Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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? _________________

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...