\(b^2-4ac\ge0\)
\(\left\{{}\begin{matrix}x_1+x_2=-\frac{b}{a}\\ax_1+bx_2=-c\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}ax_1+ax_2=-b\\ax_1+bx_2=-c\end{matrix}\right.\)
\(\Rightarrow\left(a-b\right)x_2=c-b\Rightarrow x_2=\frac{c-b}{a-b}\) (có thể dễ dàng biện luận ko thể xảy ra \(a=b=c\))
\(\Rightarrow x_1=-\frac{b}{a}-x_2=-\frac{b}{a}-\frac{c-b}{a-b}\)
\(\Rightarrow\left(\frac{c-b}{a-b}\right)\left(-\frac{b}{a}-\frac{c-b}{a-b}\right)=\frac{c}{a}\)
\(\Leftrightarrow a\left(c-b\right)^2+b\left(a-b\right)\left(c-b\right)+c\left(a-b\right)^2=0\)
\(\Leftrightarrow ac^2+b^3+a^2c-3abc=0\)
\(\Leftrightarrow ac\left(a+c-3b\right)+b^3=0\)
