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Topic review (newest first)

irspow
2006-02-06 02:43:18

Sorry, always looking for the easy way out.  The relationship that I found was obviously just applicable to this particular problem.

krassi_holmz
2006-02-05 18:05:16

and for ispow:
the full-root formula is VERY VERY larger than yours.

krassi_holmz
2006-02-05 18:04:02

but i just cant remember  where.

krassi_holmz
2006-02-05 18:02:24

The full roots formula includes non-real roots also.
i'have been posted it somewhere here.

Tigeree
2006-02-05 14:40:14

that euquled 0 hmm

irspow
2006-02-05 03:42:31

Nice, krassi_holmz!  Is this formula of the form;

1/a [ a/2 √( a/2 (c/6 - 2√c/3))] and

1/a [ a/2 √( a/2 (c/6 + 2√c/3))]

edit*

where of course ax^4 + bx^3 + cx^2 + dx + e = 0

krassi_holmz
2006-02-05 03:35:30

There's a formula, like the quadratic root formula.

RauLiTo
2006-02-04 22:23:28

i got that answers by using the excel but i want the way you get it algebraically.

krassi_holmz
2006-02-04 20:12:07

The exact roots using the root formula are:

krassi_holmz
2006-02-04 20:00:52

{
{x -> 0.33000947820757186759866712044837765639971206511454...},
{x -> 0.66999052179242813240133287955162234360028793488546...},
{x -> 0.06943184420297371238802675555359524745213731018514...},
{x -> 0.93056815579702628761197324444640475254786268981486...}
}
Interesting, all four roots are real.

Ricky
2006-02-04 08:48:24

In the end, we have developed technology which allows us to find such roots simply, there really is no need to drone away reinventing the wheel.

If you could find a root to a n'th degree polynomial in O(1) time (constant, not changing based on the values of the equation), then you could greatly speed up many different algorithms, especially for calculators and math software.

It would, indeed, be very useful.

irspow
2006-02-04 08:38:30

If you have never seen the quartic formula, then you need to see just how completely ridiculous it really is to believe it.  It is of absolutely no use to mere mortals.  Just Google the quartic formula to see what I mean.

  There are other methods out there describing "easier methods" that are derived from the quartic formula, but really they are not that much easier.

  You can even use four seesions of Newton's method with differing starting values, but again this is quite time consuming and not as exact as the above methods.

  In the end, we have developed technology which allows us to find such roots simply, there really is no need to drone away reinventing the wheel.  Maybe the algorithyms would be useful in a world destroyed, or you may just be curious as to how such a computation is possible, and there is nothing wrong with that.  But, if you are looking for a reasonable alternative to using technology for this particular purpose, don't waste your time.  There are an infinite amount of more important questions to be answered in this universe.

Ricky
2006-02-04 08:37:05

There is no easy mathimatical way to solve that equation.  You could use the quartic equation, but that's really messy.  The best way to solve those is to use a computer.

ryos
2006-02-04 08:33:43

As far as I can tell it can't be done algebraically. I'm guessing that this was assigned to teach the rational roots theorem? Unfortunately, I don't remember the theorem, just the name.

My calculator gives 4 roots between 0 and 1.

RauLiTo
2006-02-04 07:23:22

hi guys ... how you doing all of you ?
i have a equation and i want to show me how can we solve this type of equations ... i get the answers easily from the excel but i want to know how can we solve it in algebra way

70X^4-140X^3+90X^2-20X+1 = o

X = ??? dunno

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