\(\text{Δ}=\left[-2\left(m+1\right)\right]^2-4\cdot1\cdot2m\)
\(=4\left(m^2+2m+1\right)-8m\)
\(=4m^2+4>=4>0\forall m\)
=>Phương trình luôn có hai nghiệm phân biệt
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-\left[-2\left(m+1\right)\right]}{1}=2\left(m+1\right)\\x_1\cdot x_2=\dfrac{c}{a}=\dfrac{2m}{1}=2m\end{matrix}\right.\)
\(P=x_1^2+x_2\cdot2\left(m+1\right)+4x_1x_2\)
\(=x_1^2+x_2\cdot\left(x_1+x_2\right)+4\cdot2m\)
\(=x_1^2+x_2^2+x_1x_2+8m\)
\(=\left(x_1+x_2\right)^2-x_1x_2+8m\)
\(=\left(2m+2\right)^2-2m+8m\)
\(=4m^2+8m+4+6m\)
\(=4m^2+14m+4\)
\(=4\left(m^2+\dfrac{7}{2}m+1\right)\)
\(=4\left(m^2+2\cdot m\cdot\dfrac{7}{4}+\dfrac{49}{16}-\dfrac{33}{16}\right)\)
\(=4\left(m+\dfrac{7}{4}\right)^2-\dfrac{33}{4}>=-\dfrac{33}{4}\forall m\)
Dấu '=' xảy ra khi \(m+\dfrac{7}{4}=0\)
=>\(m=-\dfrac{7}{4}\)
\(\Delta'=\left(m+1\right)^2-2m=m^2+1>0;\forall m\)
Phương trình luôn có 2 nghiệm pb với mọi m
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=2m\end{matrix}\right.\)
\(P=x_1^2+2\left(m+1\right)x_2+4x_1x_2\)
\(=x_1\left(x_1+x_2\right)-x_1x_2+2\left(m+1\right)x_2+4x_1x_2\)
\(=2\left(m+1\right)x_1+2\left(m+1\right)x_2+3x_1x_2\)
\(=2\left(m+1\right)\left(x_1+x_2\right)+3x_1x_2\)
\(=4\left(m+1\right)^2+6m\)
\(=4m^2+14m+4\)
\(=4\left(m+\dfrac{7}{4}\right)^2-\dfrac{33}{4}\ge-\dfrac{33}{4}\)
\(P_{min}=-\dfrac{33}{4}\) khi \(m=-\dfrac{7}{4}\)