Cubic Equation solution

Question

I am trying to solve this cubic equation: $(b t)x^3 + (1-a t) x + q = 0$; Where $a$ and $b$ are constants, $t$= time vector and $q$= vector

Is it possible to derive an analytical solution for $x$?

Answer 1

There's the Cardano solution to the cubic]1, which works for all cubic equations. It's messy enough that in practice, many people choose not to use it, however. You'll need to apply this to each component of your vector equation.

The first step in the Cardano solution, if I remember rightly, is to remove the quadratic term by a linear change-of-variable, so because of the form of your problem, you get that step for free.

Also: it'd be a good idea to simply let $A = bt$, and $B = 1 - at$ and $C = q$, and then use Cardano on $Ax^3 + Bx + C = 0$, because that'll match the notation in almost every writeup of Cardano that you'll find.

One final point: given that your equation is parameterized by $t$, you might be hoping to find a solution that's a continuous function of $t$, but that's unlikely, for if $b = a = 1, q = 0$, then for $0 \le t \le 1$, there's exactly one solution, while for $ t = 1.5$ there are three solutions. It's hard to make an algebraic formula that "chooses" one of three possible paths.

Answer 2

Interpreting the "vector" equation like

$$bt_ix^3+(1-at_i)x+q_i=0,i\in[1,d]$$where $d=1000$, every single equation yields $1$ or $3$ real solutions in $x$ (by Cardano's formulas).

Unless there is a special degeneracy, these solutions have no reason to coincide and there's no global solution.