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

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 $x$ is $6$ and you know that two roots of the cubic are $5-i$ and $5+i$. What do you know about roots and the coefficient of the $x$ term in a cubic?

Hint, if you call the real root $m$, then what are $(5-i)(5+i)$, $m(5+i)$ and $m(5-i)$ and what should they all add up to?

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 $\alpha, \beta, \gamma$ then $\alpha \beta + \alpha \gamma + \beta \gamma = \text{coefficient of } \, x$.

So, in this case: $(5-i)(5+i) + m(5-i) + m(5+i) = 6$.

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?

Correct.




Indeed.

Correct.




Indeed.

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?

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.

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 .

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

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/AQA-MFP2-TEXTBOOK.PDF should help.

So if i did -(5+5-2) for p and -(5x5x-2) for q. That would be the quicker method?

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.