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

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1. Hi,
Can anyone help me answer this question? Im stuck on how to get the real root as the coefficients are not given
Thanks
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2. (Original post by paccap)
Hi,
Can anyone help me answer this question? Im stuck on how to get the real root as the coefficients are not given
Thanks
You're given that the coefficient of is and you know that two roots of the cubic are and . What do you know about roots and the coefficient of the term in a cubic?
3. (Original post by Zacken)
You're given that the coefficient of is and you know that two roots of the cubic are and . What do you know about roots and the coefficient of the term in a cubic?
Hint, if you call the real root , then what are , and and what should they all add up to?
4. (Original post by Zacken)
Hint, if you call the real root , then what are , and and what should they all add up to?
Should the sum of the roots add up to 6?
5. If all the coefficients of a polynomial function are real and say the complex number, is a root, this also means that it's conjugate is also a root of the polynomial.
6. (Original post by paccap)
Should the sum of the roots add up to 6?
Not the sum of the roots. But the sum of alternating two products of the roots.

You know know that if a cubic has roots then .

So, in this case: .
7. (Original post by Zacken)
Not the sum of the roots. But the sum of alternating two products of the roots.

You know know that if a cubic has roots then .

So, in this case: .
OK, I get that part now! So solving it gives me m=-2
With that can I multiply out (z-5-i)(z-5+1)(z+2) and compare coefficients to get what p and q are?
8. (Original post by paccap)
OK, I get that part now! So solving it gives me m=-2
Correct.

With that can I multiply out (z-5-i)(z-5+1)(z+2) and compare coefficients to get what p and q are?
Indeed.
9. (Original post by Zacken)
Correct.

Indeed.
10. (Original post by Zacken)
Correct.

Indeed.
But I think it would be much quicker just to take the sum and product of the roots?
11. (Original post by paccap)
OK, I get that part now! So solving it gives me m=-2
With that can I multiply out (z-5-i)(z-5+1)(z+2) and compare coefficients to get what p and q are?
I'd just do p=-(sum of roots), q=-(product of roots) for a much quicker solution.
12. (Original post by IrrationalRoot)
But I think it would be much quicker just to take the sum and product of the roots?
I agree, but the OP doesn't seem very comfortable with the Vieta formulae so I thought it best to let him go about it on his own devices rather than unsettle him more.
13. (Original post by Zacken)
I agree, but the OP doesn't seem very comfortable with the Vieta formulae so I thought it best to let him go about it on his own devices rather than unsettle him more.
Still, he needs to be comfortable with them and get used to getting the answers the quickest way/way intended by the examiners.
For now it's ok I guess but good to let him know .
14. (Original post by IrrationalRoot)
Still, he needs to be comfortable with them and get used to getting the answers the quickest way/way intended by the examiners.
For now it's ok I guess but good to let him know .
Yeah, fair enough. I agree, but I'd try and let him know after he's done the question and reported back with success.
15. (Original post by Zacken)
I agree, but the OP doesn't seem very comfortable with the Vieta formulae so I thought it best to let him go about it on his own devices rather than unsettle him more.
would it be possible to show me how? maybe i can try and understand whats going on. I did get the correct answers, q =52 and p = -8
16. (Original post by paccap)
would it be possible to show me how? maybe i can try and understand whats going on. I did get the correct answers, q =52 and p = -8
Page 27 of this http://filestore.aqa.org.uk/subjects...2-TEXTBOOK.PDF should help.
17. (Original post by paccap)
would it be possible to show me how? maybe i can try and understand whats going on. I did get the correct answers, q =52 and p = -8
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.
18. (Original post by IrrationalRoot)
Page 27 of this http://filestore.aqa.org.uk/subjects...2-TEXTBOOK.PDF should help.
So if i did -(5+5-2) for p and -(5x5x-2) for q. That would be the quicker method?
19. (Original post by paccap)
So if i did -(5+5-2) for p and -(5x5x-2) for q. That would be the quicker method?
Well not 5s but 5+i and 5-i, since those are your roots (and 2).
But yes that would be quick.
20. (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.
Awesome! Thanks for the explanation, i remember doing this but i wasnt 100% on the method. Ill be sure to practice this method

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