I am trying to solve a cubic equation as follows

Solve[2 (Sqrt[x])^3 - 1.5*x^0.5*vd == 2 (Sqrt[\[Eta]0])^3, {x}]

I want to get the solution of $x$ in terms of $\eta_0 $, with rational/irrational factors, and then in the limit $vd<<\eta_0, x$ get the solution. When I try to use solve, I get solution like this

enter image description here since I am trying to solve an analytical physical equation, I want to get the pre-factors right in terms of actual numbers (rational or irrational) and not their decimal values (like $\sqrt{2}$ instead of 1.414 and $\frac{3}{5}$ instead of 0.6 in the pre factors.


closed as off-topic by Daniel Lichtblau, Henrik Schumacher, Carl Lange, garej, m_goldberg May 21 at 17:24

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  • 3
    $\begingroup$ Replace 1.5 with 3/2 and 0.5 with 1/2 in your equation. $\endgroup$ – Lotus May 17 at 4:07

You can get a series expansion for small vd directly with AsymptoticSolve: Assuming you want the real-valued branch,

AsymptoticSolve[2 (Sqrt[x])^3 - 3/2*Sqrt[x]*vd == 2 (Sqrt[η0])^3, {x, η0}, {vd, 0, 4}]

{{x -> vd/2 - vd^3/(96 η0^2) + vd^2/(16 η0) + η0}}

  • 1
    $\begingroup$ You don't need AsymptoticSolve. Try Solve[2 (Sqrt[x])^3 - 1.5*x^0.5*vd == 2 (Sqrt[\[Eta]0])^3 // Rationalize, {x}] (see comment @Lotus) $\endgroup$ – Ulrich Neumann May 17 at 7:22
  • $\begingroup$ @UlrichNeumann as the OP is after the limiting behavior for small vd the idea here was to present the new version 12 functionality that not everybody has seen yet. $\endgroup$ – Roman May 17 at 16:28

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