+948 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?
+1926 First, let's subtract 10 from both sides. \(x^2-mx+14\)
The answer to this question lies in the number of possible multiples 14 has.
For every mutliple, there are 4 possibles.
Say the two multiples are \(a, b\)
We have \((x-a)(x-b), (x+a)(x+b), (x+a)(x-b), (x-a)(x+b)\)
For example, 2 and 7.
\((x-7)(x+2), (x-2)(x+7), (x-2)(x-7), (x+7)(x+2)\)These all give different values for m.
We have only 2 possible groups. 1 and 14, 2 and 7.
This means there are 4*2=8 different values for m
Thanks! :)
