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

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

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

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

(iv) $x^2+\dfrac{1}{x^2}$

(i) Given,

$x = \dfrac{1}{x - 5}\\[1em] ⇒ x - 5 = \dfrac{1}{x}\\[1em] ⇒ x - \dfrac{1}{x} = 5$

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

(ii) From equation (i),

$x - \dfrac{1}{x} = 5$

Squaring both sides, we get:

$⇒ \Big(x - \dfrac{1}{x}\Big)^2 = 5^2\\[1em] ⇒ x^2 + \dfrac{1}{x^2} - 2 \times x \times \dfrac{1}{x} = 25\\[1em] ⇒ x^2 + \dfrac{1}{x^2} - 2 = 25\\[1em] ⇒ x^2 + \dfrac{1}{x^2} = 25 + 2\\[1em] ⇒ x^2 + \dfrac{1}{x^2} = 27 ………. (A)$

Adding 2 on both sides, we get:

$⇒ x^2 + \dfrac{1}{x^2} + 2 = 27 + 2\\[1em] ⇒ x^2 + \dfrac{1}{x^2} + 2 \times x \times \dfrac{1}{x} = 29\\[1em] ⇒ \Big(x + \dfrac{1}{x}\Big)^2 = 29\\[1em] ⇒ x + \dfrac{1}{x} = \sqrt{29}\\[1em] ⇒ x + \dfrac{1}{x} = \sqrt{29} \text{ or } -\sqrt{29}$

Hence, $x + \dfrac{1}{x} = \sqrt{29} \text{ or } -\sqrt{29}$.

(iii) We can write, $x^2-\dfrac{1}{x^2} = \Big(x - \dfrac{1}{x}\Big) \Big(x + \dfrac{1}{x}\Big)$

From (i) and (ii),

$x^2-\dfrac{1}{x^2} = 5 \times \sqrt{29} \text{ or } 5 \times (-\sqrt{29})$

= $5\sqrt{29} \text{ or } -5\sqrt{29}$

Hence, $x^2-\dfrac{1}{x^2} = 5\sqrt{29} \text{ or } -5\sqrt{29}$.

(iv) $x^2+\dfrac{1}{x^2}$

From equation (A), $x^2 + \dfrac{1}{x^2} = 27$.

Hence, $x^2 + \dfrac{1}{x^2} = 27$.