Solve the following equation:

$$
\sqrt{4 x-7-12 \sqrt{x-4}}+\sqrt{4 x-16 \sqrt{x-4}}=1
$$
Let's eliminate the square roots in the usual way. Move the second term on the left side to the right side, then square both sides, rearrange, and simplify by 2, after which we have

$$
2 \sqrt{x-4}-4=\sqrt{4 x-16 \sqrt{x-4}}
$$

Squaring both sides again and rearranging, we get

$$
0=0
$$

Thus, it seems we have an identity, but we cannot say that every value of \( x \) satisfies our equation, because the two squaring operations may have introduced extraneous roots, and even infinitely many extraneous roots. Therefore, we need to check which values of \( x \) satisfy the original equation.

First, we need to determine that our equation is only defined if \( x \geq 4 \).

Our original equation can also be written as:

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

Let's examine how the left side changes as \( x \) increases from 4. Since the square root is generally understood to mean the positive value, the first square root represents \( 3-2 \sqrt{x-4} \) as long as \( 2 \sqrt{x-4} \leq 3 \), and \( 2 \sqrt{x-4}-3 \) otherwise. Since the inequality holds for both sides being positive or zero when \( x \geq 4 \), we can square both sides and get

$$
\sqrt{(2 \sqrt{x-4}-3)^{2}}= \begin{cases}3-2 \sqrt{x-4}, & \text { if } 4 \leq x \leq \frac{25}{4} \\ 2 \sqrt{x-4}-3, & \text { if } x \geq \frac{25}{4}\end{cases}
$$

Similarly,

$$
\sqrt{(2 \sqrt{x-4}-4)^{2}}=\left\{\begin{array}{l}
4-2 \sqrt{x-4}, & \text { if } 4 \leq x \leq 8 \\
2 \sqrt{x-4}-4, & \text { if } x \geq 8
\end{array}\right.
$$

Thus,

$$
\sqrt{(2 \sqrt{x-4}-3)^{2}}+\sqrt{(2 \sqrt{x-4}-4)^{2}}=\left\{\begin{array}{cl}
7-4 \sqrt{x-4}, & \text { if } 4 \leq x \leq \frac{25}{4} \\
1, & \text { if } \frac{25}{4} \leq x \leq 8 \\
4 \sqrt{x-4}-7, & \text { if } 8 \leq x
\end{array}\right.
$$

The behavior of the function is illustrated by the curve in the figure below.

![](https://cdn.mathpix.com/cropped/2024_05_02_619b08558108ff2e427dg-1.jpg?height=414&width=626&top_left_y=1815&top_left_x=739)

We see that equation (1) holds for every \( x \) in the interval \(\frac{25}{4}=6.25\) to 8 (including the endpoints), so for infinitely many \( x \) values. However, values of \( x \) outside this interval do not satisfy the original equation but only the equality obtained after the two squaring operations.

Note: If we incorrectly connect the expressions valid in different intervals during the solution process, we get the values \( x=\frac{25}{4} \) and \( x=8 \) as the common endpoints of the two adjacent intervals.