Để pt có 2 n0 pb thì \(\left\{{}\begin{matrix}ae0\\\Delta>0\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}me0\\\left(m^2-3\right)^2-4m^2\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}me0\\m^4-10m^2+3>0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}me0\\\left(m^2-5\right)^2-16>0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}me0\\\left(m^2-5\right)^2>16\left(2\right)\end{matrix}\right.\)
Xét (2) => \(\left[{}\begin{matrix}m^2-5< -4\\m^2-5>4\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}m^2< 1\\m^2>9\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}-1< m< 1\\\left[{}\begin{matrix}m< -3\\m>3\end{matrix}\right.\end{matrix}\right.\)
=> -1 < m < 1, m \(e\) 0 và m < -3 hoặc m > 3 (x)
Với (x) thì pt luôn có n0 . Theo Vi-ét ta có
x1 + x2 = \(\dfrac{-b}{a}=\dfrac{3-m^2}{m}\)
Theo đề bài ta có x1 + x2 = \(\dfrac{13}{4}\)
<=> \(\dfrac{3-m^2}{m}=\dfrac{13}{4}\)
<=> 12 - 4m2 = 13m
<=> 4m2 + 13m - 12 = 0
<=> (4m - 3)(m + 4) = 0
<=>\(\left[{}\begin{matrix}m=\dfrac{3}{4}\left(TM\right)\\m=-4\left(TM\right)\end{matrix}\right.\)
Vậy \(m=\dfrac{3}{4}\) hoặc m = -4
