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4.8 Quadratic formula and the discriminant 4.8 Warm up

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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!!

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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

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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

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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

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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

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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

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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

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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?

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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

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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

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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

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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

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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

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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

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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

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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

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