Question

Gọi \(x_1,x_2\) là 2 nghiệm của phương trình \(x^2-2\left(m-1\right)x-4=0\)

Tìm m để \(|x_1\left|+\right|x_2|=5\)

Answer 1

\(\Delta'=\left(m-1\right)^2+4>0\) phương trình luôn có 2 nghiệm pb

Theo Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1x_2=-4\end{matrix}\right.\)

\(\left|x_1\right|+\left|x_2\right|=5\Leftrightarrow x_1^2+x_2^2+2\left|x_1x_2\right|=25\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2+2\left|x_1x_2\right|=25\)

\(\Leftrightarrow4\left(m-1\right)^2+8+8=25\)

\(\Leftrightarrow\left(m-1\right)^2=\frac{9}{4}\Rightarrow\left[{}\begin{matrix}m-1=\frac{3}{2}\\m-1=-\frac{3}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m=\frac{5}{2}\\m=-\frac{1}{2}\end{matrix}\right.\)