Practice | Hashtable

Home > Practice > Arithmetic Aptitude > Surds and Indices > Miscellaneous

1. If $x = 5 + 2 \sqrt{6},$ then $\sqrt{x} - \frac{1}{\sqrt{x}}$ = is?

A. $2 \sqrt{2}$

B. $2 \sqrt{3}$

C. $\sqrt{3} + \sqrt{2}$

D. $\sqrt{3} - \sqrt{2}$

Solution:

\begin{aligned} {(\sqrt{x} - \frac{1}{\sqrt{x}})}^{2} \\ = x + \frac{1}{x} - 2 \\ = (5 + 2 \sqrt{6}) + \frac{1}{(5 + 2 \sqrt{6})} - 2 \\ = (5 + 2 \sqrt{6}) + \frac{1}{(5 + 2 \sqrt{6})} \times \frac{(5 - 2 \sqrt{6})}{(5 - 2 \sqrt{6})} - 2 \\ = (5 + 2 \sqrt{6}) + (5 - 2 \sqrt{6}) - 2 \\ = 10 - 2 \\ = 8 \\ ∴ (\sqrt{x} - \frac{1}{\sqrt{x}}) = \sqrt{8} = 2 \sqrt{2} \end{aligned}

You must login to add comments. Login now.