Question

cho \(x^2-\left(2m+1\right)x-3=0\),tìm m để pt có 2 nghiệm pb thõa mãn /x1/-/x2/=5

Answer 1

\(ac=-3< 0\Rightarrow\) pt có 2 nghiệm pb trái dấu

TH1: \(x_1< 0< x_2\Rightarrow\left\{{}\begin{matrix}\left|x_1\right|=-x_1\\\left|x_2\right|=x_2\end{matrix}\right.\)

\(\left|x_1\right|-\left|x_2\right|=5\Leftrightarrow-x_1-x_2=5\)

\(\Leftrightarrow x_1+x_2=-5\)

\(\Leftrightarrow2m+1=-5\Rightarrow m=-3\)

TH2: \(x_2< 0< x_1\)

\(\left|x_1\right|-\left|x_2\right|=5\Leftrightarrow x_1-\left(-x_2\right)=5\)

\(\Leftrightarrow x_1+x_2=5\)

\(\Leftrightarrow2m+1=5\Rightarrow m=2\)

Vậy \(\left[{}\begin{matrix}m=-3\\m=2\end{matrix}\right.\)