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

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nehamodak
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What is (x^2y^2)^1/2 if x < 0 and y > 0 ? [#permalink] New post   Updated on: 17 Jul 2014, 01:04
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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
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A

Originally posted by nehamodak on 16 Jul 2014, 19:57.
Last edited by Bunuel on 17 Jul 2014, 01:04, edited 1 time in total.
Edited the question.
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KarishmaB
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ? [#permalink] New post   16 Jul 2014, 22:58
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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, \(|x|*|y| = -x*y\)

Answer (A)
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Bunuel
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What is (x^2y^2)^1/2 if x < 0 and y > 0 ? [#permalink] New post   17 Jul 2014, 01:24
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PareshGmat wrote:
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, \(|x|*|y| = -x*y\)

Answer (A)


It seems there is a risk if we plugin numbers eg: -2 & 2 which may not give an undisputed answer as computed by Karishma above :)

Experts, have your say..


If you plug x=-2 and y=2, then \(\sqrt{x^2*y^2}=4\):

(A) –xy = 4. Match.
(B) xy = -4. Discard.
(C) –|xy| = -4. Discard.
(D) |y|x = -4. Discard.

So, plunging in gives the same answer.

One could discard B, C and D right away. The square root function cannot give negative result, while all these options are negative, which means that they cannot be correct answers:

(A) –xy = -(negative)(positive) = positive. OK.
(B) xy = (negative)(positive) = negative. Discard.
(C) –|xy| = -positive = negative. Discard.
(D) |y|x = positive*negative = negative. Discard.

Hope it's clear.
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PareshGmat
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ? [#permalink] New post   16 Jul 2014, 23:27
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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, \(|x|*|y| = -x*y\)

Answer (A)


It seems there is a risk if we plugin numbers eg: -2 & 2 which may not give an undisputed answer as computed by Karishma above :)

Experts, have your say..
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ? [#permalink] New post   16 Jul 2014, 23:48
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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, \(|x|*|y| = -x*y\)

Answer (A)


Karishma--

I understand the algebraic method of solving the problem, but I don't understand why plugging numbers isn't accurate on these kinds of problems. Could you explain? Thanks!
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ? [#permalink] New post   17 Jul 2014, 04:41
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napoleonsselfie wrote:
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, \(|x|*|y| = -x*y\)

Answer (A)


Karishma--

I understand the algebraic method of solving the problem, but I don't understand why plugging numbers isn't accurate on these kinds of problems. Could you explain? Thanks!


As explained by Bunuel above, plugging-in numbers isn't necessarily a problem. Sometimes, it can be problematic because multiple options may match. Then you would need to try another set of numbers on the options that matched. Usually, you will get one correct answer within two rounds.
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ? [#permalink] New post   17 Jul 2014, 09:18
Hi Bunuel and Karisma,
does that mean square root of 4 is always 2 and not -2?
I thought it had two options just like x to the power 2 = 4 gives, x = 2 and x = -2.
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ? [#permalink] New post   17 Jul 2014, 09:22
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GMatAspirerCA wrote:
Hi Bunuel and Karisma,
does that mean square root of 4 is always 2 and not -2?
I thought it had two options just like x to the power 2 = 4 gives, x = 2 and x = -2.


When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root. That is, \(\sqrt{4}=2\), NOT +2 or -2.

In contrast, the equation \(x^2=4\) has TWO solutions, +2 and -2.

Hope this helps.
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ? [#permalink] New post   15 Aug 2016, 04:24
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] New post   16 Aug 2016, 01:42
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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] New post   27 Jul 2017, 16:50
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] New post   27 Jul 2017, 21:04
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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] New post   27 Jul 2017, 21:17
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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] New post   01 Aug 2023, 09:10
KarishmaB 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, \(|x|*|y| = -x*y\)

Answer (A)


Hi KarishmaB Bunuel, can't be write (x^2y^2)^1/2 =|x*y|. Why do we have to put x and y in separate modulus?
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ? [#permalink] New post   01 Aug 2023, 13:41
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ankitapugalia wrote:
KarishmaB 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, \(|x|*|y| = -x*y\)

Answer (A)


Hi KarishmaB Bunuel, can't be write (x^2y^2)^1/2 =|x*y|. Why do we have to put x and y in separate modulus?


Both are the same: |xy| = |x|*|y|.
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ? [#permalink]
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