Solving Integration Problem: y^2=\frac{x^2-4}{4}

[bmanmcfly]

Hi, I'm new to the board, and newly returning to university after a 10 year hiatus... And for the most part I'm keeping up with the higher level math, but I'm slipping with the simpler stuff...

Without asking for you to solve my math homework for me, I just would like to ask what the next step in this would be:

I have \(\displaystyle y^2=\frac{x^2-4}{4}\)

This is where the 10 years away from a math book is killing me...

If I do the square root, do I have to do the square root of the entire side, or can I do the squares of each piece??

The other option is to have the expression to the power of 1/2... But that creates extra confusion when I do the integral and wind up with a power of 3/2...

How would you handle a situation like this?

 

[alyafey22]

Hey Bmanmcfly , welcome :)

I think your question is how to take the square root of :

\(\displaystyle y^2=\frac{x^2-4}{4}\)

Then we take it to the whole side :

\(\displaystyle y=\pm\sqrt{\left(\frac{x^2-4}{4}\right)}\)

If that is not clear , can you give me the real question ?

 

[bmanmcfly]

Hey Bmanmcfly , welcome :)

I think your question is how to take the square root of :

\(\displaystyle y^2=\frac{x^2-4}{4}\)

Then we take it to the whole side :

\(\displaystyle y=\pm\sqrt{\left(\frac{x^2-4}{4}\right)}\)

If that is not clear , can you give me the real question ?

Ya, that's what I thought... I'm minimizing questions asked because this does relate to homework.

Thanks for the nudge in the right direction.