Question

if x=3+√8 then the value of √x-1/√x is​

Answer 1

Given-

$x = 3 + \sqrt{8}$

To find -

$the \: value \: of \: \sqrt{x} - \frac{1}{ \sqrt{x} }$

Solution -

$\implies x = 3 + \sqrt{8}$

$\implies \: \sqrt{x} = \sqrt{3 + \sqrt{8} }$

$\implies \sqrt{x} = \sqrt{3 + \sqrt{2 \times 4} }$

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

$\implies \sqrt{x} = \sqrt{3 + 2 \sqrt{2} \times \sqrt{1} }$

$\implies \sqrt{x} = \sqrt{ { \sqrt{2} }^{2} + { \sqrt{1} }^{2} + 2 \times \sqrt{2} \times \sqrt{1} }$

$\implies \sqrt{x} = \sqrt{ {( \sqrt{2} + \sqrt{1} )}^{2} }$

$\implies \sqrt{x} = \sqrt{2} + \sqrt{1}$

$again$

$\frac{1}{ \sqrt{x} } = \frac{1}{ \sqrt{2 } + \sqrt{1} }$

$By \: rationalising \: the \: denominator$

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

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

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

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

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

$Now$

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

$= \sqrt{2} + \sqrt{1} - ( \sqrt{2} - \sqrt{1} )$

$= \sqrt{2} + \sqrt{1} - \sqrt{2} + \sqrt{1}$

$= 2 \sqrt{1}$