Day 3 Solving quadratics with complex solutionsmultiplying complex

Day 3: Solving quadratics with complex solutions/multiplying complex numbers Unit 3 Day 3

Day 3: Solving quadratics with complex solutions/multiplying complex numbers Quarterly Make Up Sit in last row Unit 3 Da. Y 3 Page 9

Day 3: Solving quadratics with complex solutions/multiplying complex numbers x 2 = -100 x 2 = -24 2 x 2 = -36 x 2 = -18

Day 3: Solving quadratics with complex solutions/multiplying complex numbers -10 x = 20 x = -2 12 i = 3 yi y = 4 a + bi FORM 3 + x = 6 x = 3 5 i - yi = -2 i i(5 – y) = -2 i 5 – y = -2 y = 7

Day 3: Solving quadratics with complex solutions/multiplying complex numbers 5 + 3 i – 4 i 5 - i

Day 3: Solving quadratics with complex solutions/multiplying complex numbers x 2 -5 x + 4 x - 20 x 2 - x - 20

Day 3: Solving quadratics with complex solutions/multiplying complex numbers -24 i + 4 i 2 -24 i + 4(-1) -24 i - 4 -4 - 24 i -36 + 63 i + 8 i - 14 i 2 -36 + 71 i – 14(-1) -36 + 71 i + 14 -22 + 71 i

Day 3: Solving quadratics with complex solutions/multiplying complex numbers 9 – i – 6 + 7 i 3 + 6 i -30 i 2 -30(-1) 30 3 + 7 i – 8 + 2 i -5 + 9 i 8 i-i 2 8 i-(-1) 8 i + 1 1 + 8 i -4 – 1 - i – 5 - 9 i -10 - 10 i 15 – 3 i + 5 i -i 2 15 + 2 i –(-1) 15 + 2 i + 1 16 + 2 i

Day 3: Solving quadratics with complex solutions/multiplying complex numbers Homework #3: Textbook pp. 108 #8, 12, 14, 16, 22, 37 -44 all, 49 -59 odd, 65

Day 3: Solving quadratics with complex solutions/multiplying complex numbers An object in launched directly upward at 64 feet per second (ft/s) from a platform 80 feet high. What will be the object's maximum height? When will it attain this height? s(t) = – 16 t 2 + 64 t + 80

Day 3: Solving quadratics with complex solutions/multiplying complex numbers An object is launched at 19. 6 meters per second (m/s) from a 58. 8 -meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = – 4. 9 t 2 + 19. 6 t + 58. 8, where s is in meters. When does the object strike the ground?

Day 3: Solving quadratics with complex solutions/multiplying complex numbers

Day 3: Solving quadratics with complex solutions/multiplying complex numbers The formula h (t) = -16 t 2 + 32 t + 80 gives the height h above ground, in feet, of an object thrown, at t = 0, straight upward from the top of an 80 feet building. a - What is the highest point reached by the object? b - How long does it take the object to reach its highest point? c - After how many seconds does the object hit the ground? d - For how many seconds is the hight of the object higher than 90 feet? Solution to Problem 1: • a - The height h given above is a quadratic function. The graph of h as a function of time t gives a parabolic shape and the maximum height h occur at the vertex of the parabola. For a quadratic function of the form h = a t 2 + b t + c, the vertex is located at t = - b / 2 a. Hence for h given above the vertex is at t t = -32 / 2(-16) = 1 second.

Day 3: Solving quadratics with complex solutions/multiplying complex numbers • 1 second after the object was thrown, it reaches its highest point (maximum value of h) which is given by h = -16 (1) 2 + 32 (1) + 80 = 96 feet • b - It takes 1 second for the object to reach it highest point. • c - At the ground h = 0, hence the solution of the equation h = 0 gives the time t at which the object hits the ground. -16 t 2 + 32 t + 80 = 0 • The above quadratic equation has two • d - The object is higher than 90 feet for all values of t statisfying the inequality h > 90. Hence

Day 3: Solving quadratics with complex solutions/multiplying complex numbers Homework #3: Textbook pp. 108 #8, 12, 14, 16, 22, 37 -44 all, 49 -59 odd, 65