Question 1121770
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*[illustration 3_Side_Walkway]


Original Area: *[tex \Large 60\ \times\ 80\ =\ 4800\text{ m^2}]


Double the original area:  *[tex \Large 9600\text{ m^2}]


New length:  *[tex \LARGE 2x\ +\ 80]


New width:  *[tex \LARGE x\ +\ 60]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (2x\ +\ 80)(x\ +\ 60)\ =\ 9600]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ +\ 200x\ +\ 4800\ =\ 9600]


Put the quadratic into standard form then solve for *[tex \LARGE x].  Note:  You will get two roots to the quadratic, one will be positive and the other negative.  Which one do you suppose is the correct one?  
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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