Question

If a = under root 3 - under root 2 upon under root + under root 2 and b = under root 3 + under root 2 upon under root 3 - under root 2 , Find the value of a2 + b2 -5ab.

Answer 1

Hey friend,
Here is the answer you were looking for:
$a = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ using \: the \: identities \: \\ {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{ {( \sqrt{3} )}^{2} + {( \sqrt{2}) }^{2} - 2 \times \sqrt{3} \times \sqrt{2} }{ {( \sqrt{3}) }^{2} - {( \sqrt{2} )}^{2} } \\ \\ = \frac{3 + 2 - 2 \sqrt{6} }{3 - 2} \\ \\ a = 5 - 2 \sqrt{6} \\ \\ b = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ using \: the \: identities \\ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{ {( \sqrt{3}) }^{2} + {( \sqrt{2} )}^{2} + 2 \times \sqrt{3} \times \sqrt{2} }{ {( \sqrt{3}) }^{2} - {( \sqrt{2} )}^{2} } \\ \\ = \frac{3 + 2 + 2 \sqrt{6} }{3 - 2} \\ \\ b = 5 + 2 \sqrt{6} \\ \\ {a}^{2} + {b}^{2} - 5ab \\ \\ (5 - 2 \sqrt{6} )^{2} + {(5 + 2 \sqrt{6} )}^{2} - 5(5 - 2 \sqrt{6} )(5 + 2\sqrt{6} ) \\ \\ using \: same \: identities \\ \\ = ( {(5)}^{2} + {(2 \sqrt{6} )}^{2} - 2 \times 5 \times 2 \sqrt{6} ) + ( {(5)}^{2} + {(2 \sqrt{6} )}^{2} + 2 \times 5 \times 2 \sqrt{6} ) - 5( {(5)}^{2} - {(2 \sqrt{6}) }^{2} ) \\ \\ = (25 + 24 - 20 \sqrt{6} ) + (25 + 24 + 20 \sqrt{6} ) - 5(25 - 24) \\ \\ = (49 - 20 \sqrt{6} ) + (49 + 20 \sqrt{6} ) - 5(1) \\ \\ = 49 - 20 \sqrt{6} + 49 + 20 \sqrt{6} - 5 \\ \\ = 49 + 49 - 5 \\ ( - 20 \sqrt{6} \: and \: + 20 \sqrt{6} \: got \: cancel) \\ \\ = 98 - 5 \\ \\ = 93$

Hope this helps!!!

@Mahak24

Thanks....
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