If $a -\dfrac{1}{a}= 4$, find: $a^2 +\dfrac{1}{a^2}$

Using the formula,

[∵ (x - y)² = x² - 2xy + y²]

So,

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

Putting the value $a - \dfrac{1}{a} = 4$,we get

$4^2 = a^2 - 2 + \dfrac{1}{a^2}\\[1em] ⇒ 16 = a^2 - 2 + \dfrac{1}{a^2}\\[1em] ⇒ a^2 + \dfrac{1}{a^2} = 16 + 2\\[1em] ⇒ a^2 + \dfrac{1}{a^2} = 18$

Hence, the value of $a^2 + \dfrac{1}{a^2}$ is 18.