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How does abs(x) = -x, and abs(y) = -y when x<0 and y<0? I am do not understand this concept
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[quote="Bunuel"]Official Solution:

-1-|y|=-1+y

I have not understood the above portion. Could you please tell me how to get this?

I have done it by taking numbers.

Let x=-1 and y=-1 Here i have taken integer value thinking that I will get an answer which can be in option also.

Now putting the value of x and y I am getting the value of question is -2.

Only option D gives me the value. Is my approach correct?
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Raihanuddin wrote:
Bunuel wrote:
Official Solution:

-1-|y|=-1+y

I have not understood the above portion. Could you please tell me how to get this?

I have done it by taking numbers.

Let x=-1 and y=-1 Here i have taken integer value thinking that I will get an answer which can be in option also.

Now putting the value of x and y I am getting the value of question is -2.

Only option D gives me the value. Is my approach correct?

-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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having a hard time understanding this basic rule: $$\sqrt{x^2}=|x|$$

Assuming it's an integer, shouldn't this be the positive integer of x? why |x|
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xnthic wrote:
having a hard time understanding this basic rule: $$\sqrt{x^2}=|x|$$

Assuming it's an integer, shouldn't this be the positive integer of x? why |x|

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 $$\sqrt{x^2}=\sqrt{25}=5=|-5|=|x|$$

Theory on Abolute Values: math-absolute-value-modulus-86462.html
Absolute value tips: absolute-value-tips-and-hints-175002.html

DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hard set on Abolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html
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hi,

it is not that the square root function cannot be negative. The expression inside the square root is always positive or 0.
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attrirahul456 wrote:
hi,

it is not that the square root function cannot be negative. The expression inside the square root is always positive or 0.

You are wrong. It also seems that you did not read the whole discussion.

$$\sqrt{...}$$ is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign ($$\sqrt{...}$$) always means non-negative square root.

The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

$$\sqrt{9} = 3$$, NOT +3 or -3;
$$\sqrt[4]{16} = 2$$, NOT +2 or -2;
Similarly $$\sqrt{\frac{1}{16}} = \frac{1}{4}$$, NOT +1/4 or -1/4.

Notice that in contrast, the equation $$x^2 = 9$$ has TWO solutions, +3 and -3. Because $$x^2 = 9$$ means that $$x =-\sqrt{9}=-3$$ or $$x=\sqrt{9}=3$$.
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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. $$x-y$$

HI GMATGuruNY, MentorTutoring

Can you help me with this problem?

Consider x=-2 & y= -3

$$\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}$$ ==> $$\frac{\sqrt{-2^2}}{-2}-\sqrt{-(-3)*|-3|}$$

= -1-3 ==> -4

Only option D is valid. Am'I right?
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NandishSS 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. $$x-y$$

HI GMATGuruNY, MentorTutoring

Can you help me with this problem?

Consider x=-2 & y= -3

$$\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}$$ ==> $$\frac{\sqrt{(-2)^2}}{-2}-\sqrt{-(-3)*|-3|}$$

= -1-3 ==> -4

Only option D is valid. Am I right?

Looks good!
(One small correction: the -2 under the square root should be enclosed in parentheses, as shown here.)
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NandishSS 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. $$x-y$$

HI GMATGuruNY, MentorTutoring

Can you help me with this problem?

Consider x=-2 & y= -3

$$\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}$$ ==> $$\frac{\sqrt{-2^2}}{-2}-\sqrt{-(-3)*|-3|}$$

= -1-3 ==> -4

Only option D is valid. Am'I right?

Yes, and -2 and -3 are some of my go-to numbers for such complicated looking problems, since the two, when squared or cubed, will not often lead to multiple valid answer choices when you substitute for the variables there. You might say you read my mind. I see that GMATGuruNY beat me to the parentheses tip. Make sure you understand the difference between the following:

$$-2^2=-4$$

and

$$(-2)^2=4$$

It has to do with the order of operations, and the former is telling you to take 2, square it, and make the result negative. Although you still got the answer correct in this case, you might not in another such question, one in which the number needed to be negative.

Well done. Thank you for tagging me, and good luck with your studies.

- Andrew
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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 quick and easy approach is to test some values of $$x$$ and $$y$$
If $$x = -3$$ and $$y = -2$$, we get: $$\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{\sqrt{(-3)^2}}{-3}-\sqrt{-(-2)*|-2|}$$

$$=\frac{\sqrt{9}}{-3}-\sqrt{4}$$

$$=\frac{3}{-3}-2$$

$$=(-1)-2$$

Aside: since the term with the x's turns into -1, we can eliminate answer choices A, B and E
To determine the correct answer let's rearrange what we have so far to get:

$$=(-1)+ (-2)$$

$$=(-1)+ y$$

$$=y-1$$

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hi, since there can only be a positive number under the root then, root(-y(|y|)) must be a positive number with the magnitude of y

thereby taking the whole expression to -1 - |y|

now, how can we claim this to be equal to y - 1 ?
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parvjindal13 wrote:
hi, since there can only be a positive number under the root then, root(-y(|y|)) must be a positive number with the magnitude of y

thereby taking the whole expression to -1 - |y|

now, how can we claim this to be equal to y - 1 ?

Since we are given that y is negative, then |y| = -y and -1 - |y| becomes -1 - (-y) = y - 1.

So, $$\sqrt{-y*|y|}=|y|=-y$$. Notice here that since y is negative then -y is positive and the result of the square root is still positive.
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I think this is a high-quality question and I agree with explanation.
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I think this is a high-quality question and I don't agree with the explanation. Hi,
I have a query regarding this.
I got the answer as 1 - y

Now, I think by taking values also, I am getting 1 +y.
Let x= -4 and y = -3
So, the expression will look as follows:
[{(16)^(1/2)}/(-4)] - [{-(-3)*|-3|}^(1/2)]
--> {(-4)/(-4)} - {9^(1/2)} ........(Since x<0, 16^(1/2) = -4)
--> 1- (-3) ................(Since y<0, 9^(1/2) = -3)
--> 1 - y ................(Since y=-3)

Pl let me know where I am going wrong
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SGhosh08 wrote:
I think this is a high-quality question and I don't agree with the explanation. Hi,
I have a query regarding this.
I got the answer as 1 - y

Now, I think by taking values also, I am getting 1 +y.
Let x= -4 and y = -3
So, the expression will look as follows:
[{(16)^(1/2)}/(-4)] - [{-(-3)*|-3|}^(1/2)]
--> {(-4)/(-4)} - {9^(1/2)} ........(Since x<0, 16^(1/2) = -4)
--> 1- (-3) ................(Since y<0, 9^(1/2) = -3)
--> 1 - y ................(Since y=-3)

Pl let me know where I am going wrong

If x = -4 and y = -3, then:

$$\frac{ \sqrt{x^2} } {x}-\sqrt{-y*|y|}$$

$$=\frac{ \sqrt{16} }{-4} - \sqrt{3*3}=$$

$$=\frac{4}{-4}-3=$$

$$=-1-3=$$

$$=-1+(-3)=$$

$$=-1+y$$.
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