Question 326290
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So far, so good. *[tex \Large x(x\ +\ 8)\ =\ 84].  Looks like it has the makings of a quadratic equation.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 8x\ -\ 84\ =\ 0]


Let's see, -6 times 14 is -84 and -6 plus 14 is 8, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 6)(x\ +\ 14)\ =\ 0]


So:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 6]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -14]


Discard the negative root because you are looking for a positive measure of length.


The width is 6, hence the length is 6 plus 8 = 14. 



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