SOLUTION: find three consecutive positive odd integers such that the square of the sum of the first two exceeds the square of the third by 63

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Question 243244: find three consecutive positive odd integers such that the square of the sum of the first two exceeds the square of the third by 63
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
Let x, x+2 & x+4 be the 3 odd integers.
(x+x+2)^2=(x+4)^2+63
(2x+2)^2=x^2+8x+16+63
4x^2+8x+4=x^2+8x+79
4x^2-x^2+8x-8x+4-79=0
3x^2-75=0
3(x^2-25)=0
3(x+5)(x-5)=0
x-5=0
x=5 ans.
5+2=7 ans. for x+2
5+4=9 ans. for x+4
Proof:
(5+7)^2=9^2+63
12^2=81+63
144=144