a)
[tex]-x^{4}+7x^{2}-12 = 0 \ \ /\cdot(-1)\\\\x^{4}-7x^{2}+12 = 0\\\\Niech \ x^{2}=t \ \ i \ \ t \geq 0\\\\t^{2}-7t+12 = 0\\\\\Delta = (-7)^{2}-4\cdot1\cdot12 = 49-48 = 1\\\\\sqrt{\Delta} = \sqrt{1} = 1\\\\t_1 = \frac{7-1}{2} = \frac{6}{2} = 3\\\\x^{2} = 3\\\\(x+\sqrt{3})(x-\sqrt{3}) = 0\\\\\underline{x = -\sqrt{3} \ \ \vee \ \ x = \sqrt{3}}\\\\\\t_2 = \frac{7+1}{2} = \frac{8}{2} = 4\\\\x^{2} = 4\\\\x^{2}-4 = 0\\\\(x+2)(x-2) = 0\\\\\underline{x = -2 \ \ \vee \ \ x = 2}[/tex]
b)
[tex]9x^{4}+17x^{2}-2 = 0\\\\x^{2}=t \ \ i \ \ t \geq 0\\\\9t^{2}+17t-2 = 0\\\\\Delta = 17^{2}-4\cdot9\cdot(-2)= 289+72 = 361\\\\\sqrt{\Delta} = \sqrt{361}= 19\\\\t_1 = \frac{-17-19}{18} = \frac{-36}{18} = -2 \ <0, \ odrzucamy\\\\t_2 = \frac{-17+19}{18} = \frac{2}{18} = \frac{1}{9}\\\\x^{2} = \frac{1}{9}\\\\x^{2}-\frac{1}{9} = 0\\\\(x+\frac{1}{3})(x-\frac{1}{3}) = 0\\\\\underline{x = -\frac{1}{3} \ \ \vee \ \ x = \frac{1}{3}}[/tex]
