$$\begin{gathered} If\alpha satisfiestheequation\sqrt{\frac{x}{2x+1}}+\sqrt{\frac{2x+1}{x}}=2, \\ thentherootsoftheequation{\alpha }^2{x}^2+4\alpha x+3=0are \end{gathered}$$

$\begin{array}{l}Given\sqrt{\frac{x}{2x+1}}+\sqrt{\frac{2x+1}{x}}=2\\ Let\sqrt{\frac{x}{2x+1}}=a\\ \Rightarrow a+\frac{1}{a}=2\\ \Rightarrow a^2-2a+1=0\\ \Rightarrow {(a-1)}^2=0\\ \Rightarrow a-1=0,\Rightarrow a=1\\ \therefore \sqrt{\frac{x}{2x+1}}=1\\ \Rightarrow \frac{x}{2x+1}=1\Rightarrow 2x+1=x\\ \Rightarrow x+1=0,\Rightarrow x=-1\\ \therefore \alpha =-1\\ andgivenEquation{\alpha }^2{x}^2+4\alpha x+3=0\\ \Rightarrow x^2-4x+3=0\\ \Rightarrow (x-1)(x-3)=0\\ \Rightarrow x=1,3\end{array}$