2. The product of four different numbers, which are roots of the equations

$$
x^{2}+6 b x+c=0 \text{ and } x^{2}+6 c x+b=0
$$

is equal to 1. At least one of the numbers $\mathrm{b}$ and c is an integer. Find $\mathrm{b}+\mathrm{c}$ given that b and c are positive.
2. Answer: 2.5. From the condition, it follows that $\mathrm{b}^{*} \mathrm{c}=1$. Express c, substitute into the equations and find the discriminants, which must be positive.

$36 b^{2}-\frac{4}{b}>0, \frac{36}{b^{2}}-4 b>0$, from which $b>\frac{1}{\sqrt[3]{9}}, b<\sqrt[3]{9}, \quad$ and the same constraints will be satisfied for c. One of these numbers is an integer, and therefore equals 1 or 2. But if it equals 1, then both numbers equal 1, which contradicts the condition. Therefore, one of these numbers equals 2, and the other 0.5.