(2-m)x2 -2(4-3m)x+8-m=0
có Δ' = (4-3m)2- (2-m)(8-m)= 8m2 -14m
a) Để pt có 2 nghiệm dương ⇔ \(\left\{{}\begin{matrix}\Delta'>=0\\S>0\\P>0\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}8m^2-14m>=0\\\frac{8-6m}{2-m}>0\\\frac{8-m}{2-m}>0\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}\left[{}\begin{matrix}m< =0\\m>=\frac{7}{4}\end{matrix}\right.\\\left[{}\begin{matrix}m>2\\m< \frac{4}{3}\end{matrix}\right.\\\left[{}\begin{matrix}m>8\\m< 2\end{matrix}\right.\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}m< =0\\m>8\end{matrix}\right.\)
Vậy khi m ϵ (-∞;0] \(\cap\) (8;+∞) thì pt đã cho có 2 nghiệm dương
b) Để pt có 2 nghiệm âm ⇔ \(\left\{{}\begin{matrix}\Delta'>=0\\S< 0\\P>0\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}8m^{2^{ }}-14m>=0\\\frac{8-6m}{2-m}< 0\\\frac{8-m}{2-m}>0\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}\left[{}\begin{matrix}m< =0\\m>=\frac{7}{4}\end{matrix}\right.\\\frac{4}{3}< m< 2\\\left[{}\begin{matrix}m>8\\m< 2\end{matrix}\right.\end{matrix}\right.\)⇔ \(\frac{7}{4}\)≤ m < 2
Vậy khi m ϵ [\(\frac{7}{4}\);2) thì pt đã cho có 2 nghiệm âm
