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Is (x-3)^2 =3-x ? (1) x not = 3 (2) -x|x| > 3

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Is (x-3)^2 =3-x ? (1) x not = 3 (2) -x|x| > 3 [#permalink] New post 20 Dec 2009, 06:50
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Is \sqrt{(x-3)^2}=3-x?

(1) x\neq{3}

(2) -x|x| > 3
[Reveal] Spoiler: OA
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Re: TOUGH DS Q! [#permalink] New post 20 Dec 2009, 17:25
Ans: E.

First statement is not sufficient because if x=4, it may or may not be true. LHS = +1 or -1 while RHS = -1.

Second statement leads us to -3^1/2 < x < 0. For all the values between 0 and -3^1/2, which leads to |LHS| = RHS.
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Re: TOUGH DS Q! [#permalink] New post 21 Dec 2009, 01:47
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jan4dday wrote:
Is \sqrt{(x-3)^2}=3-x?

(1) x not = 3

(2) -x|x| > 3

will give OA as soon as 1st few replies come in


Remember: \sqrt{x^2}=|x|.

\sqrt{(x-3)^2}=|x-3|. So the question becomes is |x-3|=3-x.

When x>3, then RHS is negative, but LHS (absolute value) is never negative, hence in this case equations doesn't hold true.

When x\leq{3}, then LHS=|x-3|=-x+3=3-x=RHS, hence in this case equation holds true.

Basically question asks is x\leq{3}?

(1) x is not equal to 3. Clearly insufficient.

(2) -x|x| > 3, basically this inequality implies that x<0, hence x<3. Sufficient.

Answer: B.
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Re: How to solve this Question...I have no clue... [#permalink] New post 27 Feb 2010, 20:47
is \sqrt{(x-3)^2} = 3-x?
square root of a number is positive.
if x<=3 .. 3-x is always positive
if x>3 ... 3-x is negative

st 1) x!=3 .
not sufficient as we dont know if x>3 or x<3
st 2) -x |x| > 0
|x| is always greater than 0, so x has to be negative for the expression to be > 0

if x is negative, then \sqrt{(x-3)^2} = 3-x

B
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Re: How to solve this Question...I have no clue... [#permalink] New post 27 Feb 2010, 20:58
Thank you very much.
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Re: How to solve this Question...I have no clue... [#permalink] New post 17 Mar 2011, 02:58
Just a slight correction I think...
St2 is saying x<-sqrt of 3 which is still a sufficient condition.
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Re: TOUGH DS Q! [#permalink] New post 17 Mar 2011, 03:27
I got x<-1.732 from statement 2

This satisfies the x<3 requisite
Hence B

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Re: How to solve this Question...I have no clue... [#permalink] New post 17 Mar 2011, 22:07
sqrt((x-3)^2) = |x-3| = x-3 if x > 3 or 3-x if x < 3

(1) is not enough

(2) -x|x| is positive then

So -x is also +ve and hence x is -ve

so x < 3, hence answer is B
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Re: How to solve this Question...I have no clue... [#permalink] New post 18 Mar 2011, 19:49
I'm a bit confused by this problem
Why can I not square both sides of the statement to get rid of the radical?
I will then have (x-3)^2=(3-x)^2
Statement 1 and 2 both will not apply in this case and the answer is then E.

However if I choose Bunuels method of doing it, I still dont get how B is the answer.
Statement 2 basically says that x=-2,-3....and so on and so forth will satisfy the statement. So basically all negative numbers equal to or less than 2 will satisfy the condition...
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Is square_root of (x-3)^2 = 3 - x ? [#permalink] New post 27 Jun 2012, 11:08
Is \sqrt{(x-3)^2} = 3 - x ?

(1) x is not equal to 3
(2) -x |x| > 0

According to my reasoning:
\sqrt{(x-3)^2}= |x-3|
So, the question is:
Is |x-3| = 3 -x?
In other words:
Is x - 3 < 0 ?
Is x < 3?

So, let's see the clues:
(1) x is not equal to 3
INSUFFICIENT.

(2) -x |x| > 0
Based on this info we know that x is not zero.
So,
-x > 0
x < 0
SUFFICIENT.

IMO, the OA is :
[Reveal] Spoiler:
B

Please, confirm whether I am Ok.
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Re: Is square_root of (x-3)^2 = 3 - x ? [#permalink] New post 27 Jun 2012, 11:15
Expert's post
Merging similar topics.

metallicafan wrote:
Is \sqrt{(x-3)^2} = 3 - x ?

(1) x is not equal to 3
(2) -x |x| > 0

According to my reasoning:
\sqrt{(x-3)^2}= |x-3|
So, the question is:
Is |x-3| = 3 -x?
In other words:
Is x - 3 < 0 ?
Is x < 3?

So, let's see the clues:
(1) x is not equal to 3
INSUFFICIENT.

(2) -x |x| > 0
Based on this info we know that x is not zero.
So,
-x > 0
x < 0
SUFFICIENT.

