1. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall

2. Chapter 9 Quadratic and Higher Degree Equations and Functions

3. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall 9.3 Solving Equations by Using Quadratic Methods

4. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Solving a Quadratic Equation 1.If the equation is in the form (ax + b) 2 = c, use the square root property and solve. If not, go to Step 2. 2.Write the equation in standard form: ax 2 + bx + c = 0. 3.Try to solve the equation by the factoring method. If not possible, go to Step 4. 4.Solve the equation by the quadratic formula.

5. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Solve: We need to isolate the radical and square both sides of the equation. Example Solution Square both sides. Set the equation equal to 0. continue

6. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Check: Square both sides. Set the equation equal to 0. x – 8 = 0 or x – 2 = 0 x = 8 or x = 2 True False The solution is 8 or the solution set is {8}.

7. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Solve Example Solution Simplify. continue

8. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Let a = 1, b = −9, c = − 6 We now have the equation in a standard quadratic form, so we can solve by the quadratic formula. Neither proposed solution will make the denominator 0. The solutions set is

9. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall 0 = 4x 2 – 12x + 4 0 = 4(x 2 – 3x + 1) Let a = 1, b = –3, c = 1 Example Solve 12x = 4x 2 + 4. Solution 12x = 4x 2 + 4

10. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Solve Multiply by 8. Example Solution

11. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Solve 9x 4 + 5x 2 – 4 = 0. Substitute w = x 2 into the equation. 9w 2 + 5w – 4 = 0 (9w – 4)(w + 1) = 0 Factor. Substitute the original variable back into the equation. (9x 2 – 4)(x 2 + 1) = 0 (3x – 2)(3x + 2)(x 2 + 1) = 0 Example Solution continue

12. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Now we need to set each of the preceding factors equal to 0 and solve the equations. 3x – 2 = 0 or 3x + 2 = 0 or x 2 + 1 = 0 x = ± and x = ± i So the solutions are x = ± or x = ± i.

13. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Solve3x 2/3 + 11x 1/3 = 4. Replace x 1/3 with w. 3w 2 + 11w – 4 = 0 Solve by factoring. (3w – 1)(w + 4) = 0 3w – 1= 0 w + 4 = 0 Example Solution 3x 2/3 + 11x 1/3 = 4

14. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Since w = x 1/3 x 1/3 = (x 1/3 ) 3 = ( ) 3 Example or x 1/3 = −4 (x 1/3 ) 3 = (−4) 3 x = −64 Both solutions check, so the solutions are

15. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Beach and Fargo are about 400 miles apart. A salesperson travels from Fargo to Beach one day at a certain speed. She returns to Fargo the next day and drives 10 mph faster. Her total time was hours. Find her speed to Beach and the return speed to Fargo. Example continue

16. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall 1. UNDERSTAND. Read and reread the problem. Let x = the speed to Beach, so x + 10 = the return speed to Fargo. Solution continue 2.TRANSLATE.

17. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall 3. SOLVE. continue

18. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall or 4. INTERPRET. Check: The speed is not negative, so it’s not. The number 50 does check. State: The speed to Beach was 50 mph and her return speed to Fargo was 60 mph.