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

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45% (01:19) wrong

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Is

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

(1)

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

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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
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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 Am I right In my logic.Please help

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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 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 ]
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,

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 ]
24 Sep 2013, 06:22
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 ]
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 ]
27 Jul 2014, 13:50

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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. 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.

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.

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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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. 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.

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.

Re: Is sqrt ((x-3)^2) = 3-x?
[#permalink ]
03 Sep 2014, 11:18