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sqrt (9 + x^2 -6x) =sqrt (x^2 - 6x + 9) =sqrt =sqrt = |x-3|

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sqrt (9 + x^2 -6x) =sqrt (x^2 - 6x + 9) =sqrt =sqrt = |x-3| [#permalink] New post 13 Nov 2007, 06:17
=sqrt (9 + x^2 -6x)
=sqrt (x^2 - 6x + 9)
=sqrt [(X-3)(x-3)]
=sqrt [(X-3)^2]
= |x-3|

why is the correct answer |3-x|? i've seen this somewhere before but i am not quite understanding the logic/concept of square root of squares.
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 [#permalink] New post 13 Nov 2007, 07:07
There are, at least, 2 ways to demonstrate it :)

Way 1
sqrt (9 + x^2 -6x)
= sqrt(3^2 -6*x + x^2)
= sqrt( (3-x)^2 )
= |3-x| ... by one definition, the conventional one, is sqrt( x^2 ) = |x|

Way 2
sqrt (9 + x^2 -6x)
= sqrt(3^2 -6*x + x^2)
= sqrt( (x-3)^2 )
= |x-3| .... again by definition
= |(-1)*(3-x)|
= |(-1)| * |3-x|
= |1| * |3-x|
= 1*|3-x|
= |3-x|

:)
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 [#permalink] New post 13 Nov 2007, 10:01
Fig wrote:
There are, at least, 2 ways to demonstrate it :)

Way 1
sqrt (9 + x^2 -6x)
= sqrt(3^2 -6*x + x^2)
= sqrt( (3-x)^2 )
= |3-x| ... by one definition, the conventional one, is sqrt( x^2 ) = |x|

Way 2
sqrt (9 + x^2 -6x)
= sqrt(3^2 -6*x + x^2)
= sqrt( (x-3)^2 )
= |x-3| .... again by definition
= |(-1)*(3-x)|
= |(-1)| * |3-x|
= |1| * |3-x|
= 1*|3-x|
= |3-x|

:)


how did u go from |x-3| to |(-1)*(3-x)|
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 [#permalink] New post 13 Nov 2007, 15:37
bmwhype2 wrote:
Fig wrote:
There are, at least, 2 ways to demonstrate it :)

Way 1
sqrt (9 + x^2 -6x)
= sqrt(3^2 -6*x + x^2)
= sqrt( (3-x)^2 )
= |3-x| ... by one definition, the conventional one, is sqrt( x^2 ) = |x|

Way 2
sqrt (9 + x^2 -6x)
= sqrt(3^2 -6*x + x^2)
= sqrt( (x-3)^2 )
= |x-3| .... again by definition
= |(-1)*(3-x)|
= |(-1)| * |3-x|
= |1| * |3-x|
= 1*|3-x|
= |3-x|

:)


how did u go from |x-3| to |(-1)*(3-x)|


I have factorized by -1... :)

So : x-3 = (-1)*(3-x) :)
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 [#permalink] New post 13 Nov 2007, 15:49
Fig, u r starrr in da club.

:)
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 [#permalink] New post 13 Nov 2007, 15:54
Ravshonbek wrote:
Fig, u r starrr in da club.

:)


Thanks for your kind words :).... I would take the term "star" more as bringing sometimes some ligths on some abs or so ;)
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 [#permalink] New post 13 Nov 2007, 18:20
you know what .... why is |x-3| wrong ?

The way bmwhype posted in the original post seems find to me ...
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 [#permalink] New post 14 Nov 2007, 08:15
okay i figured it out. asked a few engineer friends and found the answer.

|x-3| is the distance between 3 and x.

|x-3|*(-1) ==> |-x + 3|
OR if you want to move the numbers around in the bracket
|-x + 3| becomes |3-x|
essentially the same value.



this is what they said
===
|3-x| and |x-3|
have the same value

since in both terms the "3" and the "x" are of opposite sign,
positive/negative. The number is being changed in the same order of
direction, and because they are absolute values, will generate the
same positive value.

3-7= -4 7-3=4
===
well if it's absolute value it doesn't even matter if it's x-3 or 3-x
because after you absolute value it, it's the same answer.
  [#permalink] 14 Nov 2007, 08:15
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