Question

решить возвратное уравнение​

Answer 1

$x^{4}+3x^{3}-8x^{2}+3x+1=0|:x^{2} \\\\x^{2}+3x-8+\frac{3}{x}+\frac{1}{x^{2}} =0\\\\(x^{2}+\frac{1}{x^{2}})+3(x+\frac{1}{x})-8=0\\\\x+\frac{1}{x}=m\Rightarrow x^{2}+\frac{1}{x^{2}}=m^{2} -2\\\\m^{2} -2+3m-8=0\\\\m^{2}+3m-10=0\\\\m_{1}=2\\\\m_{2}=-5$

$1)x+\frac{1}{x}=2\\\\\frac{x^{2}-2x+1 }{x}=0 \\\\\left \{ {{x^{2}-2x+1=0 } \atop {x\neq0 }} \right. \\\\(x-1)^{2}=0\\\\x_{1}=1\\\\2)x+\frac{1}{x}=-5\\\\\frac{x^{2}+5x+1 }{x}=0\\\\\left \{ {{x^{2}+5x+1=0 } \atop {x\neq0 }} \right.\\\\x^{2}+5x+1=0\\\\D=5^{2}-4*1=25-4=21\\\\x_{2}=\frac{-5-\sqrt{21}}{2}\\\\x_{3}=\frac{-5+\sqrt{21}}{2}\\\\Otvet:\boxed{1;\frac{-5-\sqrt{21}}{2};\frac{\sqrt{21}-5}{2}}$