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 other

Question 310761: 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 given conditions...i dont get this
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
let x , x+2 , and x+4 be the integers

(x+4)^2 + 65 = x^2 + (x+2)^2

x^2 + 8x + 81 = 2x^2 + 4x + 4

0 = x^2 - 4x - 77

0 = (x-11)(x+7)

x = -7 (the given solution)

x = 11

so the numbers are 11 , 13 , 15
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