1. Warm-ups 04-11-14 Find each square root. Solve each equation. 5. –6x = –606. 7. 2x – 40 = 08. 5x = 3 611 –25 1. 2. 3. 4. x = 10x = 80 x = 20

2. Solving Quadratic Equations

3. Square Root Property

4. To Solve A Quadratic Equation When b = 0… Use the same procedures you used to solve an equation to get the “x” isolated (by itself). Instead of having an “x” left, you have an “x²”. When the “x²” is isolated, find the square root of both sides (be sure to give both the principal and the negative roots!).

5. Example 1 Solve:2x² - 18 = 0 Add 18 to both sides2x² = 18 Divide both sides by 2x² = 9 Find the square root of both sides x = ± 3

6. Example 2 Solve:2x² + 72 = 0 Subtract 72 from both sides 2x² = -72 Divide both sides by 2x² = -36 Find the square root of both sides— oops!! You can’t find the square root of a negative number (-36) so there is NO SOLUTION!

7. Example 3 Solve:2x² + 8x² + 16 = 32 Simplify:10x² + 16 = 32 Subtract 16 from both sides10x² = 16 Divide both sides by 10x² = 1.6 Find the square root of both sides X = ±1.264911064… round to the hundredth’s place X = ±1.26

8. Things to think about…. When will there be two solutions? When will there be one solution? When will there be no solutions? Look at the graphs of each type of equation listed above. Can you tell by looking a the graph?

9. Try these… 4x 2 + 1 = 17 Find the radius of a circle whose area is 125 in 2. 81x 2 - 49 = 0 3x 2 – 85 = 2x 2 – 36 4x 2 + 72 = 2x 2 - 28 x = ± 2 r = 6.31 in x = ± 7/9 x = ± 7 No solution

10. Completing the Square

11. Look at the trinomial “a” should be 1. It would be nice if “b” is even.

12. An expression in the form x 2 + bx is not a perfect square. However, you can use the relationship shown above to add a term to x 2 + bx to form a trinomial that is a perfect square. This is called completing the square.

13. To “complete” the square… Make sure that “a” is 1. Take “b”, cut it in half, and square it. Add it to both sides of the trinomial. When you do this, it creates a perfect square trinomial.

14. Using this to solve… x 2 + 8x – 9 = 0 Get the constant (“c”) on one side by itself. x 2 + 8x = 9 Take the “b”, divide by two, and square it. Add this to both sides. x 2 + 8x + 16 = 9 + 16 Simplify.

15. Using this to solve… x 2 + 8x + 16 = 25 What you have on the left is a perfect square trinomial and is easy to factor. So, factor it. (x + 4) 2 = 25 Find the square root of both sides. x + 4 = ± 5

16. Using this to solve… x + 4 = ± 5 x + 4 = 5 x + 4 = -5 Subtract 4 from both sides of both equations. x = 1 x = -9

17. Try these… x 2 - 2x – 3 = 0 x 2 – 4x = 5 x 2 – 12x = 4 x 2 + 10x = 16 x = -1, 3 x = -1, 5 x = -0.32, 12.32 x = -11.40, 1.40

18. x 2 - 14x = -15

19. x 2 + 6x – 7 = 0