1. 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

2. 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

3. 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.

4. 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.

5. 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

6. 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

7. 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 =