Let $p$ and $q$ be the two distinct solutions to the equation $$(x-3)(x+3) = 21x - 63.$$If $p > q$, what is the value of $p - q$?
I need to find the difference between the two solutions to this equation. One way to do that is to solve for $x$ by expanding the left side and moving everything to one side. That gives me $$x^2 - 9 = 21x - 63 \implies x^2 - 21x + 54 = 0.$$ This is a quadratic equation that I can factor by looking for two numbers that multiply to $54$ and add to $-21$. Those numbers are $-18$ and $-3$, so the equation factors as $$(x - 18)(x - 3) = 0.$$ Now I can use the zero product property and set each factor equal to zero and solve for $x$. I get $$x - 18 = 0 \implies x = 18$$ and $$x - 3 = 0 \implies x = 3.$$ These are the two solutions, $p$ and $q$. Since $p > q$, I can assign $p = 18$ and $q = 3$. Then the difference between them is just $$p - q = 18 - 3 = 15.$$ 