Solve the following equation:

$$
11 x-x^{2}=24-2 \sqrt{x}
$$
I. Solution: Write our equation in the form

$$
x^{2}-11 x+24=2 \sqrt{x}
$$

Notice that if we add $(x+1)$ to both sides, we will get a perfect square on both sides:

$$
x^{2}-10 x+25=x+2 \sqrt{x}+1
$$

or

$$
(x-5)^{2}=(\sqrt{x}+1)^{2}
$$

From this, it follows - taking into account the ambiguity of the square root - that either

$$
x-5=\sqrt{x}+1
$$

or

$$
x-5=-(\sqrt{x}+1)=-\sqrt{x}-1
$$

In both equations, replace $\sqrt{x}$ with $y$:

$$
\begin{aligned}
& y^{2}-y-6=0 \\
& y^{2}+y-4=0
\end{aligned}
$$

From (4),

$$
y_{1}=3, \quad y_{2}=-2 ; \quad x_{1}=9, \quad\left(x_{2}=4\right)
$$

By substituting into the given equation (1), we can verify that only the value $x_{1}=9$ satisfies our equation. From (5),

$$
y_{3}=\frac{-1+\sqrt{17}}{2}, \quad y_{4}=\frac{-1-\sqrt{17}}{2} ; \quad x_{3}=\frac{9-\sqrt{17}}{2}, \quad\left[x_{4}=\frac{9+\sqrt{17}}{2}\right]
$$

By substitution, we can verify that only $x_{3}$ is a root of the given equation (1).

Bayer Márta (Bp. XX., Bagi Ilona lg. I. o. t.)

II. Solution: The given equation can be directly reduced to an algebraic equation by the substitution $\sqrt{x}=v$. Then $x=v^{2}$ and $x^{2}=v^{4}$, and our equation takes the form

$$
v^{4}-11 v^{2}-2 v+24=0
$$

Since the coefficient of the highest degree term in our equation is 1, according to a known theorem, the rational roots of our equation can only be the divisors of the constant term - currently 24. By substitution, we can verify that among the divisors of 24, $v_{1}=-2$ and $v_{2}=3$ satisfy the equation (3).

The product of the two root factors

$$
(v+2)(v-3)=v^{2}-v-6
$$

Dividing the left side of (6) by this, we get the equation

$$
v^{2}+v-4=0
$$

which provides the other two roots, and which coincides with equation (5) of the first solution.

Based on $x=v^{2}$, we get

$$
\left[x_{1}=v_{1}^{2}=4\right], \quad x_{2}=v_{2}^{2}=9, \quad x_{3}=v_{3}^{2}=\frac{9-\sqrt{17}}{2}, \quad\left[x_{4}=v_{4}^{2}=\frac{9+\sqrt{17}}{2}\right]
$$

By substitution, we verified that only the roots $x_{2}$ and $x_{3}$ satisfy the original equation (1).