ta có : \(\Delta=\left(2m-1\right)^2-4.4=\left(2m-1\right)^2-16\)
phương trình có 2 nghiệm \(x_1;x_2\) \(\Leftrightarrow\Delta\ge0\)
\(\Leftrightarrow\left(2m-1\right)^2-16\ge0\Leftrightarrow\left(2m-1\right)^2\ge16\)
\(\Leftrightarrow\left[{}\begin{matrix}2m-1\ge4\\2m-1\le-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2m\ge5\\2m\le-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}m\ge\dfrac{5}{2}\\m\le\dfrac{-3}{2}\end{matrix}\right.\)
áp dụng hệ thức \(vi-ét\) ta có : \(\left\{{}\begin{matrix}x_1+x_2=2m-1\\x_1x_2=4\end{matrix}\right.\)
thay vào phương trình : \(x_1^2+\left(2m-1\right)x_2+8-17m=0\)
\(\Leftrightarrow x_1^2+\left(2m-1\right)\left(\left(2m-1\right)-x_1\right)+8-17m=0\)
\(\Leftrightarrow x_1^2+\left(2m-1\right)^2-\left(2m-1\right)x_1+8-17m=0\)
\(\Leftrightarrow\left(x_1-\left(2m-1\right)\right)x_1+\left(2m-1\right)^2+8-17m=0\)
\(\Leftrightarrow-x_2x_1+\left(2m-1\right)^2+8-17m=0\)
\(\Leftrightarrow-4+4m^2-4m+1+8-17m=0\)
\(\Leftrightarrow4m^2-21m+5=0\Leftrightarrow4m^2-m-20m+5=0\)
\(\Leftrightarrow m\left(4m-1\right)-5\left(4m-1\right)=0\Leftrightarrow\left(4m-1\right)\left(m-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4m-1=0\\m-5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}m=\dfrac{1}{4}\left(loại\right)\\m=5\left(tmđk\right)\end{matrix}\right.\) vậy \(m=5\)
