Task B-2.6. Determine all values of the real parameter $p \neq 0$ such that for the solutions of the equation $p x(x+2)+2 p=3$ it holds $\left|x_{1}^{3} x_{2}+x_{1} x_{2}^{3}\right| \geq 6$.
## Solution.

We write the given quadratic equation in the form $p x^{2}+2 p x+2 p-3=0$.

For its solutions, Vieta's formulas apply:

$$
x_{1}+x_{2}=-2, \quad x_{1} \cdot x_{2}=\frac{2 p-3}{p}
$$

Let's write the given inequality in a convenient form.

$$
\begin{aligned}
\left|x_{1}^{3} x_{2}+x_{1} x_{2}^{3}\right| & \geq 6 \\
\left|x_{1} x_{2}\left(x_{1}^{2}+x_{2}^{2}\right)\right| & \geq 6 \\
\left|x_{1} x_{2}\left[\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}\right]\right| & \geq 6
\end{aligned}
$$

Then, from Vieta's formulas, it follows that

$$
\begin{aligned}
\left\lvert\, \frac{2 p-3}{p} \cdot\left[(-2)^{2}-2 \cdot \frac{2 p-3}{p}\right]\right. & \geq 6 \\
\left|\frac{2 p-3}{p} \cdot \frac{6}{p}\right| & \geq 6 \\
\left|\frac{2 p-3}{p^{2}}\right| & \geq 1
\end{aligned}
$$

It follows that

$$
\frac{2 p-3}{p^{2}} \geq 1 \text { or } \frac{2 p-3}{p^{2}} \leq-1
$$

Then $p^{2}-2 p+3 \leq 0$ or $p^{2}+2 p-3 \leq 0$.

(1 point)

The inequality $p^{2}-2 p+3 \leq 0$ has no real solutions.

(1 point)

The solution to the inequality $p^{2}+2 p-3 \leq 0$, with the condition from the problem that $p \neq 0$, is the set $[-3,1] \backslash\{0\}$.

(2 points)