Solve the following equation:

$$
\sqrt{x-\sqrt{1-x}}+\sqrt{x}=1
$$
We move the $\sqrt{x}$ on the left side of the equation to the right side and then square both sides:

$$
x-\sqrt{1-x}=1+x-2 \sqrt{x}, \quad \text { i.e. } \quad-\sqrt{1-x}=1-2 \sqrt{x} \ldots
$$

Squaring again: $1-x=1-4 \sqrt{x}+4 x$.

After simplification:

$$
4 \sqrt{x}=5 x \ldots
$$

From this:

$$
16 x=25 x^{2} \quad \text { or } \quad x(25 x-16)=0 \ldots
$$

The roots of equation (3) are:

$$
x_{1}=0 \quad \text { and } \quad x_{2}=\frac{16}{25}
$$

We need to check whether these values satisfy the original equation. From the original form of the equation, it follows that $0 \leq \sqrt{x}<\frac{1}{2}, \quad x>\frac{1}{4}
$$

Thus, the requirement $\frac{1}{4}<x<1$ is only satisfied by $\frac{16}{25}$.

Indeed: substituting $x=0$ into the original equation leads to a contradiction. If $x=\frac{16}{25}$, then

$$
\begin{gathered}
\sqrt{x}=\frac{4}{5}, \quad \sqrt{1-x}=\sqrt{\frac{9}{25}}=\frac{3}{5} \\
\sqrt{x-\sqrt{1-x}}=\sqrt{\frac{16}{25}-\frac{3}{5}}=\sqrt{\frac{16}{25}-\frac{15}{25}}=\frac{1}{5} \\
\frac{1}{5}+\frac{4}{5}=1
\end{gathered}
$$

$x=0$ is a root of the equation $\sqrt{x+\sqrt{1-x}}+\sqrt{x}=1$. If $+\sqrt{1-x}$ stands on the left side of (1), we also arrive at (2) after squaring.

Márta Erőd (Koháry István g. VI. o. Gyöngyös).