1. Multiple Methods for Solving Quadratics Section P.5

2. Definition: Quadratic Equation in x A quadratic equation in x is one that can be written in the form ax + bx + c = 0 2 where a, b, and c are real numbers with a = 0.  Now, our five methods of solving for today…

3. Method #1: Factoring 1. Set the equation equal to zero 2. Completely factor the quadratic equation 3. Set each factored part separately equal to zero, solve for the unknown in each solve for the unknown in each Guided Practice: Solve for x: x – 6x + 5 = 0 2 Factored form: (x – 5)(x – 1) = 0 x – 5 = 0 OR x – 1 = 0 Zero Factor Property x = 5 OR x = 1

4. Guided Practice: Solve by factoring: x – 5x = 14 2 x = –2, 7

5. Guided Practice: Solve by factoring: 6x – 7x – 24 = 0 2 x = – 3 2, 8 3 Hint: use the “grouping method”

6. Method #2: Graphically 1. Graph the quadratic equation (set an appropriate viewing window) viewing window) 2. Calculate the zeros (x-intercepts) using your grapher Back to the first problem… Solve for x: x – 6x + 5 = 0 2 This time, use your grapher!!! 3. Note: if you use your graph, you must always include a sketch of that graph with your solution!

7. Method #3: Extracting Square Roots 1. Get the “squared” term by itself on one side of the equation the equation 2. Take the square root of both sides (remember to take either a positive or negative answer when take either a positive or negative answer when “extracting” the roots!) “extracting” the roots!) 3. Solve for the unknown (two separate equations)

8. Guided Practice: Solve by extracting square roots: (2x – 1) = 9 2 Solution: x = 2, –1

9. Guided Practice: Solve by extracting square roots: 6w – 13 = 15 – 3w 2 2 w = – 2 7 3 + Solution:

10. Method #4: Completing the Square 1. Collect the “x” terms by themselves on one side of the equation of the equation 2. Factor the “x” terms so that the x coefficient is 1 3. Add (b/2) to both sides of the equation 2 2 4. Factor the new “x” terms in the equation 5. Solve for x by extracting roots, as in the previous method method

11. Guided Practice: Solve by completing the square: 4x – 20x + 17 = 0 2 Solution:

12. So, what is this “most famous formula?”… (method 5 by the way…)

13. I’m speaking, of course, of the Quadratic Formula: The solutions of the quadratic equation ax + bx + c = 0, where a = 0, are given by the quadratic formula 2 x = – b b – 4ac 2a + 2 (note: this formula is derived via the “completing the square” method…)

14. Guided Practice: Solve for x (using the quad formula): 3x – 6x = 5 2 x = 1 + 2 6 3 = 2.633, –0.633  Can we support this answer graphically? Solution:

15. Guided Practice: Solve by using the quadratic formula: Solution:

16. Other Types of Problems… Solve the equation graphically and with a table: x = 1.325 x – x – 1 = 0 3 Not an exact answer  rounded to the thousandth

17. Other Types of Problems… Solve the equation graphically(this may be done 2 different ways graphically): x = –1.627, 2.827 5x – 6x – 23 = 0 2 Solution: Remember to show your graph as part of your solution!

18. Other Types of Problems… Solve the equation algebraically (support graphically): x = 3.5, –2.5 |2x – 1| = 6 Graphical Support??? Solution:

19. Other Types of Problems… Solve the equation graphically (think intersections of 2 separate equations): x = –3, 1 x = |2x – 3| 2 Solution:

20. Other Types of Problems… A particular rugby pitch is 30 meters longer than it is wide, and the area of the pitch is 8800 m. What are the dimensions of this particular pitch? 80 meters x 110 meters 2 Area: (length)(width) = 8800 But here, length = width + 30… Solution:

21. Whiteboard Problems: Solve by using the quadratic formula (not the program): Solution:

22. Whiteboard Problems: Solve the equation graphically: No Solution! |3x – 2| = x – 1 1 2 Why??

23. Whiteboard Problems: Solve by graphing: 3x + 2x – 9 = 0 2 x = 1.431, –2.097  How could we get exact answers???  Quadratic equation (not the program)! Solution:

24. Whiteboard Problems: Solve by completing the square: 3x – 6x – 7 =0 2 Solution:

25. Whiteboard problems… Solve the equation graphically: x = –1.942, 0.558, 1.384 x = x – 3 1 3 1 2 Solution:

26. Homework: p. 50-51 1-23 odd, 31, 43 Quiz tomorrow on sections 3, 4, and 5 !!! Remember I’m here before school if you need help on some material!