
1. Find all real numbers $p$ such that the system

$$
\begin{aligned}
& x^{2}+(p-1) x+p \leqslant 0 \\
& x^{2}-(p-1) x+p \leqslant 0
\end{aligned}
$$

of inequalities has at least one solution $x \in \mathbb{R}$.


Solution. If $p \leqslant 0$ then $x=0$ is clearly a solution. If $p>0$ then summing up we get $2 x^{2}+2 p \leqslant 0$ which doesn't hold for any real $x$.

Answer. The answer is $p \in(-\infty, 0]$.

Another solution. The graphs of functions $f(x)=x^{2}+u x+v$ and $g(x)=x^{2}-u x+v$ are symmetric about the $y$ axis, hence the solutions to the inequalities $f(x) \leqslant 0$, $g(x) \leqslant 0$ are two (possibly degenerate) intervals symmetric about 0 . The intersection of these intervals is nonempty if and only if $v=f(0)=g(0) \leqslant 0$. In our case, this happens if and only if $p \leqslant 0$.
