If a² - 5a - 1 = 0 and a ≠ 0, find :

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

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

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

(i) Given,

⇒ a² - 5a - 1 = 0

⇒ a² - 1 = 5a

Dividing above equation by a, we get :

$\Rightarrow \dfrac{a^2 - 1}{a} = \dfrac{5a}{a}$

$\Rightarrow a - \dfrac{1}{a} = 5$

Hence, $a - \dfrac{1}{a} = 5$.

(ii) By formula,

$\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - \Big(a - \dfrac{1}{a}\Big)^2 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - 5^2 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - 25 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 29 \\[1em] \Rightarrow a + \dfrac{1}{a} = \pm \sqrt{29}.$

Hence, $a + \dfrac{1}{a} = \pm \sqrt{29}$.

(iii) By formula,

$\Rightarrow \Big(a^2 - \dfrac{1}{a^2}\Big) = \Big(a + \dfrac{1}{a}\Big)\Big(a - \dfrac{1}{a}\Big) \\[1em] = \pm \sqrt{29} \times 5 \\[1em] = \pm 5\sqrt{29}.$

Hence, $a^2 - \dfrac{1}{a^2} = \pm 5\sqrt{29}$.