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Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 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 25 Jun 2019, 00:17
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 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.




Hi Bunuel, in the highlighted portion above, how can we deduce that x will be less than 3 if x is less than 0? x can be 1,2 also, right? May be I am missing something. Can you please clarify?
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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 Jun 2019, 00:20
shobhitkh 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 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.




Hi Bunuel, in the highlighted portion above, how can we deduce that x will be less than 3 if x is less than 0? x can be 1,2 also, right? May be I am missing something. Can you please clarify?


If a number is less than 0, does not it mean that it's less than 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 25 Jun 2019, 00:23
Ah! Yes, got it. I was thinking the other way round.

Thank you for clearing the doubt.
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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 Jun 2019, 01:41
gmatnub wrote:
Is \(\sqrt{(x-3)^2} = 3-x\)?

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

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

Attachment:
fasdfasdfasdfasdf.JPG


given \(\sqrt{(x-3)^2} = 3-x\)
or we can say
x-3=3-x
x=3
and x=0
#1
\(x\neq{3}\)
insufficient
#2
\(-x|x| > 0\)
sufficeint as x has to be -ve ..
IMO B
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Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

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New post 26 Jun 2019, 04:34
gmatnub wrote:
Is \(\sqrt{(x-3)^2} = 3-x\)?

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

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

Attachment:
fasdfasdfasdfasdf.JPG


Alternative Approach


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

Case 1: |x-3| > 0 => x > 3
x-3 = 3-x?
2x=6?
x=3?
x=3 is not possible ever since x > 3

Case 2: |x-3| <= 0 => x <= 3
-x + 3 = 3-x?
0=0?
LHS = RHS?
This case would always be true since it can't violate any conditions.

Rephrased Q: Is x <= 3?

Stmt 1: x != 3
Doesn't tell anything about x if it's more than or less than 3. Not sufficient.

Stmt 2: -x|x| > 0
That implies x is always negative or x < 0. Hence x < 3 is also true. Sufficient.

ANSWER: B

Bunuel EducationAisle VeritasKarishma I got this Q wrong with my initial approach (shown below) of squaring both sides. I was wondering whether we can solve this Q by squaring both sides. If not, why not? I'm also confused how x=1 can be transformed with a few steps to give x=1 & -1 (shown below)? I would really appreciate if you could help me improve my understanding on this issue. Thanks!

Initial Approach: Square both sides



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

Square both sides

(x-3)^2 = (3-x)^2?
x^2 + 9 - 6x = 9 + x^2 - 6x?
0 = 0?
LHS = RHS?

Not sure how to proceed?

x=1 transforms to x=+1,-1?
x = 1
Square both sides
x^2 = 1
Take square root of both sides
|x| = 1
x = +1, -1
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Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

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New post 27 Jun 2019, 23:34
1
dabaobao wrote:
gmatnub wrote:
Is \(\sqrt{(x-3)^2} = 3-x\)?

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

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

Attachment:
fasdfasdfasdfasdf.JPG


Alternative Approach


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

Case 1: |x-3| > 0 => x > 3
x-3 = 3-x?
2x=6?
x=3?
x=3 is not possible ever since x > 3

Case 2: |x-3| <= 0 => x <= 3
-x + 3 = 3-x?
0=0?
LHS = RHS?
This case would always be true since it can't violate any conditions.

Rephrased Q: Is x <= 3?

Stmt 1: x != 3
Doesn't tell anything about x if it's more than or less than 3. Not sufficient.

Stmt 2: -x|x| > 0
That implies x is always negative or x < 0. Hence x < 3 is also true. Sufficient.

ANSWER: B

Bunuel EducationAisle VeritasKarishma I got this Q wrong with my initial approach (shown below) of squaring both sides. I was wondering whether we can solve this Q by squaring both sides. If not, why not? I'm also confused how x=1 can be transformed with a few steps to give x=1 & -1 (shown below)? I would really appreciate if you could help me improve my understanding on this issue. Thanks!

Initial Approach: Square both sides



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

Square both sides

(x-3)^2 = (3-x)^2?
x^2 + 9 - 6x = 9 + x^2 - 6x?
0 = 0?
LHS = RHS?

Not sure how to proceed?

x=1 transforms to x=+1,-1?
x = 1
Square both sides
x^2 = 1
Take square root of both sides
|x| = 1
x = +1, -1


You need to differentiate between what is given and what is asked.
\(Is \sqrt{(x-3)^2} = 3-x\) ?
is not the same as
\(Is (x - 3)^2 = (3 - x)^2?\)
The squaring rids the expression of the negative sign and then the terms may become equal even if they are not equal

If the question is asking: Is 4 = -4?
it is not the same as Is 16 = 16?

The best way here is to realise that \(\sqrt{(x-3)^2} = |x - 3|\)

Now, we know that |x - 3| = (x - 3) when (x - 3) >= 0
and |x - 3| = (3 - x) when (x - 3) < 0

So if x < 3, then answer will be yes.
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Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0   [#permalink] 27 Jun 2019, 23:34

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