Question

plss answer ........
urgently needed..

Answer 1

Hey!!!!

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

Therefore,
$\frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} }$

$= \frac{3 - 2 \sqrt{2} }{ {3}^{2} - {2 \sqrt{2} }^{2} }$

$= \frac{3 - 2 \sqrt{2} }{9 - 8}$

$= 3 - 2 \sqrt{2}$

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

${ (\sqrt{x} + \frac{1}{ \sqrt{x} } })^{2} = x + \frac{1}{x} + 2$
$= 6 + 2$
$8$

Therefore,
$\sqrt{x} + \frac{1}{ \sqrt{x} } = \sqrt{8}$

#BE BRAINLY