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Show that the root equation (x-a)(x-b)=k^2 are always real numbers

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Henry

  1. x^2 - (a+b) x + a b - k^2 = 0

    x = [(a+b) +/- sqrt { (a+b)^2 -4(ab-k^2)} ]/2

    if
    (a+b)^2 -4(ab-k^2)
    is positive, there is no complexity

    = a^2 + 2 ab +b^2 -4ab + 4k^2

    = a^2 -2ab + b^2 + 4 k^2
    = (a-b)^2 + 4 k^2
    BOTH of those terms are positive being squares of real numbers, whew we did it !

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    Damon

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