\(x^2+2\left(m+1\right)x+2m-4=0\)
a) \(\Delta^'=b'^2-ac=\left(m+1\right)^2-1.\left(2m-4\right)\)
=\(m^2+2m+1-2m+4\)
\(=m^2+5\)
pt có 2 nghiệm phân biệt khi \(\Delta'>0\)
Ta có: \(m^2\ge0\)
\(\Leftrightarrow m^2+5>0\)
Do đó pt có 2 nghiệm phân biệt
b) Theo định lý vi ét:
\(x_1+x_2=-\dfrac{b}{a}=\dfrac{-2\left(m+1\right)}{1}=-2m-2\)
\(x_1.x_2=\dfrac{c}{a}=\dfrac{2m-4}{1}=2m-4\)
Mà \(x_1^2+x_2^2=12\)
\(\Rightarrow\left(x_1+x_2\right)^2-2x_1x_2=12\)
=>\(\left(-2m-2\right)^2-2\left(2m-4\right)=12\)
\(\Rightarrow\left(-2m\right)^2-2.2m.2+2^2-4m+8=12\)
\(\Rightarrow4m^2-8m+4-4m+8=12\)
\(\Rightarrow4m^2-12m+12=12\)
\(\Rightarrow4m^2-12m+12-12=0\)
\(\Rightarrow4m^2-12m=0\)
=>\(2m.\left(2m-6\right)=0\)
=>\(\left[{}\begin{matrix}2m=0\\m-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}m=0\\m=3\end{matrix}\right.\)
vậy với m=0, m=3 thì \(x_1^2+x^2_2=12\)
