Solve the following equation:

$\dfrac{1}{x + 2} + \dfrac{1}{x} = \dfrac{3}{4}.$

Given, $\dfrac{1}{x + 2} + \dfrac{1}{x} = \dfrac{3}{4}$.

By taking L.C.M. we get,

$\Rightarrow \dfrac{x + x + 2}{x(x + 2)} = \dfrac{3}{4}$

By cross multiplication we get,

⇒ 4(2x + 2) = 3x(x + 2)

⇒ 8x + 8 = 3x² + 6x

⇒ 3x² + 6x - 8x - 8 = 0

⇒ 3x² - 2x - 8 = 0

⇒ 3x² - 6x + 4x - 8 = 0

⇒ 3x(x - 2) + 4(x - 2) = 0

⇒ (3x + 4)(x - 2) = 0

⇒ 3x + 4 = 0 or x - 2 = 0

⇒ 3x = -4 or x = 2

⇒ x = $-\dfrac{4}{3}$ or x = 2.

Hence, roots of the given equations are 2, $-\dfrac{4}{3}$