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

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Is sqrt ((x-3)^2) = 3-x? [#permalink] New post 24 Apr 2008, 16:42
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Is \sqrt{(x-3)^2} = 3-x?

(1) x\neq{3}
(2) -x|x| > 0
[Reveal] Spoiler: OA

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Re: gmatprep DS: is sqrt ((x-3)^2) = 3-x? [#permalink] New post 24 Apr 2008, 16:50
Nevermind, I've got it.

Here is the answer, part of the clue is from http://gmatclub.com/forum/7-t61120:

sqrt ((x-3)^2) is |x-3|

if x = 1 or 2 or 0, then |x-3| = 3-x

if x = 5, etc, then |x-3| is not equal to 3-x
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Re: gmatprep DS: is sqrt ((x-3)^2) = 3-x? [#permalink] New post 24 Apr 2008, 20:45
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From the que: 3-x is always >0 --> x has to be less than 3.

Option 1: X can also be >3 when ans fails so insufficient
Option 2: -x|x|>0 implies x is always < 0 which means x is less than 3 hence sufficient.

Ans : B
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Re: gmatprep DS: is sqrt ((x-3)^2) = 3-x? [#permalink] New post 24 Apr 2008, 23:23
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gmatnub wrote:
is sqrt ((x-3)^2) = 3-x?

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

The oa is B, but why is A alone not enough?


given |x-3| can be equal to 3-x for x < 3,

1) X can be greater than 3
2) X is less than 0, i.e x < 3, for all x.

so B
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Re: gmatprep DS: is sqrt ((x-3)^2) = 3-x? [#permalink] New post 13 Jun 2010, 01:57
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I understand that 1) is insuff

But for 2) -x|x| > 0 means x cant be +ve => |x| = -x so that -x (-x) = x^2> 0

If x is -ve => (x-3)^2 = X^2+9-6x = (-ve)^2+9-6(-ve) = +ve+9-(-ve) = +ve +9 + (+ve) = +ve

=> sqrt ((x-3)^2) = +X-3

=> sqrt ( (x-3) ^2 ) is not equal to 3-x

=> Option B

Am I right In my logic.Please help
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Re: gmatprep DS: is sqrt ((x-3)^2) = 3-x? [#permalink] New post 13 Jun 2010, 02:41
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gautamsubrahmanyam wrote:
I understand that 1) is insuff

But for 2) -x|x| > 0 means x cant be +ve => |x| = -x so that -x (-x) = x^2> 0

If x is -ve => (x-3)^2 = X^2+9-6x = (-ve)^2+9-6(-ve) = +ve+9-(-ve) = +ve +9 + (+ve) = +ve

=> sqrt ((x-3)^2) = +X-3

=> sqrt ( (x-3) ^2 ) is not equal to 3-x

=> Option B

Am I right In my logic.Please help


Yes, the answer for this question is B.

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

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

Couple of things:

The point here is that square root function can not give negative result: wich means that \sqrt{some \ expression}\geq{0}.

So \sqrt{x^2}\geq{0}. But what does \sqrt{x^2} equal to?

Let's consider following examples:
If x=5 --> \sqrt{x^2}=\sqrt{25}=5=x=positive;
If x=-5 --> \sqrt{x^2}=\sqrt{25}=5=-x=positive.

So we got that:
\sqrt{x^2}=x, if x\geq{0};
\sqrt{x^2}=-x, if x<0.

What function does exactly the same thing? The absolute value function! That is why \sqrt{x^2}=|x|

Back to the original question:

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

When x>3, then RHS (right hand side) 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\neq{3}. Clearly insufficient.

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

Answer: B.

Hope it helps.
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Re: Is sqrt ((x-3)^2) = 3-x? [#permalink] New post 16 Sep 2013, 01:42
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Re: gmatprep DS: is sqrt ((x-3)^2) = 3-x? [#permalink] New post 23 Sep 2013, 07:09
Hello Bunuel,

Please help me understand where I am going wrong.

After this point..

|x-3|=3-x

Can this equation be written this way?

1)
x-3 = 3-x => x = 3

2)
-(x-3) = 3-x .. this leads to nothing

So I concluded that the question is whether x=3 and hence I chose A as answer.. but I am wrong.

