If $$\frac{{3\left( {{x^2} + 1} \right) - 7x}}{{3x}} = 6,$$    x ≠ 0 the value $$\sqrt x + \frac{1}{{\sqrt x }}$$  is:

Solution (By Examveda Team)

$$\eqalign{ & \frac{{3\left( {{x^2} + 1} \right) - 7x}}{{3x}} = 6 \cr & 3{x^2} + 3 - 7x = 18x \cr & 3{x^2} + 3 = 25x \cr & 3\left( {{x^2} + 1} \right) = 25x \cr & {\text{divided by }}'x' \cr & 3\left( {x + \frac{1}{x}} \right) = 25 \cr & x + \frac{1}{x} = 25 \cr & x + \frac{1}{x} + 2 = \frac{{25}}{3} \cr & {\left( {x + \frac{1}{x}} \right)^2} = \frac{{25}}{3} + 2 \cr & {\left( {\sqrt x + \frac{1}{{\sqrt x }}} \right)^2} = \frac{{31}}{3} \cr & \sqrt x + \frac{1}{{\sqrt x }} = \sqrt {\frac{{31}}{3}} \cr} $$