Find the value of $$(x+y)^2$$​, if $$x=\frac{\sqrt2+\sqrt3}{\sqrt2-\sqrt3}$$​ and $$y=\frac{\sqrt2-\sqrt3}{\sqrt2+\sqrt3}$$​.

Correct option is B

Given:

​$$x = \dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}}, \quad y = \dfrac{\sqrt{2} - \sqrt{3}}{\sqrt{2} + \sqrt{3}}$$​​

Formula Used:

​$$x^2 - y^2 = (x - y)(x + y)$$​​

​$$(x + y)^2 = x^2 + y^2 + 2xy$$​​

Solution:

​Now,

​$$(x + y)^2 =( (-5 - 2\sqrt{6}) + (-5 + 2\sqrt{6}))^2 = (-10)^2 = 100$$​​