The roots of the equation $x + \dfrac{1}{x}$ = 3 are:

1. $\dfrac{2 \pm \sqrt{5}}{2}$

2. $\dfrac{3 \pm \sqrt{5}}{2}$

3. $\dfrac{1 \pm \sqrt{3}}{2}$

4. none of these

Given,

$x + \dfrac{1}{x}$ = 3

Multiplying whole equation by x, we get:

$\Rightarrow \dfrac{x^2 + 1}{x} = 3$

⇒ x² + 1 = 3x

⇒ x² - 3x + 1 = 0

Comparing equation x² - 3x + 1 = 0 with ax² + bx + c = 0, we get :

a = 1, b = -3 and c = 1.

By formula,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values we get :

$\Rightarrow x = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4 \times 1 \times 1}}{2 \times 1} \\[1em] = \dfrac{3 \pm \sqrt{9 - 4}}{2} \\[1em] = \dfrac{3 \pm \sqrt{5}}{2}.$

Hence, option 2 is the correct option.