Absolute X

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Absolute X

by LSB » Tue Aug 19, 2008 3:12 pm
If X < 0, then sqrt(-x * absolute(x) ) is:

A -x
B -1
C 1
D x
E sqrt(X)
Source: — Problem Solving |

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by madeline » Tue Aug 19, 2008 7:50 pm
I wish for a whiteboard!

Here goes:

|x| > 0 no matter what x is
x < 0, therefore |x| = -x > 0
x < 0, therefore -x > 0
-x * |x| = (-x) * (-x) = x^2 > 0
sqrt( -x * |x| ) = sqrt( x^2 )

Note that in the GMAT, the square root of anything is always positive - this is the key to this question.

sqrt( x^2 ) = -x > 0

Answer = A

The quicker way I think about this (may be confusing to some):
The thing inside the sqrt is not going to be negative, since "undefined" is not an answer choice. Therefore, the two x's inside the sqrt will yield a positive number, x^2. (No need to work out the details of absolute value or negatives.) Therefore, square root of x^2 will be "x-like", ie, either x or -x. Since x<0, but sqrt(anything)>0, the answer is -x, which is greater than 0.

Hint: it's important to master this kinda question :wink:

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by LSB » Wed Aug 20, 2008 3:21 am
Thanks madeline.

I did the same thing and arrived at the last line.

sqrt( -x * |x| ) = sqrt( x^2 )

Then I selected D. How did you differentiate between D & A?

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by pepeprepa » Wed Aug 20, 2008 3:42 am
sqrt( -x * |x| ) = sqrt( x^2 )

To clarify what you can write is
sqrt( -x * |x| ) = sqrt( (absolute value of x)^2 )=absolute value of x=-x
given that x<0
What you can tell you is that x is negative, if the square gives you x there is a problem givent that square can only give you positive number.

Need to take care with this kind of question, can fastly make an error.

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by LSB » Wed Aug 20, 2008 4:04 am
I get it now. Man ... it's silly stuff that trips you up on the GMAT.

Thanks guys.