2003 Vietnamese Mathematical Olympiad

1. We have: $$P(-2)=-1;P(-1)=18;P(3/2)=-9/2\text{ and }P(\sqrt{33}/3)=(15-\sqrt{33})/9.(1)$$ $$Q(-2)=-57;Q(-1)=2;Q(1/3)=-2/9\text{ and }Q(1)=12.(2)$$ 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) $$\text{Since }\alpha \text{ is the greatest root of }P(x),\text{ from (1), we get: }\frac{\sqrt{33}}{3}>\alpha >\frac{3}{2}.(4)$$ 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: $$\begin{array}{rl}Q\left(\frac{\sqrt{3(4-{\alpha }^2)}}{3}\right)=0& \iff \frac{4}{3}(4-{\alpha }^2)\sqrt{3(4-{\alpha }^2)}+2(4-{\alpha }^2)-\frac{7}{3}\sqrt{3(4-{\alpha }^2)}+1=0\\ & \iff \left(3-\frac{4{\alpha }^2}{3}\right)\sqrt{3(4-{\alpha }^2)}+9-2{\alpha }^2=0\\ & \iff (9-2{\alpha }^2{)}^2={\left(3-\frac{4{\alpha }^2}{3}\right)}^2\cdot 3(4-{\alpha }^2)\text{(do (4))}\\ & \iff 3(81-36{\alpha }^2+4{\alpha }^4)=(81-72{\alpha }^2+16{\alpha }^4)(4-{\alpha }^2)\\ & \iff 16{\alpha }^6-124{\alpha }^4+261{\alpha }^2-81=0.\end{array}$$ (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 ${\alpha }^2+3{\beta }^2=4$.