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If xy>0, which of the following must be positive?

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If xy>0, which of the following must be positive? [#permalink]

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New post 11 May 2017, 03:03
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If xy>0, which of the following must be positive?


A. \((\sqrt{xy})^2x^3y^2\)

B. \((\sqrt[3]{x})(\sqrt[5]{y})x^3y^2\)

C. \((\sqrt{xy})x^7y^9\)

D. \((\sqrt[3]{x^2y})x^2y^4\)

E. \((\sqrt{x^3y^5})x^2y^5\)
Spoiler: OA

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Re: If xy>0, which of the following must be positive? [#permalink]

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New post Updated on: 11 May 2017, 05:04
Bunuel wrote:
If xy>0, which of the following must be positive?


A. \((\sqrt{xy})^2x^3y^2\)

B. \((\sqrt[3]{x})(\sqrt[5]{y})x^3y^2\)

C. \((\sqrt{xy})x^7y^9\)

D. \((\sqrt[3]{x2y})x^2y^4\)

E. \((\sqrt{x3y5})x^2y^5\)


xy > 0
So either x>0, y>0 OR x<0, y<0
A. \((\sqrt{xy})^2x^3y^2\) = \(x^4 * y^3\) so -ve value is possible if y<0, x<0

B. \((\sqrt[3]{x})(\sqrt[5]{y})x^3y^2\) = \((\sqrt[3]{x})(\sqrt[5]{y})(xy)^2x\)
For this option if y<0, x<0, \((\sqrt[3]{x})\) is +ve , \((\sqrt[5]{y})\) is -ve \(x^2y^2\) is -ve & x is -ve. So overall value is -ve.

C. \((\sqrt{xy})x^7y^9\) = \((\sqrt{xy})(xy)^7y^2\).. All the terms are +ve. So overall +ve.

D. \((\sqrt[3]{x2y})x^2y^4\) = \((\sqrt[3]{xy*y})x^2y^4\) . So if y <0 & x<0, \((\sqrt[3]{xy*y})\) is -ve and hence overall -ve is possible.

E. \((\sqrt{x3y5})x^2y^5\) = \((\sqrt{(xy)^3*y^2})(xy)^2y^3\) (+ve) * (+ve) *(-ve) is y<0 & x<0


Spoiler: ::
Answer C

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Originally posted by shashankism on 11 May 2017, 04:23.
Last edited by shashankism on 11 May 2017, 05:04, edited 2 times in total.
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Re: If xy>0, which of the following must be positive? [#permalink]

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New post 11 May 2017, 04:43
There are 2 possibilities when xy > 0.
1. Both x and y are positive
2. Both x and y are negative

The only possibility in the answer choices which is definitely positive is Option C
Both the exponents of x and y are odd powers and since both of them are negative,
the expression will yield a positive answer(Option C)
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Re: If xy>0, which of the following must be positive? [#permalink]

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New post 15 May 2017, 17:26
Bunuel wrote:
If xy>0, which of the following must be positive?


A. \((\sqrt{xy})^2x^3y^2\)

B. \((\sqrt[3]{x})(\sqrt[5]{y})x^3y^2\)

C. \((\sqrt{xy})x^7y^9\)

D. \((\sqrt[3]{x2y})x^2y^4\)

E. \((\sqrt{x3y5})x^2y^5\)


We are given that xy > 0, which means either x and y are both positive or both negative.

Scanning our answer choices, we see that only answer choice C MUST BE positive.

Since we know that xy is greater than zero, we know that √xy must also be greater than zero.

So, we have:

(positive)(x^7)(y^9)

Since xy is greater than zero, we see that (x^7)(y^9) = (x^7)(y^7)(y^2) = (xy)^7(y^2) must also be greater than zero.

Thus, (√xy)(x^7)(y^9) is greater than zero.

Answer: C
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Re: If xy>0, which of the following must be positive? [#permalink]

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New post 16 May 2017, 13:03
Bunuel Can you please tell me how can D be negative ,

I am stuck between C and D.
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Re: If xy>0, which of the following must be positive? [#permalink]

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New post 16 May 2017, 20:38
Bunuel wrote:
If xy>0, which of the following must be positive?


\(xy > 0 \implies x > 0 \; \text{and}\; y > 0 \quad \text{or} \quad x < 0 \;\text{and}\; y< 0 \)

A. \((\sqrt{xy})^2x^3y^2\)
\((\sqrt{xy})^2 > 0 \quad \forall xy > 0\)
\(x^3y^2=(xy)^2 \times x\)

If \(x < 0 \implies x^3y^2 < 0 \implies (\sqrt{xy})^2x^3y^2 < 0\)
Eliminated.

