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Rationalizing the Numerator

  1. Sep 8, 2009 #1
    Alright, so I've go this question where I'm to rationalize the numerator. [Grade 10 Level]
    1. The problem statement, all variables and given/known data
    (1+sqrt(2)-sqrt(3))/(sqrt(2))
    http://www.myalgebra.com/math_image.aspx?p=SMB02FSMB031+SMB02RSMB032SMB02rSMB03-SMB02RSMB033SMB02rSMB03SMB10SMB02RSMB032SMB02rSMB03SMB02fSMB03?p=74?p=42 [Broken]


    2. Relevant equations



    3. The attempt at a solution
    I'm not going to type all of my entire failed attempts. Though, I did try seperating this into two fractions, but I don't think that'd be useful. I also multiplied the numerator and denominator by the numerator's conjugate, I think. Although I'm not sure:
    (1+sqrt(2)-sqrt(3))/(sqrt(2))*((sqrt(2))+(sqrt(3)))/((sqrt(2))+(sqrt(3)))
    http://www.myalgebra.com/math_image.aspx?p=SMB02FSMB031+SMB02RSMB032SMB02rSMB03-SMB02RSMB033SMB02rSMB03SMB10SMB02RSMB032SMB02rSMB03SMB02fSMB03*SMB02FSMB03(SMB02RSMB032SMB02rSMB03)+(SMB02RSMB033SMB02rSMB03)SMB10(SMB02RSMB032SMB02rSMB03)+(SMB02RSMB033SMB02rSMB03)SMB02fSMB03?p=164?p=42 [Broken]



    I think the positive one in the equation is what's causing me to lose.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 8, 2009 #2

    berkeman

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    Are you sure they don't want you to rationalize the denominator instead? Why would you want to rationalize the numerator of a fraction?
     
    Last edited by a moderator: May 4, 2017
  4. Sep 8, 2009 #3
    The teacher just told us to... It's homework.
     
  5. Sep 8, 2009 #4

    berkeman

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    Very weird, but fair enough. I think you had the right idea to separate it into multiple fractions. I'd split it up into all 3 fractions, rationalize the numerator of each, and look for any re-combinations (if there are any) that do not re-introduce radicals in the numerator(s). Not sure that's what they want, but it's all that I see at the moment.
     
  6. Sep 8, 2009 #5
    Oh, so I seperated it into three fractions and I think I just won.

    http://www.myalgebra.com/math_image.aspx?p=2+SMB02FSMB031-SMB02RSMB033SMB02rSMB03SMB10SMB02RSMB032SMB02rSMB03SMB02fSMB03?p=66?p=42 [Broken]

    Thanks :)
     
    Last edited by a moderator: May 4, 2017
  7. Sep 8, 2009 #6

    berkeman

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    But your answer still has a radical in a numerator... (and I'm not sure your answer is equal to the original problem either... have you tried putting both forumulas into your calculator to check that they are equal?)
     
    Last edited by a moderator: May 4, 2017
  8. Sep 8, 2009 #7
    Oh... You're right. I guess that's a problem then.
     
  9. Sep 8, 2009 #8

    arildno

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    Hint:

    Start with noting:
    [tex]\frac{1+\sqrt{2}-\sqrt{3}}{\sqrt{2}}=\frac{(1+(\sqrt{2}-\sqrt{3}))(1-(\sqrt{2}-\sqrt{3}))}{\sqrt{2}(1-(\sqrt{2}-\sqrt{3}))}[/tex]

    The numerator you'll end up with will be easy to rationalize
     
  10. Sep 8, 2009 #9
    Thanks, I noticed that your method would require more steps to complete than:
    [tex]\frac{1+\sqrt{2}-\sqrt{3}}{\sqrt{2}}=\frac{((1+\sqrt{2})-\sqrt{3})((1+\sqrt{2})+\sqrt{3})}{\sqrt{2}((1-\sqrt{2})-\sqrt{3})}[/tex]

    Anyways so I've solved it, thanks for the help. :)
     
  11. Sep 9, 2009 #10

    arildno

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    As a final note, we may simplify this very much:
    [tex]\frac{1+\sqrt{2}-\sqrt{3}}{\sqrt{2}}=\frac{(1+(\sqrt{2}-\sqrt{3}))(1-(\sqrt{2}-\sqrt{3}))}{\sqrt{2}(1-(\sqrt{2}-\sqrt{3}))}=\frac{2(\sqrt{6}-2)}{\sqrt{2}(1-(\sqrt{2}-\sqrt{3}))}=\frac{4}{\sqrt{2}(1-(\sqrt{2}-\sqrt{3}))(\sqrt{6}+2)}[/tex]

    By multiplying out the denumerator, you'll end up with:
    [tex]\frac{1+\sqrt{2}-\sqrt{3}}{\sqrt{2}}=\frac{2}{1+\sqrt{2}+\sqrt{3}}[/tex]
     
  12. Sep 9, 2009 #11

    arildno

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    This won't work!!

    Unless the -signs in the denumerator really should be +signs, in which case it works nicely.
     
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