# How to understand the results of solve:ifun? [duplicate]

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I am trying to obtain an analytical solution of the following equation: My goal is to express \rho in terms of all other parameters (analytically solve RR function), so what is the result that Mathematica trying to tell me? Can anyone help me with this issue?

## marked as duplicate by Kuba♦Jan 10 '18 at 9:43

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• When you see a message like this, click the ..., then the documentation link that appears. For this message, everything is explained there. – Szabolcs Jan 10 '18 at 9:37
• @Szabolcs, thanks for your feedback. I read the content in the . . . but I'm still have trouble interpreting the results. How can \rho be mapped to a function of 0? The solution looks quite weird to me, could you help me explain it? – Joseph Martin Jan 10 '18 at 9:55
• Please post copyable code so people can experiment. – Szabolcs Jan 10 '18 at 11:57
• Thank you! The code is as follow: RR[ρ_]:= b (1+k) (2+k) ρ ((b/(1+ρ))^(1/(2+k)))^(-1-k) (-α+(b/(1+ρ))^(1/(2+k)))+(b/(1+ρ))^(1/(2+k)) (1+ρ) ((3+2 k) α-2 (1+k) (b/(1+ρ))^(1/(2+k)))+2 (1+k) (b/(1+ρ))^(1/(2+k)) (-1+ρ^(1+k)) ((2+k-2 ρ) (b/(1+ρ))^(1/(2+k))+α (-2-k+ρ)). It's essentially trying to solve out \rho in terms of all other parameters. But I guess it's unlikely to obtain analytic solution in this case. – Joseph Martin Jan 12 '18 at 0:16