If x = $5 - 2\sqrt{6}$, find the value of $x^2 + \dfrac{1}{x^2}$.

Given x = $5 - 2\sqrt{6}$

$\therefore \dfrac{1}{x} = \dfrac{1}{5 - 2\sqrt{6}} = \dfrac{1}{5 - 2\sqrt{6}} × \dfrac{5 + 2\sqrt{6}}{5 + 2\sqrt{6}} \\[1.5em] \Rightarrow \dfrac{5 + 2\sqrt{6}}{(5)^2 - (2\sqrt{6})^2} \\[1.5em] \Rightarrow \dfrac{5 + 2\sqrt{6}}{25 - 24} = \dfrac{5 + 2\sqrt{6}}{1} \\[1.5em] \Rightarrow\dfrac{1}{x} = 5+2\sqrt6 \\[1.5em] \therefore (x + \dfrac{1}{x}) = (5 - 2\sqrt6) + (5 + 2\sqrt6) = 10 \qquad \text{….(i)}$

We know that ${\Big(x + \dfrac{1}{x}\Big)}^2 = x^2 + \dfrac{1}{x^2} + 2$

$\Rightarrow x^2 + \dfrac{1}{x^2} = {\Big(x + \dfrac{1}{x}\Big)}^2 -2 \\[1.5em] \Rightarrow x^2 + \dfrac{1}{x^2} = 10^2 - 2 ….. \text{using(i)} \\[1.5em] \Rightarrow x^2 + \dfrac{1}{x^2} = 100 - 2 \\[1.5em] \bold{x^2 + \dfrac{1}{x^2} = 98}$