If x = 5 - $2\sqrt{6}$, find the value of:

1. x + $\dfrac{1}{x}$

2. $x^2 +\dfrac{1}{x^2}$

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

$\dfrac{1}{x} = \dfrac{1}{5 - 2\sqrt{6}}$

$\Rightarrow \dfrac{1}{x} = \dfrac{1 \times (5 + 2\sqrt{6})}{(5 - 2\sqrt{6}) \times (5 + 2\sqrt{6})}\\[1em] = \dfrac{5 + 2\sqrt{6}}{5^2 - (2\sqrt{6})^2}\\[1em] = \dfrac{5 + 2\sqrt{6}}{25 - 24}\\[1em] = 5 + 2\sqrt{6}$

Now, the value of $x + \dfrac{1}{x}$ = $(5 + 2\sqrt{6}) + (5 - 2\sqrt{6})$

= $5 + 2\sqrt{6} + 5 - 2\sqrt{6}$

= 10

Hence, the value of $x + \dfrac{1}{x}$ = 10.

2. $x + \dfrac{1}{x}$ = 10

Squaring both sides, we get

$\Rightarrow\Big(x + \dfrac{1}{x}\Big)^2 = 10^2\\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} + 2\times x \times \dfrac{1}{x} = 100\\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} + 2 = 100\\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 100 - 2\\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 98$

Hence, the value of $x^2 +\dfrac{1}{x^2} = 98$.