# quartic formula

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Type of Math Object:
Theorem
Major Section:
Reference
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## Mathematics Subject Classification

### computer solution

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,
Caxias do Sul, Brasil

### Re: computer solution

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 ?

Best regards,
BenoÃƒÂ®t THOUY
Grenoble, France

### Re: computer solution

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...
Regards,

### Re: computer solution

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.

### Failure at solving a quartic equation

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?

### Re: Failure at solving a quartic equation

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.