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Science Forum Index » Mathematics Forum » Solving equation with a power
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| Francogrex |
 Posted: Fri Apr 25, 2008 3:47 am |
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Hi Probably a very basic question, excuse my lack of knowledge:
Is this equation involving a power solvable algebraically at all? x^n
- x·z = 0 I want to solve for x
SOLVE(x^n - x·z = 0, x, Real)
I can't seem to find a solution or even an approximation. Thanks |
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| The Qurqirish Dragon |
| Posted: Fri Apr 25, 2008 4:01 am |
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On Apr 25, 9:47 am, Francogrex <fra...@grex.org> wrote:
Quote: Hi Probably a very basic question, excuse my lack of knowledge:
Is this equation involving a power solvable algebraically at all? x^n
- x·z = 0 I want to solve for x
SOLVE(x^n - x·z = 0, x, Real)
I can't seem to find a solution or even an approximation. Thanks
x^n - xz = x * (x^(n-1) - z), so x = 0 is a solution. Whether or not
there are other solutions depends on whether or not you allow complex
results. If z is a positive real number, then there is either 1 or 2
more real solutions (depending on n being even or odd). If z is a
negative real number, then there is either 0 or 1 more real solution
(again, depending on the parity of n). If you allow complex numbers,
then the other n-1 solutions are the (n-1)th roots of z, regardless of
the types of numbers z and n are.
In any case, they can all be found algebraically. The general n-th
order polynomial cannot be solved algebraically if n>4, but your
specific example can, for any n. |
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| bill |
| Posted: Fri Apr 25, 2008 7:10 am |
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On Apr 25, 7:01 am, The Qurqirish Dragon <qurqiri...@aol.com> wrote:
Quote: On Apr 25, 9:47 am, Francogrex <fra...@grex.org> wrote:
Hi Probably a very basic question, excuse my lack of knowledge:
Is this equation involving a power solvable algebraically at all? x^n
- x·z = 0 I want to solve for x
SOLVE(x^n - x·z = 0, x, Real)
I can't seem to find a solution or even an approximation. Thanks
x^n - xz = x * (x^(n-1) - z), so x = 0 is a solution. Whether or not
there are other solutions depends on whether or not you allow complex
results. If z is a positive real number, then there is either 1 or 2
more real solutions (depending on n being even or odd). If z is a
negative real number, then there is either 0 or 1 more real solution
(again, depending on the parity of n). If you allow complex numbers,
then the other n-1 solutions are the (n-1)th roots of z, regardless of
the types of numbers z and n are.
In any case, they can all be found algebraically. The general n-th
order polynomial cannot be solved algebraically if n>4, but your
specific example can, for any n.
Why isn't the solution just "x ^ (n-1) = z"?
Bill J |
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| bassam king karzeddin |
| Posted: Fri Apr 25, 2008 8:11 am |
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| [Mr.] Lynn Kurtz |
| Posted: Fri Apr 25, 2008 12:36 pm |
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On Fri, 25 Apr 2008 10:10:15 -0700 (PDT), bill <b92057@yahoo.com>
wrote:
Quote: On Apr 25, 7:01 am, The Qurqirish Dragon <qurqiri...@aol.com> wrote:
On Apr 25, 9:47 am, Francogrex <fra...@grex.org> wrote:
Hi Probably a very basic question, excuse my lack of knowledge:
Is this equation involving a power solvable algebraically at all? x^n
- x·z = 0 I want to solve for x
SOLVE(x^n - x·z = 0, x, Real)
I can't seem to find a solution or even an approximation. Thanks
x^n - xz = x * (x^(n-1) - z), so x = 0 is a solution. Whether or not
there are other solutions depends on whether or not you allow complex
results. If z is a positive real number, then there is either 1 or 2
more real solutions (depending on n being even or odd). If z is a
negative real number, then there is either 0 or 1 more real solution
(again, depending on the parity of n). If you allow complex numbers,
then the other n-1 solutions are the (n-1)th roots of z, regardless of
the types of numbers z and n are.
In any case, they can all be found algebraically. The general n-th
order polynomial cannot be solved algebraically if n>4, but your
specific example can, for any n.
Why isn't the solution just "x ^ (n-1) = z"?
Bill J
Aside from the fact that your expression isn't explicitly solved for
x, the original equation is satisfied for x = 0 no matter what z is,
and your equation isn't satisfied for x = 0 if z isn't zero.
--Lynn |
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| Tim Little |
| Posted: Fri Apr 25, 2008 11:12 pm |
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On 2008-04-25, Francogrex <franco@grex.org> wrote:
Quote: Hi Probably a very basic question, excuse my lack of knowledge:
Is this equation involving a power solvable algebraically at all?
x^n - x z = 0 I want to solve for x
It easily factorizes to
x (x^(n-1) - z) = 0,
which means x = 0 or x^(n-1) = z.
If x has to be real, then the second equation has 0, 1, or 2 solutions
depending upon whether n is odd/even, and whether z is positive/negative.
If x may be complex, then it will have (n-1) solutions except when
z = 0.
- Tim |
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| Francogrex |
| Posted: Sat Apr 26, 2008 4:57 am |
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Thanks all, I tried this seems to be giving me a sensible result (at
least when you try it out with some examples) with this estimation:
x=(1/z)^(1/(1 - n))
On Apr 26, 6:12 am, Tim Little <t...@soprano.little-possums.net>
wrote:
Quote: It easily factorizes to
x (x^(n-1) - z) = 0,
which means x = 0 or x^(n-1) = z.
If x has to be real, then the second equation has 0, 1, or 2 solutions
depending upon whether n is odd/even, and whether z is positive/negative.
If x may be complex, then it will have (n-1) solutions except when
z = 0.
- Tim |
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