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Re: Is square root[(y-4)^2] = 4-y? [#permalink]
28 Aug 2013, 03:05
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: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).
BACK TO THE ORIGINAL QUESTION:
Question asks: is \(\sqrt{(y-4)^2}=4-y\)?
Now, according to the above \(\sqrt{(y-4)^2}=|y-4|\)? So the question becomes: is \(|y-4|=4-y\)? --> or is \(|y-4|=-(y-4)\)? Which, again according to the properties of absolute value, is true when \(y-4\leq{0}\) or when \(y\leq{4}\).
Hope it's clear.[/quote][/quote]
Not sure understand your question. Can you please elaborate?
P.S. Note that the absolute value is ALWAYS non-negative, also I did not multiply anything by -1...[/quote]
Hello Bunnel, I have one small doubt. When you say |X| = +X, when Xgeq{0} and -X when xleq{0}, then in this question how could you compare |y-4| to be 4-y when yleq{0}, dont you think it should be when y<0. ?
Re: Is square root[(y-4)^2] = 4-y? [#permalink]
28 Aug 2013, 03:08
Reposting---------
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: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).
BACK TO THE ORIGINAL QUESTION:
Question asks: is \(\sqrt{(y-4)^2}=4-y\)?
Now, according to the above \(\sqrt{(y-4)^2}=|y-4|\)? So the question becomes: is \(|y-4|=4-y\)? --> or is \(|y-4|=-(y-4)\)? Which, again according to the properties of absolute value, is true when \(y-4\leq{0}\) or when \(y\leq{4}\).
Hope it's clear.[/quote][/quote]
Not sure understand your question. Can you please elaborate?
P.S. Note that the absolute value is ALWAYS non-negative, also I did not multiply anything by -1...[/quote]
Hello Bunnel, I have one small doubt. When you say |X| = +X, when X<eq0 and -X when <0 , then in this question how could you compare |y-4| to be 4-y when comparing y<eq4 ? Shouldnt it be y<4 only? On x=0 how does the mod function behave, its neither postive nor negative then dont you think at x= 0, we cant be sure wether F(X) is positive or negative ?
Re: Is square root[(y-4)^2] = 4-y? [#permalink]
28 Aug 2013, 08:49
Expert's post
ygdrasil24 wrote:
Reposting---------
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: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).
BACK TO THE ORIGINAL QUESTION:
Question asks: is \(\sqrt{(y-4)^2}=4-y\)?
Now, according to the above \(\sqrt{(y-4)^2}=|y-4|\)? So the question becomes: is \(|y-4|=4-y\)? --> or is \(|y-4|=-(y-4)\)? Which, again according to the properties of absolute value, is true when \(y-4\leq{0}\) or when \(y\leq{4}\).
Hope it's clear
Not sure understand your question. Can you please elaborate?
P.S. Note that the absolute value is ALWAYS non-negative, also I did not multiply anything by -1...
Hello Bunnel, I have one small doubt. When you say |X| = +X, when X<eq0 and -X when <0 , then in this question how could you compare |y-4| to be 4-y when comparing y<eq4 ? Shouldnt it be y<4 only? On x=0 how does the mod function behave, its neither postive nor negative then dont you think at x= 0, we cant be sure wether F(X) is positive or negative ?
Check your approach with numbers: if \(y=4\) --> \(\sqrt{(4-4)^2}=0\) and \(4-4=0\) --> \(0=0\).
Re: Is square root[(y-4)^2] = 4-y? [#permalink]
16 Nov 2013, 07:30
Expert's post
Bunuel wrote:
Is \(\sqrt{(y-4)^2} = 4-y\)?
Is \(\sqrt{(y-4)^2}=4-y\)? --> is \(|y-4|=4-y\)? is \(y\leq{4}\)?
(1) |y-3| less than or equal to 1 --> \(|y-3|\) is just the distance between 3 and \(y\) on the number line. We are told that this distance is less than or equal to 1: --2----3----4-- so, \(2\leq{y}\leq{4}\). Sufficient.
(2) y*|y|>0 --> just says that \(y>0\). Not sufficient.
Answer: A.
Bunuel,
\(|y-3| \leq1\) means \(-(y-3) \leq 1\) also. But could you please clarify why we're not considering the same? _________________
Re: Is square root[(y-4)^2] = 4-y? [#permalink]
16 Nov 2013, 07:34
Expert's post
bagdbmba wrote:
Bunuel wrote:
Is \(\sqrt{(y-4)^2} = 4-y\)?
