Question 649789
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Let *[tex \LARGE w] represent the width.  Then *[tex \LARGE w\ +\ 2] must represent the length.  Since the area is given by the length times the width:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w^2\ +\ 2w\ =\ 143]


Which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w^2\ +\ 2w\ -\ 143\ =\ 0]


Solve the factorable quadratic for *[tex \LARGE w] then calculate *[tex \LARGE w\ +\ 2]


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