Решите уравнение: x^4-3x^2+27/16=0.

\[x^{4} - 3x^{2} + \frac{27}{16} = 0\ \ \ \ \ \ | \cdot 16\]

\[16x^{4} - 48x^{2} + 27 = 0\]

\[t = x^{2};\ \ \ \ \ t \geq 0.\]

\[16t^{2} - 48t + 27 = 0\]

\[D = ( - 48)^{2} - 4 \cdot 16 \cdot 27 =\]

\[= 2304 - 1728 = 576;\ \ \ \ \]

\[\sqrt{D} = 24.\]

\[t_{1} = \frac{48 + 24}{2 \cdot 16} = \frac{72}{32} = \frac{9}{4};\ \ \ \ \ \]

\[\text{\ \ }t_{2} = \frac{48 - 24}{2 \cdot 16} = \frac{24}{32} = \frac{3}{4}\]

\[Ответ:\ \pm 1,5;\ \ \pm \frac{\sqrt{3}}{2}.\]