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4.

$$\begin{gathered} x=\sqrt{2-\sqrt{3}} \\ \mathit{L}\mathit{H}\mathit{S}\mathbf{:} \\ \mathbf{=}x+\frac{1}{x} \\ =\frac{x^2+1}{x} \\ =\frac{2-\sqrt{3}+1}{\sqrt{2-\sqrt{3}}} \\ k=\frac{3-\sqrt{3}}{\sqrt{2-\sqrt{3}}} \\ We need, k^2 \\ {k}^2=\frac{{\left(3-\sqrt{3}\right)}^2}{2-\sqrt{3}} \\ {k}^2=\frac{9+3-6\sqrt{3}}{2-\sqrt{3}}=\frac{6(2-\sqrt{3})}{2-\sqrt{3}} \\ {k}^2=6 \end{gathered}$$

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