Math

If x≠0, x/2=y^2 and x/4=4y, then what is the x

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asked by Bob
  1. if x/2 = y^2 ----> x = 2y^2
    if x/4 = 4y ----> x = 16y

    then 2y^2 = 16y
    2y^2 - 16y =0
    2y(y - 8 ) = 0
    y = 0 or y = 8
    then in x = 2y^2
    x = 0 or x = 128
    but x ≠ 0,
    so x = 128

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    posted by Reiny

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