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# M26-25

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Math Expert
Joined: 02 Sep 2009
Posts: 53831

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10 Dec 2016, 05:58
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..

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25 Mar 2017, 20:46
Very Good question
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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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13 May 2018, 09:02
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, $$\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y$$

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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13 May 2018, 09:48
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, $$\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y$$

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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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
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11 Jul 2018, 11:26
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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21 Aug 2018, 19:15
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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23 Aug 2018, 00:04
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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26 Aug 2018, 08:34
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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14 Sep 2018, 08:39
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, $$\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y$$

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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15 Sep 2018, 10:12
I think this is a high-quality question.
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03 Jan 2019, 11:31
1
The answer can be calculate very easily

Do 1 thing :- substitute x = -a and y = -b (as question says both x and y are negative numbers) and solve with basic math and reverse back the result in terms of "y". Answer will be "-1+y". Try your hand once.
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19 Mar 2019, 20:47
I agree It should be -1 -y since the expression after the -ve sign should be positive and would be the absolute value of y
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19 Mar 2019, 21:11
Bokaage wrote:
I agree It should be -1 -y since the expression after the -ve sign should be positive and would be the absolute value of y

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Re: M26-25   [#permalink] 19 Mar 2019, 21:11

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