Question

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

Answer 1

$t^4+8t^3+6t^2-8t+1=0|:t^2\t^2+8t+6-frac{8}{t} +frac{1}{t^2}=0\t^2+frac{1}{t^2}+8(t-frac{1}{t})+6=0\ t-frac{1}{t}=xRightarrow t^2+frac{1}{t^2}=x^2+2\x^2+2+8x+6=0\x^2+8x+8=0\D_{1}=(frac{b}{2})^2-ac=4^2-8=16-8=8,\ x_{1,2}=frac{-frac{b}{2}pmsqrt{D_1} }{a}=frac{-4pmsqrt{8}}{1}=-4pm2sqrt2$

$t-frac{1}{t}=-4pm2sqrt2\left [ {{t-frac{1}{t} =-4+2sqrt2} atop {t-frac{1}{t} =-4-2sqrt{2}}} right. \left [ {{t^2-1=(-4+2sqrt{2})t} atop {t^2-1=(-4-2sqrt{2})t}} right.\left [ {{t^2+(4-2sqrt{2})t-1=0(1)} atop {t^2+(4+2sqrt{2})t-1=0(2)}} right. \(1)D=(4-2sqrt{2})^2+4=16-16sqrt2+8+4=28-16sqrt2=4(7-4sqrt2)\t_{1,2}=frac{-(4-2sqrt2)pmsqrt{4(7-4sqrt2)} }{2} =frac{-4+2sqrt2pm2sqrt{7-4sqrt2} }{2}=-2+sqrt2pmsqrt{7-4sqrt2}.$

$(2)D=(4+2sqrt2)^2+4=16+16sqrt2+8+4=28+16sqrt2=4(7+4sqrt2)\t_{3,4}=frac{-(4+2sqrt2)pmsqrt{4(7+4sqrt2)} }{2}=frac{-4-2sqrt2pm2sqrt{7+4sqrt2}}{2} =-2-sqrt2pm sqrt{7+4sqrt2}$

ОТВЕТ: $-2+sqrt2pmsqrt{7-4sqrt2}, -2-sqrt2pmsqrt{7+4sqrt2}$