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 \(n\).
+6248 \(\text{one real solution means the discriminant is 0}\\ \text{for the quadratic }ax^2 + b x + c \text{ the discriminant }\\ D = b^2 - 4ac\\ \text{here we have}\\ D = m^2 - 4(m+n) = 0\\ m^2 -4m - 4n = 0\\ m = 2 \left(1\pm\sqrt{n+1}\right)\)
\(\text{we're told that }m \text{ is unique, so }\\ 1+\sqrt{n+1} = 1 - \sqrt{n+1}\\ 2\sqrt{n+1} = 0\\ n=-1\)
.+6248 \(\text{one real solution means the discriminant is 0}\\ \text{for the quadratic }ax^2 + b x + c \text{ the discriminant }\\ D = b^2 - 4ac\\ \text{here we have}\\ D = m^2 - 4(m+n) = 0\\ m^2 -4m - 4n = 0\\ m = 2 \left(1\pm\sqrt{n+1}\right)\)
\(\text{we're told that }m \text{ is unique, so }\\ 1+\sqrt{n+1} = 1 - \sqrt{n+1}\\ 2\sqrt{n+1} = 0\\ n=-1\)
