1. 4.8 Quadratic formula and the discriminant 4.8 Warm up

2. 4.8 Quadratic formula and the discriminant If we solve the quadratic formula for x We get an equation to find x, given any quadratic equation!!

3. 4.8 Quadratic formula and the discriminant Solve x 2 + 3x = 2. x 2 + 3x = 2 Write original equation. x 2 + 3x – 2 = 0 Write in standard form. x = – b + b 2 – 4ac 2a2a Quadratic formula x = – 3 + 3 2 – 4(1)( –2) 2(1) Identify a, b, and c a = 1, b = 3, c = –2 Simplify. x = – 3 +– 3 + 17 2 The solutions are x = – 3 +– 3 + 17 2 0.56 and x = – 3 –– 3 – 17 2 – 3.56. ANSWER

4. 4.8 Quadratic formula and the discriminant 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

5. 4.8 Quadratic formula and the discriminant Solve –x 2 + 4x = 5. –x 2 + 4x = 5 Write original equation. Write in standard form. x = – 4+ 4 2 – 4(– 1)(– 5) 2(– 1) a = –1, b = 4, c = –5 Simplify. –x 2 + 4x – 5 = 0. x = – 4+ – 4 – 2 – 4+ 2i x = – 2 Simplify. Rewrite using the imaginary unit i. x = 2 + i The solution is 2 + i and 2 – i. ANSWER

6. 4.8 Quadratic formula and the discriminant Use the quadratic formula to solve the equation. x 2 = 6x – 4 x = – b + b 2 – 4ac 2a2a x = – (– 6) + (– 6) 2 – 4(1)( 4) 2(1) x = + 3 ++ 3 + 20 2 x 2 – 6x + 4 = 0 x 2 = 6x – 4 The solutions are x = 3 + 20 2 and x = 3 – 3 – 20 2 = 3 – 5 = 3 + 5

7. 4.8 Quadratic formula and the discriminant Use the quadratic formula to solve the equation. 4x 2 – 10x = 2x – 92. x = – b + b 2 – 4ac 2a2a x = – (– 12) + (– 12) 2 – 4(4)( 9) 2(4) x = 12 + 0 8 4x 2 – 10x = 2x – 9 4x 2 – 12x + 9 = 0 SOLUTION 3 2 The solution is. = 1 2 1

8. 4.8 Quadratic formula and the discriminant Use the quadratic formula to solve the equation. 7x – 5x 2 – 4 = 2x + 33. x = – b + b 2 – 4ac 2a2a x = – (5) + (5) 2 – 4(– 5)(– 7) 2(– 5)2(– 5) –115 x = – 5 + – 5 + – 10 7x – 5x 2 – 4 = 2x + 3 – 5x 2 + 5x – 7 = 0 SOLUTION 115 x = 5 + i 10 ANSWER The solutions are and. 115 5 + i 10 115 5 – i 10

9. 4.8 Quadratic formula and the discriminant Lets look at just a part of the quadratic formula What effect does this part have on your solution if it is positive? If it is negative? If it is zero?

10. 4.8 Quadratic formula and the discriminant Find the discriminant and type of solutions for –x 2 + 4x = 5. a = –1, b = 4, c = –5 = 16 – 20 = – 4, two imaginary solutions disc. < 0two imaginary solutions disc. > 0two real solutions disc. = 0one real solution

11. 4.8 Quadratic formula and the discriminant Find the discriminant of the quadratic equation and give the number and type of solutions of the equation. a. x 2 – 8x + 17 = 0 b. x 2 – 8x + 16 = 0 c. x 2 – 8x + 15 = 0 SOLUTION Equation DiscriminantSolution(s) ax 2 + bx + c = 0b 2 – 4ac x = – b+ b 2 – 4ac 2a2a a. x 2 – 8x + 17 = 0(–8) 2 – 4(1)(17) = – 4 Two imaginary: 4 + i b. x 2 – 8x + 16 = 0(–8) 2 – 4(1)(16) = 0 One real: 4 c. x 2 – 8x + 15 = 0(–8) 2 – 4(1)(15) = 4 Two real: 3,5

12. 4.8 Quadratic formula and the discriminant Find the discriminant of the quadratic equation and give the number and type of solutions of the equation. 2x 2 + 4x – 4 = 0 SOLUTION Equation DiscriminantSolution(s) ax 2 + bx + c = 0b 2 – 4ac 2x 2 + 4x – 4 = 0 4 2 – 4(2)(– 4 ) x = – b+ b 2 – 4ac 2a2a = 48 Two real solutions

13. 4.8 Quadratic formula and the discriminant SOLUTION Equation DiscriminantSolution(s) ax 2 + bx + c = 0b 2 – 4ac 12 2 – 4(12)(3 ) x = – b+ b 2 – 4ac 2a2a = 0 One real solution 3x 2 + 12x + 12 = 0

14. 4.8 Quadratic formula and the discriminant SOLUTION Equation DiscriminantSolution(s) ax 2 + bx + c = 0b 2 – 4ac x = – b+ b 2 – 4ac 2a2a 8x 2 = 9x – 11 8x 2 – 9x + 11 = 0(– 9) 2 – 4(8)(11 ) = – 271 Two imaginary solutions

15. 4.8 Quadratic formula and the discriminant SOLUTION Equation DiscriminantSolution(s) ax 2 + bx + c = 0b 2 – 4ac x = – b+ b 2 – 4ac 2a2a 7x 2 – 2x = 5 (– 2) 2 – 4(7)(– 5 ) = 144 Two real solutions 7x 2 – 2x – 5 = 0

16. 4.8 Quadratic formula and the discriminant SOLUTION Equation DiscriminantSolution(s) ax 2 + bx + c = 0b 2 – 4ac x = – b+ b 2 – 4ac 2a2a 4x 2 + 3x + 12 = 3 – 3x (6) 2 – 4(4)(9) = – 108 4x 2 + 6x + 9 = 0 Two imaginary solutions

17. 4.8 Quadratic formula and the discriminant SOLUTION Equation DiscriminantSolution(s) ax 2 + bx + c = 0b 2 – 4ac x = – b+ b 2 – 4ac 2a2a 3x – 5x 2 + 1 = 6 – 7x (4) 2 – 4(– 5)(– 5) = 0 – 5x 2 + 4x– 5 = 0 One real solution HOMEWORK 4.8 p. 296 #3-60 EOP