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M26-25

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New post 16 Sep 2014, 01:25
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A
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C
D
E

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Question Stats:

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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
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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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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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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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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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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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New post 18 Oct 2015, 13:39
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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

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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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
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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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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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New post 05 Dec 2017, 23:09
Bunuel wrote:
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: http://gmatclub.com/forum/m26-184464.html#p1729559


Hi Bunuel, certain application of the sqrt function seem to be contradictory. If sqrt function can't give negative numbers, why do we substitute both -ve and +ve values of \(\sqrt{expression}\) while solving algebraic equations.
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New post 05 Dec 2017, 23:22
abhishek911 wrote:
Bunuel wrote:
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: http://gmatclub.com/forum/m26-184464.html#p1729559


Hi Bunuel, certain application of the sqrt function seem to be contradictory. If sqrt function can't give negative numbers, why do we substitute both -ve and +ve values of \(\sqrt{expression}\) while solving algebraic equations.



When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).
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Re: M26-25  [#permalink]

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New post 11 Jun 2018, 02:11
Hi,

Since x is negative, \sqrt{x^2}/-x - \sqrt{-y*-y}
= \(-x/-x\) - \sqrt{y^2}
= 1-(-y)
= 1+y

Can someone please explain what wrong am I doing here?
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New post 11 Jun 2018, 05:44
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New post 11 Jul 2018, 08:26
Hi,

In one of the other questions, the explanation is provided as follows

If p=−1
then (|p|!)^p=(|−1|!)^−1=1^−1=1

Here |-1| = 1

But in this problem |y| = -y since y is negative

Can you please explain the discrepancy or the misunderstanding in my thought
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Re: M26-25   [#permalink] 11 Jul 2018, 08:26

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