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Further Pure 1 - roots of polynomial equations

I was happily answering some questions from my further pure textbook earlier until I encountered a particularly challenging question; forgive me in advance for my lack of understanding, I'm a pretty average year 12 student at the moment and I really don't have as much brain power as most people on this site!

Here's the question:

The roots of the equation x3+ax+bx^3 + ax + b are α,β,γ\alpha, \beta, \gamma. Find the equation with roots βγ+γβ,γα+αγ,αβ+βα\frac{\beta}{\gamma} + \frac{\gamma}{\beta}, \frac{\gamma}{\alpha} + \frac{\alpha}{\gamma}, \frac{\alpha}{\beta} + \frac{\beta}{\alpha}.

Initially I tried to use substitution but that failed epicly. Then I just wrote out an equation showing the information I knew:

(x−(βγ+γβ))(x−(γα+αγ))(x−(αβ+βα))(x - (\frac{\beta}{\gamma} + \frac{\gamma}{\beta}))(x - (\frac{\gamma}{\alpha} + \frac{\alpha}{\gamma}))(x - (\frac{\alpha}{\beta} + \frac{\beta}{\alpha}))

Overall, I'm really confused - please could somebody hint at me?

Also, on a side note, I'm new here! Just thought you'd like to know :biggrin:
(edited 13 years ago)

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You're going about it the right way. Notice that you haven't used the 'a' or 'b' that the question has introduced. How can you get them to appear in your equation?

Note: You need an "=" for it to be an equation
Reply 2
Avatar for Femto
Femto
OP
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Right yes you are correct.

So, first of all I've probably copied down the equation wrong; it should be as you've said x3+ax+b=0x^3 + ax + b = 0.

Therefore can I assume with the simpler roots α,β,γ\alpha, \beta ,\gamma that obviously:

α+β+γ=−ba=0[br][br]αβ+αγ+βγ=ca=a[br][br]αβγ=−da=−b\alpha + \beta + \gamma = -\frac{b}{a} = 0[br][br]\alpha\beta + \alpha\gamma + \beta\gamma = \frac{c}{a} = a[br][br]\alpha\beta\gamma = -\frac{d}{a} = -b

Would this be correct? Also, how can I use this to my advantage to answer the main question? I don't really relish the task of finding the sum of those fractional roots nor multiplying them together or what not unless I'm thinking about this the completely wrong way?
(edited 13 years ago)
Reply 3
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Femto
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Would it be helpful to label each root as p,q,rp, q, r perhaps?

You see, every other preceding question was easier to tackle but seeing as this is the last question of an OCR miscellaneous exercise, there's no surprise it's quite difficult.
Reply 4
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Femto
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Infact forget what I just said.
(edited 13 years ago)
Reply 5
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Melanie-v
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What exam board is this?
Reply 6
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Femto
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Original post by Melanie-v
What exam board is this?


OCR but you would be unlucky to get a question such as this on the FP1 exam; they are generally easier.
(edited 13 years ago)
Original post by Femto
Right yes you are correct.

So, first of all I've probably copied down the equation wrong; it should be as you've said x3+ax+b=0x^3 + ax + b = 0.

Therefore can I assume with the simpler roots α,β,γ\alpha, \beta ,\gamma that obviously:

α+β+γ=−ba=0[br][br]αβ+αγ+βγ=ca=a[br][br]αβγ=−da=−b\alpha + \beta + \gamma = -\frac{b}{a} = 0[br][br]\alpha\beta + \alpha\gamma + \beta\gamma = \frac{c}{a} = a[br][br]\alpha\beta\gamma = -\frac{d}{a} = -b

Would this be correct? Also, how can I use this to my advantage to answer the main question? I don't really relish the task of finding the sum of those fractional roots nor multiplying them together or what not unless I'm thinking about this the completely wrong way?


The Sum of roots, sum of pairs, and product are all fine

and find the sum of roots, sum of pairs and products of the roots of the new equation, then you should be good to go (assuming you haven't already solved this)
Reply 8
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Melanie-v
10
Aah OK, I did Edexcel and didn't remember something like that. I'll try it.
Reply 9
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Femto
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Original post by Rational Paradox
The Sum of roots, sum of pairs, and product are all fine

and find the sum of roots, sum of pairs and products of the roots of the new equation, then you should be good to go (assuming you haven't already solved this)


Right but how do I find the sum of the pairs with the new roots? I can understand finding the sum and product of the roots but not the pair with this type of question.
Original post by Femto
Right but how do I find the sum of the pairs with the new roots? I can understand finding the sum and product of the roots but not the pair with this type of question.


