If x² - 2x + 1 = 0, the value of $x^2 + \dfrac{1}{x^2}$ is :

1. 4

2. 2

3. 1

4. 3

Given,

⇒ x² - 2x + 1 = 0

⇒ x² + 1 = 2x

Dividing above equation by x, we get :

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

Squaring both sides we get :

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 2^2 \\[1em] \Rightarrow x^2 + \Big(\dfrac{1}{x}\Big)^2 + 2 \times x \times \dfrac{1}{x} = 4 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} + 2 = 4 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 2.$

Hence, Option 2 is the correct option.