1. In the field of real numbers, solve the equation

$$
\sqrt{x+3}+\sqrt{x}=p
$$

with the unknown $x$ and the real parameter $p$.
1. For the left side of the equation to be defined, both expressions under the square roots must be non-negative, which is satisfied precisely for all $x \geq 0$. For non-negative $x$, we then have $p = \sqrt{x+3} + \sqrt{x} \geq \sqrt{3}$, so the equation can have a solution only for $p \geq \sqrt{3}$.

Now let's manipulate the given equation:

$$
\begin{aligned}
\sqrt{x} + \sqrt{x+3} & = p, \\
2x + 3 + 2 \sqrt{x(x+3)} & = p^2, \\
2 \sqrt{x(x+3)} & = p^2 - 2x - 3, \\
4x(x+3) & = (p^2 - 2x - 3)^2, \\
4x^2 + 12x & = p^4 + 4x^2 + 9 - 4p^2 x - 6p^2 + 12x, \\
x & = \frac{(p^2 - 3)^2}{4p^2}.
\end{aligned}
$$

Since we squared the given equation, it is necessary to verify by substitution that the computed $x$ is a solution to the original equation for the value of the parameter $p \geq \sqrt{3}$:

$$
\begin{aligned}
\sqrt{\frac{(p^2 - 3)^2}{4p^2} + 3} + \sqrt{\frac{(p^2 - 3)^2}{4p^2}} & = \sqrt{\frac{p^4 - 6p^2 + 9 + 12p^2}{4p^2}} + \sqrt{\frac{(p^2 - 3)^2}{4p^2}} = \\
& = \sqrt{\frac{(p^2 + 3)^2}{4p^2}} + \sqrt{\frac{(p^2 - 3)^2}{4p^2}} = \frac{p^2 + 3}{2p} + \frac{p^2 - 3}{2p} = p
\end{aligned}
$$

In the second-to-last step, we used the condition $p \geq \sqrt{3}$ (and thus also $p^2 - 3 \geq 0$ and $p > 0$), so $\sqrt{(p^2 - 3)^2} = p^2 - 3$ and $\sqrt{4p^2} = 2p$.

Note. Instead of the verification, it is sufficient to check that for the found $x$, all the squared expressions are non-negative, i.e., essentially just that

$$
p^2 - 2x - 3 = \frac{(p^2 - 3)(p^2 + 3)}{2p^2} \geq 0.
$$

For $p \geq \sqrt{3}$, this is indeed the case.

For a complete solution, award 6 points. For omitting the verification, deduct one point. If the condition $p \geq \sqrt{3}$ is not mentioned, award 3 points (regardless of whether the verification is present or not - if it is, it is incorrect).