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    (Original post by Ano123)
    It's open for anyone to have a go at.
    Ok. But, why on a thread about logarithms lol? It doesn't involve logs.
    No one will even look at this thread unless they are here to help me or are learning logs too....
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    (Original post by Chittesh14)
    Ok. But, why on a thread about logarithms lol? It doesn't involve logs.
    No one will even look at this thread unless they are here to help me or are learning logs too....
    It was the most active thread, it's all dead since exams have come and gone,
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    (Original post by Ano123)
    It was the most active thread, it's all dead since exams have come and gone,
    Lol. The Maths section is active - just for A-levels because all GCSE exams have finished. Just wait till next year when you go to A-levels lol, it'll be much more fun as you will be able to relate to the threads too. It's just too active at A-level - mainly because it's tough lool.
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    (Original post by Chittesh14)
    Lol. The Maths section is active - just for A-levels because all GCSE exams have finished. Just wait till next year when you go to A-levels lol, it'll be much more fun as you will be able to relate to the threads too. It's just too active at A-level - mainly because it's tough lool.
    I don't follow what you mean. In what world is this a GCSE question?
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    (Original post by Ano123)
    I don't follow what you mean. In what world is this a GCSE question?
    I never said it was lol?
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    (Original post by Chittesh14)
    I never said it was lol?
    I don't get what you were saying. You said wait next year until you get to A-levels.
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    (Original post by Ano123)
    I don't get what you were saying. You said wait next year until you get to A-levels.
    Aren't you in Year 11 or just finished GCSEs?
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    (Original post by Chittesh14)
    Aren't you in Year 11 or just finished GCSEs?
    Yes, year 11, but doing A-levels.
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    (Original post by Ano123)
    Yes, year 11, but doing A-levels.
    Lol what?
    Oh, as in just A-level Maths and just FM GCSE with it?
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    (Original post by Chittesh14)
    Lol what?
    Oh, as in just A-level Maths and just FM GCSE with it?
    I'm not doing GCSE further maths.
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    (Original post by Ano123)
    I'm not doing GCSE further maths.
    So you just post questions on the thread for people to attempt?
    What are you actually doing lol ?
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    I'm a troll in a sense. I just put up questions and stuff I find interesting, let others who might be interested have a go. Look up the hard integral thread, I do what others do in that thread.
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    You are wrong, book is right
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    (Original post by Ano123)
    I'm a troll in a sense. I just put up questions and stuff I find interesting, let others who might be interested have a go. Look up the hard integral thread, I do what others do in that thread.
    Lol ok. What A-level modules have you done so far?
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    (Original post by chittesh14)
    lol ok. What a-level modules have you done so far?
    c1-4, m1-2, fp1-4, s1-2
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    (Original post by Ano123)
    c1-4, m1-2, fp1-4, s1-2
    Woah lool. You must do some intense revision. Well done, you must've taken some of those modules as exams too?
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    (Original post by Chittesh14)
    Woah lool. You must do some intense revision. Well done, you must've taken some of those modules as exams too?
    All of them.
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    (Original post by Ano123)
    All of them.
    How did you do in them?
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    (Original post by Ano123)
    Find the real roots of the equation
     \displaystyle x^6+8x^5+16x^4+18x^3+16x^2+8x+1=  0 .
    It's 10pm, I'm tired, and I really hate polynomials with degree n>3 so usually I'd tell you to go **** yourself... but I gave it a shot anyway with a nice long method, enjoy. =)

    Let f(x)=x^6+8x^5+16x^4+18x^3+16x^2+  8x+1

    \rightarrow f(-1)=0 therefore x=-1 is a solution.

    ---------------------------------------------------------------------------------

    \Rightarrow f(x)=0 therefore (x^3+x^{-3})+8(x^2+x^{-2})+16(x+x^{-1})+18=0

    (x+x^{-1})^3=x^3+x^{-3}+3(x+x^{-1}) \Rightarrow (x^3+x^{-3})=(x+x^{-1})^3-3(x+x^{-1})

    (x+x^{-1})^2=x^2+x^{-2}+2 \Rightarrow (x^2+x^{-2})=(x+x^{-1})^2-2

    \Longrightarrow [(x+x^{-1})^3-3(x+x^{-1})]+8[(x+x^{-1})^2-2]+16(x+x^{-1})+18=0

    \Longrightarrow (x+x^{-1})^3+8(x+x^{-1})^2+13(x+x^{-1})+2=0

    Let a=x+x^{-1}

    \Rightarrow a^3+8a^2+13a+2=0

    Consider roots of the cubic:

    \sum(\alpha)=-8, \sum(\alpha\beta)=13, \alpha\beta\gamma=-2

    We know x=-1 is a root for f(x)=0 therefore a=(-1)+(-1)^{-1}=-2 therefore \alpha=-2

    Coefficients of cubic are real therefore if \alpha is real then \beta and \gamma are complex conjugates of each other.

    \sum(\alpha)=\alpha+\beta+\gamma  =-8 therefore \beta+\gamma=-6

    \alpha\beta\gamma=-2 therefore \gamma=\frac{1}{\beta}

    \Rightarrow \beta+\frac{1}{\beta}=-6 \Rightarrow \gamma^2+6\gamma+1=0 \Rightarrow (\gamma+3)^2=8 \Rightarrow \gamma=-3+\sqrt8 therefore \beta=-3-\sqrt8

    Therefore roots of a^3+8a^2+13a+2=0 are a=-2 / -3+\sqrt8 / -3-\sqrt8

    ---------------------------------------------------------------------------------

    Coming back to the original f(x)...

    For a=-2 ;
    x+x^{-1}=-2 \Rightarrow x^2+2x+1=0 \Rightarrow (x+1)^2=0 \Rightarrow x=-1

    For a=-3+\sqrt8 ;
    x+x^{-1}=-3+\sqrt8 \Rightarrow x^2+(3-\sqrt8)x+1=0 \Rightarrow b^2-4ac<0 therefore no real roots.

    For a=-3-\sqrt8 ;
    x+x^{-1}=-3-\sqrt8 \Rightarrow x^2+(3+\sqrt8)x+1=0 \Rightarrow b^2-4ac=13+12\sqrt2 \Rightarrow x=\frac{-(3+\sqrt8)\pm\sqrt((13+12\sqrt2)  )}{2}

    -----------------------------------------------------------------------------------

    Therefore the real solutions to f(x)=0 are:
    x=-1

    x= -\frac{(3+\sqrt8)+\sqrt{13+\sqrt2  \cdot12}}{2}

    x= -\frac{(3+\sqrt8)-\sqrt{13+\sqrt2\cdot12}}{2}
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    (Original post by Chittesh14)
    How did you do in them?
    We'll find out.
 
 
 
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