Question 1142962
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Let *[tex \Large x] represent the largest of the three integers, then the next smaller one would be *[tex \Large x\ -\ 1], and the next smaller would be *[tex \Large x\ -\ 2].


Six times the largest:  *[tex \Large 6x].  Five less than six times the largest: *[tex \Large 6x\ -\ 5].  The square of the smallest: *[tex \Large (x\ -\ 2)^2]. Two times the middle one:  *[tex \Large 2(x\ -\ 1)].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6x\ -\ 5\ =\ (x\ -\ 2)^2\ -\ 2(x\ -\ 1)]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 12\ +\ 11\ =\ 0]


Solve for *[tex \Large x].  Discard the smaller of the two roots (1) because if one is the largest of three integers they can't all be positive.
								
								
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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