Question

\(x^2-mx-m^2+m-4=0\),tìm m để pt có nghiệm x1<x2 thõa mãn \(\left|x1\right|-\left|x2\right|=2\)

Answer 1

\(ac=-m^2+m-4=-\left(m-\frac{1}{2}\right)^2-\frac{15}{4}< 0;\forall m\)

\(\Rightarrow\) Phương trình luôn có 2 nghiệm trái dấu

Mà \(x_1< x_2\Rightarrow x_1< 0< x_2\Rightarrow\left\{{}\begin{matrix}\left|x_1\right|=-x_1\\\left|x_2\right|=x_2\end{matrix}\right.\)

\(\left|x_1\right|-\left|x_2\right|=2\)

\(\Leftrightarrow-x_1-x_2=2\)

\(\Leftrightarrow x_1+x_2+2=0\)

\(\Leftrightarrow m+2=0\Rightarrow m=-2\)