Question

Cho phương trình: \(x^2+2\left(2-m\right)x-3=0\)

Tìm m để phương trình có 2 nghiệm \(x_1;x_2\) thỏa mãn \(\left|x_1x_2^2\right|+\left|x_1^2x_2\right|=18\)

Answer 1

\(ac< 0\) nên pt luôn có 2 nghiệm pb trái dấu

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

\(\left|x_1x_2.x_2\right|+\left|x_1x_2.x_1\right|=18\)

\(\Leftrightarrow\left|-3x_2\right|+\left|-3x_1\right|=18\)

\(\Leftrightarrow\left|x_1\right|+\left|x_2\right|=6\)

\(\Leftrightarrow x_1^2+x_2^2+2\left|x_1x_2\right|=36\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2+2\left|x_1x_2\right|=36\)

\(\Leftrightarrow4\left(m-2\right)^2+12=36\)

\(\Leftrightarrow\left(m-2\right)^2=6\Rightarrow m=2\pm\sqrt{6}\)