Question 184616
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You need to re-read your problem.  The problem says that the length is 2 units longer than the width, not 2 times the width.  Furthermore, the width of the box formed is 4 inches less than the original width of the cardboard, and the length of the box formed is 4 inches less than the original length of the cardboard because of the 2 inch squares cut out of the corners before the box was folded.  So width is *[tex \Large w - 4], length is *[tex \Large w + 2 - 4 = w - 2], and height is 2.



*[tex \LARGE \ \ \ \ \ \ \ \ \ \  V = lwh]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  2(w - 4)(w - 2) = 70 \ \ \Rightarrow\ \ 2w^2 - 12w + 16 - 70 = 0 \ \ \Rightarrow\ \ w^2 - 6w - 27 = 0]


Note: *[tex \Large -9 \times 3 = -27] and *[tex \Large - 9 + 3 = -6] so this factors neatly.  One of your roots will be negative.  Discard this extraneous root because it would be absurd to have a negative measure of length.  The positive root is your width of the original cardboard and the length is two inches longer.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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