For what value of $m$ are the two zeroes of $x^2+(m-3)x-12$ equal in magnitude but opposite in sign?

Zeroes p and −p add to 0, so the sum of zeroes must vanish: $-\frac{b}{a}=-(m-3)=0$, giving $m=3$. (The polynomial is then $x^2-12$, whose zeroes $2\sqrt{3}$ and $-2\sqrt{3}$ are indeed equal in magnitude and opposite in sign.) −3 solves $m+3=0$; 0 sets $m$ itself to zero instead of $m-3$; 12 confuses the condition with the constant term.