I have two solutions; I want is to find an equation that these solutions satisfy using mathematica.

The solutions are $x= \frac{{(26-k)}^2}{26}$ and $y= \frac{{k}^2}{26}$.

I know by hand computing that $x$ and $y$ satisfy $\sqrt{x}+\sqrt{y}=\sqrt{26}$; How can I show this using Mathematica?


This is just a slight alteration of the method posted by @Ajasja (which I like, and upvoted, but I'm extending to give an idea of how to get the desired relation).

 Eliminate[{x - (26 - k)^2/26, y - k^2/26, sqrtx^2 - x, 
    sqrty^2 - y} == 0, {k, x, y}], sqrtx]

(* Out[53]= {{sqrtx -> -Sqrt[26] - sqrty}, {sqrtx -> 
   Sqrt[26] - sqrty}, {sqrtx -> -Sqrt[26] + sqrty}, {sqrtx -> 
   Sqrt[26] + sqrty}} *)

From here one can substitute sqrtx->X^(1/2) and similar for y, and check to see which solutions remain valid after introducing radicals (some could be so-called "parasite" solutions).

  • $\begingroup$ Thank you so much! this is what I want :) $\endgroup$ – Farrokh Jun 9 '14 at 15:56
  • 2
    $\begingroup$ You are welcome. That said, I have this vague recollection that the prior response had been accepted. If so, that's probably the more appropriate choice since mine was a relatively small alteration of that one. (I do the "algebraic massage", in effect. Come to think of it, I could use one of those on my shoulder..) $\endgroup$ – Daniel Lichtblau Jun 10 '14 at 13:32

You can get half-way there by

eq1 = x == (26 - k)^2/26
eq2 = y == k^2/26
FullSimplify[Eliminate[{eq1, eq2}, k]]
(*x^2 + (-26 + y)^2 == 2 x (26 + y)*)

For a simpler form some more algebraic massage would be necessary...

  • $\begingroup$ Thank you so much! why we can't find the simplest one that we want? I need this one (\sqrt{x} + \sqrt{y} = \sqrt{26}) $\endgroup$ – Farrokh Jun 9 '14 at 15:21

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