If roots of the equation $x^2 - bx + 3 = 0$ are $\left( \frac{r}{\sqrt{r^2-1}}, \frac{r}{\sqrt{r^2+1}} \right)$, then what is the value of $b$ ?
$1)\pm2\sqrt6\qquad\qquad2)\pm2\sqrt3\qquad\qquad3)2\sqrt6\qquad\qquad4)2\sqrt3$
Here is my approach:
We have $\dfrac{r^2}{\sqrt{r^4-1}}=3$. Hence $\dfrac{r^4}{r^4-1}=9$ and $r^4=\dfrac98\Rightarrow r^2=\dfrac{3}{2\sqrt2}$. And $b$ is equal to sum of the roots:
$$b=\frac{r}{\sqrt{r^2-1}}+\frac{r}{\sqrt{r^2+1}}=\frac{r(\sqrt{r^2+1}+\sqrt{r^2-1})}{\sqrt{r^4-1}}=\frac{\sqrt{r^4+r^2}+\sqrt{r^4-r^2}}{\sqrt{r^4-1}}$$ $$=2\sqrt2\times(\sqrt{\frac98+\frac{3\sqrt8}{8}}+\sqrt{\frac98-\frac{3\sqrt8}{8}})=\sqrt{9+3\sqrt8}+\sqrt{9-3\sqrt8}$$ We have $b^2=24$. So $b=\pm2\sqrt6$.
My question is, can we solve this problem with other approaches?
Let $\alpha$ and $\beta $ be the roots of the equation with $\beta>\alpha$.
$${1\over\alpha^2}=1-\frac1{r^2}$$ $${1\over\beta^2}=1+\frac1{r^2}$$ $$\implies \frac{1}{\alpha^2}+\frac{1}{\beta^2}=2$$ $$\implies \alpha^2+\beta^2=18$$ $$\implies \alpha^2+\beta^2+2\alpha\beta=24$$ $$\implies b^2=24$$