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# What is (x^2y^2)^1/2 if x < 0 and y > 0 ?

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What is (x^2y^2)^1/2 if x < 0 and y > 0 ?  [#permalink]

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25 Jun 2016, 23:36
$$\sqrt{x^2}=(-x)$$ because -x > 0
$$\sqrt{y^2}=y$$ because y > 0
$$\sqrt{x^2*y^2}= \sqrt{x^2} * \sqrt{y^2} = (-x)y$$
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What is (x^2y^2)^1/2 if x < 0 and y > 0 ?  [#permalink]

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Updated on: 15 Aug 2016, 06:06
nehamodak wrote:
What is $$\sqrt{x^2*y^2}$$ if x < 0 and y > 0?

(A) –xy
(B) xy
(C) –|xy|
(D) |y|x
(E) No solution

take x as -2 and y as 2
$$\sqrt{-2^2*2^2}$$
$$\sqrt{4*4}$$
$$\sqrt{16}$$ = 4

Now see what option gives you 4
x*y= -2*2=-4 WRONG

-(x)*y = -(-2)*2 = 2*2 = 4 CORRECT
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Originally posted by LogicGuru1 on 24 Jul 2016, 22:56.
Last edited by LogicGuru1 on 15 Aug 2016, 06:06, edited 1 time in total.
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ?  [#permalink]

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15 Aug 2016, 03:24
What's the difference between A and C?

Both will give us the same answer right?

A = - (of a positive xy)
B = - (of a positive xy)
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ?  [#permalink]

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16 Aug 2016, 00:42
1
ameyaprabhu wrote:
What's the difference between A and C?

Both will give us the same answer right?

A = - (of a positive xy)
B = - (of a positive xy)

You are given that x < 0 (say x is -2) and y > 0 (say y is 5).
So xy will be -2*5 = -10.

Hence xy will actually be a negative number.

Option (A) = - xy = - (-10) = 10 (a positive number)

Option (C) = - |xy| = - |-10| = -10 (a negative number)
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ?  [#permalink]

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16 Aug 2016, 02:15
got it. Thanks

[quote="VeritasPrepKarishma"][quote="ameyaprabhu"]What's the difference between A and C?
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What is (x^2y^2)^1/2 if x < 0 and y > 0 ?  [#permalink]

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10 Sep 2016, 06:54
But isn't saying that |x| = -x contradictory?.. You are saying that "something always positive equals a negative umber"

How can this be?

Or is this something like |x| = - (X) , so "if X is negative then modulus is (obviously) positive and if you plug a negative X into a negative -> negative X you get a positive X ?

The way the choice A is written, is misleading imo. It looks as you are saying that a modulus is somehow equivalent to a negative number
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ?  [#permalink]

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10 Sep 2016, 08:35
iliavko wrote:
But isn't saying that |x| = -x contradictory?.. You are saying that "something always positive equals a negative umber"

How can this be?

Or is this something like |x| = - (X) , so "if X is negative then modulus is (obviously) positive and if you plug a negative X into a negative -> negative X you get a positive X ?

The way the choice A is written, is misleading imo. It looks as you are saying that a modulus is somehow equivalent to a negative number

|x| = -x implies what I have highlighted above.
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What is (x^2y^2)^1/2 if x < 0 and y > 0 ?  [#permalink]

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10 Sep 2016, 11:29
iliavko wrote:
But isn't saying that |x| = -x contradictory?.. You are saying that "something always positive equals a negative umber"

How can this be?

Or is this something like |x| = - (X) , so "if X is negative then modulus is (obviously) positive and if you plug a negative X into a negative -> negative X you get a positive X ?

The way the choice A is written, is misleading imo. It looks as you are saying that a modulus is somehow equivalent to a negative number

|x| actually behaves differently for values greater than 0 and less than 0.

In other words |x| has two cases
CASE 1) x>0;
for x >0 |x| = x
For example |4|= 4

Case 2) x<0;
for x<0 |x|= -x
For example |-4|= - of -4 or -(-4)= 4
so to generalise |-x| = -(x);
Now since x is a "variable" it is hiding a "Negative Polarity" inside itself in this case.

YOU HAVE TO BE ABSOLUTELY CLEAR ABOUT IT. A VARIABLE CAN LOOK POSITIVE BUT IT MAY CARRY A HIDDEN POLARITY .
For example x can be -9 but just looking at x you cannot quickly relate to -9. You will automatically see x as positive.
BUT REMEMBER X IS A VARIABLE THAT CAN HAVE ANY POLARITY AND VALUE.
THAT IS WHY SO MANY PEOPLE ARE GETTING CONFUSED IN THIS QUESTION.

