4. [5 points] Solve the equation $5 x^{4}+x^{2}+8 x-6 x^{2}|x+4|+16=0$.






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Answer: $-\frac{4}{5} ; 1 ; \frac{1 \pm \sqrt{17}}{2}$.

Solution. The equation can be rewritten as $5 x^{4}+(x+4)^{2}-6 x^{2}|x+4|=0$ or $|x+4|^{2}-6 x^{2} \mid x+$ $4 \mid+5 x^{4}=0$. To factorize the left side, note that it represents a quadratic trinomial in terms of $y=|x+4|$ with a discriminant equal to $\left(6 x^{2}\right)^{2}-4 \cdot 5 x^{4}=$ $16 x^{4}$. Therefore, the roots $y_{1,2}$ are $\frac{6 x^{2} \pm 4 x^{2}}{2}$, i.e., $y_{1}=5 x^{2}$ and $y_{2}=x^{2}$, and the equation becomes $\left(|x+4|-5 x^{2}\right)\left(|x+4|-x^{2}\right)=0$.

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

$$
\left[\begin{array}{l}
x^{2} = x + 4, \\
x^{2} = -x - 4, \\
5 x^{2} + x + 4 = 0, \\
5 x^{2} - x - 4 = 0
\end{array} \Leftrightarrow \left[\begin{array}{l}
x^{2} - x - 4 = 0, \\
5 x^{2} - x - 4 = 0
\end{array} \Leftrightarrow \left[\begin{array}{l}
x=\frac{1 \pm \sqrt{17}}{2} \\
x=-\frac{4}{5} \\
x=1
\end{array}\right.\right.\right.
$$