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Math Help - I'm so stuck yet probably so easy, what am I missing?...

  1. #1
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    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):
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  2. #2
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    Quote Originally Posted by lenfromkits View Post
    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):
    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}
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  3. #3
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    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.
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