

Define: Let x = width of a side of the box. Factoring to Solve Quadratic Equations LESSON 9-5 Additional Examples 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

Find the product (x + 2) (x + 5). Factoring to Solve Quadratic Equations LESSON 9-5 Additional Examples (continued) Equation: (x + 2) (x + 5) = 130 x2 + 7x + 10 = 130 Find the product (x + 2) (x + 5). x2 + 7x – 120 = 0 Subtract 130 from each side. (x – 8) (x + 15) = 0 Factor x2 + 7x – 120. x – 8 = 0 or x + 15 = 0 Use the Zero-Product Property. x = 8 or x = –15 Solve for x. The only reasonable solution is 8. So the dimensions of the box are 8 in.  11 in.  1 in.

1. Solve (2x – 3)(x + 2) = 0. Solve by factoring. Factoring to Solve Quadratic Equations LESSON 9-5 Lesson Quiz 1. Solve (2x – 3)(x + 2) = 0. Solve by factoring. 2. 6 = a2 – 5a 3. 12x + 4 = –9x2 4. 4y2 = 25 –2, 3 2 – 2 3 ± 5 2 –1, 6