240*. Find the real roots of the equation

$$
x^{2}+2 a x+\frac{1}{16}=-a+\sqrt{a^{2}+x-\frac{1}{16}} \quad\left(0<a<\frac{1}{4}\right) .
$$
240. First solution. Let

$$
x^{2}+2 a x+\frac{1}{16}=y, -a+\sqrt{a^{2}+x-\frac{1}{16}}=y_{1}
$$

In this case, our equation takes the form

$$
y=y_{1}
$$

Now let's express $x$ in terms of $y_{1}$. Simple calculations give

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

Thus, we see that $x$ is expressed in terms of $y_{1}$ in the same way as $y$ is expressed in terms of $x$. Therefore, if we plot the graphs of the functions

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

and

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

then these graphs (parabolas) will be symmetric with respect to the bisector of the first coordinate angle (Fig. 27; each point $x=x_{0}, y=y_{0}$ of the first graph corresponds to a point $x=y_{0}, y=x_{0}$ of the second graph, symmetric with respect to the bisector of the coordinate angle). The points of intersection of these two graphs correspond to such values of $x$ that $y=y_{1}$, i.e., to the roots of our equation; these points necessarily lie on the axis of symmetry of both curves, i.e., satisfy the condition

$$
y=x=y_{1} .
$$

Solving the equation $y=x$, i.e.,

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

we get:

$$
x_{1,2}=\frac{1-2 a}{2} \pm
$$

Fig. 27.

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

It can be verified by checking that for $0<a<\frac{1}{4}$ both these roots are real and satisfy the original equation.

Second solution. This problem can also be solved in a more conventional way. Eliminating the radical in the equation, we get:

$$
\left(x^{2}+2 a x+a+\frac{1}{16}\right)^{2}=a^{2}+x-\frac{1}{16}
$$

or after expanding the brackets and combining like terms

$$
\begin{aligned}
& x^{4}+4 a x^{3}+\left(4 a^{2}+2 a+\frac{1}{8}\right) x^{2}+\left(4 a^{2}+\frac{1}{4} a-1\right) x+ \\
&+\frac{a}{8}+\frac{1}{16}+\frac{1}{16^{2}}=0
\end{aligned}
$$

The left-hand side of the resulting equation can be factored:

$$
\begin{aligned}
& {\left[x^{4}+\left(2 a^{\prime}-1\right) x^{3}+\frac{1}{16} x^{2}\right]+} \\
& \quad+\left[(2 a+1) x^{3}+\left(4 a^{2}-1\right) x^{2}+\left(\frac{a}{8}+\frac{1}{16}\right) x\right]+ \\
& +\left[\left(2 a+\frac{17}{16}\right) x^{2}+\left(4 a^{2}+\frac{a}{8}-\frac{17}{16}\right) x+\left(\frac{a}{8}+\frac{1}{16}+\frac{1}{16^{2}}\right)\right]= \\
& \quad=\left[x^{2}+(2 a-1) x+\frac{1}{16}\right]\left[x^{2}+(2 a+1) x+\left(2 a+\frac{17}{16}\right)\right]
\end{aligned}
$$

From this, we immediately get the solutions:

$$
\begin{gathered}
x^{2}+(2 a-1) x+\frac{1}{16}=0 \\
x_{1,2}=\frac{1-2 a}{2} \pm \sqrt{\left(\frac{1-2 a}{2}\right)^{2}-\frac{1}{16}} \\
x^{2}+(2 a+1) x+2 a+\frac{17}{16}=0 \\
x_{3,4}=-\frac{1+2 a}{2} \pm \sqrt{\left(\frac{1+2 a}{2}\right)^{2}-2 a-\frac{17}{16}}
\end{gathered}
$$

For $0<a<\frac{1}{4}$, the first two roots are real and satisfy the original equation; the last two roots are complex.