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

Concept & Formula Express a surd of the form $a + 2\sqrt{b}$ as a perfect square $(\sqrt{m} + \sqrt{n})^2$ where $m + n = a$ and $m \times n = b$. $$ (a+b)^2 = a^2 + b^2 + 2ab $$ Step-by-Step Solution * Given: $x = 5 + 2\sqrt{6}$ * We need two numbers that add to 5 and multiply to 6. These are 3 and 2. * Rewrite $x$: $$ x = 3 + 2 + 2\sqrt{6} = (\sqrt{3})^2 + (\sqrt{2})^2 + 2(\sqrt{3})(\sqrt{2}) = (\sqrt{3} + \sqrt{2})^2 $$ * Taking the square root, we get: $\sqrt{x} = \sqrt{3} + \sqrt{2}$ * Next, find $\frac{1}{\sqrt{x}}$ which is $\frac{1}{\sqrt{3} + \sqrt{2}}$ * Rationalize the denominator by multiplying the numerator and denominator by the conjugate $(\sqrt{3} - \sqrt{2})$: $$ \frac{1}{\sqrt{3} + \sqrt{2}} \times \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} = \frac{\sqrt{3} - \sqrt{2}}{3 - 2} = \sqrt{3} - \sqrt{2} $$ * Substitute these values into the original expression: $$ \sqrt{x} - \frac{1}{\sqrt{x}} = (\sqrt{3} + \sqrt{2}) - (\sqrt{3} - \sqrt{2}) $$ $$ = \sqrt{3} + \sqrt{2} - \sqrt{3} + \sqrt{2} = 2\sqrt{2} $$ Exam Strategy & Shortcut Whenever you have a term like $\sqrt{m} + \sqrt{n}$ where the difference of their squares is $1$ (i.e., $m - n = 1$), its reciprocal is simply its conjugate $\sqrt{m} - \sqrt{n}$. Since $(\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1$, the reciprocal of $\sqrt{3} + \sqrt{2}$ is instantly $\sqrt{3} - \sqrt{2}$. Subtracting them in your head saves critical exam time. Common Pitfall Students often make a sign error when subtracting the reciprocal. Writing $-\sqrt{3} - \sqrt{2}$ instead of $-(\sqrt{3} - \sqrt{2})$ will lead to an incorrect intermediate step and a wrong option. Always use brackets when substituting terms with negative signs. Final Answer Therefore, the correct answer is $2\sqrt{2}$.