- #1
QED-Kasper
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Let's say we have:
A=B3
r=6B2
and we want to rewrite A in terms of r.
My way to do it is:
B2=r/6 => B=+-[tex]\sqrt{(r/6)}[/tex] => B3=+-(r/6)[tex]\sqrt{(r/6)}[/tex] => A=+-(r/6)[tex]\sqrt{(r/6)}[/tex]
In a book I found the following solution:
B2=r/6 => B3=(r/6)1.5 => A=(r/6)1.5
As you can see the book doesn't put a +-. Doesn't this mean that their answer is wrong as it only gives the positive answer? I think this can be brought back to a simpler problem:
[tex]\sqrt{4}[/tex] = 2
not +-2.
+-[tex]\sqrt{4}[/tex] = +-2Writing the 4 within the root symbol doesn't by definition imply that its either +-2, right?. It means that this is a positive root of 4. In order to include both the positive and negative root of 4 one must write +- in front of it, right? So does this mean that the books answer only includes the positive value, thus making it an incomplete and wrong answer?
I would like to add that in the book itself, the problem is stated as follows: Rewrite this system of equations as A(r) = and rearrange (its from a dutch book, I'm not sure if rearrange is the proper word in English) as far as possible.
So if A is a function of r, it means that it can't have both + or - values for the same r. So is the books answer valid in that case as only the principal value is taken to be a valid solution of the function? Or is it still wrong as the ± sign has to be used, because it symbolizes there are two functions and the answer actually includes two separate functions. Or maybe then the ± sign has to be put in front of the A aswell? (to distinguish between two functions, because A(r) by definition only defines one function so either another F(r) has to be put there for the negative equation, or one can just write ±A(r) where -A(r) = F(r))
Thank you.
A=B3
r=6B2
and we want to rewrite A in terms of r.
My way to do it is:
B2=r/6 => B=+-[tex]\sqrt{(r/6)}[/tex] => B3=+-(r/6)[tex]\sqrt{(r/6)}[/tex] => A=+-(r/6)[tex]\sqrt{(r/6)}[/tex]
In a book I found the following solution:
B2=r/6 => B3=(r/6)1.5 => A=(r/6)1.5
As you can see the book doesn't put a +-. Doesn't this mean that their answer is wrong as it only gives the positive answer? I think this can be brought back to a simpler problem:
[tex]\sqrt{4}[/tex] = 2
not +-2.
+-[tex]\sqrt{4}[/tex] = +-2Writing the 4 within the root symbol doesn't by definition imply that its either +-2, right?. It means that this is a positive root of 4. In order to include both the positive and negative root of 4 one must write +- in front of it, right? So does this mean that the books answer only includes the positive value, thus making it an incomplete and wrong answer?
I would like to add that in the book itself, the problem is stated as follows: Rewrite this system of equations as A(r) = and rearrange (its from a dutch book, I'm not sure if rearrange is the proper word in English) as far as possible.
So if A is a function of r, it means that it can't have both + or - values for the same r. So is the books answer valid in that case as only the principal value is taken to be a valid solution of the function? Or is it still wrong as the ± sign has to be used, because it symbolizes there are two functions and the answer actually includes two separate functions. Or maybe then the ± sign has to be put in front of the A aswell? (to distinguish between two functions, because A(r) by definition only defines one function so either another F(r) has to be put there for the negative equation, or one can just write ±A(r) where -A(r) = F(r))
Thank you.
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