Question

cho pt \(x^2-\left(2m+3\right)x+m^2+2m+2=0\) tìm m để pt có 2 nghiệm \(x_1,x_2\) thỏa mãn \(\left(x_1-x_2\right)^2=2x_1+x_2\)

Answer 1

\(\Delta=\left(2m+3\right)^2-4\left(m^2+2m+2\right)=4m+1\ge0\Rightarrow m\ge-\frac{1}{4}\)

\(\left(x_1+x_2\right)^2-4x_1x_2=x_1+x_2+x_1\)

\(\Leftrightarrow\left(2m+3\right)^2-4\left(m^2+2m+2\right)=2m+3+x_1\)

\(\Leftrightarrow4m+1=2m+3+x_1\)

\(\Rightarrow x_1=2m-2\Rightarrow x_2=2m+3-x_1=5\)

Mà \(x_1x_2=m^2+2m+2\)

\(\Rightarrow5\left(2m-2\right)=m^2+2m+2\)

\(\Rightarrow m^2-8m+12=0\Rightarrow\left[{}\begin{matrix}m=6\\m=2\end{matrix}\right.\)