Question

解方程：t²+（2-t）²=t²+4t+8-2×（16t+12t²+4t的三次方+t的四次方）+4+t²

Answer 1

t^2+(2-t)^2=t^2+4t+8-2(16t+12t^2+4t^3+t^4)+4+t^2
t^2+4-4t+t^2=t^2+4t+8-32t-24t^2-8t^3-2t^4+4+t^2
2t^4+8t^3+24t^2+24t-8=0
t^4+4t^3+12t^2+12t-4=0
t^4+4t^3+12t^2+12t-4=0
方程左边没办法因式分解，请核对原方程是否有误？

Answer 2

2t²-4t+4=t²+4t+8-32t-24t²-8t^3-2t^4+4+t²
t^4+4t^3+12t²+12t-4=0
(t²+2t)^2=-8t²-12t+8
(t²+2t+x)^2=2[(x-4)t²+(2x-6)t+(x²+4)/2]
令右式Delta=0,
x^3-6x^2+16x-34=0 令x=y+2
y^3+4y-18=0, y=2.12, x=4.12
(t²+2t+4.12)=+/-2^0.5[0.12t^2+2.12t+10.4872]
t(1,2)=-1.7707, 0.2597
t(3,4)= -1.245 +/- 2.674i

Answer 3

确实是无解！

Answer 4

无解