Вопрос:

Решите уравнение: 4x^4-13x^2+3=0.

Ответ:

\[4x^{4} - 13x^{2} + 3 = 0\]

\[Пусть\ t = x^{2} \geq 0:\ \ \]

\[4t^{2} - 13t + 3 = 0\]

\[t_{1} + t_{2} = \frac{13}{4};\ \ t_{1} \cdot t_{2} = \frac{3}{4}\]

\[\Longrightarrow t_{1} = 3,\ \ t_{2} = \frac{1}{4}.\]

\[x^{2} = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x^{2} = \frac{1}{4}\]

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

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

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