Let \(a\) be a real number for which there exists a unique value of \(b\) such that the quadratic equation \(x^2 + 2bx + (a-b) = 0\) has one real solution. Find \(a\) .
+6248
\(\text{one real solution means that the discriminant is zero}\\ \text{in the quadratic equation $a x^2 + b x +c$ the discriminant is given by $D=b^2-4ac$}\\ \text{here}\\ a=1,~b=2b,~c = a-b\\ D=(2b)^2 - 4(a-b) = 4b^2 - 4a+4ab\\~\\ D=0 \Rightarrow\\ 4b^2-4a+4ab=0\\ a = b+b^2 \)
.
