Show that :

$x^2 + \dfrac{1}{x^2} = 34, \text{ if x } = 3 + 2\sqrt{2}$.

Given,

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

$\therefore \dfrac{1}{x} = \dfrac{1}{3 + 2\sqrt{2}}$

Rationalizing,

$\Rightarrow \dfrac{1}{3 + 2\sqrt{2}} \times \dfrac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} \\[1em] \Rightarrow \dfrac{3 - 2\sqrt{2}}{3^2 - (2\sqrt{2})^2} \\[1em] \Rightarrow \dfrac{3 - 2\sqrt{2}}{9 - 8} \\[1em] \Rightarrow 3 - 2\sqrt{2} \\[1em] \therefore \dfrac{1}{x} = 3 - 2\sqrt{2}$

Substituting value of x and $\dfrac{1}{x} \text{ in } x^2 + \dfrac{1}{x^2}$, we get :

$\Rightarrow x^2 + \dfrac{1}{x^2} = (3 + 2\sqrt{2})^2 + (3 - 2\sqrt{2})^2 \\[1em] \Rightarrow 3^2 + (2\sqrt{2})^2 + 2 \times 3 \times 2\sqrt{2} + 3^2 + (2\sqrt{2})^2 - 2 \times 3 \times 2\sqrt{2} \\[1em] \Rightarrow 9 + 8 + 12\sqrt{2} + 9 + 8 - 12\sqrt{2} \\[1em] \Rightarrow 34.$

Hence, proved that $x^2 + \dfrac{1}{x^2} = 34.$