Let $f(x) = x^2 + 4x - 31$. For what value of $a$ is there exactly one real value of $x$ such that $f(x) = a$?
If f(x) = a then we have: x2 + 4x -31 = a
Rewrite as x2 + 4x - 31 - a =0
For this to have a single root we must have 42 - 4*(-31 - a) = 0 or a = -35
