Đáp án:
\(a = \frac{{ - 3}}{7},b = \frac{{36}}{7},c = \frac{{ - 192}}{7}\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
32a + 8b + c = 0\\
\frac{{ - b}}{{2a}} = 6\\
\frac{{4ac - {b^2}}}{{4a}} = - 12
\end{array} \right.(dk:a \ne 0) \leftrightarrow \left\{ \begin{array}{l}
c = - 32a - 8b\\
b = - 12a\\
4ac - {b^2} = - 48a
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
c = - 32a - 8b = - 32a - 8.( - 12a) = 64a\\
b = - 12a\\
4ac - {b^2} = - 48a
\end{array} \right.\\
\to 4a.64a - {( - 12a)^2} + 48a = 0\\
\leftrightarrow 112{a^2} + 48a = 0\\
\leftrightarrow \left[ \begin{array}{l}
a = 0(l)\\
a = \frac{{ - 3}}{7}(tm) \to b = \frac{{36}}{7},c = \frac{{ - 192}}{7}
\end{array} \right.
\end{array}\)
