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**RauLiTo****Member**- From: Bahrain
- Registered: 2006-01-11
- Posts: 142

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 = ???

ImPo$$!BLe = NoTH!nG

Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...

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**ryos****Member**- Registered: 2005-08-04
- Posts: 394

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.

El que pega primero pega dos veces.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

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.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

{

{x -> 0.33000947820757186759866712044837765639971206511454...},

{x -> 0.66999052179242813240133287955162234360028793488546...},

{x -> 0.06943184420297371238802675555359524745213731018514...},

{x -> 0.93056815579702628761197324444640475254786268981486...}

}

Interesting, all four roots are real.

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

The exact roots using the root formula are:

*Last edited by krassi_holmz (2006-02-03 21:20:49)*

IPBLE: Increasing Performance By Lowering Expectations.

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**RauLiTo****Member**- From: Bahrain
- Registered: 2006-01-11
- Posts: 142

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

ImPo$$!BLe = NoTH!nG

Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

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

IPBLE: Increasing Performance By Lowering Expectations.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

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

*Last edited by irspow (2006-02-04 04:44:13)*

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**Tigeree****Member**- Registered: 2005-11-19
- Posts: 13,883

that euquled 0

~ Anton Chekhov

Cheer up, emo kid.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

The full roots formula includes non-real roots also.

i'have been posted it somewhere here.

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

but i just cant remember where.

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

and for ispow:

the full-root formula is VERY VERY larger than yours.

IPBLE: Increasing Performance By Lowering Expectations.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

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

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