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Why does Solve give me an expression with root in it for this equation?

Mathematica Asked on June 20, 2021

"12.0.0 for Microsoft Windows (64-bit) (April 6, 2019)"

This can be solved by hand without too much difficulty. But

Solve[{s == 8/27 ((Sqrt[1 + (3/2 t)^2])^3 - 1), t > 0}, t, Reals]

results in

{{t -> ConditionalExpression[
    Root[-16 s - 27 s^2 + 16 #1^2 + 36 #1^4 + 27 #1^6 &, 2], s > 0]}}

One Answer

This works:

Module[{s},
 s[t_] := 8/27 ((Sqrt[1 + (3/2 t)^2])^3 - 1);
 InverseFunction[s]["t"]
 ]

Answered by Steven Thomas Hatton on June 20, 2021

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