if x = root 3 + root 2/ root 3 - root 2 nd y =3 root - root 2 /root 3 = root 2 find the value of x square y square +xy

if x = root 3 + root 2/ root 3 - root 2 nd y =3 root - root 2 /root 3

1.	if x = root 3 + root 2/ root 3 - root 2 nd y =3 root - root 2 /root 3 = root 2 find the value of x square y square +xy
Answer» HEY friend, Here is the ANSWER you were looking for: $x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$ On Rationalizing the denominator we get, $x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\$ Using the identities: ${(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2}$ $x = \frac{ {( \sqrt{3}) }^{2} + {( \sqrt{2} )}^{2} + 2( \sqrt{3})( \sqrt{2} ) }{ {( \sqrt{3} )}^{2} - {( \sqrt{2}) }^{2} } \\ \\ x = \frac{3 + 2 + 2 \sqrt{6} }{3 - 2} \\ \\ x = 5 + 2 \sqrt{6} \\ \\ y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$ On rationalizing the denominator we get, $y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\$ Using the identities: ${(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\$ $y = \frac{ {( \sqrt{3}) }^{2} + {( \sqrt{2}) }^{2} - 2( \sqrt{3})( \sqrt{2} )}{ {( \sqrt{3} )}^{2} - {( \sqrt{2} )}^{2} } \\ \\ y = \frac{3 + 2 - 2 \sqrt{6} }{3 - 2} \\ \\ y = 5 - 2 \sqrt{6}$ Now, Bro I GUESS you have forgot a sign in between x sq and y sq So, for that I can solve putting both + and - Here we go: ${x}^{2} + {y}^{2} + xy \\ \\ {(5 + 2 \sqrt{6} )}^{2} + {(5 - 2 \sqrt{6} )}^{2} + (5 + 2 \sqrt{6} )( 5 - 2 \sqrt{6} ) \\$ Using same identities: $( {(5)}^{2} + {(2 \sqrt{6}) }^{2} + 2(5)(2 \sqrt{6} )) + ( {(5)}^{2} + {(2 \sqrt{6}) }^{2} - 2(5)(2 \sqrt{6})) + (5 + 2 \sqrt{6} )(5 - 2 \sqrt{6} ) \\ \\ = (25 + 24 + 20 \sqrt{6} ) + (25 + 24 - 20 \sqrt{6} ) + ( {(5)}^{2} - {(2 \sqrt{6} )}^{2} ) \\ \\ = 49 + 20 \sqrt{6} + 49 - 20 \sqrt{6} + (25 - 24) \\ \\ = 98 + 1 \\ \\ = 99$ And if the question is, ${x}^{2} - {y}^{2} + xy \\ \\ = (5 + 2 \sqrt{6} )^{2} - (5 - 2 \sqrt{6} )^{2} + (5 + 2 \sqrt{6} )(5 - 2 \sqrt{6} ) \\ \\ = ( {(5)}^{2} + {(2 \sqrt{6} )}^{2} + 2(5)(2 \sqrt{6} )) - ( {(5)}^{2} + {(2 \sqrt{6}) }^{2} - 2(5)(2 \sqrt{6} )) - ( {(5)}^{2} - {(2 \sqrt{6}) }^{2} ) \\ \\ = (25 + 24 + 20 \sqrt{6} ) - (25 + 24 - 20 \sqrt{6} ) - (25 - 24) \\ \\ = 25 + 24 + 20 \sqrt{6} - 25 - 24 + 20 \sqrt{6} + 1 \\ \\ = 40 \sqrt{6} + 1$ Hope you have got your answer !!! If you have any doubt regarding to my answer, FEEL free to ask in the COMMENT section or inbox me if needed. @Mahak24 Thanks... ☺☺