Is it possible to solve $\frac{z}{y}=\left(1-{xca}\right)\left(1-{xcb}\right)(1-x)^{n}$ for $x$?

Question

$$\frac{z}{y}=\left(1-{xca}\right)\left(1-{xcb}\right)(1-x)^{n}$$

Can we solve this equation for $x$ such that, in the new equation we have, $x$ is the dependent variable, and $n$ is the independent variable, $c$, $a$, $b$, $y$, $z$ are all constants.

I have found it impossible to isolate $x$ on one side. I hope someone can help.

Answer 1

Start by simplifying the notation. Let $A = 1-\frac{m_1}a, B = 1-\frac{m_2}b, C = \frac{m_1m_2z}{aby}, k = n-2$ and $t = 1-x$. After multiplying through by $\frac{m_1m_2}{ab}$, the problem becomes solving $$(t-A)(t-B)t^k = C$$ If $C = 0$, we would be in luck. The solutions would be $t = A, t = B, t = 0$, or $x = 1-A, x = 1-B, x = 1$.

Alas, when $C \ne 0$, it is not so easy. You have a polynomial equation of degree $n = k+2$ in $t$. There is a famous result that there is no closed form solution for $t$ involving just addition, subtraction, multiplication, division, and the taking of roots. You can come up with closed form solutions using esoteric functions, but those esoteric functions are merely a mathematical relabeling of the problem. They offer no new insight into it, and do not give easier methods of calculating $t$.

So what can we do? If your need is theoretical, just the existence of a solution is usually enough, without having some specific formula for calculating it. If your need is practical, finding a numeric value for a specific set of parameter values, then there are very good means of approximating the answer to any accuracy. (And if you think you need an exact value, not an approximation - as many mathematical neophytes often think - you are very much mistaken. Any formula would have given you an approximate value only. It is impossible to ever have an exact value of an irrational number.)

The case for $k = 0$ is just a quadratic equation that is solvable with the quadratic formula Using a little calculus (yes, I see the precalculus tag, but I'll take care of the calculus and you will just have to trust that what I tell you from it is accurate), for $k > 0$ the turning points of the polynomial $P(t) = (t-A)(t-B)t^k - C$ are at the solutions to the equation $$t^{k-1}[t(t - B) + t(t - A) + k(t-A)(t-B)] = 0$$ which are

• $r_0 = 0$ for $k > 1$ only,
• $r_1 = \dfrac{(k+1)(A+B) - \sqrt{(k+1)^2(A+B)^2 - 4(k+2)kAB}}{2k+4}$,
• $r_2 = \dfrac{(k+1)(A+B) + \sqrt{(k+1)^2(A+B)^2 - 4(k+2)kAB}}{2k+4}$.

When $k$ is even, all three will be turning points. These turning points divide the real line into four intervals, one from $-\infty$ to the lowest root, the next from the lowest root to the middle root, and so on. When $k$ is odd, $r_0$ will not be a turning point. Instead the curve will level off there, but then continue on in the same direction, much like $y = x^3$ does around $x = 0$. So for odd $k, P(t)$ has only two turning poimts at $r_1, r_2$. These divide the real line into three intervals.

Let $y_0 = P(r_0) = P(0) = -C, y_1 = P(r_1), y_2 = P(r_2)$. For $k$ even,

• if all three $y$ values are $< 0$, the equation $P(t) = 0$ will have two roots, one below the lowest turning point, and the other above the greatest turning point.
• If two of the $y$ values is $< 0$, the equation will have four roots, one in each of the intervals defined by the turning points.
• If one of the $y$ values is $<0$, the equation will have two roots, one in each of the intervals on either side of the negative turning point.
• If none of the $y$ values is $\le 0$, then the equation has no real roots.

For $k$ odd, ignore $y_0$, and consider only $y_1,y_2$.

• If both $y$ values are $< 0$, the equation has a single root, greater than the larger of $r_1, r_2$.
• If one $y$ value is negative and the other positive, the equation has three roots, one in each of the intervals.
• If neither $y$ value is $\le 0$, the equation has a single root, less than the lower of $r_1, r_2$.

How can you find the roots? For each interval where a root resides per the conditions above, pick an arbitrary point $t_0$ in the interval (not the end points). Then for each $i$, calculate $$t_{i+1} = t_i - \dfrac{(t_i-A)(t_i-B)t_i^k - C}{t_i^{k-1}[t_i(2t_i - A - B) + k(t_i-A)(t_i-B)]}$$ The $t_i$ will converge to the root of the equation in that interval, and will do so very quickly. Keep calculating new $t_i$ until the difference between consecutive values is below the level of tolerance your application needs. Use the last value as your root.