Question

FOR THOSE WHO Said THEY ARE THE TEACHER OF MATHS

$x = \frac{ \sqrt{3} + 1}{ \sqrt{3} - 1} and \: y = \frac{ \sqrt{3} - 1}{ \sqrt{3} + 1 }$
Find value

${x}^{2} + xy + {y}^{2}$
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Answer 1

$\huge \color{tomato}{Hey} \\ \\ \huge \boxed{Here \: is \: your \: answer \: in \: pic \: attached} \\ \\ \huge \boxed{Thank \: you} \:$

Answer 2

$<b >hey here is your answer$

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$x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \: \times \: \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ = \frac{( \sqrt{3} + \sqrt{2} {)}^{2} }{( \sqrt{3} {)}^{2} - ( \sqrt{2} {)}^{2} } \\ \\ = \frac{( \sqrt{3} {)}^{2} + ( \sqrt{2} {)}^{2} + 2 \times \sqrt{3} \times \sqrt{2} }{3 - 2}\\ \\ = \frac{3 + 2 + 2 \sqrt{6} }{1} \\ \\ = 5 + 2 \sqrt{6} \\ \\ \\y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ = \frac{( \sqrt{3} - \sqrt{2} {)}^{2} }{( \sqrt{3} {)}^{2} - ( \sqrt{2} {)}^{2} } \\ \\ = \frac{( \sqrt{3} {)}^{2} + ( \sqrt{2} {)}^{2} - 2 \times \sqrt{3} \times \sqrt{2} }{3 - 2} \\ \\ = \frac{3 + 2 - 2 \sqrt{6} }{1} \\ \\ = 5 - 2 \sqrt{6}$

now ,

${x}^{2} + {y}^{2} + xy = (5 + 2 \sqrt{6} {)}^{2}\\ \\ + (5 - 2 \sqrt{6} {)}^{2}+ (5+2 \sqrt{6} )(5-2 \sqrt{6} )\\ \\ = ( {5})^{2} + (2 \sqrt{6} {)}^{2} + 2 \times 5 \times 2 \sqrt{6} +\\ \\ (5 {)}^{2} + (2 \sqrt{6} {)}^{2} - 2 \times 5 \times 2 \sqrt{6} + [ 5^2 -( 2 \sqrt{6})^2] \\ \\ = 25 + 24 + 20 \sqrt{6} + 25 + 24 - 20 \sqrt{6}+ (25-24) \\ \\ = 25 + 24 + 25 + 24 -1\\ \\ = 99$

99 answer

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hope this helps you

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