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 $n$.
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
