If $a^2 - \dfrac{1}{a^2} = 7$, then $a^4 + \dfrac{1}{a^4}$ =

1. 49

2. 51

3. 53

4. 60

Given,

$a^2 - \dfrac{1}{a^2} = 7$

Upon squaring both sides,

$\Rightarrow \Big(a^2 - \dfrac{1}{a^2}\Big)^2 = a^4 + \dfrac{1}{a^4} - 2 \\[1em] \Rightarrow (7)^2 = \Big(a^4 + \dfrac{1}{a^4}\Big) - 2 \\[1em] \Rightarrow 49 = \Big(a^4 + \dfrac{1}{a^4}\Big) - 2 \\[1em] \Rightarrow \Big(a^4 + \dfrac{1}{a^4}\Big) = 49 + 2 \\[1em] \Rightarrow \Big(a^4 + \dfrac{1}{a^4}\Big) = 51 \\[1em]$

Hence, Option 2 is the correct option.