Question 265728
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Let *[tex \Large x] represent April's age.  Then *[tex \Large x\ -\ 1] is Mae, *[tex \Large x\ -\ 2] is June, and *[tex \Large x\ -\ 3] is Julie.


The sum of the squares of their ages:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ \left(x\ -\ 1\right)^2\ +\ \left(x\ -\ 2\right)^2\ +\ \left(x\ -\ 3\right)^2\ =\ 534]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ \left(x^2\ -\ 2x\ -\ 1\right)\ +\ \left(x^2\ -\ 4x\ +\ 4\right)\ +\ \left(x^2\ -\ 6x\ +\ 9\right)\ =\ 534]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x^2\ -\ 12x\ +\ 14\ =\ 534]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x^2\ -\ 12x\ -\ 520\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 3x\ -\ 130\ =\ 0]


Giving you an easily factorable quadratic.  Discard the negative root because ages of people are positive numbers.  The positive root is April's age and you just count down for the other three.


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