IMO, the OA is :
[Reveal] Spoiler:
B

Please, confirm whether I am Ok.


You are right OA is B.
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Re: TOUGH DS Q! [#permalink] New post 28 Jun 2012, 13:53
Bunuel wrote:
jan4dday wrote:
Is \sqrt{(x-3)^2}=3-x?

(1) x not = 3

(2) -x|x| > 3

will give OA as soon as 1st few replies come in


Remember: \sqrt{x^2}=|x|.

\sqrt{(x-3)^2}=|x-3|. So the question becomes is |x-3|=3-x.

When x>3, then RHS is negative, but LHS (absolute value) is never negative, hence in this case equations doesn't hold true.

When x\leq{3}, then LHS=|x-3|=-x+3=3-x=RHS, hence in this case equation holds true.

Basically question asks is x\leq{3}?

(1) x is not equal to 3. Clearly insufficient.

(2) -x|x| > 3, basically this inequality implies that x<0, hence x<3. Sufficient.

Answer: B.


Dear Bunuel,
is it possible in this question to solve by taking conditions x>0 and x<0 for |x-3| = 3-x instead of x>3 ?
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Re: Is (x-3)^2 =3-x ? (1) x not = 3 (2) -x|x| > 3 [#permalink] New post 01 Jul 2012, 12:07
B it is!!Good explanation though!
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DS Question: Is ((x-3)^2)^(1/2) = 3-x? (1) x<>3 (2) -x|x|>0 [#permalink] New post 21 Nov 2012, 19:11
I have no idea how to solve this. Aren't the two of them always equal? I thought LHS = RHS without (1) and (2). VERY confused. Please help
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Re: DS Question: Is ((x-3)^2)^(1/2) = 3-x? (1) x<>3 (2) -x|x|>0 [#permalink] New post 21 Nov 2012, 21:02
arvindbhat1887 wrote:
I have no idea how to solve this. Aren't the two of them always equal? I thought LHS = RHS without (1) and (2). VERY confused. Please help


The question is

Is \sqrt{(x-3)^2}= 3-x?

(1) x<>3

(2) -x|x|>0

\sqrt{y} can only be positive or 0.

So,The question is basically asking whether 3-x >= 0

1) Insufficient. If x=1, true, If x=10, false.

2) From this we get that x is negative. So -x is postivie. So, 3 + (-x) will always be a positive number. Sufficient.

Answer is hence B.

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Re: DS Question: Is ((x-3)^2)^(1/2) = 3-x? (1) x<>3 (2) -x|x|>0 [#permalink] New post 22 Nov 2012, 04:48
Expert's post
arvindbhat1887 wrote:
I have no idea how to solve this. Aren't the two of them always equal? I thought LHS = RHS without (1) and (2). VERY confused. Please help


Merging similar questions. Please refer to the solutions above.

Also, please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html
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Re: TOUGH DS Q! [#permalink] New post 09 Jan 2013, 18:14
Bunuel wrote:
jan4dday wrote:
Is \sqrt{(x-3)^2}=3-x?

(1) x not = 3

(2) -x|x| > 3

will give OA as soon as 1st few replies come in


Remember: \sqrt{x^2}=|x|.

\sqrt{(x-3)^2}=|x-3|. So the question becomes is |x-3|=3-x.

When x>3, then RHS is negative, but LHS (absolute value) is never negative, hence in this case equations doesn't hold true.

When x\leq{3}, then LHS=|x-3|=-x+3=3-x=RHS, hence in this case equation holds true.

Basically question asks is x\leq{3}?

(1) x is not equal to 3. Clearly insufficient.

(2) -x|x| > 3, basically this inequality implies that x<0, hence x<3. Sufficient.

Answer: B.




(2) Actually -x|x| > 3, implies that x<-sqrt(3), but still, as you mentioned, consequently implies that x<3. Sufficient.
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Re: TOUGH DS Q! [#permalink] New post 31 Mar 2014, 11:55
Hi Bunuel,

In the solution provided by you, why not choosing two conditions x>=0 and x<0? Also how did you come to this conclusion that Basically question is asking x\leq{3}?

Please help me understand this concept.

Regards,
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Re: TOUGH DS Q! [#permalink] New post 31 Mar 2014, 12:22
email2vm wrote:
Hi Bunuel,

In the solution provided by you, why not choosing two conditions x>=0 and x<0? Also how did you come to this conclusion that Basically question is asking x\leq{3}?

Please help me understand this concept.

Regards,
Ravi


I got it now!

|x-3| = (3-x) ===> |3-x| = (3-x) ===>which is only possible when (3-x) >=0 (i.e +ve)

(like |x| =x only when x>=0)

so we have 3-x >=0 ====> x<=3


Also for option B

-x|x| > 3
==> x|x| < 3
so |x| is always positive but so make it less than 3 , x must be negative. i.e. x<0
Re: TOUGH DS Q!   [#permalink] 31 Mar 2014, 12:22
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