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

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Manager
Joined: 07 Apr 2012
Posts: 126
Location: United States
Concentration: Entrepreneurship, Operations
Schools: ISB '15
GMAT 1: 590 Q48 V23
GPA: 3.9
WE: Operations (Manufacturing)
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Kudos [?]: 11 [1] , given: 45

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

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28 Aug 2013, 04:05
1
KUDOS
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: 126
Location: United States
Concentration: Entrepreneurship, Operations
Schools: ISB '15
GMAT 1: 590 Q48 V23
GPA: 3.9
WE: Operations (Manufacturing)
Followers: 0

Kudos [?]: 11 [0], given: 45

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

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28 Aug 2013, 04:08
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
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Posts: 38989
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Kudos [?]: 106448 [0], given: 11626

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

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28 Aug 2013, 09:49
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.
_________________
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Joined: 27 Aug 2012
Posts: 1194
Followers: 134

Kudos [?]: 1618 [0], given: 147

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

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16 Nov 2013, 08:30
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?
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Kudos [?]: 106448 [0], given: 11626

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

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16 Nov 2013, 08:34
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 $$-1\leq{y-3}\leq{1}$$ --> $$2\leq{y}\leq{4}$$.
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Re: Is square root[(y-4)^2] = 4-y? [#permalink]

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16 Nov 2013, 08:47
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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Kudos [?]: 106448 [0], given: 11626

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

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16 Nov 2013, 08:53
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 $$-1\leq{x}\leq{1}$$.
_________________
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Re: Is square root[(y-4)^2] = 4-y? [#permalink]

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16 Nov 2013, 08:58
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?
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Is square root[(y-4)^2] = 4-y? [#permalink]

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28 Jul 2015, 12:05
Bunuel wrote:
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.

Bunuel chetan2u

I have a question here

I am clear with the concept of absolute function. |x| = +x when x>=0
-x when x<0

Here, you have considered the case as |y-4| = y-4, when y>4
4-y, when y<=4

My question is at y =4, the case can be either y-4 OR 4-y. Both yield 0 as the answer. Then why have you only considered y<=4 in (4-y) case

Thanking in anticipation

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

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

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19 Oct 2016, 06:54
Keats wrote:
Bunuel wrote:
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.

Bunuel chetan2u

I have a question here

I am clear with the concept of absolute function. |x| = +x when x>=0
-x when x<0

Here, you have considered the case as |y-4| = y-4, when y>4
4-y, when y<=4

My question is at y =4, the case can be either y-4 OR 4-y. Both yield 0 as the answer. Then why have you only considered y<=4 in (4-y) case

Thanking in anticipation

Cheers

Hi Keats,

It doesn't matter where you take EQUAL sign, with less than or greater than.. only point is it should be catered in any one.....
You are having TWO ranges..
1) y is LESS than or EQUAL to 4 and y is more than 4.... so this completes the range
2) y is less than 4 and y is GREATER than or EQUAL to 4... this again completes the range

But you cannot have EQUAL to in both as it is duplicated while you combine the two to find the different ranges..

The MOD in my signature below may be of some use here
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Re: Is square root[(y-4)^2] = 4-y?   [#permalink] 19 Oct 2016, 06:54

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