9.3 Two parabolas with different vertices are the graphs of quadratic trinomials with leading coefficients $p$ and $q$. It is known that the vertex of each parabola lies on the other parabola. What can $p+q$ be?
9.3 Two parabolas with different vertices are the graphs of quadratic trinomials with leading coefficients $p$ and $q$. It is known that the vertex of each parabola lies on the other parabola. What can $p+q$ be equal to?

Answer: 0.

Solution: Let $\left(x_{1}, y_{1}\right)$ be the coordinates of the vertex of one parabola, $\left(x_{2}, y_{2}\right)$ - the other. Then the equations of the parabolas can be represented as

$$
y=p\left(x-x_{1}\right)^{2}+y_{1} \quad \text { and } y=q\left(x-x_{2}\right)^{2}+y_{2} \text {. }
$$

The point ( $x_{2}, y_{2}$ ) lies on the first parabola:

$$
y_{2}=p\left(x_{2}-x_{1}\right)^{2}+y_{1},
$$

and the point ( $x_{1}, y_{1}$ ) lies on the second parabola:

$$
y_{1}=q\left(x_{1}-x_{2}\right)^{2}+y_{2} \text {. }
$$

Adding the obtained equations, we find

$$
y_{2}+y_{1}=p\left(x_{2}-x_{1}\right)^{2}+y_{1}+q\left(x_{1}-x_{2}\right)^{2}+y_{2}
$$

From which (since $\left.\left(x_{2}-x_{1}\right)^{2}=\left(x_{1}-x_{2}\right)^{2}\right)$

$$
(p+q)\left(x_{1}-x_{2}\right)^{2}=0
$$

Since $x_{1} \neq x_{2}$ (if $x_{1}=x_{2}$, then from (*) $y_{2}=y_{1}$ and the vertices of the parabolas coincide), then $p+q=0$.