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I have to solve the following system: $$\begin{cases} 16x^4-40ax^3+(15a^2+24b)x^2- 18abx+3b^2 = 0 \\ 5ax-4x^2-b>0 \\ 15ax-20x^2-3b<0 \\ 4x^3-8 c x^2+5acx-cb <0\end{cases},$$ where $c<0$, and $a<0$, which implies $x<0$. I first ran this in Mathematica

Solve[
  16 x^4 - 40 a x^3 + (15 a^2 + 24 b) x^2 - 18 a b x + 3 b^2 == 0 && 
  5 a x - 4 x^2 - b > 0 && 15 a x - 20 x^2 - 3 b < 0 && 
  4 x^3 - 8 c x^2 + 5 a c x - c b < 0 && a < 0 && x < 0 && c < 0, x]

This gives me solutions with Root Objects. Now, I'm only interested in the solutions $x$ over the reals. Furthermore, I'm especially interested in the constraints on $a,b$ and $c$ such that the system admits real solutions. To that end, I just added 'Reals' at the end of the code:

Solve[
  16 x^4 - 40 a x^3 + (15 a^2 + 24 b) x^2 - 18 a b x + 3 b^2 == 0 && 
  5 a x - 4 x^2 - b > 0 && 15 a x - 20 x^2 - 3 b < 0 && 
  4 x^3 - 8 c x^2 + 5 a c x - c b < 0 && a < 0 && x < 0 && c < 0, x, Reals]

However, Mathematica does not display anything, which is quite odd? (I'm using the online version on wolframcloud.com)

Therefore, I wanted to ask the following: is it necessary to add 'Reals' to guarantee the existence of real roots when I already have added a constraint $x<0$? I mean, it would be weird that Mathematica displays complex solutions with the constraint $x<0$? I think I can trust that, without explicitly adding 'Reals', my first code displays the real roots only. Furthermore, could someone perhaps run my second code (with 'Reals') run? Then I can compare the results.

Finally, one of the constraints from the Output is for instance $$\mbox{Root}[f(x),1] < c \leq \mbox{Root}[f(x),2],$$ where $f(x)$ is a certain quartic polynomial. I interpret the above as: $c$ lies between the smallest real root of $f(x)$ and the greatest real root of $f(x)$. Is this the correct way?

Thanks! Cheers

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closed as off-topic by MarcoB, m_goldberg, Jens, Yves Klett, Kuba Jul 1 '16 at 12:08

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If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ "1:eJztWNFqwjAU9VMyBKm0QpLWto4+TbYP0MeuD43zQRhWWgeC9m/\ 3IYuNL3FQQ8xAxnlpWhoONzfnnpObJ1Etlt/DweB4PLw/v005n1fbj81+\ U23Lz9fDrl43jXzPF1W1z0MipyRz1j1n3TNdfu3WdbOqN7v9S3XI1S/\ RDTRQ82dcfRZqmLCUlESQIfM9NjWFLPsgfR4RMTaFGrLe8CIqw7OECvWwWGwLFGlAo4AVg\ eciU5nI1FYmtplP9WyFMaWWSLGGdOZDStPUEkxPmCMqxDPGnRDUEZ84jYzzc4tRvBiNOir\ wxLKsr4LzzIHKPiAiJizm5jXYKzbjTmYmkS3b9az5nLsRLCL8xJhZNxbohOrnHNnWsb59c\ vM6fbcO7JpW5pTv9wgpxvZB/dbjbJWhfFA+\ KB9b8ykzOj6dcKrBqQanGsgyZPkhZFmeaqiS5qINcCVwCe8hrwQ4rgRgnjBPmCfM0zLlf2\ eesq+BpkJT1UWRyFysDdIMaYY0o6/5931NVAToSeCfFzYw9CQwPhjf/\ cZXtO0PvDQfYA==" $\endgroup$ – Sektor Mar 18 '16 at 13:17
  • $\begingroup$ Uncompress the string above $\endgroup$ – Sektor Mar 18 '16 at 13:17
  • $\begingroup$ What happens if you use Reduce[] instead? $\endgroup$ – J. M. will be back soon Mar 18 '16 at 13:20
  • 1
    $\begingroup$ What @Sektor was trying to say (rather obscurely, I'm afraid), is that Solve[yourequation, x, Reals] returns solutions just fine (I'm on the desktop MMA 10.4). I also agree with you that setting x < 0 automatically restricts $x$ to be real: indeed the solutions one obtains with and without the Reals domain restriction seem identical. $\endgroup$ – MarcoB Mar 18 '16 at 14:28
  • 3
    $\begingroup$ I'm voting to close this question as off-topic because it is too localized; i.e, it applies only to the local situation and needs of its poster and answers will not benefit others. $\endgroup$ – m_goldberg Jun 30 '16 at 0:51

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