Re: polynomial to solve with Maple
From: M. A. Fajjal (H2MAF_at_se.com.sa)
Date: 08/11/04
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Date: Wed, 11 Aug 2004 12:02:07 +0000 (UTC)
On 10 Aug 04 00:07:14 -0400 (EDT), M. A. Fajjal wrote:
>On Mon, 9 Aug 2004 20:50:18 -0400, Leo Harten wrote:
>>
>>"Edwin Clark" <eclark@math.usf.edu> wrote in message
>><a
>href="news://%KURc.38778$DZ.5543443@twister.tampabay.rr.com...">news://%KURc.38778$DZ.5543443@twister.tampabay.rr.com...>
>
>>> What the radical solution for
>>>
>>> 5 / 35 1/2\ / 1/2\
>>> X + |10 - I ---- * 2 |*X + |52 - I 5 * 2 | = 0
>>> \ 4 / \ /
>>>
>>>
>>> Where I = (-1)^(1/2)
>>
>>The format does not survive the posting. Could you type it in an
>input form
>>such as
>>
>>x^5+(10-sqrt(35)*%i/4)*x+(52-%i*sqrt(5)*2)=0
>>
>>so that we can all read it as intended?
>
>
>The polynomial to be solved is:
>x^5+(10-(35/4)*sqrt(2)*I)*x+52-5*sqrt(2)*I=0
>Where I = (-1)^(1/2)
>
>M. A. Fajjal
I have got the following solution
Root1=
((4*m+3)/(m^2+1)/5)^(1/5)*((1-2*m+2*(m^2+1)^(1/2)+(5-4*m+8*m^2+(500*(m^2+1)*(1-2*m)+(4*m+3)^3)/125/(1+m^2)^(1/2))^(1/2))^(1/5)+(1-2*m-2*(m^2+1)^(1/2)-(5-4*m+8*m^2-(500*(m^2+1)*(1-2*m)+(4*m+3)^3)/125/(1+m^2)^(1/2))^(1/2))^(1/5)*w^3+(1-2*m-2*(m^2+1)^(1/2)+(5-4*m+8*m^2-(500*(m^2+1)*(1-2*m)+(4*m+3)^3)/125/(1+m^2)^(1/2))^(1/2))^(1/5)*w^2+(1-2*m+2*(m^2+1)^(1/2)-(5-4*m+8*m^2+(500*(m^2+1)*(1-2*m)+(4*m+3)^3)/125/(1+m^2)^(1/2))^(1/2))^(1/5)*w^3)
m=1+sqrt(2)*I
w=(1/4)*(-1+sqrt(5)+sqrt(10+2*sqrt(5))*I)
Can anyone verify this solution by using Maple
M. A. Fajjal
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