What have you done so far? if you got some working, post it up and i can go through it
Reply 11
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Femto
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Original post by Rational Paradox
What have you done so far? if you got some working, post it up and i can go through it


Right well I've found the sum of the roots (I hope) as:

α(β2+γ2)+β(α2+γ2)+γ(α2+β2)αβγ\frac{\alpha(\beta^2 + \gamma^2) + \beta(\alpha^2 + \gamma^2) + \gamma(\alpha^2 + \beta^2)}{\alpha\beta\gamma}

That's what I've done so far.
(edited 13 years ago)
Original post by Femto
x


Do you have an answer for this question btw? I'm trying to do it as well and so far I have:

Spoiler

Reply 13
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Femto
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Original post by electriic_ink
Do you have an answer for this question btw? I'm trying to do it as well and so far I have:

Spoiler



Yes I do however I'm trying not to look at it at the moment because I'm trying to solve it but when you solve it I'll tell you the answer and then perhaps you could help out with my dire attempt.
Original post by Femto
Yes I do however I'm trying not to look at it at the moment because I'm trying to solve it but when you solve it I'll tell you the answer and then perhaps you could help out with my dire attempt.


OK. If I were you I'd have looked at it ages ago :p:

That is the sum of the roots and also the co-eff of x^2, which is the only one I've done. You need to use α+β+γ=0 \alpha + \beta + \gamma = 0 to simplify it.
Reply 15
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Femto
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Original post by electriic_ink
OK. If I were you I'd have looked at it ages ago :p:

That is the sum of the roots and also the co-eff of x^2, which is the only one I've done. You need to use α+β+γ=0 \alpha + \beta + \gamma = 0 to simplify it.


Ok I've looked and the answer is extremely strange; contrary to the other answers of this question format from the whole of FP1, this answer has no mention of the variable xx. But I think they've used substitution instead.
(edited 13 years ago)
Reply 16
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Femto
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Nearly given up with this question.
Reply 17
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ilyking
Original post by Femto
Right but how do I find the sum of the pairs with the new roots? I can understand finding the sum and product of the roots but not the pair with this type of question.


on edexcel we do this, but instead with imaginary numbers.

So I'm afraid I can't be of any help! :angry:
Reply 18
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Femto
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Original post by ilyking
on edexcel we do this, but instead with imaginary numbers.

So I'm afraid I can't be of any help! :angry:


No worries :biggrin:
Reply 19
Avatar for nkvuong
nkvuong
Original post by Femto
No worries :biggrin:


I think I could give this one a try
We have
α=−(β+γ)\alpha=-(\beta+\gamma)
And from
αβ+αγ+βγ=a[br]⇒βγ+α(β+γ)=a\alpha\beta+\alpha\gamma+\beta \gamma = a[br]\Rightarrow\beta\gamma+\alpha( \beta+ \gamma)=a
βγ=a+α2(1) \beta\gamma=a+\alpha^2 (1)

From

αβγ=−b⇒βγ=−bα(2)\alpha \beta \gamma = -b \Rightarrow \beta \gamma = -\frac{b}{\alpha} (2)

From (1) and (2)

α3+aα=−b\alpha^3+a\alpha = -b
so α3+β3+γ3+a(α+β+γ)=α3+β3+γ3=−3b\alpha^3+\beta^3+\gamma^3+a( \alpha +\beta+\gamma) = \alpha^3+\beta^3+\gamma^3 = -3b

Now consider the new root

βγ+γβ=(β+γ)2βγ−2=α2βγ−2=−α3b−2[br]\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]

So the sum of the 3 roots of the new equation will be
−α3+β3+γ3b−6=3−6=−3-\frac{\alpha^3+\beta^3+\gamma^3}{b}-6 = 3-6 =-3

You can repeat with the other two coefficients.
(edited 13 years ago)

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