Chapter 5: Exercises with Solutions

Chapter 5Quadratic Functions0 < D0 < 9 − 20kSubtracting 9 from each side,−9 < 20kNow divide both sides by -20 (reverse the direction of the inequality because we’redividing by a negative number):9/20 > kSo, any k less than 9/20 will work. Our solution set is therefore {k : k < 9/20}39. The graph of a quadratic function has no x-intercepts when the discriminant isnegative. For this function f(x) = 2x 2 − x + 5k, a = 2, b = −1 and c = 5k, so thediscriminant is D = b 2 − 4ac = (−1) 2 − 4(2)(5k) = 1 − 40k. In order for there to be nox-intercepts, this must be negative. So, set it less than zero and solve for the value ofk.Subtracting 1 from each side,0 > D0 > 1 − 40k−1 > −40kNow divide both sides by -40 (reverse the direction of the inequality because we’redividing by a negative number):1/40 < kSo, any k greater than 1/40 will work. Our solution set is therefore {k : k > 1/40}41. Use factoring and the principle of zero products:63x 2 + 74x − 1 = 8=⇒ 63x 2 + 74x − 9 = 0=⇒ (7x + 9)(9x − 1) = 0=⇒ x = − 9 7 , 1 9Alternatively, use the quadratic formula x = −b ± √ b 2 − 4ac.2a43. x 2 − x − 5 = 2 =⇒ x 2 − x − 7 = 0. The polynomial on the left side does notfactor, so use the quadratic formula:Version: Fall 2007