If x² - 2x + 1 = 0; the value of $x^4 + \dfrac{1}{x^4}$ is

1. 9

2. 7

3. 3

4. 2

Given,

⇒ x² - 2x + 1 = 0

⇒ x² + 1 = 2x

Dividing above equation by x, we get :

⇒ $\dfrac{x^2 + 1}{x} = \dfrac{2x}{x}$

⇒ $x + \dfrac{1}{x}$ = 2.

Squaring both sides we get :

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

Squaring both sides we get :

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

Hence, Option 4 is the correct option.