Let $x^2-mx+24$ be a quadratic with roots $x_1$ and $x_2$. If $x_1$ and $x_2$ are integers, how many different values of $m$ are possible?
I'm assuming we can use Vieta's Formula.
We can say $x_1+x_2=m,$ and $x_1\cdot x_2=24.$
$16$ values satisfy both of these conditions, so I think our solution would be $\boxed{16}.$ Did I go wrong somewhere in my process, or am I correct? Thank you in advance.
Best Answer
You are on the right track, but consider that the pairs $(3/8)$ and $(8/3)$ , for example, give the same sum. Now, you should be able to solve the problem.