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Section 10-7 & 8 Quadratic Formula and Discriminant SPI 23E: Find the solution to a quadratic equation given in standard form SPI 23H: select the discriminant of a quadratic equation Objectives: Use the Quadratic formula to solve any quadratic equation Use the discriminant to determine the number of solutions Quadratic Equation: equation written in the form ax 2 + bx + c = 0 Solutions of a quadratic equation: the x-intercepts since they are on the real number line a quadratic equation can have 1, 2 or no solutions

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Find Solutions using the Quadratic Formula Use to find all real number solutions ax + by = c Two Methods for Solving Quadratic Equations: Factoring and using the Zero-Product Property Using the Quadratic Formula

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Solve x 2 + 6 = 5x using the quadratic formula. x 2 – 5x + 6 = 0 x = 5 + 1 2 x = 5 – 1 2 or Write two solutions. x = 3orx = 2Simplify. Find Solutions using the Quadratic Formula Step 1: Write equation in standard form. Step 2: Substitute values into the quadratic formula. a = 1 1 b = -5 c = 6 -5 6 1 Step 3. Simplify and write as two separate equations.

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Find Solutions using the Quadratic Formula Solve x 2 + 2 = –3x using the quadratic formula. x 2 + 3x +2 = 0 x = -3 + 1 2 x = -3 – 1 2 or Write two solutions. x = -1orx = -2Simplify. Step 1: Write equation in standard form. Step 2: Substitute values into the quadratic formula. a = 1 1 b = 3 c = 2 3 2 1 3 Step 3. Simplify and write as two separate equations.

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Discriminant Determines number of real number solutions Discriminant is the expression under the radical sign in the quadratic formula Property of the Discriminant For a quadratic equation, if: b 2 – 4ac > 0 (2 solutions) b 2 – 4ac = 0 (1 solution) b 2 – 4ac < 0 (No solutions) b 2 – 4ac x 2 – 6x + 39 = 0 b 2 – 4ac = 0 x 2 – 6x + 3 = 0x 2 – 6x + 12 = 0 b 2 – 4ac = -12 b 2 – 4ac = 24

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A child throws a ball upward with an initial upward velocity of 15 ft/s from a height of 2 ft. If no one catches the ball, after how many seconds will it land? Use the vertical motion formula h = –16t 2 + vt + c, where h = 0, v = velocity, c = starting height, and t = time to land. Round to the nearest hundredth of a second. Step 1: Use the vertical motion formula. h = –16t 2 + vt + c 0 = –16t 2 + 15t + 2 Substitute 0 for h, 15 for v, and 2 for c. Step 2: Use the quadratic formula. x = –b ± b 2 – 4ac 2a Relate Quadratic Equations to Physical Science

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t = –15 ± 15 2 – 4(–16)(2) 2(–16) Substitute –16 for a, 15 for b, 2 for c, and t for x. t = –15 + 18.79 –32 or t = –15 – 18.79 –32 Write two solutions. t–0.12ort1.06 Simplify. Use the positive answer because it is the only reasonable answer in this situation. The ball will land in about 1.06 seconds. –15 ± 225 + 128 –32 t = Simplify. –15 ± 353 –32 t =

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