jmc

The discriminant of the given quadratic equation is ${\left(b+\frac{1}{b}\right)}^2-4c$. For the quadratic to have one root, it follows that the discriminant must be equal to zero, so $b^2+2-4c+\frac{1}{b^2}=0$. We are also given that there should be exactly one positive value $b$ satisfying this equation. Multiplying through by $b^2$ (for we know that $b\ne 0$) yields that $b^4+(2-4c)b^2+1=0$; this is a quadratic equation in $b^2$ that has discriminant $(2-4c{)}^2-4$. Again, this discriminant must be equal to zero, so $(2-4c{)}^2=4⟹2-4c=\pm 2$. The non-zero value of $c$ satisfying this equation is $c=\boxed{1}$.