$\begin{align}  & {{\left( \frac{2-x}{1+3x} \right)}^{\tfrac{1}{2}}}={{(2-x)}^{\tfrac{1}{2}}}{{(1+3x)}^{-\tfrac{1}{2}}} \\ & {{\left( \frac{2-x}{1+3x} \right)}^{\tfrac{1}{2}}}=\left[ {{2}^{\tfrac{1}{2}}}{{(1-\tfrac{x}{2})}^{\tfrac{1}{2}}} \right]{{(1+3x)}^{-\tfrac{1}{2}}} \\ & {{2}^{\tfrac{1}{2}}}{{(1-\tfrac{x}{2})}^{\tfrac{1}{2}}}=\sqrt{2}\left( 1+\tfrac{1}{2}(-\tfrac{x}{2})+\frac{(\tfrac{1}{2})(-\tfrac{1}{2})}{2}{{(-\tfrac{x}{2})}^{2}}+--- \right) \\ & {{2}^{\tfrac{1}{2}}}{{(1-\tfrac{x}{2})}^{\tfrac{1}{2}}}=\sqrt{2}\left( 1-\frac{x}{4}-\frac{{{x}^{2}}}{32}+--- \right) \\ & {{(1+3x)}^{-\tfrac{1}{2}}}=1-\frac{1}{2}(3x)+\frac{(-\tfrac{1}{2})(-\tfrac{3}{2})}{2!}{{(3x)}^{2}}+--- \\ & {{(1+3x)}^{-\tfrac{1}{2}}}=1-\frac{3x}{2}+\frac{27{{x}^{2}}}{8}+--- \\ & {{2}^{\tfrac{1}{2}}}{{(1-\tfrac{x}{2})}^{\tfrac{1}{2}}}{{(1+3x)}^{-\tfrac{1}{2}}}=\sqrt{2}\left( 1-\frac{x}{4}-\frac{{{x}^{2}}}{32}+--- \right)\left( 1-\frac{3x}{2}+\frac{27{{x}^{2}}}{8} \right) \\ & {{\left( \frac{2-x}{1+3x} \right)}^{\tfrac{1}{2}}}={{2}^{\tfrac{1}{2}}}{{(1-\tfrac{x}{2})}^{\tfrac{1}{2}}}{{(1+3x)}^{-\tfrac{1}{2}}}=\sqrt{2}\left( 1-\frac{3x}{2}+\frac{27{{x}^{2}}}{8}-\frac{x}{4}-\frac{3{{x}^{2}}}{8}-\frac{{{x}^{2}}}{32} \right) \\ & {{\left( \frac{2-x}{1+3x} \right)}^{\tfrac{1}{2}}}={{2}^{\tfrac{1}{2}}}{{(1-\tfrac{x}{2})}^{\tfrac{1}{2}}}{{(1+3x)}^{-\tfrac{1}{2}}}=\sqrt{2}\left( 1-\frac{7x}{4}+\frac{96{{x}^{2}}}{32} \right) \\ & \text{The value of }x\text{ to obtain }\sqrt{3}\text{ can be calculated} \\ & \frac{2-x}{1+3x}=3 \\ & 2-x=3(1+3x) \\ & x=-\frac{1}{10} \\\end{align}$