Example 33. Solve the system

$$
\left\{\begin{array}{l}
\sqrt{5 x(x+2)}+\sqrt{5 y(x+2)}=2 \sqrt{2(x+y)(x+2)} \\
x y=9
\end{array}\right.
$$
The solution is as follows. We replace system (24) with the system

$$
\begin{cases}
\sqrt{5|x|} \sqrt{|x+2|} + \sqrt{5|y|} \sqrt{|x+2|} = 2 \sqrt{2|x+y|} \sqrt{|x+2|} \\
xy = 9
\end{cases}
$$

which is a consequence of system (24). System (25) is equivalent to the combination of two systems:

$$
\left\{\begin{array}{l}
|x + 2| = 0, \\
xy = 9
\end{array} \quad \left\{\begin{array}{l}
\sqrt{5|x|} + \sqrt{5|y|} = 2 \sqrt{2|x+y|} \\
xy = 9
\end{array}\right.\right.
$$

The first system in the combination (26) is equivalent to the system

$$
\left\{\begin{array}{l}
x = -2 \\
y = -9 / 2
\end{array}\right.
$$

The second system in the combination (26) is equivalent to the system

$$
\left\{\begin{array}{l}
5|x| + 5|y| + 10 \sqrt{|x||y|} = 8|x+y| \\
xy = 9
\end{array}\right.
$$

From the condition $xy = 9$, it follows that $x$ and $y$ have the same sign; therefore, the following equalities hold:

$$
|x||y| = |xy| = xy = 9, \quad |x+y| = |x| + |y|
$$

Thus, system (27) is equivalent to the system

$$
\left\{\begin{array}{l}
|x| + |y| = 10 \\
|x||y| = 9 \\
xy > 0
\end{array}\right.
$$

The equation $a^2 - 10a + 9 = 0$ has roots $a_1 = 1$ and $a_2 = 9$. According to Vieta's theorem, system (28) is equivalent to the combination of two systems:

$$
\left\{\begin{array}{r}
|x| = 1, \\
|y| = 9 \\
xy > 0
\end{array} \quad \left\{\begin{array}{r}
|x| = 9 \\
|y| = 1 \\
xy > 0
\end{array}\right.\right.
$$

which is equivalent to the combination of four systems:

$$
\left\{\begin{array}{l}
x = -1, \\
y = -9
\end{array}, \left\{\begin{array}{l}
x = 1, \\
y = 9
\end{array}, \left\{\begin{array}{l}
x = 9, \\
y = 1
\end{array}, \left\{\begin{array}{l}
x = -9 \\
y = -1
\end{array}\right.\right.\right.\right.
$$

Thus, the set of all solutions to system (24) is contained among the set of ordered pairs

$$
\{(-1, -9); \quad(-9, -1); \quad(-2, -9/2); (1, 9); (9, 1)\}
$$

corresponding to the variable set $(x, y)$. By substituting each of these pairs into system (24), we verify that the pair $(-1, -9)$ does not satisfy this system, while the other pairs do satisfy it.

Therefore, the set of all solutions to system (24) consists of four ordered pairs: $(-9, -1); (-2, -9/2); (1, 9); (9, 1)$, corresponding to the variable set $(x, y)$.

Example 34. Solve the equation for each value of $a$:

$$
\sqrt{a + \sqrt{a + x}} = x
$$

Solution. Equation (29) is equivalent to the system

$$
\left\{\begin{array}{l}
a + \sqrt{a + x} = x^2 \\
x \geq 0
\end{array}\right.
$$

which in turn is equivalent to the system

$$
\left\{\begin{array}{l}
a + x = (x^2 - a)^2 \\
x^2 - a \geq 0 \\
x \geq 0
\end{array}\right.
$$

The equation in system (30) is a fourth-degree equation in $x$ and a second-degree equation in $a$. Rewriting it as

$$
a^2 - (2x^2 + 1)a + (x^4 - x) = 0
$$

we factor the left-hand side. The discriminant of the quadratic trinomial in $a$ is

$$
(2x^2 + 1)^2 - 4(x^4 - x) = 4x^2 + 4x + 1 = (2x + 1)^2
$$

and thus,

$$
\begin{aligned}
a^2 - (2x^2 + 1)a + x^4 - x &= \\
&= \left(a - \frac{(2x^2 + 1) + (2x + 1)}{2}\right)\left(a - \frac{(2x^2 + 1) - (2x + 1)}{2}\right)
\end{aligned}
$$

Therefore, equation (29) is equivalent to the combination of two mixed systems:

$$
\left\{\begin{array}{l}
x^2 + x + 1 = a, \\
x^2 - a \geq 0, \\
x \geq 0
\end{array} \quad \left\{\begin{array}{l}
x^2 - x - a = 0 \\
x^2 - a \geq 0 \\
x \geq 0
\end{array}\right.\right.
$$

i.e., the combination of systems

$$
\left\{\begin{array}{l}
x^2 - a = -x - 1, \\
-x - 1 \geq 0, \\
x \geq 0
\end{array} \quad \left\{\begin{array}{l}
x^2 - a = x \\
x \geq 0
\end{array}\right.\right.
$$

The first system in this combination has no solutions. The second system in the combination is equivalent to the system

$$
\left\{\begin{array}{l}
\left(x - \frac{1 + \sqrt{1 + 4a}}{2}\right)\left(x - \frac{1 - \sqrt{1 + 4a}}{2}\right) = 0 \\
x \geq 0 \\
a \geq -1/4
\end{array}\right.
$$

which is equivalent to the combination of systems:

$$
\left\{\begin{array}{l}
-1/4 \leq a \leq 0 \\
\left[\begin{array}{l}
x = \frac{1 - \sqrt{1 + 4a}}{2} \\
x = \frac{1 + \sqrt{1 + 4a}}{2}
\end{array}\right.
\end{array} \quad \left\{\begin{array}{l}
a > 0 \\
x = \frac{1 + \sqrt{1 + 4a}}{2}
\end{array}\right.\right.
$$

Thus, for equation (29), we obtain:

- For $a > 0$, there is a unique root $x = \frac{1 + \sqrt{1 + 4a}}{2}$.
- For $-1/4 < a \leq 0$, there are two roots: $x_1 = \frac{1 - \sqrt{1 + 4a}}{2}$ and $x_2 = \frac{1 + \sqrt{1 + 4a}}{2}$.