Question

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

Answer 1

$x=3+2\sqrt{2}$

Then, $\frac{1}{x}=\frac{1}{3+2\sqrt{2}}=\frac{1\times (3-2\sqrt{2})}{(3+2\sqrt{2})(3-2\sqrt{2})}$

$=\frac{3-2\sqrt{2}}{(3{)}^2-(2\sqrt{2}{)}^2}=\frac{3-2\sqrt{2}}{9-8}=\frac{3-2\sqrt{2}}{1}$

$\therefore x+\frac{1}{x}=3+2\sqrt{2}+3-2\sqrt{2}=6$

Now, ${(\sqrt{x}-\frac{1}{\sqrt{x}})}^2=x+\frac{1}{x}-2$

$=6-2=4=(2{)}^2$

$\therefore \sqrt{x}-\frac{1}{\sqrt{x}}=2$