How to solve this equation with square roots and fractions?

Question

A problem in Optics comes down to this equation:

Given that $A$, $B$, $C$ and $D$ are all real positive numbers, $-1 \lt x \lt 1$ and $-1 \lt \sqrt{C} x \lt 1$

$$\frac{Ax}{\sqrt{1 - x^2}} + \frac{Bx}{\sqrt{1 - Cx^2}} = D$$

How to solve this kind of equation?

To use a dot for multiplication, use \cdot, so A\cdot x gives $A \cdot x$, or you can use nothing. — Ross Millikan, Aug 11 '16 at 21:10
@NgChungTak yes definitely, that is exactly what I am looking for — ielyamani, Aug 12 '16 at 15:32
@ielyamani: I just wanted to mention that you actually were on the right track with your answer here: https://math.stackexchange.com/a/3028365/42969, but I did not not recognize the connection at that point. You are welcome to undelete your answer, so that I can honor it properly. — Martin R, Mar 06 '20 at 13:13

score 1 · Answer 1 · answered Aug 11 '16 at 21:09

1

You multiply by the product of the denominators and square the result. That will get you down to one square root, which you can isolate and square again. Unfortunately, that gives you an eighth degree equation unless there is cancellation. You can do numeric root finding on the original.

answered Aug 11 '16 at 21:09

Ross Millikan

383,099

The eighth degree equation will be a quartic equation in $x^2$, so theoretically you could end up with eight exact though messy solutions, some of which may be spurious – Henry Aug 11 '16 at 21:13

score 1 · Answer 2 · answered Aug 11 '16 at 21:25

Firstly I think that either the second denominator is $\sqrt{1-(Cx)^2}$ or the condition with $C$ is $-1 \leq \sqrt{C}x \leq 1$ with the additional $C \geq 0$.

A possible approach to the equation is the substitution $x = \cos{y}$ ot $x = \sin{y}$ (since $|x|<1$). With the first substitution, the equation becomes $ \frac{B\cos{y}}{\sqrt{1-C\sin^2{y}}} = D - A\cot{y}$. Now a necessary condition for the equation to have a solution is $(D-A\cot{y})(B\cos{y}) \geq 0$. For all such $y$ rise to the power of two and you will get rid of the square root. However, the left trigonometric equation could be difficult to solve explicitly for arbitrary parameters $A,B,C,D$. You can use universal substitution to make the equation polynomial. If you obtain some roots remember check if for them holds $(D-A\cot{y})(B\cos{y}) \geq 0$. It is possible that the problem must be solved numerically after the last step with the universal substitution.

How to solve this equation with square roots and fractions?

2 Answers2