Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. 13.6 Solving Quadratic Equations by Factoring.

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Solving Quadratic Equations using Factoring.  has the form: ax 2 + bx + c = 0 If necessary, we will need to rearrange into this form before we solve!
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Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Solving Quadratic Equations by Factoring

Martin-Gay, Developmental Mathematics, 2e 22 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Quadratic Equation A quadratic equation is one that can be written in the form ax 2 + bx + c = 0, where a, b and c are real numbers and a ≠ 0. The form ax 2 + bx + c = 0 is referred to as standard form.

Martin-Gay, Developmental Mathematics, 2e 33 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Zero-Factor Property If a and b are real numbers and if ab = 0, then a = 0 or b = 0.

Martin-Gay, Developmental Mathematics, 2e 44 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solve: (x – 4)(x + 2) = 0 x – 4 = 0 or x + 2 = 0 x = 4 or x = –2 You can check the solutions on your own. Example

Martin-Gay, Developmental Mathematics, 2e 55 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. To Solve Quadratic Equations by Factoring Step 1:Write the equation in standard form so that one side of the equation is 0. Step 2:Factor the quadratic completely. Step 3:Set each factor containing a variable equal to 0. Step 4:Solve the resulting equations. Step 5:Check each solution in the original equation. Solving Quadratic Equations

Martin-Gay, Developmental Mathematics, 2e 66 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solve: x 2 – 5x = 24 First write the quadratic equation in standard form. x 2 – 5x – 24 = 0 Now factor. x 2 – 5x – 24 = (x – 8)(x + 3) = 0 Set each factor equal to 0. x – 8 = 0 or x + 3 = 0, which will simplify to x = 8 or x = –3 Example continued

Martin-Gay, Developmental Mathematics, 2e 77 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Check both possible answers in the original equation. 8 2 – 5(8) = 64 – 40 = 24 True (–3) 2 – 5(–3) = 9 – (–15) = 24 True So the solutions are 8 or –3. continued ? ?

Martin-Gay, Developmental Mathematics, 2e 88 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solve: 4x(8x + 9) = 5 4x(8x + 9) = 5 32x x = 5 32x x – 5 = 0 (8x – 1)(4x + 5) = 0 8x – 1 = 0 or 4x + 5 = 0 8x = 1 4x = – 5 Example continued and x = The solutions are

Martin-Gay, Developmental Mathematics, 2e 99 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Check both possible answers in the original equation. True Example ? ? andThe solutions are

Martin-Gay, Developmental Mathematics, 2e 10 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solve: Replace x with each solution in the original equation. The solutions all check. Example