$\sqrt{\frac{x}{2 x+1}}+\sqrt{\frac{2 x+1}{x}}=2$
Let $\quad \sqrt{\frac{x}{2 x+1}}=y$
Then,
$$
y+\frac{1}{y}=2
$$
$$
\begin{aligned}
& \Rightarrow \quad y^2-2 y+1=0 \\
& \Rightarrow \quad(y-1)(y-1)=0 \\
& \Rightarrow \quad(y-1)^2=0 \\
& \Rightarrow \quad y-1=0 \\
& \Rightarrow \quad y=1 \\
& y=1 \\
& \Rightarrow \quad \sqrt{\frac{x}{2 x+1}}=1 \\
& \Rightarrow \quad \frac{x}{2 x+1}=1 \\
& \Rightarrow \quad 2 x+1=x \\
& \Rightarrow \quad x=-1 \\
&
\end{aligned}
$$
$\because \alpha$ satisfies the Eq. (i), therefore $\alpha=-1$
Now,
$$
\alpha^2 x^2+4 \alpha x+3=0
$$
Put the value of $\alpha$
$$
\begin{aligned}
(-1)^2 x^2+4(-1) x+3 & =0 \\
x^2-4 x+3 & =0 \\
(x-1)(x-3) & =0 \\
x & =1,3
\end{aligned}
$$
Hence, the roots of the equation $\alpha^2 x^2+4 \alpha x+3=0$ are 1,3