Question

Cho \(x^2-2\left(m-1\right)x+m^2-m=0\). Tìm m để pt có 2 nghiệm thỏa mãn: \(\left(1+x_1\right)^2+\left(1+x_2\right)^2=6\)

Answer 1

\(\Delta'=\left(m-1\right)^2-m^2+m=-m+1\ge0\Rightarrow m\le1\)

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

\(\left(1+x_1\right)^2+\left(1+x_2\right)^2-6=0\)

\(\Leftrightarrow x_1^2+2x_1+x^2_2+2x_2-4=0\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2+2\left(x_1+x_2\right)-4=0\)

\(\Leftrightarrow4m^2-8m+4-2m^2+2m+4m-4-4=0\)

\(\Leftrightarrow2m^2-2m-4=0\)

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