Để pt có hai nghiệm \(\Leftrightarrow\Delta\ge0\)
\(\Leftrightarrow4\left(m+1\right)^2-4\left(9m-5\right)\ge0\)
\(\Leftrightarrow4m^2-28m+24\ge0\)\(\Leftrightarrow\left[{}\begin{matrix}m\le1\\m\ge6\end{matrix}\right.\)
Theo viet có: \(\left\{{}\begin{matrix}x_1+x_2=-2\left(m+1\right)\\x_1x_2=9m-5\end{matrix}\right.\)
\(P=-2x_1x_2+9\left(x_1^2+x_2^2\right)+1945\)
\(=-2x_1x_2+9\left(x_1+x_2\right)^2-18x_1x_2+1945\)
\(=-20\left(9m-5\right)+9.4\left(m+1\right)^2+1945\)
\(=36m^2-108m+2081\)
BBT f(m)=36m2-108m+2081 , \(m\in\left(-\infty;1\right)\cup\left(6;+\infty\right)\)
=> minP=minf(m) với m\(\in R\backslash\left(1;6\right)\)
=>minP=2009 <=> m=1
