ty asked in 科學及數學數學 · 1 decade ago

math question

using the method of taking square roots(leave the radical sign "開方根" in the answers)

Q1) 4(1-1/3x)的二次-5=0

2 Answers

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  • 1 decade ago
    Favorite Answer

    4 ( 1 - x / 3 )^2 - 5 = 0

    4 ( 1 - x / 3 )^2 = 5

    ( 1 - x / 3 )^2 = 5 / 4

    1 - x / 3 = ± √5 / 2

    1 ± √5 / 2 = x / 3

    x = 3 ± 3√5 / 2

    2007-09-08 00:40:04 補充:

    But if yr Q is 4 ( 1 - 1 / ( 3x ))^2 - 5 = 0, then( 1 - 1 / (3x ))^2 = 5 / 41 - 1 /( 3x ) = ± √5 / 21± √5 / 2 = 1 / ( 3x )6x± 3√5 x = 2 ( 6 ± 3√5 ) x = 2 x = 2 / ( 6 ± 3√5 )

    Source(s): My Maths Knowledge
  • 1 decade ago

    4(1-1/3x)^2 - 5 = 0

    => (1-1/3x)^2 = 5/4

    => 1-1/3x = (5/4)^(1/2)

    => 1/3x = 1 - (5/4)^(1/2)

    => 3x = 1 / [ 1-(5/4)^(1/2) ]

    => x = 1 / 3 [ 1 - (5/4)^(1/2) ]

    ^(1/2) represents square root

    Source(s): my own self
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