A8. How many different (real) solutions does the equation $\left(\left(x^{2}-2\right)^{2}-5\right)^{2}=1 \quad$ have?
A) 4
B) 5
C) 6
D) 7
E) 8

## B-questions
A8. B) 5 The equation is equivalent to

$$
\left(x^{2}-2\right)^{2}-5=1 \quad \text { or } \quad\left(x^{2}-2\right)^{2}-5=-1 .
$$

The first part is equivalent to $x^{2}-2=\sqrt{6}$ or $x^{2}-2=-\sqrt{6}$, with 2 and 0 solutions respectively (the latter due to $-\sqrt{6}+2<0$).

The second part is equivalent to $x^{2}-2=2$ or $x^{2}-2=-2$, with 2 and 1 solution(s) respectively. Therefore, in total $2+0+2+1=5$ solutions.