Question

If x = `((sqrt(3) + 1))/((sqrt(3) - 1)` and y = `((sqrt(3) - 1))/((sqrt(3) - 1)`, find the values of 

x² - y² + xy

Answer 1

x² - y² + xy

x² - y² + xy = (x + y) (x - y) + xy   ----(1)

∴ (x + y) = `((sqrt(3) + 1))/((sqrt(3) - 1)) + ((sqrt(3) - 1))/((sqrt(3) + 1)`

= `((sqrt(3) + 1)^2 + (sqrt(3) - 1)^2)/(3 - 1)`

= `(3 + 1 + 2sqrt(3) + 3 + 1 - 2sqrt(3))/(2)`

= `(8)/(2)`

= 4

(x - y) = `((sqrt(3) + 1))/((sqrt(3) - 1)) xx ((sqrt(3) - 1))/((sqrt(3) + 1)`

= `((sqrt(3) + 1)^2 - (sqrt(3) - 1)^2)/(3 - 1)`

= `(3 + 1 + 2sqrt(3) - 3 - 1 + 2sqrt(3))/(2)`

= `2sqrt(3)`

and xy = `((sqrt(3) + 1))/((sqrt(3) - 1)) xx ((sqrt(3) - 1))/((sqrt(3) + 1)`

= `(3 - 1)/(3 - 1)`

= 1
substitutingin (1), we get
x2 - y2 + xy
= (x+ y) (x - y) + xy
= `4 xx 2sqrt(3) + 1`
= `8sqrt(3) + 1`