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.
+128408 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
