Question

if x = √3+√2/√3-√2 and y = √3-√2/√3+√2 find x^2+y^2+xy

Answer 1

Heya , here .

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Solution .
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$x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ = > \frac{( { \sqrt{3} + \sqrt{2} ) }^{2} }{ {( \sqrt{3}) }^{2} - ( \sqrt{2} {)}^{2} } \\ \\ \: \: \: \: \: \: \: \: \: (after \: \: rationalisation \: \: ) \\ \\ = > \frac{( \sqrt{3} {)}^{2} + ( \sqrt{2} {)}^{2} + 2. \sqrt{3}. \sqrt{2} }{3 - 2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: |(a + b {)}^{2} = {a}^{2} + 2ab + {b}^{2} | \\ \: \: \: \: \: \: \: \: \: \: \: |(a + b)(a - b) = {a}^{2} - {b}^{2} | \\ \\ = > 3 + 2 + 2 \sqrt{6} \\ x = 5 + 2 \sqrt{6} \\ = > {x}^{2} = (5 + 2 \sqrt{6} {)}^{2} \\ = > ( {5})^{2} + (2 \sqrt{6} {)}^{2} + 2.5.2 \sqrt{6} \\ = > {x}^{2} = 25 + 24 + 20 \sqrt{6} \\ = > 49 + 20 \sqrt{6} \\ = > {x}^{2} = 49 + 20 \sqrt{6} \\ \\ similarly \: \: \\ {y}^{2} = 49 - 20 \sqrt{6} \\ \\ now \: \\ {x}^{2} + {y}^{2} = 49 + 20 \sqrt{6} + 49 - 20 \sqrt{6} \\ \\ = > 98 \: \: \: \: \: \: \: \: \: \: answer$
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