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Thread: Writing Math Paper 1 on Friday - Help Please

  1. #31
    Bar0n janvdl's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    And what are $\displaystyle p\mbox{ and }q$.
    $\displaystyle p = -5$
    $\displaystyle q = -4$

    Quote Originally Posted by ThePerfectHacker View Post
    I relized something interesting,
    Given $\displaystyle x^3+ax^2+bx+c=0$ which has exactly 1 root.
    It is necessary (thought maybe not sufficient) that $\displaystyle b^3=a^3c$.
    Can you prove that?
    It might help you.
    But $\displaystyle b^3$ is not equal to $\displaystyle a^3c$

    I really have no idea how to do this. I've never seen a problem like this before, and i dont know how to simplify it. And on top of it all, it only counts 2 marks.
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  2. #32
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    Quote Originally Posted by janvdl View Post
    But $\displaystyle b^3$ is not equal to $\displaystyle a^3c$
    That is because you do not have it in the right form.

    $\displaystyle 2x^3 - 5x^2 -4x+3=k$
    Thus,
    $\displaystyle 2x^3 - 5x^2 - 4x + (3-k)=0$
    Thus,
    $\displaystyle x^3 - \frac{5}{2}x^2 - 2x+\frac{3-k}{2}=0$
    So we require that,
    $\displaystyle \left( -\frac{5}{2} \right)\left( \frac{3-k}{2} \right) = (-2)^3$
    This gives us an equation for $\displaystyle k$.

    And hence the only possible number. You still have to check whether or not it gives you exactly one zero.
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  3. #33
    Bar0n janvdl's Avatar
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    Hey guys, just thought i'd let you know it went great with my first math paper.

    Thank you to everyone who helped.

    Oh and TPH, i still owe you thanks. I'll give it to you as soon as i get to my pc.
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  4. #34
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    Quote Originally Posted by janvdl View Post
    Oh and TPH, i still owe you thanks. .
    I want your soul instead.
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  5. #35
    Bar0n janvdl's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    I want your soul instead.
    Nope
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