The sum of squares of all the roots of the equation:

\((x + \frac{1}{x})^2 - 4 = \frac{3}{2}(x - \frac{1}{x}), x \neq 0\) is:

Given:

The equation is (x + 1/x)² - 4 = 3/2(x - 1/x), x ≠ 0

Formula used:

(a + b)² = a² + b² + 2ab

(a - b)² = a² + b² - 2ab

We know that (x + 1/x)² = x² + 1/x² + 2

And (x - 1/x)² = x² + 1/x² - 2

Also, (x + 1/x)² - (x - 1/x)² = 4

Calculation:

Let y = x - 1/x.

Then (x + 1/x)² = (x - 1/x)² + 4 = y² + 4.

Substitute these into the given equation:

(y² + 4) - 4 = 3/2 y

⇒ y² = 3/2 y

⇒ 2y² = 3y

⇒ 2y² - 3y = 0

⇒ y(2y - 3) = 0

This gives two possible values for y:

1. y = 0

2. 2y - 3 = 0 ⇒ y = 3/2

Case 1: x - 1/x = 0

⇒ (x² - 1)/x = 0

⇒ x² - 1 = 0

⇒ x² = 1

⇒ x = ± 1

So, the roots are 1 and -1.

Case 2: x - 1/x = 3/2

⇒ (x² - 1)/x = 3/2

⇒ 2(x² - 1) = 3x

⇒ 2x² - 2 = 3x

⇒ 2x² - 3x - 2 = 0

⇒ 2x² - 4x + x - 2 = 0

⇒ 2x(x - 2) + 1(x - 2) = 0

⇒ (2x + 1)(x - 2) = 0

⇒ x - 2 = 0 ⇒ x = 2

⇒ 2x + 1 = 0 ⇒ x = -1/2

The roots of the equation are 1, -1, 2, and -1/2.

Now, calculate the sum of the squares of all the roots:

Sum of squares = (1)² + (-1)² + (2)² + (-1/2)²

= 1 + 1 + 4 + 1/4

= 6 + 1/4

= (24 + 1)/4

= 25/4 = \(6\frac{1}{4}\)

∴ The correct answer is option 3.