Równanie wielomianowe
Veves: Rozwiąż równanie
x4−5x3+6x2−5x+1=0
16 lis 21:53
Adamm: t
2−2−5t+6=0
t
2−5t+4=0
t=1 lub t=4
x
2−x+1=0 lub x
2−4x+1=0
x=2+
√3 lub x=2−
√3
16 lis 21:58
Jack:
x
4 + 1 − 5x
3 − 5x + 6x
2 = 0
| 1 | | 1 | |
x2[(x2+ |
| ) − 5(x+ |
| ) + 6] = 0 |
| x2 | | x | |
| 1 | | 1 | |
x2[(x+ |
| )2 − 2 − 5*(x+ |
| )+6] = 0 |
| x | | x | |
| 1 | | 1 | |
x2[(x+ |
| )2 − 5(x+ |
| ) + 4] = 0 |
| x | | x | |
x
2[t
2 − 5t + 4] = 0
x
2[(t−4)(t−1)]
| 1 | | 1 | |
x2[(x+ |
| −4)(x+ |
| −1)]=0 |
| x | | x | |
(x
2+1−4x)(x
2+1−x)=0
(x
2−4x+1)(x
2−x+1)=0
x
2−4x+1 = 0 lub x
2−x+1=0
delta...
16 lis 22:01
Adamm: ha, szybszy
16 lis 22:02