700+ GMATPrep question

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700+ GMATPrep question

by aleph777 » Tue Jan 11, 2011 8:55 am
If x ≠ 0, then (√x^2)/x =

a. -1
b. 0
c. 1
d. x
e. |x|/x

OA: E

Can someone please explain this one to me? I solved for C, assuming that the square root and square cancel each other out, leaving x/x. And whether x was positive or negative, that would give us a positive result.
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by GMATGuruNY » Tue Jan 11, 2011 9:04 am
aleph777 wrote:If x ≠ 0, then (√x^2)/x =

a. -1
b. 0
c. 1
d. x
e. |x|/x

OA: E

Can someone please explain this one to me? I solved for C, assuming that the square root and square cancel each other out, leaving x/x. And whether x was positive or negative, that would give us a positive result.
Plug in multiple times until only 1 answer choice remains:

Let x=2.
(√x^2)/x = (√2^2)/2 = (√4)/2 = 2/2 = 1.
Eliminate A, B and D.

Let x = -2.
(√x^2)/x = (√(-2)^2)/(-2) = (√4)/(-2) = 2/(-2) = -1.
Eliminate C.

The correct answer is E.
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by aleph777 » Tue Jan 11, 2011 9:19 am
Thanks, Mitch.

I guess what still confuses me is that squared terms always disguise the sign, and since we know outright that x is a negative, I thought you need to resolve the sq rt of x sqrd accurately.

Am I just crossing my DS wires with honest square-roots resolve positively logic?

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by Geva@EconomistGMAT » Tue Jan 11, 2011 9:26 am
aleph777 wrote:Thanks, Mitch.

I guess what still confuses me is that squared terms always disguise the sign, and since we know outright that x is a negative, I thought you need to resolve the sq rt of x sqrd accurately.

Am I just crossing my DS wires with honest square-roots resolve positively logic?
You are right that x^2 will always be non-negative, regardless of whether x itself is positive or negative. If x is 2 or -2, the result of x^2 is a uniform 4.
But the expression the question is asking for is dependent on the x in the denominator as well. The numerator will always be positive, but the denominator can be positive or negative, affecting the value of the entire fraction.

Having said that, I would not really expect you (or any other GMAT student) to immediately recognize the underlying mathematical complexity above. Instead, get used to plugging numbers and eliminating trap answer choices as a general method to check that your mathematical understanding does not ignore specific end cases such as above. If you had taken Mitch's advice and plugged in numbers, instead of listening to your own inner algebraian, you would've chosen E without even understanding why, or, in fact, without needing to waste time understanding: all the other answer choices are eliminated, and one answer choice in 5 must be the correct one.
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by GMATGuruNY » Tue Jan 11, 2011 9:27 am
aleph777 wrote:Thanks, Mitch.

I guess what still confuses me is that squared terms always disguise the sign, and since we know outright that x is a negative, I thought you need to resolve the sq rt of x sqrd accurately.

Am I just crossing my DS wires with honest square-roots resolve positively logic?
√ means the positive root only:

√4 = 2.

However, if we're told that x^2 = 4, then x=2 or x= -2.

So in the problem above, while the numerator of the fraction must be positive, the denominator could be positive or negative. Hence, answer choice C can be eliminated.
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by Night reader » Tue Jan 11, 2011 9:27 am
aleph777 wrote:If x ≠ 0, then (√x^2)/x =

a. -1
b. 0
c. 1
d. x
e. |x|/x

OA: E

Can someone please explain this one to me? I solved for C, assuming that the square root and square cancel each other out, leaving x/x. And whether x was positive or negative, that would give us a positive result.
even number^Sqr(negative number) IS NOT defined for GMAT quantitative testing purpose. Therefore 2^Sqr(x) as well as 4^Sqr(x) where x<0, and so on ... are not defined. But we are given even number^Sqr(number^2) which means that regardless of even number^Sqr(x) rule we need to consider case when x is positive OR negative and squared is always positive to satisfy the rule.

So we want to rewrite this as |x|/x which means that x can be negative OR positive.

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by shashank.ism » Tue Jan 11, 2011 9:59 am
see √x will always be + ve
hence √(x^2) = |x|
now √(x^2) / x = |x|/x

you cant write |x|/x =1 as
|x| = x when x >0
and |x| = -x when x <0......because |x| is always +ve
[spoiler]so ans is E |x|/x.[/spoiler]
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