Đặt \(\sqrt{x}=a\ge0\) ta được:
\(a^4-a^3-2a^2-2a+4=0\)
\(\Leftrightarrow a^4+2a^3+2a^2-3a^3-6a^2-6a+2a^2+4a+4=0\)
\(\Leftrightarrow a^2\left(a^2+2a+2\right)-3a\left(a^2+2a+2\right)+2\left(a^2+2a+2\right)=0\)
\(\Leftrightarrow\left(a^2-3a+2\right)\left(a^2+2a+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a^2-3a+2=0\\a^2+2a+2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a=1\\a=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
Vậy pt có 2 nghiệm \(x=1;x=4\)