Question

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

Answer 1

Given: $x=2+\sqrt{3}$
$$\begin{gathered} \frac{1}{x}=\frac{1}{2+\sqrt{3}} \\ =\frac{2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)} \\ =\frac{2-\sqrt{3}}{2^2-{\left(\sqrt{3}\right)}^2}\left[\because \left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3} \end{gathered}$$
Now
$$\begin{gathered} x-\frac{1}{x}=2+\sqrt{3}-\left(2-\sqrt{3}\right) \\ =2+\sqrt{3}-2+\sqrt{3} \\ =2\sqrt{3} \end{gathered}$$
Hence, $x-\frac{1}{x}=2\sqrt{3}$.