GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Q02-38 Question # 08 Section # 09
[#permalink]
Updated on: 04 Jul 2019, 23:42
Updated on: 04 Jul 2019, 23:42
1
1
4
Bookmarks
00:00
Question Stats:
18% (01:58) correct
81% (01:45) wrong based on 48 sessions
Hide Show timer Statistics
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
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
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?
