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

Given,

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

So,

$x + \dfrac{1}{x} = 5 - 2\sqrt{6} + 5 + 2\sqrt{6} = 10.$

We know that,

$\Rightarrow \Big(\sqrt{x} + \dfrac{1}{\sqrt{x}}\Big)^2 = x + \dfrac{1}{x} + 2 \times \sqrt{x} \times \dfrac{1}{\sqrt{x}} \\[1em] \Rightarrow \Big(\sqrt{x} + \dfrac{1}{\sqrt{x}}\Big)^2 = x + \dfrac{1}{x} + 2 \\[1em] \Rightarrow \sqrt{x} + \dfrac{1}{\sqrt{x}} = \sqrt{x + \dfrac{1}{x} + 2} = \sqrt{10 + 2} = \sqrt{12} = \pm 2\sqrt{3}.$

Hence, $\sqrt{x} + \dfrac{1}{\sqrt{x}} = \pm 2\sqrt{3}.$