+0  
 
0
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avatar+1359 

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?

 Apr 2, 2024
 #3
avatar+129690 
+1

Simplify as 

 

x^2 - mx + 14 =  0

 

Roots             m

-1  -14            15

1    14           - 15

-2   -7               9

2     7              -9 

 

cool cool cool  

 Apr 2, 2024
 #4
avatar+945 
+1

 

Hi Chris, 

So, I blundered into the right answer but for the wrong reason. 

All the wrong reasons.  Thanks for posting the correct answer. 

Ron  

.

Bosco  Apr 2, 2024
 #5
avatar+129690 
+1

No prob.....I have those days when I should have just stayed in  bed, too....

 

 

cool cool cool

CPhill  Apr 2, 2024

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