1. Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. Section 6.7 Solving Quadratic Equations by Factoring

2. 2 Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. Solving Quadratic Equations by Factoring A quadratic equation is a polynomial equation in one variable that contains a variable term of degree 2 and no terms of higher degree. z 2 + 3z + 7 = 0 4k 2 – 5 = 0

3. 3 Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. Standard Form of a Quadratic Equation ax 2 + bx + c = 0, where a, b, and c are real numbers, and a  0. Zero Factor Property If a · b = 0, then a = 0 or b = 0. Quadratic Equations

4. 4 Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. 1. Make sure the equation is set equal to zero. 2. Factor, if possible, the quadratic expression that equals 0. 3. Set each factor containing a variable equal to 0. 4. Solve the resulting equations to find each root. 5. Check each root. Solving a Quadratic Equation

5. 5 Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. Example Solve the equation to find the two roots. x 2 – x – 20 = 0 x 2 – x – 20 = 0 The equation is in standard form. (x – 5)(x + 4) = 0 Factor the quadratic equation. x – 5 = 0 x + 4 = 0 Set each factor equal to 0. Solve the equations to find the roots. x =  4 x = 5 The two solutions are –4 and 5.

6. 6 Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. Example (cont) Solve the equation to find the two roots. x 2 – x – 20 = 0 Check each solution 5 and –4. (5) 2 – (5) = 20 25 – 5 = 20 (  4) 2 – (  4) = 20 16 + 4 = 20 Both solutions check.

7. 7 Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. Example Solve the equation to find the two roots. Both solutions check.

8. 8 Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. Example Solve the equation to find the two roots. Both solutions check.

9. 9 Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. Example Solve the equation to find the two roots. Both solutions check.

10. 10 Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. Example The area of a rectangle is 84 square inches. Determine the length and width if the length is 2 inches less than twice the width. Let w = width Let 2w – 2 = length Area = l · w 84 = (2w – 2)w

11. 11 Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. Example (cont) 84 = (2w – 2)w 84 = 2w 2 – 2w 2w 2 – 2w – 84 = 0 2(w 2 – w – 42) = 0 2(w – 7)(w + 6) = 0 2 = 0 w – 7 = 0 w + 6 = 0 w = 7 w = –6 The width of the rectangle is 7 inches. The length is 2w – 2 = 2(7) – 2 = 12 inches

12. 12 Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. Example The top of a local cable television tower has several small triangular reflectors. The area of each triangle is 49 square centimeters. The altitude of each triangle is 7 centimeters longer than the base. Find the altitude and the base of one of the triangles. Let b = length of the base Let b + 7 = length of the altitude

13. 13 Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. Example (cont) Let b = length of the base Let b + 7 = length of the altitude The triangular reflector has a base of 7 centimeters and an altitude of 7 + 7 = 14 cm.