If $$\sqrt x + \frac{1}{\sqrt x} = 2\sqrt2,  then  x^2 + \frac{1}{x^2}$$ is equal to:

Given that,

$$\sqrt x + \frac{1}{\sqrt x} = 2\sqrt2$$

Taking the square of the both side,

$$\Rightarrow (\sqrt x + \frac{1}{\sqrt x})^2 = (2\sqrt2)^2$$

$$\Rightarrow (\sqrt x)^2 +( \frac{1}{\sqrt x})^2+2 = (2\sqrt2)^2$$

$$\Rightarrow x +\dfrac{1}{x}+2 = 8$$

$$\Rightarrow x +\dfrac{1}{x} = 6$$-------------------(i)

Again taking square of the equation (i)

$$\Rightarrow (x +\dfrac{1}{x})^2 = 6^2$$

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

$$\Rightarrow x^2+\dfrac{1}{x^2}=36-2=34$$

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