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From: David Bernier on 31 Jul 2010 07:04
Wood wrote:
> I am trying to learn how to solve a quartic equation. In
> "Summary of Ferrari's method" of page
> http://en.wikipedia.org/wiki/Quartic_function
> a root in the most simple case is
> x = -B/4/A + sqrt((-alfa + sqrt(alfa^2 - 4*gama))/2)
>
> My question is how to know the express inside sqrt is not negative:
> how to know
> alfa^2 - 4*gama>= 0
> and
> (-alfa + sqrt(alfa^2 - 4*gama))/2>= 0

I think it might be helpful to look at a concrete example, if
you get stuck.


You might want to look at the Web page
<
http://mathdl.maa.org/mathDL/46/?pa=content&sa=viewDocument&nodeId=3312&bodyId=3621>

which solves one problem.

I think this can be pretty long to do in closed form.
Ludovico Ferrari solved the quartic
x^4 + 6 x^2 + 36 = 60x around 1540.

Starting from the discussion here:
http://mathdl.maa.org/mathDL/46/?pa=content&sa=viewDocument&nodeId=3312&bodyId=3621


Following their calculations, I get the series of
explicit formulas, with numerical values on the next line:

u = cuberoot(190+sqrt(33903)) + cuberoot(190 - sqrt(33903)) - 5
so u ~= 4.0097912285348773231421386699582591 .

B = -sqrt(2*u + 6)
so B ~= -3.7442732882456316033499879583779239 .

C = 6 + u + 30/B
so C~= 1.9975555580834789833719521920884714 .

x = (-B - sqrt(B*B - 4*C))/2
so x~= 0.64439886422681550176229551792942104 .

[ or: x = (-B + sqrt(B*B - 4*C))/2 ].

result = x^4 + 6*x^2 - 60*x + 36
result ~= 2.06887 E-36, i.e. 0.


David Bernier
From: Wood on 31 Jul 2010 10:22
Thanks a lot for your help!

I checked the example in the link, I still got problem in the value inside sqrt.
I start in

"Summary of Ferrari's method" of page
http://en.wikipedia.org/wiki/Quartic_function
the root in the most simple case
x = -B/4/A + sqrt((-alfa + sqrt(alfa^2 - 4*gama))/2)

If (-alfa + sqrt(alfa^2 - 4*gama))/2 is a complex number, sqrt((-alfa + sqrt(alfa^2 - 4*gama))/2)
will have two roots and the whole quartic equation will have more than four roots!!??
From: Ray Vickson on 31 Jul 2010 17:56
On Jul 31, 11:22 am, Wood <wood0...(a)sohu.com> wrote:
> Thanks a lot for your help!
>
> I checked the example in the link, I still got problem in the value inside sqrt.
> I start in
>
> "Summary of Ferrari's method" of pagehttp://en.wikipedia.org/wiki/Quartic_function
> the root in the most simple case
> x = -B/4/A + sqrt((-alfa + sqrt(alfa^2 - 4*gama))/2)
>
> If (-alfa + sqrt(alfa^2 - 4*gama))/2 is a complex number, sqrt((-alfa + sqrt(alfa^2 - 4*gama))/2)
> will have two roots and the whole quartic equation will have more than four roots!!??

What is stopping you from trying it yourself on some numerical
examples, to see what is happening?

R.G. Vickson
From: Wood on 31 Jul 2010 20:39
Thank you. Please check what I was doing wrong:

Acooding to "Summary of Ferrari's method" of page
http://en.wikipedia.org/wiki/Quartic_function
when beta = 0, the root is
x = -B/4/A +-s sqrt((-alfa +-t sqrt(alfa^2 - 4*gama))/2)

I think that means there are 4 expressions:
x = -B/4/A + sqrt((-alfa + sqrt(alfa^2 - 4*gama))/2)
x = -B/4/A + sqrt((-alfa - sqrt(alfa^2 - 4*gama))/2)
x = -B/4/A - sqrt((-alfa + sqrt(alfa^2 - 4*gama))/2)
x = -B/4/A - sqrt((-alfa - sqrt(alfa^2 - 4*gama))/2)

