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Math Help - [SOLVED] roots

  1. #1
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    [SOLVED] roots

    here's the problem.
    i don't know how to solve this , can you help me? thanks.
    \frac{(^4\sqrt{4b^4})+(^4\sqrt{9b^4}-(\sqrt{12a^2}-(^4\sqrt{64a^4}+(\sqrt{20a^2}-(\sqrt{5b^2})}{b-2a}^{-1}
    Attached Thumbnails Attached Thumbnails [SOLVED] roots-picture-49.jpg   [SOLVED] roots-picture-51.jpg  
    Last edited by princess_21; March 28th 2009 at 03:30 AM.
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    Quote Originally Posted by princess_21 View Post
    here's the problem.
    i don't know how to solve this , can you help me? thanks.
    The first exemple must be an olympic exercise:

    \left(\dfrac{\sqrt[4]{4b^4} + \sqrt[4]{9b^4}-\sqrt{12a^2}-\sqrt[4]{64a^4} + \sqrt{20a^2} - \sqrt{5b^2}}{b-2a} \right)^{-1} =

    \left(\dfrac{b\sqrt{2} + b\sqrt{3}-2a\sqrt{3}-2a\sqrt{2} + 2a\sqrt{5} - b\sqrt{5}}{b-2a} \right)^{-1} =

    \left(\dfrac{b \left(\sqrt{2} + \sqrt{3}-\sqrt{5}\right)-2a \left(\sqrt{3}+\sqrt{2} - \sqrt{5}\right) }{b-2a} \right)^{-1} =

    \left(\dfrac{ \left(\sqrt{2} + \sqrt{3}-\sqrt{5}\right)(b-2a)  }{b-2a} \right)^{-1} = \dfrac1{\sqrt{2} + \sqrt{3}-\sqrt{5}}

    Now eliminate the square roots in the denominator:

    \dfrac1{\sqrt{2} + \sqrt{3}-\sqrt{5}} \cdot \dfrac{\sqrt{2} + \sqrt{3}+\sqrt{5}}{\sqrt{2} + \sqrt{3}+\sqrt{5}} = \dfrac{\sqrt{2} + \sqrt{3}+\sqrt{5}}{2+2\sqrt{6}+3-5} =

    \dfrac{\sqrt{2} + \sqrt{3}+\sqrt{5}}{2\sqrt{6}} = \dfrac1{12} \cdot \left(2\sqrt{3}+3\sqrt{2} + \sqrt{30}  \right)


    phooo!

    The next exemple is much easier. Show your work and I'M going to help you if you don't succeed. (You should get 2\sqrt{6})
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    Question

    Quote Originally Posted by earboth View Post
    The first exemple must be an olympic exercise:

    \left(\dfrac{\sqrt[4]{4b^4} + \sqrt[4]{9b^4}-\sqrt{12a^2}-\sqrt[4]{64a^4} + \sqrt{20a^2} - \sqrt{5b^2}}{b-2a} \right)^{-1} =

    \left(\dfrac{b\sqrt{2} + b\sqrt{3}-2a\sqrt{3}-2a\sqrt{2} + 2a\sqrt{5} - b\sqrt{5}}{b-2a} \right)^{-1} = how did it become like this?

    \left(\dfrac{b \left(\sqrt{2} + \sqrt{3}-\sqrt{5}\right)-2a \left(\sqrt{3}+\sqrt{2} - \sqrt{5}\right) }{b-2a} \right)^{-1} =

    \left(\dfrac{ \left(\sqrt{2} + \sqrt{3}-\sqrt{5}\right)(b-2a)  }{b-2a} \right)^{-1} = \dfrac1{\sqrt{2} + \sqrt{3}-\sqrt{5}}

    Now eliminate the square roots in the denominator:

    \dfrac1{\sqrt{2} + \sqrt{3}-\sqrt{5}} \cdot \dfrac{\sqrt{2} + \sqrt{3}+\sqrt{5}}{\sqrt{2} + \sqrt{3}+\sqrt{5}} = \dfrac{\sqrt{2} + \sqrt{3}+\sqrt{5}}{2+2\sqrt{6}+3-5} =

    \dfrac{\sqrt{2} + \sqrt{3}+\sqrt{5}}{2\sqrt{6}} = \dfrac1{12} \cdot \left(2\sqrt{3}+3\sqrt{2} + \sqrt{30}  \right)


    phooo!

    The next exemple is much easier. Show your work and I'M going to help you if you don't succeed. (You should get 2\sqrt{6})
    i'm still confused... can you help me again?
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  4. #4
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    Quote Originally Posted by princess_21 View Post
    i'm still confused... can you help me again?
    I'll show you how to do the second question, but if I shall help you I must know what exactly is confusing you.

    Ooooh, I finally found your question in the quoted calculations.

