If $$x = 4 + \sqrt{15}$$, then what is the value of $$[x^2 + (\frac{1}{x^2})]$$ ?

Given : $$x = 4 + \sqrt{15}$$ -------------(i)

=> $$\frac{1}{x} = \frac{1}{4 + \sqrt{15}}$$

=> $$\frac{1}{x} = \frac{1}{4 + \sqrt{15}}\times(\frac{4-\sqrt{15}}{4-\sqrt{15}})$$

=> $$\frac{1}{x}=\frac{4-\sqrt{15}}{(16-15)}=4-\sqrt{15}$$ -------------(ii)

To find : $$[x^2 + (\frac{1}{x^2})]$$ 

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

Substituting values from equations (i) and (ii), we get :

= $$[(4+\sqrt{15})+(4-\sqrt{15})]^2-2$$

= $$(8)^2-2=64-2=62$$

=> Ans - (A)