Finding Solutions to a Cubic Equation with Multiple Roots

[Big-Daddy]

I'm finding it hard to find the solutions to this cubic equation:

1/2 x^3 - 2.025647693*10^14 x^2 + 8.102590772*10^11 x - 8.102590772*10^8 = 0

I'm looking for the smallest real positive solution but no matter what solver I use I keep getting only one root (the one of order of magnitude 10^14). There should be a root of the order of 10^-3 but I don't know how to find it, nor why my usually trusty solvers are giving only 1 root.

 

[micromass]

You are probably using the Newton-Rhapson algorithm. This algorithm is used to solve $f(x)=0$ for various functions. It works by taking an initial guess and then recursively refine the guess.

Not all guesses work however. So the question is to find out what the best guess is and whether there are good guesses at all!

We have the following: http://en.wikipedia.org/wiki/Newton-Rhapson#Analysis

So if $\alpha$ is your root and if $f^\prime(\alpha)$ is nonzero, then there are good initial guesses. However, this is exactly what goes wrong here! The root $\alpha$ that you seek does have $f^\prime(\alpha)=0$. This is why no initial guess will work and why you didn't find the solution.

There are other ways to find the solution though. It is hardly practical in many cases, but the rational root theorem works here. Either way, wolfram alpha could find the root: http://www.wolframalpha.com/input/?...x^2+++8.102590772*10^11+x+-+8.102590772*10^8+
and it's a very pretty root too!

 

[lurflurf]

^Yes
one thing you can do without changing to a different method is to solve
f'(x)=0
3/2 x^2 - 2*2.025647693*10^14 x + 8.102590772*10^11 = 0
if f'(x)=0 and f(x)=0 then x is a double root of f

 

[Chestermiller]

If you substitute x = 10^-3y into the equation, you get:

y²-4y+4 = 0. This neglects the cubic term, which is going to be minute compared to the other terms. So there is a double root at y = 2, or, equivalently, at x = 2 x 10^-3.

Chet

 

[CellCoree]

I would recommend approaching this problem by first simplifying the equation and then using various methods such as factoring, the rational root theorem, or the cubic formula to find the solutions. It is also important to check for any errors in the input of the equation or the use of the solvers. Additionally, it may be helpful to plot the equation and visually identify any potential solutions. If all else fails, consulting with a colleague or seeking assistance from a mathematics expert may provide further insight and solutions to the problem.