Question 1120771
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Let *[tex \Large x] represent an integer.  Then the next consecutive integer is, perforce, *[tex \Large x\ +\ 1].  The product of these two integers is *[tex \Large x^2\ +\ x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2\ +\ x\ =\ 1122]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2\ +\ x\ -\ 1122\ =\ 0]


Solve the factorable quadratic.  There will be a positive root and a negative root, so that gives rise to two distinct answers to the question because negative numbers are included in the integers.  Factoring hint:  Since you multiplied two numbers that differ by only 1 to get the product 1122, the two numbers are going to be very close to the square root of 1122.
								
								
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">
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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