Đáp án: $a = \pm \sqrt {14} ;b = - 5$
Giải thích các bước giải:
$\begin{array}{l}
\Delta > 0\\
\Rightarrow {a^2} - 4\left( {b + 2} \right) > 0\left( * \right)\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = - a\\
{x_1}.{x_2} = b + 2
\end{array} \right.\\
Do:\left\{ \begin{array}{l}
{x_1} - {x_2} = 4\\
x_1^3 - x_2^3 = 28
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} = 16\\
\left( {{x_1} - {x_2}} \right)\left( {x_1^2 + {x_1}{x_2} + x_2^2} \right) = 28
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{a^2} - 4.\left( {b + 2} \right) = 16\\
4.\left[ {{{\left( {{x_1} + {x_2}} \right)}^2} - {x_1}{x_2}} \right] = 28
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{a^2} - 4b = 24\\
{a^2} - b - 2 = 7
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{a^2} - 4b = 24\\
{a^2} - b = 9
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
b = - 5\\
{a^2} = 14\left( {tm} \right)
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
b = - 5\\
a = \pm \sqrt {14}
\end{array} \right.
\end{array}$
