# $\left( \frac{r}{\sqrt{r^2-1}}, \frac{r}{\sqrt{r^2+1}} \right)$ are roots of the equation $x^2 - bx + 3 = 0$. What is the value of $b$?

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?

• Just out of curiosity why do you need another approach?
– user876009
Commented Jul 11, 2021 at 14:45
• @JitendraSingh The approach has lots of calculations and one may make some mistakes if not be so careful. So I'm looking for some elegant methods. And learning some new ideas. Commented Jul 11, 2021 at 14:47

Let $$\alpha$$ and $$\beta$$ be the roots of the equation with $$\beta>\alpha$$.
$${1\over\alpha^2}=1-\frac1{r^2}$$ $${1\over\beta^2}=1+\frac1{r^2}$$ $$\implies \frac{1}{\alpha^2}+\frac{1}{\beta^2}=2$$ $$\implies \alpha^2+\beta^2=18$$ $$\implies \alpha^2+\beta^2+2\alpha\beta=24$$ $$\implies b^2=24$$