If a = $\dfrac{1}{a - 5}$, find

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

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

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

Given,

$\Rightarrow a = \dfrac{1}{a - 5} \\[1em] \therefore a(a - 5) = 1 \\[1em] \Rightarrow a^2 - 5a = 1 \\[1em] \Rightarrow a^2 - 5a - 1 = 0$

Now,

(i) Dividing above equation by a we get,

$\Rightarrow a - 5 - \dfrac{1}{a} = 0 \\[1em] \Rightarrow a - \dfrac{1}{a} = 5.$.

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

(ii) We know that,

$\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} + 2 \space …..(i) \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} - 2 \space …..(ii) \\[1em]$

Subtracting eq. (ii) from (i) we get,

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

Substituting values we get,

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

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

(iii) We know that,

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

Substituting values we get,

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

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