Question

if x is equals to root 3 + 1 / root 3 - 1, Y is equals to root 3 minus 1 by root 3+1 find the value of x2 +xy-y2​

Answer 1

$\sf x = \frac{ \sqrt{3} + 1 }{ \sqrt{3} - 1} \\ \\ \sf x = \frac{ \sqrt{3} + 1 }{ \sqrt{3} - 1} \times \frac{ \sqrt{3 + 1} }{ \sqrt{3} + 1 } \\ \\ \sf x = \frac{ (\sqrt{3} + 1)^{2}}{(\sqrt{3}) {}^{2} - (1)^{2} } \\ \\ \sf x = \frac{( \sqrt{3})^{2} + 1^{2} + 2 \times \sqrt{3} \times 1 }{3 - 1} \\ \\ \sf x = \frac{3 + 1 + 2 \sqrt{3} }{2} \\ \\ \sf x = \frac{4 + 2 \sqrt{3} }{2} \\ \\ \sf x = \frac{2(2 + \sqrt{3})}{2} \\ \\ \boxed{\bf x = 2 + \sqrt{3}}$

$\sf{Now, \: again \: find \: value \: of \: y}$

$\sf y = \frac{ \sqrt{3} - 1 }{ \sqrt{3} + 1} \\ \\ \sf y = \frac{ \sqrt{3} - 1 }{ \sqrt{3} + 1} \times \frac{ \sqrt{3} - 1}{ \sqrt{3} - 1 } \\ \\ \sf y = \frac{ (\sqrt{3} - 1)^{2}}{(\sqrt{3}) {}^{2} - (1)^{2} } \\ \\ \sf y = \frac{( \sqrt{3})^{2} + 1^{2} - 2 \times \sqrt{3} \times 1 }{3 - 1} \\ \\ \sf y = \frac{3 + 1 - 2 \sqrt{3} }{2} \\ \\ \sf y = \frac{4 - 2 \sqrt{3} }{2} \\ \\ \sf y = \frac{2(2 - \sqrt{3})}{2} \\ \\ \boxed{\bf y = 2 - \sqrt{3}}$

$\underline{\bf To \: find \: value \: of \: x^{2} + xy - y^{2} ;}$

$\bf x^{2} +xy - y^{2} \\ \\ \sf (2 + \sqrt{3} ){}^{2} + (2 + \sqrt{3})(2 - \sqrt{3}) - (2 - \sqrt{3} ) {}^{2} \\ \\ \sf 4 + 3 + 4 \sqrt{3} + 4 - 3 - 4 - 3 + 4 \sqrt{3} \\ \\ \sf 11 + 4 \sqrt{3} - 10 + 4 \sqrt{3} \\ \\ \boxed{\underline{\red{ \bf 1 + 8 \sqrt{3}}}}$

$\sf{Hence,\: the \: answer \: is \: found.}$