Solve the following equation:

$\dfrac{x}{x + 1} + \dfrac{x + 1}{x} = 2\dfrac{1}{6}$

By taking L.C.M. we get,

$\Rightarrow \dfrac{x^2 + (x + 1)^2}{x(x + 1)} = \dfrac{13}{6} \\[1em] \Rightarrow \dfrac{x^2 + x^2 + 1 + 2x}{x^2 + x} = \dfrac{13}{6} \\[1em] \Rightarrow 6(2x^2 + 2x + 1) = 13(x^2 + x)$

⇒ 12x² + 12x + 6 = 13x² + 13x

⇒ 13x² - 12x² + 13x - 12x - 6 = 0

⇒ x² + x - 6 = 0

⇒ x² + 3x - 2x - 6 = 0

⇒ x(x + 3) - 2(x + 3) = 0

⇒ (x + 3)(x - 2) = 0

⇒ x + 3 = 0 or x - 2 = 0

⇒ x = -3 or x = 2.

Hence, roots of the given equations are -3, 2.