\(\Delta=\left(-3\right)^2-4.\left(-1\right).4=9+16=25>0\)
\(\Delta>0\) thì pt có 2 nghiệm phân biệt \(\left\{{}\begin{matrix}x_1=\dfrac{3-5}{8}=-\dfrac{1}{4}\\x_2=\dfrac{3+5}{8}=1\end{matrix}\right.\)
\(Ta\) \(có\) \(4x^2-3x-1=0\)
\(\Rightarrow\) \(4x^2-4x+x-1=0\)
\(\Rightarrow\) \(4x.\left(x-1\right)+\left(x-1\right)=0\)
\(\Rightarrow\) \(\left(x-1\right).\left(4x+1\right)=0\)
\(\Rightarrow\) \(\left[{}\begin{matrix}x-1=0\\4x+1=0\end{matrix}\right.\) \(\Rightarrow\) \(\left[{}\begin{matrix}x=1\\x=\dfrac{-1}{4}\end{matrix}\right.\)
\(Vậy\) \(đa\) \(thức\) \(có\) \(nghiệm\) \(x=1\) \(hoặc\) \(x=\dfrac{-1}{4}\)
