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**A cubic...**

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**God****Member**- Registered: 2005-08-25
- Posts: 59

Can the cubic 0 = x^3 - x - 1 be solved by means of simple algebra (without the cubic formula)?

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 17,009

x³ - x - 1 = 0

x³ - x = 1

x(x²-1)=1

x = 1/(x²-1)

Does this cubic equation have a real solution?

Character is who you are when no one is looking.

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

No, it can't be factored. The only real factor is a real number x~1.32. So any attempts to factor it would be futile.

"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: 1,906

It has one real root, but I used the cubic formula:

x≈1.32471795724474602596090885448...

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

So what is the exact root of it, krassi?

"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: 1,906

No. Do you REALLY want it?

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

If you use the cubic formula, shouldn't you come up with the exact root?

"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: 1,906

The roots of quadratic equation ax^2+bx+c:

*Last edited by krassi_holmz (2006-01-04 22:08:31)*

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,906

Cubic equation:

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,906

And quadric equation (useless):

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,906

For you question we get:

*Last edited by krassi_holmz (2006-01-04 22:12:44)*

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,906

Equations >4 are unsolvable exactin radicals. (Galoa)

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**krisper****Member**- Registered: 2005-12-20
- Posts: 19

I am sure you could find this without using the cubic formula.

x^3 - x - 1 = 0

x(x^2-1) = 1

if x = 0, the equasion has no meaning, so x must differ from 0 (x<>0); Now we can devide by x.

x^2 - 1 = 1/x.

We draw the graphics of these two expressions - x^2 - 1 and 1/x. After that we check where they cross eachother. This happens only in I quadrant which means that x^3 - x - 1 = 0 has only one real root. Afterwards doing some calculas we find that the root is somewhere between 1 and √2. Now we have to use the tangents and make some calculations and I am sure we will get the exact value of this root.

*Last edited by krisper (2006-01-04 22:16:42)*

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,906

this is the exact root!

krassi_holmz wrote:

For you question we get:

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 17,009

Very good posts and images uploaded, krassi_holmz. Those would be very useful. BTW, did you know the insolubility of the quintic equation was shown first by Neils Henrik Abel, the Norwegain Mathematician?

Character is who you are when no one is looking.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,906

Yes, I've read very much about Abel.

And the Galous theory simplifies the result.

The solubility of an equation depends of the structude if its Galous group.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,906

The general quintic can be solved in terms of Jacobi theta functions, as was first done by Hermite in 1858. Kronecker subsequently obtained the same solution more simply, and Brioschi also derived the equation.

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**God****Member**- Registered: 2005-08-25
- Posts: 59

Lol.thanks... wow... that looks so complicated lol

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,906

That's why I prefer the approximated result.

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**God****Member**- Registered: 2005-08-25
- Posts: 59

I guess you might as well just use Newton's recursion for these things...

*Last edited by God (2006-01-08 09:41:58)*

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