## Task A-2.2.

Solve the equation $\sqrt{9-5 x}-\sqrt{3-x}=\frac{6}{\sqrt{3-x}}$ in the set of real numbers.
## Solution.

Let's group the expressions that include $\sqrt{3-x}$ on one side of the equation, then square it:

$$
\begin{array}{rlrl}
\sqrt{9-5 x} & =\sqrt{3-x}+\frac{6}{\sqrt{3-x}}, \quad /^{2} & \\
9-5 x & =3-x+12+\frac{36}{3-x}, & & 1 \text { point } \\
-6-4 x & =\frac{36}{3-x}, \quad / \cdot(3-x) & \\
(-6-4 x)(3-x)=36, & 1 \text { point } \\
4 x^{2}-6 x-54=0 . & 1 \text { point }
\end{array}
$$

The solutions to the resulting quadratic equation are $x=\frac{9}{2}$ and $x=-3$. 1 point

Substituting into the original equation, we see that for $x=\frac{9}{2}$ the equation is not defined, 1 point while $x=-3$ is a solution.

Thus, the only solution to the equation is $x=-3$.

Note: Instead of substituting the value $x=\frac{9}{2}$, it is possible to verify that the original equation is defined for values of $x$ that satisfy $9-5 x \geqslant 0$ and $3-x>0$, i.e., $x \leqslant \frac{9}{5}$. This analysis changes the second-to-last 1 point in the scoring scheme. The number $x=-3$ must be substituted regardless of this analysis.

The first steps of the solution can also be derived in a different order: it is possible to first multiply the equation by $\sqrt{3-x}$, and then square the rearranged equation. Each of these steps also carries 1 point.

Reducing the problem to a quadratic equation with the process totals 3 points. A different squaring method (such as squaring the original equation without prior rearrangement) can lead to a fourth-degree equation. In such a solution, the correct higher-degree equation is worth 2 points, while 1 point is awarded for justifying that finding the remaining roots of the polynomial and justifying that they are not solutions to the original equation. The last 3 points in the above scoring scheme are awarded in the same way.