Question

Решите пожалуйста уравнение.
t^4+8t^3+6t^2-8t+1=0​

Answer 1

$t^4+8t^3+6t^2-8t+1=0|:t^2\\t^2+8t+6-\frac{8}{t} +\frac{1}{t^2}=0\\t^2+\frac{1}{t^2}+8(t-\frac{1}{t})+6=0\\ t-\frac{1}{t}=x\Rightarrow t^2+\frac{1}{t^2}=x^2+2\\x^2+2+8x+6=0\\x^2+8x+8=0\\D_{1}=(\frac{b}{2})^2-ac=4^2-8=16-8=8,\\ x_{1,2}=\frac{-\frac{b}{2}\pm\sqrt{D_1} }{a}=\frac{-4\pm\sqrt{8}}{1}=-4\pm2\sqrt2$

$t-\frac{1}{t}=-4\pm2\sqrt2\\\left \ [ {{t-\frac{1}{t} =-4+2\sqrt2} \atop {t-\frac{1}{t} =-4-2\sqrt{2}}} \right. \\\left \ [ {{t^2-1=(-4+2\sqrt{2})t} \atop {t^2-1=(-4-2\sqrt{2})t}} \right.\\\left \ [ {{t^2+(4-2\sqrt{2})t-1=0(1)} \atop {t^2+(4+2\sqrt{2})t-1=0(2)}} \right. \\(1)D=(4-2\sqrt{2})^2+4=16-16\sqrt2+8+4=28-16\sqrt2=4(7-4\sqrt2)\\t_{1,2}=\frac{-(4-2\sqrt2)\pm\sqrt{4(7-4\sqrt2)} }{2} =\frac{-4+2\sqrt2\pm2\sqrt{7-4\sqrt2} }{2}=-2+\sqrt2\pm\sqrt{7-4\sqrt2}.$

$(2)D=(4+2\sqrt2)^2+4=16+16\sqrt2+8+4=28+16\sqrt2=4(7+4\sqrt2)\\t_{3,4}=\frac{-(4+2\sqrt2)\pm\sqrt{4(7+4\sqrt2)} }{2}=\frac{-4-2\sqrt2\pm2\sqrt{7+4\sqrt2}}{2} =-2-\sqrt2\pm \sqrt{7+4\sqrt2}$

ОТВЕТ: $-2+\sqrt2\pm\sqrt{7-4\sqrt2}, -2-\sqrt2\pm\sqrt{7+4\sqrt2}$