SOLUTION: Three consecutive odd integers are such that the square of the third integer is 65 less than the sum of the squares of the first two. One solution is -7, -5, and -3. Find three oth

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Question 899720: Three consecutive odd integers are such that the square of the third integer is 65 less than the sum of the squares of the first two. One solution is -7, -5, and -3. Find three other consecutive odd integers that also satisfy the five conditions.
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
let the integers be x,x+2,x+4
(x+4)^2= x^2+(x+2)^2-65
x^2+8x+16= x^2+x^2+4x+4-65
x^2-4x-77=0
x^2-11x+7x-77=0
x(x-11)+7(x-11)=0
(x-11)(x+7)=0
x= 11 Or -7
11,13,15