SOLUTION: The product of two consecutive integers is 71 more than the square of the preceding smaller integer. Find the numbers.

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Question 778871: The product of two consecutive integers is 71 more than the square of the preceding smaller integer. Find the numbers.
Answer by ramkikk66(644) About Me  (Show Source):
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The product of two consecutive integers is 71 more than the square of the preceding smaller integer. Find the numbers.
Ans:
Let the integers be n and n+1. Then the preceding smaller integer is (n-1).
It is given that
n%2A%28n%2B1%29+=+%28n-1%29%5E2+%2B+71
n%5E2+%2B+n+=+n%5E2+-+2%2An+%2B+1+%2B+71+=+n%5E2+-+2%2An+%2B+72
Cancelling out n^2 and simplifying
3%2An+=+72
n+=+24
The numbers are 24 and 25.
Hope you got it :)