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March 1, 2015

March 1, 2015

Posted by **Jess** on Thursday, April 26, 2012 at 3:19pm.

- algebra 2 -
**Reiny**, Thursday, April 26, 2012 at 3:37pmI will assume you want the area to be

(1-x)/(x^2 +1)

area = l x w

so (1-x)/(x^2 + 1 ) = w(x-1)

w = [ (1-x)/(x^2 + 1) ] / (x-1)

= -1/(x^2 + 1)

What I would say is "There is no real number for the width"

-since x^2 is always positive, then x^2 + 1 is always positive

Thus -1/(a positive) = a negative

We cannot have a negative width

unless you want your typed version to have a different meaning besides my interpretation.

the way you typed it it would have a value of 2 - 1/x

and I am pretty sure you didn't mean that.

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