Problem 11.2. Let $\alpha$ and $\beta$ be the real roots of the equation $x^{2}-x-2021=0$, with $\alpha>\beta$. Denote

$$
A=\alpha^{2}-2 \beta^{2}+2 \alpha \beta+3 \beta+7
$$

Find the greatest integer not exceeding $A$.
Answer: -6055.

Solution. By Vieta's theorem, we have $\alpha+\beta=1$ and $\alpha \beta=-2021$. Also, $\beta^{2}-\beta-2021=0$, since $\beta$ is a root of the equation. Therefore,

$A=\alpha^{2}-2 \beta^{2}+2 \alpha \beta+3 \beta+7=(\alpha+\beta)^{2}-3\left(\beta^{2}-\beta-2021\right)-6063+7=1^{2}-3 \cdot 0-6063+7=-6055$, so the answer to the problem is the number -6055.

The fact that $\alpha>\beta$ was not used.