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Math Expert V
Joined: 02 Sep 2009
Posts: 58434

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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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Manager  S
Joined: 20 Jul 2018
Posts: 87
GPA: 2.87

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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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Hasnain Afzal

"When you wanna succeed as bad as you wanna breathe, then you will be successful." -Eric Thomas
Intern  B
Joined: 12 Jan 2017
Posts: 34
Location: United States (NY)
Schools: Booth '21 (D)
GMAT 1: 710 Q47 V41 GPA: 3.48

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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!
Intern  B
Joined: 13 Dec 2018
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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.
Intern  B
Joined: 22 May 2018
Posts: 7
Location: India
Schools: LBS '22
GMAT 1: 700 Q49 V35 Show Tags

hi,

it is not that the square root function cannot be negative. The expression inside the square root is always positive or 0.
Math Expert V
Joined: 02 Sep 2009
Posts: 58434

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attrirahul456 wrote:
hi,

it is not that the square root function cannot be negative. The expression inside the square root is always positive or 0.

You are wrong. It also seems that you did not read the whole discussion.

$$\sqrt{...}$$ is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign ($$\sqrt{...}$$) always means non-negative square root. The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
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 non-negative root. That is:

$$\sqrt{9} = 3$$, NOT +3 or -3;
$$\sqrt{16} = 2$$, NOT +2 or -2;
Similarly $$\sqrt{\frac{1}{16}} = \frac{1}{4}$$, NOT +1/4 or -1/4.

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 $$x=\sqrt{9}=3$$.
_________________ Re: M26-25   [#permalink] 29 May 2019, 04:35

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