\section*{Problem 5 - 191045}

Determine all real numbers \(x\) for which \(\sqrt{x^{2}+5 x+28}\) (as a real number) is defined and which satisfy the equation

\[
x^{2}+5 x+4=5 \sqrt{x^{2}+5 x+28}
\]
}

Assume a real number \(x\) has the mentioned properties. Then \(x^{2}+5 x+28 \geq 0\), and if we set \(y=\sqrt{x^{2}+5 x+28}\), we have \(y^{2}-24=5 y\).

This quadratic equation has the solutions \(y=8\) and \(y=-3\), so one of the equations

\[
x^{2}+5 x+4=40 \quad \text { or } \quad x^{2}+5 x+4=-15
\]

follows. Of these two equations, only the first has real solutions, namely \(x=-9\) and \(x=4\). Therefore, at most these two numbers can have the required properties.

These numbers do indeed have these properties, since

\[
(-9)^{2}+5(-9)+28=64 \geq 0 \quad \text { and } \quad 4^{2}+5 \cdot 4+28=64 \geq 0
\]

so for these numbers \(x\), the square root \(\sqrt{x^{2}+5 x+28}\) exists and we have

\[
\begin{gathered}
(-9)^{2}+5(-9)+4=40=5 \cdot 8=5 \sqrt{(-9)^{2}+5(-9)+28} \\
4^{2}+5 \cdot 4+4=40=5 \cdot 8=5 \sqrt{4^{2}+5 \cdot 4+28}
\end{gathered}
\]

Therefore, exactly the numbers \(x=-9\) and \(x=4\) satisfy the conditions of the problem.