Question 152498: Three consecutive odd integers are such that the square of the third is 264 more than the square of the second. Find the three integers.
Answer by mangopeeler07(462)   (Show Source):
You can put this solution on YOUR website! consecutive odd integers= x, x+2, x+4
(x+4)^2=(x+2)^2+264
Simplify
x^2+8x+16=x^2+4x+4+264
Combine like terms on each side
x^2+8x+16=x^2+4x+268
Subtract 4x from both sides
x^2+4x+16=x^2+268
Subtract 268 from both sides
x^2+4x-252=x^2
Subtract x^2 from both sides
4x-252=0
Add 252 to both sides
4x=252
Divide both sides by 4
x=63
consecutive odd integers= x, x+2, x+4
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Answer: 63, 65, and 67
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