A quick glance through all the option, we will see that

option D which is \(u + 3v > 0\) is not possible, so might be the answer, however lets look at the remaing option

option A :: \(\frac{u}{3} < -v\) - This is possible if we divide by 3 on both sides

Option B:: \(\frac{u}{v} > -3\) - This is possible, since -3v > 0 so v < 0

Option C:: \(\sqrt{\frac{u}{-v}} < \sqrt{3}\) - It is possible by dividing \(\sqrt{-v}\) on both sides

Option E:: \(u < -3v\) - It is also possible by squaring on both sides.

** When the GRE writes a root sign, they are indicating a nonnegative root only**
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here, u>0, and -3v>0
So, u+(-3v)>0
U-3v >0
But, option D is not as above . So, D.

If \(\sqrt{u} < \sqrt{-3v}\), then it must also be true that u < -3v

Scan the answer choices....

Hmmm, D says that u + 3v > 0
However, if we take u < -3v and add 3v to both sides, we get: u + 3v < 0, which is the OPPOSITE of what answer choice D is saying.

Answer: D

Cheers,
Brent
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** When the GRE writes a root sign, they are indicating a nonnegative root only**


Can someone explain what is the exact meaning of this with an example?
Thanks in advance!

Every positive number has two square roots.
For example, the square root of 25 is 5, since 5² = 25 AND -5, since (-5)² = 25

However, the square root NOTATION tells us to indicate the non-negative root only.

So, for example:
\(\sqrt{25}=5\)
\(\sqrt{49}=7\)
\(\sqrt{0.25}=0.5\)
\(\sqrt{0}=0\)
etc

Does that help?

Cheers,
Brent
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