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Intern
Joined: 29 Jul 2009
Posts: 17

Kudos [?]: 5 [0], given: 6

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06 Oct 2009, 22:49
00:00

Difficulty:

(N/A)

Question Stats:

50% (00:01) correct 50% (00:31) wrong based on 4 sessions

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What is √ (x^^2 -6x + 9) + √(2-x) + (x-3) if each expression under the square root is greater than or equal to 0?

A√(2-x)
B √(2-x) + 2x -6
C √(2-x) + x-3
D √(x-2) + 2x -6
E √(x-2)

OA is
[Reveal] Spoiler:
A

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Manager
Joined: 08 Nov 2005
Posts: 180

Kudos [?]: 22 [0], given: 10

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06 Oct 2009, 23:13
This is strange. Where did you find the OA?
where is this problem from?
You can argue both ways
\sqrt{(x-3)^2} = x-3 or 3-x
so both A and B could be the answer.
something is missing.

Kudos [?]: 22 [0], given: 10

Intern
Joined: 01 Jan 2009
Posts: 29

Kudos [?]: 2 [2], given: 1

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07 Oct 2009, 05:08
2
KUDOS
Its not a Radical problem.. in fact just requires little reasoning.

sqrt{(x-3)^2} = x-3 or 3-x

If X-3 is positive, then X>3 ; So 2-X will be a -ve no

So , the expression must be = 3-x + √(2-x) + (x-3) = √(2-x) ( Answer A )

Kudos [?]: 2 [2], given: 1

Founder
Joined: 04 Dec 2002
Posts: 16108

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Location: United States (WA)
GMAT 1: 750 Q49 V42

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07 Oct 2009, 20:18
 ! Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/No posting of PS/DS questions is allowed in the main Math forum.

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Kudos [?]: 29415 [0], given: 5348

Manager
Joined: 08 Nov 2005
Posts: 180

Kudos [?]: 22 [0], given: 10

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08 Oct 2009, 12:46
good job. LAKSH0328. I didnt think about the 2-x.

Kudos [?]: 22 [0], given: 10

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