Solve the equation

$$
x^{2}-8(x+3) \sqrt{x-1}+22 x-7=0
$$
Solution. Clearly $x \geq 1$.

We assume that the left side of the equation can be written as the square of the sum of two terms. In this case, the double product can only be $8(x+3) \sqrt{x-1}$, so the two terms could be $(x+3)$ and $4 \sqrt{x-1}$. Checking:

$$
\begin{aligned}
{[(x+3)-4 \sqrt{x-1}]^{2} } & =x^{2}+6 x+9-8(x+3) \sqrt{x-1}+16 x-16= \\
& =x^{2}-8(x+3) \sqrt{x-1}+22 x-7
\end{aligned}
$$

Thus, our assumption is correct, and our equation is

$$
[(x+3)-4 \sqrt{x-1}]^{2}=0
$$

This is true if and only if $(x+3)-4 \sqrt{x-1}=0$, or $(x+3)=4 \sqrt{x-1}$. Squaring both sides and rearranging the equation:

$$
x^{2}+6 x+9=16 x-16, \quad x^{2}-10 x+25=0, \quad \text { or } \quad(x-5)^{2}=0
$$

Thus, $x=5$.

Substituting the solution back into the original equation, we get a true statement, so the only solution to the equation is 5.