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

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Reply 20
Avatar for Zacken
Zacken
22
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
Reply 21
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paccap
OP
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
Reply 22
Avatar for Zacken
Zacken
22
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)≠5×5(5-i)(5+i) \neq 5 \times 5.
Reply 23
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paccap
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Original post by Zacken
They don't cancel out in the multiplication though. (5−i)(5+i)≠5×5(5-i)(5+i) \neq 5 \times 5.


That would become (25+5i-5i-i2) which simpflies to 26?
Reply 24
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Zacken
22
Original post by paccap
That would become (25+5i-5i-i2) which simpflies to 26?


Indeed!
Reply 25
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paccap
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Original post by Zacken
Indeed!

Ok, good to know!
Reply 26
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Zacken
22
Original post by paccap
Ok, good to know!


Just to check, your final answer should be x3−8x2+6x+52x^3 -8x^2 + 6x + 52.
Original post by Zacken
So, if a cubic equation x3+ax2+bx+cx^3 + ax^2 + bx + c has roots α,β,γ\alpha, \beta, \gamma then we can say three things:

c=−αβγc = -\alpha \beta \gamma, and b=αβ+αγ+γβb = \alpha \beta + \alpha \gamma + \gamma \beta and a=−(α+β+γ)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−α)(x−β)(x−γ)=x3+ax2+bx+c(x-\alpha)(x-\beta)(x-\gamma) = x^3 + ax^2 + bx + c and comparing coefficients.

In this case, we have α=5+i\alpha = 5+i, β=5−i\beta = 5-i and γ=−2\gamma = -2 so −αβγ=q-\alpha \beta \gamma = q is easy to work out and −(α+β+γ)=p-(\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.
Reply 28
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Zacken
22
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. :-)
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.
Reply 30
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paccap
OP
Original post by Zacken
Just to check, your final answer should be x3−8x2+6x+52x^3 -8x^2 + 6x + 52.


Yep thats what i got.
Reply 31
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Zacken
22
Original post by paccap
Yep thats what i got.


Awesome. Well done! :biggrin:

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