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how is that sqrt ( (x-a)^2 ) = |x-a| Thanks

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how is that sqrt ( (x-a)^2 ) = |x-a| Thanks [#permalink]

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New post 28 Oct 2008, 14:32
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

how is that

sqrt ( (x-a)^2 ) = |x-a|

Thanks
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kris

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Re: basic question regarding modulus [#permalink]

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New post 28 Oct 2008, 16:58
is so because, sqrt is basically the same as saying (....)^1/2, so if you write the question again,
((x-a)^2)^1/2, thus the 2's cancel out and you get | x-a |, its absolute because sqrt can never be negative.

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Re: basic question regarding modulus [#permalink]

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New post 28 Oct 2008, 21:56
krishan wrote:
how is that
sqrt ((x-a)^2) = |x-a|
Thanks


This is basic but very important. the concept is that any square has 2 roots. for ex: if x^2 = 4
x could be 2 and -2. it can also be written as x = sqrt(4) or -sqrt(4). if x = sqrt(4), x = 2. if x = -sqrt(4), x = -2.

so if we apply the same rule in the above expression, we also get 2 roots for (x-a)^2: (1) sqrt {(x-a)^2} and (2) -[sqrt {(x-a)^2}]. so

(1) sqrt {(x-a)^2} = lx-al
(2) - [sqrt {(x-a)^2}] = - lx-al
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Re: basic question regarding modulus [#permalink]

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New post 29 Oct 2008, 00:56
krishan wrote:
how is that

sqrt ( (x-a)^2 ) = |x-a|

Thanks


sqrt((x-a)^2) can have two values, (x-a) or -(x-a) and hence |x-a|.

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Re: basic question regarding modulus [#permalink]

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New post 29 Oct 2008, 11:29
thanks for the clarification .
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Re: basic question regarding modulus   [#permalink] 29 Oct 2008, 11:29
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