Question

Tìm m để pt x²+2(m+2)x+3m+10=0 có 2 nghiệm x₁,x₂ thỏa mãn /x₁-x₂/≤4

Answer 1

\(\Delta'=\left(m+2\right)^2-3m-10=m^2+m-6\ge0\) \(\Rightarrow\left[{}\begin{matrix}m\le-3\\m\ge2\end{matrix}\right.\)

Theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=-2\left(m+2\right)\\x_1x_2=3m+10\end{matrix}\right.\)

\(\left|x_1-x_2\right|\le4\)

\(\Leftrightarrow\left(x_1-x_2\right)^2\le16\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2-16\le0\)

\(\Leftrightarrow4\left(m+2\right)^2-4\left(3m+10\right)-16\le0\)

\(\Leftrightarrow m^2+m-10\le0\) \(\Rightarrow\frac{-1-\sqrt{41}}{2}\le m\le\frac{-1+\sqrt{41}}{2}\)

Vậy \(\left[{}\begin{matrix}\frac{-1-\sqrt{41}}{2}\le m\le-3\\2\le m\le\frac{-1+\sqrt{41}}{2}\end{matrix}\right.\)