All of us have solved many problems on cubic polynomials who have their roots in Arithmetic Progression and Geometric Progression.So I have found a relationship between the coefficients of these polynomials which makes these roots special.

So first , let's start with the roots in Arithmetic Progression.

Consider a polynomial : \(\alpha_1x^3+\alpha_2x^2+\alpha_3x+\alpha_4\).

Let its roots be \(\beta , \gamma , \lambda\) such that they form an arithmetic progression.So we have a relation :

\[\Rightarrow 2\gamma = \beta + \lambda \dots (1)\]

Now our favorite polynomial tool cum friend that is , Vieta's formula will help us to have some more relations:

\[\sigma_1=\beta +\gamma + \lambda = \dfrac{-\alpha_2}{\alpha_1} \dots (2) \\ \sigma_2=\beta\gamma + \beta\lambda + \lambda\gamma = \dfrac{\alpha_3}{\alpha_1} \dots (3) \\ \sigma_3= \beta\gamma\lambda = \dfrac{-\alpha_4}{\alpha_1} \dots (4)\]

Using \((1),(2)\) we have :

\[\beta +\gamma + \lambda = 2\gamma+ \gamma = 3\gamma= \dfrac{-\alpha_2}{\alpha_1} \Rightarrow \gamma = \dfrac{-\alpha_2}{3\alpha_1} \dots (5)\]

Using \((1),(3),(5)\) we have :

\[\beta\gamma + \beta\lambda + \lambda\gamma = \dfrac{\alpha_3}{\alpha_1} \\ \Rightarrow \gamma(\beta+\lambda)+\beta\lambda = \dfrac{\alpha_3}{\alpha_1} \dots(6) \\ \Rightarrow 2\gamma^2 + \beta\lambda = \dfrac{\alpha_3}{\alpha_1} \\ \Rightarrow 2\left( \dfrac{-\alpha_2}{3\alpha_1}\right)^2 + \beta\lambda = \dfrac{\alpha_3}{\alpha_1} \\ \Rightarrow \beta\lambda = \dfrac{\alpha_3}{\alpha_1} - \dfrac{2\alpha_2^2}{9\alpha_1^2} \\ \Rightarrow \beta\lambda=\dfrac{9\alpha_1\alpha_3-2\alpha_2^2}{9\alpha_1^2} \dots (7)\]

Using \((4) , (7)\) we have :

\[\dfrac{\sigma_3}{\beta\lambda} = \gamma \\ \Rightarrow \dfrac{\dfrac{-\alpha_4}{\alpha_1}}{\dfrac{9\alpha_1\alpha_3-2\alpha_2^2}{9\alpha_1^2}} = \dfrac{-\alpha_2}{3\alpha_1} \\ \Rightarrow \left(\dfrac{-\alpha_4}{\alpha_1}\right) \times \left(\dfrac{9\alpha_1^2}{9\alpha_1\alpha_3-2\alpha_2^2}\right)=\dfrac{-\alpha_2}{3\alpha_1} \\ \Rightarrow \dfrac{-9\alpha_1^2\alpha_4}{9\alpha_1\alpha_3-2\alpha_2^2} \\ \Rightarrow 27\alpha_1^2\alpha_4=9\alpha_1\alpha_2\alpha_3 - 2\alpha_2^3 \]

Thus , we can write the relation between coefficients as :

\[\Large\boxed{\alpha_4(2\alpha_2^3+27\alpha_1^2\alpha_4) = 9\displaystyle\prod_{i=1}^4 \alpha_i}\]

Now let the roots \(\beta , \gamma , \lambda\) form a Geometric progression.Thus we have the relation :

\[\gamma^2=\beta\lambda \dots (8)\]

Using \((1)\) we have :

\[\beta+\lambda = \dfrac{-\alpha_2}{\alpha_1} - \gamma \Rightarrow \beta+\lambda=-\left(\dfrac{\alpha_2}{\alpha_1} + \gamma\right) \dots (9)\]

Using \((4),(8)\) we have :

\[\beta\gamma\lambda = \dfrac{-\alpha_4}{\alpha_1} \Rightarrow \gamma^3= \dfrac{-\alpha_4}{\alpha_1} \Rightarrow \gamma= \sqrt[3]{ \dfrac{-\alpha_4}{\alpha_1}} \dots (10)\]

