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The sum of squares of all the roots of the equation:​(x+1x)2−4=32(x−1x),x≠0\left(x + \frac{1}{x}\right)^2 - 4 = \frac{3}{2} \left(x - \frac{1}{x}\righ
Question

The sum of squares of all the roots of the equation:
​(x+1x)2−4=32(x−1x),x≠0\left(x + \frac{1}{x}\right)^2 - 4 = \frac{3}{2} \left(x - \frac{1}{x}\right), x \neq 0 is:

A.

6

B.

​6146\frac{1}{4}​​

C.

​8148\frac{1}{4}​​

D.

4

Correct option is B

​Given:(x+1x)2−4=32(x−1x), x≠0Concept Used:Algebraic identities and polynomial equationsFormula Used:(x+1x)2=x2+1x2+2Solution:x2+1x2−2=32(x−1x)2x2+2x2−4=3x−3x2x4−4x2+2=3x3−3x2x4−3x3−4x2+3x+2=0(x−1)(x+1)(2x2−3x−2)=0x=1, −1, −12, 2Sum of squares=1+1+14+4=254=614\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​​

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