# Solving a cubic equation

### Question

As part of a program I'm writing, I need to solve a cubic equation exactly (rather than using a numerical root finder):

``````a*x**3 + b*x**2 + c*x + d = 0.
``````

I'm trying to use the equations from here. However, consider the following code (this is Python but it's pretty generic code):

``````a =  1.0
b =  0.0
c =  0.2 - 1.0
d = -0.7 * 0.2

q = (3*a*c - b**2) / (9 * a**2)
r = (9*a*b*c - 27*a**2*d - 2*b**3) / (54*a**3)

print "q = ",q
print "r = ",r

delta = q**3 + r**2

print "delta = ",delta

# here delta is less than zero so we use the second set of equations from the article:

rho = (-q**3)**0.5

# For x1 the imaginary part is unimportant since it cancels out
s_real = rho**(1./3.)
t_real = rho**(1./3.)

print "s [real] = ",s_real
print "t [real] = ",t_real

x1 = s_real + t_real - b / (3. * a)

print "x1 = ", x1

print "should be zero: ",a*x1**3+b*x1**2+c*x1+d
``````

But the output is:

``````q =  -0.266666666667
r =  0.07
delta =  -0.014062962963
s [real] =  0.516397779494
t [real] =  0.516397779494
x1 =  1.03279555899
should be zero:  0.135412149064
``````

so the output is not zero, and so x1 isn't actually a solution. Is there a mistake in the Wikipedia article?

ps: I know that numpy.roots will solve this kind of equation but I need to do this for millions of equations and so I need to implement this to work on arrays of coefficients.

1
9
7/20/2015 12:32:39 PM

Wikipedia's notation `(rho^(1/3), theta/3)` does not mean that `rho^(1/3)` is the real part and `theta/3` is the imaginary part. Rather, this is in polar coordinates. Thus, if you want the real part, you would take `rho^(1/3) * cos(theta/3)`.

I made these changes to your code and it worked for me:

``````theta = arccos(r/rho)
s_real = rho**(1./3.) * cos( theta/3)
t_real = rho**(1./3.) * cos(-theta/3)
``````

(Of course, `s_real = t_real` here because `cos` is even.)

24
12/1/2009 10:44:54 PM

I've looked at the Wikipedia article and your program.

I also solved the equation using Wolfram Alpha and the results there don't match what you get.

I'd just go through your program at each step, use a lot of print statements, and get each intermediate result. Then go through with a calculator and do it yourself.

I can't find what's happening, but where your hand calculations and the program diverge is a good place to look.