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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?

 Jan 13, 2024
 #1
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Simplify as

 

x^2 -mx + 14 =  0

 

(x +1) (x + 14)  = 0    x^2 + 15x + 14 =  0     m =  15

 

So proceeding likewise....

 

(x -14) ( x -1)   = 0            m = -15

 

(x + 2) ( x + 7)  =  0         m =  9

 

(x -2) (x -7)  =  0             m  =  - 9

 

cool cool cool

 Jan 13, 2024

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