## Task A-2.1.

Determine the value of the real parameter $p$ so that the solutions of the equation

$$
(p-3) x^{2}+\left(p^{2}+1\right) x-11 p+18=0
$$

are the lengths of the legs of a right triangle with a hypotenuse of length $\sqrt{17}$.
## Solution.

Let $x_{1}$ and $x_{2}$ be the solutions of the given equation. The condition given in the problem then states $x_{1}^{2}+x_{2}^{2}=17$. According to Viète's formulas, we have

$$
x_{1}+x_{2}=-\frac{p^{2}+1}{p-3} \quad \text { and } \quad x_{1} x_{2}=\frac{-11 p+18}{p-3}
$$

so

$$
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=\left(\frac{p^{2}+1}{p-3}\right)^{2}-2 \cdot \frac{-11 p+18}{p-3}
$$

Therefore, we seek all real numbers $p$ such that

$$
\begin{gathered}
\left(\frac{p^{2}+1}{p-3}\right)^{2}-2 \cdot \frac{-11 p+18}{p-3}=17 \\
\Longleftrightarrow\left(p^{2}+1\right)^{2}-2(-11 p+18)(p-3)=17(p-3)^{2} \\
\Longleftrightarrow p^{4}+7 p^{2}-44=0
\end{gathered}
$$

By substituting $t=p^{2}$, we arrive at the quadratic equation

$$
t^{2}+7 t-44=0
$$

whose solutions are $t=4$ and $t=-11$.

Since the equation $p^{2}=-11$ has no solutions, we discard the second option. The first option leads to $p^{2}=4$, i.e., $p= \pm 2$.

Solving the initial equation for $p=-2$ yields the solutions $x_{1,2}=\frac{1-\sqrt{33}}{2}$, one of which is not a positive number, so we discard the option $p=-2$.

Solving the equation for $p=2$ yields $x_{1}=1$ and $x_{2}=4$, which are indeed positive real numbers. Therefore, the only solution is $p=2$.