+608 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?
+1952 First, let's simplify the equation so that we have a quadratic equal to 0. We have
x2–mx+24=10x2–mx+14=0
The coefficient of x is the sum of the factors of 14
The integer factors of 14 are (2, 7) and (–2, –7)
and (1, 14) and (–1, –14)
So m could equal 9 or –9 or 15 or –15
Therefore, m could have 4 possible values
Thanks! :)
+1952 First, let's simplify the equation so that we have a quadratic equal to 0. We have
x2–mx+24=10x2–mx+14=0
The coefficient of x is the sum of the factors of 14
The integer factors of 14 are (2, 7) and (–2, –7)
and (1, 14) and (–1, –14)
So m could equal 9 or –9 or 15 or –15
Therefore, m could have 4 possible values
Thanks! :)
