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The sum of the squares of two consecutive positive even integers is one hundred sixty-four. Find the two integers.

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2 answers

  1. n^2 + (n+2)^2 = 164
    2n^2 + 4n - 160 = 0
    n^2 + 2n - 80 = 0
    (n+10)(n-8) = 0
    ...

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  2. x 1st + even integer
    x+2 2nd + even integer

    x^2+(x+2)^2= 164
    x^2+x^2+4x+4-164=0
    1/2(2x^2+4x-160=0)
    x^2+2x-80=0
    (x+10)(x-8)=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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