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# Is sqrt[(x-5)^2] = 5-x?

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Director
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05 May 2005, 17:37
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This question is from OG.

Is sqrt[(x-5)^2] = 5-x?

i) -x|x|>0
ii) 5-x>0

Please correct my approach if it is wrong. I am having some confusion with this problem.
From 1)

since |x| is always positive -x|x| > 0 only if x<0

==>> sqrt[(x-5)^2] = 5-x means (x-5)<0

from 2) 5-x>0 means (x-5)<0

So we can get from 1 and 2 .

Is this approach correct??
VP
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05 May 2005, 20:56
let me do this way. first simplify the question. remember this is yes/no question.

given that sqrt[(x-5)^2] = 5-x is equivalant to x=5? because sqrt [(x-5)^2] = x-5 (because sqrt of something is always positive. so it is x-5).
so the enequality is x-5 = 5-x......after simplification, x=5.

i) -x|x|>0
from this, x is -ve. so x is definitely not 5. sufficient...............

from ii, 5>x. this also means x is not equal to 5. it is also sufficient............
Senior Manager
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06 May 2005, 00:13
MA, your approach is easy fine.

I got the same answer (D) by picking up numbers.

1) From statement 1, we get x to be negative. I tried substituting the original formulas, for various numbers, and the base statement was true. Hence statement 1 is sufficient.

2) From statement 2, we get X < 5. I substituted x to be a positive number less than 5, and a negative number, and the statement still holds good.

But having looked at your post, felt there was a better way of doing it. Thanks.
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06 May 2005, 00:29
MA wrote:
because sqrt [(x-5)^2] = x-5 (because sqrt of something is always positive. so it is x-5).
..

I didn't understand what u mean by that

Sqrt (9) = +- 3 rt?

Kindly explain

Saurabh Malpani
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06 May 2005, 03:05
sqrt[(x-5)^2] = 5-x

what if we rewrite it as (x-5)^2=(5-x)^2, and then it would mean (x-5)^2=(-(x-5))^2, right? (x-5)^2=(x-5)^2 thus, it might be concluded that x can be any number, which makes the answer D since it does not really matter what they put in st1 or st2.
Director
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06 May 2005, 03:11
Thanks guys ....The OA is D
VP
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06 May 2005, 06:18
saurabhmalpani wrote:
MA wrote:
because sqrt [(x-5)^2] = x-5 (because sqrt of something is always positive. so it is x-5).
..

I didn't understand what u mean by that
Sqrt (9) = +- 3 rt?
Kindly explain. Saurabh Malpani

saurabh,
Sqrt (9) = always and only +3 not -3.
Director
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06 May 2005, 08:37
MA wrote:
saurabhmalpani wrote:
MA wrote:
because sqrt [(x-5)^2] = x-5 (because sqrt of something is always positive. so it is x-5).
..

I didn't understand what u mean by that
Sqrt (9) = +- 3 rt?
Kindly explain. Saurabh Malpani

saurabh,
Sqrt (9) = always and only +3 not -3.

Can you figure out if there are any flaws in my reasoning? That would be of great help to me.....Also suggestions on where to find more "Inequalities" problems....
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06 May 2005, 14:13
D for me

the key here is that we need to make sure X is less than 5!

if that is the case then we can solve the problem...

I and II both say X is less than 5, therefore sufficient!
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06 May 2005, 16:30
D for me too !

From the question LHS is always = + ( x-5)

A. -x|x| > 0

=> -x>0 because |x| is always +ve
=> 5-x>5 or 5-x>0

LHS is not = RHS .........A is suff

B.

LHS = +(x-5)
5-x > 0 implies (x-5)<0
LHS not = RHS
B is sufficient.

ans is D
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ash
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I'm crossing the bridge.........

VP
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06 May 2005, 16:48
gmat2me2 wrote:
This question is from OG.
Is sqrt[(x-5)^2] = 5-x?
i) -x|x|>0
ii) 5-x>0

Please correct my approach if it is wrong. I am having some confusion with this problem.
From 1) since |x| is always positive -x|x| > 0 only if x<0
==>> sqrt[(x-5)^2] = 5-x means (x-5)<0
from 2) 5-x>0 means (x-5)<0
So we can get from 1 and 2 . Is this approach correct??

i think, there is nothing wrong with your approach. you only did not simplified the given inequality in the question. if the given inequality is complex and multiple variables, it better to simplify to the possible extent. if you simplifiy the question, you can get the answer.
Director
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06 May 2005, 18:21
MA wrote:
gmat2me2 wrote:
This question is from OG.
Is sqrt[(x-5)^2] = 5-x?
i) -x|x|>0
ii) 5-x>0

Please correct my approach if it is wrong. I am having some confusion with this problem.
From 1) since |x| is always positive -x|x| > 0 only if x<0
==>> sqrt[(x-5)^2] = 5-x means (x-5)<0
from 2) 5-x>0 means (x-5)<0
So we can get from 1 and 2 . Is this approach correct??

i think, there is nothing wrong with your approach. you only did not simplified the given inequality in the question. if the given inequality is complex and multiple variables, it better to simplify to the possible extent. if you simplifiy the question, you can get the answer.

Thanks a bunch MA for your review
Re: DS: Inequalities   [#permalink] 06 May 2005, 18:21
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