Further Pure 1 - roots of polynomial equations

Now consider the new root

$\frac{\beta}{\gamma}+\frac{ \gamma}{\beta}=\frac{(\beta+ \gamma)^2}{\beta\gamma}-2=\frac{\alpha^2}{\beta\gamma}-2=\frac{\alpha^3}{b}-2[br]$

Wow that was impressive; but how did you get -2? Another thing - when you cross multiply $\frac{\beta}{\gamma}+\frac{ \gamma}{\beta}$ do you not get $\alpha^2 + \beta^2$ in the numerator?

I'm going to provide the answer anyway for those interested:

$b^2(1 + u)^3 + a^3(u + 2) = 0$

Where $u$ is simply a variable. I still don't understand how you arrive at that answer.

This question seems to be a bit more complicated than what is normally in an FP1 paper, so don't worry too much. May I suggest using the "finding the new roots method", as you were doing previously. So we need to find the sum of the roots, the sum of the product of pairs of roots, and the product of all the roots.



Sum of roots = $\displaystyle \frac{\beta}{\gamma} + \frac{\gamma}{\beta} + \frac{\gamma}{\alpha} + \frac{\alpha}{\gamma} + \frac{\alpha}{\beta} + \frac{\beta}{\alpha}$

and you got to $\displaystyle\frac{\alpha(\beta^2 + \gamma^2) + \beta(\alpha^2 + \gamma^2) + \gamma(\alpha^2 + \beta^2)}{\alpha\beta\gamma}$.

Note that this simplifies further to (You might want to check to see if that's correct :P ). You know the values of all of the things in that fraction from the above equations, so you can obtain the new sum of the roots.

Now you will need to repeat this for the product of pairs and the product of the roots.

Thank you very much - I'm too daft to realise the simplfied form of that expression, so sorry about that

I will get to work on it once I've done some worthwile revision for my January exams; this FP1 question isn't important right now, I just encountered it for a bit of fun.

It turns out it wasn't fun at all!