quartic formula

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quartic formula

The four roots $r_{1},r_{2},r_{3},r_{4}$ of a quartic polynomial equation $x^{4}+ax^{3}+bx^{2}+cx+d=0$ are given by

$\displaystyle r_{1}$	$\displaystyle=$	$\displaystyle{\frac{-a}{4}-\frac{1}{2}{\sqrt{\frac{a^{2}}{4}-\frac{2b}{3}+% \frac{2^{{\frac{1}{3}}}\left(b^{2}-3ac+12d\right)}{3{\left(2b^{3}-9abc+27c^{2}% +27a^{2}d-72bd+{\sqrt{-4{\left(b^{2}-3ac+12d\right)}^{3}+{\left(2b^{3}-9abc+27% c^{2}+27a^{2}d-72bd\right)}^{2}}}\right)}^{{\frac{1}{3}}}}+\left(\frac{{2b^{3}% -9abc+27c^{2}+27a^{2}d-72bd+{\sqrt{-4{\left(b^{2}-3ac+12d\right)}^{3}+{\left(2% b^{3}-9abc+27c^{2}+27a^{2}d-72bd\right)}^{2}}}}}{54}\right)^{\frac{1}{3}}}}-% \frac{1}{2}{\sqrt{\frac{a^{2}}{2}-\frac{4b}{3}-\frac{2^{{\frac{1}{3}}}\left(b^% {2}-3ac+12d\right)}{3{\left(2b^{3}-9abc+27c^{2}+27a^{2}d-72bd+{\sqrt{-4{\left(% b^{2}-3ac+12d\right)}^{3}+{\left(2b^{3}-9abc+27c^{2}+27a^{2}d-72bd\right)}^{2}% }}\right)}^{{\frac{1}{3}}}}-\left(\frac{{2b^{3}-9abc+27c^{2}+27a^{2}d-72bd+{% \sqrt{-4{\left(b^{2}-3ac+12d\right)}^{3}+{\left(2b^{3}-9abc+27c^{2}+27a^{2}d-7% 2bd\right)}^{2}}}}}{54}\right)^{\frac{1}{3}}-\frac{-a^{3}+4ab-8c}{4{\sqrt{% \frac{a^{2}}{4}-\frac{2b}{3}+\frac{2^{{\frac{1}{3}}}\left(b^{2}-3ac+12d\right)% }{3{\left(2b^{3}-9abc+27c^{2}+27a^{2}d-72bd+{\sqrt{-4{\left(b^{2}-3ac+12d% \right)}^{3}+{\left(2b^{3}-9abc+27c^{2}+27a^{2}d-72bd\right)}^{2}}}\right)}^{{% \frac{1}{3}}}}+\left(\frac{{2b^{3}-9abc+27c^{2}+27a^{2}d-72bd+{\sqrt{-4{\left(% b^{2}-3ac+12d\right)}^{3}+{\left(2b^{3}-9abc+27c^{2}+27a^{2}d-72bd\right)}^{2}% }}}}{54}\right)^{\frac{1}{3}}}}}}}}$
$\displaystyle r_{2}$	$\displaystyle=$	$\displaystyle{\frac{-a}{4}-\frac{1}{2}{\sqrt{\frac{a^{2}}{4}-\frac{2b}{3}+% \frac{2^{{\frac{1}{3}}}\left(b^{2}-3ac+12d\right)}{3{\left(2b^{3}-9abc+27c^{2}% +27a^{2}d-72bd+{\sqrt{-4{\left(b^{2}-3ac+12d\right)}^{3}+{\left(2b^{3}-9abc+27% c^{2}+27a^{2}d-72bd\right)}^{2}}}\right)}^{{\frac{1}{3}}}}+\left