The two zeroes of $x^2-13x+c$ differ by 5. What is the value of $c$?

$(\alpha -\beta {)}^2=(\alpha +\beta {)}^2-4\alpha \beta =169-4c$. The difference is 5, so $169-4c=25$, giving $4c=144$ and $c=36$. (The zeroes are 9 and 4, which differ by 5 and multiply to 36.) 41 sets the expression equal to the difference 5 instead of its square; 72 uses the ${\alpha }^2+{\beta }^2$ formula $169-2c$ by mistake; 144 stops at $4c$ without dividing.