If a is uniformly distributed over [-12,16] , what is the probability that the roots of the equation
x^2+ax+a+3=0
are both real?

 1. x^2 + ax + a + 48 = 0 Discriminant D = a^2 - 4 * 1 * (a+48) D = a^2 - 4a - 192 For the roots to be real, D >= 0 a^2 - 4a - 192 >= 0 a(a+12) - 16(a+12) >= 0 (a+12)(a-16) >= 0 a <= -12, a >= 16 The required probability P(roots are real) = P(a <= -12) + P(a >= 16) The interval is [-19, 22] Therefore P(roots are real) = P(a E [-19, -12]) + P (a E [16, 22]) = [(-12)-(-19)]/[22-(-19)] + (22-16)/[22-(-19)] = 7/41 + 6/41 = 13/41 Ans: 13/41