SOLUTION: find the two consecutive integers such that the square of the sum of the two integers is 11 more than the first integer.

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Question 111627This question is from textbook Beginning & intermediate algebra
: find the two consecutive integers such that the square of the sum of the two integers is 11 more than the first integer. This question is from textbook Beginning & intermediate algebra

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
First integer - N
Next consecutive integer - N+1
N%5E2%2B%28N%2B1%29%5E2=11%2BN
N%5E2%2B%28N%5E2%2B2N%2B1%29=11%2BN
2N%5E2%2B2N%2B1=11%2BN
2N%5E2%2BN-10=0
%282N%2B5%29%28N-2%29=0
The solutions are N=-5/2 and N=2
Since you require the solution to be an integer, we only accept the second solution (N=2).
Check the answer.
N%5E2%2B%28N%2B1%29%5E2=11%2BN
2%5E2%2B3%5E2=11%2B2
4%2B9=13
13=13
True statement.
Good answer.
The integers are 2 and 3.