If $$\sqrt x + \frac{1}{\sqrt x} = \sqrt6$$, then $$x^2 + \frac{1}{x^2}$$ is equal to:

Given that,

$$\sqrt x + \frac{1}{\sqrt x} = \sqrt6 -------(i)$$

taking square of both side of equation (i)

$$(\sqrt x + \frac{1}{\sqrt x})^2 =( \sqrt6)^2 $$

$$(\sqrt x)^2 + (\frac{1}{\sqrt x})^2 +\dfrac{2\sqrt x}{\sqrt x} =( \sqrt6)^2 $$

$$x + \frac{1}{x} +2 =6 $$

$$x + \frac{1}{x}=4 -----------(ii)$$

Now squaring of equation(ii)

$$x^2 + \frac{1}{ x}^2 +\dfrac{2 x}{ x} =( 4)^2 $$

$$x^2 + \frac{1}{ x^2} +2 =16$$

$$x^2 + \frac{1}{ x^2}  =14$$