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# M26-25

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 01:25
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65% (hard)

Question Stats:

45% (01:13) correct 55% (01:18) wrong based on 214 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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16 Sep 2014, 01:25
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5
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, 05:44
1
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, 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

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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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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

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, 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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28 Nov 2014, 05:38
2
2
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, 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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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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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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31 Aug 2016, 07:48
2
1
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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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

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

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07 Dec 2016, 11:46
1
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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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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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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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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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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11 Jun 2018, 05:44
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?

Why are you changing x to -x there? The stem just says that x is negative, so x stands for some negative number, changing x to -x does not make sense.
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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
Re: M26-25   [#permalink] 11 Jul 2018, 08:26

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# M26-25

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