Đặt \(x^2=t\left(t\ge0\right)\)
\(\Rightarrow4t^2+7t-2=0\\ \Rightarrow4t^2+8t-t-2=0\\ \Rightarrow4t\left(t+2\right)-\left(t+2\right)=0\\ \Leftrightarrow\left(4t-1\right)\left(t+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}4t-1=0\\t+2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}t=\dfrac{1}{4}\left(t/m\right)\\t=-2\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow x^2=\dfrac{1}{4}\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Đặt t = x2 (\(t\ge0\))
Phương trình trở thành:
\(4t^2+7t-2=0\)
\(\Delta=7^2-4.4.\left(-2\right)=49+32=81>0\)
=> Phương trình có 2 nghiệm phân biệt.
\(t_1=\dfrac{-7+\sqrt{81}}{2.4}=\dfrac{-7+9}{8}=\dfrac{1}{4}>0\Rightarrow x=\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\)
\(t_2=\dfrac{-7-\sqrt{81}}{2.4}=\dfrac{-7-9}{8}=-2\) (loại vì điều kiện \(t\ge0\))
Đặt \(t=x^2\left(t\ge0\right)\)
Pt trở thành : \(4t^2+7t-2=0\)
\(\Delta=b^2-4ac=7^2-4.4.\left(-2\right)=81>0\)
\(\Rightarrow\) Pt có 2 nghiệm pb \(t_1,t_2\)
\(\left\{{}\begin{matrix}t_1=\dfrac{-7+\sqrt{81}}{2.4}=\dfrac{1}{4}\left(tm\right)\\t_2=\dfrac{-7-\sqrt{81}}{2.4}=-2\left(ktm\right)\end{matrix}\right.\)
Ta có : \(x^2=\dfrac{1}{4}\Leftrightarrow x=\pm\dfrac{1}{2}\)
Vậy \(S=\left\{-\dfrac{1}{2};\dfrac{1}{2}\right\}\)
