If $$x = 7 - 4 \sqrt{3}$$, then the value of $$x^2 + \frac{1}{x^2}$$ is

Given : $$x = 7 - 4 \sqrt{3}$$ ------------(i)

=> $$\frac{1}{x}=\frac{1}{7-4\sqrt3}$$

Rationalizing the denominator, we get

=> $$\frac{1}{x}=\frac{1}{7-4\sqrt3}\times(\frac{7+4\sqrt3}{7+4\sqrt3})$$

=> $$\frac{1}{x}=\frac{(7+4\sqrt3)}{(7)^2-(4\sqrt3)^2}$$

=> $$\frac{1}{x}=\frac{7+4\sqrt3}{49-48}=7+4\sqrt3$$ ------------(ii)

Adding equations (i) and (ii), we get : $$(x+\frac{1}{x})=14$$

Squaring both sides, $$(x+\frac{1}{x})^2=(14)^2$$

=> $$x^2+\frac{1}{x^2}+2(x)(\frac{1}{x})=196$$

=> $$x^2+\frac{1}{x^2}=196-2=194$$

=> Ans - (D)