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Get no real solutions in Solve?

Posted 8 years ago

Hello!

In one of my previous posts I asked how to solve the a system of equations with 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]

Mathematica displays lengthy solutions with root objects. Now, I'm only interested in solutions over the reals. Hence I adjusted the previous 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 only displays x == 0. Maybe I did something wrong? Or should I derive from this output that the only real solution is in fact x == 0? Can somebody recheck this?

Many thanks! Cheers

POSTED BY: Cédric Cavents
3 Replies

That is not my experience. Even solving on the reals I get conditional expressions, and not just x==0. Have you tried Reduce instead of Solve? What version are you using?

POSTED BY: Gianluca Gorni

I'm using the online version on wolframcloud.com. I noticed that I added x<0. I tested with some easy examples and adding something x<0 or x>0 obviously works to only extracts real solutions. So, since I do not have a legitimate version of Mathematica I guess I should hold on to that?

Could you perhaps show me what you obtain for: 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 && x<0 && c < 0, x, Reals]

POSTED BY: Cédric Cavents

EDIT: Even, when I write

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, x, Reals]

meaning that I do not impose any conditions on a,c or x then it does not give any solutions as well. This is quite strange?

POSTED BY: Cédric Cavents
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