Question 228684
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Let *[tex \Large w] represent the width.  Twice the width is then *[tex \Large 2w].  200 yards more than that is *[tex \Large 2w + 200], which is the length.  The length times the width is the area:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2w^2\ +\ 200w\ =\ 40000]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2w^2\ +\ 200w\ -\ 40000\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w^2\ +\ 100w\ -\ 20000\ =\ 0]


Just solve the quadratic for *[tex \Large w] to get the width, then calculate the length from that.  <b>Hint:</b>  This factors.  You will get two real number roots but one of them will be negative.  Exclude the extraneous negative root because you are looking for a positive measure of length.


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