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

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Is sqrt ((x-3)^2) = 3-x? [#permalink]  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]  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]  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]  24 Apr 2008, 23:23
2
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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]  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

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Re: gmatprep DS: is sqrt ((x-3)^2) = 3-x? [#permalink]  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

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.

(1) x\neq{3}. Clearly insufficient.

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

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

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.

(1) x\neq{3}. Clearly insufficient.

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

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

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.

(1) x\neq{3}. Clearly insufficient.

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

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]  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

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]  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

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 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]  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)!
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Is sqrt ((x-3)^2) = 3-x? [#permalink]  01 Sep 2014, 23:24
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.

(1) x\neq{3}. Clearly insufficient.

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

Hope it helps.

Hi Bunuel,
Can't we rephrase the question like:
Is \sqrt{(x-3)^2}=3-x?
Or : (x-3)^2=(3-x)^2
Or : x-3=3-x
Or : x=3?

Please tell me where I am doing wrong?
Thanks.
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Last edited by scofield1521 on 01 Sep 2014, 23:37, edited 1 time in total.
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Re: Is sqrt ((x-3)^2) = 3-x? [#permalink]  01 Sep 2014, 23:33
scofield1521 wrote:
Hi Bunuel,
Can't we rephrase the question like:
Is \sqrt{(x-3)^2}=3-x?
Or : (x-3)^2=3-x
Or : x-3=3-x
Or : x=3?

Please tell me where I am doing wrong?
Thanks.

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

Squaring Both sides...

(x-3)^{2}=(3-x)^{2}
x^{2}+9-6x=9+x^{2}-6x=0

Now this expression will always be equal..for any value of x...so this approach leads you to no where
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Re: Is sqrt ((x-3)^2) = 3-x? [#permalink]  02 Sep 2014, 06:08
Hi Bunuel,

Thanks for the great explanations and the GMAT Club Math Book as well! Great resource.

A question regarding the above question and square roots/absolute values in general. In the GMAT Club Math Book you write "That is, SQRT (25)=5￼ , NOT +5 or -5. In contrast, the equation x^2=25￼ has TWO solutions, +5 and -5. Even roots ￼have only a positive value on the GMAT."

What you are basically saying here is that whenever we have an equation with a positive and even square root we shall utilize the absolute value. BUT when we have just a random figure, we shall use only the positive root. Right?

Thanks again!
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Thank you very much for reading this post till the end! Kudos?

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Re: Is sqrt ((x-3)^2) = 3-x? [#permalink]  03 Sep 2014, 04:20
Thanks Bunnel ! This was one brilliant explanation !

scofield1521 wrote:
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.

(1) x\neq{3}. Clearly insufficient.

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

Hope it helps.

Hi Bunuel,
Can't we rephrase the question like:
Is \sqrt{(x-3)^2}=3-x?
Or : (x-3)^2=(3-x)^2
Or : x-3=3-x
Or : x=3?

Please tell me where I am doing wrong?
Thanks.
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Re: Is sqrt ((x-3)^2) = 3-x? [#permalink]  03 Sep 2014, 11:18
gmatnub wrote:
Is \sqrt{(x-3)^2} = 3-x?

(1) x\neq{3}
(2) -x|x| > 0

mod(x-3)=3-x?

mod(y) = -y when y is -ve => mod(x-3) can be equal to -(x-3) only when x-3 is negative i.e x-3<0 => x<3

1) x not equal to 3=> which means x can be greater than 3 or less than 3
2) -x|x|>0=> this is possible only when x is -ve i.e x<0

statement 2 gives x<0, so statement 2 alone solves the problem.
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Re: Is sqrt ((x-3)^2) = 3-x? [#permalink]  30 Sep 2014, 10:25
Bunuel wrote:
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

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.

(1) x\neq{3}. Clearly insufficient.

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

Hope it helps.

Hi Bunuel,

I am ok with X<=3 but statement B says x < 0 . Lets say X=2 in that case also the LHS = RHS but B says X<0 . I am confused where am i going wrong in my approach. Please help
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Re: Is sqrt ((x-3)^2) = 3-x? [#permalink]  30 Sep 2014, 10:34
Expert's post
snehamd1309 wrote:
Bunuel wrote:
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

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.

(1) x\neq{3}. Clearly insufficient.

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

Hope it helps.

Hi Bunuel,

I am ok with X<=3 but statement B says x < 0 . Lets say X=2 in that case also the LHS = RHS but B says X<0 . I am confused where am i going wrong in my approach. Please help

The second statement says that x<0. So, the answer to the question is yes.

Also, if we know that x < 0, then HOW can x be 2?
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Re: Is sqrt ((x-3)^2) = 3-x?   [#permalink] 30 Sep 2014, 10:34
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