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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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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]

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04 Jun 2019, 10:14
tamalmallick wrote:
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.

A lot of people ask this question - you're not alone. I'm not Bunuel, but I did write a short article about it once that should clear things up:

https://www.manhattanprep.com/gmat/blog ... -the-gmat/
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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, 00:17
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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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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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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Re: Is (x-3)^2 =3-x ? (1) x not = 3 (2) -x|x| > 3  [#permalink]

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30 Aug 2019, 22:20
jan4dday wrote:
Is $$\sqrt{(x-3)^2}=3-x$$?

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

(2) -x|x| > 3

$$\sqrt{(x-3)^2}=3-x$$
This will be true only when x = 3 or x= 2

Statement 1
$$x\neq{3}$$

It might be equal to 2, 4, anything.

Insufficient.

Statement 2
$$-x|x| > 3$$
$$|x|$$ is always +ve
if $$-x|x| > 3$$, then $$-x > 0$$
this means that $$x$$ is -ve
if $$x$$ is -ve, it cannot equal either $$3$$ or $$2$$.

Sufficient.

Hence, B.
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Re: Is (x-3)^2 =3-x ? (1) x not = 3 (2) -x|x| > 3  [#permalink]

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04 Sep 2019, 21:59
@Bunel - can you please explain how to approach this?
I am also confused of how to simplify the equation given in the question stem.
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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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04 Sep 2019, 22:35
pzgupta wrote:
@Bunel - can you please explain how to approach this?
I am also confused of how to simplify the equation given in the question stem.

My solution is on the first page: https://gmatclub.com/forum/is-x-3-2-1-2 ... ml#p737280
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Re: Is ((x-3)^2)^(1/2) = 3-x? (1) x ≠ 3 (2) -x|x| > 0   [#permalink] 04 Sep 2019, 22:35

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