\(\text{Δ}=\left[-2\left(k-1\right)\right]^2-4\cdot1\cdot\left(-4k\right)\)
\(=\left(2k-2\right)^2+16k\)
\(=4k^2+8k+4=\left(2k+2\right)^2\)
Để phương trình có hai nghiệm phân biệt thì Δ>0
=>\(\left(2k+2\right)^2>0\)
=>\(2k+2\ne0\)
=>\(k\ne-1\)
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(k-1\right)\\x_1x_2=\dfrac{c}{a}=-4k\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x_1+x_2=2k-2\\3x_1-x_2=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x_1=2k\\x_1+x_2=2k-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=0,5k\\x_2=2k-2-0,5k=1,5k-2\end{matrix}\right.\)
\(x_1\cdot x_2=-4k\)
=>\(0,5k\left(1,5k-2\right)=-4k\)
=>\(0,75k^2-k+4k=0\)
=>\(0,75k^2+3k=0\)
=>\(0,75\left(k^2+4k\right)=0\)
=>k(k+4)=0
=>\(\left[{}\begin{matrix}k=0\left(nhận\right)\\k=-4\left(nhận\right)\end{matrix}\right.\)
