If n is a constant and if there exists a unique value of m for which the quadratic equation $x^2 + mx + (2m+n) = 0$ has one real solution, then find n.
Find n.
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\(x^2 + mx + (2m+n) = 0\\ x=-\frac{m}{2}\pm \sqrt{(\frac{m}{2})^2-2m-n}\\ n\le m( \frac{m}{4}-2)\)
if m = 8 then n = 0
so
x = - 4
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