+817 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 $x^2 + (b + 4c)x + c^2 = 0$.
+121 Let's consider the quadratic equation \(x^2 + (b + 4c)x + c^2 = 0\).
For a quadratic equation \(ax^2 + bx + c = 0\) to have exactly one solution, its discriminant (\(b^2 - 4ac\)) must be equal to \(0\).
In this case, the discriminant is \((b + 4c)^2 - 4 \cdot 1 \cdot c^2 = b^2 + 8bc + 16c^2 - 4c^2\).
Simplify the discriminant:
\[b^2 + 8bc + 12c^2.\]
For the quadratic equation to have exactly one solution, the discriminant must be equal to \(0\):
\[b^2 + 8bc + 12c^2 = 0.\]
Now, we want to find the non-zero value of \(c\) for which there is exactly one positive value of \(b\) that satisfies this equation.
Notice that the quadratic equation above is a quadratic in \(b\), and we want it to have a single solution. This means the discriminant of this quadratic in \(b\) must be equal to \(0\):
\[8^2 - 4 \cdot 12c^2 = 0.\]
Solve for \(c\):
\[64 - 48c^2 = 0 \Rightarrow 48c^2 = 64 \Rightarrow c^2 = \frac{64}{48} = \frac{4}{3}.\]
Take the positive square root:
\[c = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}.\]
So, the non-zero value of \(c\) that satisfies the given conditions is \(\boxed{\frac{2\sqrt{3}}{3}}\).
