+0

Algebra

0
2
1
+221

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?

Jul 25, 2024

#1
+1657
+1

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! :)

Jul 25, 2024
edited by NotThatSmart  Jul 25, 2024

#1
+1657
+1

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! :)

NotThatSmart Jul 25, 2024
edited by NotThatSmart  Jul 25, 2024