4. [5 points] Solve the equation $2 x^{4}+x^{2}-6 x-3 x^{2}|x-3|+9=0$.

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Answer: $-\frac{3}{2} ; 1 ; \frac{-1 \pm \sqrt{13}}{2}$.

Solution. The equation can be rewritten as $2 x^{4}+(x-3)^{2}-3 x^{2}|x-3|=0$ or $|x-3|^{2}-$ $3 x^{2}|x-3|+2 x^{4}=0$. To factor the left side, note that it represents a quadratic trinomial in terms of $y=|x-3|$ with a discriminant equal to $\left(3 x^{2}\right)^{2}-4$. $2 x^{4}=x^{4}$. Therefore, the roots $y_{1,2}$ are $\frac{3 x^{2}+x^{2}}{2}$, i.e., $y_{1}=2 x^{2}$ and $y_{2}=x^{2}$, and the equation becomes $\left(|x-3|-2 x^{2}\right)\left(|x-3|-x^{2}\right)=0$.

The product is zero if and only if one of the factors is zero. From this, we get

$$
\left[\begin{array}{l}
x^{2} = x - 3, \\
x^{2} = -x + 3, \\
2 x^{2} + x - 3 = 0, \\
2 x^{2} - x + 3 = 0
\end{array} \Leftrightarrow \left[\begin{array}{l}
x^{2} + x - 3 = 0, \\
2 x^{2} + x - 3 = 0
\end{array} \Leftrightarrow \left[\begin{array}{l}
x=\frac{-1 \pm \sqrt{13}}{2} \\
x=-1.5 \\
x=1
\end{array}\right.\right.\right.
$$