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FP1- January 2010 Question 6 help! (complex and real roots) watch

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    (Original post by paccap)
    So if i did -(5+5-2) for p and -(5x5x-2) for q. That would be the quicker method?
    Not quite, remember that you need to do -(5 + i + 5 - i - 2) = -(8) = -8 = p which is the same as what you've written, luckily. But you need to do q = -(5-i)(5+i)(-2) = 2(5^2 + 1) = 52
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    (Original post by Zacken)
    Not quite, remember that you need to do -(5 + i + 5 - i - 2) = -(8) = -8 = p which is the same as what you've written, luckily. But you need to do q = -(5-i)(5+i)(-2) = 2(5^2 + 1) = 52
    Oh, i forgot to write that i simplified it down as the imaginary parts cancel out
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    (Original post by paccap)
    Oh, i forgot to write that i simplified it down as the imaginary parts cancel out
    They don't cancel out in the multiplication though. (5-i)(5+i) \neq 5 \times 5.
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    (Original post by Zacken)
    They don't cancel out in the multiplication though. (5-i)(5+i) \neq 5 \times 5.
    That would become (25+5i-5i-i2) which simpflies to 26?
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    (Original post by paccap)
    That would become (25+5i-5i-i2) which simpflies to 26?
    Indeed!
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    (Original post by Zacken)
    Indeed!
    Ok, good to know!
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    (Original post by paccap)
    Ok, good to know!
    Just to check, your final answer should be x^3 -8x^2 + 6x + 52.
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    (Original post by Zacken)
    So, if a cubic equation x^3 + ax^2 + bx + c has roots \alpha, \beta, \gamma then we can say three things:

    c = -\alpha \beta \gamma, and b = \alpha \beta + \alpha \gamma + \gamma \beta and a = -(\alpha + \beta + \gamma).

    If you think this looks like magic, it's not. You can easily derive them yourself (and I do, since I always forget) by setting (x-\alpha)(x-\beta)(x-\gamma) = x^3 + ax^2 + bx + c and comparing coefficients.

    In this case, we have \alpha = 5+i, \beta = 5-i and \gamma = -2 so -\alpha \beta \gamma = q is easy to work out and -(\alpha + \beta + \gamma) = p is also easy to work out.
    Lol. The first method you suggested was unnecessarily long. The second one requires derivation/memorisation. I recommend just expanding the quadratic with the complex roots, multiplying it by ax+b then comparing coefficients. They already tell you the first coefficient is 1. So no prob at all.
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    (Original post by DrownedDeity)
    Lol. The first method you suggested was unnecessarily long. The second one requires derivation/memorisation. I recommend just expanding the quadratic with the complex roots, multiplying it by ax+b then comparing coefficients. They already tell you the first coefficient is 1. So no prob at all.
    I don't think you understand the first method if that's what you're saying. What about it confuses you? I can try and clear it up for you. :-)
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    (Original post by Zacken)
    I don't think you understand the first method if that's what you're saying. What about it confuses you? I can try and clear it up for you. :-)
    Solving for the real root, then expanding to later compare coefficients is unnecessary.

    When it sufficed to expand and compare coefficients in the first place.
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    (Original post by Zacken)
    Just to check, your final answer should be x^3 -8x^2 + 6x + 52.
    Yep thats what i got.
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    (Original post by paccap)
    Yep thats what i got.
    Awesome. Well done!
 
 
 
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