91. Let's solve the equation

$$
(3 x-1) \sqrt{x-4}=17 \sqrt{2}
$$
Solution. Method 1. The domain of expression (3) is the interval $[4 ;+\infty)$. Squaring both sides of the equation, we obtain the equivalent equation

$$
9 x^{3}-42 x^{2}+25 x-582=0
$$

We find the critical points of the function $f(x)=9 x^{3}-42 x^{2}+25 x-582$:

$$
f^{\prime}(x)=27 x^{2}-84 x+25=0
$$

from which $x=\frac{1}{3}$ and $x=\frac{25}{9}$. Both critical points do not belong to the domain of equation (3). Therefore, we look for the zero of the function in the interval $[4 ;+\infty)$, where $f^{\prime}(x)>0$. In this interval, the function is increasing, and it has a unique zero. Since

$$
\begin{aligned}
& f(4)=9 \cdot 4^{3}-42 \cdot 4^{2}+25 \cdot 4-5820
\end{aligned}
$$

by the Bolzano-Cauchy theorem, the zero of the function belongs to the interval $[4 ; 10]$. By trial, we find the root of the equation—it is equal to 6.

Method 2. Suppose that the factors containing radicals are equal. Then the system holds

$$
\left\{\begin{array}{l}
3 x-1=17 \\
\sqrt{x-4}=\sqrt{2}
\end{array}\right.
$$

Solving it, we get $x=6$. It is easy to verify that the number 6 is a root of the given equation. We will prove that the equation has no other roots.

Indeed, for $x>6$, the system of inequalities holds

$$
\left\{\begin{array}{l}
3 x-1>17 \\
\sqrt{x-4}>\sqrt{2}
\end{array}\right.
$$

Then

$$
(3 x-1) \cdot \sqrt{x-4}>17 \sqrt{2}
$$

If $4 \leqslant x<6$, then the system of inequalities holds

$$
\left\{\begin{array}{l}
3 x-1<17 \\
\sqrt{x-4}<\sqrt{2}
\end{array}\right.
$$

i.e.

$$
(3 x-1) \cdot \sqrt{x-4}<17 \sqrt{2}
$$

Therefore, the original equation has a unique root $x=6$.