Question 188544
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Integer, not inte<b>r</b>ger, please.


Let <i>x</i> be the first integer, then <i>x</i> + 1 is the second, and <i>x</i> + 2 is the third, so:


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


Simplify:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2 + x = 5x + 10 - 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2 - 4x - 5 = 0]


Factor:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  (x + 1)(x - 5) = 0]


so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x = -1]


and the three integers are -1, 0, and 1


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x = 5]


and the three integers are 5, 6, and 7


Check:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  (-1)(0) =^? 5(1) - 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  (5)(6) =^? 5(7) - 5]


You get to do the arithmetic.


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