Question 245435
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Let *[tex \Large w] represent the width of the rectangle.  Then, since the width is one foot less than the length, the length must be represented by *[tex \Large w\ +\ 1].  The area of a rectangle is the length times the width and the area of this rectangle is given as 2, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w(w\ +\ 1)\ =\ 2]


Distribute and put into standard form:


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


Now solve the factorable quadratic and discard the negative root.  The positive root is the value of the width that you seek.


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