If $a - \dfrac{1}{a} = 8$ and a ≠ 0; find :

(i) $a + \dfrac{1}{a}$

(ii) $a^2 - \dfrac{1}{a^2}$

(i) By formula,

$\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - \Big(a - \dfrac{1}{a}\Big)^2$ = 4

Substituting values we get :

$\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - 8^2 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - 64 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 68 \\[1em] \Rightarrow a + \dfrac{1}{a} = \sqrt{68} = \pm 2\sqrt{17}.$

Hence, $a + \dfrac{1}{a} = \pm 2\sqrt{17}.$

(ii) By formula,

$\Rightarrow a^2 - \dfrac{1}{a^2} = \Big(a + \dfrac{1}{a}\Big)\Big(a - \dfrac{1}{a}\Big)$

Substituting values we get :

$\Rightarrow a^2 - \dfrac{1}{a^2} = \pm 2\sqrt{17} \times 8 \\[1em] = \pm 16\sqrt{17}.$

Hence, $a^2 - \dfrac{1}{a^2} = \pm 16\sqrt{17}$