9.2. The numbers $x, y$ satisfy the system of equations

$$
\left\{\begin{array}{l}
x+y=a \\
x^{2}+y^{2}=-a^{2}+2
\end{array}\right.
$$

What is the greatest and the least value that the product $x y$ can take?
Answer: The maximum value is $1 / 3$, the minimum is -1.

Solution. We have $2 x y=(x+y)^{2}-\left(x^{2}+y^{2}\right)=a^{2}-\left(-a^{2}+2\right)=2\left(a^{2}-1\right)$, i.e., $x y=a^{2}-1$. The system $\left\{\begin{array}{l}x+y=a \\ x y=a^{2}-1\end{array}\right.$ is equivalent to the original system (since from it, using the above transformations, the second equation of the original system is obtained). The solution of the obtained system is the roots of the quadratic equation $t^{2}-a t+\left(a^{2}-1\right)=0$ (by the inverse Vieta's theorem). The discriminant of this equation $D=a^{2}-4\left(a^{2}-1\right)=4-3 a^{2}$ must be non-negative, i.e., $0 \leq a^{2} \leq \frac{4}{3}$. Therefore, $x y=a^{2}-1 \in\left[-1 ; \frac{1}{3}\right]$.