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M26-25

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Re: M26-25  [#permalink]

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New post 11 Jul 2018, 11:26
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Re: M26-25  [#permalink]

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New post 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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Re: M26-25  [#permalink]

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New post 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, \(\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y\)


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 03 Jan 2019, 11:31
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The answer can be calculate very easily :cool:

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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Re: M26-25  [#permalink]

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New post 29 May 2019, 04:27
hi,

it is not that the square root function cannot be negative. The expression inside the square root is always positive or 0.
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New post 29 May 2019, 04:35
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.

Image
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;
\(\sqrt[4]{16} = 2\), 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 \(x=\sqrt{9}=3\).
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Re: M26-25   [#permalink] 29 May 2019, 04:35

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