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?
Last question of the day for me.
First, let's simplify the equation. Bringing all terms to on side and combining all like terms, we get
\(x^2 - mx + 14 = 0 \)
Now, we simply search for factors of 14, both negative and positive. We have
Factors of 14 m
(1 , 14) (x + 1) (x + 14) -14
(-1, -14) (x - 1) (x - 14) 14
(2,7) (x + 2) ( x + 7) -9
(-2, -7) (x - 2) ( x - 7) 9
4 should be the answer.
Thanks! :)
Last question of the day for me.
First, let's simplify the equation. Bringing all terms to on side and combining all like terms, we get
\(x^2 - mx + 14 = 0 \)
Now, we simply search for factors of 14, both negative and positive. We have
Factors of 14 m
(1 , 14) (x + 1) (x + 14) -14
(-1, -14) (x - 1) (x - 14) 14
(2,7) (x + 2) ( x + 7) -9
(-2, -7) (x - 2) ( x - 7) 9
4 should be the answer.
Thanks! :)