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Thread: Quadratics

  1. #1
    Member Furyan's Avatar
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    Quadratics

    Hello

    I've been looking at a question for some time, but haven't been able to find a way to start. The question is:

    'If x^2 - px - q = 0, where p and q are positive integers, which of the following could not equal x^3?'

    4x + 3, 8x + 5, 8x + 7, 10x + 3, 26x + 5

    I know the equation has one positive and one negative root. I've completed the square and found the discriminant, but I can't find a way into the question. Could someone please give me a pointer.

    Thank you.
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  2. #2
    Super Member Quacky's Avatar
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    Re: Quadratics

    I have a solution, but I'm not sure it's the neatest approach.

    $\displaystyle x^2=px+q$

    $\displaystyle x^3=px^2+qx$

    But as $\displaystyle x^2=px+q$,

    $\displaystyle x^3=p(px+q)+qx$

    $\displaystyle x^3=p^2x+qx+pq$

    $\displaystyle x^3=x(p^2+q)+pq$

    We have these options:
    $\displaystyle 4x + 3$, $\displaystyle 8x + 5$, $\displaystyle 8x + 7$, $\displaystyle 10x + 3$, $\displaystyle 26x + 5$

    First, I checked $\displaystyle 26x+5$

    p and q are integers and $\displaystyle pq=5$
    This means $\displaystyle p$ and $\displaystyle q$ are $\displaystyle 1$ and $\displaystyle 5$, although I don't know which letter corresponds to which number.

    Can I get $\displaystyle p^2+q=26$?
    Yes! If $\displaystyle p=5$ and $\displaystyle q=1$! So I can rule this out. Check the others: I found one which didn't work.
    Last edited by Quacky; Sep 24th 2011 at 07:05 PM.
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  3. #3
    Member Furyan's Avatar
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    Re: Quadratics

    Quacky, thank you for taking the time to reply. I see now how I should have approached the problem.
    Last edited by Furyan; Sep 25th 2011 at 12:24 AM. Reason: Spelling
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