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If x > 0, then x = A. 3x B. 1 C. x2 D. x E. 1+x2 [#permalink]
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11 Jun 2007, 07:22
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If x > 0, then x =
A. 3x B. 1 C. x2 D. x E. 1+x2
Last edited by nick_sun on 28 Jun 2008, 02:30, edited 1 time in total.



Manager
Joined: 23 Dec 2006
Posts: 134

A, unless there is some subtlety that I'm missing.



Director
Joined: 19 Mar 2007
Posts: 522

sludge wrote: A, unless there is some subtlety that I'm missing.
It is not "A". The unofficial answer is D. Could anybody please explain why it is D.



SVP
Joined: 01 May 2006
Posts: 1796

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



Director
Joined: 06 Sep 2006
Posts: 736

Guys, what happens if 'x > 0' not provided. what 'x > 0' means here?



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Joined: 01 May 2006
Posts: 1796

asaf wrote: Guys, what happens if 'x > 0' not provided. what 'x > 0' means here?
We need it because sqrt(x) is in the expression : x has to be positive or nul to make srqt() existing. Also, if x=0, we have 1/0, impossible for the GMAT context



GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore

For a moment, I got A because that square root notation appears like a letter 'v' on my screen. (With a variable like 'v', it works out to be A)



Manager
Joined: 28 Aug 2006
Posts: 160

Fig,
You solved the problem as if it were an Theorem and you knew where to arrive at the end.
This is an ambiguous question. Thanks for the info



Director
Joined: 13 Mar 2007
Posts: 543
Schools: MIT Sloan

LOL yes, the sqrt appears as V which left me blinking !!
its a simple conjucate problem !



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Joined: 01 May 2006
Posts: 1796

vijay2001 wrote: Fig,
You solved the problem as if it were an Theorem and you knew where to arrive at the end.
This is an ambiguous question. Thanks for the info
On such question, I try to simplify the expression by:
> factorizing what I can (ie: 1/sqrt(x))
> pulling up from the denominator to the numerator the constants containing sqrt() (Note that it could be to move a sqrt(x) up also)



Manager
Joined: 17 Oct 2006
Posts: 52

these kind of questions (i have forgot their real name) were there in my 9th grade and loved to do them back then. Now when i have seen one such question, i dont know what like after 15 years or so, it took me a few moments to figure out what to do.
anyway, the rules that my teacher told us were, 1)always try to get rid of sqrt from the denominators and 2) use a^2b^2=(a+b)(ab) formula.
SO by multiplying and dividing the fraction with sqrt2x+sqrtx, we can easily solve the problem.
and yes, the answer is D.



Director
Joined: 30 Nov 2006
Posts: 591
Location: Kuwait

Using x is positive, 1/[v(2x)+vx] = 1/v[2x+x] = 1/v(3x)
Answer: A
This question is too easy to be on the GMAT.



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Joined: 01 May 2006
Posts: 1796

Mishari wrote: Using x is positive, 1/[v(2x)+vx] = 1/v[2x+x] = 1/v(3x)
Answer: A
This question is too easy to be on the GMAT.
v is actually sqrt()... Hardening the question & making it a GMAT question?



Manager
Joined: 17 Oct 2006
Posts: 52




Director
Joined: 30 Nov 2006
Posts: 591
Location: Kuwait

Now this is a GMAT question, a good one actually:
sqrt(2x) = sqrt(2) * sqrt(x)
1/[sqrt(2x)+sqrt(x)] = 1/[(sqrt(2)*sqrt(x)) + sqrt(x)]
= 1/[sqrt(x)*(sqrt(2)+1)]
= 1/[sqrt(x)*(sqrt(2)+sqrt(1)]
= 1/sqrt(x)*sqrt(3)
= 1/sqrt(3x)
ANSWER: A
Thanks Fig



SVP
Joined: 01 May 2006
Posts: 1796

Mishari wrote: Now this is a GMAT question, a good one actually: sqrt(2x) = sqrt(2) * sqrt(x) 1/[sqrt(2x)+sqrt(x)] = 1/[(sqrt(2)*sqrt(x)) + sqrt(x)] = 1/[sqrt(x)*(sqrt(2)+1)] = 1/[sqrt(x)*(sqrt(2)+sqrt(1)] = 1/sqrt(x)*sqrt(3)= 1/sqrt(3x) ANSWER: AThanks Fig
U are weclome
Also, sqrt(2)+sqrt(1) != sqrt(3)... Similarly sqrt(4) = 2 != sqrt(2) + sqrt(2) = 1,7 + 1,7 = 3,4



Director
Joined: 30 Nov 2006
Posts: 591
Location: Kuwait

Okay then .. I dunno how to solve .. I give up










