Question 522260
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Let *[tex \Large x] represent "another" number.  Then "one" number is *[tex \Large 3x\ +\ 1]


then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x(3x\ +\ 1)\ =\ 154]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x^2\ +\ x\ -\ 154\ =\ 0]


Solve the quadratic for the two values of "another" number then calculate *[tex \Large 3x\ +\ 1] for each.  Check both results to ensure you didn't introduce an extraneous root.


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