Toán Lớp 9: PT : $x^{2}$ – ( m+5 )x -m + 6 = 0
Tìm m để pt có 2 nghiệm phân biệt cùng dương
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Toán Lớp 9: PT : $x^{2}$ – ( m+5 )x -m + 6 = 0
Tìm m để pt có 2 nghiệm phân biệt cùng dương
Comments ( 1 )
\Leftrightarrow \left\{ \begin{array}{l}
{m^2} + 10m + 25 – 4\left( { – m + 6} \right) > 0\\
m + 5 > 0\\
– m + 6 > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{m^2} + 10m + 25 + 4m – 24 > 0\\
6 > m > – 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{m^2} + 14m + 1 > 0\\
6 > m > – 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{m^2} + 14m + 49 > 48\\
6 > m > – 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{\left( {m + 7} \right)^2} > 48\\
6 > m > – 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m + 7 > 4\sqrt 3 \\
m + 7 < – 4\sqrt 3
\end{array} \right.\\
6 > m > – 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m > – 7 + 4\sqrt 3 \\
m < – 7 – 4\sqrt 3
\end{array} \right.\\
6 > m > – 5
\end{array} \right.\\
\to 6 > m > – 7 + 4\sqrt 3
\end{array}\)