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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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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. $$(\sqrt{x^3y^5})x^2y^5$$

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

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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

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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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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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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.

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

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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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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.

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

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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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20 May 2017, 22:38
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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21 May 2017, 01:44
sayiurway wrote:
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? &nbs [#permalink] 21 May 2017, 01:44
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