1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simplifying Radicals with varibles?

  1. Nov 27, 2008 #1
    I understand that the square root of (for example)j^16 is j^8. But when you have an odd squared root, like the square root of j^19, would it be j^9square root of j?

    Another quick question 12 radical 36 would be 72?

    Sorry, I know this is basic algebra, but I really have forgotten how to do it...
     
  2. jcsd
  3. Nov 28, 2008 #2

    Mentallic

    User Avatar
    Homework Helper

    Delete - Multipost.
     
  4. Nov 28, 2008 #3

    Mentallic

    User Avatar
    Homework Helper

    Yes that is exactly what it would be.

    [tex]\sqrt{x}=x^{\frac{1}{2}}[/tex]
    and
    [tex]\sqrt{x^a}=x^{\frac{a}{2}[/tex]

    So for your question [tex]\sqrt{j^{16}}=j^{\frac{16}{2}}=j^8[/tex]

    Thus, [tex]\sqrt{j^19}=j^{\frac{19}{2}}=j^{9\frac{1}{2}}=j^9\sqrt{j}[/tex]


    For the second one, [tex]12\sqrt{36}=72[/tex] because [tex]\sqrt{36}=6[/tex] therefore [tex]12x6=72[/tex]

    If theres something you still don't understand, just ask :smile:
     
  5. Nov 28, 2008 #4
    I would add a plus or minus to every result :tongue:
     
  6. Nov 28, 2008 #5

    Mentallic

    User Avatar
    Homework Helper

    Well if you want to be wrong then go ahead.
     
    Last edited: Nov 28, 2008
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Simplifying Radicals with varibles?
  1. Simplifying radicals (Replies: 8)

Loading...