Question

Answer 1

Theo đề, ta có hệ phương trình:

\(\left\{{}\begin{matrix}\dfrac{-b}{2a}=-2\\-\dfrac{b^2-4ac}{4a}=5\\a\cdot1^2+b\cdot1+c=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=4a\\b^2-4ac=-20a\\a+b+c=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=4a\\16a^2-4ac+20a=0\\a+b+c=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=4a\\4a\left(4a-c+5\right)=0\\a+b+c=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}b=4a\\4a-c+5=0\\a+b+c=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=4a\\c=4a+5\\4a+4a+5+a=-1\end{matrix}\right.\)

=>a=-2/3; b=-8/3; c=-8/3+5=7/3