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Math Help - Simplifying help

  1. #1
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    Simplifying help

    Hey there I need some help simplifying this problem
    hope you guys can see it

    Simplifying help-simplifying.jpg

    oh and that is an 8th root if you guys are wondering
    also I'm looking for the steps to go about simplifying this problem not just the answer thanks
    Last edited by gfbrd; November 24th 2012 at 09:27 AM.
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  2. #2
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    Re: Simplifying help

    \sqrt[8]{\frac{2207-987\sqrt{5}}{2}}

    what makes you think the expression can be simplified further than the form you've written?
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  3. #3
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    Re: Simplifying help

    because i know the simplified form

    i just dont know how to go about simplifying it thats all
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  4. #4
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    Re: Simplifying help

    care to share that "simplified form" ... because my CAS can't do it.
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  5. #5
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    Re: Simplifying help

    it should be
    (3 + sqrt(5))/ 2
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    Re: Simplifying help

    uhh ...

    \sqrt[8]{\frac{2207-987\sqrt{5}}{2}} \approx 0.381966010851...

    \frac{3+\sqrt{5}}{2} \approx 2.61803398875...

    ???
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  7. #7
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    Re: Simplifying help

    oh sorry i made a mistake in typing the problem
    it should be + 987sqrt(5) not subtract
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    Re: Simplifying help

    That looks better to the calculator, but I can't see a quick simplification at this time. If it doesn't come to me, hopefully someone else can help.
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  9. #9
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    Re: Simplifying help

    oh alright thanks for helping
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  10. #10
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    Re: Simplifying help

    Quote Originally Posted by gfbrd View Post
    Hey there I need some help simplifying this problem
    hope you guys can see it

    Click image for larger version. 

Name:	simplifying.jpg 
Views:	8 
Size:	19.5 KB 
ID:	25901

    oh and that is an 8th root if you guys are wondering
    also I'm looking for the steps to go about simplifying this problem not just the answer thanks
    A word of advice, ALL "roots" of fractions are considered unsimplified. You are expected to always rationalise denominators at the very least.

    \displaystyle \begin{align*} \sqrt[8]{\frac{2207 + 987\sqrt{5}}{2}} &= \frac{\sqrt[8]{2207 + 987\sqrt{5}}}{\sqrt[8]{2}} \\ &= \frac{\sqrt[8]{2207 + 987\sqrt{5}}\cdot \left( \sqrt[8]{2} \right)^7}{\sqrt[8]{2} \cdot \left(\sqrt[8]{2}\right)^7} \\ &= \frac{\sqrt[8]{2207 + 987\sqrt{5}} \cdot \sqrt[8]{2^7}}{2} \end{align*}

    I'll have to think about how the top simplifies to \displaystyle \begin{align*} 3 + \sqrt{5} \end{align*}...
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  11. #11
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    Re: Simplifying help

    Hello, gfbrd!

    \text{It should be }\:\frac{3 + \sqrt{5}}{2} . . . . no

    The original expression has a "minus": . \sqrt[8]{\frac{2207 \:{\color{red}-}\: 987\sqrt{5}}{2}}

    You must mean: . \frac{3\:{\color{red}-}\:\sqrt{5}}{2}
    . . which happens to equal: . \frac{1}{\phi^2}
    . . where \phi \,=\,\tfrac{1+\sqrt{5}}{2}, the Golden Ratio.

    It can be shown that: . \frac{1}{\phi^{16}} \:=\:\frac{2207 - 987\sqrt{5}}{2}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    I'd like to share a "trick" for raising \phi to higher powers.

    One of the root of the quadratic equation: . x^2 - x - 1 \:=\:0 .is \phi.

