Re: "Solve" on polynomial equality results in expressions

*To*: mathgroup at smc.vnet.net*Subject*: [mg128522] Re: "Solve" on polynomial equality results in expressions*From*: Bob Hanlon <hanlonr357 at gmail.com>*Date*: Sun, 28 Oct 2012 01:55:19 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-newout@smc.vnet.net*Delivered-to*: mathgroup-newsend@smc.vnet.net*References*: <20121027034831.2AEC16893@smc.vnet.net>

x is strictly not a function of y since it is not single-valued. The conditional expressions break the result into regions where it is single-valued. The function is real in each of these regions. sol = x /. Solve[y == 3 x^2 - 2 x^3, x, Reals] // ToRadicals {ConditionalExpression[ (1/2)*(1 + 1/(1 - 2*y + 2*Sqrt[-y + y^2])^(1/3) + (1 - 2*y + 2*Sqrt[-y + y^2])^ (1/3)), 0 < y < 1 || y > 1 || y < 0], ConditionalExpression[ 1/2 - (1 + I*Sqrt[3])/ (4*(1 - 2*y + 2*Sqrt[-y + y^2])^(1/3)) - (1/4)*(1 - I*Sqrt[3])* (1 - 2*y + 2*Sqrt[-y + y^2])^ (1/3), 0 < y < 1], ConditionalExpression[ 1/2 - (1 - I*Sqrt[3])/ (4*(1 - 2*y + 2*Sqrt[-y + y^2])^(1/3)) - (1/4)*(1 + I*Sqrt[3])* (1 - 2*y + 2*Sqrt[-y + y^2])^ (1/3), 0 < y < 1]} Plot[Evaluate[sol], {y, -3, 2}, Frame -> True, Axes -> False, PlotStyle -> Evaluate[{#, AbsoluteThickness[2]} & /@ {Red, Blue, Green}], FrameLabel -> {"y", "x"}] ContourPlot[y == 3 x^2 - 2 x^3, {y, -3, 2}, {x, -0.5, 2}, AspectRatio -> 1/GoldenRatio, FrameLabel -> {"y", "x"}] ParametricPlot[{3 x^2 - 2 x^3, x }, {x, -0.5, 2}, PlotRange -> {{-3, 2}, {-0.5, 2}}, Frame -> True, Axes -> False, AspectRatio -> 1/GoldenRatio, FrameLabel -> {"y", "x"}] BobHanlon On Fri, Oct 26, 2012 at 11:48 PM, <coder0xff at gmail.com> wrote: > If I do Solve[y==3x^2-2x^3,x] I get expressions containing imaginary parts. If I set the domain to the Reals (Solve[y==3x^2-2x^3,x,Reals]) then I got stuff like Root[y - 3 #1^2 + 2 #1^3 &, 2]. I know there's a solution using only reals (since the range and domain are both real, and complex math is performed as operations on reals after all) How can I get it to give me the actual solution using only reals? >

**References**:**"Solve" on polynomial equality results in expressions containing "Root"***From:*coder0xff@gmail.com