It is useless to rote or memorize the formula that |x|= x until and unless how it works in case of x>0 and x<0

Again to revise

IF x>0 then |x|= x {plain and simple to remember}
If x<0 then |x|=-x {You need to understand the concept of hidden polarity inside a variable}.
In this case particular example -x will ultimately be a positive numerical value but if the question does not involve substitution of numerical values to x and -x then you have to be solve the question very carefuly.

Hope its clear to all now .
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ?  [#permalink]

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19 Apr 2017, 20:45
VeritasPrepKarishma, I don't understand your explanation:

$$\sqrt{(x^2*y^2)}$$ = |x|*|y| --> I understand this part, but how do you then go here: "Now, if x < 0, |x|=−x"

Let me rephrase:
- if I told you that I know x=-2, when you take |-2|, you get 2.
- if, on the other hand, I told you that the value of x changed, and is now 25, then when you take |25|, you get 25.
* The point here being: the absolute value will give you the positive value of whatever is inside the brackets, regardless of what sign the content inside the bracket is

So I'm having a hard time rationalizing this: |x|=−x. How can the absolute value of a variable ever be negative?
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ?  [#permalink]

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19 Apr 2017, 23:30
1
LakerFan24 wrote:
VeritasPrepKarishma, I don't understand your explanation:

$$\sqrt{(x^2*y^2)}$$ = |x|*|y| --> I understand this part, but how do you then go here: "Now, if x < 0, |x|=−x"

Let me rephrase:
- if I told you that I know x=-2, when you take |-2|, you get 2.
- if, on the other hand, I told you that the value of x changed, and is now 25, then when you take |25|, you get 25.
* The point here being: the absolute value will give you the positive value of whatever is inside the brackets, regardless of what sign the content inside the bracket is

So I'm having a hard time rationalizing this: |x|=−x. How can the absolute value of a variable ever be negative?

What makes you say that -x is a negative quantity?
Note that the absolute value is -x when x itself is negative. So x itself will have a negative sign and the two negatives will cancel off each other to give you a positive value.
-x is a way of expressing a positive quantity when x itself is negative.

Does this make sense?
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ?  [#permalink]

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20 Apr 2017, 18:41
VeritasPrepKarishma wrote:
LakerFan24 wrote:
VeritasPrepKarishma, I don't understand your explanation:

$$\sqrt{(x^2*y^2)}$$ = |x|*|y| --> I understand this part, but how do you then go here: "Now, if x < 0, |x|=−x"

Let me rephrase:
- if I told you that I know x=-2, when you take |-2|, you get 2.
- if, on the other hand, I told you that the value of x changed, and is now 25, then when you take |25|, you get 25.
* The point here being: the absolute value will give you the positive value of whatever is inside the brackets, regardless of what sign the content inside the bracket is

So I'm having a hard time rationalizing this: |x|=−x. How can the absolute value of a variable ever be negative?

What makes you say that -x is a negative quantity?
Note that the absolute value is -x when x itself is negative. So x itself will have a negative sign and the two negatives will cancel off each other to give you a positive value.
-x is a way of expressing a positive quantity when x itself is negative.

Does this make sense?

Let me see if I understand this correctly: |x| is always positive, and since we know that x is a negative number, in order to express a positive quantity, we need to add a negative to "cancel out" the negatives to make positive? This way, |x| = -(negative value,x). If I were to substitute values for x (let's say, -2), then: |2| = -(-2)
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What is (x^2y^2)^1/2 if x < 0 and y > 0 ?  [#permalink]

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06 Jul 2017, 20:16
VeritasPrepKarishma wrote:
nehamodak wrote:
What is \sqrt{x^2*y^2} if x < 0 and y > 0?
(A) –xy
(B) xy
(C) –|xy|
(D) |y|x
(E) No solution

Please explain how do we solve this

Note that $$\sqrt{x^2} = |x|$$. It is not x, it is |x|. When we talk about square root, it implies the principal square root i.e. just the positive square root.

$$\sqrt{x^2*y^2} = |x|*|y|$$

Now, if x < 0, $$|x| = -x$$
If y > 0, $$|y| = y$$

Hence, $$|x|*|y| = -x*y$$

$$\sqrt{x^2} = |x|$$

^ in the above equation, if the X value is not specified, can you know for certain whether X is positive or negative?

i.e. - the only way we know |X| is negative is because the stem tells us x<0. Without this information, are we unable to determine the sign of X?
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What is (x^2y^2)^1/2 if x < 0 and y > 0 ?  [#permalink]

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06 Jul 2017, 20:19
LogicGuru1 wrote:
iliavko wrote:
But isn't saying that |x| = -x contradictory?.. You are saying that "something always positive equals a negative umber"

How can this be?

