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Homework Help: Square root problem

  1. Feb 24, 2010 #1
    1. The problem statement, all variables and given/known data[/b]

    i have this thing:


    is this expression the same as:


    2. Relevant equations

    3. The attempt at a solution
    Last edited: Feb 24, 2010
  2. jcsd
  3. Feb 24, 2010 #2
    it means [tex] 2u^{3/2}m^{1/2} [/tex] Where is m in your expression.
  4. Feb 24, 2010 #3
    great! i am just bad with latex :)
  5. Feb 24, 2010 #4
    so later on i get this

    [tex]\lambda[/tex] = [tex]3u^{1/2}m^{1/2}[/tex]

    and [tex]\lambda[/tex] = [tex]u^{3/2}/m^{1/2}[/tex]

    i am supposed to solve this as an equation:

    [tex]3u^{1/2}m^{1/2}[/tex] = [tex]u^{3/2}/m^{1/2}[/tex]

    but how?
    Last edited: Feb 24, 2010
  6. Feb 24, 2010 #5
    [tex] 3u^{1/2}.m^{1/2} [/tex] = [tex]\frac{u^{3/2}}{m^{1/2}} [/tex]

    that is equal to u = 3m
  7. Feb 24, 2010 #6
    how did you do that? did you cross multiply or something?
  8. Feb 24, 2010 #7
    yes, I've done cross multiplication. I'm giving you some more detail.

    take [tex]m^{1/2}[/tex] from denominator on right hand side to numerator at left hand side and similarly take [tex]3u^{1/2}[/tex] from numerator at right hand side to denominator at right hand side. Now you have to know that whenever me multiply powers add up and whenever we divide power subtract up. Do you have any doubt now.
  9. Feb 24, 2010 #8
    i am stil having some issues with this.

    so is it correct that i would end up with:

    [tex]m^{1/2}*3u^{1/2}m^{1/2}[/tex] = [tex]u^{3/2}/m^{1/2}*3u^{1/2}[/tex]
  10. Feb 24, 2010 #9
    no, you have moved [tex] m^{1/2}[/tex] from right side to left hand side and its correct but now you have to delete it from the right hand side, and similar with the other one.
  11. Feb 24, 2010 #10
    in computer language you don't have to copy and paste actually you have to cut and paste.
  12. Feb 24, 2010 #11
    okey so then i end up with:

    [tex]m^{1/2}m^{1/2}[/tex] = [tex]u^{3/2}3u^{1/2}[/tex]

    and so:

    [tex]m[/tex] = [tex]3u^{4/2}[/tex]

    am i right?
  13. Feb 24, 2010 #12
    no, some other members may help you. I'm tiered using latex. If you still confused, write your all the problem, if you can scan then scan and post it to www.monbattle.com [Broken]. I'll solve each and every problem for you in a detailed manner, scan it and post it there so that you can copy and understand.

    last thing I should add for your problem is that when you cross multiply, take digits from top of one side to bottom of other side and bottom on one side to top of other side.
    Last edited by a moderator: May 4, 2017
  14. Feb 24, 2010 #13
    okey, thank you very much for your help :)
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