Official Solution:

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 \(x \lt 0\) and \(y \lt 0\) 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
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Below links on absolute value should help you:

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First of all, read this: rules-for-posting-please-read-this-before-posting-133935.html#p1096628 (Writing Mathematical Formulas on the Forum)

Next, the point is that the square root function cannot give negative result (\(\sqrt{some \ expression}\geq{0}\)) and \(\sqrt{x^2}=|x|\), which means that \(\sqrt{x^2}/ x=\frac{|x|}{x}=\frac{-x}{x}=-1\).

Please follow the links in my previous post for more on absolute values.
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-1 - |y| = -1 + y because we are given that y is negative. When y is negative |y| = -y. Hence -1 - |y| = -1 - (-y) = -1 + y.
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That's wrong. \(\sqrt{-y*|y|}=\sqrt{-(negative)*positive}=\sqrt{positive*positive}\)
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MUST KNOW: \(\sqrt{x^2}=|x|\):

The point here is that since square root function cannot give negative result then \(\sqrt{some \ expression}\geq{0}\).

So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?

Let's consider following examples:
If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\);
If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).

So we got that:
\(\sqrt{x^2}=x\), if \(x\geq{0}\);
\(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).

For example if x = -5, then \(\sqrt{x^2}=\sqrt{25}=5=|-5|=|x|\)


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Hi,

Having trouble wrapping my head around why \(\sqrt{x^2}\) is equal to |x|

I looked at the post on this page, but did not see any explanation.

For example, how is the equation \(\sqrt{4}\) = {-2, 2} any different from \(\sqrt{2^2}\) = ? Is it because the solution of \(\sqrt{x^2}\) must be equal to the solution of \((\sqrt{x})^2\)? if so, why?

Thanks in advance for any help!

(with regard to bunnel's explanation to xnthic, throughout the explanation bunnel writes "\(\sqrt{25}\) = 5" as though this is given, but my understanding is that \(\sqrt{25}\) could be either positive or negative 5, as either positive or negative 5 squared result in 25. If I am incorrect, then that is most likely the root of my confusion)

Hi Bunuel, certain application of the sqrt function seem to be contradictory. If sqrt function can't give negative numbers, why do we substitute both -ve and +ve values of \(\sqrt{expression}\) while solving algebraic equations.
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Hi,

Since x is negative, \sqrt{x^2}/-x - \sqrt{-y*-y}
= \(-x/-x\) - \sqrt{y^2}
= 1-(-y)
= 1+y

Can someone please explain what wrong am I doing here?

Hi,

In one of the other questions, the explanation is provided as follows

If p=−1
then (|p|!)^p=(|−1|!)^−1=1^−1=1

Here |-1| = 1

But in this problem |y| = -y since y is negative

Can you please explain the discrepancy or the misunderstanding in my thought