Problem 3. The equations $x^{2}+2019 a x+b=0$ and $x^{2}+2019 b x+a=0$ have one common root. What can this root be, given that $a \neq b$?
Answer: $\frac{1}{2019}$.

Solution. Let the common root of the given equations be $r$. Then

$$
r^{2}+2019 a r+b=0=r^{2}+2019 b r+a
$$

From this, we obtain that $2019 r(a-b)=(a-b)$. Since $a \neq b$, it follows that $r=\frac{1}{2019}$

Criteria

4 p. A complete and justified solution is provided.

In the absence of a correct solution, the highest applicable criterion from those listed below is used:

3 6. A system of equations equivalent to the one considered in the solution is set up, but there is no further progress

1 6. The correct answer is provided, but there is no justification

0 6. The problem is not solved or is solved incorrectly.