Mathematica does not display real solutions?

[Siron]

Hello!

I let Mathematica run the following command:

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]

It displays a solutions in function of Roots Objects. However, I'm only interested in solutions over the reals. Therefore, I thought to just add 'Reals' at the end:

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]

and Mathematica returns

x == 0

It looks like it can't find any reals, which is quite strange? Is there someone who can tell what I'm doing wrong here? Furthermore, can someone perhaps recheck this with Mathematica?

Thanks in advance!

 

[topsquark]

Hello!

I let Mathematica run the following command:

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]

It displays a solutions in function of Roots Objects. However, I'm only interested in solutions over the reals. Therefore, I thought to just add 'Reals' at the end:

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]

and Mathematica returns

x == 0

It looks like it can't find any reals, which is quite strange? Is there someone who can tell what I'm doing wrong here? Furthermore, can someone perhaps recheck this with Mathematica?

Thanks in advance!

See here for W|A's solution.

If you look carefully there are no complex numbers listed so I'm guessing that means this is your solution. It's a quartic equation so the results can be predicted to be messy. Were you expecting any particular form for the result?

-Dan

 

[Siron]

Thanks for the answer!

The expressions are indeed quite messy to work so its difficult to conclude if a solution is indeed real. I did some small tests with some easy examples. Apparently, adding 'Reals' or 'Complexes' does not matter to Mathematica, it displays the same solutions. So it looks like this is not the way to guarantee the existence of real solutions. On the other hand, what works better is to add '&& x <0' or '&& x > 0'. In this case, it seems that Mathematica only displays real solutions.

After some investigation of the quartic, I found that $a<0$ implies $x<0$. Since in my calculations $a<0$ is always satisfied, adding '&& x<0' should lead to real solutions only.