Question

Please answer this question 14

Answer 1

$x = \frac{ \sqrt{3} + 1 }{ \sqrt{3} - 1 } \: \: and \: y = \frac{ \sqrt{3} - 1}{ \sqrt{3} + 1} \\ \\ x = \frac{ \sqrt{3} + 1 }{ \sqrt{3} - 1} \times \frac{ \sqrt{3} + 1}{ \sqrt{3} + 1 } = \frac{4 + 2 \sqrt{3} }{2} = 2 + \sqrt{3} \\ \\ y = \frac{ \sqrt{3} - 1}{ \sqrt{3} + 1} \times \frac{ \sqrt{3} - 1 }{ \sqrt{3} - 1 } = \frac{4 - 2 \sqrt{3} }{2} = 2 - \sqrt{3} \\ \\ now \\ {x}^{2} - {y}^{2} + xy = ({2 + \sqrt{3} })^{2} - ( {2 - \sqrt{3} })^{2} + (2 + \sqrt{3} )(2 - \sqrt{3} ) \\ = 7 + 4 \sqrt{3} - 7 + 4 \sqrt{3} + 4 \\ = 4 + 8 \sqrt{3}$