Question

$\huge \bf x = 3+2\sqrt{2}$

$\huge \bf \sqrt{x}-\dfrac{1}{\sqrt{x}}$ = ¿?

Answer 1

$\displaystyle\bf x = 3+2\sqrt{2} \\ \displaystyle\sf \sqrt{x}-\dfrac{1}{\sqrt{x}} \\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ \\ \begin{gathered} \begin{gathered}\displaystyle\sf :\implies \dfrac{1}{x} = \dfrac{1}{3+2\sqrt{2}}\\\end{gathered}\end{gathered} \\ \displaystyle\sf :\implies \dfrac{1}{3+2\sqrt{2}}\times \dfrac{3-2\sqrt{2}}{3-2\sqrt{2}} \\ \displaystyle\sf :\implies \dfrac{3-2\sqrt{2}}{3^2-(2\sqrt{2}^2)} \\ \displaystyle\sf :\implies \dfrac{3-2\sqrt{2}}{9-8} \\ \displaystyle\sf :\implies \dfrac{1}{x} = 3-2\sqrt{2} \\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ \\ \displaystyle\sf :\implies x+\dfrac{1}{x} = (3+2\sqrt{2}) + (3-2\sqrt{2}) \\ \displaystyle\sf :\implies x+\dfrac{1}{x} = 3+2\sqrt{2} + 3 - 2\sqrt{2} \\ \displaystyle\sf :\implies x+\dfrac{1}{x} \\ \\ \sf \small{So \: here \: we \: know \: that \: we \: may \: split \: the \: number \: 6 \: into \: 4+2 } \\ \\ \displaystyle\sf :\implies x+\dfrac{1}{x} = 4 \\ \displaystyle\sf :\implies x+\dfrac{1}{x}-2 = 4 \\ \displaystyle\sf :\implies \bigg\lgroup \sqrt{x}-\dfrac{1}{\sqrt{x}}\bigg\rgroup^2 = 4 \\ \displaystyle\sf :\implies \sqrt{x}-\dfrac{1}{\sqrt{x}} = \sqrt{4} \\ \displaystyle\sf :\implies \sqrt{x}-\dfrac{1}{\sqrt{x}} = \pm 2 \\ \\ \displaystyle\therefore\:{\textbf{ \blue{The value of \sqrt{ \sf x}-\dfrac{\sf 1}{\sqrt{\sf x}} is \textbf{ \pm 2 }}}}$

Answer 2

The value is to be 2

The square root x-1/root x is 2

Refer the attachment