If x = 5 + $2\sqrt{6}$, then $x^2 + \dfrac{1}{x^2}$ =

1. 98

2. 142

3. 49

4. 138

Given,

x = $(5 + 2\sqrt{6})$

$\therefore \dfrac{1}{x} = \dfrac{1}{(5 + 2\sqrt{6})}$

Rationalizing,

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

By formula,

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

Substituting values we get :

$\Rightarrow x^2 + \dfrac{1}{x^2} = (5 + 2\sqrt{6} + 5 - 2\sqrt{6})^2 - 2 \\[1em] = 10^2 - 2 \\[1em] = 100 - 2 \\[1em] = 98.$

Hence, Option 1 is the correct option.