Chapter 5: Exercises with Solutions

Chapter 5Quadratic Functions63. To find the x-intercepts, solve the equation f(x) = 0:− 56x 2 + 47x + 18 = 0=⇒ (−8x + 9)(7x + 2) = 0=⇒ x = 9 8 , −2 7Alternatively, use the quadratic formula x = −b ± √ b 2 − 4ac.2a65. To find the x-intercepts, solve the equation f(x) = 0. Note that each term isdivisible by 4, then factor the resulting perfect square trinomial.36x 2 + 96x + 64 = 0=⇒ 9x 2 + 24x + 16 = 0=⇒ (3x + 4) 2 = 0=⇒ x = − 4 3Alternatively, use the quadratic formula x = −b ± √ b 2 − 4ac.2a67. Compute the discriminant b 2 − 4ac to determine the number of real solutions.If b 2 − 4ac > 0, then the equation has two real solutions.If b 2 − 4ac = 0, then the equation has one real solution.If b 2 − 4ac < 0, then the equation has no real solutions.In this case, b 2 − 4ac = 6 2 − 4(9)(1) = 0, so the equation has one real solution.69. Compute the discriminant b 2 − 4ac to determine the number of real solutions.If b 2 − 4ac > 0, then the equation has two real solutions.If b 2 − 4ac = 0, then the equation has one real solution.If b 2 − 4ac < 0, then the equation has no real solutions.In this case, b 2 − 4ac = 4 2 − 4(−6)(−7) = −152 < 0, so the equation has no realsolutions.71. Compute the discriminant b 2 − 4ac to determine the number of real solutions.If b 2 − 4ac > 0, then the equation has two real solutions.If b 2 − 4ac = 0, then the equation has one real solution.If b 2 − 4ac < 0, then the equation has no real solutions.In this case, b 2 − 4ac = (−10) 2 − 4(−5)(−5) = 0, so the equation has one real solution.Version: Fall 2007