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

 

angryangelsad

 May 8, 2021
edited by Guest  May 8, 2021
 #1
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Note  that if  we  only have  one real solution , then  the discriminant  must =   0 

 

So......

 

m^2  - 4 (m + n)   =  0

 

m^2  - 4m  - 4n   =   0

 

m^2  - 4m  =  4n        complete  the  square  on   m

 

m^2 - 4m  + 4  =  4n  + 4

 

(m - 2)^2  =  4( n + 1)

 

Note  that  if   m  =  4    and  n  =  0, this is  true

 

So.....n =  0

 

 

cool cool cool

 May 8, 2021

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