SOLUTION: the square root of the sum of two consecutive integers is eleven. find the larger integer.

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Question 246756: the square root of the sum of two consecutive integers is eleven. find the larger integer.
Found 2 solutions by suan, drk:
Answer by suan(1) About Me  (Show Source):
You can put this solution on YOUR website!
Let x and (x+1) be the consecutive integers.
sqrt%28x%2B%28x%2B1%29%29=11
(x+(x+1))=121
2x+1=121
2x=120
x=60
largest integer is x+1. So 61.

Answer by drk(1908) About Me  (Show Source):
You can put this solution on YOUR website!
Let n and n+1 be two consecutive integers. Now SQRT%28n%2Bn%2B1%29 = 11 gives us SQRT%282n%2B1%29 = 11. To get rid of the square root, you square both sides to get 2n+1 = 121. Solving for n, we get n = 60. So our two consecutive integers are 60 and 61.