Để phương trình có hai nghiệm trái dấu thì a*c<0
=>12-5m^2<0
=>5m^2>12
=>m^2>12/5
=>\(\left[{}\begin{matrix}m>\sqrt{\dfrac{12}{5}}\\m< -\sqrt{\dfrac{12}{5}}\end{matrix}\right.\)
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=3m-1\\x_1x_2=\dfrac{c}{a}=12-5m^2\end{matrix}\right.\)
\(x_1^2+x_1+2x_1x_2=8x_2^2+2x_2\)
=>\(\left(x_1^2+2x_1x_2-8x_2^2\right)+\left(x_1-2x_2\right)=0\)
=>\(\left(x_1^2-2x_1x_2+4x_1x_2-8x_2^2\right)+\left(x_1-2x_2\right)=0\)
=>\(\left[x_1\left(x_1-2x_2\right)+4x_2\left(x_1-2x_2\right)\right]+\left(x_1-2x_2\right)=0\)
=>\(\left(x_1-2x_2\right)\left(x_1+4x_2+1\right)=0\)
=>\(\left[{}\begin{matrix}x_1-2x_2=0\\x_1+4x_2=-1\end{matrix}\right.\)
TH1: \(x_1-2x_2=0\)
Theo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}x_1-2x_2=0\\x_1+x_2=3m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-3x_2=-3m+1\\x_1=2x_2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_2=m-\dfrac{1}{3}\\x_1=2m-\dfrac{2}{3}\end{matrix}\right.\)
\(x_1x_2=12-5m^2\)
=>\(12-5m^2=\left(2m-\dfrac{2}{3}\right)\left(m-\dfrac{1}{3}\right)\)
=>\(12-5m^2=2m^2-\dfrac{4}{3}m+\dfrac{2}{9}\)
=>\(-7m^2+\dfrac{4}{3}m+\dfrac{106}{9}=0\)
=>\(\left[{}\begin{matrix}m=\dfrac{2+\sqrt{746}}{21}\left(loại\right)\\m=\dfrac{2-\sqrt{746}}{21}\left(nhận\right)\end{matrix}\right.\)
TH2: \(x_1+4x_2=-1\)
Theo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}x_1+4x_2=-1\\x_1+x_2=3m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x_2=-1-3m+1=-3m\\x_1+4x_2=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x_2=-m\\x_1=-1-4x_2=-1-4\cdot\left(-m\right)=4m-1\end{matrix}\right.\)
\(x_1x_2=12-5m^2\)
=>\(12-5m^2=-m\left(4m-1\right)=-4m^2+m\)
=>\(12-5m^2+4m^2-m=0\)
=>\(-m^2-m+12=0\)
=>\(m^2+m-12=0\)
=>(m+4)(m-3)=0
=>\(\left[{}\begin{matrix}m=-4\left(nhận\right)\\m=3\left(nhận\right)\end{matrix}\right.\)
