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Math Help - Rationalizing radicals in the denominator

  1. #1
    Lil
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    Post Rationalizing radicals in the denominator

    It's little difficult for me...

    \frac{2}{\sqrt[5]{4}}
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    MHF Contributor harish21's Avatar
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    \frac{2}{\sqrt[5]{4}} = \frac{2}{4^{(1/5)}}= \frac{2}{(2)^{2(1/5)}}  = \frac{2}{2^{(2/5)}}

    NOw complete it
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  3. #3
    Lil
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    I don't understand. Correct answer is \sqrt[5]{8}
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  4. #4
    MHF Contributor harish21's Avatar
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    YOu need to look at your books/notes:

     \frac{2}{\sqrt[5]{4}} = \frac{2}{4^{(1/5)}}= \frac{2}{(2)^{2(1/5)}} = \frac{2}{2^{(2/5)}} = 2^{1-\frac{2}{5}} = 2^{(3/5)}=2^{3(1/5)} = 8^{(1/5)} = \sqrt[5]{8}


    Clear??
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  5. #5
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    \dfrac{2}{{\sqrt[5]{4}}} = \dfrac{2}{{\sqrt[5]{{2^2 }}}}\dfrac{{\sqrt[5]{{2^3 }}}}{{\sqrt[5]{{2^3 }}}} = \sqrt[5]{8}
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  6. #6
    Lil
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    I see. I just think you leave answer in that first post. I miss to read complete it. Sorry. I'm absent-minded today.
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  7. #7
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    Quote Originally Posted by Lil View Post
    It's little difficult for me...

    \frac{2}{\sqrt[5]{4}}
    Express the numerator as a fifth-root also...

    \displaystyle\frac{2}{\sqrt[5]4}}=\frac{\sqrt[5]{2^5}}{\sqrt[5]4}=\sqrt[5]{\frac{2^5}{2^2}}
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  8. #8
    Lil
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    I solve this now. It's correct or no?
    \frac{4}{3\sqrt{5}-4}=\frac{4(3\sqrt{5}+4)}{(3\sqrt{5}-4){(3\sqrt{5}+4)}}=\frac{4(3\sqrt{5}+4)}{(3\sqrt{5  })^2-(4)^{2}}=\frac{4(3\sqrt{5}+4)}{29}.
    Last edited by Lil; October 9th 2010 at 06:40 AM.
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  9. #9
    MHF Contributor harish21's Avatar
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    Yes!

    It can also be expressed as:

     \frac{4(3\sqrt{5}+4)}{29} = \frac{12\sqrt{5}}{29} + \frac{16}{29}
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  10. #10
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    Quote Originally Posted by Lil View Post
    I solve this now. It's correct or no?

    \frac{4}{3\sqrt{5}-4}=\frac{4(3\sqrt{5}+4)}{(3\sqrt{5}-4){(3\sqrt{5}+4)}}=\frac{4(3\sqrt{5}+4)}{(3\sqrt{5  }-4)^{2}}=\frac{4(3\sqrt{5}+4)}{29}.
    You have the right idea (surd conjugate) and your final answer has rationalised the denominator and is correct.

    However (3\sqrt{5}-4)(3\sqrt{5}+4)=(3\sqrt{5})^2-4^2=29

    (3\sqrt{5}-4)(3\sqrt{5}+4)\ \ne\ (3\sqrt{5}-4)^2
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  11. #11
    Lil
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    Quote Originally Posted by Archie Meade View Post
    You have the right idea (surd conjugate) and your final answer has rationalised the denominator and is correct.

    However (3\sqrt{5}-4)(3\sqrt{5}+4)=(3\sqrt{5})^2-4^2=29

    (3\sqrt{5}-4)(3\sqrt{5}+4)\ \ne\ (3\sqrt{5}-4)^2
    What a stupid mistake!!! I know this rule. My head don't work today. Thanks.
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  12. #12
    Lil
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    Question

    I won't make new topic. Just continue this, because my work is with roots.
    Stuck with this.
    Just calculate expression:

    2*\left(\frac{2}{\sqrt{6}+3} +\frac{3}{\sqrt{6}-2}-\frac{5}{\sqrt{6}}\right)
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  13. #13
    MHF Contributor harish21's Avatar
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    2*\left(\frac{2}{\sqrt{6}+3} +\frac{3}{\sqrt{6}-2}-\frac{5}{\sqrt{6}}\right)

     =  2*\left( \left[\frac{2}{\sqrt{6}+3} \times \frac{\sqrt{6}-3}{\sqrt{6}-3}\right]+ \left[\frac{3}{\sqrt{6}-2} \times \frac{\sqrt{6}+2}{\sqrt{6}+2}\right]-\left[\frac{5}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}}\right]\right)

    Now Simplify.....
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  14. #14
    Lil
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    I won't write all solution... It's clear now. Answer I get 10. I think it's correct.
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  15. #15
    MHF Contributor harish21's Avatar
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    Yes!!
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