Question

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

• A. 1
• B. $2\sqrt{2}$
• C. 2
• D. $3\sqrt{3}$

Answer 1

The correct option is C 2

$x=3+2\sqrt{2}$ so $\sqrt{x}-\frac{1}{\sqrt{x}}=\sqrt{3+2\sqrt{2}}-\frac{1}{\sqrt{3+2\sqrt{2}}}$
$\Rightarrow \sqrt{3+2\sqrt{2}}-\frac{1}{\sqrt{3+2\sqrt{2}}}\times \frac{\sqrt{3-2\sqrt{2}}}{\sqrt{3-2\sqrt{2}}}$
$\Rightarrow \sqrt{3+2\sqrt{2}}-\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{\left(\sqrt{3+2\sqrt{2}}\right)\left(\sqrt{3-2\sqrt{2}}\right)}}$
$\Rightarrow \sqrt{3+2\sqrt{2}}-\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{(3^2-{\left(2\sqrt{2}\right)}^2)}}$
$\Rightarrow \sqrt{3+2\sqrt{2}}-\frac{\sqrt{3-2\sqrt{2}}}{9-8}$
$\Rightarrow \sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}$
$\Rightarrow \sqrt{2+1+2\sqrt{2}}-\sqrt{2+1-2\sqrt{2}}$
$\Rightarrow \sqrt{{(\sqrt{2}+1)}^2}-\sqrt{{\left(\sqrt{2-1}\right)}^2}$
$\Rightarrow \sqrt{2}+1-\sqrt{2}+1\Rightarrow 2$