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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
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x³ - x - 1 = 0

x³ - x = 1

x(x²-1)=1

x = 1/(x²-1)

Does this cubic equation have a real solution?

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**Ricky****Moderator**- Registered: 2005-12-04
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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: 1,905

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

x≈1.32471795724474602596090885448...

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**Ricky****Moderator**- Registered: 2005-12-04
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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
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No. Do you REALLY want it?

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**Ricky****Moderator**- Registered: 2005-12-04
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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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**ganesh****Moderator**- Registered: 2005-06-28
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I did not know the roots of a cubic equation and looked for it on the net. I got to this pdf. Ricky, you can find the root of the incomplete cubic equation using the formulae.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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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
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Cubic equation:

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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And quadric equation (useless):

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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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
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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
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this is the exact root!

krassi_holmz wrote:

For you question we get:

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**ganesh****Moderator**- Registered: 2005-06-28
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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?

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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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
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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
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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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