Question 287723
<font face="Garamond" size="+2">


Let *[tex \Large x] represent the first odd integer.  Since every odd integer is 2 larger than the previous one, the next odd integer must be *[tex \Large x\ +\ 2] and the one after that, *[tex \Large x\ +\ 2\ +\ 2\ =\ x\ +\ 4]


The square of the first one is:  *[tex \Large x^2]


The square of the second one is:  *[tex \Large (x\ +\ 2)^2\ =\ x^2\ +\ 4x\ + 4]


The square of the third one is:  *[tex \Large (x\ +\ 4)^2\ =\ x^2\ +\ 8x\ + 16]


The sum of the squares, after collecting like terms (verification left as an exercise for the student) is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x^2\ +\ 12x\ +\ 20]


And this sum is equal to 155:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x^2\ +\ 12x\ +\ 20\ =\ 155]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x^2\ +\ 12x\ -\ 135\ =\ 0]


Factor and solve the quadratic.  Quadratics have two solutions.  Check both of them to see if they both satisfy the conditions of the problem.



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