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

Expression : $$x = 3 + 2\sqrt{2}$$

=> $$\frac{1}{x} = \frac{1}{3 + 2\sqrt{2}}$$

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

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

$$\therefore$$ $$x + \frac{1}{x} = 3 + 2\sqrt{2} + 3 - 2\sqrt{2} = 6$$

Squaring both sides, we get :

=> $$(x + \frac{1}{x})^2 = 6^2$$

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

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