Question

If x = √3 -√2/ √3 + √2 and y = √3 + √2 / √3 - √2 then find x² + xy + y²

Answer 1

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Answer 2

Answer:-

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

$to \: find : \: \: \: {x}^{2} +xy + {y}^{2}$

⇒$( \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} })^{2} +( \frac{\sqrt{3} - \sqrt{2} }{\sqrt{3} + \sqrt{2}})( \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2}}) + ( \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2}})^{2}$

⇒$\frac{3 - 2}{3 + 2} + \frac{( \sqrt{3} ) ^{2} - ( \sqrt{2})^{2} }{( \sqrt{3}) ^{2} - ( \sqrt{2})^{2}} + \frac{3 + 2}{3 - 2}$

⇒$\frac{1}{5} + \frac{( \sqrt{3} - \sqrt{2})^{2} - 2 \times \sqrt{3} \times \sqrt{2} }{( \sqrt{3} - \sqrt{2})^{2} - 2 \times \sqrt{3} \times \sqrt{2} } + \frac{5}{1}$

∴[(a - b)^{2} - 2× a× b)]

.⇒$\frac{1}{5} + \frac{3 - 2 - 2 \sqrt{5} }{3 - 2 - 2 \sqrt{5}} + \frac{5}{1}$

⇒$\frac{1}{5} + \frac{1 - 2 \sqrt{5} }{1 - 2 \sqrt{5}} + \frac{5}{1}$

⇒$\frac{1}{5} + \frac{1}{1} + \frac{5}{1}$

⇒$\frac{1 + 5 + 25}{5} = \frac{31}{5}$