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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
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. _________________
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.
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)!
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! _________________
Thank you very much for reading this post till the end! Kudos?
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
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? _________________
Re: Is sqrt ((x-3)^2) = 3-x? [#permalink]
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. ?
Re: Is sqrt ((x-3)^2) = 3-x? [#permalink]
24 Oct 2014, 02:21
1
This post received KUDOS
Expert's post
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 \(x=\sqrt{9}=3\).
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