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.

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- May 5th 2011, 03:53 AManthonyetranpose 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 AMUnknown008
This is because

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

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

$\displaystyle y = \pm\sqrt{a^2w^2 - v^2} / w.$ - May 5th 2011, 04:03 AManthonye
What I thought those two things would be equal please elaborate

- May 5th 2011, 04:05 AManthonye
Oh think I just got it parenthesis done first of course