## Task B-2.1.

Determine the value of the expression $\frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}$ if $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $2 x^{2}+3=6 x$.
## First Solution.

For the quadratic equation $2 x^{2}-6 x+3=0$, Vieta's formulas apply, that is,

$$
x_{1}+x_{2}=3 \text{ and } x_{1} \cdot x_{2}=\frac{3}{2}
$$

Then, we proceed as follows:

$$
\begin{array}{rlrl}
\frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}} & =\frac{x_{1}^{2}+x_{2}^{2}}{x_{1}^{2} x_{2}^{2}} & & 1 \text{ point} \\
& =\frac{\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}}{\left(x_{1} x_{2}\right)^{2}} & & 2 \text{ points } \\
& =\frac{3^{2}-2 \cdot \frac{3}{2}}{\left(\frac{3}{2}\right)^{2}}=\frac{6}{\frac{9}{4}}=\frac{8}{3} . & 1 \text{ point}
\end{array}
$$