Chapter 5: Exercises with Solutions

Section 5.4The Quadratic Formula11. If (x + 2) 2 = 0, then there is only one possibility for x + 2, namelyx + 2 = 0.To solve for x, subtract 2 from both sides of this last equation.So the solution is x=-2.x = −2.13. If x is a real number, then so is x + 6. It’s not possible to square the real numberx + 6 and get a negative number like -81. Hence, the equation (x + 6) 2 = −81 has noreal solutions.15. If (x − 8) 2 = 15, then there are two possibilities for x − 8, namelyx − 8 = ± √ 15.To solve for x, add 8 to both sides of this last equation.x = 8 ± √ 15.Our TI83 gives the following approximations.(a)(b)17. First, complete the square:f(x) = x 2 − 4x − 8= x 2 − 4x + 4 − 4 − 8= ( x 2 − 4x + 4 ) − 4 − 8= (x − 2) 2 − 12Read off the vertex as (h, k) = (2, −12). The axis of symmetry is a vertical line throughthe vertex with equation x = 2.To find the x-intercepts algebraically, set x 2 −4x−8 = 0 and use the quadratic formulawith a = 1, b = −4 and c = −8:Version: Fall 2007