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

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

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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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.

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

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.

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

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 \((\sqrt[3]{x})\) 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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