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# If x > 0, then 1/ = 1/sqrt3x 1/ 1/(xsqrt2)

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Manager
Joined: 10 Oct 2005
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If x > 0, then 1/ = 1/sqrt3x 1/ 1/(xsqrt2) [#permalink]  18 Oct 2005, 19:24
If x > 0, then 1/[sqrt(2x)+sqrt(x)]=

1/sqrt3x

1/[2sqrt(2x)]

1/(xsqrt2)

(sqrt2-1)/sqrtx

(1+2)/sqrtx
Director
Joined: 14 Oct 2003
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Location: On Vacation at My Crawford, Texas Ranch
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Re: p/s sqrt [#permalink]  18 Oct 2005, 19:28
trickygmat wrote:
If x > 0, then 1/sqrt(2x)+sqrtx=

1/sqrt3x

1/[2sqrt(2x)]

1/(xsqrt2)

(sqrt2-1)/sqrtx

(1+2)/sqrtx

Director
Joined: 21 Aug 2005
Posts: 793
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Kudos [?]: 10 [0], given: 0

D (sqrt2-1)/sqrtx

(1/sq(2x)+sq(x))*([sq(2x)-sq(x)]/[sq(2x)+sq(x)])
= [sq(2x)-sq(x)]/x = sq(x)*[sq(2)-1]/x = [sq(2)-1]/sq(x)
Director
Joined: 14 Oct 2003
Posts: 588
Location: On Vacation at My Crawford, Texas Ranch
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gsr wrote:
D (sqrt2-1)/sqrtx

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

can you mutiply by

(sq(2x) - sq(x))
(sq(2x) +sq(x)) ?

how do you get from

sq(x)[sq(2) - 1] -------> [sq(2)-1]
--------- X ------------------- Sq(X)

Director
Joined: 21 Aug 2005
Posts: 793
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Kudos [?]: 10 [0], given: 0

Titleist wrote:
gsr wrote:
D (sqrt2-1)/sqrtx

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

can you mutiply by

(sq(2x) - sq(x))
(sq(2x) +sq(x)) ?

how do you get from

sq(x)[sq(2) - 1] -------> [sq(2)-1]
--------- X ------------------- Sq(X)

sorry typo...multiply Numerator and Den. by sqrt(2x)-sqrtx

on the other one -

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

= [sq(2)-1]/sq(x)
Director
Joined: 14 Oct 2003
Posts: 588
Location: On Vacation at My Crawford, Texas Ranch
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Kudos [?]: 10 [0], given: 0

gsr wrote:
Titleist wrote:
gsr wrote:
D (sqrt2-1)/sqrtx

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

can you mutiply by

(sq(2x) - sq(x))
(sq(2x) +sq(x)) ?

how do you get from

sq(x)[sq(2) - 1] -------> [sq(2)-1]
--------- X ------------------- Sq(X)

sorry typo...multiply Numerator and Den. by sqrt(2x)-sqrtx

on the other one -

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

= [sq(2)-1]/sq(x)

ahhh yes. this can mean only two things; i'm getting tired and my new avatar is distracting the hell out of me. i'm going to bed goodnight all! see you tomorrow. :
Manager
Joined: 03 Aug 2005
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Finally I got it!!! This question was driving me crazy.

There is another typo:

Stem is 1/(sq(2x)+sq(x)) not (1/sq(2x)+sq(x)),
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5068
Location: Singapore
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I have edited the question for trickygmat so the parentheses are correctly placed.

By the way, D is my answer as well. Same method as the rest.
VP
Joined: 20 Sep 2005
Posts: 1021
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1/[sqrt(2x)+sqrt(x)]= 1/ ( (sqrt(2) + 1)sqrt(x) ) .

Since (sqrt(2) -1 )(sqrt(2) + 1 ) = 2 - 1 = 1. Put that in the numerator and we get (sqrt(2) -1 )/sqrt(x).

All this sqrt text intead of the symbol makes typing miserable .
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