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30 July, 07:21

Is there a solution to 16^2-8x+1=0

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Answers (2)
  1. 30 July, 07:22
    0
    Yes X = 32, here is the procedure
  2. 30 July, 07:37
    0
    If ya meant 16x²-8x+1=0 then

    complete square

    group x

    (16x²-8x) + 1=0

    undistribute 16

    16 (x²-0.5x) + 1=0

    take 1/2 of linear coefient and squre positive and negaitve of it inside parentheasees

    -0.5/2=-0.25, (-0.25) ^2=0.0625

    16 (x²-0.5x+0.0625-0.0625) + 1=0

    factor perfect square trinomial

    16 ((x-0.25) ²-0.0625) + 1=0

    expand

    16 (x-0.25) ²-1+1=0

    16 (x-0.25) ²=0

    (x-0.25) ²=0

    x-0.25=0

    x=0.25

    or, if you mean

    16²-8x+1=0

    256-8x+1=0

    257-8x=0

    add 8x both sides

    257=8x

    divide both sides by 8

    257/8=x
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