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# M26-25

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:25
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Difficulty:

75% (hard)

Question Stats:

41% (00:47) correct 59% (00:44) wrong based on 314 sessions

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If $$x$$ and $$y$$ are negative numbers, what is the value of $$\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}$$?

A. $$1+y$$
B. $$1-y$$
C. $$-1-y$$
D. $$y-1$$
E. $$x-y$$

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Joined: 02 Sep 2009
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16 Sep 2014, 00:25
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Official Solution:

If $$x$$ and $$y$$ are negative numbers, what is the value of $$\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}$$?

A. $$1+y$$
B. $$1-y$$
C. $$-1-y$$
D. $$y-1$$
E. $$x-y$$

Note that $$\sqrt{a^2}=|a|$$. Next, since $$x \lt 0$$ and $$y \lt 0$$ then $$|x|=-x$$ and $$|y|=-y$$.

So, $$\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y$$

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25 Oct 2014, 04:44
How does abs(x) = -x, and abs(y) = -y when x<0 and y<0? I am do not understand this concept
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25 Oct 2014, 04:47
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delm2010 wrote:
How does abs(x) = -x, and abs(y) = -y when x<0 and y<0? I am do not understand this concept

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25 Oct 2014, 04:51
Got it now, thanks!
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25 Oct 2014, 22:33
Hi Bunuel,
I have a query.... wouldn't \sqrt{x^2}/ x would be equal to +1?

I say so since it's already given in the question that x is a -ve number, so by \sqrt{x^2} we know that we need to consider only the -ve Root.

But in the denominator the variable x would be a -ve number. Hence after the resolution of \sqrt{x^2} one part of the question would be |x|/x, where |x| would give us -x(since the -ve root needs to be considered) and the denominator in isolation is again a -ve number... so the fraction becomes \frac{-x}{-x} = 1.

Kindly clarify
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Joined: 02 Sep 2009
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26 Oct 2014, 05:08
avaneeshvyas wrote:
Hi Bunuel,
I have a query.... wouldn't \sqrt{x^2}/ x would be equal to +1?

I say so since it's already given in the question that x is a -ve number, so by \sqrt{x^2} we know that we need to consider only the -ve Root.

But in the denominator the variable x would be a -ve number. Hence after the resolution of \sqrt{x^2} one part of the question would be |x|/x, where |x| would give us -x(since the -ve root needs to be considered) and the denominator in isolation is again a -ve number... so the fraction becomes \frac{-x}{-x} = 1.

Kindly clarify

Next, the point is that the square root function cannot give negative result ($$\sqrt{some \ expression}\geq{0}$$) and $$\sqrt{x^2}=|x|$$, which means that $$\sqrt{x^2}/ x=\frac{|x|}{x}=\frac{-x}{x}=-1$$.

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27 Nov 2014, 12:37
[quote="Bunuel"]Official Solution:

-1-|y|=-1+y

I have not understood the above portion. Could you please tell me how to get this?

I have done it by taking numbers.

Let x=-1 and y=-1 Here i have taken integer value thinking that I will get an answer which can be in option also.

Now putting the value of x and y I am getting the value of question is -2.

Only option D gives me the value. Is my approach correct?
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28 Nov 2014, 04:38
2
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Raihanuddin wrote:
Bunuel wrote:
Official Solution:

-1-|y|=-1+y

I have not understood the above portion. Could you please tell me how to get this?

I have done it by taking numbers.

Let x=-1 and y=-1 Here i have taken integer value thinking that I will get an answer which can be in option also.

Now putting the value of x and y I am getting the value of question is -2.

Only option D gives me the value. Is my approach correct?

-1 - |y| = -1 + y because we are given that y is negative. When y is negative |y| = -y. Hence -1 - |y| = -1 - (-y) = -1 + y.
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16 Oct 2015, 06:57
There is a step that is off in this question. If you substitute negative y from the get go -(-y) = +y * (-y) leaves you with a negative inside the square root and this cannot happen. Should be written as square root of y*lyl
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18 Oct 2015, 12:39
danjbon wrote:
There is a step that is off in this question. If you substitute negative y from the get go -(-y) = +y * (-y) leaves you with a negative inside the square root and this cannot happen. Should be written as square root of y*lyl

That's wrong. $$\sqrt{-y*|y|}=\sqrt{-(negative)*positive}=\sqrt{positive*positive}$$
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26 Aug 2016, 07:32
I think this is a high-quality question and I agree with explanation.
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Joined: 26 Aug 2015
Posts: 34
Concentration: Strategy, Economics
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26 Aug 2016, 08:24
2
Bunuel wrote:
Raihanuddin wrote:
Bunuel wrote:
Official Solution:

-1-|y|=-1+y

I have not understood the above portion. Could you please tell me how to get this?

I have done it by taking numbers.

Let x=-1 and y=-1 Here i have taken integer value thinking that I will get an answer which can be in option also.

Now putting the value of x and y I am getting the value of question is -2.

Only option D gives me the value. Is my approach correct?

