, 2003

1. We have:
$$P(-2) = -1;\ P(-1) = 18;\ P(3/2) = -9/2 \text{ and } P(\sqrt{33}/3) = (15 - \sqrt{33})/9. \quad (1)$$
$$Q(-2) = -57;\ Q(-1) = 2;\ Q(1/3) = -2/9 \text{ and } Q(1) = 12. \quad (2)$$
From (1) it follows that $P(x)$ has three distinct real roots, and from (2) it follows that $Q(x)$ has three distinct real roots.

2. As $\alpha$ is a root of $P(x)$,
$$4\alpha^3 - 2\alpha^2 - 15\alpha + 9 = 0.$$
Therefore $4\alpha^3 - 15\alpha = 2\alpha^2 - 9$, so
$$16\alpha^6 - 120\alpha^4 + 225\alpha^2 = 4\alpha^4 - 36\alpha^2 + 81$$
i.e.
$$16\alpha^6 - 124\alpha^4 + 261\alpha^2 - 81 = 0.$$
(3)
$$\text{Since } \alpha \text{ is the greatest root of } P(x), \text{ from (1), we get: } \frac{\sqrt{33}}{3} > \alpha > \frac{3}{2}. \quad (4)$$
and so $4 - \alpha^2 > 0$. We shall prove that $\frac{\sqrt{3(4 - \alpha^2)}}{3}$ is a root of $Q(x)$.
Indeed, we have:
$$\begin{align*} Q\left(\frac{\sqrt{3(4 - \alpha^2)}}{3}\right) = 0 &\Leftrightarrow \frac{4}{3}(4 - \alpha^2)\sqrt{3(4 - \alpha^2)} + 2(4 - \alpha^2) - \frac{7}{3}\sqrt{3(4 - \alpha^2)} + 1 = 0 \\ &\Leftrightarrow \left(3 - \frac{4\alpha^2}{3}\right)\sqrt{3(4 - \alpha^2)} + 9 - 2\alpha^2 = 0 \\ &\Leftrightarrow (9 - 2\alpha^2)^2 = \left(3 - \frac{4\alpha^2}{3}\right)^2 \cdot 3(4 - \alpha^2) \quad \text{(do (4))} \\ &\Leftrightarrow 3(81 - 36\alpha^2 + 4\alpha^4) = (81 - 72\alpha^2 + 16\alpha^4)(4 - \alpha^2) \\ &\Leftrightarrow 16\alpha^6 - 124\alpha^4 + 261\alpha^2 - 81 = 0. \end{align*}$$
(3) shows that the last equality is true, so $x_0 = \frac{\sqrt{3(4 - \alpha^2)}}{3}$ is a root of $Q(x)$.
Moreover, from (4) it is easy to see that $x_0 \in (\frac{1}{3}; \frac{\sqrt{21}}{6}) \subset (\frac{1}{3}; 1)$.
On the other hand, since $\beta$ is the greatest root of $Q(x)$, (2) shows that $\beta$ is the unique root of $Q(x)$ in the interval $(\frac{1}{3}; 1)$.
From these results, it follows that $\frac{\sqrt{3(4 - \alpha^2)}}{3} = \beta$, thus $\alpha^2 + 3\beta^2 = 4$.