Question

1) 4x⁴ - 5x² + 1 = 0
2) 9x⁴ - 9x² + 2 = 0
3) (x² + x - 3)² - 12 (x² + x - 3) + 27 = 0​

Answer 1

$\displaystyle\bf\\1)\\\\4x^{4} -5x^{2} +1=0\\\\x^{2} =m \ \ , \ \ m\geq 0\\\\4m^{2} -5m+1=0\\\\D=(-5)^{2} -4\cdot4\cdot 1=25-16=9=3^{2} \\\\\\m_{1} =\frac{5+3}{8} =1\\\\m_{2}=\frac{5-3}{8} =\frac{1}{4} =0,25 \\\\x^{2} =1\\\\x_{1} =-1 \ \ ; \ \ x_{2} =1\\\\\\x^{2} =0,25\\\\x_{3}=-0,5 \ \ ; \ \ x_{4} =0,5\\\\Otvet: \ -1 \ ; \ 1 \ ; \ -0,5 \ ; \ 0,5$

$\displaystyle\bf\\2)\\\\9x^{4} -9x^{2} +2=0\\\\x^{2} =m \ \ , \ \ m\geq 0\\\\9m^{2} -9m+2=0\\\\D=(-9)^{2} -4\cdot9\cdot 2=81-72=9=3^{2} \\\\\\m_{1} =\frac{9+3}{18} =\frac{2}{3} \\\\m_{2}=\frac{9-3}{18} =\frac{1}{3} \\\\\\x^{2} =\frac{2}{3} \\\\x_{1} =-\sqrt{\frac{2}{3} } =-\frac{\sqrt{6} }{3} \\\\x_{2} =\sqrt{\frac{2}{3} } =\frac{\sqrt{6} }{3} \\\\\\x^{2} =\frac{1}{3} \\\\x_{3} =-\sqrt{\frac{1}{3} } =-\frac{\sqrt{3} }{3} \\\\x_{4} =\sqrt{\frac{1}{3} } =\frac{\sqrt{3} }{3}$

$\displaystyle\bf\\Otvet: \ -\frac{\sqrt{6} }{3} \ ; \ \frac{\sqrt{6} }{3} \ ; \ -\frac{\sqrt{3} }{3} \ ; \ \frac{\sqrt{3} }{3} \\\\\\3)\\\\(x^{2} +x-3)^{2} -12(x^{2} +x-3)+27=0\\\\x^{2} +x-3=m\\\\m^{2}-12m+27=0\\\\Teorema \ Vieta:\\\\m_{1}+ m_{2} =12\\\\m_{1}\cdot m_{2} =27\\\\m_{1}=3\\\\ m_{2} =9\\\\\\x^{2} +x-3=3\\\\x^{2} +x-6=0\\\\Teorema \ Vieta:\\x_{1}+ x_{2} =-1\\\\x_{1}\cdot x_{2} =-6\\\\x_{1} =-3\\\\x_{2} =2\\\\\\x^{2} +x-3=9\\\\x^{2} +x-12=0$

$\displaystyle\bf\\Teorema \ Vieta:\\\\x_{1}+ x_{2} =-1\\\\x_{1}\cdot x_{2} =-12\\\\x_{1} =-4\\\\x_{2} =3\\\\\\Otvet: \ -3 \ ; \ 2 \ ; \ -4 \ ; \ 3$