# 2. CONDITION

Find all pairs $(a ; b)$ of real numbers $a$ and $b$ such that the equations $x^{2}+a x+b^{2}=0$ and $x^{2}+b x+a^{2}=0$ have at least one common root.
Solution. Suppose that $a \neq b$, and let $x_{0}$ be a common root of the equations. Substituting $x_{0}$ into the equations and subtracting one from the other, we get

$$
\begin{gathered}
(a-b) x_{0}+b^{2}-a^{2}=0, \\
\text { hence } x_{0}=a+b . \text { Therefore, } \\
(a+b)^{2}+a(a+b)+b^{2}=0, \text { which means, } \\
2 a^{2}+3 a b+2 b^{2}=2(a+3 b / 4)^{2}+7 b^{2} / 8=0, \text { and thus } a=b=0-
\end{gathered}
$$

$$
\text { contradiction. }
$$

Let $a=b$. Then both equations take the form $x^{2}+a x+a^{2}=0$. The discriminant of this equation is $-3 a^{2} \leq 0$. Therefore, the only possibility is: $a=b=0$. It is clear that this pair satisfies the condition.

Answer. $(0 ; 0)$.