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Bunuel
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. \(x-y\)


Hello Bunuel

Really amazing question and explanation to understand the relation between absolute values and square roots.

The square root of y^2 will be |y|.

I just had one query here, how would we simplify the following expression:

Square root of (-y * y): would it be |y| only? Or square root of -x^2 ? Would it be |x| only?

Thank you Bunuel

First of all, \(\sqrt{-y * y}=\sqrt{-y^2}\), so it's equivalent to \(\sqrt{-x^2}\).

All numbers on the GMAT are real numbers by default. Hence, for the square root (or any even root) to be defined, the expression under it must be non-negative. Therefore, \(\sqrt{-x^2}\) is defined only when x = 0, (resulting in \(\sqrt{-x^2}=0\)). For any other value of x, \(\sqrt{-x^2}\) results in \(\sqrt{negative}\), making it undefined for GMAT purpose.
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­An easy way to solve it is plugging values.
For example x=-1 y=-2
Then solve the equation:
 1/(-1) - sqr(2*2) =
-1-2=-3
Using the same values:
y-1 =-2-1 =-3

CORRECT: D
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i solve questions with such negative variables by substitutuing them with a negative sign variable:
let
x=-a
y=-b
after substituting the values we get the answer simply by manipulating.
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I like the solution - it’s helpful.
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I don’t quite agree with the solution. -1-Y
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I don’t quite agree with the solution. -1-Y

Both the question and the solution are absolutely correct and fully aligned with the highest GMAT standards.

I suggest spending a bit more time carefully reviewing both the solution and the discussion above - that should help clarify things.

If something still seems unclear, feel free to ask a specific follow-up. As it stands, your feedback is too vague to address or elaborate on meaningfully.
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Hi Bunuel.

Thanks for the answer. This is a tricky question from my point of view, because one has to really remember the absolute value properties.

though in the this step:

(|x|/x)−sqrt[(−y)∗(−y)]

then the inside the square root then is y ^2. If 2 negative get multiplied they become posistive. How come sqrt( y ^2) is not equal to y, so the answer would be -1-y?

Thanks in advance.
Bunuel
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 both \(x < 0\) and \(y < 0\), then \(|x| = -x\) and \(|y| = -y\).

Hence, we can express the given expression as follows:

\(\frac{\sqrt{x^2} }{x}-\sqrt{-y*|y|}=\)

\(=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\)

\(=\frac{-x}{x}-\sqrt{y^2}=\)

\(=-1-|y|=\)

\(=-1-(-y)=\)

\(=-1+y\)


Answer: D
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Hi Bunuel.

Thanks for the answer. This is a tricky question from my point of view, because one has to really remember the absolute value properties.

though in the this step:

(|x|/x)−sqrt[(−y)∗(−y)]

then the inside the square root then is y ^2. If 2 negative get multiplied they become posistive. How come sqrt( y ^2) is not equal to y, so the answer would be -1-y?

Thanks in advance.
Bunuel
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 both \(x < 0\) and \(y < 0\), then \(|x| = -x\) and \(|y| = -y\).

Hence, we can express the given expression as follows:

\(\frac{\sqrt{x^2} }{x}-\sqrt{-y*|y|}=\)

\(=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\)

\(=\frac{-x}{x}-\sqrt{y^2}=\)

\(=-1-|y|=\)

\(=-1-(-y)=\)

\(=-1+y\)


Answer: D

Your doubt is addressed here: https://gmatclub.com/forum/m26-184464.html#p1729559 Hope it helps.
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