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What is (x^2y^2)^1/2 if x < 0 and y > 0 ?
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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) yx (E) No solution
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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.
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ?
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16 Jul 2014, 22:58
nehamodak wrote: What is \sqrt{x^2*y^2} if x < 0 and y > 0? (A) –xy (B) xy (C) –xy (D) yx (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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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ?
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16 Jul 2014, 20:53
shouldn't it be just xy, since for sqrt of any value we need to take the absolute positive value.



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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ?
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16 Jul 2014, 21:44
plugging in values x = 2 y = 2
sqrt of \sqrt{4*4} = 4
answer can be +4 or 4
xy = 4 satisfies one option
xy = 4 satisfies other option.
Which one is correct here?



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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ?
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16 Jul 2014, 22:26
Answer should be B..
sqrt(x^2) = x



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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ?
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16 Jul 2014, 23:27
VeritasPrepKarishma wrote: nehamodak wrote: What is \sqrt{x^2*y^2} if x < 0 and y > 0? (A) –xy (B) xy (C) –xy (D) yx (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 ?
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16 Jul 2014, 23:48
VeritasPrepKarishma wrote: nehamodak wrote: What is \sqrt{x^2*y^2} if x < 0 and y > 0? (A) –xy (B) xy (C) –xy (D) yx (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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What is (x^2y^2)^1/2 if x < 0 and y > 0 ?
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17 Jul 2014, 01:24
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) yx (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) yx = 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) yx = positive*negative = negative. Discard. Hope it's clear.
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ?
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17 Jul 2014, 02:32
nehamodak wrote: What is \(\sqrt{x^2*y^2}\) if x < 0 and y > 0?
(A) –xy (B) xy (C) –xy (D) yx (E) No solution Let us say that x = 2 and y = 3 then the value is 6 (A) (2)(3) = 6 (B) 6 (ELIMINATED) (C) 6 (ELIMINATED) (D) 6 (ELIMINATED) (E) There is a solution Hence the answer is A
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ?
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17 Jul 2014, 04:41
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) yx (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, pluggingin 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 ?
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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 ?
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17 Jul 2014, 09:22
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 ?
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17 Jul 2014, 09:59
Thanks Bunuel. But this contradicts my math concepts on square root. according to Princeton review radical can never be negative. that is x can not be negative. will have to revisit this section on square roots once. Thanks



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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ?
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17 Jul 2014, 10:03



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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ?
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17 Jul 2014, 11:11
my doubt is this. \sqrt{4}=2, NOT +2 or 2. i had understood that the possibities of sq root are both + or  2.



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What is (x^2y^2)^1/2 if x < 0 and y > 0 ?
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17 Jul 2014, 11:13
GMatAspirerCA wrote: my doubt is this. \sqrt{4}=2, NOT +2 or 2. i had understood that the possibities of sq root are both + or  2. NO. \(\sqrt{4}=2\) ONLY. Please read again: 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, \(\sqrt{4}=2\) ONLY, NOT +2 or 2. In contrast, the equation \(x^2=4\) has TWO solutions, +2 and 2.
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ?
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15 Sep 2014, 02:27
VeritasPrepKarishma wrote: nehamodak wrote: What is \sqrt{x^2*y^2} if x < 0 and y > 0? (A) –xy (B) xy (C) –xy (D) yx (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) Quote: Hey I have a doubt regarding the above, about the interpretation of the question . you have looked at the question as X raised to the power 2 multiplied into Y raised to the power 2 . But the question is (What is (x^2y^2)^1/2 if x < 0 and y > 0 ? ) so shouldnt we consider it as X raised to the power 2ysquare ?
In actual GMAT, the formatting will be very clear so you will never have this problem. Also, I think that if it were meant to be "X raised to the power 2ysquare", it would be written as x^{2y^2} if we were to write it without proper formatting. Anyway, as I said, this problem will not arise once it's properly formatted as will be the case in actual GMAT.
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ?
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16 Apr 2016, 10:17
nehamodak wrote: What is \(\sqrt{x^2*y^2}\) if x < 0 and y > 0?
(A) –xy (B) xy (C) –xy (D) yx (E) No solution \(\sqrt{x^2*y^2}\) is simply \(\sqrt{(xy)^2}\) i.e. xy. product of x and y is negative when x and y has opposite sign. it is already given. hence xy = xy Choice A is correct.
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Re: What is (x^2y^2)^1/2 if x < 0 and y > 0 ?
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17 Apr 2016, 04:47
Plugging in with easy numbers can help X=2 Y=4 Sqrt((2^2)*(4^2)) = Sqrt(4*16) = Sqrt(64) = 8 8 is 2*4. To get 2 instead of 2, x needs to be noted as negative, so that it becomes (2)*4. Therefore it is x*y.
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What is (x^2y^2)^1/2 if x < 0 and y > 0 ?
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Updated on: 26 Jun 2016, 01:28
x * y = x * y (since y>0) = (x)*y = xy (since x<0)
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Originally posted by rishi02 on 25 Jun 2016, 10:40.
Last edited by rishi02 on 26 Jun 2016, 01:28, edited 1 time in total.




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