Task B-2.2. Rowing evenly, a fisherman traveled upstream as many kilometers as the solution to the equation $\sqrt{5 x}-\frac{4}{\sqrt{3 x+1}}=\sqrt{3 x+1}$, and returned to the harbor without stopping in 1 hour and 20 minutes. At what speed did he row if the river's speed is $2 \mathrm{~km} / \mathrm{h}$?
## Solution.

First, let's solve the given equation

$$
\sqrt{5 x}-\frac{4}{\sqrt{3 x+1}}=\sqrt{3 x+1}
$$

Multiply both sides by the denominator.

$$
\begin{aligned}
\sqrt{5 x(3 x+1)}-4 & =3 x+1 \\
\sqrt{5 x(3 x+1)} & =3 x+5
\end{aligned}
$$

If we square both sides, we get

$$
\begin{aligned}
15 x^{2}+5 x & =9 x^{2}+30 x+25 \\
6 x^{2}-25 x-25 & =0
\end{aligned}
$$

The only positive solution to this equation that satisfies the original equation is $x=5$ $\mathrm{km}$.

The fisherman walked a distance of $5 \mathrm{~km}$ upstream and the same distance downstream in a total of $1 \mathrm{~h}$ and $20 \mathrm{~min}$, which is $\frac{4}{3} \mathrm{~h}$.

If the speed at which the fisherman rows is $v$, then

$$
\begin{aligned}
\frac{5}{v-2}+\frac{5}{v+2} & =\frac{4}{3} / \cdot 3(v-2)(v+2) \\
15(v+2)+15(v-2) & =4\left(v^{2}-4\right) \\
2 v^{2}-15 v-8 & =0
\end{aligned}
$$

The positive solution to this equation is $v=8$. The fisherman rowed at a speed of $8 \mathrm{~km} / \mathrm{h}$.