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If n is a constant and if there exists a unique value of m for which the quadratic equation x^2 + mx + (m+2n) = 0 has one real solution, then find n.

 Jul 14, 2021
 #1
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If we   have one real solution  then we  have a double root......which means that the  discriminant = 0

 

So

 

m^2  - 4(m + 2n)   =  0

 

m^2  - 4m  -  8n  =  0

 

Maybe  many solutions for  this but  one possible solution  is   m = 8   and  n = 4

 

So  we  have

 

x^2   +  8x   +   16 =  0

 

(x + 4) ^2   = 0

 

x  = -4   is  the  only solution

 

 

cool cool cool

 Jul 14, 2021

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