Task B-2.6. For which values of the positive real parameter $m$ does the equation

$$
\sqrt{3+x}+\sqrt{5+x}=m
$$

have solutions?
## First Solution.

For the equation to make sense, it must be that $5+x \geq 0$ and $3+x \geq 0$, which means $x \geq -3$. (1 point) If we write the equation in the form

$$
\sqrt{5+x}=m-\sqrt{3+x}
$$

the right-hand side must satisfy the following condition

$$
m-\sqrt{3+x} \geq 0
$$

In this case, squaring both sides gives

$$
5+x=m^{2}-2 m \sqrt{3+x}+3+x
$$

This simplifies to

$$
\sqrt{3+x}=\frac{m^{2}-2}{2 m} \quad(*)
$$

For the equation to make sense, it must be that

$$
\frac{m^{2}-2}{2 m} \geq 0
$$

Given that $m$ is positive, this implies $m \geq \sqrt{2}$.

We also need to check the condition

$$
\begin{gathered}
m-\sqrt{3+x} \geq 0 \\
m-\frac{m^{2}-2}{2 m} \geq 0, \quad \frac{m^{2}+2}{2 m} \geq 0
\end{gathered}
$$

which holds for all positive real numbers $m$.

From $(*)$ it follows that

$$
x=\left(\frac{m^{2}-2}{2 m}\right)^{2}-3
$$

which is certainly $\geq -3$, so the initial condition $x \geq -3$ is also satisfied.

(1 point)

Therefore, the solution to the problem is $m \in [\sqrt{2}, +\infty)$.