1. Zadatak

Prvo rješenje.

Za kvadratnu jednadžbu $2x^2-6x+3=0$ vrijede Vieteove formule, odnosno

$$x_1+x_2=3 \quad \text{i} \quad x_1\cdot x_2=\frac32.$$

Tada redom slijedi:

$$\begin{aligned} \frac{1}{x_1^2}+\frac{1}{x_2^2} &=\frac{x_1^2+x_2^2}{x_1^2x_2^2}\\ &=\frac{(x_1+x_2)^2-2x_1x_2}{(x_1x_2)^2}\\ &=\frac{3^2-2\cdot\frac32}{(\frac32)^2} =\frac{6}{\frac94} =\frac83. \end{aligned}$$

Drugo rješenje.

Rješenja kvadratne jednadžbe $2x^2-6x+3=0$ su $x_{1,2}=\dfrac{3\pm\sqrt3}{2}$.

Tada redom slijedi

$$x_{1,2}^2=\frac{9\pm6\sqrt3+3}{4}=\frac{12\pm6\sqrt3}{4}=\frac{6\pm3\sqrt3}{2},$$ $$\frac{1}{x_{1,2}^2}=\frac{2}{6\pm3\sqrt3}\cdot\frac{6\mp3\sqrt3}{6\mp3\sqrt3}=\frac{2\cdot(6\mp3\sqrt3)}{36-27}=\frac{4\mp2\sqrt3}{3},$$ $$\frac{1}{x_1^2}+\frac{1}{x_2^2}=\frac{4-2\sqrt3}{3}+\frac{4+2\sqrt3}{3}=\frac83.$$