Why the complicated Solve output?

Questions
1

Hi to all’s,

the problem:

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The second is ok while the first … :frowning:

Why?
Regards

Stefan

2

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}}
3

Thanks,

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

Stefan

4

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
5

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