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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

Best Answer 

 #1
avatar+1926 
+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
avatar+1926 
+1
Best Answer

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

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