## Task A-1.2.

Determine all ordered triples $(x, y, z)$ of real numbers for which

$$
\left\{\begin{array}{l}
x+y=\sqrt{28} \\
x y-2 z^{2}=7
\end{array}\right.
$$
## First solution.

Substitute into the second equation $y=2 \sqrt{7}-x$ and transform the obtained equation:

$$
\begin{gathered}
x(2 \sqrt{7}-x)-2 z^{2}=7 \\
x^{2}-2 x \sqrt{7}+7+2 z^{2}=0 \\
(x-\sqrt{7})^{2}+2 z^{2}=0
\end{gathered}
$$

Since the squares of real numbers cannot be negative,

Then $y=\sqrt{7}$.

Check that for $(x, y, z)=(\sqrt{7}, \sqrt{7}, 0)$ both equations are satisfied.

We have $\sqrt{7}+\sqrt{7}=2 \sqrt{7}=\sqrt{28}$ and $\sqrt{7} \cdot \sqrt{7}-2 \cdot 0^{2}=7$.