$\begin{cases} x_2=x_1+\dfrac{2A}{\sqrt3}\\ x_1^2+\left(x_1+ \dfrac{2A}{\sqrt3}\right)^2=A^2\end{cases}$
$\to \begin{cases} x_2=x_1+\dfrac{2A}{\sqrt3}\\ x_1^2+x_1^2+\dfrac{4Ax_1}{\sqrt3}+\dfrac{4}{3}A^2-A^2=0\end{cases}$
$\to \begin{cases} x_2=x_1+\dfrac{2A}{\sqrt3}\\ 2x_1^2+\dfrac{4x_1.A\sqrt3}{3}+\dfrac{A^2}{3}=0(*)\end{cases}$
$\Delta'_{(*)}=A^2\left( (\dfrac{2}{\sqrt3})^2-2.\dfrac{1}{3}\right)=\dfrac{2A^2}{3}$
Suy ra:
$x_1=\dfrac{-\dfrac{2A\sqrt3}{3}\pm A\dfrac{\sqrt6}{3}}{2}= \dfrac{\pm \sqrt{6}-2\sqrt3}{6}A$
$\to x_1^2=A^2\dfrac{3\pm 2\sqrt2}{6}$
$\to \begin{cases} x_1=\dfrac{3-2\sqrt2}{6}A^2\\ x_2=\dfrac{3+2\sqrt2}{6}A^2\end{cases}$
