\section*{Problem 5 - 071035}

For which real \(a\) does the sum of the squares of the solutions of the equation \(x^{2}+a x+a-2=0\) attain its smallest value?
}

The discriminant of the polynomial \(x^{2}+a x+a-2\) is \(a^{2}-4(a-2)=(a-2)^{2}+4\) which is positive for every real \(a\), so there are exactly two real solutions \(x_{1}, x_{2} \in \mathbb{R}\) of the equation \(x^{2}+a x+a-2=0\) for every real \(a\), and these are not double roots.

By Vieta's formulas, \(x_{1}+x_{2}=-a\) and \(x_{1} x_{2}=a-2\). Therefore,

\[
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=a^{2}-2(a-2)=a^{2}-2 a+4=(a-1)^{2}+3
\]

The sum of the squares of the solutions is minimized exactly when \(a=1\).

\section*{Solution adopted from \([5]\)}