# I'm so stuck yet probably so easy, what am I missing?...

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• Jun 7th 2010, 09:35 PM
lenfromkits
I'm so stuck yet probably so easy, what am I missing?...
Hi. I have been trying to solve this for weeks add can't figure it out.

I need to solve for 'b' in this equation:
(ie, I need to be able to reverse the equation so if given 'z' then I can find b. 'a' is just a constant).

$z = ab / sqrt (a^2 - b^2)$

Please if anyone can save me and show me how to do this. Thank you so much in advance.

(not sure if this image will show):
http://24.82.3.191/MyPublic/publicImages/z.JPG
• Jun 8th 2010, 01:32 AM
Isomorphism
Quote:

Originally Posted by lenfromkits
Hi. I have been trying to solve this for weeks add can't figure it out.

I need to solve for 'b' in this equation:
(ie, I need to be able to reverse the equation so if given 'z' then I can find b. 'a' is just a constant).

$z = ab / sqrt (a^2 - b^2)$

Please if anyone can save me and show me how to do this. Thank you so much in advance.

(not sure if this image will show):
http://24.82.3.191/MyPublic/publicImages/z.JPG

Square both sides:
$z = \frac{ab}{\sqrt{a^2 - b^2}} \implies z^2(a^2 - b^2) = (ab)^2$

Group terms with $b^2$:

$z^2(a^2 - b^2) = (ab)^2 \implies z^2a^2 = (z^2+a^2)b^2$

Finally,
$z^2a^2 = (z^2+a^2)b^2 \implies b^2 = \frac{a^2z^2}{a^2 + z^2}$
• Jun 8th 2010, 02:12 AM
lenfromkits
see, just like I said, easy....
...for a pro. Thank you so much! I get it now. I couldn't see the way but you make it seem easy.