Question

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PLS SOLVE ABOVE QUESTION
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Answer 1

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If $x = \: \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$and$\:\:y = \: \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

Find the value of x² + y² + xy

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$\implies\:\:$$\tt x = \: \frac{ \sqrt{3 \:} - \sqrt{2} }{\sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3 - \sqrt{2} } } = \frac{3 + 2 - 2 \sqrt{6} }{1}$

$\implies\:\:$$\tt 5 - 2 \sqrt{6}$

$\implies\:\:$$\tt y = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{2} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } = \frac{3 + 2 + 2 \sqrt{6} }{3 - 2 }$

$\implies\:\:$$\tt 5 + 2 \sqrt{6}$

$\implies\:\:$$\tt {x}^{2} + {y}^{2} + xy \: = \: (5 - 2 \sqrt{6} ) ^{2} + (5 + 2 \sqrt{6} ) {}^{2} + (5 - 2 \sqrt{6} )(5 + 2 \sqrt{6} )$

$\implies\:\:$$\tt 25 + 24 - \cancel{20 \sqrt{6}} + 25 - 24 + 25 + 24 + \cancel{20 \sqrt{6}}$

$\implies\:\:$$\tt 98 + 1$

$\implies\:\:$$\tt 99$

Answer 2

Answer:

hope this will help you