Question

$\text{If}x=\frac{\sqrt{5}-2}{\sqrt{5}+2}$, then find the value of $x+\frac{1}{x}$.

Answer 1

Solving through rationalization:
Given, $x=\frac{\sqrt{5}-2}{\sqrt{5}+2}$

Rationalizing the denominator,

$x=\frac{\sqrt{5}-2}{\sqrt{5}+2}\times \frac{\sqrt{5}-2}{\sqrt{5}-2}$

$x=$$\frac{(\sqrt{5}-2{)}^2}{(\sqrt{5}{)}^2-(2{)}^2}$

$x=$$\frac{(\sqrt{5}{)}^2-2(\sqrt{5}\left)\right(2)+(2{)}^2}{5-4}$

$x=$$\frac{5-4\sqrt{5}+4}{1}$ =$9-4\sqrt{5}...\left(i\right)$
Take $\left(i\right)$ as equation 1.

Now,

$\frac{1}{x}=\frac{1}{9-4\sqrt{5}}\times \frac{9+4\sqrt{5}}{9+4\sqrt{5}}$

$\frac{1}{x}=$$\frac{9+4\sqrt{5}}{(9{)}^2-(4\sqrt{5}{)}^2}$

$\frac{1}{x}=$$\frac{9+4\sqrt{5}}{81-80}$

$\frac{1}{x}=$$9+4\sqrt{5}...\left(ii\right)$
Take $\left(ii\right)$ as equation 2.

Now, from $\left(i\right)$ and $\left(ii\right)$, we get that

$x+\frac{1}{x}=9-4\sqrt{5}+9+4\sqrt{5}$

$\therefore \text{Answer}=18$