# tranpose issue 3

• May 5th 2011, 03:53 AM
anthonye
tranpose issue 3
Hi;
I need to take the sqrt of both sides to find y. y^2 = a^2w^2 - v^2 / w^2

Ok got that and it becomes y = sqrt(a^2w^2 - v^2) / w.

why has the denominator been reduces to w and not the numerator to aw -v.

Thanks.
• May 5th 2011, 03:57 AM
Unknown008
This is because

$a^2w^2 - v^2 \neq (aw-v)^2$

If you had $(aw-v)^2$ in the numerator, then you would be able to put aw - v
• May 5th 2011, 04:00 AM
Ackbeet
Also, don't forget the dual sign possibilities:

$y = \pm\sqrt{a^2w^2 - v^2} / w.$
• May 5th 2011, 04:03 AM
anthonye
What I thought those two things would be equal please elaborate
• May 5th 2011, 04:05 AM
anthonye
Oh think I just got it parenthesis done first of course