Message Boards Message Boards

0
|
1489 Views
|
4 Replies
|
0 Total Likes
View groups...
Share
Share this post:

Why the complicated Solve output?

Posted 10 years ago
Hi to all's,

the problem:

The second is ok while the first ... emoticon

Why?
Regards

Stefan
POSTED BY: Stefan AS
4 Replies
Posted 10 years ago
Thanks again!
Mathematica 9.0.1 on XP 32 bit

Now
k = 2
In[233]:= Reduce[(x + k)^2/3 + (3 x + k)/2 < (2 x - 7)/2 - (x - 5)/2 + 3 k , x]

Out[233]= > -(1/8) && -(7/2) - (3 Sqrt[1 + 8])/2 < x < -(7/2) + (3 Sqrt[1 + 8])/2
or
In[232]:= Reduce[(x + k)^2/3 + (3 x + k)/2 - (2 x - 7)/2 + (x - 5)/2 - 3 k < 0 , x]

Out[232]= -8 < x < 1

Stefan
POSTED BY: Stefan AS
Mathematica 9.0.1 on Windows 7 64 bit does it
In[5]:= Reduce[(2 x + 1)^2 - (x - 3)^2 >= 2 (x - 1) (x + 2) - 11, x]
Out[5]= x <= -7 || x >= -1

ordering an inequality forces x to be an integer, rational number or a real one. What happens if you state the tautology in your version
In[8]:= Reduce[(2 x + 1)^2 - (x - 3)^2 >= 2 (x - 1) (x + 2) - 11, x, Reals]
Out[8]= x <= -7 || x >= -1
POSTED BY: Udo Krause
Posted 10 years ago
Thanks,

I must solve too an inequality and there is same problem!

Stefan
Attachments:
POSTED BY: Stefan AS
The first is exactly the same as the second by the way, if you are concerned with the representation, use Simplify
In[3]:= Solve[(2 x + 1)^2 - (x - 3)^2 == 2 (x - 1) (x + 2) - 11, x] // Simplify
Out[3]= {{x -> -7}, {x -> -1}}
POSTED BY: Udo Krause
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Reply Preview
Attachments
Remove
or Discard

Group Abstract Group Abstract