+279 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?
+302 The quadratic equation takes the form of x^2 - mx + 24 = 0. Given that x_1 and x_2 are its roots and are integers, it means that the sum of these roots (x_1 + x_2) equals m (based on Vieta's formulas), and the product of these roots equals 24 (x_1*x_2=24). In this case, we need to factor the constant term (24), in other words, list all possible pairs of integers that multiply to give 24. These include {1,24}, {-1,-24}, {2,12}, {-2,-12}, {3,8}, {-3,-8}, {4,6}, {-4,-6}, and their negative counterparts. As both roots are integers, and each pair of factors correspond to one unique value of m (the sum of those two integers), there are altogether 16 different values for m, considering both positive and negative pairs.
Answer is: 16
