Solve the following equation:

$$
\sqrt{x+2+4 \sqrt{x-2}}+\sqrt{x+7-6 \sqrt{x-2}}=5
$$
The solution can only be sought among the values $x \geq 2$, otherwise the inner roots are not real. The two expressions under the outer square roots can be written as follows:

$$
\begin{aligned}
& x+2+4 \sqrt{x-2}=(x-2)+4 \sqrt{x-2}+4=(\sqrt{x-2}+2)^{2}, \\
& x+7-6 \sqrt{x-2}=(x-2)-6 \sqrt{x-2}+9=(\sqrt{x-2}-3)^{2},
\end{aligned}
$$

thus the two terms on the left side are $|\sqrt{x-2}+2|$ and $|\sqrt{x-2}-3|$. The former expression is equal to $\sqrt{x-2}+2$ because this is positive, as $\sqrt{x-2}$ cannot be negative. On the other hand, for $\sqrt{x-2}-3$, we need to consider two cases based on its sign.

a) If $\sqrt{x-2}-3 \leq 0$, or $x-2 \leq 3^{2}, x \leq 11$, then (1) becomes:

$$
(\sqrt{x-2}+2)+(3-\sqrt{x-2})=5,
$$

so (1) holds for all $x$ in the range $2 \leq x \leq 11$.

If, however, $\sqrt{x-2}-3>0$, or $x-2>3^{2}, x>11$, then

$$
(\sqrt{x-2}+2)+(\sqrt{x-2}-3)=5
$$

and from this, $\sqrt{x-2}=3, x=11$, which contradicts $x>11$, so there is no solution in this case.

In summary: the given equation is satisfied by the numbers between 2 and 11, and only by these, including both endpoints. We can also say that (1) is an identity in the interval $2 \leq x \leq 11$.

Judit Gombás (Budapest, Ságvári E. Gyak. Ig. III. o. t.)