Cho pt x^2 -2x + m -3=0. Tìm m để pt có 2 nghiệm x1;x2 thỏa mãn x1^2 + 3.×2^2 =4x1x2

Đáp án:

\(m = \dfrac{{15 + \sqrt {33} }}{8}\)

Giải thích các bước giải:

 Để phương trình có 2 nghiệm

\(\begin{array}{l}
 \to \Delta ‘ \ge 0\\
 \to 1 – m + 3 \ge 0\\
 \to 4 \ge m\\
 \to \left[ \begin{array}{l}
x = 1 + \sqrt {4 – m} \\
x = 1 – \sqrt {4 – m} 
\end{array} \right.\\
Vi – et:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\\
{x_1}{x_2} = m – 3
\end{array} \right.\\
{x_1}^2 + 3{x_2}^2 = 4{x_1}{x_2}\\
 \to {x_1}^2 + {x_2}^2 + 2{x_2}^2 = 4{x_1}{x_2}\\
 \to \left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2}} \right) + 2{x_2}^2 = 6{x_1}{x_2}\\
 \to {\left( {{x_1} + {x_2}} \right)^2} + 2{x_2}^2 = 6{x_1}{x_2}\\
 \to \left[ \begin{array}{l}
4 + 2.{\left( {1 + \sqrt {4 – m} } \right)^2} = 6\left( {m – 3} \right)\\
4 + 2.{\left( {1 – \sqrt {4 – m} } \right)^2} = 6\left( {m – 3} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
4 + 2\left( {1 + 2\sqrt {4 – m}  + 4 – m} \right) = 6m – 18\\
4 + 2\left( {1 – 2\sqrt {4 – m}  + 4 – m} \right) = 6m – 18
\end{array} \right.\\
 \to \left[ \begin{array}{l}
14 – 2m + 4\sqrt {4 – m}  = 6m – 18\\
14 – 2m – 4\sqrt {4 – m}  = 6m – 18
\end{array} \right.\\
 \to \left[ \begin{array}{l}
4\sqrt {4 – m}  = 8m – 32\\
4\sqrt {4 – m}  =  – 8m + 32
\end{array} \right.\\
 \to \sqrt {4 – m}  = 2m – 4\\
 \to 4 – m = 4{m^2} – 16m + 16\left( {DK:4 \ge m \ge 2} \right)\\
 \to 4{m^2} – 15m + 12 = 0\\
 \to \left[ \begin{array}{l}
m = \dfrac{{15 + \sqrt {33} }}{8}\left( {TM} \right)\\
m = \dfrac{{15 – \sqrt {33} }}{8}\left( l \right)
\end{array} \right.
\end{array}\)