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M26-25

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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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Re M26-25 [#permalink]

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New post 16 Sep 2014, 01:25
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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, \(\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y\)


Answer: D
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Re: M26-25 [#permalink]

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New post 25 Oct 2014, 05:44
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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New post 25 Oct 2014, 05:47
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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


Below links on absolute value should help you:

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Re: M26-25 [#permalink]

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New post 25 Oct 2014, 05:51
Got it now, thanks!
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Re: M26-25 [#permalink]

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New post 25 Oct 2014, 23:33
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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Re: M26-25 [#permalink]

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New post 26 Oct 2014, 06:08
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


First of all, read this: rules-for-posting-please-read-this-before-posting-133935.html#p1096628 (Writing Mathematical Formulas on the Forum)

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 \(\sqrt{x^2}/ x=\frac{|x|}{x}=\frac{-x}{x}=-1\).

Please follow the links in my previous post for more on absolute values.
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M26-25 [#permalink]

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New post 27 Nov 2014, 13:37
[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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Re: M26-25 [#permalink]

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New post 28 Nov 2014, 05:38
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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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Re: M26-25 [#permalink]

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New post 16 Oct 2015, 07:57
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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Re: M26-25 [#permalink]

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New post 18 Oct 2015, 13:39
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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Re M26-25 [#permalink]

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New post 26 Aug 2016, 08:32
I think this is a high-quality question and I agree with explanation.
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M26-25 [#permalink]

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New post 26 Aug 2016, 09:24
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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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M26-25 [#permalink]

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New post 31 Aug 2016, 07:25
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|
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New post 31 Aug 2016, 07:48
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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

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New post 31 Aug 2016, 19:04
Thanks Bunuel ! amazing as always !
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Re: M26-25 [#permalink]

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New post 01 Sep 2016, 00:12
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

Checking answer choices

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


Answer: D
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M26-25 [#permalink]

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New post 07 Dec 2016, 11:46
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)
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Re: M26-25 [#permalink]

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New post 08 Dec 2016, 13:53
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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M26-25 [#permalink]

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New post 09 Dec 2016, 18:51
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   [#permalink] 09 Dec 2016, 18:51

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