Tanisha2819 wrote:
Bunuel wrote:
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. \(x-y\)
A. \(1+y\)
B. \(1-y\)
C. \(-1-y\)
D. \(y-1\)
E. \(x-y\)
HelloBunuel
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 youBunuel
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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Statistics : Posted by Bunuel • on 16 Sep 2014, 01:25 • Replies 23 • Views 52679
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