## Task B-2.2.

Determine all real numbers $p$ such that for the solutions $x_{1}, x_{2}$ of the quadratic equation

$$
x(2 x-3 p)-3 p(x-3 p)=x-1
$$

it holds that $\left|x_{1}^{2}+x_{2}^{2}\right|>1$
## Solution.

Let's write the given equation in the general form: $2 x^{2}-(6 p+1) x+9 p^{2}+1=0$. 1 point

It is true that $\left|x_{1}^{2}+x_{2}^{2}\right|=\left|\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}\right|$.

1 point

Apply Vieta's formulas

$x_{1}+x_{2}=-\frac{b}{a}=\frac{6 p+1}{2}, x_{1} \cdot x_{2}=\frac{c}{a}=\frac{9 p^{2}+1}{2}$.

Then we have

$\left|x_{1}^{2}+x_{2}^{2}\right|=\left|\frac{(6 p+1)^{2}}{4}-\left(9 p^{2}+1\right)\right|=\left|\frac{12 p-3}{4}\right|$.

From $\left|\frac{12 p-3}{4}\right|>1$ we get $p \in\left\langle-\infty,-\frac{1}{12}\right\rangle \cup\left\langle\frac{7}{12},+\infty\right\rangle$.