1. Determine all real numbers $p$ for which the system of inequalities

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

has at least one solution in the set of real numbers.
Another solution. The graphs of the functions $f(x)=x^{2}+u x+v$ and $g(x)=x^{2}-u x+v$ (parabolas opening upwards) are symmetric with respect to the $y$-axis, since $f(x)=g(-x)$ for every $x$. Therefore, the possible solution sets of the individual inequalities $f(x) \leqq 0$, resp. $g(x) \leqq 0$, which are known to be closed intervals that may degenerate into single-point sets, are two connected sets symmetrically associated with respect to the origin on the $x$-axis. Their intersection is non-empty precisely when they have a common point $x=0$, which occurs precisely when $v=f(0)=$ $=g(0) \leqq 0$. For the given system, this is the inequality $p \leqq 0$. (As is clear from the previous solution, the result does not depend on the form of the coefficient $u=p-1$ at all.)