Expanding the quadratic on the left gives
\[x^2-mx+14=0.\]The discriminant is $\sqrt{m^2-56}.$ To have real solutions for $x$, the discriminant must be nonnegative, so $m^2-56\ge 0$. Applying the difference of squares, $(m - 8)(m + 8) \ge 0$, so $m \le -8$ or $m \ge 8$. If $m = -8k$ with $k = 0,$ $1$, $2,$ or $3$, then $m^2-56 = 64k^2-56$ is positive. If $m = 8k$ with $k = 1$, $2, \dots$, $7$, then $m^2-56 = 64k^2-56$ is again positive. Thus, the answer is $3 + 7 = \boxed{10}$.
The quadratic equation $x^2-mx+24 = 10$ has roots $x_1$ and $x_2$. If $x_1$ and $x_2$ are integers, how many different values of $m$ are possible?
1 answer