Solve the following equation by factorization:

$2\Big(\dfrac{x}{x + 1}\Big)^2 - 5\Big(\dfrac{x}{x + 1}\Big) + 2 = 0$, x ≠ -1

Let us consider y = $\dfrac{x}{x + 1}$.

Substituting y = $\dfrac{x}{x + 1}$ in equation $2\Big(\dfrac{x}{x + 1}\Big)^2 - 5\Big(\dfrac{x}{x + 1}\Big) + 2 = 0$, we get :

⇒ 2y² - 5y + 2 = 0

⇒ 2y² - 4y - y + 2 = 0

⇒ 2y(y - 2) - 1(y - 2) = 0

⇒ (2y - 1)(y - 2) = 0

⇒ (2y - 1) = 0 or (y - 2) = 0      [Using Zero-product rule]

⇒ 2y = 1 or y = 2

⇒ y = $\dfrac{1}{2}$ or y = 2.

Now we have,

Case 1 : y = $\dfrac{1}{2}$

$\Rightarrow y = \dfrac{x}{x + 1} \\[1em] \Rightarrow \dfrac{1}{2} = \dfrac{x}{x + 1} \\[1em] \Rightarrow x + 1 = 2x \\[1em] \Rightarrow 2x - x = 1 \\[1em] \Rightarrow x = 1.$

Case 2 : y = 2

⇒ y = $\dfrac{x}{x + 1}$

⇒ 2 = $\dfrac{x}{x + 1}$

⇒ 2(x + 1) = x

⇒ 2x + 2 = x

⇒ 2x - x = -2

⇒ x = -2.

Hence, x = {-2, 1}.