B. \((\sqrt[3]{x})(\sqrt[5]{y})x^3y^2\)

\((\sqrt[3]{x})(\sqrt[5]{y})x^3y^2=(x)^{\frac{1}{3}}(y)^{\frac{1}{5}}x^3y^2=(x)^{\frac{10}{3}}(y)^{\frac{11}{5}} \)
\(=(x)^{\frac{10}{3}}(y)^{\frac{10}{5}} \times (y)^{\frac{1}{5}} \)
\(= \Big ( (x)^{\frac{1}{3}}(y)^{\frac{1}{5}} \Big ) ^{10} \times (y)^{\frac{1}{5}}\)

If \(y < 0 \implies (y)^{\frac{1}{5}} < 0 \implies \Big ( (x)^{\frac{1}{3}}(y)^{\frac{1}{5}} \Big ) ^{10} \times (y)^{\frac{1}{5}} < 0\)

Eliminated.

C. \((\sqrt{xy})x^7y^9\)
\((\sqrt{xy})x^7y^9 = (xy)^{\frac{1}{2}}(xy)^7y^2 \)
We have \(\sqrt{xy} > 0\)
\((xy)^7 > 0\) because \(xy>0\)
\(y^2 > 0\) because \(y \neq 0\)

Hence \((\sqrt{xy})x^7y^9 > 0\). This is the correct answer.

D. \((\sqrt[3]{x^2y})x^2y^4\)
\((\sqrt[3]{x^2y})x^2y^4=\sqrt[3]{x^2}\times\sqrt[3]{y}\times x^2y^4\)

If \(y < 0 \implies \sqrt[3]{y} < 0 \implies \sqrt[3]{x^2}\times\sqrt[3]{y}\times x^2y^4 < 0\).

Eliminated.

E. \((\sqrt{x^3y^5})x^2y^5\)
\(\sqrt{x^3y^5} \times x^2y^5 = \sqrt{(xy)^3 \times y^2}\times (xy)^2y^3 = \sqrt{(xy)^3} \times \sqrt{y^2} \times (xy)^2y^3 = \sqrt{(xy)^3} \times |y| \times (xy)^2y^3\)

Note that \(\sqrt{(xy)^3} > 0\) and \((xy)^2 > 0\) and \(|y| > 0\).

If \(y < 0 \implies y^3 < 0 \). Hence this choice is wrong.

The answer is C.
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Re: If xy>0, which of the following must be positive? [#permalink]

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New post 17 May 2017, 02:40
xy > 0
So either x>0, y>0 OR x<0, y<0
A. if y<0, x<0
(xy−−√)2x3y2(xy)2x3y2 = x4∗y3x4∗y3 so -ve value is possible

B. (x√3)(y√5)x3y2(x3)(y5)x3y2 = (x√3)(y√5)(xy)2x(x3)(y5)(xy)2x
For this option if y<0, x<0, (x√3)(x3) is +ve , (y√5)(y5) is -ve x2y2x2y2 is -ve & x is -ve. So overall value is -ve.

C. All the terms are +ve.

D. (x2y−−−√3)x2y4(x2y3)x2y4 = (xy∗y−−−−−√3)x2y4(xy∗y3)x2y4 . So if y <0 & x<0, (xy∗y−−−−−√3)(xy∗y3) is -ve and
hence -ve is possible.

E. (x3y5−−−−√)x2y5(x3y5)x2y5 = ((xy)3∗y2−−−−−−−−√)(xy)2y3((xy)3∗y2)(xy)2y3 (+ve) * (+ve) *(-ve) is y<0 & x<0 -ve is possible


Option C is correct.
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Re: If xy>0, which of the following must be positive? [#permalink]

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New post 20 May 2017, 22:38
I wrongly chose answer E.
My explanation is: squareroot(x^3*y^5)*x^2*y*5= x^3/2*y^5/2*x^2*y^5= x^7/2*y*15/2=squareroot(x^7*y^15)=squareroot(x^7*y^7*y^8).
Since xy>0 => (xy)^7>0 and y^8> 0
=> E has to be positive.
Can you pls explain?
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Re: If xy>0, which of the following must be positive? [#permalink]

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New post 21 May 2017, 01:44
sayiurway wrote:
I wrongly chose answer E.
My explanation is: squareroot(x^3*y^5)*x^2*y*5= x^3/2*y^5/2*x^2*y^5= x^7/2*y*15/2=squareroot(x^7*y^15)=squareroot(x^7*y^7*y^8).
Since xy>0 => (xy)^7>0 and y^8> 0
=> E has to be positive.
Can you pls explain?


Say x = y = -1, then E. \((\sqrt{x^3y^5})x^2y^5=-1=negative\).

Hope it helps.
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Re: If xy>0, which of the following must be positive?   [#permalink] 21 May 2017, 01:44
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