x-3

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x-3

by pathaniaus » Mon Aug 17, 2009 11:10 am
sqrt((x-3)^2) = 3 - x

1) x NOT EQUAL TO 3

2) -x|x| > 0


Please explain. I've gotten a headache just trying to figure this thing out.

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by capnx » Mon Aug 17, 2009 1:15 pm
simplify the expression and you'll find X can take on 3 values: -3, 0, and 3

(x-3)^2 = (3-x)^2
so: x-3 = +/- (3-x) and +/-(x-3) = 3-x
either way plugging in +/-3 and 0 is easiest way lol

I is not enough, can be -3 or 0
II: -x|x| > 0 --> x|x| < 0 , so x has to be negative

B

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Re: x-3

by chetanojha » Mon Aug 17, 2009 1:21 pm
pathaniaus wrote:sqrt((x-3)^2) = 3 - x

1) x NOT EQUAL TO 3

2) -x|x| > 0


Please explain. I've gotten a headache just trying to figure this thing out.
Can you please post the complete question clearly?

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Re: x-3

by tohellandback » Tue Aug 18, 2009 7:57 pm
pathaniaus wrote:sqrt((x-3)^2) = 3 - x

1) x NOT EQUAL TO 3

2) -x|x| > 0


Please explain. I've gotten a headache just trying to figure this thing out.
sqrt(x^2)=|x|
sqrt((x-3)^2)=|x-3|
we are asked whether x-3<0 or x<3
1) NOT SUFF
2)-x|x| > 0 means x is negative
so of course x<3
SUFFICIENT
answer B
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Re: x-3

by adilka » Wed Aug 19, 2009 7:39 am
tohellandback wrote:sqrt(x^2)=|x|
is that so... I thought that only when X^2=Y, then x=|sqrt(Y)|
But whenever you're taking a square root of a number it can only be positive, i.e. SQRT(X) can only be a positive number.
What's GMAT's stance?

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Re: x-3

by tohellandback » Wed Aug 19, 2009 8:03 am
adilka wrote:
tohellandback wrote:sqrt(x^2)=|x|
is that so... I thought that only when X^2=Y, then x=|sqrt(Y)|
But whenever you're taking a square root of a number it can only be positive, i.e. SQRT(X) can only be a positive number.
What's GMAT's stance?
if you say x^2=y, then x can be +sqrt(y) or -sqrt(y).
but if it is said that sqrt(x^2), then it is given as |x|.
it's just a rule or convention. whatever you say. I am yet to see a "proof" for it. I have seen Ian post the same thing. If you do a little search, you will find it.
But it is a rule and just remember it.

not a very reliable source, but still:
https://en.wikipedia.org/wiki/Square_root
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