If $\Big(x^2 + \dfrac{1}{x^2}\Big) = 7$, find the values of :

(i) $\Big(x + \dfrac{1}{x}\Big)$

(ii) $\Big(x - \dfrac{1}{x}\Big)$

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

(i) Given,

$\Big(x^2 + \dfrac{1}{x^2}\Big) = 7$

Using identity,

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} + 2 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 7 + 2 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 9 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big) = \pm \sqrt{9} \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big) = \pm 3.$

Hence, $\Big(x + \dfrac{1}{x}\Big) = \pm 3$.

(ii) Given,

$\Big(x^2 + \dfrac{1}{x^2}\Big) = 7$

Using identity,

$\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 7 - 2 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 5 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big) = \pm \sqrt{5} \\[1em]$

Hence, $\Big(x - \dfrac{1}{x}\Big) = \pm \sqrt{5}$.

(iii) Given,

$\Big(x^2 + \dfrac{1}{x^2}\Big) = 7$

From part (i) and (ii),

$\Big(x + \dfrac{1}{x}\Big) = \pm 3 \text{ and } \Big(x - \dfrac{1}{x}\Big) = \pm \sqrt{5}$

Using identity,

$\Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = \Big(x + \dfrac{1}{x}\Big)\Big(x - \dfrac{1}{x}\Big) \\[1em] \Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = (\pm 3) \times (\pm \sqrt{5}) \\[1em] \Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = \pm 3\sqrt{5} \\[1em] \Rightarrow 2\Big(x^2 - \dfrac{1}{x^2}\Big) = 2 \times \pm 3\sqrt{5} \\[1em] \Rightarrow \Big(2x^2 - \dfrac{2}{x^2}\Big) = 2 \times (\pm 3\sqrt{5}) \\[1em] \Rightarrow \Big(2x^2 - \dfrac{2}{x^2}\Big) = \pm 6\sqrt{5}$

Hence, $\Big(2x^2 - \dfrac{2}{x^2}\Big) = \pm 6\sqrt{5}$.