Is \(\sqrt{(y-4)^2}=4-y\)? --> is \(|y-4|=4-y\)? is \(y\leq{4}\)?
(1) |y-3| less than or equal to 1 --> \(|y-3|\) is just the distance between 3 and \(y\) on the number line. We are told that this distance is less than or equal to 1: --2----3----4-- so, \(2\leq{y}\leq{4}\). Sufficient.
(2) y*|y|>0 --> just says that \(y>0\). Not sufficient.
Answer: A.
Bunuel,
\(|y-3| \leq1\) means \(-(y-3) \leq 1\) also. But could you please clarify why we're not considering the same?
\(|y-3|\leq{1}\) means that \(-1\leq{y-3}\leq{1}\) --> \(2\leq{y}\leq{4}\). _________________
Re: Is square root[(y-4)^2] = 4-y? [#permalink]
16 Nov 2013, 07:47
Expert's post
Bunuel wrote:
\(|y-3|\leq{1}\) means that \(-1\leq{y-3}\leq{1}\) --> \(2\leq{y}\leq{4}\).
There I'm getting bit confused...For any such inequality(\(|x|\leq{1}\)) it can have two values i.e. either \(x\leq{1}\) or \(-x\leq{1}\) dependning upon whether \(x>=0\) or \(x<0\). Simultaneously both can't hold good...!
Now my question is in case of 'must be true' problems should we consider the range \(-1\leq{x}\leq{1}\) or two individual possibilities separately? _________________
Re: Is square root[(y-4)^2] = 4-y? [#permalink]
16 Nov 2013, 07:53
Expert's post
bagdbmba wrote:
Bunuel wrote:
\(|y-3|\leq{1}\) means that \(-1\leq{y-3}\leq{1}\) --> \(2\leq{y}\leq{4}\).
There I'm getting bit confused...For any such inequality(\(|x|\leq{1}\)) it can have two values i.e. either \(x\leq{1}\) or \(-x\leq{1}\) dependning upon whether \(x>=0\) or \(x<0\). Simultaneously both can't hold good...!
Now my question is in case of 'must be true' problems should we consider the range \(-1\leq{x}\leq{1}\) or two individual possibilities separately?
You are overthinking it: \(|x|\leq{1}\) means \(-1\leq{x}\leq{1}\). _________________
Re: Is square root[(y-4)^2] = 4-y? [#permalink]
16 Nov 2013, 07:58
Expert's post
Bunuel wrote:
bagdbmba wrote:
Bunuel wrote:
\(|y-3|\leq{1}\) means that \(-1\leq{y-3}\leq{1}\) --> \(2\leq{y}\leq{4}\).
There I'm getting bit confused...For any such inequality(\(|x|\leq{1}\)) it can have two values i.e. either \(x\leq{1}\) or \(-x\leq{1}\) depending upon whether \(x>=0\) or \(x<0\). Simultaneously both can't hold good...!
Now my question is in case of 'must be true' problems should we consider the range \(-1\leq{x}\leq{1}\) or two individual possibilities separately?
You are overthinking it: \(|x|\leq{1}\) means \(-1\leq{x}\leq{1}\).
Okay...So.even in case of 'must be true' questions we'll consider that \(|x|\leq{1}\) means \(-1\leq{x}\leq{1}\) always, unless otherwise stated explicitly that \(x>=0\) or \(x<0\). Right? _________________
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: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).
BACK TO THE ORIGINAL QUESTION:
Question asks: is \(\sqrt{(y-4)^2}=4-y\)?
Now, according to the above \(\sqrt{(y-4)^2}=|y-4|\)? So the question becomes: is \(|y-4|=4-y\)? --> or is \(|y-4|=-(y-4)\)? Which, again according to the properties of absolute value, is true when \(y-4\leq{0}\) or when \(y\leq{4}\).
Hope it's clear
Not sure understand your question. Can you please elaborate?
P.S. Note that the absolute value is ALWAYS non-negative, also I did not multiply anything by -1...
Hello Bunnel, I have one small doubt. When you say |X| = +X, when X<eq0 and -X when <0 , then in this question how could you compare |y-4| to be 4-y when comparing y<eq4 ? Shouldnt it be y<4 only? On x=0 how does the mod function behave, its neither postive nor negative then dont you think at x= 0, we cant be sure wether F(X) is positive or negative ?
Check your approach with numbers: if \(y=4\) --> \(\sqrt{(4-4)^2}=0\) and \(4-4=0\) --> \(0=0\).
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