What is that I doing wrong here?

Thanks
C23678


Bunuel wrote:
Yes, the answer for this question is B.

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

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

Couple of things:

The point here is that square root function can not give negative result: wich means that \sqrt{some \ expression}\geq{0}.

So \sqrt{x^2}\geq{0}. But what does \sqrt{x^2} equal to?

Let's consider following examples:
If x=5 --> \sqrt{x^2}=\sqrt{25}=5=x=positive;
If x=-5 --> \sqrt{x^2}=\sqrt{25}=5=-x=positive.

So we got that:
\sqrt{x^2}=x, if x\geq{0};
\sqrt{x^2}=-x, if x<0.

What function does exactly the same thing? The absolute value function! That is why \sqrt{x^2}=|x|

Back to the original question:

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

When x>3, then RHS (right hand side) 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\neq{3}. Clearly insufficient.

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

Answer: B.

Hope it helps.
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Re: gmatprep DS: is sqrt ((x-3)^2) = 3-x? [#permalink] New post 24 Sep 2013, 06:22
Expert's post
c23678 wrote:
Hello Bunuel,

Please help me understand where I am going wrong.

After this point..

|x-3|=3-x

Can this equation be written this way?

1)
x-3 = 3-x => x = 3

2)
-(x-3) = 3-x .. this leads to nothing

So I concluded that the question is whether x=3 and hence I chose A as answer.. but I am wrong.

What is that I doing wrong here?

Thanks
C23678


Bunuel wrote:
Yes, the answer for this question is B.

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

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

Couple of things:

The point here is that square root function can not give negative result: wich means that \sqrt{some \ expression}\geq{0}.

So \sqrt{x^2}\geq{0}. But what does \sqrt{x^2} equal to?

Let's consider following examples:
If x=5 --> \sqrt{x^2}=\sqrt{25}=5=x=positive;
If x=-5 --> \sqrt{x^2}=\sqrt{25}=5=-x=positive.

So we got that:
\sqrt{x^2}=x, if x\geq{0};
\sqrt{x^2}=-x, if x<0.

What function does exactly the same thing? The absolute value function! That is why \sqrt{x^2}=|x|

Back to the original question:

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

When x>3, then RHS (right hand side) 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\neq{3}. Clearly insufficient.

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

Answer: B.

Hope it helps.


|x-3|=-(x-3) when x\leq{3}. In this case we'd have -(x-3)=3-x --> 3=3 --> true. This means that when x\leq{3}, then the equation holds true.

Try numbers less than or equal to 3 to check.
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Re: Is sqrt ((x-3)^2) = 3-x? [#permalink] New post 27 Jul 2014, 09:29
Please clarify a doubt which i have in this question :

If we have a question, Is x<=5,
A. X<0
B. X<=0

What will be the answer?

In the original question, I am confused because 0, which satisfies the equation, doesn't appear in x|x| < 0. And hence the solution is incomplete.
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Re: Is sqrt ((x-3)^2) = 3-x? [#permalink] New post 27 Jul 2014, 13:50
Expert's post
vibsaxena wrote:
Please clarify a doubt which i have in this question :

If we have a question, Is x<=5,
A. X<0
B. X<=0

What will be the answer?

In the original question, I am confused because 0, which satisfies the equation, doesn't appear in x|x| < 0. And hence the solution is incomplete.


The answer would be D.

The original question asks whether x\leq{3}: the answer would be YES if x is 3 or less than 3. (2) says that x<0, so the answer is clearly YES.

Does this make sense?
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Re: Is sqrt ((x-3)^2) = 3-x? [#permalink] New post 30 Jul 2014, 12:31
Thanks Bunnel,

I get it now. Actually, I didn't at first, then stumbled upon another question, a geometry one this time, and could draw the parallels. The question here doesn't ask if x<=0, it asks if x<3.

Got it, thanks. The geometry question I am referring to is this - a-circle-is-drawn-on-a-coordinate-plane-if-a-line-is-drawn-161692.html (the question asks if slope is less than 1, not zero, not anything else)!
Re: Is sqrt ((x-3)^2) = 3-x?   [#permalink] 30 Jul 2014, 12:31
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