Question 921690
The square root of the sum of 2 consecutive odd integers is 1 less than 3 times the smaller integer. Find the integers
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{{{sqrt(n + n+2) + 1 = 3n}}}
{{{sqrt(2n+2) = 3n-1}}}
Square both sides
{{{2n+2 = 9n^2 - 6n + 1}}}
Now it's a quadratic
{{{9n^2 - 8n - 1 = 0}}}
(9n + 1)*(n - 1) = 0
n = 1
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