\(\Delta=\left(m-2\right)^2+8m=m^2+4m+4=\left(m+2\right)^2\ge0\) \(\forall m\)
\(\Rightarrow\) Phương trình luôn có 2 nghiệm
Theo Viet ta có \(\left\{{}\begin{matrix}x_1+x_2=m-2\\x_1x_2=-2m\end{matrix}\right.\)
\(A=x_1^2+x_2^2=x_1^2+x_2^2+2x_1x_2-2x_1x_2\)
\(A=\left(x_1+x_2\right)^2-2x_1x_2=\left(m-2\right)^2+4m\)
\(A=m^2+4\ge4\)
\(\Rightarrow A_{min}=4\) khi \(m=0\)