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

Given that,

$$\sqrt x + \frac{1}{\sqrt x} = 2\sqrt2  ---------------(i)$$

Taking square of both side,

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

$$\Rightarrow (\sqrt x )^2+ (\frac{1}{\sqrt x})^2 +2 \times \sqrt x \times \frac{1}{\sqrt x}= 8$$

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

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

Now, again taking square of both sides.

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

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

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

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