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there are three consecutive positive integers such that the sum of the squares of the smallest two is 221.
write and equation to find the three consecutive positive integers let x= the smallest integer

  • math -

    let the 3 consecutive positive integers be
    x , x+1, and x+2

    x^2 + (x+1)^2 = 221
    x^2 + x^2 + 2x + 1 = 221
    2x^2 + 2x - 220 = 0
    x^2 + x - 110 = 0
    (x+11)(x-10) = 0
    x = -11 or x = 10
    but x must be positive

    so the 3 numbers are 10,11, and 12

    check: is 10^2 + 11^2 = 221 ?
    YES

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