Using \((6),(9),(10)\) we will have :

\[\gamma(\beta+\lambda)+\beta\lambda = \dfrac{\alpha_3}{\alpha_1} \\ \Rightarrow -\gamma\left(\dfrac{\alpha_2}{\alpha_1} + \gamma\right) + \gamma^2 = \dfrac{\alpha_3}{\alpha_1} \\ \Rightarrow \dfrac{-\alpha_2\gamma}{\alpha_1} - \gamma^2+\gamma^2 = \dfrac{\alpha_3}{\alpha_1} \\ \Rightarrow \dfrac{-\alpha_2\gamma}{\alpha_1} = \dfrac{\alpha_3}{\alpha_1} \\ \Rightarrow \gamma = \dfrac{-\alpha_3}{\alpha_2} \\ \Rightarrow \sqrt[3]{ \dfrac{-\alpha_4}{\alpha_1}} = \dfrac{-\alpha_3}{\alpha_2} \\ \Rightarrow \dfrac{-\alpha_4}{\alpha_1} = \dfrac{-\alpha_3^3}{\alpha_2^3} \\ \alpha_4\alpha_2^3 = \alpha_1\alpha_3^3 \]

This when rearranged properly we get the following relationship between the coefficients:

\[\Large\boxed{\dfrac{\alpha_1}{\alpha_4}=\left(\dfrac{\alpha_2}{\alpha_3}\right)^3}\]

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## Comments

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TopNewestI assume you're only solving for cubic polynomials.

Now try for Harmonic progression as well! And post some challenges!

Nice read! Great work!

Next stop: Apply Newton's sum! ｡◕‿◕｡ – Pi Han Goh · 2 years, 4 months ago

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– Archit Boobna · 2 years, 4 months ago

Hey, I observed that even \[\left\lceil { \pi }^{ \pi }\ln { \pi } \right\rceil \] is the meaning of life.Log in to reply

– Nihar Mahajan · 2 years, 4 months ago

Thanks a lot. I will for sure try it for Harmonic progression. And yeah , I will not make Newton's sums feel lonely. \(\ddot\smile\).Log in to reply

This is really cool. Good work, keep it up! :) – Prasun Biswas · 2 years, 4 months ago

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– Nihar Mahajan · 2 years, 4 months ago

Thanks a lot :)Log in to reply

@Calvin Lin I will be happy if you read this. :) – Nihar Mahajan · 2 years, 4 months ago

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For example, let's take the Geometric progression case. Here is a nice simple solution:

Now, I have still hidden the beauty of the solution. If you truly understand it, the following problem has a "one line" solution.

– Calvin Lin Staff · 2 years, 3 months agoLog in to reply

My thought:Its a good idea to assume the roots as \(1,r,r^2\).But one can argue that we get the relationship \(\dfrac{\alpha_4}{\alpha_1} = \left(\dfrac{\alpha_3}{\alpha_2}\right)^3\) only if one of the roots is \(1\).What about when it is not equal to \(1\)? There is no guarantee that the relationship remains the same for all possible roots.For a polynomial whose \(n\) roots are in geometric progression, I think \(\dfrac{\alpha_{n+1}}{\alpha_1} = \left(\dfrac{\alpha_n}{\alpha_2}\right)^n\). Am I right?

Also I didn't understand what do you want me to find about \(\dfrac{\alpha_{n-i}}{\alpha_{2+i}}\). – Nihar Mahajan · 2 years, 3 months ago

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Of course, another approach would be to check that

\[ (x-a)(x-ar)(x-ar^2) = x^3 - (a +ar+ar^2) x^2 + (a^2r + a^2r^2 + a^2r^3) - ar^3. \]

from which the identity follows. However, this is still "tedious magical algebra", instead of getting to the "actual reason why this relationship should exist" which would allow us to easily generalize to \(a_{n-i} / a_{2+i} \). – Calvin Lin Staff · 2 years, 3 months ago

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And even the GP case can be done easily with a similar analysis. – Ankit Kumar Jain · 5 months, 2 weeks ago

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@Nihar Mahajan @Pi Han Goh What about this?

Let the roots be \(x-y,x,x+y\) , by Vieta's relations we have

\(3x=\dfrac{-a_2}{a_1}\)

But since \(x\) is a root of the equation , therefore substitute this value of \(x\) in the equation and simplify by to get the desired result. – Ankit Kumar Jain · 5 months, 2 weeks ago

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You're on the right track, yes, if you let "p-q, p, p + q" be the 3 roots that follow an AP, then yes, you can prove all the results that Nihar has shown with less work. – Pi Han Goh · 5 months, 2 weeks ago

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– Ankit Kumar Jain · 5 months, 2 weeks ago

Oops...I should have used the same variable..Sorry!!Log in to reply

– Ankit Kumar Jain · 5 months, 2 weeks ago

I mean ...A different variableLog in to reply

Here are some relevant articles: Cardano's method and Cubic discriminant. – Pi Han Goh · 5 months, 2 weeks ago

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Thanks a lot sir.! – Ankit Kumar Jain · 5 months, 2 weeks ago

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gud work nihar – Idris Muhammed · 2 years, 3 months ago

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Excellent work! – Swapnil Das · 2 years, 3 months ago

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Gud... Thanks... – Abhay Agnihotri · 2 years, 4 months ago

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– Nihar Mahajan · 2 years, 4 months ago

Welcome buddy!Log in to reply