-1 - |y| = -1 + y because we are given that y is negative. When y is negative |y| = -y. Hence -1 - |y| = -1 - (-y) = -1 + y.

Hello Bunuel,

I had the same mistake. Could you clarify something for me please?

I understand that $$|y| = -y$$ given that it is negative. Let's say $$y=-2$$ so $$|y| = -(-2) = 2$$

when I substitute, I get this:

$$= -1 - \sqrt{-(-2)(2)} = -1 - \sqrt{4} = -1 -2$$ .... OH!! I see it now. $$-2 = + y$$

= - 1 + y.

Lol, I'm still going to post it in case someone has the same doubt.
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31 Aug 2016, 06:25
having a hard time understanding this basic rule: $$\sqrt{x^2}=|x|$$

Assuming it's an integer, shouldn't this be the positive integer of x? why |x|
Math Expert
Joined: 02 Sep 2009
Posts: 51280

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31 Aug 2016, 06:48
1
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xnthic wrote:
having a hard time understanding this basic rule: $$\sqrt{x^2}=|x|$$

Assuming it's an integer, shouldn't this be the positive integer of x? why |x|

MUST KNOW: $$\sqrt{x^2}=|x|$$:

The point here is that since square root function cannot give negative result then $$\sqrt{some \ expression}\geq{0}$$.

So $$\sqrt{x^2}\geq{0}$$. But what does $$\sqrt{x^2}$$ equal to?

Let's consider following examples:
If $$x=5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=x=positive$$;
If $$x=-5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=-x=positive$$.

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 $$\sqrt{x^2}=|x|$$.

For example if x = -5, then $$\sqrt{x^2}=\sqrt{25}=5=|-5|=|x|$$

Theory on Abolute Values: math-absolute-value-modulus-86462.html
Absolute value tips: absolute-value-tips-and-hints-175002.html

DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hard set on Abolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html

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31 Aug 2016, 18:04
Thanks Bunuel ! amazing as always !
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31 Aug 2016, 23:12
Bunuel wrote:
If $$x$$ and $$y$$ are negative numbers, what is the value of $$\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}$$?

A. $$1+y$$
B. $$1-y$$
C. $$-1-y$$
D. $$y-1$$
E. $$x-y$$

Alternate approach is to testing some easy values:

As per question stem, x & y are negative number..

Let x= -1 & y=-1

Substituting the in equation

$$\frac{\sqrt{1}}{-1}-\sqrt{-(-1)*|-1|}$$ = -2

y-1=-1-1=-2 only answer that matches what we get above

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07 Dec 2016, 10:46
Hi,

Having trouble wrapping my head around why $$\sqrt{x^2}$$ is equal to |x|

I looked at the post on this page, but did not see any explanation.

For example, how is the equation $$\sqrt{4}$$ = {-2, 2} any different from $$\sqrt{2^2}$$ = ? Is it because the solution of $$\sqrt{x^2}$$ must be equal to the solution of $$(\sqrt{x})^2$$? if so, why?

Thanks in advance for any help!

(with regard to bunnel's explanation to xnthic, throughout the explanation bunnel writes "$$\sqrt{25}$$ = 5" as though this is given, but my understanding is that $$\sqrt{25}$$ could be either positive or negative 5, as either positive or negative 5 squared result in 25. If I am incorrect, then that is most likely the root of my confusion)
Math Expert
Joined: 02 Sep 2009
Posts: 51280

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08 Dec 2016, 12:53
blitheclyde wrote:
Hi,

Having trouble wrapping my head around why $$\sqrt{x^2}$$ is equal to |x|

I looked at the post on this page, but did not see any explanation.

For example, how is the equation $$\sqrt{4}$$ = {-2, 2} any different from $$\sqrt{2^2}$$ = ? Is it because the solution of $$\sqrt{x^2}$$ must be equal to the solution of $$(\sqrt{x})^2$$? if so, why?

Thanks in advance for any help!

(with regard to bunnel's explanation to xnthic, throughout the explanation bunnel writes "$$\sqrt{25}$$ = 5" as though this is given, but my understanding is that $$\sqrt{25}$$ could be either positive or negative 5, as either positive or negative 5 squared result in 25. If I am incorrect, then that is most likely the root of my confusion)

$$\sqrt{4}=2$$ ONLY, not 2 or -2. The square root function cannot give negative result. Check here: m26-184464.html#p1729559
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09 Dec 2016, 17:51
Doubt for this q:

$$\sqrt{-y * |y|}$$

= $$\sqrt{y^2}$$

and Square root of Y has to be the positive root.. Thus = y

Thus the answer must be -1-y.

I do understand that $$\sqrt{x^2}$$= |x| ... and if X is -ve then mod x = -x, but we don't have to simplify$$\sqrt{y^2}$$ to mod y.
We can just take the positive root.. since per GMAT anything under the square root sign is the positive root ?

What am I missing here?

I guess my q is on the GMAT, square root of numbers are ONLY positive..
M26-25 &nbs [#permalink] 09 Dec 2016, 17:51

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# M26-25

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