<PRE><B>Question 890821</B>
find three consecutive odd positive integers
x, (x+2), (x+4), their sum: (3x+6)
;
such that 2 times the sum of all three is 105 less than the product of the first and second integers
2(3x+6) = x(x+2) - 105
6x + 12 = x^2 + 2x - 105
Form a quadratic equation on the right
0 = x^2 + 2x - 6x - 105 - 12
x^2 - 4x - 117 = 0
you can use the quadratic formula, but this will factor to:
(x+9)(x-13) = 0
The positive solution
x = 13 is the first odd number, followed by 15 and 17
:
:
See if this checks out
2(13+15+17) = 13*15 - 105
2(45) = 195 - 105

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