3-ча 3. Find the real roots of the equation:

\[
x^{2}+2 a x+\frac{1}{16}=-a+\sqrt{a^{2}+x-\frac{1}{16}} \quad\left(0<a<\frac{1}{4}\right)
\]
Solution 3. Let \(f(x)\) be the left-hand side of the given equation. The graph of the function \(y=f(x)\) is a parabola. If \(00\). Therefore, the original equation is equivalent to the equation

\[
x^{2}+2 a x+\frac{1}{16}=-a \pm \sqrt{a^{2}+x-\frac{1}{16}}
\]

The right-hand side of equation (1) represents the inverse function \(f^{-1}(x)\). Indeed, if \(y^{2}+2 a y+\) \(\frac{1}{16}=x\), then \(y=-a \pm \sqrt{a^{2}+x-\frac{1}{16}}\). To find the real roots of the given equation, we need to find the points of intersection of the graphs \(y=f(x)\) and \(y=f^{-1}(x)\). The graphs of mutually inverse functions are symmetric with respect to the line \(y=x\), so the real roots of the given equation are precisely the real roots of the equation

\[
x^{2}+2 a x+\frac{1}{16}=x
\]

Solving this equation, we find

\[
x_{1,2}=\frac{1-2 a}{2} \pm \sqrt{\left(\frac{1-2 a}{2}\right)^{2}-\frac{1}{16}}
\]