By the quadratic formula, the solutions are \[\frac{-b - 4c \pm \sqrt{(b + 4c)^2 - 4c^2}}{2},\]which gives us the discriminant $(b + 4c)^2 - 4c^2 = b^2 + 8bc + 16c^2 - 4c^2 = b^2 + 8bc + 12c^2$.
We can rewrite this as \[b^2 + 8bc + 12c^2 = (b^2 + 4bc + 4c^2) + 4bc + 8c^2 = (b + 2c)^2 + 4bc + 8c^2.\]Therefore, we need $4bc + 8c^2 = 0$ and $(b + 2c)^2 = 0$ to coincide, meaning $b = -2c$.
Since only one positive $b$ exists, $-2c > 0$, or $c < 0$. Furthermore, we need to discard the case $c = 0$ because then the given equation becomes $x^2 = 0$, which has only one solution. Therefore, $c$ must be negative and nonzero. We can set $c = -1$, so the non-zero value of $c$ is $\boxed{-1}$.
Find the non-zero value of $c$ for which there is exactly one positive value of $b$ for which there is one solution to the equation $x^2 + (b + 4c)x + c^2 = 0$.
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