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

Given,  $$x=3+2\sqrt{2}$$

$$\Rightarrow$$  $$\frac{1}{x}=\frac{1}{3+2\sqrt{2}}\times\frac{3-2\sqrt{2}}{3-2\sqrt{2}}$$

$$\Rightarrow$$  $$\frac{1}{x}=\frac{3-2\sqrt{2}}{9-8}$$

$$\Rightarrow$$  $$\frac{1}{x}=3-2\sqrt{2}$$

$$\left(x+\frac{1}{x}\right)^2=\left(3+2\sqrt{2}+3-2\sqrt{2}\right)^2$$

$$\Rightarrow$$  $$x^2+\frac{1}{x^2}+2=6^2$$

$$\Rightarrow$$  $$x^2+\frac{1}{x^2}+2=36$$

$$\Rightarrow$$  $$x^2+\frac{1}{x^2}=34$$

Hence, the correct answer is Option D