Distinct real roots of quadratic in (x + 1/x)

Let $y = x + \frac{1}{x}$. The equation becomes: $y^2 - 3y + 2 = 0 \implies (y - 1)(y - 2) = 0$ So $y = 1$ or $y = 2$.

• Case 1: $x + \frac{1}{x} = 1 \implies x^2 - x + 1 = 0$. Discriminant $D = 1 - 4 = -3 < 0$, so no real roots.

• Case 2: $x + \frac{1}{x} = 2 \implies x^2 - 2x + 1 = 0 \implies (x - 1)^2 = 0$. This gives $x = 1$.

Thus, there is only 1 distinct real root.