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Question 669649: Find three consecutive odd integers such that the square of the largest integer is 25 more than three times the sum of the smaller numbers. Give your answers as comma-separated numbers in ascending order.
I have tried everything I learned to answer this question, but I just seem to keep getting stuck around here(if I did it right):X^2-6x-15=0
Please help if possible it it my last question and it is due at midnight! Thank you!!
Answer by DrBeeee(684)   (Show Source):
You can put this solution on YOUR website! You must have had an error. The correct quadratic is
(1) x^2 + 2*x - 15 = 0, which has roots at
(2) x = {-5,3}, where we can get two solution
sequences
(3) (-5,-3,-1) and (3,5,7)
To check use
(4) (x+4)^2 = 25 + 3*(x + x + 2)
Is ((-1)^2 = 25+3*(-5-3))?
Is (1 = 25 + 3*(-8))?
Is (1 = 25 - 24)?
Is (1 = 1)? Yes
Is (7^2 = 25+3*(3+5))?
Is (49 = 25+3*8)?
Is (49 = 49)? Yes
Answer: There are two sequences of consecutive odd intergers that satisfy the problem statement - (-5,-3,-1) and (3,5,7).
Let me derive (1) so you may find your error.
Let x = the first odd integer
Then the next odd integer is x + 2 and the third is x + 4.
The problem statement has
(5) (x+4)^2 = 25 + 3*(x + x+2) or
(6) x^2 + 8*x + 16 = 25 + 3*(x + x+2) or
(7) x^2 + 8*x + 16 = 25 + 6*x + 6 or
(8) x^2 + 2*x - 15 = 0 which agrees with (1)
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