Solve the following equation:

$$
\sqrt{x-a}+\sqrt{x-2 a}=\sqrt{x-3 a} .
$$
By squaring the equation, only one radical expression remains, which we leave on the left side, then square again:

$$
2 x-3 a+2 \sqrt{(x-a)(x-2 a)}=x-3 a
$$

$$
\begin{gathered}
2 \sqrt{(x-a)(x-2 a)}=-x \\
4(x-a)(x-2 a)=x^{2}, \quad 3 x^{2}-12 a x+8 a^{2}=0
\end{gathered}
$$

Accordingly, the solution of (1) can only be

$$
x_{1}=2 a\left(1+\frac{1}{\sqrt{3}}\right) \quad(\approx a \cdot 3.15), \quad x_{2}=2 a\left(1-\frac{1}{\sqrt{3}}\right) \quad(\approx a \cdot 0.85)
$$

Since we performed non-equivalent transformations, we must check whether $x_{1}$ and $x_{2}$ satisfy (1).

In the case of $x_{1}$, under all three square root signs in (1), a product of $a$ with a positive number stands, so the square roots are only real for $a \geq 0$. For $a>0$, $x_{1}>0$ results, which is impossible due to (2), since the left side of (2) cannot be negative. For $a=0$, $x_{1}=0$, and this satisfies (1).

In the case of $x_{2}$, under all three square root signs in (1), a product of $a$ with a negative number stands, so the square roots are only real for $a \leq 0$ (and in this case, $x_{2}$ is also negative or 0). In this case, $x_{2}$ satisfies (1), indeed - always taking the positive value of the square root -

$$
\begin{gathered}
\sqrt{x_{2}-a}=\sqrt{-a\left(-1+\frac{2}{\sqrt{3}}\right)}=\sqrt{-a \cdot \frac{4-2 \sqrt{3}}{\sqrt{12}}}=\sqrt{-a} \cdot \frac{\sqrt{3}-1}{\sqrt[4]{12}} \\
\sqrt{x_{2}-2 a}=\sqrt{-a \cdot \frac{2}{\sqrt{3}}}=\sqrt{-a \cdot \frac{4}{\sqrt{12}}}=\sqrt{-a} \cdot \frac{2}{\sqrt[4]{12}} \\
\sqrt{x_{2}-3 a}=\sqrt{-a\left(1+\frac{2}{\sqrt{3}}\right)}=\sqrt{-a \cdot \frac{4+2 \sqrt{3}}{\sqrt{12}}}=\sqrt{-a} \cdot \frac{\sqrt{3}+1}{\sqrt[4]{12}}
\end{gathered}
$$

and the sum of the first two is indeed equal to the third.

The above case $a=0, x_{1}=0$ is also given by this expression (and in the case of an irrational equation, we do not speak of a double root), therefore the solution of (1) is

$$
x=2 a\left(1-\frac{1}{\sqrt{3}}\right), \quad \text { if } \quad a \leq 0
$$

Katalin Péli (Makó, József A. g. II. o. t.)

Remark. Similarly, it can be shown that the above expression for $x_{1}$ is the solution of the equation $\sqrt{x-a}-\sqrt{x-2 a}=\sqrt{x-3 a}$ for $a \geq 0$.