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zetafunc.

Guest

quadratic, cubic, quartic, nth degree... ? help

the solution for k^2 - k - 1 = 0 is k = phi and k = (1-phi).

How do I solve k^3 - k^2 - k - 1 = 0? How many roots of this equation are there? Is there a better way of solving this equation without using the cubic formula?

and what about k^4 - k^3 - k^2 - k - 1 = 0? or k^7 - k^6 - k^5 - k^4 - k^3 - k^2 - k - 1 = 0?

How to solve these?

Is there a pattern for each root of each equation?

Help.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: quadratic, cubic, quartic, nth degree... ? help

Hi zetafunc;

Yes, there are better methods to solve all polynomials higher than quadratics. You need some background. Go here.

http://www.mathsisfun.com/algebra/polynomials.html

http://www.mathsisfun.com/algebra/polyn … lving.html

Is there a pattern for each root of each equation?

Yes, I believe there is: The dominant real root approaches 2.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

zetafunc.

Guest

Re: quadratic, cubic, quartic, nth degree... ? help

how do I write that the dominant real root approaches 2 algebraically?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: quadratic, cubic, quartic, nth degree... ? help

Hi zetafunc;

Before we write it don't you think we should prove it? It is just an offhand conjecture of mine.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: quadratic, cubic, quartic, nth degree... ? help

bobby:

This fact once becomes obvious once you think of numbers in binary form:

1111111 + 1 = 10000000

The immediate corollary is your proposition.

zetafunc: p(k) = k^7 - k^6 - k^5 - k^4 - k^3 - k^2 - k - 1 is an irreducible polynomial.  This means that if there are two polynomials f(k) and g(k) with integer coefficients such that f * g = p, then either f = ±1 or g = ±1.

Further, p(k) has only one real root, easily verified by graphing.  An elementary result from these two facts from Galois theory is that the roots of p(x) can not be expressed in terms of +, -, ×, ÷, and nth roots alone.  Such roots are called insolvable.

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"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..."

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: quadratic, cubic, quartic, nth degree... ? help

Hi Rick;

Thanks, for providing that.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.