Or is this something like |x| = - (X) , so "if X is negative then modulus is (obviously) positive and if you plug a negative X into a negative -> negative X you get a positive X ?

The way the choice A is written, is misleading imo. It looks as you are saying that a modulus is somehow equivalent to a negative number

|x| actually behaves differently for values greater than 0 and less than 0.

In other words |x| has two cases
CASE 1) x>0;
for x >0 |x| = x
For example |4|= 4

Case 2) x<0;
for x<0 |x|= -x
For example |-4|= - of -4 or -(-4)= 4
so to generalise |-x| = -(x);
Now since x is a "variable" it is hiding a "Negative Polarity" inside itself in this case.

YOU HAVE TO BE ABSOLUTELY CLEAR ABOUT IT. A VARIABLE CAN LOOK POSITIVE BUT IT MAY CARRY A HIDDEN POLARITY .
For example x can be -9 but just looking at x you cannot quickly relate to -9. You will automatically see x as positive.
BUT REMEMBER X IS A VARIABLE THAT CAN HAVE ANY POLARITY AND VALUE.
THAT IS WHY SO MANY PEOPLE ARE GETTING CONFUSED IN THIS QUESTION.

It is useless to rote or memorize the formula that |x|= x until and unless how it works in case of x>0 and x<0

Again to revise

IF x>0 then |x|= x {plain and simple to remember}
If x<0 then |x|=-x {You need to understand the concept of hidden polarity inside a variable}.
In this case particular example -x will ultimately be a positive numerical value but if the question does not involve substitution of numerical values to x and -x then you have to be solve the question very carefuly.

Hope its clear to all now .

This helps and I just want to clarify that without the added information that x<0, there would be no way of determining whether X is positive or negative? Correct?
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ?  [#permalink]

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06 Jul 2017, 22:42
ak1802 wrote:

This helps and I just want to clarify that without the added information that x<0, there would be no way of determining whether X is positive or negative? Correct?

Hi ak1802 ,

Yes, that's true. We cannot find the nature of x unless we are given such condition.
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ?  [#permalink]

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27 Jul 2017, 15:50
VeritasPrepKarishma

am i correct in saying that $$\sqrt{x^{2}}$$ is NOT the same as writing $$\sqrt{(x^{2})}$$, so is this the reason here that $$\sqrt{x^{2}}$$ = (-x)?

is this what the problem is testing? because otherwise, i cannot understand how an absolute value is negative -- an absolute value can never be negative, because it is the measure of how far away numbers are on a number line
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ?  [#permalink]

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27 Jul 2017, 20:04
1
LakerFan24 wrote:
VeritasPrepKarishma

am i correct in saying that $$\sqrt{x^{2}}$$ is NOT the same as writing $$\sqrt{(x^{2})}$$, so is this the reason here that $$\sqrt{x^{2}}$$ = (-x)?

is this what the problem is testing? because otherwise, i cannot understand how an absolute value is negative -- an absolute value can never be negative, because it is the measure of how far away numbers are on a number line

$$\sqrt{x^{2}}$$ is the same as $$\sqrt{(x^{2})}$$.

$$\sqrt{x^{2}}=|x|$$. We are given that x is negative. When x < 0, we know that |x| = -x, so $$\sqrt{x^{2}}=|x|=-x$$. Notice that since x is negative, -x = -negative = positive, thus both the square root and the absolute value return positive result.
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ?  [#permalink]

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27 Jul 2017, 20:17
LakerFan24 wrote:
VeritasPrepKarishma

am i correct in saying that $$\sqrt{x^{2}}$$ is NOT the same as writing $$\sqrt{(x^{2})}$$, so is this the reason here that $$\sqrt{x^{2}}$$ = (-x)?

is this what the problem is testing? because otherwise, i cannot understand how an absolute value is negative -- an absolute value can never be negative, because it is the measure of how far away numbers are on a number line

In addition to what Bunuel said above, let me also add a line on how I explain this in words: You are right. An absolute value can never be negative. So |x| will never be negative. But it can be equal to -x. When? When x itself is negative. Note that a variable x CAN stand for a negative value too. So if x itself is negative, -x becomes POSITIVE. And that is the case in which |x| is equal to -x (which is positive).
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ?  [#permalink]

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