It is currently 20 Nov 2017, 01:18

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Is (x-3)^2 = 3 - x ? (1) x 3 (2) x | x | > 0 Some

Author Message
Intern
Joined: 01 May 2009
Posts: 41

Kudos [?]: 42 [0], given: 0

Is (x-3)^2 = 3 - x ? (1) x 3 (2) x | x | > 0 Some [#permalink]

### Show Tags

20 Jul 2009, 01:49
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Is \sqrt{(x-3)^2} = 3 - x ?
(1) x ≠ 3
(2) – x | x | > 0

Some help & explanations needed please.

Many thanks!

Kudos [?]: 42 [0], given: 0

Current Student
Joined: 12 Jun 2009
Posts: 1836

Kudos [?]: 277 [0], given: 52

Location: United States (NC)
Concentration: Strategy, Finance
Schools: UNC (Kenan-Flagler) - Class of 2013
GMAT 1: 720 Q49 V39
WE: Programming (Computer Software)

### Show Tags

20 Jul 2009, 08:22
cialit0506 wrote:
Is \sqrt{(x-3)^2} = 3 - x ?
(1) x ≠ 3
(2) – x | x | > 0

Some help & explanations needed please.

Many thanks!

the square root of the (x-3)^2 basically makes whatever's in it positive all the time. on the left it is different though..
(1). INSUFFICIENT. Since the left side will ALWAYS be positive if I plug in 4 on the left i get 1 and the right get -1. If i plug in 1 i get 2 on the left and 2 on the right as well so no good.
(2). only way to have that statement true is to have a NEGATIVE x (so the - negates it..). Now if you plug in any negative number on the left you will still get a positive result and on the right you will always get positive result as well and they always equal to eachother so SUFFICIENT.

B. kudos
_________________

Kudos [?]: 277 [0], given: 52

Manager
Joined: 07 Apr 2009
Posts: 145

Kudos [?]: 13 [0], given: 3

### Show Tags

20 Jul 2009, 08:30
lets attack the option B.

-x|x| > 0

$$-x *(x) >0$$ or $$-x*(-x) > 0$$
$$x^2 < 0$$or $$x^2 > 0$$

$$x^2$$ can never be less than 0, so x cannot be positive. so from the above statement we have to conclude that x is negative.

for any negative value of X the given statement will hold good. so B is SUFF.

A is not SUFF because when x > 3, the given statement doesnt hold good.

hence 'B'

Kudos [?]: 13 [0], given: 3

Senior Manager
Joined: 01 Mar 2009
Posts: 367

Kudos [?]: 96 [0], given: 24

Location: PDX

### Show Tags

20 Jul 2009, 08:37
Well ... for option B if x is -ve :

sqrt ( x - 3) ^ 2 = 3 - x ?

let's say x=-1
so sqrt (-4)^2 = -4 .. which need not be true because sqrt of 16 can be -4 or +4 .. isn't that right ? or am I missing something ?
_________________

In the land of the night, the chariot of the sun is drawn by the grateful dead

Kudos [?]: 96 [0], given: 24

Manager
Joined: 07 Apr 2009
Posts: 145

Kudos [?]: 13 [1], given: 3

### Show Tags

20 Jul 2009, 10:24
1
KUDOS
Quote:
let's say x=-1
so sqrt (-4)^2 = -4 .. which need not be true because sqrt of 16 can be -4 or +4 .. isn't that right ? or am I missing something ?
Well ... for option B if x is -ve :

sqrt ( x - 3) ^ 2 = 3 - x ?

let's say x=-1
so sqrt (-4)^2 = -4 .. which need not be true because sqrt of 16 can be -4 or +4 .. isn't that right ? or am I missing something ?

IMO, while considering square root , we consider only positive square root.

16 = $$(+4)^2$$ or $$(-4)^2$$
$$sqrt (16) = 4$$

Kudos [?]: 13 [1], given: 3

Senior Manager
Joined: 04 Jun 2008
Posts: 286

Kudos [?]: 160 [1], given: 15

### Show Tags

20 Jul 2009, 10:40
1
KUDOS
pleonasm wrote:
Well ... for option B if x is -ve :

sqrt ( x - 3) ^ 2 = 3 - x ?

let's say x=-1
so sqrt (-4)^2 = -4 .. which need not be true because sqrt of 16 can be -4 or +4 .. isn't that right ? or am I missing
something ?

I was thinking the same

Actually its a confusion betwn sqrt x and x^2

if X^2 = 16; then x can be +4 or -4
but if asked to find out sqrt, sqrt 16 will always be 4

Kudos [?]: 160 [1], given: 15

Senior Manager
Joined: 01 Mar 2009
Posts: 367

Kudos [?]: 96 [0], given: 24

Location: PDX

### Show Tags

20 Jul 2009, 13:20
So sqrt of a number is always a positive value ... hmmmm .. ok I guess I now get it..

sqrt (16 ) = sqrt (4 *4) = sqrt(4) * sqrt(4) = 2*2 = 4
sqrt(16) = sqrt(-4*-4) = sqrt (-4) * sqrt (-4) .. oh well there's no square root for a -ve number .. Thanks guys
_________________

In the land of the night, the chariot of the sun is drawn by the grateful dead

Kudos [?]: 96 [0], given: 24

Intern
Joined: 01 May 2009
Posts: 41

Kudos [?]: 42 [0], given: 0

### Show Tags

23 Jul 2009, 08:16
Dear ppl,

Thanks!

Kudos [?]: 42 [0], given: 0

Re: Inequalities - Get Answers by Filling Up the Blanks?   [#permalink] 23 Jul 2009, 08:16
Display posts from previous: Sort by