Question

2 + √8 ÷ √2 + √3 = a + b√2
$\frac{2 + \sqrt{8} }{ \sqrt{2 + \sqrt{3} } } = a \: + b \sqrt{2}$
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Answer 1

Question

$: \implies\rm \dfrac{2 + \sqrt{8} }{ \sqrt{2} + \sqrt{ 3} } = a + b \sqrt{2}$

Solution:-

$: \implies\rm \dfrac{2 + \sqrt{8} }{ \sqrt{2} + \sqrt{ 3} }$

Using rationalization method

$: \implies\rm \dfrac{2 + \sqrt{8} }{ \sqrt{2} + \sqrt{ 3} } \times \dfrac{ \sqrt{2} - \sqrt{3} }{ \sqrt{2} - \sqrt{3} }$

$: \implies\rm \dfrac{(2 + \sqrt{8} )( \sqrt{2} - \sqrt{3}) }{ (\sqrt{2} + \sqrt{ 3} )( \sqrt{2} - \sqrt{3}) }$

Using this identity

$\rm : \implies(a {}^{2} - {b}^{2} ) = (a + b)(a - b)$

We get

$\rm : \implies \dfrac{2 \times \sqrt{2} - 2 \times \sqrt{3} + \sqrt{8} \times \sqrt{2} - \sqrt{8} \times \sqrt{3} }{( \sqrt{2}) {}^{2} - ( \sqrt{3}) {}^{2} }$

$: \implies \dfrac{2 \sqrt{2} - 2 \sqrt{3} + \sqrt{16} - \sqrt{24} }{2 - 3}$

$: \implies - (2 \sqrt{2} - 2 \sqrt{3} + 4 - 2 \sqrt{6} )$

$: \implies \: 2 \sqrt{6} + 2 \sqrt{3} - 4 - 2 \sqrt{2}$

where

$\rm \: a = 2 \sqrt{6} + 2 \sqrt{3 } - 4 \: \: \: and \: \: b \: = - 2$