Let S be the set of all integer values of k for which the graph of the parabola y = (k+2)x^2 - x + 2 lies strictly above the line y = -(2k+1)x - 1 for all real x . The sum of the elements in S is:

Step-by-step solution

For the parabola to lie strictly above the line for all x R , we must have: (k+2)x^2 - x + 2 > -(2k+1)x - 1 (k+2)x^2 + 2kx + 3 > 0 for all x R . For a quadratic expression to be strictly positive for all real x , its leading coefficient must be positive and its discriminant must be negative. Condition 1: Leading coefficient > 0 k + 2 > 0 k > -2 Condition 2: Discriminant D (2k)^2 - 4(k+2)(3) 4k^2 - 12k - 24 k^2 - 3k - 6 The roots of the equation k^2 - 3k - 6 = 0 are k = 3 9 - 4(1)(-6) 2 = 3 33 2 . Since 33 is approx