(\frac{{2b^{3}% -9abc+27c^{2}+27a^{2}d-72bd+{\sqrt{-4{\left(b^{2}-3ac+12d\right)}^{3}+{\left(2% b^{3}-9abc+27c^{2}+27a^{2}d-72bd\right)}^{2}}}}}{54}\right)^{\frac{1}{3}}}}+% \frac{1}{2}{\sqrt{\frac{a^{2}}{2}-\frac{4b}{3}-\frac{2^{{\frac{1}{3}}}\left(b^% {2}-3ac+12d\right)}{3{\left(2b^{3}-9abc+27c^{2}+27a^{2}d-72bd+{\sqrt{-4{\left(% b^{2}-3ac+12d\right)}^{3}+{\left(2b^{3}-9abc+27c^{2}+27a^{2}d-72bd\right)}^{2}% }}\right)}^{{\frac{1}{3}}}}-\left(\frac{{2b^{3}-9abc+27c^{2}+27a^{2}d-72bd+{% \sqrt{-4{\left(b^{2}-3ac+12d\right)}^{3}+{\left(2b^{3}-9abc+27c^{2}+27a^{2}d-7% 2bd\right)}^{2}}}}}{54}\right)^{\frac{1}{3}}-\frac{-a^{3}+4ab-8c}{4{\sqrt{% \frac{a^{2}}{4}-\frac{2b}{3}+\frac{2^{{\frac{1}{3}}}\left(b^{2}-3ac+12d\right)% }{3{\left(2b^{3}-9abc+27c^{2}+27a^{2}d-72bd+{\sqrt{-4{\left(b^{2}-3ac+12d% \right)}^{3}+{\left(2b^{3}-9abc+27c^{2}+27a^{2}d-72bd\right)}^{2}}}\right)}^{{% \frac{1}{3}}}}+\left(\frac{{2b^{3}-9abc+27c^{2}+27a^{2}d-72bd+{\sqrt{-4{\left(% b^{2}-3ac+12d\right)}^{3}+{\left(2b^{3}-9abc+27c^{2}+27a^{2}d-72bd\right)}^{2}% }}}}{54}\right)^{\frac{1}{3}}}}}}}}$
$\displaystyle r_{3}$	$\displaystyle=$	$\displaystyle{\frac{-a}{4}+\frac{1}{2}{\sqrt{\frac{a^{2}}{4}-\frac{2b}{3}+% \frac{2^{{\frac{1}{3}}}\left(b^{2}-3ac+12d\right)}{3{\left(2b^{3}-9abc+27c^{2}% +27a^{2}d-72bd+{\sqrt{-4{\left(b^{2}-3ac+12d\right)}^{3}+{\left(2b^{3}-9abc+27% c^{2}+27a^{2}d-72bd\right)}^{2}}}\right)}^{{\frac{1}{3}}}}+\left(\frac{{2b^{3}% -9abc+27c^{2}+27a^{2}d-72bd+{\sqrt{-4{\left(b^{2}-3ac+12d\right)}^{3}+{\left(2% b^{3}-9abc+27c^{2}+27a^{2}d-72bd\right)}^{2}}}}}{54}\right)^{\frac{1}{3}}}}-% \frac{1}{2}{\sqrt{\frac{a^{2}}{2}-\frac{4b}{3}-\frac{2^{{\frac{1}{3}}}\left(b^% {2}-3ac+12d\right)}{3{\left(2b^{3}-9abc+27c^{2}+27a^{2}d-72bd+{\sqrt{-4{\left(% b^{2}-3ac+12d\right)}^{3}+{\left(2b^{3}-9abc+27c^{2}+27a^{2}d-72bd\right)}^{2}% }}\right)}^{{\frac{1}{3}}}}-\left(\frac{{2b^{3}-9abc+27c^{2}+27a^{2}d-72bd+{% \sqrt{-4{\left(b^{2}-3ac+12d\right)}^{3}+{\left(2b^{3}-9abc+27c^{2}+27a^{2}d-7% 