Question

If $x=\sqrt{13}+2\sqrt{3}$, find the value of $x-\frac{1}{x}$.

Answer 1

$$\begin{gathered} x=\sqrt{13}+2\sqrt{3}.....\left(1\right) \\ \Rightarrow \frac{1}{x}=\frac{1}{\sqrt{13}+2\sqrt{3}} \\ \Rightarrow \frac{1}{x}=\frac{1}{\sqrt{13}+2\sqrt{3}}\times \frac{\sqrt{13}-2\sqrt{3}}{\sqrt{13}-2\sqrt{3}} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{1}{x}=\frac{\sqrt{13}-2\sqrt{3}}{{\left(\sqrt{13}\right)}^2-{\left(2\sqrt{3}\right)}^2} \\ \Rightarrow \frac{1}{x}=\frac{\sqrt{13}-2\sqrt{3}}{13-12} \\ \Rightarrow \frac{1}{x}=\sqrt{13}-2\sqrt{3}.....\left(2\right) \end{gathered}$$
Subtracting (2) from (1), we get
$$\begin{gathered} x-\frac{1}{x}=\left(\sqrt{13}+2\sqrt{3}\right)-\left(\sqrt{13}-2\sqrt{3}\right) \\ \Rightarrow x-\frac{1}{x}=\sqrt{13}+2\sqrt{3}-\sqrt{13}+2\sqrt{3} \\ \Rightarrow x-\frac{1}{x}=4\sqrt{3} \end{gathered}$$
Thus, the value of $x-\frac{1}{x}$ is $4\sqrt{3}$.