+972 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?
+1950 First, let's subtract 10 from both sides. x2−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! :)
