Solved. 1

$\text{When }ax^2 + bx + c = 0\text{ has only 1 solution, }b^2 = 4ac$

So 

$\left(b+\dfrac 1 b\right)^2 = 4c\\ b^2 + \dfrac 1 {b^2} - 4c + 2 = 0\\ \left(b^2\right)^2 - (4c - 2)b^2 +1 = 0\\ $

And there is only 1 positive value for b, so there are only 1 possible value for b.

Use that again and you get: 

$\left(-(4c-2)\right)^2 = 4\\ 2c - 1 = 1\\ c = 1$

I am not sure about what the question is talking about, but I guess you meant that.