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I have the following equation:

$$D=\frac{1}{64} \pi A ^3 B \sin \left(C\right)-\frac{1}{2} \pi A B \sin \left(C\right)$$

which I want to solve for $A$. The equation is cubic in $A$ so this should give me 3 answers and, potentially, imaginary parts to the answers. I know, from the physical meaning of the parameters, that all parameters ($A$,$B$,$C$ and $D$) are positive and real.

My question is: how can I use Solve on this equation and make sure that there will be no imaginary parts popping up in the solution?

I have used the suggestion by Chris and tried:

Solve[d == 1/64 π a^3 b Sin[c] - 1/2 π a b Sin[c], a, Reals]

but I still see that there is a $\imath\sqrt{3}$ term in the answer. Why does this still appear?

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  • $\begingroup$ related? mathematica.stackexchange.com/questions/3369/… $\endgroup$
    – chris
    Commented Feb 26, 2013 at 20:34
  • $\begingroup$ Not exactly. I found this post but I would like Solve to work with the assumptions already 'in mind'. Not `change' the answer afterwards $\endgroup$
    – Michiel
    Commented Feb 26, 2013 at 20:37
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    $\begingroup$ Solve[F[x],x,Reals] as suggested in reference.wolfram.com/mathematica/ref/Solve.html ? e.g. 'Solve[x^3 + 2 x^2 + 3 x + 4 == 0, x, Reals] // N' $\endgroup$
    – chris
    Commented Feb 26, 2013 at 20:38
  • $\begingroup$ That is actually a good point. I didn't know that Solve had that option. I tried using $Assumptions, but that didn't help. Thanks! $\endgroup$
    – Michiel
    Commented Feb 26, 2013 at 20:41
  • $\begingroup$ `Solve[x^3 + 2 x^2 + 3 x + 4 == 0, x, Reals] // ToRadicals // First // First' gives you the formal unique real root $\endgroup$
    – chris
    Commented Feb 26, 2013 at 20:44

1 Answer 1

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You can use the Reals Domain option in Solve as in 

 Solve[x^3 + 2 a x^2 + 4 == 0, x, Reals] // ToRadicals // First // First

Mathematica graphics

For the example above, calling d=2D/Pi/B/Sin[C]

 Solve[d == 1/32  A^3  -  A , A, Reals] // ToRadicals // First //First // FullSimplify

Mathematica graphics

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  • $\begingroup$ I have done what you mentioned, except for the //First//First part, because that will give me the wrong 1 of the 3 answers. $\endgroup$
    – Michiel
    Commented Feb 26, 2013 at 20:56

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