$ax^{13}+bx^{12}+c=0$ by hand. Is there any chance?

Question

I'd like to know if is there any way to get an approximation for the roots of the equation below by hand.

$$ax^{13}+bx^{12}+c=0.$$

You are allowed to use calculator to calculate powers, logarithms, roots, etc. (for example, $\text{(some number)}^{13}$, $\text{(some number)}^{1/12}$, etc.).

This problem came from the equation

$$5328.31=50000\frac{(1+i)^{13}\cdot i}{(1+i)^{13}-1}\cdot \frac{1}{1+i}$$

from where I have to calculate the value of $i$ (interest rate). If we write $x=1+i$, then the equation becomes

$$-8.3838x^{13}+9.3838x^{12}-1=0.$$

Answer 1

The problem of finding an algebraic formula is closed since longtime ago: it is not possible. Therefore you need an approximation and the particular values of the coefficients are very important in each case of course.

For your equation $8.3838x^{13}-9.3838x^{12}+1=0$ you have the equivalent one $$8.3838(x^{13}-x^{12})=x^{12}-1\iff 8.3838=\frac{x^{12}-1}{x^{13}-x^{12}}=\frac{1-1/x^{12}}{x-1}$$

Consider separately the function $f(x)=\frac{1-1/x^{12}}{x-1}$.

You have f(-1)=0. $f(1.06)\approx 8.384$ and $f(-0.794)\approx 8.34$ and other approximated values. You can see the special value $x=1.06$ gives a "good" approximated root.

Here the graph of $f$

Answer 2

Iterate:

$$x_{n+1}=\frac {-c}{a(x_n)^{12}}-\frac{b}{a} $$

$$x_0=\frac {-b}{a} $$

I will try to edit my answer and put bounds that indicate the rate of convergence

Answer 3

Inspired by this question of mine, we can approximate the solution "quite" easily using Padé approximants.

Let the equation be $$\frac{(1+i)^{13}\cdot i}{(1+i)^{13}-1}\cdot \frac{1}{1+i}-r=0$$ Building the simplest $[1,1]$ Padé approximant around $i=0$, we have $$0=\frac{\frac{2}{39} i (26 r+7)+\frac{1}{13} (1-13 r)}{1-\frac{4 i}{3}}$$ Canceling the numerator gives $$i=\frac{3 (13 r-1)}{2 (26 r+7)}$$ So, using your numbers $$i\approx 0.0591605$$ while the exact solution would be $0.0600001$ (I suppose that the true solution is $0.06$).

More difficult would be to build the $[1,2]$ Padé approximant, but it is doable. Canceling the numerator gives $$i=\frac{4 \left(338 r^2+65 r-7\right)}{3 \left(-845 r^2+546 r+35\right)}$$ and, using your numbers $$i\approx 0.0600597$$ which is much better.

Simpler (but less accurate) would be to develop the expression as a Taylor series around $i=0$; this would give $$r=\frac{1}{13}+\frac{6 i}{13}+\frac{8 i^2}{13}+O\left(i^3\right)$$ Using the first two terms would lead to $i\approx 0.0642268$ (we know that this is an overestimate of the solution by Darboux theorem since it corresponds to the first iterate of Newton method using $i_0=0$). Using the three terms implies solving a quadratic for which the positive root is $i\approx 0.0595056$.