6.369 Solve the equation $2 x^{5}-x^{4}-2 x^{3}+x^{2}-4 x+2=0$, given that it has three roots, two of which differ only in sign.
Solution. Let $x_{1}, x_{2}, x_{3}$ be the roots, $x_{1}=-x_{2}$. Then

$\left\{\begin{array}{l}2 x_{1}^{5}-x_{1}^{4}-2 x_{1}^{3}+x_{1}^{2}-4 x_{1}+2=0, \\ -2 x_{1}^{5}-x_{1}^{4}+2 x_{1}^{3}+x_{1}^{2}+4 x_{1}+2=0\end{array} \Rightarrow 4 x_{1}^{5}-4 x_{1}^{3}-8 x_{1}=0 \Leftrightarrow\right.$ $x_{1}^{4}-x_{1}^{2}-2=0 \Leftrightarrow\left(x_{1}^{2}+1\right)\left(x_{1}^{2}-2\right)=0 \Leftrightarrow x_{1,2}= \pm \sqrt{2}$. In addition, $2 x^{5}-x^{4}-2 x^{3}+x^{2}-4 x+2=\left(x^{4}-x^{2}-2\right)(2 x-1)=0 \Rightarrow x_{3}=\frac{1}{2}$.

Answer: The roots of the equation are: $x_{1,2}= \pm \sqrt{2} ; x_{3}=\frac{1}{2}$.