Roots of polynomials in an arithmetic sequence proof problem

[rinik]

Prove that if the roots of x^3 - ax^2 + bx -c=0 are in an arithmetic sequence
then 2a^3 -9ab + 27c=0.
Hence find x : x^3 - 12x^2 +39x -28=0
I tried making equation with the sum and products of the terms as well as making an equation with the sequence but had no luck so far.

[Guest]

My initial guess is that you should write the cubic as: (x-p)(x-q)(x-r)=x^3 -ax^2 + bx -c =0

Then you can express a, b and c in terms of the zeros and possibly vice versa.

As they are in an arithmetic sequence: p-q=q-r -> 2q - p - r = 0.

Maybe you can find a way to substitute p,q and r and confirm the 2a^3 - 9ab - 27c=0

I'm not sure about it but maybe it helps

[flinquinnster]

I actually managed to do this question! I was so intrigued by it that I had to solve it through until an answer came up. I used sum/product of roots to solve it, so maybe you just had a mistake in your method. Bear with me through the notation here.

Let your roots be e-f, e, e+f (they are in AS).

Given your cubic equation of x^3-ax^2+bx-c=0, we can make several equations.

a = sum of roots taken one at a time = (e-f)+e+(e+f) = 3e

b = sum of roots taken two at a time = e(e-f)+e(e+f)+(e-f)(e+f)=3e^2-f^2

c = sum of roots taken three at a time = e(e-f)(e+f) = e(e^2-f^2)

Therefore, that means we can substitute into 2a^3-9ab+27c=0, and prove that LHS=RHS=0.

LHS = 2a^3-9ab+27c = 2(3e)^3-9(3e)(3e^2-f^2)+27(e(e^2-f^2))

= 54e^3 - 81e^3 + 27ef^2 + 27e^3-27ef^2

= 0 = RHS

That means that for these particular circumstances, 2a^3-9ab+27c=0 is true.

Now onto the second part of the question - solving x^3-12x^2+39x-28=0.

We can compare this with the form x^3-ax^2+bx-c=0. Here, a=12, b=39, c=28.

We know that a = 3e. Therefore, a=12=3e --> e=4

We know that c = e(e^2-f^2). Substituting e=4, c=28=4(16-f^2). Solving gives you f = 3.

As the roots are e-f, e, e+f, that menas that the roots of x^3-12x^2+39x-28=0 are x=1,4,7. Which can be confirmed using a GDC.

Hope that helped! Perhaps there is a quicker/more elegant way to do this, but if you do everything correctly, this should get you there. This question is actually good revision for me, as I have been woefully negligent in practising maths questions lately.

[rinik]

I actually managed to do this question! I was so intrigued by it that I had to solve it through until an answer came up. I used sum/product of roots to solve it, so maybe you just had a mistake in your method. Bear with me through the notation here.

Let your roots be e-f, e, e+f (they are in AS).

Given your cubic equation of x^3-ax^2+bx-c=0, we can make several equations.

a = sum of roots taken one at a time = (e-f)+e+(e+f) = 3e

b = sum of roots taken two at a time = e(e-f)+e(e+f)+(e-f)(e+f)=3e^2-f^2

c = sum of roots taken three at a time = e(e-f)(e+f) = e(e^2-f^2)

Therefore, that means we can substitute into 2a^3-9ab+27c=0, and prove that LHS=RHS=0.

LHS = 2a^3-9ab+27c = 2(3e)^3-9(3e)(3e^2-f^2)+27(e(e^2-f^2))

= 54e^3 - 81e^3 + 27ef^2 + 27e^3-27ef^2

= 0 = RHS

That means that for these particular circumstances, 2a^3-9ab+27c=0 is true.

Now onto the second part of the question - solving x^3-12x^2+39x-28=0.

We can compare this with the form x^3-ax^2+bx-c=0. Here, a=12, b=39, c=28.

We know that a = 3e. Therefore, a=12=3e --> e=4

We know that c = e(e^2-f^2). Substituting e=4, c=28=4(16-f^2). Solving gives you f = 3.

As the roots are e-f, e, e+f, that menas that the roots of x^3-12x^2+39x-28=0 are x=1,4,7. Which can be confirmed using a GDC.

Hope that helped! Perhaps there is a quicker/more elegant way to do this, but if you do everything correctly, this should get you there. This question is actually good revision for me, as I have been woefully negligent in practising maths questions lately.

Thanks!

I solved it yesterday after posting( mine was a little more complex but it gave the same results ). I left it here hoping for a quicker solution and this is just what I wanted.

Maybe there is an even quicker solution but when it comes to polynomials the proof is always long compared to some others problems IMO