1. Let $P(x), Q(x)$ be quadratic trinomials such that three of the roots of the equation $P(Q(x))=0$ are the numbers $-22, 7, 13$. Determine the fourth root of this equation.
SOLUTION. Given that the equation $P(Q(x))=0$ has a real root, the quadratic equation $P(x)=0$ has two real roots $r_{1}, r_{2}$ (we do not exclude the possibility that $r_{1}=r_{2}$). The polynomial $P(Q(x))$ can therefore be written in the form

$$
P(Q(x))=a\left(Q(x)-r_{1}\right)\left(Q(x)-r_{2}\right),
$$

where $a$ is a real number $a \neq 0$. The equation $P(Q(x))=0$ has, according to the problem, four real roots, so each of the quadratic equations $Q(x)-r_{1}=0, Q(x)-r_{2}=0$ must have two real roots. From Vieta's formulas, it follows that the sum of the roots in both quadratic equations is the same, since both equations have the same coefficient for the linear term. Furthermore, three of the four real roots of both quadratic equations $Q(x)-r_{1}=0, Q(x)-r_{2}=0$ are the numbers $-22,7,13$, and the fourth root we denote as $q$. There are three possible cases:

(i) One of the quadratic equations has roots $-22,7$, the other has roots 13 and $q$. Then $-22+7=13+q$, so $q=-28$.

(ii) One of the quadratic equations has roots $-22,13$, the other has roots 7 and $q$. Then $-22+13=7+q$, so $q=-16$.

(iii) One of the quadratic equations has roots 13,7, the other has roots -22 and $q$.

Then $13+7=-22+q$, so $q=42$.

It is clear that in each of the cases (i), (ii), (iii), there exist corresponding quadratic trinomials $P(x)$ and $Q(x)$. If one of the quadratic equations $Q(x)-r_{1}=0$, $Q(x)-r_{2}=0$ has roots $-22,7$ and the other has roots $13,-28$, we set $Q(x)=x^{2}+15 x$, $r_{1}=(-22) \cdot 7=-154, r_{2}=13 \cdot(-28)=-364, P(x)=(x+154)(x+364)=$ $=x^{2}+518 x+56056$. Similarly, we can proceed in the remaining cases.

The fourth root of the equation $P(Q(x))=0$ can be any of the numbers -28, $-16,42$.