Question

State True or False.
If $x=3-2\sqrt{2}$, then the value of $\sqrt{x}+\frac{1}{\sqrt{x}}$$=\pm 2\sqrt{2}$.

• A. True
• B. False

Answer 1

The correct option is A True

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