Question

Rationalise the denominator ouestion 1 . 1/(root 3+ root 5)- root 2

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{3} + \sqrt{5} - \sqrt{2} }$

can be re-arranged as

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{3} - \sqrt{2} + \sqrt{5} }$

$\rm \:  =  \: \dfrac{1}{( \sqrt{3} - \sqrt{2} ) + \sqrt{5} }$

On rationalizing the denominator, we get

$\rm \:  =  \: \dfrac{1}{( \sqrt{3} - \sqrt{2} ) + \sqrt{5} } \times \dfrac{( \sqrt{3} - \sqrt{2} ) - \sqrt{5} }{( \sqrt{3} - \sqrt{2} ) - \sqrt{5} }$

We know,

$\rm :\longmapsto\:\boxed{\tt{ (x + y)(x - y) = {x}^{2} - {y}^{2} \: }}$

So, using this, we get

$\rm \:  =  \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{ {( \sqrt{3} - \sqrt{2} )}^{2} - {( \sqrt{5}) }^{2} }$

$\rm \:  =  \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{3 + 2 - 2 \sqrt{6} - 5}$

$\rm \:  =  \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{5 - 2 \sqrt{6} - 5}$

$\rm \:  =  \: \dfrac{ \sqrt{3} - \sqrt{2} - \sqrt{5} }{ - 2 \sqrt{6}}$

$\rm \:  =  \: \dfrac{ - ( - \sqrt{3} + \sqrt{2} + \sqrt{5}) }{ - 2 \sqrt{6}}$

$\rm \:  =  \: \dfrac{- \sqrt{3} + \sqrt{2} + \sqrt{5}}{2 \sqrt{6}}$

On rationalizing the denominator, we get

$\rm \:  =  \: \dfrac{- \sqrt{3} + \sqrt{2} + \sqrt{5}}{2 \sqrt{6}} \times \dfrac{ \sqrt{6} }{ \sqrt{6} }$

$\rm \:  =  \: \dfrac{- \sqrt{18} + \sqrt{12} + \sqrt{30}}{2 \times 6}$

$\rm \:  =  \: \dfrac{- \sqrt{3 \times 3 \times 2} + \sqrt{2 \times 2 \times 3} + \sqrt{30}}{12}$

$\rm \:  =  \: \dfrac{- 3\sqrt{2} + 2 \sqrt{3} + \sqrt{30}}{12}$

Hence,

$\boxed{\tt{ \rm \dfrac{1}{ \sqrt{3} + \sqrt{5} - \sqrt{2} } =\dfrac{- \sqrt{3 \times 3 \times 2} + \sqrt{2 \times 2 \times 3} + \sqrt{30}}{12}}}$

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More Identities to know

$\purple{\boxed{\tt{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2}}}}$

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