Question

cho pt\(x^2-\left(2m+1\right)x+m^2+m=0\)

Tìm M để pt có 2 nghiệm \(x_1,x_2\)thỏa mãn -2<\(x_1\)<\(x_2\)<4

Answer 1

\(\Delta=\left(2m+1\right)^2-4\left(m^2+m\right)=1>0;\forall m\)

Đặt \(f\left(x\right)=x^2-\left(2m+1\right)x+m^2+m\)

Để pt có 2 nghiệm thỏa mãn \(-2< x_1< x_2< 4\)

\(\Leftrightarrow\left\{{}\begin{matrix}f\left(-2\right)>0\\f\left(4\right)>0\\-2< \frac{x_1+x_2}{2}< 4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m^2+m+2\left(2m+1\right)+4>0\\m^2+m-4\left(2m+1\right)+16>0\\-4< 2m+1< 8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m^2+5m+6>0\\m^2-7m+12>0\\-\frac{5}{2}< m< \frac{7}{2}\end{matrix}\right.\) \(\Rightarrow-2< m< 3\)