Phương trình hoành độ giao điểm:
\(x^2=mx-m^2+6\Leftrightarrow x^2-mx+m^2-6=0\)
\(\Delta=m^2-4\left(m^2-6\right)=24-3m^2\ge0\Rightarrow-2\sqrt{2}\le m\le2\sqrt{2}\)
Khi đó, theo Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=m^2-6\end{matrix}\right.\)
Theo bài ra ta có: \(x_1=2x_2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2=m\\x_1=2x_2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x_2+x_2=m\\x_1=2x_2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_2=\frac{m}{3}\\x_1=\frac{2m}{3}\end{matrix}\right.\)
\(\Rightarrow\left(\frac{m}{3}\right)\left(\frac{2m}{3}\right)=m^2-6\Leftrightarrow2m^2=9\left(m^2-6\right)\)
\(\Leftrightarrow7m^2=54\Rightarrow m=\pm\frac{3\sqrt{42}}{7}\)
