2. Determine all pairs $(a, b)$ of real numbers for which the equations

$$
x^{2}+(3 a+b) x+4 a=0, \quad x^{2}+(3 b+a) x+4 b=0
$$

have a common real root.
2. If $x_{0}$ is a common root of both equations, then

$$
x_{0}^{2}+(3 a+b) x_{0}+4 a=0, \quad x_{0}^{2}+(3 b+a) x_{0}+4 b=0
$$

By subtracting these equations, we get $(2 a-2 b) x_{0}+4(a-b)=0$, which simplifies to $(a-b)\left(x_{0}+2\right)=0$.

We will consider two cases:

The number $x_{0}=-2$ is a common root of both equations. Substituting into either of them gives $4-2 a-2 b=0$, hence $b=2-a$. For such $b$, both equations have the common root -2 for any value of the parameter $a$.

If $a=b$, both given equations have the same form $x^{2}+4 a x+4 a=0$. At least one root (necessarily common) exists precisely when the discriminant $16 a^{2}-16 a$ is non-negative, i.e., precisely when $a \notin(0,1)$.

Conclusion: The given equations have at least one common root for all pairs of the form $(a, 2-a)$, where $a$ is arbitrary, and for all pairs $(a, a)$, where $a \notin(0,1)$.

For a complete solution, award 6 points. For deriving the condition $(a-b)\left(x_{0}+2\right)=0$, award 2 points, 2 points for correctly analyzing the case $a=b$, and 2 points for finding the solution $b=2-a$.