Question

x= 2- root3/2+root 3 & y = 2+root 3 / 2-root 3 find x^2 - y^2

Answer 1

$x = \frac{2 - \sqrt{3} }{2 + \sqrt{3} } \\ y = \frac{2 + \sqrt{3} }{2 - \sqrt{3} } \\ x + y = \frac{2 - \sqrt{3} }{2 + \sqrt{3} } + \frac{2 + \sqrt{3} }{2 - \sqrt{3} } \\ = \frac{( {2 - \sqrt{3} })^{2} + ( {2 + \sqrt{3} })^{2} }{ {2}^{2} - ( { \sqrt{3} })^{2} } \\ = \frac{2( {2}^{2} + ( { \sqrt{3} })^{2} ) }{4 - 3} \\ = 2(4 + 3) \\ = 2 \times 7 = 14 \\ x - y = \frac{2 - \sqrt{3} }{2 + \sqrt{3} } - \frac{2 + \sqrt{3} }{2 - \sqrt{3} } \\ = \frac{( {2 - \sqrt{3} })^{2} - ( {2 + \sqrt{3} })^{2} }{4 - 3} \\ = - 2 \times 2 \times \sqrt{3} - 2 \times 2 \times \sqrt{3} \\ = - 8 \sqrt{3} \\ now \: \: \: {x}^{2} - {y}^{2} = (x - y)(x + y) \\ = ( - 8 \sqrt{3} )(14) \\ = - 112 \sqrt{3}$