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Is root ((x-5)^2)=5-x?

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Senior Manager
Joined: 19 Oct 2004
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Location: Missouri, USA

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10 Nov 2004, 12:52
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Is root ((x-5)^2)=5-x?

1)-x|x|>0
2)5-x>0

can we please have some explanation here???
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Director
Joined: 16 Jun 2004
Posts: 891

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10 Nov 2004, 13:01
Iam unsure if I am reading this right..

Is root ((x-5)^2)=5-x?
=> x-5 = 5-x
or x=5

question is, Is x=5?

1. doesnt say anything much..
2. says x<5. so x is not equal to 5. suff.

B it is. I am unsure if I am looking at it very naively..
Director
Joined: 07 Nov 2004
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Re: DS " a real terror" [#permalink]

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10 Nov 2004, 13:31
ruhi160184 wrote:
Is root ((x-5)^2)=5-x?

1)-x|x|>0

Ruhi, I don't understand statement 1. How can a -ve multiplied by a poitive be greater than zero?
Senior Manager
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10 Nov 2004, 13:38
YEs thats the way the statement is... even i cant understand the head or tail of this one....
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Joined: 19 May 2004
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10 Nov 2004, 13:56
D. Nasty one

Question is: Is |x-5| = 5-x ?
Or Is |x-5| = - (x-5) ?

And the answer is: Yes, if (x-5)<0,
So the question is now: is x<5 ?

Statement 1: Sufficient. x is negative so it's surely <5.
Statement 2: x<5.

D.
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Joined: 15 Dec 2003
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10 Nov 2004, 17:21
yes, D it is. root of a square is equal to the abs value of the term squared. So if you have: Sqrt[(a-b)^2], this is equivalent to la-bl
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Paul

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Joined: 31 Aug 2004
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10 Nov 2004, 17:24
Agree with Dookie and Paul...

This one could have been a gmatclub challenge question...
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Joined: 07 Nov 2004
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10 Nov 2004, 21:02
Paul/ Dookie can you please explain statement 1...
Intern
Joined: 08 Nov 2004
Posts: 46
Location: Montreal

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10 Nov 2004, 22:41
Statement 1:

-x|x|>0

Can be divided into 2 inequalities:

1. If x>0:

-x (x) > 0
-x^2>0 (Impossible except if we consider imaginary #'s)

2. If x<0:

-x (-x) > 0
x^2 > 0 (True)

So, we can conclude that x is -ve
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11 Nov 2004, 01:11
gayathri,

You know that |x| is positive.
Only thing left to make sure the equation is > 0
is to know that x<0. This way (-x) turns positive.
And finally, if x<0 it is surely <5.
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11 Nov 2004, 07:45
Thanks Dookie & Complex Vision, got it now.
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12 Nov 2004, 00:44
thanks dookie and complex vision. real nice explanations. OA is infact D.

*happy that the terror is over*
CIO
Joined: 09 Mar 2003
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12 Nov 2004, 09:05
D. let me be very clear about something. This is a terrible problem, but very doable. It comes from the OG, and it's actually one I use in my lessons.

The reason is this. You need to know the GMAT VERY well to do this one right, and quickly. And there's exactly one great gmat trick imbedded here which is a red flag: x-5 and 5-x are in the same problem.

Whenever you see the same numbers twice in a problem, reversed around the minus sign, don't freak out. Just remember, this is only a positive/negative question.

Why? Take a look at this example: 10-7 = 3, 7-10=-3. Do you see it? The answer of these two subtraction problems is the same, but one is negative. That will always be true. So it can be said that |10-7|=|7-10|

That's huge. Knowing this means we can forget any algebra crap we are now freaking out about and approach the problem conceptually.

How can the sqaure root of a number squared be equal to its negative? Well, only if the number itself is negative to begin with. Look at -5. If we square it, it's 25, and then if we square root that, we're down to 5! Bingo. If we started with 5, and then squared it, we'd get 25, and then square root, we're back to the same positive 5. So it must be that the number under the radical is negative and the number on the other side of the equal is positive!

Now we're ready to make the final call. If all that is true, then x-5 must be negative, or, more specifically, x must be less than 5.

THAT'S ALL WE NEED TO FIGURE OUT!

Now, to the statements.

1)-x|x|>0
How can this be true? If the product here is positive, then they ultimately must both be positive. The |x| is definately positive, so the -x must also be. That means x must be negative. If negative, then less than 0, so less than 5. ie, enough info.

2)5-x>0
Could it get any better? If 5-x is positive, then x-5 is negative, and this whole thing works out! ie, enough info.

D

ps - i've talked about this in more detail in some of my other posts, if you're interested...
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12 Nov 2004, 18:11
Thanks for your detailed explanation ian, I learned a lot from your teaching! Keep it up .
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Paul

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Joined: 19 Oct 2004
Posts: 315
Location: Missouri, USA

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13 Nov 2004, 10:16
Anonymous wrote:
thanks dookie and complex vision. real nice explanations. OA is infact D.

*happy that the terror is over*

That was me forgot to logiin again while posting. Thanks everyone for their explanations. Really got this problem now.
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13 Nov 2004, 10:16
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