Đáp án:
b) \(\left[ \begin{array}{l}
m = - 1\\
m = - \dfrac{2}{3}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:m = 3\\
Pt \to {x^2} - 4x - 8 = 0\\
\to {x^2} - 4x + 4 = 12\\
\to {\left( {x - 2} \right)^2} = 12\\
\to \left| {x - 2} \right| = 2\sqrt 3 \\
\to \left[ \begin{array}{l}
x - 2 = 2\sqrt 3 \\
x - 2 = - 2\sqrt 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2 + 2\sqrt 3 \\
x = 2 - 2\sqrt 3
\end{array} \right.
\end{array}\)
b) Để phương trình có 2 nghiệm
\(\begin{array}{l}
\to \Delta ' \ge 0\\
\to 4 + 3m - 1 \ge 0\\
\to m \ge - 1\\
\to \left[ \begin{array}{l}
x = 2 + \sqrt {3m + 3} \\
x = 2 - \sqrt {3m + 3}
\end{array} \right.\\
Có:{x_1}^2 + 2{x_2}^2 - {x_1}{x_2} = 8\\
\to {x_1}^2 + {x_2}^2 + {x_2}^2 - {x_1}{x_2} = 8\\
\to {x_1}^2 + {x_2}^2 + {x_2}^2 + 2{x_1}{x_2} - 3{x_1}{x_2} = 8\\
\to {\left( {{x_1} + {x_2}} \right)^2} + {x_2}^2 - 3{x_1}{x_2} = 8\\
\to \left[ \begin{array}{l}
16 + {\left( {2 + \sqrt {3m + 3} } \right)^2} - 3\left( {1 - 3m} \right) = 8\\
16 + {\left( {2 - \sqrt {3m + 3} } \right)^2} - 3\left( {1 - 3m} \right) = 8
\end{array} \right.\\
\to \left[ \begin{array}{l}
16 + 4 + 4\sqrt {3m + 3} + 3m + 3 - 3 + 9m = 8\\
16 + 4 - 4\sqrt {3m + 3} + 3m + 3 - 3 + 9m = 8
\end{array} \right.\\
\to \left[ \begin{array}{l}
4\sqrt {3m + 3} = - 12m - 12\\
4\sqrt {3m + 3} = 12m + 12
\end{array} \right.\\
\to \sqrt {3m + 3} = 3m + 3\\
\to \sqrt {3m + 3} \left( {1 - \sqrt {3m + 3} } \right) = 0\\
\to \left[ \begin{array}{l}
m = - 1\\
3m + 3 = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
m = - 1\\
m = - \dfrac{2}{3}
\end{array} \right.\left( {TM} \right)
\end{array}\)
