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If x > 0, then 1/ = A. 1/sqrt(3x) B. 1/ C. 1/(xsqrt(2))

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If x > 0, then 1/ = A. 1/sqrt(3x) B. 1/ C. 1/(xsqrt(2)) [#permalink] New post 06 Feb 2006, 21:32
If x > 0, then 1/[sqrt(2x)+sqrt(x)] =

A. 1/sqrt(3x)
B. 1/[2sqrt(2x)]
C. 1/(xsqrt(2))
D. (sqrt(2)-1)/sqrt(x)
E. (1+sqrt(2))/sqrt(x)
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Re: square roots [#permalink] New post 06 Feb 2006, 22:25
joemama142000 wrote:
If x > 0, then 1/[sqrt(2x)+sqrt(x)] =

A. 1/sqrt(3x)
B. 1/[2sqrt(2x)]
C. 1/(xsqrt(2))
D. (sqrt(2)-1)/sqrt(x)
E. (1+sqrt(2))/sqrt(x)


I don't think its A, B, C

So I'm gonna go with D?

I simplified it down to

1/[(sqrt x)*(sqrt(2) + 1)]
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Re: square roots [#permalink] New post 06 Feb 2006, 22:31
TeHCM wrote:
joemama142000 wrote:
If x > 0, then 1/[sqrt(2x)+sqrt(x)] =

A. 1/sqrt(3x)
B. 1/[2sqrt(2x)]
C. 1/(xsqrt(2))
D. (sqrt(2)-1)/sqrt(x)
E. (1+sqrt(2))/sqrt(x)


I don't think its A, B, C

So I'm gonna go with D?

I simplified it down to

1/[(sqrt x)*(sqrt(2) + 1)]


After that if you multiply and divide by sqrt(2)-1, you will get D
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Re: square roots [#permalink] New post 06 Feb 2006, 22:33
joemama142000 wrote:
If x > 0, then 1/[sqrt(2x)+sqrt(x)] =

A. 1/sqrt(3x)
B. 1/[2sqrt(2x)]
C. 1/(xsqrt(2))
D. (sqrt(2)-1)/sqrt(x)
E. (1+sqrt(2))/sqrt(x)


By rationalisation

mulitply and divide the expression by [sqrt(2x)-sqrt(x)]. This gives

[sqrt(2x)-sqrt(x)]/x => (sqrt(2)-1)/sqrt(x). Hence D
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 [#permalink] New post 06 Feb 2006, 22:38
the oa is D,

can you guys show the earlier steps? im bad at this
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 [#permalink] New post 06 Feb 2006, 22:50
1/(sqrt(2x) + sqrt(x))

= 1/[sqrt(x) * ( sqrt(2) +1)]

= (sqrt(2) -1 ) /[sqrt(x) * ( sqrt(2) +1)* (sqrt(2) -1 ) ]

= (sqrt(2) -1 ) /[sqrt(x) * ( 2-1) ]

= (sqrt(2) -1 ) /[sqrt(x) ]
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 [#permalink] New post 06 Feb 2006, 22:52
joemama142000 wrote:
the oa is D,

can you guys show the earlier steps? im bad at this


Multiply and divide by sqrt(2)-1
{1/sqrt(2)*[sqrt(2)+1]}*{sqrt(2)-1/sqrt(2)-1}

put sqrt(2) = a
1 = b

Denominator = {sqrt(2)+1}*{sqrt(2)-1} = (a+b)(a-b) = a^2-b^2
= 2 -1 = 1

Hence sqrt(2)-1/sqrt(x)
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 [#permalink] New post 06 Feb 2006, 22:56
thanks guys :-D
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Re: square roots [#permalink] New post 07 Feb 2006, 17:19
prash_c wrote:
joemama142000 wrote:
If x > 0, then 1/[sqrt(2x)+sqrt(x)] =

A. 1/sqrt(3x)
B. 1/[2sqrt(2x)]
C. 1/(xsqrt(2))
D. (sqrt(2)-1)/sqrt(x)
E. (1+sqrt(2))/sqrt(x)


By rationalisation

mulitply and divide the expression by [sqrt(2x)-sqrt(x)]. This gives

[sqrt(2x)-sqrt(x)]/x => (sqrt(2)-1)/sqrt(x). Hence D


prash, how did you simplify this part?
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Re: square roots [#permalink] New post 11 Feb 2006, 04:15
giddis method , got D
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Re: square roots   [#permalink] 11 Feb 2006, 04:15
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