Question

Cho phương trình \(\left(m+1\right)x^2-2\left(m+1\right)x+m-3=0\) Tìm m để phương tình có hai nghiệm thỏa mãn \(4\left(x_1+1\right)\left(4x_2+1\right)=18\)

Answer 1

Để pt có 2 nghiệm

\(\Leftrightarrow\left\{{}\begin{matrix}m\ne-1\\\Delta'=\left(m+1\right)^2-\left(m+1\right)\left(m-3\right)\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m\ne-1\\4m+4\ge0\end{matrix}\right.\) \(\Rightarrow m>-1\)

Theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=\frac{m-3}{m+1}\end{matrix}\right.\)

Sửa đề: \(\left(4x_1+1\right)\left(4x_2+1\right)=18\)

\(\Leftrightarrow16x_1x_2+4\left(x_1+x_2\right)+1=18\)

\(\Leftrightarrow\frac{16\left(m-3\right)}{m+1}+9=18\)

\(\Leftrightarrow16\left(m-3\right)=9\left(m+1\right)\Rightarrow m=\frac{57}{7}\)