# find the value

Question:

If $x=\frac{5-\sqrt{21}}{2}$, find the value of $x+\frac{1}{x}$

Solution:

$x=\frac{5-\sqrt{21}}{2} \quad \ldots \ldots(1)$

$\Rightarrow \frac{1}{x}=\frac{1}{\frac{5-\sqrt{21}}{2}}$

$\Rightarrow \frac{1}{x}=\frac{2}{5-\sqrt{21}}$

$\Rightarrow \frac{1}{x}=\frac{2}{5-\sqrt{21}} \times \frac{5+\sqrt{21}}{5+\sqrt{21}}$

$\Rightarrow \frac{1}{x}=\frac{2(5+\sqrt{21})}{5^{2}-(\sqrt{21})^{2}}$

$\Rightarrow \frac{1}{x}=\frac{2(5+\sqrt{21})}{25-21}$

$\Rightarrow \frac{1}{x}=\frac{2(5+\sqrt{21})}{4}$

$\Rightarrow \frac{1}{x}=\frac{5+\sqrt{21}}{2}$ ............(2)

Adding (1) and (2), we get

$x+\frac{1}{x}=\frac{5-\sqrt{21}}{2}+\frac{5+\sqrt{21}}{2}$

$\Rightarrow x+\frac{1}{x}=\frac{5-\sqrt{21}+5+\sqrt{21}}{2}$

$\Rightarrow x+\frac{1}{x}=\frac{10}{2}=5$

Thus, the value of $x+\frac{1}{x}$ is 5 .