Question

`x^2 -(2+m)x+m+1=0`

Tìm m để pt có hai nghiệm thỏa mãn \(2\left|x_1\right|+3\left|x_2\right|=5\)

Answer 1

\(a+b+c=1-\left(2+m\right)+m+1=0\)

\(\Rightarrow\) Pt luôn có 2 nghiệm \(x=\left\{1;m+1\right\}\)

TH1: \(\left\{{}\begin{matrix}x_1=1\\x_2=m+1\end{matrix}\right.\)

\(\Rightarrow2.1+3\left|m+1\right|=5\)

\(\Rightarrow\left|m+1\right|=1\Rightarrow\left[{}\begin{matrix}m=0\\m=-2\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x_1=m+1\\x_2=1\end{matrix}\right.\)

\(\Rightarrow2\left|m+1\right|+3=5\Rightarrow\left|m+1\right|=1\)

\(\Rightarrow\left[{}\begin{matrix}m=0\\m=-2\end{matrix}\right.\)