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Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0

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

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

(1) \(x\neq{3}\)

(2) \(-x|x| > 0\)

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Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

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New post 13 Jun 2010, 03:41
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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 cannot 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 ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

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New post 24 Apr 2008, 21:45
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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: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

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New post 25 Apr 2008, 00:23
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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: gmat prep_inequality  [#permalink]

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New post 28 Mar 2010, 12:38
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One important thing to know about the square root of a square term:
(1) If it's a number under the square root sign, the answer is only the positive root: sqrt[(-3)^2] = sqrt[(3)^2] = sqrt[9] = 3.
(2) If there are variables under the square root sign, you must consider the fact that the squared "thing" may have been either pos or neg to begin with: sqrt[(-x)^2] = sqrt[(x)^2] = |x|. We are still looking for the positive root, but we don't know whether +x or -x is actually greater than zero--that depends on the sign of x.

So for this question: Is sqrt[(x-3)^2] = 3-x?

Rephrase to: Is |x-3| = 3-x?

If stuff in the absolute value sign is positive or zero, this becomes:
Is x-3 = 3-x?
Is 2x = 6?
Is x = 3?

If stuff in the absolute value sign is negative, this becomes:
Is -(x-3) = 3-x?
Is -x+3 = 3-x?
The answer is yes for all x, but remember this was just "all x" such that x-3 was negative, or x < 3: So we ask "Is x < 3?"

Put the two cases together for the final rephrase: "Is x =<3 ? "

(1) x is not 3, but no info on whether it is less than or greater than 3. INSUFF.
(2) -x|x| > 0.
|x| is positive, so -x would have to be positive too (pos*pos > 0, but neg*pos < 0). Thus, x is negative. If x is negative, it is definitely less than 3. The answer is definitely Yes. SUFF.

The answer is B.
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Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

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New post 13 Jun 2010, 02:57
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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: Is [m][square_root](x - 3)[/square_root]^2[/m] = 3 - x  [#permalink]

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New post 24 Jul 2011, 23:47
Lets disect the question stem first.

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

Carrying out the squaring and then the square root operation, we get:

|(x-3)| = 3 - x

To solve the modulus sign, we will have two cases -
Case 1: (x-3)>=0
=> x = 3

Case 2: (x-3) < 0
=> Every value of x less than 3 satisfies the equation

Therefore sqrt((x-3)^2) = 3 - x
for all values of x<=3, but this equation is not satisfied for values of x>3

Now consider the statements of the DS.

Statement (1): x is not equal to 3
If x is not equal to 3, it could be greater than 3 or less than 3. If x is less than 3, the equation is satisfied. If x is greater than 3, the equation is not satisfied. Therefore we cannot say if the equation will be satisfied if x is not equal to 3.
Statement (1) alone is therefore insufficient to answer the question.

Next lets consider statement (2):
Statement (2): -x|x|>0
Again to remove the modulus sign make two cases.
Case 1: x>=0 => -x(x) >0 => -(x^2) > 0, which is impossible because x^2 is always >=0, so -(x^2) cannot be >0 for any values of x. Therefore to satisfy this inequality, x cannot be >0
Case 2: x<0 => -x (-x) > 0
=> x^2 > 0 , which is true for all values of x except 0.
Therefore the inequality is satisfied for all values of x<0

Therefore from statement (2) we know that x<0. And from the stem of the question (after we solved it), we know that the equation given in the stem is satisfied for all values of x<=3. Therefore the equation will hold for all values of x specified by statement (2). Therefore statement (2) alone is enough to solve the question.

