# [?] Solve cubic equation w.r.t X ?

Posted 1 year ago
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 Hi,I have partially differentiated an equation w.r.t. x, see picture below, and want to isolate x in the equation to solve for a local maximum. This equation is part of my master's thesis and quite crucial. I know that, if rewritten, it will be cubic in x, thus, there can be at least three different solutions. However, at this moment, I am not able to obtain any solutions using Mathematica. Can you help me?Attached there is a picture of the equation as it looks in the paper and also a picture of how I've written it into Mathematica. Ideally, I would like to simplify it and find values of x that solves the equation.I hope you can help.   Answer
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Posted 1 year ago
 Hi Mads,Please post code, not images.One obvious problem is that solve should be Solve. Functions in WL are case sensitive and all built-in functions start with a capital letter. Answer
Posted 1 year ago
 Sorry for the inconvenience. Please find my code below.Ideally, I would like the expression on a cubic form, if Wolfram can provide it, and then isolate x in the expression. x is the only variable, the others are constants. Solve[[Alpha](ph - pl)(R1 + R2[Gamma][Lambda]*( { {x(ph)^2 + (1 - x)(pl)^2}, {xph + (1 - x)pl} } - { {x(1 - ph)ph + (1 - x)(1 - pl)pl}, {x(1 - ph) + (1 - x)(1 - pl)} }) + R2*( { {(1 - [Gamma]*[Lambda])}, {1 + r} } )) - 2b[Phi]*x == 0, x] Answer
Posted 1 year ago
 Did you mean this? Solve[\[Alpha] (ph - pl) (R1 + R2\[Gamma]*\[Lambda]*(((x (ph)^2 + (1 - x) (pl)^2)/(x* ph + (1 - x) pl)) - ((x (1 - ph) ph + (1 - x) (1 - pl) pl)/(x (1 - ph) + (1 - x) (1 - pl)))) + R2*((((1 - \[Gamma]*\[Lambda]))/(1 + r)))) - 2 b*\[Phi]*x == 0, x] Answer
Posted 1 year ago
 Thank you Gianluca!I am new to Mathematica. The code you provided gives me three different solutions, as expected. However, the solutions are not that meaningful. Is there any way I can simplify these in some way?and what do the grey rounded boxes with numbers mean? i.e. (--- 1 ---) etc.  Answer
Posted 1 year ago
 The expression is very complicated, and Mathematica gives an oultine of the structure. Click on "show more", "show all" to explore. Answer