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Math Help - Rearranging formula for z

  1. #1
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    Cool Rearranging formula for z

    (z+h)+[(Q/b*(z+h))^2/(2*g)]=z

    Could someone please solve for z.

    We know all the variables apart from z.

    Cheers
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  2. #2
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    Quote Originally Posted by Haris View Post
    (z+h)+[(Q/b*(z+h))^2/(2*g)]=z

    Could someone please solve for z.

    We know all the variables apart from z.

    Cheers

    What you wrote is, apparently, \displaystyle{z+h+\left[\frac{Q}{b}\cdot\frac{(z+h)^2}{2g}\right]}=z\Longrightarrow \displaystyle{h+\left[\frac{Q}{b}\frac{(z+h)^2}{2g}\right]}=0\Longrightarrow \displaystyle{(z+h)^2=-\frac{2bgh}{Q}} ...

    Yet the above is so simple that perhaps you meant another first expression...

    Tonio
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  3. #3
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    z+h+(\frac{Q}{b(z+h)})^2*\frac{1}{2*g}=z

    sorry, it was meant to look like this.
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  4. #4
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    \displaystyle z+h+\left(\frac{Q}{b(z+h)}\right)^2\times\frac{1}{  2g}=z

    Subtract z from both sides

    \displaystyle h+\left(\frac{Q}{b(z+h)}\right)^2\times\frac{1}{2g  }=0

    Take h from both sides

    \displaystyle \left(\frac{Q}{b(z+h)}\right)^2\times\frac{1}{2g}=-h

    Now lets simplify this a bit

    \displaystyle \frac{Q^2}{b^2(z+h)^2}\times\frac{1}{2g}=-h

    \displaystyle \frac{Q^2}{2gb^2(z+h)^2}=-h

    Now your turn, multiply both sides through by \displaystyle (z+h)^2
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  5. #5
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    Well, a "z" cancels out from each side (why even bother showing those?!),
    leaving a less wieldy:
    Q^2 / [b(z + h)]^2 = -2gh
    Surely you can handle that...
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