Đáp án:
Xét pt hoành độ giao điểm
$\begin{array}{l}
- {x^2} + 2x + 3 = \left( {m + 4} \right).x + m + 2\\
\Rightarrow {x^2} + \left( {m + 4} \right).x - 2x + m + 2 - 3 = 0\\
\Rightarrow {x^2} + \left( {m + 2} \right).x + m - 1 = 0
\end{array}$
=> pt có 2 nghiệm trái dấu
$\begin{array}{l}
\Rightarrow a.c < 0\\
\Rightarrow 1.\left( {m - 1} \right) < 0\\
\Rightarrow m < 1\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = - m - 2\\
{x_1}{x_2} = m - 1
\end{array} \right.\\
Do:\left| {{x_2}} \right| = 2\left| {{x_1}} \right|\\
\Rightarrow {x_2} = - 2{x_1}\\
\Rightarrow {x_1} - 2{x_1} = - m - 2\\
\Rightarrow - {x_1} = - m - 2\\
\Rightarrow {x_1} = m + 2\\
\Rightarrow {x_2} = - 2m - 4\\
\Rightarrow \left( {m + 2} \right).\left( { - 2m - 4} \right) = m - 1\\
\Rightarrow - 2{\left( {m + 2} \right)^2} = m - 1\\
\Rightarrow 2{m^2} + 8m + 8 = 1 - m\\
\Rightarrow 2{m^2} + 9m + 7 = 0\\
\Rightarrow \left( {2m + 7} \right)\left( {m + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
m = - \frac{7}{2}\left( {tm} \right)\\
m = - 1\left( {tm} \right)
\end{array} \right.
\end{array}$
