If $x + \dfrac{1}{x} = 2$, the value of $x^2 + \dfrac{1}{x^2} + 5$ is :

1. -1

2. 2

3. 9

4. 7

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

Given,

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

$\therefore x^2 + \dfrac{1}{x^2} = 2^2 - 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 4 - 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 2$

Adding 5 on both sides,

$x^2 + \dfrac{1}{x^2} + 5 = 2 + 5 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} + 5 = 7$

Hence, Option 4 is the correct option.