The answer is therefore (B).
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Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

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New post 23 Sep 2013, 08:09
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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: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

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New post 24 Sep 2013, 07:22
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 ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

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New post 27 Jul 2014, 10:29
1
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 ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

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New post 27 Jul 2014, 14:50
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 ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

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New post 03 Sep 2014, 12:18
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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 ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

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New post 23 Oct 2014, 23:40
Hi Bunuel
I am still confused about this.Please help me out.
As a^2 = 25 has two solutions -------------------------> a=5 and a= -5
therefore a= sqrt 25 should also have two solutions-----> a=5 and a= -5

Then why do we say that square root of a positive no. is always positive?
Shouldn't sqrt 25 have two possible values +5 and -5. ?
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Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

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New post 24 Oct 2014, 03:21
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arpitsharms wrote:
Hi Bunuel
I am still confused about this.Please help me out.
As a^2 = 25 has two solutions -------------------------> a=5 and a= -5
therefore a= sqrt 25 should also have two solutions-----> a=5 and a= -5

Then why do we say that square root of a positive no. is always positive?
Shouldn't sqrt 25 have two possible values +5 and -5. ?


NO!

When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root.

That is:
\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).

Hope it's clear.
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Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

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New post 22 Nov 2014, 03:02
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

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.




can u pls help in decoding option B where , you say x <0
i mean -x|x|> 0 , could you elaborate on this .

thanks.
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Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

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New post 22 Nov 2014, 07:20
2
hanschris5 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

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.




can u pls help in decoding option B where , you say x <0
i mean -x|x|> 0 , could you elaborate on this .

thanks.


Sure. We have that \(-x|x| >0\), so the product of two multiples -x and |x| is positive. Now, we know that |x| must be positive, hence -x must also be positive for the product to be positive, thus -x > 0, which is the same as x < 0.

Hope it's clear.
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Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

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New post 26 Jan 2017, 19:55
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Be definition:
√(x²) = |x|.
|x-y| is the DISTANCE between x and y.
The DISTANCE between two numbers must be greater than or equal to 0.

The question stem above, rephrased: Is |x-3| = 3-x?
In words:
Is the DISTANCE between x and 3 equal to the DIFFERENCE of 3 and x?
The answer will be YES if the DIFFERENCE of 3 and x is greater than or equal to 0:
3-x≥0
x≤3.

The question stem rephrased: Is x≤3?

Statement 1: x is not equal to 3.
It is possible that x<3 or that x>3.
INSUFFICIENT.

Statement 2: -x*|x| > 0 .
Thus, the left-hand side must be positive*positive or negative*negative.
Since |x| cannot be negative, both factors on the left-hand side must be positive.
Thus:
-x>0
x<0.
Since x<0, we know that x≤3.
SUFFICIENT.

The correct answer is B.
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Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

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New post 02 Oct 2018, 11:30
Bunuel, Thanks for the explanation, but i had one question.
You mentioned : "Couple of things:
The point here is that square root function can not give negative result: wich means that some expression−−−−−−−−−−−−−−√≥0some expression≥0."

Can you please let me know why? Because in general square root gives two values, one positive and one negative.
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Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

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New post 02 Oct 2018, 20:35
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PriyankaGehlawat wrote:
Bunuel, Thanks for the explanation, but i had one question.
You mentioned : "Couple of things:
The point here is that square root function can not give negative result: wich means that some expression−−−−−−−−−−−−−−√≥0some expression≥0."

Can you please let me know why? Because in general square root gives two values, one positive and one negative.



\(\sqrt{...}\) is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign (\(\sqrt{...}\)) always means non-negative square root.

Image
The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).
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Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

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New post 04 Jun 2019, 10:05
Dear Brunel,

You said-

"Is (x−3)2−−−−−−−√=3−x(x−3)2=3−x?

Remember: x2−−√=|x|x2=|x|. Why?

Couple of things:

The point here is that square root function can not give negative result: wich means that some expression−−−−−−−−−−−−−−√≥0some expression≥0.

So x2−−√≥0x2≥0. But what does x2−−√x2 equal to?

Let's consider following examples:
If x=5x=5 --> x2−−√=25−−√=5=x=positivex2=25=5=x=positive;
If x=−5x=−5 --> x2−−√=25−−√=5=−x=positivex2=25=5=−x=positive."

My doubt is as follows-

All we know that sqrt of a number can be positive or negative results both, how you are saying "square root function can not give negative result"? If you kindly answer this question it would be a great help for me. Looking forward to hear you from.
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Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0   [#permalink] 04 Jun 2019, 10:05

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