\(\Delta=2^2-4m=4-4m\)
Để pt có 2 nghiệm phân biệt thì \(4-4m>0\Rightarrow m< 1\)
Theo hệ thức Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=-2\\x_1.x_2=m\end{matrix}\right.\)
a.\(x_1=2x_2\)
\(\Leftrightarrow x_1+x_2=3x_2\)
\(\Leftrightarrow3x_2=-2\) \(\Leftrightarrow x_2=-\dfrac{2}{3}\)
Ta có: \(x_1+x_2=-2\)
\(\Leftrightarrow x_1=-2-x_2\Leftrightarrow x_1=-2+\dfrac{2}{3}=-\dfrac{4}{3}\)
Thế \(x_1=-\dfrac{4}{3};x_2=-\dfrac{2}{3}\) vào \(x_1.x_2=m\)
\(\Leftrightarrow m=-\dfrac{4}{3}.-\dfrac{2}{3}=\dfrac{8}{9}\left(n\right)\)
b.\(x_1^2+x_2^2=0\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1.x_2=0\)
\(\Leftrightarrow\left(-2\right)^2-4m=0\)
\(\Leftrightarrow4-4m=0\)
\(\Leftrightarrow m=1\left(l\right)\) ( vì m<1 )
c.\(x_1-x_2=14\)
\(\Leftrightarrow\left(x_1-x_2\right)^2=14^2\)
\(\Leftrightarrow x_1^2+x_2^2-2x_1.x_2=196\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-4x_1.x_2=196\)
\(\Leftrightarrow\left(-2\right)^2-4m=196\)
\(\Leftrightarrow-4m=192\)
\(\Leftrightarrow m=-48\left(n\right)\)
