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

**yoman360** · June 13th 2009, 08:06 PM

The set of real numbers z such that $(z-1)^2 < 1$ is an open interval in the form (a,b). Express a and b as integers.

**apcalculus** · June 13th 2009, 09:47 PM

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!

**yoman360** · June 13th 2009, 10:51 PM

Originally Posted by apcalculus

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
$(z-1)^2 < 1$
$z^2-2z+1 < 1$
$z^2-2z < 0$
z(z-2) < 0
z<0
z-2<0 -----> z<2
(0,2)

**Raoh** · June 14th 2009, 03:25 AM

hi
$z(z-2)< 0$ needs $z > 0$ and $z-2 < 0$
the interval is indeed $]0,2[$

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