Detailed solutions to # 26 and 27 on study guide for exam 1 26. Solve: x 2 3 4x 1 3 12 0 This is an equation of the quadratic type: Set u = x 1 3 The rewrite the equation as u 2 4u 12 0 Factor to get (u + 6) (u – 2) = 0. Solve for u: u = -6 or u = 2 Plug in x 1 3 for u : 1 x 3 = -6 or x 1 Cube both sides of x 3 = -6 1 (x 3 ) 3 = (-6 )3 or 1 3 and x =2 1 3 =2 1 (x 3 ) 3 = 23 Then x = - 216 or x = 8 Put answers in a set: {-216, 8} 27. A bullet is fired into the air with an initial upward velocity of 112 feet per second from the top of a building 128 feet high. The equation that gives the height of the bullet at any time t is. h 128 112t 16t 2 (This is correct, original equation was not!) Find the times at which the bullet will be 288 feet in the air. Since h gives the height of the bullet at any time t, we need to replace h by 288 and solve the resulting equation for t. 288 128 112t 16t 2 Set the equation equal to 0: Divide both sides by 16: t 2 7t 10 0 Factor: (t – 5) (t – 2) = 0 , so t = 5 or t = 2 Answer: 2 seconds and 5 seconds. See next page for 2 examples. 16t 2 112t 160 0 Example: ( Like # 26) Solve . The substitution puts the equation in a form that is more obviously quadratic, , which we can factor. Then or But recall that , so corresponds to , corresponds to . To eliminate the fifth-roots in the two equations, raise the equations to the fifth powers. or or The solution set is { -1, 32} . Example (like # 27) A projectile is launched into the air with an initial upward velocity of 48 feet per second from a building 32 feet high. The height (h) in feet of the projectile can be modeled by h = - 16t2+ 48t + 32, where t is the time in seconds after it was launched. How much time, in seconds, does it take the object to reach 64 feet above the ground? Plug in 64 for h in h = - 16t2+ 48t + 32 then 64 = - 16t2+ 48t + 32 Set the quadratic equation equal to 0. 16t2 - 48t + 32 = 0 Divide the equation by 16. t 2- 3t + 2 = 0. Factor: (t – 2) (t – 1) = 0 so t = 1 or t= 2 Solution: After 1 second and after 2 seconds.