I am experimenting with quartics because of a computer program. I wanted to see if there is a simpler general way than using the quartic formula, or descartes, or ferrari methods to find the smallest positive real root of a quartic if I know that it is of the form:
$\left(at^2+bt+c\right)^2+\left(dt^2+gt+h\right)^2-k^2=0$ where $t\in R$ is what I am trying to solve for, and $a,b,c,d,g,h,k\in R$ are already defined constants.
Note: A computer needs to be able to compute this very fast so no algebraic manipulation except plugging variables into functions. This also means that newtons method is not the right tool because it wouldn't be a fast enough to guess where the starting value is such that it converges to the correct root without becoming unstable.
Anyways, I decided to try the $a^2+b^2=\left(a+bi\right)\left(a-bi\right)$ equation on the $\left(at^2+bt+c\right)^2+\left(dt^2+gt+h\right)^2$ part of the equation and I got $\left(\left(a+di\right)t^2+\left(b+gi\right)t+\left(c+hi\right)\right)\left(\left(a-di\right)t^2+\left(b-gi\right)t+\left(c-hi\right)\right)-k^2=0$.
So let $A=a+di$, $B=b+gi$, $C=c+hi$, $D=a-di$, $G=b-gi$, and $H=c-hi$
This is how I got to the equation with the six complex coefficients from the question title. I also noticed that half of them are conjugates of the other half which is expected.
If equation didn't have this $-k^2$ term, this would be easy because I could just solve both complex quadratics for zero using the quadratic formula, but the $-k^2$ term is creating a dead end in this experiment. I am not sure how the roots of a quartic change when subtracting a constant term.
Is there a way that I can solve this with the $-k^2$ term? Or is there a proof that this cannot be solved with the $-k^2$ term, without using any of the methods I have rejected in this post?