kensafw
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I got the question:

The roots of x^4+ax^3+bx^2+cx+d=0 are alpha, beta, gamma, and delta. Given that alpha + beta = gamma + delta , show that a^3+8c=4ab.

I know that I can expand out the roots but I'm not sure if that'd be too helpful. Could you please give me some pointers on where to start and go?
Many thanks for your help!
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DFranklin
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I've spoilered the answer because I thought this was actually one of the nicest questions of this form I've ever seen and thought some other people might like to give it a try.

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Consider factorising into a pair of quadratics (x^2 + Ax + B)(x^2 + Cx + D) with roots alpha,beta and gamma,delta respectively. It comes out pretty quickly from there.


ghostwalker RDKGames mqb2766 davros
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mqb2766
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(Original post by DFranklin)
I've spoilered the answer because I thought this was actually one of the nicest questions of this form I've ever seen and thought some other people might like to give it a try.

Spoiler:
Show
Consider factorising into a pair of quadratics (x^2 + Ax + B)(x^2 + Cx + D) with roots alpha,beta and gamma,delta respectively. It comes out pretty quickly from there.


ghostwalker RDKGames mqb2766 davros
Yes, it does fall out nicely. The first step is fairly obvious, and going down the quadratic route gives enough encouragement to keep going for the other two steps.
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ghostwalker
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(Original post by DFranklin)
I've spoilered the answer....
Too late - I'd seen your initial response prior to spoilering (if that's at word) and had worked it through. Nice method.
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kensafw
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(Original post by DFranklin)
I've spoilered the answer because I thought this was actually one of the nicest questions of this form I've ever seen and thought some other people might like to give it a try.

Spoiler:
Show
Consider factorising into a pair of quadratics (x^2 + Ax + B)(x^2 + Cx + D) with roots alpha,beta and gamma,delta respectively. It comes out pretty quickly from there.


ghostwalker RDKGames mqb2766 davros
I'm afraid I'm still not sure what to do after this. The others in my class have the same issue.
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mqb2766
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(Original post by kensafw)
I'm afraid I'm still not sure what to do after this. The others in my class have the same issue.
The answer has a, b and c in it (no d), so that suggests relating the roots to the cubic, quadratic and linear coefficients.
To start with, what is the relationship between (combinations of the) roots and "a", the cubic coefficient?

If you go with the previuos hint, make two quadratics out of the alpha and beta, and the gamma and delta roots. Then the overall quartic is the product of these two quadratics (but don't expand at the moment).
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RDKGames
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(Original post by mqb2766)
The answer has a, b and c in it (no d), so that suggests relating the roots to the cubic, quadratic and linear coefficients.
To start with, what is the relationship between (combinations of the) roots and "a", the cubic coefficient?
From that hint, you know that

\alpha+\beta = -A and \delta + \gamma = -C.

Since \alpha + \beta = \delta + \gamma we have A=C.

When you multiply out these brackets for the x^3 term, you have a = C+A = 2A.

When you multiply out these brackets for the x^2 term, you have b = ..

When you multiply out these brackets for the x term, you have c = ..

From here you can show the result holds.
Last edited by RDKGames; 3 weeks ago
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DFranklin
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(Original post by RDKGames)
~snip~
A somewhat bigger hint than I'd consider reasonable...
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kensafw
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RDKGamesmqb2766DFranklin
Thank you for all your help but unfortunately, I just cannot get my head round it! I have a Maths lesson tomorrow in which I'll ask my teacher to go over this question but hopefully, this can be of use to someone else in the future.
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RDKGames
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(Original post by kensafw)
RDKGamesmqb2766DFranklin
Thank you for all your help but unfortunately, I just cannot get my head round it! I have a Maths lesson tomorrow in which I'll ask my teacher to go over this question but hopefully, this can be of use to someone else in the future.
Since x^4 + ax^3 + bx^2 + cx + d has four roots \alpha,\beta,\gamma,\delta then we can rewrite as a product of two quadratics


\begin{aligned} x^4 + ax^3 + bx^2 + cx + d & = (x^2 + Ax + B)(x^2 + Cx + D) \\ & = x^4 + (A+C)x^3 + (AC + B + D)x^2 + (AD + BC)x + BD


where (x^2 + Ax+B) has roots \alpha,\beta and (x^2 + Cx+D) has roots \gamma , \delta.

Considering the first quadratic, we know that \alpha + \beta = -A, and if we consider the second quadratic we know that \gamma + \delta = -C.

The question tells us that \alpha + \beta = \gamma + \delta which means that A=C via these two relations we have found.

Now, equate coefficients between the quartics above. We get

a=A+C
b = AC + B + D
c=AD+BC
d=BD

The result you seek does not involve d so we can forget about it and just consider the first three results. Since A=C we can reduce them to

a=2A
b=A^2 + (B+D)
c=A(B+D)

what you can now proceed to do is eliminate A,B,D between these three relations.

E.g. from first relation we know that A = \dfrac{a}{2} which means our other two become

b=\dfrac{a^2}{4} + (B+D)

c = \dfrac{a}{2}(B+D)

What can you do with these two equations to eliminate (B+D) terms? You should hence obtain a result entirely in terms of a,b,c which corresponds to the result you seek.
Last edited by RDKGames; 3 weeks ago
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kensafw
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(Original post by RDKGames)
Since x^4 + ax^3 + bx^2 + cx + d has four roots \alpha,\beta,\gamma,\delta then we can rewrite as a product of two quadratics


\begin{aligned} x^4 + ax^3 + bx^2 + cx + d & = (x^2 + Ax + B)(x^2 + Cx + D) \\ & = x^4 + (A+C)x^3 + (AC + B + D)x^2 + (AD + BC)x + BD


where (x^2 + Ax+B) has roots \alpha,\beta and (x^2 + Cx+D) has roots \gamma , \delta.

Considering the first quadratic, we know that \alpha + \beta = -A, and if we consider the second quadratic we know that \gamma + \delta = -C.

The question tells us that \alpha + \beta = \gamma + \delta which means that A=C via these two relations we have found.

Now, equate coefficients between the quartics above. We get

a=A+C
b = AC + B + D
c=AD+BC
d=BD

The result you seek does not involve d so we can forget about it and just consider the first three results. Since A=C we can reduce them to

a=2A
b=A^2 + (B+D)
c=A(B+D)

what you can now proceed to do is eliminate A,B,D between these three relations.

E.g. from first relation we know that A = \dfrac{a}{2} which means our other two become

b=\dfrac{a^2}{4} + (B+D)

c = \dfrac{a}{2}(B+D)

What can you do with these two equations to eliminate (B+D) terms? You should hence obtain a result entirely in terms of a,b,c which corresponds to the result you seek.
Thank you so much! This is really helpful and I do now understand it. I'm just curious as to why you can just disregard d without having to cancel it out or anything?
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RDKGames
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(Original post by kensafw)
Thank you so much! This is really helpful and I do now understand it. I'm just curious as to why you can just disregard d without having to cancel it out or anything?
As I said:
(Original post by RDKGames)
... The result you seek does not involve d so we can forget about it ...
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