From: Jon on
The root to ax^5+bx+c=0 is approximately,

x = { -c/({a^2+b^2}^(1/2)) }^(1/5)

Using this formula, the roots to,

x^5+x-34=0 x=1.888 should be 2
32x^5+4x-3=0 x=0.623 should be 1/2
x^5+x-0.10001 x=0.5887 should be 1/10


From: KY on
cf;
http://www.wolframalpha.com/input/?i=32x%5E5%2B4x-3%3D0

etc
From: Sam Wormley on
Jon wrote:
> The root to ax^5+bx+c=0 is approximately,
>
> x = { -c/({a^2+b^2}^(1/2)) }^(1/5)
>
> Using this formula, the roots to,
>
> x^5+x-34=0 x=1.888 should be 2
> 32x^5+4x-3=0 x=0.623 should be 1/2
> x^5+x-0.10001 x=0.5887 should be 1/10
>
>

http://www.wolframalpha.com/input/?i=roots+ax%5E5%2Bbx%2Bc%3D0
From: master1729 on
Sam Wormley wrote :

> Jon wrote:
> > The root to ax^5+bx+c=0 is approximately,
> >
> > x = { -c/({a^2+b^2}^(1/2)) }^(1/5)
> >
> > Using this formula, the roots to,
> >
> > x^5+x-34=0 x=1.888 should be 2
> > 32x^5+4x-3=0 x=0.623 should be 1/2
> > x^5+x-0.10001 x=0.5887 should be 1/10
> >
> >
>
>
>
> http://www.wolframalpha.com/input/?i=roots+ax%5E5%2Bb
> x%2Bc%3D0

hahaha LMFAO

this is the ' result ' of wolfram alpha = the link given above :

Results:

x = root[#1^5a + #1 b + c&,1] and a =/= 0
x = root[#1^5a + #1 b + c&,2] and a =/= 0
x = root[#1^5a + #1 b + c&,3] and a =/= 0
x = root[#1^5a + #1 b + c&,4] and a =/= 0
x = root[#1^5a + #1 b + c&,5] and a =/= 0

a = 0 b = 0 c = 0

a = 0 x = -c/b b =/= 0

hahaha so trivial.

is this the best math sci.math (posters) has to offer ?

some link to some lame result ???

LMFAO

wolfram alpha = " computational knowledge engine "

hahahahaha

no offense , but this is hilaric.


tommy1729
From: hagman on
On 6 Dez., 03:20, "Jon" <jon8...(a)peoplepc.com> wrote:
> The root to ax^5+bx+c=0 is approximately,
>
> x = { -c/({a^2+b^2}^(1/2)) }^(1/5)
>
> Using this formula, the roots to,
>
> x^5+x-34=0  x=1.888  should be 2
> 32x^5+4x-3=0 x=0.623 should be 1/2
> x^5+x-0.10001 x=0.5887 should be 1/10

That approximation is exact if either b=0 or c=0, but as your examples
show of doubtful quality in the general case