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Solve[beta* phi *((1 + (1 - m)*alpha*theta ((phi^(-0.5) - 1)/((1 - theta)*phi^(-0.5) + theta)))) - delta == 0, phi]

I run it and get nothing showing as results. Thanks so much for your help!!! Appreciated!

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closed as off-topic by Michael E2, garej, Chris K, Henrik Schumacher, rhermans Jul 5 at 16:10

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Michael E2, garej, Chris K, Henrik Schumacher, rhermans
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Use Solve (not solve)? $\endgroup$ – kglr Jul 3 at 5:07
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    $\begingroup$ Also, use -1/2 instead of -0.5 in the exponent to allow for exact solving. $\endgroup$ – Roman Jul 3 at 6:31
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    $\begingroup$ This is a 3rd-order equation. In general, such equations cannot be solved in a simple form. However, a nice solution might exist if you set some parameters to zero. $\endgroup$ – yarchik Jul 3 at 9:10
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The most compact way of writing these solutions would be as squares of Root objects:

{Root[#^2*β-δ+(#-1)*(#^2*(1+(m-1)*α)*β-δ)*θ &, 1]^2,
 Root[#^2*β-δ+(#-1)*(#^2*(1+(m-1)*α)*β-δ)*θ &, 2]^2,
 Root[#^2*β-δ+(#-1)*(#^2*(1+(m-1)*α)*β-δ)*θ &, 3]^2}
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So the Mathematica syntax for what you show above is:

Solve[\[Beta] \[Phi] ((1 + (1 - m) \[Alpha] \[Theta] ((\[Phi]^(-1/2) - 1)/((1 - \[Theta]) \[Phi]^(-1/2) + \[Theta])))) - \[Delta] == 0, \[Phi]]

While this does give solutions, they are not pretty and quite complex. But you might be able to work with them.

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