11.2. Solve the system of equations in real numbers:

$$
x^{2}+x y+y^{2}=4, x^{4}+x^{2} y^{2}+y^{4}=8
$$
Answer. $\pm\left(\sqrt{\frac{3 \pm \sqrt{5}}{2}}, \sqrt{\frac{3 \mp \sqrt{5}}{2}}\right)$ - a total of 4 solutions.

## Solution. Consider the cases.

1) $x=y$, then $3 x^{2}=4,3 x^{4}=8$, from which $x^{2}=2$, which clearly does not satisfy both equations. No solutions.
2) $x=-y$, then $x^{2}=4,3 x^{4}=8$, from which $x^{2}=\frac{2}{3}$ which also clearly does not satisfy both equations. No solutions.
3) $x \neq \pm y$. Multiply the first equation by $x-y$. We get $x^{3}-y^{3}=4(x-y)$. Multiply the second equation by $x^{2}-y^{2}$, we get: $x^{6}-y^{6}=8\left(x^{2}-y^{2}\right)$. Divide the second equation by the first, we get $x^{3}+y^{3}=2(x+y)$, from which $x^{2}-x y+y^{2}=2$. Considering the first equation, $x y=1, x^{2} + y^{2}=3$. Replacing $y=\frac{1}{x}$, we get the biquadratic equation $x^{4}-3 x^{2}+1=0$, from which $x= \pm \sqrt{\frac{3 \pm \sqrt{5}}{2}}, y= \pm \sqrt{\frac{3 \mp \sqrt{5}}{2}}-$ a total of 4 solutions.

Grading criteria. Loss of part of the solutions or acquisition of extra solutions: minus 2-3 points, depending on the number of lost/acquired solutions.