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http://prntscr.com/l8rzbo

 Oct 21, 2018
 #1
avatar+5800 
+3

\(x^2 + m x + n = 0,~(m,n)\in \mathbb{Z} \\ \text{the only possible value for x is }x = -3\\ x^2 + m x + n = (x+3)^2 = x^2 + 6x + 9\\ m = 6\)

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 Oct 21, 2018
 #2
avatar+103148 
+2

Since   x  = -3 is the only value that makes the quadratic  = 0.....then this has a "double" root

 

So...the sum  of the roots  =  -3 + -3  =  -6

 

And the sum of the roots  in the form ax^2 + bx + c  =  -b/a 

 

So...in our case, the sum of the roots are  =  - m / 1  =   -m

 

So

 

-m   = -6

 

m  = 6

 

 

cool cool cool

 Oct 22, 2018

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