    This means: . \phi^2 - \phi - 1 \:=\:0 \quad\Rightarrow\quad \boxed{\phi^2 \:=\:\phi + 1}

    Divide by \phi\!:\;\phi \:=\:1 + \frac{1}{\phi} \quad\Rightarrow\quad \boxed{\frac{1}{\phi} \:=\:\phi - 1}


    We have: . \frac{1}{\phi} \;=\;\phi - 1

    Square: . \frac{1}{\phi^2} \;=\;(\phi - 1)^2 \;=\;\phi^2 - 2\phi + 1 \;=\;(\phi+1) - 2\phi + 1 \;=\;2 - \phi

    Square: . \frac{1}{\phi^4} \;=\;(2-\phi)^2 \;=\;4 - 4\phi + \phi^2 \;=\;4 - 4\phi + (\phi + 1) \;=\;5-3\phi

    Square: . \frac{1}{\phi^8} \;=\;(5-3\phi)^2 \;=\;25 - 30\phi + 9\phi^2 \;=\;25 - 30\phi + 9(\phi+1) \;=\;34 - 21\phi

    Square: . \frac{1}{\phi^{16}} \;=\;(34-21\phi)^2 \;=\;1156 - 1428\phi + 441\phi^2 \;=\; 1597 - 987\phi

    Hence: . \frac{1}{\phi^{16}} \;=\;1597 - 987\left(\frac{1+\sqrt{5}}{2}\right) \;=\;\frac{2207 - 987\sqrt{5}}{2}

    Therefore: . \sqrt[8]{\frac{2207 - 987\sqrt{5}}{2}} \;=\;\frac{1}{\phi^2} \;=\;\frac{3 - \sqrt{5}}{2}
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  12. #12
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    Re: Simplifying help

    Quote Originally Posted by Soroban View Post
    Hello, gfbrd!


    The original expression has a "minus": . \sqrt[8]{\frac{2207 \:{\color{red}-}\: 987\sqrt{5}}{2}}

    You must mean: . \frac{3\:{\color{red}-}\:\sqrt{5}}{2}
    . . which happens to equal: . \frac{1}{\phi^2}
    . . where \phi \,=\,\tfrac{1+\sqrt{5}}{2}, the Golden Ratio.

    It can be shown that: . \frac{1}{\phi^{16}} \:=\:\frac{2207 - 987\sqrt{5}}{2}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    I'd like to share a "trick" for raising \phi to higher powers.

    One of the root of the quadratic equation: . x^2 - x - 1 \:=\:0 .is \phi.

    This means: . \phi^2 - \phi - 1 \:=\:0 \quad\Rightarrow\quad \boxed{\phi^2 \:=\:\phi + 1}

    Divide by \phi\!:\;\phi \:=\:1 + \frac{1}{\phi} \quad\Rightarrow\quad \boxed{\frac{1}{\phi} \:=\:\phi - 1}


    We have: . \frac{1}{\phi} \;=\;\phi - 1

    Square: . \frac{1}{\phi^2} \;=\;(\phi - 1)^2 \;=\;\phi^2 - 2\phi + 1 \;=\;(\phi+1) - 2\phi + 1 \;=\;2 - \phi

    Square: . \frac{1}{\phi^4} \;=\;(2-\phi)^2 \;=\;4 - 4\phi + \phi^2 \;=\;4 - 4\phi + (\phi + 1) \;=\;5-3\phi

    Square: . \frac{1}{\phi^8} \;=\;(5-3\phi)^2 \;=\;25 - 30\phi + 9\phi^2 \;=\;25 - 30\phi + 9(\phi+1) \;=\;34 - 21\phi

    Square: . \frac{1}{\phi^{16}} \;=\;(34-21\phi)^2 \;=\;1156 - 1428\phi + 441\phi^2 \;=\; 1597 - 987\phi

    Hence: . \frac{1}{\phi^{16}} \;=\;1597 - 987\left(\frac{1+\sqrt{5}}{2}\right) \;=\;\frac{2207 - 987\sqrt{5}}{2}

    Therefore: . \sqrt[8]{\frac{2207 - 987\sqrt{5}}{2}} \;=\;\frac{1}{\phi^2} \;=\;\frac{3 - \sqrt{5}}{2}
    The OP already made it clear that the original post had a typo and should have been \displaystyle \begin{align*} \sqrt[8]{\frac{2207 \mathbf{+} 987\sqrt{5}}{2}} \end{align*} giving the correct result.
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  13. #13
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    Re: Simplifying help

    Thanks for your help everyone, I'll see if I can figure something out from what soroban did
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