2bd\right)}^{2}}}}}{54}\right)^{\frac{1}{3}}+\frac{-a^{3}+4ab-8c}{4{\sqrt{% \frac{a^{2}}{4}-\frac{2b}{3}+\frac{2^{{\frac{1}{3}}}\left(b^{2}-3ac+12d\right)% }{3{\left(2b^{3}-9abc+27c^{2}+27a^{2}d-72bd+{\sqrt{-4{\left(b^{2}-3ac+12d% \right)}^{3}+{\left(2b^{3}-9abc+27c^{2}+27a^{2}d-72bd\right)}^{2}}}\right)}^{{% \frac{1}{3}}}}+\left(\frac{{2b^{3}-9abc+27c^{2}+27a^{2}d-72bd+{\sqrt{-4{\left(% b^{2}-3ac+12d\right)}^{3}+{\left(2b^{3}-9abc+27c^{2}+27a^{2}d-72bd\right)}^{2}% }}}}{54}\right)^{\frac{1}{3}}}}}}}}$
$\displaystyle r_{4}$	$\displaystyle=$	$\displaystyle{\frac{-a}{4}+\frac{1}{2}{\sqrt{\frac{a^{2}}{4}-\frac{2b}{3}+% \frac{2^{{\frac{1}{3}}}\left(b^{2}-3ac+12d\right)}{3{\left(2b^{3}-9abc+27c^{2}% +27a^{2}d-72bd+{\sqrt{-4{\left(b^{2}-3ac+12d\right)}^{3}+{\left(2b^{3}-9abc+27% c^{2}+27a^{2}d-72bd\right)}^{2}}}\right)}^{{\frac{1}{3}}}}+\left(\frac{{2b^{3}% -9abc+27c^{2}+27a^{2}d-72bd+{\sqrt{-4{\left(b^{2}-3ac+12d\right)}^{3}+{\left(2% b^{3}-9abc+27c^{2}+27a^{2}d-72bd\right)}^{2}}}}}{54}\right)^{\frac{1}{3}}}}+% \frac{1}{2}{\sqrt{\frac{a^{2}}{2}-\frac{4b}{3}-\frac{2^{{\frac{1}{3}}}\left(b^% {2}-3ac+12d\right)}{3{\left(2b^{3}-9abc+27c^{2}+27a^{2}d-72bd+{\sqrt{-4{\left(% b^{2}-3ac+12d\right)}^{3}+{\left(2b^{3}-9abc+27c^{2}+27a^{2}d-72bd\right)}^{2}% }}\right)}^{{\frac{1}{3}}}}-\left(\frac{{2b^{3}-9abc+27c^{2}+27a^{2}d-72bd+{% \sqrt{-4{\left(b^{2}-3ac+12d\right)}^{3}+{\left(2b^{3}-9abc+27c^{2}+27a^{2}d-7% 2bd\right)}^{2}}}}}{54}\right)^{\frac{1}{3}}+\frac{-a^{3}+4ab-8c}{4{\sqrt{% \frac{a^{2}}{4}-\frac{2b}{3}+\frac{2^{{\frac{1}{3}}}\left(b^{2}-3ac+12d\right)% }{3{\left(2b^{3}-9abc+27c^{2}+27a^{2}d-72bd+{\sqrt{-4{\left(b^{2}-3ac+12d% \right)}^{3}+{\left(2b^{3}-9abc+27c^{2}+27a^{2}d-72bd\right)}^{2}}}\right)}^{{% \frac{1}{3}}}}+\left(\frac{{2b^{3}-9abc+27c^{2}+27a^{2}d-72bd+{\sqrt{-4{\left(% b^{2}-3ac+12d\right)}^{3}+{\left(2b^{3}-9abc+27c^{2}+27a^{2}d-72bd\right)}^{2}% }}}}{54}\right)^{\frac{1}{3}}}}}}}}$

