If 3 ( x^2 + 1 ) - 7x3x = 6, x ≠ 0 the value x + 1 x is:

Answer: Option B

$$\begin{array}{rl}& \frac{3\left(x^2+1\right)-7x}{3x}=6\\ & 3x^2+3-7x=18x\\ & 3x^2+3=25x\\ & 3\left(x^2+1\right)=25x\\ & {\text{divided by }}^{\prime }{x}^{\prime }\\ & 3\left(x+\frac{1}{x}\right)=25\\ & x+\frac{1}{x}=25\\ & x+\frac{1}{x}+2=\frac{25}{3}\\ & {\left(x+\frac{1}{x}\right)}^2=\frac{25}{3}+2\\ & {\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)}^2=\frac{31}{3}\\ & \sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt{\frac{31}{3}}\end{array}$$