Question

If x=√3+√2\√3-√2 and y=√3-√2\√3+√2, find the value of x^2-y^2+xy, if √6=2.4

Answer 1

here's your solution..

Answer 2

$x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ \\ x + y = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } + \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ x + y = \frac{( { \sqrt{3} + \sqrt{2} })^{2} + ( { \sqrt{3} - \sqrt{2} })^{2} }{3 - 2} \\ \\ x + y = 2(3 + 2) = 10 \\ \\ \\ x - y = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } - \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ x - y = \frac{( { \sqrt{3} + \sqrt{2} })^{2} - ( { \sqrt{3} - \sqrt{2} })^{2} }{3 - 2} \\ \\ x - y = 4 \sqrt{3} \sqrt{2} = 4 \sqrt{6} \\ \\ \\ {x}^{2} - {y}^{2} = (x + y)(x - y) \\ = 10 \times 4 \sqrt{6} \\ = 40 \sqrt{6} \\ = 40 \times 2.4 \\ = 96 \\ \\ xy = 1 \\ \\ \\ {x}^{2} - {y}^{2} + xy = 96 + 1 \\ = 97$