CHAPTER 9 Quadratic Equations Slide 2Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. 9.1Introduction to Quadratic Equations 9.2Solving Quadratic Equations by Completing the Square 9.3The Quadratic Formula 9.4Formulas 9.5Applications and Problem Solving 9.6Graphs of Quadratic Equations 9.7Functions

OBJECTIVES 9.5 Applications and Problem Solving Slide 3Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. aSolve applied problems using quadratic equations.

EXAMPLE The park board wants to increase the size of the children’s play area. The addition will be in the shape of a rectangle and have an area of 270 square feet. The width of the park is to be 3 feet less than the length. Find the dimensions of the addition. 9.5 Applications and Problem Solving a Solve applied problems using quadratic equations. AApplications of Quadratic Equations (continued) Slide 4Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE Solution 1. Familiarize. We first make a drawing and label it. We let l = length of one side and l – 3 = the width. Area is the length times the width 2. Translate. A = lw 270 = l(l – 3) 270 sq. ft l l  Applications and Problem Solving a Solve applied problems using quadratic equations. AApplications of Quadratic Equations (continued) Slide 5Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE 3. Solve. 270 = l(l – 3) 270 = l2 – 3l 0 = l 2 – 3l – 270 a = 1, b = –3, c = – Applications and Problem Solving a Solve applied problems using quadratic equations. AApplications of Quadratic Equations (continued) Slide 6Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE 9.5 Applications and Problem Solving a Solve applied problems using quadratic equations. AApplications of Quadratic Equations (continued) Slide 7Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE 4. Check. Length cannot be negative, so –15 does not check. A = lw = (18)(15) = State. The length is 18 feet and the width is 15 feet. 9.5 Applications and Problem Solving a Solve applied problems using quadratic equations. AApplications of Quadratic Equations Slide 8Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE s + 28 s 52 Solution 1. Familiarize. We first make a drawing and label it. We let s = length, in yards, of one leg. Then s + 28 = the length, in yards, of the other leg. 9.5 Applications and Problem Solving a Solve applied problems using quadratic equations. BThe hypotenuse of a right triangle is 52 yards long. One leg is 28 yards longer than the other. Find the lengths of the legs. (continued) Slide 9Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE 2. Translate. We use the Pythagorean theorem: s 2 + (s + 28) 2 = 52 2 s + 28 s Applications and Problem Solving a Solve applied problems using quadratic equations. BThe hypotenuse of a right triangle is 52 yards long. One leg is 28 yards longer than the other. Find the lengths of the legs. (continued) Slide 10Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE 3. Solve. s 2 + (s + 28) 2 = 52 2 s 2 + s s = s s – 1920 = 0 s s – 960 = 0 a = 1, b = 28, c = – Applications and Problem Solving a Solve applied problems using quadratic equations. BApplications of Quadratic Equations (continued) Slide 11Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE 9.5 Applications and Problem Solving a Solve applied problems using quadratic equations. BApplications of Quadratic Equations (continued) Slide 12Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE 9.5 Applications and Problem Solving a Solve applied problems using quadratic equations. BApplications of Quadratic Equations (continued) Slide 13Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE 4. Check. Length cannot be negative, so –48 does not check. 5. State. One leg is 20 yards and the other leg is 48 yards. 9.5 Applications and Problem Solving a Solve applied problems using quadratic equations. BApplications of Quadratic Equations Slide 14Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.