SOLUTION: The square of the first of three consecutive odd integers is 9 more than 6 times the sum of the second and the first. Find the numbers.

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Question 265347: The square of the first of three consecutive odd integers is 9 more than 6 times the sum of the second and the first. Find the numbers.
Answer by mananth(16946) About Me  (Show Source):
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The square of the first of three consecutive odd integers is 9 more than 6 times the sum of the second and the first. Find the numbers.
let the numbers be x , x+2 , x+4
(x+x+2+x+4)^2 = 6(x+x+2) +9
(3x+6)^2=12x=12+9
9x^2+36x+36=12x+3
9x^2+36x-12x+36-3=0
9x^2+24x+15=0 ( divide equation by3
3x^2+8x+5=0
3x^2+3x+5x+5=0
3x(x+1)+5(x+1)=0
(x+1)(3x+5)=0
x=-1
the numbers are -1 , +1, +3