SOLUTION: find two consecutive integers whose sum is 16 less than the square of the larger number

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Question 489417: find two consecutive integers whose sum is 16 less than the square of the larger number
Answer by jorel1380(3719)   (Show Source): You can put this solution on YOUR website!
n+(n+1)+16=(n+1)2
2n+17=n2+2n+1
n2-16=0
n2=16
n=4 or -4
The two possible answers for n are 4 and -4, therefore the two possible values for n+1 are 5 and -3. ☺☺☺☺

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