If f(x) = x^2 - 3x + 2, find the sum of all real values of x for which f(f(x)) = 0.
Correct: D
First, solve f(x) = 0: x^2 - 3x + 2 = (x-1)(x-2) = 0, so x = 1 or x = 2. Now, we need to solve f(x) = 1 and f(x) = 2. For f(x) = 1: x^2 - 3x + 2 = 1, so x^2 - 3x + 1 = 0. The solutions are x = (3 +/- sqrt(5))/2. For f(x) = 2: x^2 - 3x + 2 = 2, so x^2 - 3x = 0, which gives x = 0 or x = 3. The sum of all solutions is (3 + sqrt(5))/2 + (3 - sqrt(5))/2 + 0 + 3 = 3 + 3 = 6.