Problem 9.1. Find all pairs $(a, b)$ of non-negative real numbers such that the equations $x^{2}+a^{2} x+b^{3}=0$ and $x^{2}+b^{2} x+a^{3}=0$ have a common real root.

Peter Boyvalenkov
9.1. If $x_{0}$ is a common root of the equations, then

$$
x_{0}\left(a^{2}-b^{2}\right)=a^{3}-b^{3}
$$

Case 1. For $a \neq b$ we have $x_{0}=\frac{a^{2}+a b+b^{2}}{a+b}$. Since $x_{0}>0$, it follows by the first equation that $x_{0}^{2}+a^{2} x_{0}+b^{3}>0$, a contradiction.

Case 2. If $a=b$, then the equations coincide. They have a real root when $D=a^{4}-4 a^{3}=a^{3}(a-4) \geq 0$. Since $a \geq 0$, we conclude that the solutions of the problem are the pairs $(a, a)$, where $a \in\{0\} \cup[4,+\infty)$.