Question 497679
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Just off the top of my head, I would guess 5 feet by 6 feet.  But let's check on my intuition.


Let *[tex \Large w] represent the width.  Then *[tex \Large w\ +\ 1] must represent the length.  And since the area is 30 square feet:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w(w\ +\ 1)\ =\ 30] 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w^2\ +\ w\ -\ 30\ =\ 0]


Solve the quadratic for *[tex \Large w], then calculate *[tex \Large w\ +\ 1].


Was my guess correct?


Extra Credit.  Solve the same problem using *[tex \Large l] to represent the length and *[tex \Large l\ -\ 1] to represent the width.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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