M15#37 : Retired Discussions [Locked]
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# M15#37

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Manager
Joined: 29 Sep 2008
Posts: 146
Followers: 3

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25 May 2012, 08:28
there is a question in test m15

a^2-b^2=b^2-c^2.Is a=|b|

1)b=|c|
2)b=|a|

from 1 we get a^2=b^2

a=root b^2

or a=|b|

a^2=b^2
so |a|=|b|

i have doubt regarding mod

|a|=root a^2

so above how can we write a^2=|a|
Math Expert
Joined: 02 Sep 2009
Posts: 37098
Followers: 7249

Kudos [?]: 96415 [2] , given: 10738

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25 May 2012, 11:45
2
KUDOS
Expert's post
mrinal2100 wrote:
there is a question in test m15

a^2-b^2=b^2-c^2.Is a=|b|

1)b=|c|
2)b=|a|

from 1 we get a^2=b^2

a=root b^2

or a=|b|

a^2=b^2
so |a|=|b|

i have doubt regarding mod

|a|=root a^2

so above how can we write a^2=|a|

I guess your main question is about the following issue is why is $$\sqrt{x^2}=|x|$$ true.

The point here is that as square root function can not give negative result then $$\sqrt{some \ expression}\geq{0}$$.

So $$\sqrt{x^2}\geq{0}$$. But what does $$\sqrt{x^2}$$ equal to?

Let's consider following examples:
If $$x=5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=x=positive$$;
If $$x=-5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=-x=positive$$.

So we got that:
$$\sqrt{x^2}=x$$, if $$x\geq{0}$$;
$$\sqrt{x^2}=-x$$, if $$x<0$$.

What function does exactly the same thing? The absolute value function: $$|x|=x$$, if $$x\geq{0}$$ and $$|x|=-x$$, if $$x<0$$. That is why $$\sqrt{x^2}=|x|$$.

Hope it helps.
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Senior Manager
Joined: 01 Nov 2010
Posts: 295
Location: India
Concentration: Technology, Marketing
GMAT Date: 08-27-2012
GPA: 3.8
WE: Marketing (Manufacturing)
Followers: 10

Kudos [?]: 76 [0], given: 44

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27 May 2012, 01:17
mrinal2100 wrote:
there is a question in test m15

a^2-b^2=b^2-c^2.Is a=|b|

1)b=|c|
2)b=|a|

from 1 we get a^2=b^2

a=root b^2

or a=|b|

a^2=b^2
so |a|=|b|

i have doubt regarding mod

|a|=root a^2

so above how can we write a^2=|a|

here,
a^2=b^2
==> |a|=|b|
because here a &b both are variable.

|a|=root a^2; in all cases, but when you are equating a variable to variable a^2=b^2,
==> |a|=|b|, you have to right it this way.
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Re: M15#37   [#permalink] 27 May 2012, 01:17
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# M15#37

Moderator: Bunuel

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