SOLUTION: The square of the largest of 3 positive consecutive integers is 140 less than the sum of the squares of the two smaller integers. What are the integers
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Question 468039: The square of the largest of 3 positive consecutive integers is 140 less than the sum of the squares of the two smaller integers. What are the integers Answer by jorel1380(3719) (Show Source):
You can put this solution on YOUR website! x2+(x+1)2=(x+2)2+140
x2+x2+2x+1=x2+4x+4+140
2x2+2x+1=x2+4x+144
x2-2x-143=0
(x-13)(x+11)=0
x=13 or -11
Throwing out the negative answer, we get the consecutive integers to be 13,14, and 15..
169+196-140=225
365-140=225
225=225..
