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Is sqrt ((x3)^2) = 3x?
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24 Apr 2008, 17:42
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Is \(\sqrt{(x3)^2} = 3x\)? (1) \(x\neq{3}\) (2) \(xx > 0\)
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Is sqrt ((x3)^2) = 3x?
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13 Jun 2010, 03:41
gautamsubrahmanyam wrote: I understand that 1) is insuff
But for 2) xx > 0 means x cant be +ve => x = x so that x (x) = x^2> 0
If x is ve => (x3)^2 = X^2+96x = (ve)^2+96(ve) = +ve+9(ve) = +ve +9 + (+ve) = +ve
=> sqrt ((x3)^2) = +X3
=> sqrt ( (x3) ^2 ) is not equal to 3x
=> Option B
Am I right In my logic.Please help Yes, the answer for this question is B. Is \(\sqrt{(x3)^2}=3x\)? 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 \(\sqrt{(x3)^2}=x3\) and the question becomes is: \(x3=3x\)? 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 \(LHS=x3=x+3=3x=RHS\), hence in this case equation holds true. Basically question asks is \(x\leq{3}\)? (1) \(x\neq{3}\). Clearly insufficient. (2) \(xx >0\), basically this inequality implies that \(x<0\), hence \(x<3\). Sufficient. Answer: B. Hope it helps.
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Re: Is sqrt ((x3)^2) = 3x?
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24 Apr 2008, 21:45
From the que: 3x is always >0 > x has to be less than 3.
Option 1: X can also be >3 when ans fails so insufficient Option 2: xx>0 implies x is always < 0 which means x is less than 3 hence sufficient.
Ans : B




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Re: Is sqrt ((x3)^2) = 3x?
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25 Apr 2008, 00:23
gmatnub wrote: is sqrt ((x3)^2) = 3x?
1) x not equal to 3 2) xx > 0
The oa is B, but why is A alone not enough? given x3 can be equal to 3x 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 ((x3)^2) = 3x?
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13 Jun 2010, 02:57
I understand that 1) is insuff
But for 2) xx > 0 means x cant be +ve => x = x so that x (x) = x^2> 0
If x is ve => (x3)^2 = X^2+96x = (ve)^2+96(ve) = +ve+9(ve) = +ve +9 + (+ve) = +ve
=> sqrt ((x3)^2) = +X3
=> sqrt ( (x3) ^2 ) is not equal to 3x
=> Option B
Am I right In my logic.Please help



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Re: Is sqrt ((x3)^2) = 3x?
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23 Sep 2013, 08:09
Hello Bunuel, Please help me understand where I am going wrong. After this point.. \(x3=3x\) Can this equation be written this way? 1) x3 = 3x => x = 3 2) (x3) = 3x .. 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 \(\sqrt{(x3)^2}=3x\)?
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 \(\sqrt{(x3)^2}=x3\) and the question becomes is: \(x3=3x\)?
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 \(LHS=x3=x+3=3x=RHS\), hence in this case equation holds true.
Basically question asks is \(x\leq{3}\)?
(1) \(x\neq{3}\). Clearly insufficient.
(2) \(xx >0\), basically this inequality implies that \(x<0\), hence \(x<3\). Sufficient.
Answer: B.
Hope it helps.



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Re: Is sqrt ((x3)^2) = 3x?
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24 Sep 2013, 07:22
c23678 wrote: Hello Bunuel, Please help me understand where I am going wrong. After this point.. \(x3=3x\) Can this equation be written this way? 1) x3 = 3x => x = 3 2) (x3) = 3x .. 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 \(\sqrt{(x3)^2}=3x\)?
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 \(\sqrt{(x3)^2}=x3\) and the question becomes is: \(x3=3x\)?
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 \(LHS=x3=x+3=3x=RHS\), hence in this case equation holds true.
Basically question asks is \(x\leq{3}\)?
(1) \(x\neq{3}\). Clearly insufficient.
(2) \(xx >0\), basically this inequality implies that \(x<0\), hence \(x<3\). Sufficient.
Answer: B.
Hope it helps. \(x3=(x3)\) when \(x\leq{3}\). In this case we'd have \((x3)=3x\) > 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 ((x3)^2) = 3x?
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27 Jul 2014, 10:29
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 xx < 0. And hence the solution is incomplete.



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Re: Is sqrt ((x3)^2) = 3x?
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27 Jul 2014, 14:50



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Re: Is sqrt ((x3)^2) = 3x?
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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  acircleisdrawnonacoordinateplaneifalineisdrawn161692.html (the question asks if slope is less than 1, not zero, not anything else)!



