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quartic formula

biquadratic formula, quartic equation, biquadratic equation
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12D10 no label found



"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));
v = u^(1/3);
%%%% end MATLAB code %%%

I think this may be helpful.

Adalberto Ayjara Dornelles Filho
Caxias do Sul, Brasil

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,
Benoît THOUY
Grenoble, France

Dear Benoit,

I think the problem is in R6 computation.
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
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.

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:

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

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.

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.

Note that the .pdf version is not fully readable. Can this be rectified? Thanks.

super long lines don’t display very well in the HTML version either, I see.

Well, you can download the source and try to compile it yourself into a poster of some kind… is that sufficient?

Long term, I can look for a strategy to wrap lines in the PDFs.

Thanks for the report!


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