Solve the following system of equations.

$$
\begin{gathered}
\sqrt{x}+\sqrt{y}=z \\
2 x+2 y+p=0 \\
z^{4}+p z^{2}+q=0
\end{gathered}
$$

Under what conditions do the real values of $x, y$, and $z$ satisfy the system of equations?
$$
\begin{gathered}
\sqrt{x}+\sqrt{y}=z \\
2 x+2 y+p=0 \\
z^{4}+p z^{2}+q=0
\end{gathered}
$$

If we square both sides of the first equation, we get:

$$
x+2 \sqrt{x y}+y=z^{2}
$$

Furthermore, if we multiply both sides of this by 2 and subtract the right or left side of (2) from them, we arrive at the following equation:

$$
4 \sqrt{x y}=2 z^{2}+p
$$

or squaring both sides:

$$
x y=\left(\frac{2 z^{2}+p}{4}\right)^{2}
$$

while from (2):

$$
x+y=-\frac{p}{2}
$$

From equations (6) and (7), it follows that $x$ and $y$ are the roots of the following quadratic equation:

$$
u^{2}+\frac{p}{2} u+\left(\frac{2 z^{2}+p}{4}\right)^{2}=0
$$

The roots of this equation, $u'$ and $u''$, or $x$ and $y$, are:

$$
-\frac{p}{4}+\frac{1}{2} \sqrt{\frac{p^{2}}{4}-4\left(\frac{2 z^{2}+p}{4}\right)^{2}}
$$

or

$$
-\frac{p}{4}+\frac{1}{2} \sqrt{-2^{4}-z^{2} p}
$$

But this expression, considering (3), takes the following form:

$$
-\frac{p}{4}+\frac{1}{2} \sqrt{q}
$$

and this is the common form of $x$ and $y$, that is,

$$
\begin{aligned}
x & =-\frac{p}{4}+\frac{1}{2} \sqrt{q} \\
y & =-\frac{p}{4}-\frac{1}{2} \sqrt{q}
\end{aligned}
$$

while the value of $z$ from (3) is:

$$
z=\sqrt{-\frac{p}{2}+\sqrt{\frac{p^{2}}{4}-q}}
$$

For $x, y$, and $z$ to be real simultaneously, it is necessary that:

$$
\begin{gathered}
q \geq 0 \\
p^{2}-4 q \geq 0 \\
-p+\sqrt{p^{2}-4 q} \geq 0
\end{gathered}
$$

Since $\sqrt{p^{2}-4 q} \leq p$, the last inequality only holds if $p<0$ and the necessary and sufficient conditions in their simplest form are:

$$
p \leq 0, \quad q \geq 0 \text { and } p^{2}-4 q \geq 0
$$

(Suták Sándor, 8th grade student, Nyíregyháza).