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Re: Is sqrt ((x3)^2) = 3x?
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Updated on: 02 Sep 2014, 00:37
Bunuel wrote: Yes, the answer for this question is B.
Is \(\sqrt{(x3)^2}=3x\)?
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 \(\sqrt{(x3)^2}=x3\) and the question becomes is: \(x3=3x\)?
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 \(LHS=x3=x+3=3x=RHS\), hence in this case equation holds true.
Basically question asks is \(x\leq{3}\)?
(1) \(x\neq{3}\). Clearly insufficient.
(2) \(xx >0\), 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 \(\sqrt{(x3)^2}=3x\)? Or : \((x3)^2=(3x)^2\) Or : \(x3=3x\) 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 ((x3)^2) = 3x?
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02 Sep 2014, 00:33
scofield1521 wrote: Hi Bunuel, Can't we rephrase the question like: Is \(\sqrt{(x3)^2}=3x\)? Or : \((x3)^2=3x\) Or : \(x3=3x\) Or : \(x=3\)?
Please tell me where I am doing wrong? Thanks. \(\sqrt{(x3)^2}=3x\) Squaring Both sides... \((x3)^{2}=(3x)^{2}\) \(x^{2}+96x=9+x^{2}6x=0\) 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 ((x3)^2) = 3x?
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02 Sep 2014, 07:08
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 ((x3)^2) = 3x?
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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 \(\sqrt{(x3)^2}=3x\)?
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 \(\sqrt{(x3)^2}=x3\) and the question becomes is: \(x3=3x\)?
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 \(LHS=x3=x+3=3x=RHS\), hence in this case equation holds true.
Basically question asks is \(x\leq{3}\)?
(1) \(x\neq{3}\). Clearly insufficient.
(2) \(xx >0\), 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 \(\sqrt{(x3)^2}=3x\)? Or : \((x3)^2=(3x)^2\) Or : \(x3=3x\) Or : \(x=3\)? Please tell me where I am doing wrong? Thanks.



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Re: Is sqrt ((x3)^2) = 3x?
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03 Sep 2014, 12:18
gmatnub wrote: Is \(\sqrt{(x3)^2} = 3x\)?
(1) \(x\neq{3}\) (2) xx > 0 mod(x3)=3x? mod(y) = y when y is ve => mod(x3) can be equal to (x3) only when x3 is negative i.e x3<0 => x<3 1) x not equal to 3=> which means x can be greater than 3 or less than 3 2) xx>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 ((x3)^2) = 3x?
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30 Sep 2014, 11:25
Bunuel wrote: gautamsubrahmanyam wrote: I understand that 1) is insuff
But for 2) xx > 0 means x cant be +ve => x = x so that x (x) = x^2> 0
If x is ve => (x3)^2 = X^2+96x = (ve)^2+96(ve) = +ve+9(ve) = +ve +9 + (+ve) = +ve
=> sqrt ((x3)^2) = +X3
=> sqrt ( (x3) ^2 ) is not equal to 3x
=> Option B
Am I right In my logic.Please help Yes, the answer for this question is B. Is \(\sqrt{(x3)^2}=3x\)? 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 \(\sqrt{(x3)^2}=x3\) and the question becomes is: \(x3=3x\)? 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 \(LHS=x3=x+3=3x=RHS\), hence in this case equation holds true. Basically question asks is \(x\leq{3}\)? (1) \(x\neq{3}\). Clearly insufficient. (2) \(xx >0\), 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 ((x3)^2) = 3x?
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30 Sep 2014, 11:34
snehamd1309 wrote: Bunuel wrote: gautamsubrahmanyam wrote: I understand that 1) is insuff
But for 2) xx > 0 means x cant be +ve => x = x so that x (x) = x^2> 0
If x is ve => (x3)^2 = X^2+96x = (ve)^2+96(ve) = +ve+9(ve) = +ve +9 + (+ve) = +ve
=> sqrt ((x3)^2) = +X3
=> sqrt ( (x3) ^2 ) is not equal to 3x
=> Option B
Am I right In my logic.Please help Yes, the answer for this question is B. Is \(\sqrt{(x3)^2}=3x\)? 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 \(\sqrt{(x3)^2}=x3\) and the question becomes is: \(x3=3x\)? 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 \(LHS=x3=x+3=3x=RHS\), hence in this case equation holds true. Basically question asks is \(x\leq{3}\)? (1) \(x\neq{3}\). Clearly insufficient. (2) \(xx >0\), 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 ((x3)^2) = 3x?
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23 Oct 2014, 23:40
Hi BunuelI 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 ((x3)^2) = 3x?
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24 Oct 2014, 03:21
arpitsharms wrote: Hi BunuelI 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 ((x3)^2) = 3x?
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22 Nov 2014, 03:02
Bunuel wrote: gautamsubrahmanyam wrote: I understand that 1) is insuff
But for 2) xx > 0 means x cant be +ve => x = x so that x (x) = x^2> 0
If x is ve => (x3)^2 = X^2+96x = (ve)^2+96(ve) = +ve+9(ve) = +ve +9 + (+ve) = +ve
=> sqrt ((x3)^2) = +X3
=> sqrt ( (x3) ^2 ) is not equal to 3x
=> Option B
Am I right In my logic.Please help Yes, the answer for this question is B. Is \(\sqrt{(x3)^2}=3x\)? 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 \(\sqrt{(x3)^2}=x3\) and the question becomes is: \(x3=3x\)? 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 \(LHS=x3=x+3=3x=RHS\), hence in this case equation holds true. Basically question asks is \(x\leq{3}\)? (1) \(x\neq{3}\). Clearly insufficient. (2) \(xx >0\), 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 xx> 0 , could you elaborate on this . thanks.




Re: Is sqrt ((x3)^2) = 3x? &nbs
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22 Nov 2014, 03:02



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