This equation: $(\frac{-a}{x})^2=\sqrt{\frac{1}{x}}$ at $a > 0$ and $x > 0$ has a clear solution $x=a^{4/3}$, doesn't it? However,

Reduce[(-(a/x))^2 == (1/x)^(1/2) && x > 0 && a > 0] // ToRadicals


a > 0 && x == (-1)^(2/3) a^(4/3)


-(1/2) + (I Sqrt[3])/2

is a complex number, numerically N @ % yields -0.5 + 0.866025 I.


  • 2
    $\begingroup$ Quoting from Help > Documentation Center > ToRadicals > Possible Issues: "The result may not be equal to the Root object for some values of the parameter" $\endgroup$ – Daniel Lichtblau May 25 '14 at 17:56

The issue we encounter here is closely related to the problem exposed more extensively here:
Finding parameters making real part of eigenvalues vanish, however in this case we have to tackle with a bit more harmful problem. This is an undesired feature of the system.
Namely Root[-a^4 + #1^3 &, 1] has been pointed out as a solution, nevertheless since a is symbolic the system doesn't decide at an appropriate order which one of the three solutions Root[-a^4 + #1^3 &, k] where k ∈ { 1, 2, 3} it is and then ToRadicals interprets it arbitrarily:

ToRadicals @ Table[ Root[-a^4 + #1^3 &, k], {k, 3}]
 {(-1)^(2/3) a^(4/3), -(-1)^(1/3) a^(4/3), a^(4/3)}

One can use sometimes options in Reduce: Quartics and Cubics nonetheless they don't help for the problem at hand.

Reduce[(-(a/x))^2 == (1/x)^(1/2) && x > 0 && a > 0, {a, x}, Cubics -> True]
a > 0 && x == Root[-a^4 + #1^3 &, 1]

Since Mathematica allows for various ways of solving problems one can get the proper result e.g. with this:

FullSimplify[ Reduce[ (-(a/x))^2 == (1/x)^(1/2) && x > 0, {x}], a > 0]
a^(4/3) == x


FullSimplify[ x /. Solve[(-(a/x))^2 == (1/x)^(1/2) && x > 0 && a > 0,
                          {a, x}] // Quiet, a > 0]
  • $\begingroup$ "since a is symbolic it doesn't know which one of three solutions...". This means that the fact that I fixed a>0 in the Reduce function does not help. Did I get you right? $\endgroup$ – Alexei Boulbitch May 26 '14 at 7:03
  • $\begingroup$ @AlexeiBoulbitch In case you think that it doesn't answer your question, try to expose your problem more clearly. As Daniel has written the documentation mentions about this issue with ToRadicals, so in general this is not adjustable, at least in Mathematica 9. $\endgroup$ – Artes May 29 '14 at 11:26
  • $\begingroup$ @ Artes No, the question is closed. I have simply been on leave. The only thing is that I always prefer to put things as clearly as possible. The clearest explanation here is as follows: Mma may make here a mistake and may not. It is worse than when it always makes here a mistake. OK, good to know. For me that means that this function is better to avoid, if possible. $\endgroup$ – Alexei Boulbitch Jun 2 '14 at 7:43

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