\(\Delta'=\left(m+1\right)^2-\left(m^2+2\right)=2m-1\)
Pt có 2 nghiệm pb khi \(2m-1>0\Rightarrow m>\dfrac{1}{2}\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=m^2+2\end{matrix}\right.\)
Do \(x_1\) là nghiệm pt nên:
\(x_1^2-2\left(m+1\right)x_1+m^2+2=0\Rightarrow x_1^2=2\left(m+1\right)x_1-m^2-2\)
Từ đó ta có:
\(x_1^2+2\left(m+1\right)x_2=12m+2\)
\(\Leftrightarrow2\left(m+1\right)x_1-m^2-2+2\left(m+1\right)x_2=12m+2\)
\(\Leftrightarrow2\left(m+1\right)\left(x_1+x_2\right)-m^2-12m-4=0\)
\(\Leftrightarrow4\left(m+1\right)^2-m^2-12m-4=0\)
\(\Leftrightarrow3m^2-4m=0\Rightarrow\left[{}\begin{matrix}m=0< \dfrac{1}{2}\left(loại\right)\\m=\dfrac{4}{3}\end{matrix}\right.\)
