If x = 3 + $2\sqrt{2}$, then $x + \dfrac{1}{x}$ equals to :

1. $4\sqrt{2}$

2. $6\sqrt{2}$

3. 6

4. 4

Given,

⇒ x = 3 + $2\sqrt{2}$

$\Rightarrow \dfrac{1}{x} = \dfrac{1}{(3 + 2\sqrt{2})}$

Rationalizing,

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

Adding x and $\dfrac{1}{x}$ we get

$\Rightarrow x + \dfrac{1}{x} = 3 + 2\sqrt{2} + 3 - 2\sqrt{2} \\[1em] \Rightarrow 6$

Hence, Option 3 is the correct option.