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Formula for quintics or higher

Name: Anonymous 2006-05-29 19:23

Show that you can't get a formula to find the roots of quintics or higher.

Name: Anonymous 2006-05-30 10:35

If there were, it wouild've been found by now. It hasn't been. Therefore there is none. qed.

Name: Anonymous 2006-05-30 10:37

use group theory and GG

Name: Anonymous 2006-05-30 10:46

Use Galois theory to solve quintics. Or, altneratively

Difficulty of finding solution to:

Linear: Practice
Quadratic: Normal
Cubic: Nightmare
Quartic: Hell

There is no difficulty higher than Hell, therefore quintics cannot be solved using a formula. QED.

Name: Anonymous 2006-05-30 12:09 (sage)

oh lawd is dat sum Abel–Ruffini theorem

Name: Anonymous 2006-05-30 12:14

>>1
That could be solved if we assume 2=1

Name: Anonymous 2006-06-02 21:14

Is OP asking question or just showing that he paid attention to the introduction of sum algebra course?

Name: Anonymous 2006-06-03 9:16

Keep differentiating. Look for turning points.

Name: Anonymous 2006-06-03 10:59

technichally if it's a perfect ...quint(?)... all the roots should be a the same place, so you could use an easy first and second derivative test.  if this isn't the case, yeah, try >>8

Name: Anonymous 2006-06-03 13:29

loli haet Abel-Ruffini

My algebra textbook has an entire chapter (nearly a hundred pages) devoted to this proof. It's called "A First Course in Abstract Algebra (second edition)" by Joseph J. Rotman. It's a great book; if you really want to know, get your ass to your library and look it up.

Name: ciel- 2006-06-03 22:30

While we're here, anyone mind going through Hell and figuring out the roots for "(x^4)-(3x^3)-(22x^2)-(48x)-(32)=0"?

Name: Anonymous 2006-06-03 23:00

>>11
3±√17 and (-3±i√7)/2.

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