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Q02-38 Question # 08 Section # 09
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Updated on: 04 Jul 2019, 23:42

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Question Stats:

Given \(x\), \(y<0\), what is the value of \(\frac{\sqrt{x^2}}{x} - \sqrt{\frac{-y}{|y|}}\) ?

A. \(1+y\)

B. \(1-y\)

C. \(-1-y\)

D. \(y-1\)

E. \(x-y\)

Hello,

I was looking at the explanation of this question (ID Q02-38, Question2 section 9) and it quite does not make any sense in the very first step (or maybe I am overlooking something). Kindly check (or possibly explain it if possible/and-is-correct).

Thanks

Arsh

A. \(1+y\)

B. \(1-y\)

C. \(-1-y\)

D. \(y-1\)

E. \(x-y\)

Hello,

I was looking at the explanation of this question (ID Q02-38, Question2 section 9) and it quite does not make any sense in the very first step (or maybe I am overlooking something). Kindly check (or possibly explain it if possible/and-is-correct).

Thanks

Arsh

Re: Q02-38 Question # 08 Section # 09
[#permalink]
25 Jun 2016, 01:16

1

Expert Reply

Hi,

well I have to admit that the explanation is not the top-notch but it is not wrong.

Starting from what we do know: X could be positive or negative: we do not know at the moment AND y is negative.

Now look at the first part of the equation \(\sqrt{x^2}\) is equal to |x|. From this we also know that |x| = -x. Which means that on the left hand side X is positive, always. On the right hand side for X to be always positive X must be negative. This are math rules that is better for you to know as cold during the exam.

As such, actually we do have that X/X is = 1 and is negative. So -1

Going to the second square root |y| = -y so -y * - y = y^2 that under square root becomes |y|.

At this point we have -1 - |y| AND we already know that |y| = -y .

-1 - (-y) = -1 +y

Hope is clear this. Waiting, though, math expert for further clarification.

well I have to admit that the explanation is not the top-notch but it is not wrong.

Starting from what we do know: X could be positive or negative: we do not know at the moment AND y is negative.

Now look at the first part of the equation \(\sqrt{x^2}\) is equal to |x|. From this we also know that |x| = -x. Which means that on the left hand side X is positive, always. On the right hand side for X to be always positive X must be negative. This are math rules that is better for you to know as cold during the exam.

As such, actually we do have that X/X is = 1 and is negative. So -1

Going to the second square root |y| = -y so -y * - y = y^2 that under square root becomes |y|.

At this point we have -1 - |y| AND we already know that |y| = -y .

-1 - (-y) = -1 +y

Hope is clear this. Waiting, though, math expert for further clarification.

Re: Q02-38 Question # 08 Section # 09
[#permalink]
28 Feb 2018, 21:33

Carcass wrote:

Hi,

well I have to admit that the explanation is not the top-notch but it is not wrong.

Starting from what we do know: X could be positive or negative: we do not know at the moment AND y is negative.

Now look at the first part of the equation \(\sqrt{x^2}\) is equal to |x|. From this we also know that |x| = -x. Which means that on the left hand side X is positive, always. On the right hand side for X to be always positive X must be negative. This are math rules that is better for you to know as cold during the exam.

As such, actually we do have that X/X is = 1 and is negative. So -1

Going to the second square root |y| = -y so -y * - y = y^2 that under square root becomes |y|.

At this point we have -1 - |y| AND we already know that |y| = -y .

-1 - (-y) = -1 +y

Hope is clear this. Waiting, though, math expert for further clarification.

well I have to admit that the explanation is not the top-notch but it is not wrong.

Starting from what we do know: X could be positive or negative: we do not know at the moment AND y is negative.

Now look at the first part of the equation \(\sqrt{x^2}\) is equal to |x|. From this we also know that |x| = -x. Which means that on the left hand side X is positive, always. On the right hand side for X to be always positive X must be negative. This are math rules that is better for you to know as cold during the exam.

As such, actually we do have that X/X is = 1 and is negative. So -1

Going to the second square root |y| = -y so -y * - y = y^2 that under square root becomes |y|.

At this point we have -1 - |y| AND we already know that |y| = -y .

-1 - (-y) = -1 +y

Hope is clear this. Waiting, though, math expert for further clarification.

I still don't know why \(\sqrt{\frac{-y}{|y|}}\) =-y NOT -1?

Re: Q02-38 Question # 08 Section # 09
[#permalink]
24 May 2018, 11:07

1

Peter wrote:

Carcass wrote:

Hi,

well I have to admit that the explanation is not the top-notch but it is not wrong.

Starting from what we do know: X could be positive or negative: we do not know at the moment AND y is negative.

Now look at the first part of the equation \(\sqrt{x^2}\) is equal to |x|. From this we also know that |x| = -x. Which means that on the left hand side X is positive, always. On the right hand side for X to be always positive X must be negative. This are math rules that is better for you to know as cold during the exam.

As such, actually we do have that X/X is = 1 and is negative. So -1

Going to the second square root |y| = -y so -y * - y = y^2 that under square root becomes |y|.

At this point we have -1 - |y| AND we already know that |y| = -y .

-1 - (-y) = -1 +y

Hope is clear this. Waiting, though, math expert for further clarification.

well I have to admit that the explanation is not the top-notch but it is not wrong.

Starting from what we do know: X could be positive or negative: we do not know at the moment AND y is negative.

Now look at the first part of the equation \(\sqrt{x^2}\) is equal to |x|. From this we also know that |x| = -x. Which means that on the left hand side X is positive, always. On the right hand side for X to be always positive X must be negative. This are math rules that is better for you to know as cold during the exam.

As such, actually we do have that X/X is = 1 and is negative. So -1

Going to the second square root |y| = -y so -y * - y = y^2 that under square root becomes |y|.

At this point we have -1 - |y| AND we already know that |y| = -y .

-1 - (-y) = -1 +y

Hope is clear this. Waiting, though, math expert for further clarification.

I still don't know why \(\sqrt{\frac{-y}{|y|}}\) =-y NOT -1?

Consider y = (-2), as y<0. hence -y = -(-2) = 2.

|y| = |-2| = 2.

Re: Q02-38 Question # 08 Section # 09
[#permalink]
04 Jul 2019, 15:54

1

I was thinking another way. Let me correct if I am horribly wrong! Let x and y both equal -1. from the first, sqrt of (x square)/x we get -1 and also form second part we get -1. So adding, -2. Consider all options!. Option D fits!

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Re: Q02-38 Question # 08 Section # 09
[#permalink]
05 Jul 2019, 04:49

@AlaminMolla is correct. For ALL negative values of x and y, the expression will ALWAYS evaluate to be -2

So, the question is flawed. At the very least, the question should read "Which of the following COULD BE the value of ...", in which case A, D and E COULD equal -2

@arsh are you sure you transcribed the question correctly?

Cheers,

Brent

So, the question is flawed. At the very least, the question should read "Which of the following COULD BE the value of ...", in which case A, D and E COULD equal -2

@arsh are you sure you transcribed the question correctly?

Cheers,

Brent

Re: Q02-38 Question # 08 Section # 09
[#permalink]
28 Jul 2021, 14:25

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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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