Question

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

Answer 1

$$\begin{gathered} x=\frac{5-\sqrt{21}}{2}.....\left(1\right) \\ \Rightarrow \frac{1}{x}=\frac{1}{{\displaystyle \frac{5-\sqrt{21}}{2}}} \\ \Rightarrow \frac{1}{x}=\frac{2}{5-\sqrt{21}} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{1}{x}=\frac{2}{5-\sqrt{21}}\times \frac{5+\sqrt{21}}{5+\sqrt{21}} \\ \Rightarrow \frac{1}{x}=\frac{2\left(5+\sqrt{21}\right)}{5^2-{\left(\sqrt{21}\right)}^2} \\ \Rightarrow \frac{1}{x}=\frac{2\left(5+\sqrt{21}\right)}{25-21} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{1}{x}=\frac{2\left(5+\sqrt{21}\right)}{4} \\ \Rightarrow \frac{1}{x}=\frac{5+\sqrt{21}}{2}.....\left(2\right) \end{gathered}$$
Adding (1) and (2), we get
$$\begin{gathered} x+\frac{1}{x}=\frac{5-\sqrt{21}}{2}+\frac{5+\sqrt{21}}{2} \\ \Rightarrow x+\frac{1}{x}=\frac{5-\sqrt{21}+5+\sqrt{21}}{2} \\ \Rightarrow x+\frac{1}{x}=\frac{10}{2}=5 \end{gathered}$$
Thus, the value of $x+\frac{1}{x}$ is 5.