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

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Joined: 20 Feb 2018
Posts: 44
Location: India
Schools: ISB '20
Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

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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

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.

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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Posts: 57131
Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0  [#permalink]

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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

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.

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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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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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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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.

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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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.

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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