The sum of squares of all the roots of the equation:
​$$\left(x + \frac{1}{x}\right)^2 - 4 = \frac{3}{2} \left(x - \frac{1}{x}\right), x \neq 0$$ is:

Correct option is B

​$$\textbf{Given:} \\\left(x+\frac{1}{x}\right)^2 - 4 = \frac{3}{2}\left(x-\frac{1}{x}\right),\; x \neq 0 \\\textbf{Concept Used:} \\\text{Algebraic identities and polynomial equations} \\\textbf{Formula Used:} \\\left(x+\frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2 \\\textbf{Solution:} \\x^2 + \frac{1}{x^2} - 2 = \frac{3}{2}\left(x-\frac{1}{x}\right) \\2x^2 + \frac{2}{x^2} - 4 = 3x - \frac{3}{x} \\2x^4 - 4x^2 + 2 = 3x^3 - 3x \\2x^4 - 3x^3 - 4x^2 + 3x + 2 = 0 \\(x-1)(x+1)(2x^2-3x-2)=0 \\x = 1,\,-1,\,-\frac{1}{2},\,2 \\\text{Sum of squares} = 1 + 1 + \frac{1}{4} + 4 = \frac{25}{4} =6\frac 14$$​​