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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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24 Apr 2008, 16:42
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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 sqrt ((x-3)^2) = 3-x?  [#permalink]

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13 Jun 2010, 02:41
38
50
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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24 Apr 2008, 20:45
9
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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24 Apr 2008, 23:23
3
2
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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13 Jun 2010, 01:57
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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23 Sep 2013, 07:09
1
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.
Math Expert
Joined: 02 Sep 2009
Posts: 52902
Re: Is sqrt ((x-3)^2) = 3-x?  [#permalink]

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24 Sep 2013, 06:22
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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27 Jul 2014, 09:29
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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Posts: 52902
Re: Is sqrt ((x-3)^2) = 3-x?  [#permalink]

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27 Jul 2014, 13:50
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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30 Jul 2014, 12:31
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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Updated on: 01 Sep 2014, 23:37
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 01 Sep 2014, 23:24.
Last edited by scofield1521 on 01 Sep 2014, 23:37, edited 1 time in total.
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Re: Is sqrt ((x-3)^2) = 3-x?  [#permalink]

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01 Sep 2014, 23:33
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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02 Sep 2014, 06:08
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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03 Sep 2014, 04:20
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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Joined: 28 Jan 2013
Posts: 29
Re: Is sqrt ((x-3)^2) = 3-x?  [#permalink]

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03 Sep 2014, 11:18
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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30 Sep 2014, 10:25
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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30 Sep 2014, 10:34
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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23 Oct 2014, 22:40
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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24 Oct 2014, 02:21
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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22 Nov 2014, 02:02
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?   [#permalink] 22 Nov 2014, 02:02

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