If x+y=4, $$x^{2}+y^{2}$$= 14 and x > y, Then the correct value of x and y is:

Given : $$x+y=4$$ and $$x^2+y^2=14$$ ----------------(i)

Squaring both sides, we get :

=> $$(x+y)^2=(4)^2$$

=> $$x^2+y^2+2xy=16$$

=> $$14+2xy=16$$

=> $$2xy=16-14=2$$

=> $$xy=1$$

=> $$y=\frac{1}{x}$$

Substituting it in equation (i), => $$x+\frac{1}{x}=4$$

=> $$x^2-4x+1=0$$

=> $$x=\frac{4\pm\sqrt{(-4)^2-4(1)(1)}}{2}$$

=> $$x=\frac{4\pm\sqrt{12}}{2}$$

=> $$x=\frac{4\pm2\sqrt{3}}{2}$$

=> $$x=2\pm\sqrt3$$

$$\because$$ $$x>y$$ => $$x=2+\sqrt3$$ and $$y=2-\sqrt3$$

=> Ans - (C)