Example 23. Solve the equation

$$
(x-3)(x+1)+3(x-3) \sqrt{\frac{x+1}{x-3}}-28=0
$$
Solution. Solving the inequality $\frac{x+1}{x-3} \geqslant 0$, we find the domain of the equation (9): $x \leqslant-1$ and $x>3$.

Within the domain of the equation, we have:

$$
3(x-3) \sqrt{\frac{x+1}{x-3}}=3 \sqrt{(x+1)(x-3)} \text { for } x>3
$$

And

$$
3(x-3) \sqrt{\frac{x+1}{x-3}}=-3 \sqrt{(x+1)(x-3)} \text { for } x \leqslant-1
$$

therefore, equation (9) is equivalent to the following system of systems

$$
\left[\begin{array}{l}
\left\{\begin{array}{l}
x \leqslant-1 \\
(x-3)(x+1)-3 \sqrt{(x+1)(x-3)}-28=0
\end{array}\right. \\
\left\{\begin{array}{l}
x>3 \\
(x-3)(x+1)+3 \sqrt{(x-3)(x+1)}-28=0
\end{array}\right.
\end{array}\right.
$$

Let $t=\sqrt{(x-3)(x+1)}$. Since

$$
t^{2}-3 t-28=(t-7)(t+4)
$$

and

$$
t^{2}+3 t-28=(t+7)(t-4)
$$

the obtained system of systems is equivalent to the following system

![](https://cdn.mathpix.com/cropped/2024_05_21_aca6fef227914c05e0acg-066.jpg?height=252&width=1106&top_left_y=525&top_left_x=472)

Within the domain of the original equation $\sqrt{(x-3)(x+1)}+7>0 \quad$ and $\sqrt{(x-3)(x+1)}+4>0$; therefore, from the last system of systems, it follows that

![](https://cdn.mathpix.com/cropped/2024_05_21_aca6fef227914c05e0acg-066.jpg?height=251&width=1340&top_left_y=934&top_left_x=358)

$$
\Leftrightarrow\left[\begin{array}{l}
\left\{\begin{array}{l}
x \leqslant-1 \\
x^{2}-2 x-52=0 \\
x>3 \\
x^{2}-2 x-19=0
\end{array}\right.
\end{array}\right.
$$

The equation $x^{2}-2 x-52=0$ has two roots: $x_{1}=1+\sqrt{53}$ and $x_{2}=1-\sqrt{53}$, of which only $x_{2}$ satisfies the condition $x \leqslant -1$. The equation $x^{2}-2 x-19=0$ also has two roots: $x_{3}=1+2 \sqrt{5}$ and $x_{4}=1-2 \sqrt{5}$, of which only $x_{3}$ satisfies the condition $x>3$.

Thus, the set of all solutions to equation (9) consists of two numbers: $1-\sqrt{53}$ and $1+2 \sqrt{5}$.

Let's provide another solution to equation (9).

The domain of equation (9) is given by the union of two inequalities: $x \leqslant-1$ and $x>3$. Let $y=(x-3) \sqrt{\frac{x+1}{x-3}}$; then $y^{2}=(x-3)(x+1)$, and equation (9) can be written as

$$
y^{2}+3 y-28=0
$$

Since $y^{2}+3 y-28=(y+7)(y-4)$, the obtained equation has two roots: $y_{1}=-7$ and $y_{2}=4$. Therefore, equation (9) is equivalent to the union of two equations:

$$
(x-3) \sqrt{\frac{x+1}{x-3}}=-7, \quad(x-3) \sqrt{\frac{x+1}{x-3}}=4
$$

The expression $\sqrt{\frac{x+1}{x-3}}$ is non-negative; within the domain of the equation, we have: $x-3>0$ for $x>3$ and $x-3<0$ for $x \leqslant-1$. Therefore, the equation $(x-3) \sqrt{\frac{x+1}{x-3}}=-7$ has no solutions, and the equation $(x-3) \sqrt{\frac{x+1}{x-3}}=4$ is equivalent to the system

$$
\left\{\begin{array}{l}
x-3>0 \\
(x-3)^{2} \frac{x+1}{x-3}=16
\end{array}\right.
$$

Solving the obtained system of systems, we find the set of all roots of equation (9) - the numbers $1-\sqrt{53}$ and $1+2 \sqrt{5}$.

66 CH. 2. EQUATIONS WITH ONE UNKNOWN