3. (3 points) Solve the equation of the form $f(f(x))=x$, given that $f(x)=x^{2}+2 x-4$

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# Solution:

It is clear that some solutions to the equation $f(f(x))=x$ can be found by solving the simpler equation $f(x)=x$, which in this case is:

$$
x^{2}+2 x-4=x \Leftrightarrow x^{2}+2 x-5=0 \Leftrightarrow x=\frac{1}{2}(-1 \pm \sqrt{17})
$$

Since the roots of the polynomial $f(x)-x=x^{2}+2 x-5$ are also roots of the polynomial $f(f(x))-x=x^{4}+4 x^{3}-2 x^{2}-13 x+4$, the first polynomial divides the second. The result of the division is $x^{2}+3 x-1$. Therefore, the original equation is equivalent to the system of equations:

$$
\left[\begin{array} { c } 
{ x ^ { 2 } + 2 x - 5 = 0 } \\
{ x ^ { 2 } + 3 x - 1 = 0 }
\end{array} \Leftrightarrow \left[\begin{array}{l}
\frac{1}{2}(-1 \pm \sqrt{17}) \\
\frac{1}{2}(-3 \pm \sqrt{13})
\end{array}\right.\right.
$$

Answer: $\frac{1}{2}(-1 \pm \sqrt{17}), \frac{1}{2}(-3 \pm \sqrt{13})$.