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