Question 1025873
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Let *[tex \Large x] be an odd integer.  Then the next consecutive odd integer is *[tex \Large x\ +\ 2].  The product of these two integers is *[tex \Large x^2\ +\ 2x], and their sum is *[tex \Large 2x\ +\ 2].  Six times their sum is *[tex \Large 12x\ +\ 12].  We are given that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2\ +\ 2x\ =\ 12x\ +\ 12\ + 107]


In standard form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2\ -\ 10x\ -\ 119\ =\ 0]


Solve the quadratic for BOTH values of *[tex \Large x].  Check both of the solutions; it is possible that both of them satisfy the constraints of the problem.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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