If xy0, which of the following must be positive?

Question

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

A.  (\sqrt{xy})^2x^3y^2 
 
B.  (\sqrt {x})(\sqrt {y})x^3y^2 
 
C.  (\sqrt{xy})x^7y^9 
 
D.  (\sqrt {x^2y})x^2y^4

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

shashankism · Answer

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.

pushpitkc · Answer

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)

JeffTargetTestPrep · Answer

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

broall · Answer

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

praveenmittal95605 · Answer

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.

sayiurway · Answer

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?

Bunuel · Answer

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

Hope it helps.

SUV0508 · Answer

in C doesnt sq.root of a number give a positive and a negative number?

Noida · Answer

Have a basic doubt here:

Option C gives us  x^{15/2}y^{19/2}

Option E gives us  x^{7/2}y^{15/2}

Here, both the powers of x and y are odd.

How to deduce the answer from this point?

Please guide  Bunuel .

Bunuel · Answer

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

A.  (\sqrt{xy})^2x^3y^2 
 
B.  (\sqrt {x})(\sqrt {y})x^3y^2 
 
C.  (\sqrt{xy})x^7y^9 
 
D.  (\sqrt {x^2y})x^2y^4

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

Here is a trick how to solve such questions easily. Simplify!!!

1. xy > 0 means that  x and y have the same sign  and  none of them is 0 .

2. Since the square root sign for an even root, such as a square root, fourth root, cannot given negative result then we can  get rid of all the even roots  from the options.

3.  Get rid of all the unknowns in even powers  - they will be positive for sure.

4.  Get rid of all odd powers  (not the unknowns themselves just their odd powers!). For example, rewrite x^7 as x (divide by x^6), this will not change the sign.

5.  Get rid of all odd roots  (not the unknowns themselves just their odd roots!). For example, rewrite  (\sqrt {x})  as x (cube it), this will not change the sign.

For example, in E:

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

First of all get rid of the square root, next get rid x^2 and finally rewrite y^5 as y and you'll get y alone!

We'll get:

A.  x . This can be negative as well as positive.
 
B.  x*y*x = x^2y . here we can further drop x^2 and we'd get y. And y be negative as well as positive.
 
C.  xy . This is given to be positive.
 
D.  y . This can be negative as well as positive.

E.  y . This can be negative as well as positive.

Answer: C.

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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?

Prep Toolkit

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