\(\text{Δ}=\left(2m-6\right)^2-4\left(m^2+3m+2\right)\)
\(=4m^2-24m+36-4m^2-12m-8=-36m+28\)
Để phương trình có hai nghiệm thì -36m+28>=0
=>-36m>=-28
hay m<=7/9
Theo đề, ta có:
\(\left(x_1+x_2\right)^2-2x_1x_2=100\)
\(\Leftrightarrow\left(\dfrac{2m-6}{m+1}\right)^2-2\cdot\dfrac{m+2}{m+1}=100\)
\(\Leftrightarrow\dfrac{\left(2m-6\right)^2-2\left(m^2+3m+2\right)}{\left(m+1\right)^2}=100\)
\(\Leftrightarrow4m^2-24m+36-2m^2-6m-4=100\left(m+1\right)^2\)
\(\Leftrightarrow50\left(m+1\right)^2=m^2-15m+16\)
\(\Leftrightarrow50m^2+100m+50-m^2+15m-16=0\)
\(\Leftrightarrow49m^2+115m+34=0\)
\(\text{Δ}=115^2-4\cdot49\cdot34=6561\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}m_1=\dfrac{-115-81}{2\cdot49}=-2\left(nhận\right)\\m_2=\dfrac{-115+81}{2\cdot49}=-\dfrac{17}{49}\left(nhận\right)\end{matrix}\right.\)
