Đặt \(k=x^2\left(k\ge0\right)\)
Phương trình trở thành \(4k^2+7k-2=0\)
Ta có: \(\Delta=7^2+4.4.2=81,\sqrt{\Delta}=9\)
\(\Rightarrow\orbr{\begin{cases}k=\frac{-7+9}{8}=\frac{1}{4}\\k=\frac{-7-9}{8}=-2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x^2=\frac{1}{4}\\x^2=-2\left(VL\right)\end{cases}}\)
Vậy phương trình có 2 nghiệm \(\left\{\pm\frac{1}{2}\right\}\)
\(x=\pm\frac{1}{2}\)
\(4x^4+7x^2-2=0\)
\(\Leftrightarrow\left(4x^4+8x^2\right)-\left(x^2+2\right)=0\)
\(\Leftrightarrow4x^2\left(x^2+2\right)-\left(x^2+2\right)=0\)
\(\Leftrightarrow\left(x^2+2\right)\left(4x^2-1\right)=0\)
Vì \(x^2+2>0\forall x\inℝ\)
\(\Rightarrow4x^2-1=0\)
\(\Leftrightarrow4x^2=1\)
\(\Leftrightarrow x^2=\frac{1}{4}\)
\(\Leftrightarrow x=\pm\frac{1}{2}\)