Để phương trình có hai nghiệm phân biệt:
\(\Delta'>0\)
\(\Leftrightarrow\left[-\left(k-1\right)\right]^2+4k>0\)
\(\Leftrightarrow k^2-2k+1+4k>0\)
\(\Leftrightarrow k^2+2k+1>0\)
\(\Leftrightarrow k\ne-1\)
Theo định lý viet:
\(\left\{{}\begin{matrix}x_1+x_2=2\left(k-1\right)\\x_1.x_2=-4k\left(1\right)\end{matrix}\right.\)
Ta có \(\left\{{}\begin{matrix}x_1+x_2=2k-2\\3x_1-x_2=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x_1=2k\\x_1+x_2=2k-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{k}{2}\\\dfrac{k}{2}+x_2=2k-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{k}{2}\\x_2=\dfrac{3k}{2}-2\end{matrix}\right.\)
\(x_1.x_2=-4k\)
\(\Leftrightarrow\dfrac{k}{2}.\left(\dfrac{3k}{2}-2\right)=-4k\)
\(\Leftrightarrow\dfrac{3k^2}{4}-k=-4k\)
\(\Leftrightarrow\dfrac{3k^2}{4}+3k=0\Leftrightarrow3k\left(\dfrac{k}{4}+1\right)=0\)
Tới đây e tính được k=0 hoặc k = -4
