How to approximate the answer of a polynomial of degree 4?

Question

I want to solve this polynomial analytically. I know the useful answer is between 0 and 1. Is there any way I can write the answer based on a, b, and c? $$ 6\cdot a \cdot x^4 + 2 \cdot b \cdot x^3-b \cdot c=0 $$ Also, an approximate answer is acceptable, for example, an answer with 2% error. I will appreciate if someone can help me on this subject.

Answer 1

You could use Ferrari's method for solve in general. https://proofwiki.org/wiki/Ferrari%27s_Method this is an easy algorithmic way to do it.

Answer 2

Since you say an approximate answer is alright, up to 2% tolerance, it might be a good idea to use a numerical approximation like Newton's method.

The derivative works out to be $$ f'(x) = 24ax^3 + 6bx^2 = 6x^2 (4ax + b) $$

So you can take an $x_0$ in the range $ (0, 1) $ and try approximating your missing solution, if you know it is there. Just linearize to $ y = f'(x_0) x_1 + f(x_0) $ and solve for $ x_1 $ when $ y = 0 $, and repeat until you have the right confidence.

Newton's method can sometimes instead get you the wrong root. So you would likely have to have some kind of assured interval for the space you are looking at.

If your values are in the wrong range, you won't get a solution in $ (0, 1) $ at all, so there may be some other conditions you are looking at. For instance, suppose for a given triple $ (a, b, c) $, we have a solution to $ f $ in the range $(0, 1)$. Taking the forward image $ f( \, (0, 1) \, ) = (x , y) $ will give us an upper bound that the function attains on the interval, so if we re-pick $ c $ as $$ c' = c - \frac{y}{b} - 1 $$ then a new function $ g $ with triple $ (a, b, c') $ has no solutions in the range $ (0, 1) $ because of the vertical shift. So not every function with real parameters $a, b, c$ will have the desired property.

Answer 3

The exact solution would turn into a "complicated" expression in $a,b,c$.

If you are looking for an approximated solution, and you know that a real root is near to $1$ (and in fact it is for "normal" positive value of the parameters) , then replace $x$ with $1+y$, retain only the terms of degree $\le 2$ and solve for $y$. The shift to $x=1$ is because in $x=0$ the polynomial is quite flat (1st and 2nd derivative null).

Depending on the parameters you might find a better approximation developing instead at $x=1/2$.