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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]  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)
VP
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Location: CA
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Re: square roots [#permalink]  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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VP
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Re: square roots [#permalink]  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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- Bernard Edmonds

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Re: square roots [#permalink]  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
Director
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Kudos [?]: 62 [0], given: 0

the oa is D,

can you guys show the earlier steps? im bad at this
Senior Manager
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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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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)
_________________

"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
Joined: 17 Oct 2005
Posts: 940
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Kudos [?]: 62 [0], given: 0

thanks guys
Director
Joined: 17 Oct 2005
Posts: 940
Followers: 1

Kudos [?]: 62 [0], given: 0

Re: square roots [#permalink]  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?
Senior Manager
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Re: square roots [#permalink]  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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