Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 07 Jul 2015, 21:00

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:
Manager
Joined: 07 Apr 2012
Posts: 127
Location: United States
Concentration: Entrepreneurship, Operations
Schools: ISB '15
GMAT 1: 590 Q48 V23
GPA: 3.9
WE: Operations (Manufacturing)
Followers: 0

Kudos [?]: 9 [0], given: 45

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]

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. ?
Manager
Joined: 07 Apr 2012
Posts: 127
Location: United States
Concentration: Entrepreneurship, Operations
Schools: ISB '15
GMAT 1: 590 Q48 V23
GPA: 3.9
WE: Operations (Manufacturing)
Followers: 0

Kudos [?]: 9 [0], given: 45

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]

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 ?
Math Expert
Joined: 02 Sep 2009
Posts: 28352
Followers: 4487

Kudos [?]: 45496 [0], given: 6762

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

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.
_________________
BSchool Forum Moderator
Joined: 27 Aug 2012
Posts: 1167
Followers: 104

Kudos [?]: 794 [0], given: 126

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.

Bunuel,

$$|y-3| \leq1$$ means $$-(y-3) \leq 1$$ also. But could you please clarify why we're not considering the same?
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 28352
Followers: 4487

Kudos [?]: 45496 [0], given: 6762

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.

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 $$-1\leq{y-3}\leq{1}$$ --> $$2\leq{y}\leq{4}$$.
_________________
BSchool Forum Moderator
Joined: 27 Aug 2012
Posts: 1167
Followers: 104

Kudos [?]: 794 [0], given: 126

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?
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 28352
Followers: 4487

Kudos [?]: 45496 [0], given: 6762

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 $$-1\leq{x}\leq{1}$$.
_________________
BSchool Forum Moderator
Joined: 27 Aug 2012
Posts: 1167
Followers: 104

Kudos [?]: 794 [0], given: 126

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

Go to page   Previous    1   2   [ 28 posts ]

Similar topics Replies Last post
Similar
Topics:
3 What is the value of x^4y^2 - x^4y^2 ? 7 24 Mar 2015, 03:48
1 Given 3x + 4y = 18, what is x? 2 25 Nov 2011, 14:29
[(x^4) + (y^4)] > z^4 1 24 Dec 2008, 03:10
5 Is x^4 + y^4 > z^4 ? 4 20 Mar 2008, 09:19
What is the value of y? (1) 3|x2(readx square) 4| = y 2 (2) 8 03 Jul 2007, 17:21
Display posts from previous: Sort by