Question: Ordering Powers of -0.5

Comment on Ordering Powers of -0.5

Isnt root of a negative number an imaginary number? (which is neither negative nor positive)
greenlight-admin's picture

The SQUARE root of a negative number an imaginary number. However, this question features a CUBE root, and we CAN find cube roots of negative numbers.
For example, the cube root of -8 equals -2, since (-2)^3 = -8

Quick question, not just a sq rt but any even root of a negative number is imaginary right?
greenlight-admin's picture

That's correct.
If n is an EVEN integer, then the nth root of a negative number will be an IMAGINARY (aka COMPLEX) number.

ASIDE: For students who are wondering what this is all about, please note that imaginary (complex) numbers are not tested on the GRE.

Hi Brent,
Please if you can check for the ques

OA is A , but can I try this way

√{x²y²}= √{(-xy)²}= -xy = negative value i.e QTY A = QTY B

Let x = -5 and y = 4
Then √{(-5)²(2)²}= √{(-5*2)²} = -10
However, √{x² y²} = positive value as the sq root of positive numbers will always be positive i.e QTY A > QTY B

greenlight-admin's picture

Question link:

I'm not sure what you're doing when you write the following:
√{x²y²} = √{(-xy)²} = -xy = negative value

How did xy become -xy?
Is it because we're told x < 0?
If so, then -x represents a POSITIVE value.
For example, if x = -2, then -x = -(-2) = 2

So, that's one issue with your solution.

There's also a problem when you write: √{(-5)²(2)²}= √{(-5*2)²} = -10
The √ notation tell us to take the POSITIVE square root.
For example √25 = 5 and √49 = 7

So, in your example: √{(-5*2)²} = √{(-10)²} = √100 = 10 (not -10)

Here's my full solution:


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