2. Find all solutions to the equation: $\sqrt{x+\sqrt{2 x-1}}+\sqrt{x-\sqrt{2 x-1}}=\sqrt{2}$.
Answer: the interval $\left[\frac{1}{2} ; 1\right]$

## Solution:

Let's find the domain of admissible values of the variable $x$. From the non-negativity of the expressions under the square roots, we have: $2 x-1 \geq 0$ (i.e., $x \geq \frac{1}{2}$) and $x-\sqrt{2 x-1} \geq 0$, from which it immediately follows that $x \geq 0$ (since we have already shown that $x \geq \frac{1}{2}$), and after moving the square root to the other side of the inequality and squaring, we get an equivalent inequality: $x^{2} \geq 2 x-1$, which is the true inequality $(x-1)^{2} \geq 0$. It remains to note that for $x \geq \frac{1}{2}$, the inequality $x+\sqrt{2 x-1} \geq 0$ is also valid. Thus, the domain of admissible values are all $x \geq \frac{1}{2}$. Both sides of the original inequality are non-negative, so we can square everything and simplify:

\[
\begin{gathered}
x+\sqrt{2 x-1}+2 \sqrt{x+\sqrt{2 x-1}} \sqrt{x-\sqrt{2 x-1}}+x-\sqrt{2 x-1}=2 \\
2 x+2 \sqrt{(x+\sqrt{2 x-1})(x+\sqrt{2 x-1})}=2 \\
2 x+2 \sqrt{x^{2}-(2 x-1)}=2 \\
2 x+2 \sqrt{(x-1)^{2}}=2 \\
x+|x-1|=1
\end{gathered}
\]

If $\frac{1}{2} \leq x<1$, then upon opening the modulus, we get the identity: $x+(1-x)=1$ (so the entire interval $\frac{1}{2} \leq x<1$ is part of the solution). For $x \geq 1$, we open the modulus and get $x+(x-1)=1$, from which $x=1$ (also included in the solution).

## Criteria:

Correctly found domain of admissible values: 1 point.

When squaring, the non-negativity of the sides of the equation is not mentioned: deduct 1 point.

Acquisition of extraneous solutions: the final score is no higher than 2 points (in particular, if extraneous solutions are obtained due to an unconsidered domain of admissible values - no more than 1 point).

Loss of one value from the set of solutions - deduct 2 points, loss of more than one deduct 4 points.