I am trying to solve 0.125 + 0.5 (1-x)^3 - (12.5/y)==0 for x, when y is real and y>0. I thus want to find x= 1- 0.63 ((100-y)/y)^(1/3), so that if y=100, x=1. Mathematica's Solve yields 3 roots:(adsbygoogle = window.adsbygoogle || []).push({});

sol=Solve[0.125 + 0.5 (1 - x)^3 - (12.5/y) == 0, x]

Root 1:

1.+((0.+0. I) y^(1/3))/(-100.+1. y)^(1/3)-((0.31498+0.545562 I) (-100.+1. y)^(1/3))/y^(1/3),

Root 2:

1.+((0.+0. I) y^(1/3))/(-100.+1. y)^(1/3)-((0.31498-0.545562 I) (-100.+1. y)^(1/3))/y^(1/3),

Root 3:

1.+(0. y^(1/3))/(-100.+1. y)^(1/3)+(0.629961 (-100.+1. y)^(1/3))/y^(1/3)

If I now evaluate root 1 at y=100, I get an Infy error because of division by zero. How can I drop the imaginary part of this function? I have tried (without succes) a bunch of things such as:

xroot1=x/.sol[[1]]

*ComplexExpand[xroot1]

*Assuming[p > 0, {Simplify[xroot1]}]

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# Mathematica Real Part of Solve output

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