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

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Director
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If x > 0, then 1/ = A. 1/sqrt(3x) B. 1/ C. 1/(xsqrt(2)) [#permalink]

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06 Feb 2006, 22:32
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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)
VP
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06 Feb 2006, 23:25
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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VP
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06 Feb 2006, 23:31
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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- Bernard Edmonds

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06 Feb 2006, 23:33
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
Director
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06 Feb 2006, 23:38
the oa is D,

can you guys show the earlier steps? im bad at this
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06 Feb 2006, 23:50
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) ]
VP
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06 Feb 2006, 23:52
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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"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

- Bernard Edmonds

Director
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06 Feb 2006, 23:56
thanks guys
Director
Joined: 17 Oct 2005
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07 Feb 2006, 18:19
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?
Senior Manager
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11 Feb 2006, 05:15
giddis method , got D
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Re: square roots   [#permalink] 11 Feb 2006, 05:15
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