2. 1.8 Quadratic Formula & Discriminant p. 58 How do you solve a quadratic equation if you cannot solve it by factoring, square roots or completing the square? Is there a fast way to decide?

3. Quadratic Formula If you take a quadratic equation in standard form (ax 2 +bx+c=0), and you complete the square, you will get the quadratic formula!

4. When to use the Quadratic Formula Use the quadratic formula when you can’t factor to solve a quadratic equation. (or when you’re stuck on how to factor the equation.)

5. Quadratic Formula : the one with the song! It’s a way to remember the formula…

6. To use the quadratic formula...

7. Example – Two real solutions 1. 3x 2 +8x=35 3x 2 +8x-35=0 a=3, b=8, c= -35 OR

8. Two real solutions What does an answer with two real solutions tell you about the graph of the equation? It tells you the two places that the graph crosses the x axis (x-intercepts)

9. Example – One real solution Solve 25x 2 – 18x = 12x – 9. 25x 2 – 18x = 12x – 9. Write original equation. Write in standard form. x = 30 + (–30) 2 – 4(25)(9) 2(25) a = 25, b = –30, c = 9 Simplify. 25x 2 – 30x + 9 = 0. x = 30 + 0 50 x = 3 5 Simplify. 3 5 The solution is ANSWER

10. One solution CHECK Graph y = –5x 2 – 30x + 9 and note that the only x-intercept is 0.6 =. 3 5 

11. Example - Imaginary solutions 2. -2x 2 =-2x+3 -2x 2 +2x-3=0 a=-2, b=2, c= -3

12. Imaginary Solutions What does an answer with imaginary solutions tell you about the graph of the equation? The graph will not go through or touch the x- axis.

13. Discriminant: b 2 -4ac The discriminant tells you how many solutions and what type you will have. If the discrim: Is positive – 2 real solutions Is negative – 2 imaginary solutions Is zero – 1 real solution

14. Examples Find the discriminant and give the number and type of solutions. a. 9x 2 +6x+1=0 a=9, b=6, c=1 b 2 -4ac=(6) 2 -4(9)(1) =36-36=0 1 real solution b. 9x 2 +6x-4=0 a=9, b=6, c=-4 b 2 -4ac=(6) 2 -4(9)(-4) =36+144=180 2 real solutions c. 9x 2 +6x+5=0 a=9, b=6, c=5 b 2 -4ac=(6) 2 -4(9)(5) =36-180=-144 2 imaginary solutions

15. h = –16t 2 + v 0 t + h 0 3 = –16t 2 + 40t + 4 Write height model. Substitute 3 for h, 40 for v 0, and 4 for h 0. 0 = –16t 2 + 40t + 1 Write in standard form. t = – 40+ 40 2 – 4(– 16)(1) 2(– 16) Quadratic formula t = – 40+ 1664 – 32 Simplify. t – 0.025 or t 2.5 A juggler tosses a ball into the air. The ball leaves the juggler’s hand 4 feet above the ground and has an initial vertical velocity of 40 feet per second. The juggler catches the ball when it falls back to a height of 3 feet. How long is the ball in the air? Because the ball is thrown, use the model h = –16t 2 + v 0 t + h 0. To find how long the ball is in the air, solve for t when h = 3. Reject the solution – 0.025 because the ball’s time in the air cannot be negative. So, the ball is in the air for about 2.5 seconds.

16. AAAA ssss ssss iiii gggg nnnn mmmm eeee nnnn tttt Page 58, 3-48 every third problem, 52-54