1. CHAPTER 8 Quadratic Equations EQ: What is a quadratic equation?

2. INTRODUCTION TO QUADRATIC EQUATIONS 8-1 EQ: How are quadratic equations solved by completing the square?

3. Lesson—Activation Solve by factoring: x(3x + 4) = 0 3x 2 – 6 = 0 14x 2 – 11x = -2

4. Lesson— Quadratic Equation—a second degree equation Standard Form— ax 2 + bx + c = 0 where a ≠ 0 All quadratic equations must be equal to zero to solve

5. Lesson—activation 2 Square each of the following: (x + 4) 2 (x – 3) 2 Do you see a pattern?

6. Try to go in reverse using the rule: x 2 + 10x + 25 Divide the middle term by 2 (x + 5) 2 Try: x 2 – 18x + 81

7. Lesson What would you need to complete this square: x 2 – 12x + ( x – ) 2

8. Lesson General rule for completing the square: x 2 + bx + make sure the leading coef =1 x 2 + bx + take half the x term and square it

9. Lesson How can completing the square assist in solving an equation that can’t be factored? Work with the person next to you to try this problem x 2 + 6x + 7 = 0 x 2 + 6x + = -7+

10. Lesson—Completing the Square EXAMPLE: 4x 2 – 12x – 7 = 0 standard form 4x 2 – 12x = 7 move the constant x 2 – 3x = 7 divide by the leading 4 coefficient then follow the procedure 9 4 9 4 + +

11. HOMEWORK: PAGE(S): 345 – 346 NUMBERS: 3 to 42 by 3’s use factoring up to 16 sq. roots up to 28 completing the square through the end

12. USING QUADRATIC EQUATIONS 8-2 EQ: How are problems solved by translating into quadratic equations?

13. Lesson--Activation USE ROPES: which means???? R O P E S Read the problem Organize your thoughts in a chart Plan the equations that will work Evaluate the Solution Summarize your findings

14. Lesson A rectangular lawn is 60m by 80m. Part of the lawn is torn up to install a pool, leaving a strip of lawn of uniform width around the pool. The area of the pool is 1/6 the area of the old lawn area. How wide is the strip of lawn? x 80 60 A rect = l w A lawn = 60 x 80 = 4800 A pool = 1/6 A lawn Note: don’t forget to check for extraneous solutions L = ? W =?

15. Lesson An open box is to be made from 10cm by 20cm rectangular pieces of cardboard by cutting a square from each corner. The area of the bottom of the box is 96 cm 2 What is the length of the sides of the squares that are cut from the corners? 20 10 x A = lw

16. HOMEWORK: PAGE(S): 349 NUMBERS: 1,4,5,6, 7,8,9

17. THE QUADRATIC FORMULA 8-3 EQ: How are quadratic equations solved using the quadratic formula?

18. Activation: Work with a partner to solve by completing the square: ax 2 + bx + c = 0 ax 2 + bx + = -c x 2 + x + =

19. Lesson Solve with the quadratic equation: 3x 2 + x - 2 = 0 x 2 – 6x – 20 = 0

20. Lesson Solve with the quadratic equation: 9x 2 + 12x + 4 = 0 3x 2 + x + 2 = 0

21. HOMEWORK: Use the quadratic Formula for all problems PAGE(S): 352 NUMBERS: 3 – 30 by 3’s

22. SOLUTIONS OF QUADRATIC EQUATIONS 8-4 EQ: How are the nature of the solutions determined for a quadratic equation?

23. ACTIVATION: Look at the homework and compare the problems does there seem to be a pattern that helps you determine the number of solutions?

24. Lesson The discriminant—(b 2 – 4ac) determines how many solutions and of what type How many solutions do each of the following have? x 2 – 4x + 4 = 0x 2 - 3x – 2 = 0

25. Lesson Write an equation from the solution If x = 3 or x = -2 what is a possible equation? x – 3 = 0 x + 2 = 0

26. HOMEWORK: PAGE(S): 357 NUMBERS: 3 – 15 by 3’s and 32, 35, 36

27. EQUATIONS REDUCIBLE TO QUADRATIC FORM 8-5 EQ: How is substitution used to reduce equations to quadratic form?

28. ACTIVATION: Given a = 2 b = 3 and c = 5 Find a 2 - 2b + c What procedure did you use?

29. Lesson U substitution—a method to simplify various equations by placing them in a familiar format example: x 4 – 5x 2 + 6 = 0 Let u = __________

30. Lesson let u = __________ Then

31. Lesson—you try (x – 1) 2 – 4(x -1) – 5 = 0 Let u = ________

32. Lesson Let u = _______

33. HOMEWORK: PAGE(S): 361 NUMBERS: 3 – 24 by 3

34. FORMULAS AND PROBLEM SOLVING 8-6 EQ: What are quadratic and joint variation?

35. ACTIVATION: When a ball is on the floor what is its height?

36. Lesson Given h(t) = -16t 2 + v 0 t + h 0 where h(t) = the height at time t v 0 = the initial velocity h 0 = the initial height When will an object launched at 15 ft/sec from a height of 12 feet will land? How high is it at 1 seconds? The formula is h(t) = -4.9t 2 + v 0 t + h 0 if measurement is given in meters

37. Lesson Solve for the indicated variable: P = 4s 2 for sA = πr 2 + 2πrh for h

38. HOMEWORK: PAGE(S): 364 NUMBERS: 3 – 21 by 3’s