SOLUTION: 3 consecutive integers are such that the square of the third is 100 more than the square of the second. Find the three integers.

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Question 178138: 3 consecutive integers are such that the square of the third is 100 more than the square of the second. Find the three integers.
Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
lets call our integers a,b=a+1,and c=a+2
:
%28a%2B2%29%5E2=%28a%2B1%29%5E2%2B100
:
a%5E2%2B4a%2B4=a%5E2%2B2a%2B1%2B100 multiplied terms
:
4a%2B4=2a%2B101cancelled squared terms and combined like terms on each side
:
2a=97
:
a=97%2F2
:
there are NO integers for which these stipulations are true
:
had the 2nd number been squared plus 101 instead of 100 then the integers would have been 49,50, and 51. I suspect someone just made a mistake when writing the problem or there was a mis-typed number
:
the rational number answers under these stipulations would be
:
system%2897%2F2%2C99%2F2%2C101%2F2%29