Now check the first expression:
a, b, c, d, e = 1, 0, 2, 0, 5
alfa, beta, gama = 2, 0, 5
x = -B/4/A + sqrt((-alfa + sqrt(alfa^2 - 4*gama))/2)
= sqrt((-alfa + sqrt(alfa^2 - 4*gama))/2)
= sqrt(-1 + 2i)
= 0.786151377814533 + 1.27201964947877i , or -0.786151377928752 - 1.27201964940818i

1. substituting into ax^4 + bx^3 + cx^2 + dx + e, got
-3.59172247499373E-10 + 5.00000000071835i or -1.07752073930101E-09 + 5.00000000215505i
not equal zero !
2. Moreover, as you see, I got 2 solutions for 1 expression. So total 8 solutions!
From: Ray Vickson on 1 Aug 2010 03:35
On Jul 31, 9:39 pm, Wood <wood0...(a)sohu.com> wrote:
> Thank you. Please check what I was doing wrong:
>
> Acooding to "Summary of Ferrari's method" of pagehttp://en.wikipedia.org/wiki/Quartic_function
> when beta = 0, the root is
> x = -B/4/A +-s sqrt((-alfa +-t sqrt(alfa^2 - 4*gama))/2)
>
> I think that means there are 4 expressions:
> x = -B/4/A + sqrt((-alfa + sqrt(alfa^2 - 4*gama))/2)
> x = -B/4/A + sqrt((-alfa - sqrt(alfa^2 - 4*gama))/2)
> x = -B/4/A - sqrt((-alfa + sqrt(alfa^2 - 4*gama))/2)
> x = -B/4/A - sqrt((-alfa - sqrt(alfa^2 - 4*gama))/2)
>
> Now check the first expression:
> a, b, c, d, e = 1, 0, 2, 0, 5
> alfa, beta, gama = 2, 0, 5
> x = -B/4/A + sqrt((-alfa + sqrt(alfa^2 - 4*gama))/2)
>   = sqrt((-alfa + sqrt(alfa^2 - 4*gama))/2)
>   = sqrt(-1 + 2i)
>   = 0.786151377814533 + 1.27201964947877i , or  -0.786151377928752 - 1.27201964940818i
>
> 1. substituting into ax^4 + bx^3 + cx^2 + dx + e, got
> -3.59172247499373E-10 + 5.00000000071835i or -1.07752073930101E-09 + 5.00000000215505i
> not equal zero !  

I get zero (using Maple). Actually, I get zero exactly if I use the
exact expression for sqrt(-1 + 2i) = 1/2*sqrt(-2+2*5^(1/2))
+1/2*I*sqrt(2+2*5^(1/2)), but get -.1e-8 -.7e-8*i using 10-digit
floating-point computations.


> 2. Moreover, as you see, I got 2 solutions for 1 expression. So total 8 solutions!

NO!. Usually by writing sqrt(..) we mean the "principal" square root.
That is, the equation y^2 = x has two solutions: y = +sqrt(x) and y = -
sqrt(x), but there is just one value for sqrt(x). I think this is what
the author of the Wiki article had in mind, so that is why they wrote
+-sqrt(...). In your example all -B/4A terms = 0 so you have the
roots
x = + sqrt((-alfa + sqrt(alfa^2 - 4*gama))/2)
x = + sqrt((-alfa - sqrt(alfa^2 - 4*gama))/2)
x = - sqrt((-alfa + sqrt(alfa^2 - 4*gama))/2)
x = - sqrt((-alfa - sqrt(alfa^2 - 4*gama))/2)
You took x = + sqrt(-1 + 2*i) to have two values, namely +-
{1/2*sqrt(-2+2*5^(1/2)) +1/2*I*sqrt(2+2*5^(1/2))}. If you did this
consistently, your "third" root(s) -sqrt(...) would be -+
{1/2*sqrt(-2+2*5^(1/2)) +1/2*I*sqrt(2+2*5^(1/2))}, so are just the
same as the first. You really do get just _four_ roots, and the
simplest way to be sure of what you are doing is to use the principal
square root wherever you see the "sqrt" symbol.

R.G. Vickson

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