Question

if x=root 3 + root 3\ root 3 - root 2 and y=root 3- root 2\ root 3+root 2 find the value of x square + y square + xy

Answer 1

hope this helps you ☺

Answer 2

Heya there !!!
Here is the answer you were looking for:

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,

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

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

On rationalizing the denominator we get,

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

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

Now putting the values in x^2 + y^2 + xy

$= (49 + 20 \sqrt{6} ) + (49 - 20 \sqrt{6} ) + (5 + 2 \sqrt{6} )(5 - 2\sqrt{6} ) \\ \\ = 49 + 20 \sqrt{6} + 49 - 20 \sqrt{6} + ( {(5)}^{2} - {(2 \sqrt{6}) }^{2} ) \\ \\ = 98 + (25 - 24) \\ \\ = 98 + (1) \\ \\ = 99$

Hope you understood !!!

If you have any doubt regarding to my answer, then feel free to ask in the comment section ^_^

@Mahak24

Thanks...
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