Question

if x is equal to under root 3 minus under root 2 upon under root 3 + under root 2 and Y is equal to under root 3 + under root 2 upon under root 3 minus under root 2 then the value of x square + xy + Y square is

Answer 1

Heya !!!

Identities used :

${( x+ y)}^{2} = {x}^{2} + {y}^{2} + 2xy \\ \\ {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \\ \\ (x + y)(x - y) = {x}^{2} - {y}^{2}$

$x = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

On rationalizing the denominator we get,

$x = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ x = \frac{ {( \sqrt{3}) }^{2} + {( \sqrt{2}) }^{2} - 2( \sqrt{3} )( \sqrt{2} ) }{ {( \sqrt{3} )}^{2} - {( \sqrt{2} )}^{2} } \\ \\ x = \frac{3 + 2 - 2 \sqrt{6} }{3 - 2} \\ \\ x = 5 - 2 \sqrt{6}$


Now,

$y = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

On rationalizing the denominator we get,

$y = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ y = \frac{ {( \sqrt{3} )}^{2} + {( \sqrt{2}) }^{2} + 2( \sqrt{3} )( \sqrt{2} ) }{ {( \sqrt{3}) }^{2} - {( \sqrt{2} )}^{2} } \\ \\ y = \frac{3 + 2 + 2 \sqrt{6} }{3 - 2} \\ \\ y = 5 + 2 \sqrt{6}$

Now putting the value in

${x}^{2} + xy + {y}^{2}$

$\\ = {(5 - 2 \sqrt{6}) }^{2} + (5 - 2 \sqrt{6} )(5 + 2 \sqrt{6} ) + {(5 + 2 \sqrt{6} )}^{2} \\ \\ = ( {(5)}^{2} + {(2 \sqrt{6} )}^{2} - 2(5)(2 \sqrt{6} )) + ( {(5)}^{2} - {(2 \sqrt{6} )}^{2} ) + ( {(5)}^{2} + {(2 \sqrt{6}) }^{2} + 2(5)(2 \sqrt{6} )) \\ \\ = (25 + 24 - 20 \sqrt{6} ) + (25 - 24) + (25 + 24 + 20 \sqrt{6} ) \\ \\ = 49 - 20 \sqrt{6} + 1 + 49 + 20 \sqrt{6} \\ \\ = 49 + 49 + 1 \\ \\ = 99$

Hope this helps ☺