The formulas for the roots are much too unwieldy to be used for solving quartic equations by radicals, even with the help of a computer. A practical algorithm for solving quartic equations by radicals is given in the concluding paragraph of the Galois-theoretic derivation of the quartic formula.

Synonym:

biquadratic formula, quartic equation, biquadratic equation

Type of Math Object:

Theorem

Major Section:

Reference

Groups audience:

Buddy list for djao

Mathematics Subject Classification

12D10 Polynomials: location of zeros (algebraic theorems)

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computer solution

Permalink Submitted by ayjara on Wed, 04/25/2007 - 14:04

Hi,

"The formulas for the roots are much too unwieldy to be used for solving quartic equations by radicals, even with the help of a computer."

I think we could compute the roots almost "easily" which this procedure:

%%%% begin MATLAB code %%%%
a3 = a;
a2 = b;
a1 = c;
a0 = d;

T1 = -a3/4;
T2 = a2^2 - 3*a3*a1 + 12*a0;
T3 = (2*a2^3 - 9*a3*a2*a1 + 27*a1^2 + 27*a3^2*a0 - 72*a2*a0)/2;
T4 = (-a3^3 + 4*a3*a2 - 8*a1)/32;
T5 = (3*a3^2 - 8*a2)/48;

R1 = sqrt(T3^2 - T2^3);
R2 = cubic_root(T3 + R1);
R3 = (1/12)*(T2/R2 + R2);
R4 = sqrt(T5 + R3);
R5 = 2*T5 - R3;
R6 = T4/R4;

r1 = T1 - R4 - sqrt(R5 - R6);
r2 = T1 - R4 + sqrt(R5 - R6);
r3 = T1 + R4 - sqrt(R5 + R6);
r4 = T1 + R4 + sqrt(R5 + R6);
%%%% end MATLAB code %%%

I does some tests and I think this code is correct.
My (not ease) job was to extract the Ts and Rs from your set of equations...

The cubic_root function must be

%%%% MATLAB code %%%
function v = cubic_root(u)
if imag(u) == 0 & real(u) < 0
v = -(abs(u)^(1/3));
else
v = u^(1/3);
end
%%%% end MATLAB code %%%

I think this may be helpful.

Regards,
Adalberto Ayjara Dornelles Filho
Caxias do Sul, Brasil

Re: computer solution

Permalink Submitted by Benoit on Fri, 11/16/2007 - 14:55

Hi Adalberto,
I've done a very similar work on Matlab and I find the same results as yours with my script.
But I have serious doubts about the programs (or formulas ???) : when I try the parameters (0,0,0,-1) for the equations x^4-1=0, I find only one (false) root : 0.
Where is the problem ?
Thanks for reply me.

Best regards,
BenoÃƒÂ®t THOUY
Grenoble, France

Re: computer solution

Permalink Submitted by ayjara on Mon, 11/19/2007 - 13:26

Dear Benoit,

I think the problem is in R6 computation.
With
a3 = 0, a2 = 0, a1 = 0, a0 = -1,
we have
T1 = -a3/4 = 0
T2 = a2^2 - 3*a3*a1 + 12*a0 = -12
T3 = (2*a2^3 - 9*a3*a2*a1 + 27*a1^2 + 27*a3^2*a0 - 72*a2*a0)/2 = 0
T4 = (-a3^3 + 4*a3*a2 - 8*a1)/32 = 0
T5 = (3*a3^2 - 8*a2)/48 = 0
and
R1 = sqrt(T3^2 - T2^3) = 24 sqrt(3)
R2 = root[3](T3 + R1) = 2 sqrt(3)
R3 = (1/12)*(T2/R2 + R2) = 0
R4 = sqrt(T5 + R3) = 0
R5 = 2*T5 - R3 =
R6 = T4/R4 = 0/0 (problem!)

I don't know how to fix it. Maybe some algebraic manipulation...
If you have some idea, please contact.
Regards,
Adalberto

Re: computer solution

Permalink Submitted by DavidB52 on Mon, 06/28/2010 - 00:24

There are also online solvers that offer immediate-for-use solutions. For example, the program at the following URL is a very good Quartic Equation Solver:

http://www.akiti.ca/Quad4Deg.html

It is free, quick, easy to use--and it works.

Failure at solving a quartic equation

Permalink Submitted by potato_jack on Mon, 03/19/2012 - 02:46

Good evening.

Using the formula, I have tried to solve the following quartic equation:

0=x^4-2x^3+4x^2-3x-10 (i.e: a=-2 , b=4 , c=-3 , d=-10), which has 2 real solutions and 2 complex ones.

However, when I came to the last "part" of the formula - the last fraction with the numerator of (-a^3+4ab-8c) - I was surprised to see that, after having applied the required calculations, the denominator of this fraction ended up being equal to 0, making any attempt to solve the equation using the formula vain.

Have I missed something important?

Please give feedback.

Re: Failure at solving a quartic equation

Permalink Submitted by lvoyster on Mon, 03/19/2012 - 18:31

Let x = u + 1/2 and you get

u^4 + (5/2)u^2 - 171/16 = 0

I don't know what formula you were using, but I suspect it assumed that the depressed quartic polynomial wasn't biquadratic.

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