Algebra
posted by LaShawn .
The sum of the squares of two consecutive positive even integers is one hundred sixtyfour. Find the two integers.

n^2 + (n+2)^2 = 164
2n^2 + 4n  160 = 0
n^2 + 2n  80 = 0
(n+10)(n8) = 0
... 
x 1st + even integer
x+2 2nd + even integer
x^2+(x+2)^2= 164
x^2+x^2+4x+4164=0
1/2(2x^2+4x160=0)
x^2+2x80=0
(x+10)(x8)=0
x=10 & x=8, disregard 10 since we are looking for positive integers...
so, x=8 (1st even)
x+2= 10 (2nd even)
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