Find the real roots of the equation:

$$
x^{2}+2 a x+\frac{1}{16}=-a+\sqrt{a^{2}+x-\frac{1}{16}} \quad\left(0<a<\frac{1}{4}\right) .
$$
Let $\mathrm{f}(\mathbf{x})$ be the left-hand side of the given equation. The graph of the function $y=\mathrm{f}(\mathbf{x})$ is a parabola. If $0\mathbf{0}$. 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 $\mathrm{f}-\mathbf{- 1}(\mathbf{x})$. 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 $y=\mathrm{f}(\mathbf{x})$ and $y=\mathrm{f}^{-\mathbf{1}}(\mathbf{x})$. The graphs of inverse functions are symmetric with respect to the line $y=x$, so the real roots of the given equation are exactly the real roots of the equation

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

Solving this equation, we find

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

| [ Decimal number system ] |  |  |
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| Problem 78631 Topics: | Processes and operations | $\begin{aligned} Difficulty: \\ Classes. 910\end{aligned}$ |
|  | [ Proof by contradiction ] | Class |