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footmath
Apr22-11, 12:17 PM
how can i solve this equation :
x^5+ax+b

tiny-tim
Apr22-11, 12:32 PM
hi footmath! :wink:
how can i solve this equation :
x^5+ax+b

guessin, numerical approximation, or bribing the TA …

otherwise, you can't! :smile:

LCKurtz
Apr22-11, 12:35 PM
how can i solve this equation :
x^5+ax+b

That isn't an equation; it is just an expression. Perhaps you mean find the roots of it or solve the equation x5+ax+b = 0.

Either way you are pretty much out of luck. The quintic equation is not generally solvable although some special cases are. Whether this one is solvable generally or not, I don't know. But if you have a particular equation in mind so you know a and b, you might be able to solve it if you are lucky or you could solve it numerically.

JJacquelin
Apr23-11, 12:30 AM
The solutions (real and/or complex) of the quintic equation x5+ax+b = 0 are computed thank to a number of methods of numerical calculus.
Analytical solving is solving is possible in case of some particular values of corfficients a and b.
In the general case, the solutions cannot be expressed in terms of a conbination of a finite number of elementary or usal functions. Special functions are necessary : the solutions can be expessed in terms of Jacobi theta function (which is of no use in practice).

footmath
Apr24-11, 10:23 AM
thank you .
what do you think about this equation : x^6+x^2+x=y
if you believe that it can not be Solvable prove it please .

JJacquelin
Apr24-11, 12:57 PM
First, would you mind give your precise definition of "Solvable".

gb7nash
Apr24-11, 01:17 PM
Also, does the OP have any experience in group theory? To show a certain quintic is unsolvable requires use of Galois Theory.

LCKurtz
Apr24-11, 02:36 PM
thank you .
what do you think about this equation : x^6+x^2+x=y
if you believe that it can not be Solvable prove it please .

If what you mean by "solveable" is to find the roots, then yes, because that particular polynomial factors.

x6+x2+x = x(x2+x+1)(x3-x2+1)

footmath
May5-11, 11:51 AM
I want inverse of f(x)=x^5+ax+b

LCKurtz
May5-11, 12:44 PM
I want inverse of f(x)=x^5+ax+b

If a < 0 the function is not 1-1 and has no single valued inverse. If a > 0, at least it is increasing and has an inverse. But good luck with finding a formula for it.

JJacquelin
May9-11, 09:51 AM
I want inverse of f(x)=x^5+ax+b
The solutions (real and/or complex) of the quintic equation x5+ax+b = 0 are computed thank to a number of methods of numerical calculus.
Analytical solving is solving is possible in case of some particular values of corfficients a and b.
In the general case, the solutions cannot be expressed in terms of a conbination of a finite number of elementary or usal functions. Special functions are necessary : the solutions can be expessed in terms of Jacobi theta function (which is of no use in practice).

JJacquelin
May9-11, 10:12 AM
If you really want an analytical expression for the inverse quintic function, you can find it in :
http://mathworld.wolfram.com/QuinticEquation.html
(Very complicated : Requiring knowledge of the Jacobi theta functions)

footmath
May16-11, 09:02 AM
if x^5+ax+b cannot solve by radical you can see brings method
http://en.wikipedia.org/wiki/Bring_radical#Bring_radicals