SOLUTION: find three consecutive odd integers such that the sum of all three is 42 less than the product of the second and third integers.

Question 531676: find three consecutive odd integers such that the sum of all three is 42 less than the product of the second and third integers.
Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
Define the consecutive odd integers than:
x
x+2
x+4
.
3x+6 +42 = (x+2)(x+4)
.
3x + 48 = x^2 +6x + 8
.
x^2 +6x +8 = 3x +48
.
x^2 +6x +8 -3x -48 = 0
.
x^2 +3x -40 = 0
.
(x+8)(x-5) = 0
.
x = -8 or 5
.
Since -8 is not odd, we choose x = 5.
.
Answer: 5, 7, and 9.
.
Check the answer to determine if this is the correct answer.
Sum of all 3 = 21.
Product of 7*9 = 63.
63-21 = 42.
Correct.
.
Done.
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