Question 332644
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Let *[tex \Large x] represent the width.  Then *[tex \Large x\ +\ 4] represents the length.  Since area is length times width and the area is 21 square meters, in the first place it should be casual to the most obvious observer that the length and width have to be 7 and 3 meters, but you can always say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x(x\ +\ 4)\ =\ 21]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 4x\ -\ 21\ =\ 0]


And just solve the quadratic for *[tex \Large x]


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