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Is square root[(y-4)^2] = 4-y?

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Re: Is square root[(y-4)^2] = 4-y? [#permalink] New post 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. ?
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Re: Is square root[(y-4)^2] = 4-y? [#permalink] New post 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 ?
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Re: Is square root[(y-4)^2] = 4-y? [#permalink] New post 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.

Hope it's clear.
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Re: Is square root[(y-4)^2] = 4-y? [#permalink] New post 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?
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Re: Is square root[(y-4)^2] = 4-y? [#permalink] New post 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}.
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Re: Is square root[(y-4)^2] = 4-y? [#permalink] New post 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?
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Re: Is square root[(y-4)^2] = 4-y? [#permalink] New post 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}.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Is square root[(y-4)^2] = 4-y? [#permalink] New post 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}.


:roll: 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?
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Re: Is square root[(y-4)^2] = 4-y?   [#permalink] 16 Nov 2013, 07:58
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