Given that $ a = ((k-3)\sqrt{v})/s$ where $k$ and $v$ are known. I have to solve the following equation for $s$: $$q(a,b) = s \sqrt{v},$$ where $b = 1.08148a^2+\epsilon$ and $\epsilon>0$ a very small number (for instance $\epsilon = 10^{-6}$). Furthermore, $q(a,b)$ is the greatest real root (which exists) of the quartic polynomial: $$(48a^2+16b)x^4 - (40a^3+168 ab)x^3+(-45a^4+225 a^2 b+ 72 b^2)x^2+(27a^3b - 162 ab^2)x + 27b^3$$ Since $b$ is written in function of $a$ this $q$ shall be a function of $a$ and hence a function of $s$. Theoretically speaking, it should be possible to solve the equation. However the algebraic expression for $q(a,b)$ is quite ugly. So I would like to not use it explicitly. I guess I need some kind of numerical algorithm?
Any suggestions? Thanks in advance! Cheers
Thank your for your answer! I can not give the whole context because that would take quite some time.
I now see that it should be $b = 1.08148a^2 - \epsilon$. It comes from the fact that I verified with Mathematica that the quartic polynomial from my original post has two real roots if $$0<b<1.08148a^2$$ (there are perhaps other possible values for $b$ as well, but $0<b<1.08148a^2$ is the only region I need in my calculations).
However, when I compute $b$ in my calculations, it is often that $b>1.08148a^2$ implying that I do not always have real roots. The explicit expression for $b = (v/s)(w-10s)$. In order to make $b<1.08148a^2$ (when its not), I replace $w$ by a new value $w'$ such that $b = 1.08148a^2$ and next I extract $\epsilon$ to guarantee that the new $b$-value satisfies the admission domain to obtain two real roots.
The link with the equation $q(a,b) = s \sqrt{v}$ is that this equation also needs to be satisfied (in my calculations). Therefore, I have to solve $$q(a,1.08148a^2-\epsilon) = s \sqrt{v}$$ for $s$. I hope this makes my question more clear.
Thanks!