Question

if solve it for me please

Answer 1

x= (√3+√2)(√3+√2)/3-2
=3+2+2√6/1
5+2√6=x
x²= (5+2√6)²
= 25+24+20√6
=49+20√6
similarly y²= 49-20√6
x²+y²=
49+20√6+49-20√6
98
the required answer is 98.

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,

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

Now,

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

On rationalizing the denominator we get,

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


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

Putting the values we get,

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

Hope this helps ^_^

If you have any doubt regarding to my answer, feel free to ask me in the comment section.

@Mahak24

Thanks...
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