1. 5.8 – Quadratic Formula

2. Ready to SING? X = -b + 2a What does it all mean??? ax 2 + bx + c So, we plug into the formula and simplify the arithmetic!

3. Quadratic Formula Great for solving ax² + bx + c = 0 Solve for x : x² + 5x + 4 = 0 Here a = 1, b = 5, c = 4 And

4. Let’s try one out Use the quadratic formula to solve for x 3x 2 – x - 6 a = 3, b = -1, c = -6

5. Let’s try it out Use the quadratic formula to solve for x 3x 2 -x -6

6. Let’s try one with a complex root 2x 2 + 2x + 5

7. Let’s try one with a complex root 2x 2 + 2x + 5

8. Let’s go back to our equation X = -b + 2a Discriminant If we look at the discrininant, we can figure out how many roots the equation will give, without even solving the equation

9. Test the discrininant If b 2 - 4ac > 0 If b 2 - 4ac < 0 If b 2 - 4ac = 0 2 real solutions No real solution; 2 imaginary solutions 1 real solutions Sample graph

10. Sample Problem Given the equation x 2 + 3x – 4, how many solutions will the graph have? Real or imaginary? a = 1, b = 3, c = -4 1 2 – 4(1)(-4) 1 + 16 > 0 2 Real Solutions

11. WORD PROBLEM #1 A pool measuring 12 meters by 16 meters is to have a pedestrian pathway installed all around it, increasing the total area to 247 square meters. What will be the width of the pathway?

12. A pool measuring 12 meters by 16 meters is to have a pedestrian pathway installed all around it, increasing the total area to 285 square meters. What will be the width of the pathway? SET UP We know length x width = area Therefore (2x+16) (2x + 12) = 285 16 m 12 m x x 16 + 2x 12+ 2x SOLVE (2x+16) (2x + 12) = 285 4x² + 56x + 192 = 285 4x² + 56x – 93 = 0 (2x + 31)(2x – 3)= 0 Therefore x = -15.5 and 1.5 Which solution is it??? It must be 1.5, since -15.5 doesn’t make sense ANSWER THE QUESTION  What are the dimensions of the walkway ?  Since x = 1.5, the walkway is 1.5 m.

13. WORD PROBLEM # 2 A rocket is launched directly upward at 64 feet per second (ft/s) from a platform 80 feet high. –What will be the object's maximum height? –When will it attain this height? Use the formula, where and

14. A rocket is launched directly upward at 64 feet per second (ft/s) from a platform 80 feet high. What will be the object's maximum height? When will it attain this height? Use the formula, where and ANSWER QUESTION What will be the object's maximum height? Max height happens at vertex. VERTEX = ( -b / 2a, f(-b / 2a) ) Remember a = -16, b = 64, c = 80 So – b /2a = -64/2(-16) = 2 Since t= time, it will take 2 seconds to reach the max height To find the actual max height, plug in t = 2. -16(2)² + 64(2) + 80 = 144 So the object will reach the max height of 144 feet at 2 seconds after launch FOOTHIL L SET UP We know h = -16t² + v o t + h o and that Vo = 64 and h o = 80 therefore h = -16t² + 64 t + 80