Roots of Polynomials HW Help

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.



@ghostwalker @RDKGames @mqb2766 @davros

I've spoilered the answer....

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.



@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.

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?

~snip~

@RDKGames@mqb2766@DFranklin
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





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?

... The result you seek does not involve $d$ so we can forget about it ...