2. Quadratic trinomials $P(x)=x^{2}+\frac{x}{2}+b$ and $Q(x)=x^{2}+c x+d$ with real coefficients are such that $P(x) Q(x)=Q(P(x))$ for all $x$. Find all real roots of the equation $P(Q(x))=0$.
2. Quadratic trinomials $P(x)=x^{2}+\frac{x}{2}+b$ and $Q(x)=x^{2}+c x+d$ with real coefficients are such that $P(x) Q(x)=Q(P(x))$ for all $x$. Find all real roots of the equation $P(Q(x))=0$.

Answer: $x \in\left\{-1, \frac{1}{2}\right\}$.

In the equation $P(x) Q(x)=Q(P(x))$, expand the brackets and equate the coefficients of all powers. We get the system of equations

$$
\left\{\begin{array}{l}
c+\frac{1}{2}=1 \\
d+b+\frac{c}{2}=\frac{1}{4}+2 b+c \\
\frac{d}{2}+b c=b+\frac{c}{2} \\
b d=b c+b^{2}+d
\end{array}\right.
$$

the only solution of which is $b=-\frac{1}{2}, c=\frac{1}{2}, d=0$. Then the trinomial $P(x)$ can be represented as $P(x)=x^{2}+\frac{x}{2}-\frac{1}{2}=(x+1)\left(x-\frac{1}{2}\right)$. Further, the equation $P(Q(x))=0=(Q(x)+1)\left(Q(x)-\frac{1}{2}\right)$ reduces to either $Q(x)=-1$ or $Q(x)=\frac{1}{2}$, the first of which has no solutions, while the second has two solutions $x \in\left\{-1, \frac{1}{2}\right\}$.

Another solution. Notice that by the Remainder Theorem, the polynomial $Q(P(x))-Q(0)$ is divisible by the polynomial $P(x)$, hence, by the condition, $Q(0)$ is divisible by $P(x)$ as a polynomial, which is possible only if $d=Q(0)=0$. In this case, $Q(P(x))=P(x)(P(x)+c)$, and thus $P(x)+c=Q(x)$, from which we derive the equalities $b=-\frac{1}{2}, c=\frac{1}{2}$. After finding the unknown coefficients, the solution ends the same way as the first one.

# Criteria

+ Correct solution - 7 points.

$\pm$ Correctly found all coefficients $a, b, c$, but made arithmetic errors in solving equations $P(y)=0$ or $Q(x)=y$ - 5 points.

$\mp$ Correctly found two of the coefficients $b, c, d-3$ points

- Correctly found one of the coefficients $b, c, d-1$ point.