For what values of $j$ does the equation $(2x + 7)(x - 4) = -31 + jx$ have exactly one real solution? Express your answer as a list of numbers, separated by commas.

Answer 1

When the equation is expanded, we get \[(2x + 7)(x - 4) = -31 + jx \Rightarrow 2x^2 - x - 8 = -31 + jx \Rightarrow 2x^2 - (j + 1)x + 23 = 0.\] For there to be exactly one real solution, the discriminant of this quadratic must be equal to 0. The discriminant is given by $b^2 - 4ac$ where $a=2$, $b=-(j+1)$, and $c=23$. Thus we have \[(-1 - j)^2 - 4(2)(23) = 0 \Rightarrow j^2 + 2j - 91 = 0.\] Factoring the quadratic gives $(j + 13)(j - 7) = 0$, so the possible values of $j$ are $j = \boxed{-13,7}$.