I am currently doing some investigations on Sylvester's 4 Point Problem Probability of 4 Points being in Convex Configuration and repeatedly face the problem of solving equations between sums of euclidean distances between points, of which a typical example is $$\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}+\sqrt{(x_D-x_C)^2+(y_D-y_C)^2}=\sqrt{(x_C-x_A)^2+(y_C-y_A)^2}+\sqrt{(x_D-x_B)^2+(y_D-y_B)^2}$$
where the subscripts denote the point, to which the subscripted coordinate refers.

while it is possible to remove the square roots in that example manually by repeated squaring, it is tedious and error prone and, if there are more than two square roots on each side, then I think repeated squaring won't work.


are there any free online tools, that convert an equation with radicals into an algebraic one?

My searches in the internet only brought up a haystack of solvers for trivial instances (from the perspective of what I need).

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    $\begingroup$ Repeated squaring can always be made to work: for an equation involving $k$ square roots, one can get to an algebraic equation involving no square roots after squaring at most $k$ times. Basically, you can just collect all the terms which have an odd power of one particular square root from the original equation to one side, and after squaring both sides that original square root never shows up with an odd power again. $\endgroup$ – zeb Jul 20 '14 at 15:01
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    $\begingroup$ A similar trick works for radicals with larger exponents, using the fact that $a_1 + \cdots + a_k$ divides the determinant of a circulant matrix with top row $a_1, ..., a_k$. One collects all of the terms involving one particular $k$th root from the original equation raised to a power congruent to $i$ modulo $k$ into $a_i$, and then the circulant determinant will only involve that $k$th root raised to powers congruent to $0$ modulo $k$. $\endgroup$ – zeb Jul 20 '14 at 15:37
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    $\begingroup$ The answers at the math StackExchange question Rationalizing radicals describe algebraic procedures you can use. It can be a bit tedious, but you can use these methods along with various on-line devices that multiply-out and combine like terms of factored algebraic expressions. In the case of the expression you gave, you can rewrite with a $0$ on one side and then multiply both sides by an appropriate rationalizing factor (see André Nicolas's answer and my answer in that StackExchange question). $\endgroup$ – Dave L Renfro Jul 21 '14 at 16:37
  • $\begingroup$ There is now a similar question mathoverflow.net/questions/177765/… in which the details of removing the radicals are explained $\endgroup$ – Manfred Weis Aug 5 '14 at 5:00

I don't know about "free", but the general solution is to introduce a variable for each radical, together with an equation $y^2 = f(x),$ and then solve the resulting system. Since this is easy to do in Mathematica, I assume WolframAlpha will do it for you for free, but I never use WA, so can't promise.


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