Question

Решить возвратное уравнение: х4 – 2х3 – 22х2 – 2х +1=0;

Answer 1

$\displaystyle\bf\\x^{4} -2x^{3} -22x^{2} -2x+1=0 \ | \ :x^{2} \\\\x^{2} -2x-22-\frac{2}{x} +\frac{1}{x^{2} } =0\\\\\Big(x^{2}+\frac{1}{x^{2} } \Big)-2 \Big(x+\frac{1}{x} \Big)-22=0\\\\x+\frac{1}{x} =m \ \ \Rightarrow \ \ x^{2} +\frac{1}{x^{2} } =m^{2} -2\\\\m^{2} -2-2m-22=0\\\\m^{2} -2m-24=0\\\\Teorema \ Vieta:\\\\m_{1} =-4 \ ; \ m_{2} =6$

$\displaystyle\bf\\1)\\\\x+\frac{1}{x} =-4\\\\x^{2} +4x+1=0 \ ; \ x\neq 0\\\\D=4^{2} -4\cdot 1=16-4=12=(2\sqrt{3} )^{2} \\\\x_{1} =\frac{-4-2\sqrt{3} }{2} =-2-\sqrt{3} \\\\x_{2} =\frac{-4+2\sqrt{3} }{2}=\sqrt{3} -2\\\\2)\\\\x+\frac{1}{x}=6 \\\\x^{2} -6x+1=0\\\\D=(-6)^{2} -4\cdot1=36-4=32=(4\sqrt{2} )^{2} \\\\x_{3} =\frac{6-4\sqrt{2} }{2} =3-2\sqrt{2}\\\\x_{4}=\frac{6+4\sqrt{2} }{2} =3+2\sqrt{2}$