a: \(\text{Δ}=\left[-2\left(m+1\right)\right]^2-4\cdot1\left(m^2+2\right)\)
\(=4\left(m^2+2m+1\right)-4\left(m^2+2\right)\)
\(=4\left(m^2+2m+1-m^2-2\right)=4\left(2m-1\right)\)
Để phương trình có hai nghiệm phân biệt thì Δ>0
=>2m-1>0
=>2m>1
=>\(m>\dfrac{1}{2}\)
b: Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(m+1\right)\\x_1x_2=\dfrac{c}{a}=m^2+2\end{matrix}\right.\)
\(x_1^2+2\left(m+1\right)x_2=12m+2\)
=>\(x_1^2+x_2\left(x_1+x_2\right)=12m+2\)
=>\(\left(x_1^2+x_2^2\right)+x_1x_2=12m+2\)
=>\(\left(x_1+x_2\right)^2-x_1x_2=12m+2\)
=>\(\left(2m+2\right)^2-m^2-2=12m+2\)
=>\(4m^2+8m+4-m^2-2-12m-2=0\)
=>\(3m^2-4m=0\)
=>m(3m-4)=0
=>\(\left[{}\begin{matrix}m=0\left(loại\right)\\m=\dfrac{4}{3}\left(nhận\right)\end{matrix}\right.\)