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Thread: (z-1)^2 < 1 is..

  1. #1
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    (z-1)^2 < 1 is..

    The set of real numbers z such that $\displaystyle (z-1)^2 < 1$ is an open interval in the form (a,b). Express a and b as integers.
    Last edited by mr fantastic; Jun 13th 2009 at 11:08 PM. Reason: Corrected the inequation
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  2. #2
    Senior Member apcalculus's Avatar
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    move 1 to the left side and factor to
    (x-2)x < 0
    (0,2) is the interval. see the other post for reasoning. g luck!
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  3. #3
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    Quote Originally Posted by apcalculus View Post
    move 1 to the left side and factor to
    (x-2)x < 0
    (0,2) is the interval. see the other post for reasoning. g luck!
    Thanks I got it now
    $\displaystyle (z-1)^2 < 1$
    $\displaystyle z^2-2z+1 < 1$
    $\displaystyle z^2-2z < 0$
    z(z-2) < 0
    z<0
    z-2<0 -----> z<2
    (0,2)
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  4. #4
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    hi
    $\displaystyle z(z-2)< 0 $ needs $\displaystyle z > 0$ and $\displaystyle z-2 < 0$
    the interval is indeed $\displaystyle ]0,2[$
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