Bài 2:
Theo đề, ta có hệ:
\(\left\{{}\begin{matrix}\dfrac{-b}{2a}=2\\-\dfrac{b^2-4ac}{4a}=1\\a+b+c=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-2a\\b^2-4ac=-4a\\a+b+c=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=-2a\\\left(-2a\right)^2-4ac=-4a\\a+b+c=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-2a\\4a^2-4ac=-4a\\a+b+c=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=-2a\\a-c=-1\\a+b+c=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-2a\\c=a+1\\a-2a+a+1=-1\end{matrix}\right.\)
=>1=-1(loại)