Question

rationalize the denominator of (√3+√2) by (√3-√2)​

Answer 1

$\bf\small\green{\underline{ }} \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ \bf\small\green{\underline{ }}multiplying \: numerator \: and \: denominator \\ \bf\small\green{\underline{ }}by \: ( \sqrt{3} + \sqrt{2} ) \\ \\ = > \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ \bf\small\green{\underline{ }} = > \frac{ {( \sqrt{3} + \sqrt{2} ) }^{2} }{( \sqrt{3} - \sqrt{2} )( \sqrt{3} + \sqrt{2} ) } \\ \\ \bf\small{\underline{ (a + b)(a - b) = {a}^{2} - {b}^{2} }} \\ .....a = \sqrt{3} \: \: \: \: \: \: \: \: \: b = \sqrt{2} \\ \\ \bf\small{\underline{( {a + b)}^{2} = {a}^{2} + 2ab + {b}^{2} }} \\ .....a = \sqrt{3} \: \: \: \: \: \: \: \: b = \sqrt{2} \\ \\ = > \bf\small\green{\underline{ }} \frac{ ( { \sqrt{3}) }^{2} + 2( \sqrt{3})( \sqrt{2} ) + ( { \sqrt{2}) }^{2} }{ { (\sqrt{3}) }^{2} - { (\sqrt{2} )}^{2} } \\ \\ = > \bf\small\green{\underline{ }} \frac{3 + 2 \sqrt{6} + 2}{3 - 2} \\ \\ \bf\small\green{ = > 5 + 2 \sqrt{6} }$