smart manipulation of roots

Question

Let $r_1$, $r_2$ and $r_3$ be roots of the equation $2x^3-x^2+3x-6$. Find the value of $$\sqrt{(r_1^2+3)(r_2^2+3)(r_3^2+3)}.$$

So my approach is to multiply the bracket out and use newton sum afterwards. However it keeps giving me a value of 0 which is not correct. Can someone give me a hint or a smart way of doing this? Thank you!

Answer 1

$r_1+r_2+r_3=\frac{1}{2}$, $r_1r_1+r_1r_3+r_2r_3=\frac{3}{2}$ and $r_1r_2r_3=3$.

Thus, $$\sqrt{(r_1^2+3)(r_2^2+3)(r_3^2+3)}=$$ $$=\sqrt{x_1^2x_2^2x_3^2+3(x_1^2x_2^2+x_1^2x_3^2+x_2^2x_3^2)+9(x_1^2+x_2^2+x_3^2)+27}=$$ $$=\sqrt{r_1^2r_2^2r_3^2+3((r_1r_1+r_1r_3+r_2r_3)^2-2r_1r_2r_3(r_1+r_2+r_3))+9((r_1+r_2+r_3)^2-2(r_1r_1+r_1r_3+r_2r_3))+27}=$$ $$=\sqrt{9+3\left(\frac{9}{4}-2\cdot3\cdot\frac{1}{2}\right)+9\left(\frac{1}{4}-3\right)+27}=3.$$ Done!

Answer 2

Hint: Let $P(x) = 2x^3-x^2+3x-6$. Then $P(x) = -2(r_1-x)(r_2-x)(r_3-x)$. Since $r_1^2+3 = (r_1+\sqrt{3}i)(r_1-\sqrt{3}i)$, and similarly for $r_2,r_3$, it follows that $$ (r_1^2+3)(r_2^2+3)(r_3^2+3) = \left(-\frac{1}{2}P(-\sqrt{3}i)\right)\left(-\frac{1}{2}P(\sqrt{3}i)\right). $$ Take it from there...