If $x^2 + \dfrac{1}{x^2} = 9$, the value of $x^4 + \dfrac{1}{x^4} + 5$ is :

1. 78

2. 86

3. 84

4. 81

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

Adding 5 on both sides,

$x^4 + \dfrac{1}{x^4} + 5 = \Big(x^2 + \dfrac{1}{x^2}\Big)^2 - 2 + 5\\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} + 5 = 9^2 + 3 \\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} + 5 = 84.$

Hence, Option 3 is the correct option.