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Our deepest fear is not that we are inadequate. Our deepest fear is that we are powerful beyond measure. It is our light not our darkness that most frightens us.

Your playing small does not serve the world. There's nothing enlightened about shrinking so that other people won't feel insecure around you.

It's not just in some of us; it's in everyone. And as we let our own light shine, we unconsciously give other people permission to do the same.

As we are liberated from our own fear, our presence automatically liberates others.

Re: M26-25 If x and y are negative numbers [#permalink]
16 Sep 2013, 00:36

Expert's post

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. _________________

Re: M26-25 If x and y are negative numbers [#permalink]
03 Jan 2014, 09:53

"Next, since x<0 and y<0 then |x|=-x and |y|=-y."

Hi Bunuel, can you help me with this explanation...I thought absolute values are ALWAYS positive? Using real numbers for x & y, say x = -1 and y = -2, then |x| = |-1| = 1 and |y|=|-2| = 2. How could |x| = -x?

Re: M26-25 If x and y are negative numbers [#permalink]
09 Jan 2014, 05:50

1

This post received KUDOS

neehow wrote:

"Next, since x<0 and y<0 then |x|=-x and |y|=-y."

Hi Bunuel, can you help me with this explanation...I thought absolute values are ALWAYS positive? Using real numbers for x & y, say x = -1 and y = -2, then |x| = |-1| = 1 and |y|=|-2| = 2. How could |x| = -x?

Thanks.

Hi Neehow,

In the context of the question, we are given that x and y are both negative numbers.

The absolute value of a negative number is of course positive.

Re: M26-25 If x and y are negative numbers [#permalink]
10 Jan 2014, 12:11

Spoontrick wrote:

neehow wrote:

"Next, since x<0 and y<0 then |x|=-x and |y|=-y."

Hi Bunuel, can you help me with this explanation...I thought absolute values are ALWAYS positive? Using real numbers for x & y, say x = -1 and y = -2, then |x| = |-1| = 1 and |y|=|-2| = 2. How could |x| = -x?

Thanks.

Hi Neehow,

In the context of the question, we are given that x and y are both negative numbers.

The absolute value of a negative number is of course positive.

Re: M26-25 If x and y are negative numbers [#permalink]
27 Feb 2014, 07:09

Regarding the question above in Bunnuel, I just don't understand why the y turns positive if the subtract sign is outside the absolute signs? Thanks always for providing the best descriptive answers online.

Re: M26-25 If x and y are negative numbers [#permalink]
27 Feb 2014, 07:13

Expert's post

demart78 wrote:

Regarding the question above in Bunnuel, I just don't understand why the y turns positive if the subtract sign is outside the absolute signs? Thanks always for providing the best descriptive answers online.

We are told that y is a negative number, thus |y|=-y, so -1-y=-1-(-y)=-1+y.

Re: M26-25 If x and y are negative numbers [#permalink]
27 Feb 2014, 07:20

Bunuel wrote:

demart78 wrote:

Regarding the question above in Bunnuel, I just don't understand why the y turns positive if the subtract sign is outside the absolute signs? Thanks always for providing the best descriptive answers online.

We are told that y is a negative number, thus |y|=-y, so -1-y=-1-(-y)=-1+y.

Hope it's clear.

I still do not understand. If y is a negative number, once the absolute brackets go up around it, shouldn't it than be converted to a positive number?

Or because we are told that it is negative, we automatically assume that y<0 and hit it with the negative sign?

Re: M26-25 If x and y are negative numbers [#permalink]
27 Feb 2014, 07:30

Expert's post

demart78 wrote:

Bunuel wrote:

demart78 wrote:

Regarding the question above in Bunnuel, I just don't understand why the y turns positive if the subtract sign is outside the absolute signs? Thanks always for providing the best descriptive answers online.

We are told that y is a negative number, thus |y|=-y, so -1-y=-1-(-y)=-1+y.

Hope it's clear.

I still do not understand. If y is a negative number, once the absolute brackets go up around it, shouldn't it than be converted to a positive number?

Or because we are told that it is negative, we automatically assume that y<0 and hit it with the negative sign?

It seems that you need to brush up fundamentals on absolute values.

When x\leq{0} then |x|=-x, or more generally when some \ expression\leq{0} then |some \ expression|={-(some \ expression)}. For example: |-5|=5=-(-5);

When x\geq{0} then |x|=x, or more generally when some \ expression\geq{0} then |some \ expression|={some \ expression}. For example: |5|=5.

Re: M26-25 If x and y are negative numbers [#permalink]
05 Mar 2014, 12:18

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?

Re: M26-25 If x and y are negative numbers [#permalink]
06 Mar 2014, 02:19

Expert's post

damamikus wrote:

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?

The point here is that the square root function cannot give negative result: \sqrt{some \ expression}\geq{0}. So, \sqrt{(-1)^2}=\sqrt{1}=1.

Thus if x=-1, then \frac{\sqrt{x^2}}{x}=\frac{\sqrt{(-1)^2}}{-1}=\frac{\sqrt{1}}{-1}=\frac{1}{-1}=-1.

Re: M26-25 If x and y are negative numbers [#permalink]
05 May 2014, 21:11

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.

Hi Bunnel,

I have one doubt on this

here x and y is <0 so we are considering |x| and |y| as -x and -y

so I just want to know in denominator we have |x|/x = -x/x. why we are not considering x in denominator as -x

Re: M26-25 If x and y are negative numbers [#permalink]
06 May 2014, 02:13

Expert's post

PathFinder007 wrote:

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.

Hi Bunnel,

I have one doubt on this

here x and y is <0 so we are considering |x| and |y| as -x and -y

so I just want to know in denominator we have |x|/x = -x/x. why we are not considering x in denominator as -x

Please clarify

Thanks.

x is a negative number, so -x is a positive number. So, knowing that x is negative doesn't mean that you should write -x instead of x. _________________

Re: M26-25 If x and y are negative numbers [#permalink]
05 Aug 2014, 17:48

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.

Hi Bunuel,

Can you please explain one small point in this question. If the expression was √x^2/x −√-y∗y will the answer change? I am only trying to understand whether value will change between sqrt (-y^2) and sqrt(-y*|y|), given y is negative.

M26-25 If x and y are negative numbers [#permalink]
12 Aug 2014, 00:58

1

This post received KUDOS

Expert's post

hubahuba wrote:

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.

Hi Bunuel,

Can you please explain one small point in this question. If the expression was √x^2/x −√-y∗y will the answer change? I am only trying to understand whether value will change between sqrt (-y^2) and sqrt(-y*|y|), given y is negative.

In this case the question would be flawed: \sqrt{(-y)*y}=\sqrt{positive*negative}=\sqrt{negative} and the square roots from negative numbers are undefined for the GMAT. _________________