Решите уравнение: 9x^4-19x^2+2=0.

\[9x^{4} - 19x^{2} + 2 = 0\]

\[Пусть\ \ t = x^{2} \geq 0:\]

\[9t^{2} - 19t + 2 = 0\]

\[D = 361 - 72 = 289 = 17^{2}\]

\[t_{1} = \frac{19 + 17}{18} = 2;\ \ \]

\[t_{2} = \frac{19 - 17}{18} = \frac{2}{18} = \frac{1}{9}.\]

\[Подставим:\ \]

\[x^{2} = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x^{2} = \frac{1}{9}\]

\[x = \pm \sqrt{2}\ \ \ \ \ \ \ \ \ \ \ \ x = \pm \frac{1}{3}\]

\[Ответ:\ x = \pm \sqrt{2};\ x = \pm \frac{1}{3}.\]