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# 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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- Gabriella MontezLv 71 decade agoFavorite 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 - bertbertchanLv 51 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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