If $$2[x^2+\frac{1}{x^2}]-2[x-\frac{1}{x}]-8=0$$, then what are the two values of $$x-\frac{1}{x}$$ ?

Given : $$2[x^2+\frac{1}{x^2}]-2[x-\frac{1}{x}]-8=0$$ ------------(i)

Let $$x-\frac{1}{x}=y$$

Squaring both sides,

=> $$(x-\frac{1}{x})^2=y^2$$

=> $$x^2+\frac{1}{x^2}-2(x)(\frac{1}{x})=y^2$$

=> $$x^2+\frac{1}{x^2}=y^2+2$$

Substituting it in equation (i), we get :

=> $$2(y^2+2)-2(y)-8=0$$

=> $$2y^2-2y-4=0$$

=> $$y^2-y-2=0$$

=> $$y^2+y-2y-2=0$$

=> $$y(y+1)-2(y+1)=0$$

=> $$(y+1)(y-2)=0$$

=> $$y=-1,2$$

=> Ans - (A)