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# Q02-38 Question # 08 Section # 09

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Q02-38 Question # 08 Section # 09 [#permalink]  23 Jun 2016, 12:50
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37% (01:59) correct 62% (02:59) wrong based on 8 sessions
Given the question

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
[Reveal] Spoiler: OA

Last edited by Carcass on 25 Jun 2016, 00:50, edited 1 time in total.
Editing the post and adding the tags
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Re: Q02-38 Question # 08 Section # 09 [#permalink]  25 Jun 2016, 01:16
Expert's post
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.
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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.

I still don't know why $$\sqrt{\frac{-y}{|y|}}$$ =-y NOT -1?
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Re: Q02-38 Question # 08 Section # 09 [#permalink]  24 May 2018, 11:07
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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.

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

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