[Date Index] [Thread Index] [Author Index]

Solving the Quintic: Correction to x^5 + 20 x + 32 == 0 root example

• To: mathgroup at smc.vnet.net
• Subject: [mg95311] Solving the Quintic: Correction to x^5 + 20 x + 32 == 0 root example
• From: "Q.E.D." <aoe at netzero.net>
• Date: Thu, 15 Jan 2009 06:10:40 -0500 (EST)
• Organization: Adtech Computers, Inc.

There is an error on one of the "Solving the Quintic"
pages. It is necessary to change a minus sign to a plus
sign to get a correct N[%] value -1.363962165089939`
The change is marked below with the comment (*fixed*).

The "Solving the Quintic" page in error is
http://library.wolfram.com/examples/quintic/main.html

The page shows the following:

There are polynomials of degree greater than four which
do not factor over the rationals, but that can still be
solved in radicals. An example is the quintic:

x^5 + 20 x + 32 == 0

one of whose roots is

1/5 (-(2500 Sqrt[5] + 250 Sqrt[50 - 10 Sqrt[5]] -
     750 Sqrt[50 + 10 Sqrt[5]])^(1/5) -
(-2500 Sqrt[5] + 750 Sqrt[50 - 10 Sqrt[5]] +
     250 Sqrt[50 + 10 Sqrt[5]])^(1/5) +(*fixed*)
(2500 Sqrt[5] + 750 Sqrt[50 - 10 Sqrt[5]] +
     250 Sqrt[50 + 10 Sqrt[5]])^(1/5) -
(2500 Sqrt[5] - 250 Sqrt[50 - 10 Sqrt[5]] +
     750 Sqrt[50 + 10 Sqrt[5]])^(1/5))

The corrected result can be checked using:

N[#^5 + 20 # + 32 == 0 &[%], 100]

Can anyone who has the "Solving the Quintic" poster see
if it is correct on there?

All of the roots are given by evaluating this:

Module[{prim1, prim2, prim3, prim4, r1, r2, r3, r4},
 r1 = -(2500*5^(1/2) + 250* (50 - 10*5^(1/2))^(1/2) -
      750 *(50 + 10*5^(1/2))^(1/2))^(1/5);
 r2 = (2500*5^(1/2) + 750 *(50 - 10*5^(1/2))^(1/2) +
     250 *(50 + 10*5^(1/2))^(1/2))^(1/5);
 r3 = -(-2500*5^(1/2) + 750 *(50 - 10*5^(1/2))^(1/2) +
      250* (50 + 10*5^(1/2))^(1/2))^(1/5);
 r4 = -(2500*5^(1/2) - 250* (50 - 10*5^(1/2))^(1/2) +
      750 *(50 + 10*5^(1/2))^(1/2))^(1/5);
 prim1 = (-1 - 5^(1/2) + 2^(1/2)*I*(5 - 5^(1/2))^(1/2))/4;
 prim2 = (-1 + 5^(1/2) - 2^(1/2)*I*(5 + 5^(1/2))^(1/2))/4;
 prim3 = (-1 + 5^(1/2) + 2^(1/2)*I*(5 + 5^(1/2))^(1/2))/4;
 prim4 = (-1 - 5^(1/2) - 2^(1/2)*I*(5 - 5^(1/2))^(1/2))/4;
 {x -> #} & /@ ({{r1, r2, r3, r4}.{1, 1, 1, 1},
     {r1, r2, r3, r4}.{prim2, prim4, prim1, prim3},
     {r1, r2, r3, r4}.{prim3, prim1, prim4, prim2},
     {r1, r2, r3, r4}.{prim1, prim2, prim3, prim4},
     {r1, r2, r3, r4}.{prim4, prim3, prim2, prim1}}/5)]

Check the roots using:

N[Solve[x^5 + 20 x + 32 == 0, x], 100] == N[%, 100]

The "Solving the Quintic" poster notebooks are available at
http://library.wolfram.com/infocenter/TechNotes/158/

The Radical.nb notebook is what was used here. Change
the simp function to do Expand[Simplify[%]] on its result
and discard the Discriminant definition which is built into
versions 6 & 7 to get the results in a form like what is
shown here. Also change Solve::rad to SolveQuintic::rad
in two places before evaluating all cells.

Q.E.D.