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

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1. (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
2. (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
3. (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. .
4. (Original post by Zacken)
They don't cancel out in the multiplication though. .
That would become (25+5i-5i-i2) which simpflies to 26?
5. (Original post by paccap)
That would become (25+5i-5i-i2) which simpflies to 26?
Indeed!
6. (Original post by Zacken)
Indeed!
Ok, good to know!
7. (Original post by paccap)
Ok, good to know!
8. (Original post by Zacken)
So, if a cubic equation has roots then we can say three things:

, and and .

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 and comparing coefficients.

In this case, we have , and so is easy to work out and 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.
9. (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. :-)
10. (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.
11. (Original post by Zacken)
Yep thats what i got.
12. (Original post by paccap)
Yep thats what i got.
Awesome. Well done!

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