Bunuel wrote:
vaishnogmat wrote:
If x and y are negative numbers, what is the value of √x^2/x −√-y∗|y|?
a) 1+y
b) 1−y
c) −1−y
d) y−1
e) x−y
I don't understand why do I get a different answer when I use numbers. Bunuel please show the explanation using numbers, not just pure theory. Thanks.
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
Note that \(\sqrt{a^2}=|a|\). Next, since \(x<0\) and \(y<0\) then \(|x|=-x\) and \(|y|=-y\).
So, \(\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y\)
Answer: D.Number plugging:Since given that x and y are negative numbers, say x=-1 and y=-2.
\(\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=-1-2=-3\). Now, plug x=-1 and y=-2 into the options to see which one yields -3. Only D fits.
Absolute value questions and I don't seem to be getting close friends anytime soon
Here's what still doesn't make sense to me: if x=-1, then why does \(\frac{\sqrt{x^2}}{x}\) become -1?? i understand the rule that when x<0 then |x|=-x and that the square root of x is equal to |x| but in this case we already know that x is negative. So when taking the square root out of (-1)² then why do we not get x=-1 (as we initially said) and hence x/x=(-1)/(-1)=1? it's not quite intuitive for me to go from \(\frac{\sqrt{(-1)^2}}{-1}\) to -1. I thought we are taking the square root out of x² in order to determine x, which we already know to be negative, hence it should be -1 in this example. i just don't seem to get it ...
or is it just a convention/ rule that when using square roots in value expressions we only consider the absolute values of such expressions and that therefore \(\frac{\sqrt{(-1)^2}}{-1}\) becomes -1 because we are dividing the absolute value of x=|x|=|-1|=1 by x=-1?