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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 28, 2025
 #1
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Simplify as

 

x^2  - mx  + 14  =  0  

 

Possible linear factors

( x - 2) ( x -7)      m =  9

(x + 2) (x + 7)     m =  -9

(x - 1) ( x -14)    m  = 15

(x + 1) ( x + 14)  m = -15

 

cool cool cool

 Jan 28, 2025
edited by CPhill  Jan 28, 2025

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