Message Boards Message Boards

GROUPS:

Get a real-valued root?

Posted 7 months ago
1151 Views
|
7 Replies
|
6 Total Likes
|

enter image description here

enter image description here

enter image description here

As you can see,I want to find the real-valued root in Mathematica, what should i do? I tried so hard to search it up online, but in vein.

7 Replies

Maybe try Surd instead of Sqrt.

Posted 7 months ago

Where do you mean I should try Surd with? I mean I use 1/3 power, isn't that the same?

Posted 7 months ago

Another way, using CubeRoot

CubeRoot[7 + Sqrt[50]] + CubeRoot[7 - Sqrt[50]] // N
Posted 7 months ago

Thanks, it works, but is that really the only way to solve it? What if there are something like 1/5 power and the answer still come up with imaginary number?

Posted 7 months ago

Then you can use Surd, as Henrik suggested

Surd[7 + Sqrt[50], 5] + Surd[7 - Sqrt[50], 5] // N

CubeRoot is just a special form of Surd. The following will give the same result as if CubeRoot was used, in case of power 1/3:

Surd[7 + Sqrt[50], 3] + Surd[7 - Sqrt[50], 3] // N

The 1/5 power in Mathematica is defined as x^(1/5) == Exp[(1/5) Log[x]]. When x is negative, the principal value of the logarithm has I Pi as imaginary part, so that

In[9]:= (-1)^(1/5) == Exp[(1/5) Log[-1]]

Out[9]= True

and

In[10]:= Exp[(1/5) Log[-1]] // N

Out[10]= 0.809017 + 0.587785 I

This issue is an endless source of confusion and frustration for beginners, which had been raised in a real-centered theory of roots.

Posted 6 months ago

Thanks everyone for helping.

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