Question 374906
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Let *[tex \Large x] represent the width of the rectangle.  Then *[tex \Large x\ +\ 4] must represent the length.  The area of a rectangle is the length times the width, hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x(x\ +\ 4)\ =\ 96]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 4x\ -\ 96\ =\ 0]


For any quadratic equation in the form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2\ +\ bx\ +\ c\ =\ 0]


the solutions are given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-b\ \pm\ \sqrt{b^2\ -\ 4ac}}{2a}]


In your case:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 1\ \ ], *[tex \LARGE b\ =\ 4\ \ ], and *[tex \LARGE c\ =\ -96]


Make the appropriate substitutions and do the indicated arithmetic.  Properly performed, the operation will result in a positive and a negative root.  Since you are attempting to determine a positive measure of length, discard the negative root.  The positive root is the width and the width plus 4 is the length.


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