    If you have  \left(\dfrac{\sqrt[4]{4b^4} + \sqrt[4]{9b^4}-\sqrt{12a^2}-\sqrt[4]{64a^4} + \sqrt{20a^2} - \sqrt{5b^2}}{b-2a} \right)^{-1} = then you can simplify

    \sqrt[4]{4b^4} = (2^2 \cdot b^4)^{\frac14} = 2^{\frac24} \cdot b^{\frac44} = b\cdot 2^{\frac12} = b\cdot \sqrt{2}

    Apply similar transformations with all summands in the numerator and you'll get the next line of my calculations.

    And now the 2nd question:

    \left(\dfrac{a}{\sqrt{2}} \div (2a\sqrt{3})\right)^{-1} = \left(\dfrac{a}{\sqrt{2}} \cdot \dfrac1{2a\sqrt{3}}\right)^{-1} = \left(\dfrac1{2\sqrt{6}} \right)^{-1} = 2\sqrt{6}
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    yes understand the second equation. I don't understand the first equation. My teacher told me that I need to change all the index so they can be combined. I change them all as fourth root by raising all square root to 2. is that correct?
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  6. #6
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    Quote Originally Posted by princess_21 View Post
    yes understand the second equation. I don't understand the first equation. My teacher told me that I need to change all the index so they can be combined. I change them all as fourth root by raising all square root to 2. is that correct?
    That is possible, of course. But you'll deal with quite large numbers. For instance:

    \sqrt{12a^2} = \sqrt[4]{144a^4}

    while I prefer the transform:

    \sqrt{12a^2} = 2a\sqrt{3}

    To me this form is much easier to handle.
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    Ok I got it. but I don't understand why \frac{\sqrt{2} + \sqrt{3} +\sqrt{5}}{2\sqrt{6}}= \frac{1}{12}
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    Quote Originally Posted by princess_21 View Post


    Ok I got it. but I don't understand why \frac{\sqrt{2} + \sqrt{3} +\sqrt{5}}{2\sqrt{6}}= \frac{1}{12}
    Unfortunately you quoted me wrong:

    <br />
\dfrac{\sqrt{2} + \sqrt{3}+\sqrt{5}}{2\sqrt{6}} = \dfrac{{\color{red}\bold{\sqrt{6}}} (\sqrt{2} + \sqrt{3}+\sqrt{5})}{2\sqrt{6} \cdot {\color{red}\bold{\sqrt{6}}}} = \dfrac{\sqrt{12}+\sqrt{18} + \sqrt{30}}{12} =  \dfrac1{12} \cdot \left(2\sqrt{3}+3\sqrt{2} + \sqrt{30} \right)<br />
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    yeah I got it thanks
    I'll try to continue from here

    \frac{2\sqrt{3}+3\sqrt{2}+\sqrt{30}}{12} * (2\sqrt{6})

    \frac{4\sqrt{18}+6\sqrt{12}+2\sqrt{180}}{12}

    \frac{12\sqrt{2}+12\sqrt{3}+12\sqrt{5}}{12}

    is this correct?
    Last edited by princess_21; March 28th 2009 at 07:13 AM.
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  10. #10
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    Quote Originally Posted by princess_21 View Post
    yeah I got it thanks
    I'll try to continue from here

    \frac{2\sqrt{3}+3\sqrt{2}+\sqrt{30}}{12} * (2\sqrt{6})

    \frac{4\sqrt{18}+6\sqrt{12}+2{180}}{12}

    \frac{12\sqrt{2}+12\sqrt{3}+6\sqrt{5}}{12}

    is this correct?
    Honestly, I don't know what you did and why?

    All my calculations were necessary to get a rational denominator. Thus I multiplied numerator and denominator by \sqrt{6} which I marked in red. That means I didn't change the value of the quotient.

    2\cdot \sqrt{6}\cdot \sqrt{6} = 12

    The product of the numerator and \sqrt{6} is

    \sqrt{12}+\sqrt{18} + \sqrt{30} = \sqrt{4 \cdot 3}+\sqrt{9 \cdot 2} + \sqrt{30} which I simplified to

    2\sqrt{3}+3\sqrt{2} + \sqrt{30}

    That's all.
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    the two attachments are just one question the first one multiplied by the second one.
    this is the original equation. I multiplied the answers.
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  12. #12
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    Quote Originally Posted by princess_21 View Post
    yeah I got it thanks
    I'll try to continue from here

    \frac{2\sqrt{3}+3\sqrt{2}+\sqrt{30}}{12} * (2\sqrt{6})

    \frac{4\sqrt{18}+6\sqrt{12}+2{180}}{12}

    \frac{12\sqrt{2}+12\sqrt{3}+6\sqrt{5}}{12}

    is this correct?
    Got it - better late then never

    Simplify this expression a little bit:

    \frac{12\sqrt{2}+12\sqrt{3}+6\sqrt{5}}{12} = \sqrt{2} + \sqrt{3} + \dfrac{1}{2}\sqrt{5}
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