SOLUTION: Find three consecutive integers such that the sum of the squares of the second and third exceeds twice the square of the first by 41.

SOLUTION: Find three consecutive integers such that the sum of the squares of the second and third exceeds twice the square of the first by 41.

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Question 338996: Find three consecutive integers such that the sum of the squares of the second and third exceeds twice the square of the first by 41.
Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
Let the numbers be x, x+1 and x+2
(x+1)^2 + (x+2)^2 = 2x^2+41
x^2+2x+1+x^2+4x+4=2x^2+41
6x=36
x=6
.
Ed
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