2. Aim: Quadratics with Complex Roots Course: Adv. Alg. & Trig. Aim: How do we handle quadratic equations that result in complex roots? Do Now: Solve the following quadratic: x 2 – 8x + 17 = 0

3. Aim: Quadratics with Complex Roots Course: Adv. Alg. & Trig. Complex Roots x 2 – 8x + 17 = 0 Quadratic Formula determine a, b, and c a = 1, b = -8, c = 17 substitute into quadratic formula evaluate and simplify standard form y = ax 2 + bx + c

4. Aim: Quadratics with Complex Roots Course: Adv. Alg. & Trig. x = 4 – ix = 4 + i Checking Complex Roots x 2 – 8x + 17 = 0 check both roots (4 + i) 2 – 8(4 + i) + 17 = 0 (4 – i) 2 – 8(4 – i) + 17 = 0 16 + 8i + i 2 – 32 – 8 i + 17 = 016 – 8i + i 2 – 32 + 8 i + 17 = 0 16 + i 2 – 32 + 17 = 0 16 – 1 – 32 + 17 = 0 0 = 0 Solution: x = 4 ± i or {4 + i, 4 – i)

5. Aim: Quadratics with Complex Roots Course: Adv. Alg. & Trig. 0 = ax 2 + bx + c the roots of the parabola - where its crosses the x-axis The Graph, the Roots, & the x-axis y = x 2 y = x 2 – 18x + 82y = x 2 + 14x + 45 0 = x 2 – 18x + 820 = x 2 + 14x + 450 = x 2 y = ax 2 + bx + cEquation of parabola y = 0 2 real roots 2 real equal roots NO real roots, -complex

6. Aim: Quadratics with Complex Roots Course: Adv. Alg. & Trig. Model Problem determine a, b, and c a = 1, b = -2, c = 10 substitute into quadratic formula Solve the equation and express its roots in the form a + bi. put in standard form x 2 – 2x + 10 = 0 evaluate and simplify

7. Aim: Quadratics with Complex Roots Course: Adv. Alg. & Trig. x = 1 – 3ix = 1 + 3i Check check both roots in original equation Solution: x = 1 ± 3i or {1 + 3i, 1 – 3i) 1 – 6i + 9i 2 = -8 – 6i 1 + 6i + 9i 2 = -8 + 6i 1 – 6i – 9 = -8 – 6i 1 + 6i – 9 = -8 + 6i -8 – 6i = -8 – 6i -8 + 6i = -8 + 6i

8. Aim: Quadratics with Complex Roots Course: Adv. Alg. & Trig. Model Problem A scoop is a hockey pass that propels the puck from the ice into the air. Suppose a player makes a scoop that releases the puck with an upward velocity of 34 ft/s. The equation h = -16t 2 + 34t models the height h in feet of the puck at time t in seconds. Will the puck ever reach a height of 20 feet? If so, how many seconds will it take?

9. Aim: Quadratics with Complex Roots Course: Adv. Alg. & Trig. h = 20 ft. Model Problem When an object is dropped, thrown, or launched either up or down, you can use the vertical motion formula to find the height of the object. h is height of object, t is time is takes the object to rise or fall to a given height, v is the starting velocity of the object, s is the starting height. h = -16t 2 + vt + s quadratic equationrecall:

10. Aim: Quadratics with Complex Roots Course: Adv. Alg. & Trig. Model Problem Substitute: 20 = -16t 2 + 34t h = 20 ft Standard form: 0 = -16t 2 + 34t – 20 Use quad. form.: a = -16, b = 34, c = -20 h = -16t 2 + 34t DOES NOT REACH HEIGHT OF 20

11. Aim: Quadratics with Complex Roots Course: Adv. Alg. & Trig. Model Problem h = -16t 2 + 34t y = 20 Graph: h = -16t 2 + 34t - 20

12. Aim: Quadratics with Complex Roots Course: Adv. Alg. & Trig. Model Problems Solve the equations and express their roots in a + bi form.