4. In the set of real numbers, solve the equation

$$
(3 x+1)(4 x+1)(6 x+1)(12 x+1)=2
$$
Solution. We have:

$$
8 \cdot(3 x+1) \cdot 6 \cdot(4 x+1) \cdot 4 \cdot(6 x+1) \cdot 2 \cdot(12 x+1)=2 \cdot 8 \cdot 6 \cdot 4 \cdot 2
$$

which is equivalent to

$$
(24 x+8)(24 x+6)(24 x+4)(24 x+2)=2 \cdot 8 \cdot 6 \cdot 4 \cdot 2
$$

Let $24 x+5=y$, then

$$
(y+3)(y+1)(y-1)(y-3)=2 \cdot 8 \cdot 6 \cdot 4 \cdot 2
$$

and from there

$$
\left(y^{2}-9\right)\left(y^{2}-1\right)=2 \cdot 8 \cdot 6 \cdot 4 \cdot 2, \text{ i.e. } y^{4}-10 y^{2}+9-2 \cdot 8 \cdot 6 \cdot 4 \cdot 2=0
$$

In the set of real numbers, the solutions to the last equation are

$$
y_{1,2}= \pm \sqrt{5+\sqrt{25-9+2 \cdot 8 \cdot 6 \cdot 4 \cdot 2}}= \pm \sqrt{5+\sqrt{16(1+48)}}= \pm \sqrt{33}
$$

so therefore $x_{1,2}=\frac{-5 \pm \sqrt{33}}{24}$.

## III year