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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?

 Jun 27, 2024
 #1
avatar+129725 
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Simplify as

 

x^2 -mx + 14 =  0

 

Note that  the possible factorizations are

 

( x - 7) (x - 2)      m =  9

( x + 7) (x + 2)    m = -9

(x - 14) ( x -1)    m = 14

(x + 14) ( x + 1)  m = -14

 

cool cool cool

 Jun 27, 2024

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