\(\text{Δ}=\left(3m+1\right)^2-4\cdot2m\left(m+1\right)\)
\(=9m^2+6m+1-8m^2-8m\)
\(=m^2-2m+1=\left(m-1\right)^2\)>=0
=>(*) luôn có hai nghiệm
Theo Vi-et, ta có:
\(x_1+x_2=-\dfrac{b}{a}=3m+1;x_1x_2=\dfrac{c}{a}=2m\left(m+1\right)=2m^2+2m\)
\(\left|x_1-x_2\right|=2\)
=>\(\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}=2\)
=>\(\sqrt{\left(3m+1\right)^2-4\left(2m^2+2m\right)}=2\)
=>\(\sqrt{9m^2+6m+1-8m^2-8m}=2\)
=>\(\sqrt{m^2-2m+1}=2\)
=>\(\sqrt{\left(m-1\right)^2}=2\)
=>|m-1|=2
=>\(\left[{}\begin{matrix}m-1=2\\m-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=3\\m=-1\end{matrix}\right.\)
\(\Delta=\left(3m+1\right)^2-8m\left(m+1\right)=\left(m-1\right)^2\ge0;\forall m\)
Pt luôn có nghiệm với mọi m
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=3m+1\\x_1x_2=2m\left(m+1\right)\end{matrix}\right.\)
\(\left|x_1-x_2\right|=2\)
\(\Leftrightarrow\left(x_1-x_2\right)^2=4\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=4\)
\(\Leftrightarrow\left(3m+1\right)^2-8m\left(m+1\right)=4\)
\(\Leftrightarrow m^2-2m-3=0\Rightarrow\left[{}\begin{matrix}m=-1\\m=3\end{matrix}\right.\)
