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Absolute value and square root

  1. Feb 3, 2007 #1
    what is /squared root sign4x+1 /
    what does this equal because i'm confused when it has the square root sign on it all and it's absolute value
     
  2. jcsd
  3. Feb 3, 2007 #2

    cristo

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    Do you mean [itex]|\sqrt{4x+1}|[/itex]? Here, the absolute value sign just means take the positive square root.
     
  4. Feb 3, 2007 #3
    yup that's what i mean
     
  5. Feb 3, 2007 #4
    so when i a take the absolute value of it, it looks the same or the square root is gone
     
  6. Feb 3, 2007 #5

    cristo

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    The square root is still there, since that is the operation being applied to (4x+1). When we take the square root of a number, we get a positive number, and a negative number (consider the simple example: [itex]\sqrt{4}=\pm 2[/itex], since (-2)2=4=22). By putting the absolute value around the square root is the same as saying that we are taking the positive square root (so, in our example[itex]|\sqrt{4}|=+\sqrt{4}=2[/itex]).

    Your expression above can be written [itex]+\sqrt{4x+1}[/itex].
     
  7. Feb 3, 2007 #6
    what happens after i take the absolut value of square root sign4x+1
     
  8. Feb 3, 2007 #7
    so that's what it'll look like
     
  9. Feb 3, 2007 #8
    Actually, his expression can be written [itex]\sqrt{4x+1}[/itex]. Square roots are defined to be a FUNCTION, which means they CAN'T give you more than one result for any number in their domain (i.e. we can't have [itex]\sqrt{4}=\pm 2[/itex]). By convention, [itex]\sqrt{x} \ge 0[/itex] for all [itex]x \in [0,\infty)[/itex].
     
  10. Feb 3, 2007 #9

    cristo

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    Good point; thanks for spotting that, Moo!
     
  11. Feb 4, 2007 #10

    HallsofIvy

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    Lets be clear that in general |f(x)| can't be written as simply +f(x)! Here that was true because [itex]\sqrt{x}[/itex] is by definition non-negative. If f(x)= x and x= -4 then |f(x)|= |-4|= 4 while +f(x)= +(-4)= -4.
     
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