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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 11, 2024
 #1
avatar+128732 
+1

Simplify as

 

x^2 - mx  + 14  = 0

 

Possible factorizations

 

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

 

(x + 2) (x + 7)  = 0          m  =  9

 

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

 

(x + 14) ( x + 1)  =  0      m  =  15

 

cool cool cool

 Jan 12, 2024

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