Question

Rozwiąż równania /załącznik/

Answer 1

b)

[tex]x^{2}(x+5)(2x^{2}+3) = 0\\\\x^{2}=0 \ \vee \ x+5 = 0 \ \vee \ 2x^{2}+3 > 0\\\\x = 0 \ \vee \ x = -5\\\\\boxed{x \in\{-5,0\}}[/tex]

c)

[tex]-x^{4}-9x^{2}+10 = 0 \ \ \ |\cdot(-1)\\\\x^{4}+9x^{2}-10 = 0\\\\Niech \ \ t = x^{2}\\\\t^{2}+9t-10 = 0\\\\\Delta = b^{2}-4ac = 9^{2}-4\cdot1\cdot(-10) = 81+40 = 121\\\\\sqrt{\Delta} = \sqrt{121} = 11\\\\t_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{-9-11}{2\cdot1} = \frac{-20}{2} = -10\\\\t_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{-9+11}{2} = \frac{2}{2} = 1\\\\t_1 = x^{2}\neq -10\\\\t_2 = x^{2}=1\\\\x = -1 \ \vee \ x = 1\\\\\boxed{x\in\{-1,1\}}[/tex]