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

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Is sqrt ((x-3)^2) = 3-x?  [#permalink]

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

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

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

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Is sqrt ((x-3)^2) = 3-x?  [#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 sqrt ((x-3)^2) = 3-x?  [#permalink]

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New post 24 Apr 2008, 21:45
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3
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 sqrt ((x-3)^2) = 3-x?  [#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: Is sqrt ((x-3)^2) = 3-x?  [#permalink]

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New post 13 Jun 2010, 02:57
1
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 sqrt ((x-3)^2) = 3-x?  [#permalink]

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New post 23 Sep 2013, 08:09
2
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 sqrt ((x-3)^2) = 3-x?  [#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 sqrt ((x-3)^2) = 3-x?  [#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 sqrt ((x-3)^2) = 3-x?  [#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 sqrt ((x-3)^2) = 3-x?  [#permalink]

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New post 30 Jul 2014, 13:31
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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Re: Is sqrt ((x-3)^2) = 3-x?  [#permalink]

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New post Updated on: 02 Sep 2014, 00:37
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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Originally posted by scofield1521 on 02 Sep 2014, 00:24.
Last edited by scofield1521 on 02 Sep 2014, 00:37, edited 1 time in total.
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Re: Is sqrt ((x-3)^2) = 3-x?  [#permalink]

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New post 02 Sep 2014, 00:33
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]

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New post 02 Sep 2014, 07:08
1
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]

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New post 03 Sep 2014, 05:20
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]

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New post 03 Sep 2014, 12:18
2
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 sqrt ((x-3)^2) = 3-x?  [#permalink]

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New post 30 Sep 2014, 11:25
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.


Hi Bunuel,

I am ok with X<=3 but statement B says x < 0 . Lets say X=2 in that case also the LHS = RHS but B says X<0 . I am confused where am i going wrong in my approach. Please help
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Re: Is sqrt ((x-3)^2) = 3-x?  [#permalink]

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New post 30 Sep 2014, 11:34
snehamd1309 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.


Hi Bunuel,

I am ok with X<=3 but statement B says x < 0 . Lets say X=2 in that case also the LHS = RHS but B says X<0 . I am confused where am i going wrong in my approach. Please help


The question asks whether \(x\leq{3}\)?
The second statement says that \(x<0\). So, the answer to the question is yes.

Also, if we know that x < 0, then HOW can x be 2?
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Re: Is sqrt ((x-3)^2) = 3-x?  [#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 sqrt ((x-3)^2) = 3-x?  [#permalink]

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New post 24 Oct 2014, 03:21
1
1
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 sqrt ((x-3)^2) = 3-x?  [#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.
Re: Is sqrt ((x-3)^2) = 3-x? &nbs [#permalink] 22 Nov 2014, 03:02

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