If one of the roots of the equation x² - Kx+27=0 is the square root of the other, find the other root and hence the value of K.?

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  • khalil
    Lv 7
    2 years ago
    Favourite answer

    ax² + bx + c = 0

    the roots ... m ....√m

    the product of the roots = c/a

    m*√m = 27

    m^3/2 = 27

    m = 9 and m' = 3

    the sum of the roots = -b/a

    9 + 3 = k

    k = 12

  • sepia
    Lv 7
    2 years ago

    If one of the roots of the equation x² - Kx + 27 = 0 is the square root of the other,

    find the other root and hence the value of K.

    K = 12

    x² - Kx + 27 = x² - 12x + 27 = (x - 3)(x - 9)

  • 2 years ago

    Factor

    ( x - 9)(x - 3) = 0

    x = 9

    &

    x = 3

    '3' is the square root of '9'

    x^2 - 9x - 3x + 27 = 0

    x^2 - 12x + 27 = 0

    Hence K = 12

  • David
    Lv 7
    2 years ago

    Using the discriminant the value of k is +/- the square root of 108

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