[tex]4x^{4}+9 = 37x^{2}\\\\4x^{4}-37x^{2}+9 = 0\\\\Niech \ \ t = x^{2}\\\\4x^{2}-37x+9 = 0\\\\\Delta = b^{2}-4ac = (-37)^{2} - 4\cdot4\cdot9 = 1369 - 144 = 1225\\\\\sqrt{\Delta} = \sqrt{1225} = 35\\\\t_1 = \frac{37-35}{8} = \frac{2}{8} = \frac{1}{4}\\\\t_2 = \frac{37+35}{8} = \frac{72}{8} = 9\\\\\\x^{2} = \frac{1}{4}\\\\x^{2}-\frac{1}{4} = 0\\\\(x+\frac{1}{2})(x-\frac{1}{2}) = 0\\\\x = -\frac{1}{2}\vee \ x = \frac{1}{2}\\\\\\x^{2} = 9\\\\x^{2}-9 = 0\\\\(x+3)(x-3) = 0\\\\x = -3 \ \vee \ x = 3[/tex]
Równanie ma cztery rozwiązania:
[tex]\underline{x \in\{-3, -\frac{1}{2}, \frac{1}{2}, 3\}}[/tex]
