Question

if a= √3+√2 / √3-√2 , b= √3-√2 / √3+√2. Find the value of a square + b square

Answer 1

Hey friend,
Here is the answer you were looking for:
$a = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ using \: the \: identities \\ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{ {( \sqrt{3}) }^{2} + {( \sqrt{2} )}^{2} + 2( \sqrt{3})( \sqrt{2} )}{ {( \sqrt{3}) }^{2} - {( \sqrt{2}) }^{2} } \\ \\ = \frac{3 + 2 + 2 \sqrt{6} }{3 - 2} \\ \\ = 5 + 2 \sqrt{6} \\ \\ b = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ using \: the \: identities \\ {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{ {( \sqrt{3}) }^{2} + {( \sqrt{2} )}^{2} - 2( \sqrt{3})( \sqrt{2} )}{ {( \sqrt{3}) }^{2} - {( \sqrt{2}) }^{2} } \\ \\ = \frac{3 + 2 - 2 \sqrt{6} }{3 - 2} \\ \\ = 5 - 2 \sqrt{6} \\ \\ {a}^{2} + {b}^{2} \\ \\ = {(5 + 2 \sqrt{6}) }^{2} + {(5 - 2 \sqrt{6} )}^{2} \\ \\ = ( {(5)}^{2} + {(2 \sqrt{6}) }^{2} + 2(5)(2 \sqrt{6} )) + ( {(5)}^{2} + {(2 \sqrt{6}) }^{2} - 2(5)(2 \sqrt{6} )) \\ \\ = (25 + 24 + 20 \sqrt{6} ) + (25 + 24 - 20 \sqrt{6} ) \\ \\ = 49 + 20 \sqrt{6} + 49 - 20 \sqrt{6} \\ \\ = 49 + 49 \\ \\ = 98$

Hope this helps!!!

@Mahak24

Thanks...
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