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Homework Help: Simplifying Absolute Values

  1. Oct 17, 2012 #1
    1. The problem statement, all variables and given/known data
    Simplify.

    a) [tex]\sqrt{x^6}[/tex]
    b) [tex]8 \sqrt{x^7y^{10}} - 10 \sqrt{x^7y^{10}}[/tex]

    For b, it's y^10. I can't make it look right for some reason.
    Mod note: Fixed the exponent.

    2. Relevant equations



    3. The attempt at a solution
    I can simplify all of them but I don't know when or where I need to put in absolute value symbols to the solution. I know the solution because my book shows me, but I don't understand why the absolute values are where they are.

    For instance, the solution to a is [tex]|x^3|[/tex] but I can only get to [tex]x^3[/tex] without becoming confused.
    For b, i can get to [tex]-2x^3y^5 \sqrt{x}[/tex] but the solution is [tex]-2x^3|y^5| \sqrt{x}[/tex]
    So how do you know when an absolute value is required?
     
    Last edited by a moderator: Oct 17, 2012
  2. jcsd
  3. Oct 17, 2012 #2

    SammyS

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    Use y^{10}
    [itex]\displaystyle \sqrt{u^2}=|u|[/itex]

    Also, remember that if n is a positive integer, then [itex]u^{2n}\ge0\,,[/itex] so there is no need to use absolute value.
     
  4. Oct 17, 2012 #3
    So in my scenario, x has to be positive because it's still underneath the square root so I don't need to put an absolute value around x^3, whereas y can be either positive or negative because it's not still beneath the square root so an absolute value is required?
     
  5. Oct 17, 2012 #4

    Mark44

    Staff: Mentor

    No. x can be any real number.

    Like SammyS said,
    $$ \sqrt{x^2} = |x|$$

    so
    $$ \sqrt{x^6} = \sqrt{(x^3)^2} = |x^3|$$

    This is also the same as |x|3.
     
  6. Oct 17, 2012 #5

    SammyS

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    Which scenario ?

    If x is to an even power, that result is non-negative. So that can be under a radical -- actually one signifying a square root -- no matter what value x has.
     
  7. Oct 17, 2012 #6
    I was referring to question b. Do you mean if x has an even power in the initial equation or in the solution?
     
  8. Oct 17, 2012 #7
    Oh wow, I think it makes sense now. Seeing the step between x^6 and |x^3| was really helpful. I wasn't writing that step down, I was simply skipping to the final step. Thanks a bunch.
     
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