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M26-25

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Re: M26-25  [#permalink]

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New post 10 Dec 2016, 04:58
cuhmoon wrote:
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..


Please check the discussion above.
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Re: M26-25  [#permalink]

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New post 25 Mar 2017, 19:46
Very Good question
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Re: M26-25  [#permalink]

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New post 05 Dec 2017, 22:09
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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Re: M26-25  [#permalink]

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New post 05 Dec 2017, 22:22
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 13 May 2018, 08:02
Bunuel wrote:
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



Hi Brunel,

If x <0 . then (Sqrt (X^2))/ X should be equal to 1

Let us take consider X= -2 , then Sqrt ( -2 *-2) should be -2 . so (Sqrt (X^2))/ X = -2/-2 = 1

Can you please explain what mistake am I doing?
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Re: M26-25  [#permalink]

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New post 13 May 2018, 08:48
seevishal@gmail.com wrote:
Bunuel wrote:
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



Hi Brunel,

If x <0 . then (Sqrt (X^2))/ X should be equal to 1

Let us take consider X= -2 , then Sqrt ( -2 *-2) should be -2 . so (Sqrt (X^2))/ X = -2/-2 = 1

Can you please explain what mistake am I doing?


I think this is explained several times in this topic. For example, in the post above yours: https://gmatclub.com/forum/m26-184464-20.html#p1973777

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.

\(\sqrt{(-2)^2}=\sqrt{4}=2\), not -2.
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Re: M26-25  [#permalink]

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New post 11 Jun 2018, 01:11
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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Re: M26-25  [#permalink]

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New post 11 Jun 2018, 04:44
dishadesai wrote:
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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Re: M26-25  [#permalink]

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New post 11 Jul 2018, 07:26
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]

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New post 11 Jul 2018, 10:26
1
Fervids77 wrote:
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


When x < 0, then |x| = -x.

Say x = -1, then |-1| = -(-1) = 1.
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Re: M26-25  [#permalink]

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New post 21 Aug 2018, 18:15
I used smart numbers, where x = -2 and y = -3. The given formula will get you to -4 and by plugging in numbers, only D ends up with -4.
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Re: M26-25  [#permalink]

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New post 22 Aug 2018, 23:04
delm2010 wrote:
How does abs(x) = -x, and abs(y) = -y when x<0 and y<0? I am do not understand this concept


It's given in the question.
If x and y are negative numbers, what is the value.......
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Re: M26-25  [#permalink]

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New post 26 Aug 2018, 07:34
as x and y are negative so
\(\frac{-x}{x}\) - \(\sqrt{(-y)(-y)}\)
\((-1)\) - \(\sqrt{y^2}\)
\(-1-(-y)\)
\(-1+y\)
\(y-1\)
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Re: M26-25  [#permalink]

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New post 14 Sep 2018, 07:39
Bunuel wrote:
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


For anyone that picked C over D, like I did, plug in numbers:
Let X = -2, and Let Y = -3 (also let -X = 2, and Let -Y = 3)

You end up with:
\(\frac{2}{-2} - \sqrt{3*3}\) = \(-1 - 3\)

Now look back - 3 = -Y, so 1- -Y = 1 + Y. D'oh!
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Re M26-25  [#permalink]

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New post 15 Sep 2018, 09:12
I think this is a high-quality question.
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Re M26-25 &nbs [#permalink] 15 Sep 2018, 09:12

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