If x² - 3x + 1 = 0, the value of

$x^2 + \dfrac{1}{x^2} + 1$ is :

1. 8

2. 10

3. 5

4. $\dfrac{10}{9}$

Given,

⇒ x² - 3x + 1 = 0

⇒ x² + 1 = 3x

Dividing above equation by x, we get :

$\Rightarrow \dfrac{x^2 + 1}{x} = \dfrac{3x}{x} \\[1em] \Rightarrow x + \dfrac{1}{x} = 3$

Squaring both sides we get :

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 3^2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} + 2 \times x \times \dfrac{1}{x} = 9 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} + 2 = 9 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} + 1 + 1 = 9 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} + 1 = 9 - 1 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} + 1 = 8.$

Hence, Option 1 is the correct option.