Copyright © 2014, 2010, 2007 Pearson Education, Inc. 1 1 Chapter 9 Quadratic Equations and Functions.

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Presentation transcript:

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 1 1 Chapter 9 Quadratic Equations and Functions

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 2 Quadratic Equations and Functions CHAPTER 9.1The Square Root Principle and Completing the Square 9.2Solving Quadratic Functions Using the Quadratic Formula 9.3Solving Equations That Are Quadratic in Form 9.4Graphing Quadratic Functions 9 Copyright © 2014, 2010, 2007 Pearson Education, Inc. 2

3 Solving Equations That Are Quadratic in Form Solve equations by rewriting them in quadratic form. 2.Solve equations that are quadratic in form by using substitution.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 4 An equation is quadratic in form if it can be rewritten as a quadratic equation au 2 + bu + c = 0, where a  0 and u is a variable or an expression.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 5 Example Solve. Solution

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 6 continued Check x = 10 x = –12 The solutions are –12 and 10.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 7 Example Solve. Solution The solution is 3/4 (5 is extraneous).

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 8 Using Substitution to Solve Equations That Are Quadratic in Form To solve equations that are quadratic in form using substitution, 1. Rewrite the equation so that it is in the form au 2 + bu + c = Solve the quadratic equation for u. 3. Substitute for u and solve. 4. Check the solutions.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 9 Example Solution Solve x 4 – 5x = 0. u 2 – 5u + 4 = 0 Let u = x 2. Then we solve by substituting u for x 2 and u 2 for x 4. (u – 1)(u – 4) = 0 u = 1 or u = 4 u – 1 = 0 or u – 4 = 0 Factor Use zero factor theorem x 2 = 1 or x 2 = 4 Substitute x 2 for u. Solve by taking the square root of both sides. The solution set is {1,  1, 2,  2}.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 10 Example Solution Solve u 2 – 8u – 9 = 0 (u – 9)(u +1) = 0 u = 9 or u = –1 u – 9 = 0 or u + 1 = 0 Let u =. Then we solve by substituting u for and u 2 for x:

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 11 Check: x = 81:x = 1: The solution is {81}. FALSE TRUE