Roots of $g'(x)$ in AP: Proving the Theory

[anemone]

The roots of a fourth degree polynomial $g(x)=0$ are in an AP (arithmetic progression). Prove that the roots of $g'(x)=0$ must also form an AP.

 

[MarkFL]

My solution:

Spoiler

Without loss of generality, let us horizontally translate the function $g$ such that it is even in our coordinate system:

\(\displaystyle g(x)=(x+a)(x+3a)(x-a)(x-3a)=\left(x^2-a^2\right)\left(x^2-9a^2\right)\)

Hence:

\(\displaystyle g'(x)=2x\left(x^2-9a^2\right)+2x\left(x^2-a^2\right)=4x(x+\sqrt{5}a)(x-\sqrt{5}a)=0\)

We see that the roots of $g'$ are in fact in an AP.

 

[anemone]

My solution:

Spoiler

Without loss of generality, let us horizontally translate the function $g$ such that it is even in our coordinate system:

\(\displaystyle g(x)=(x+a)(x+3a)(x-a)(x-3a)=\left(x^2-a^2\right)\left(x^2-9a^2\right)\)

Hence:

\(\displaystyle g'(x)=2x\left(x^2-9a^2\right)+2x\left(x^2-a^2\right)=4x(x+\sqrt{5}a)(x-\sqrt{5}a)=0\)

We see that the roots of $g'$ are in fact in an AP.

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[anemone]

I also want to share a solution that I saw online, and here it goes:

Spoiler

Let's consider only the case with four real distinct roots $a,\,a+r,\,a+2r,\,a+3r$ with $r>0$.

Then $g(x)=k(x-a)(x-a-r)(x-a-2r)(x-a-3r)$ for some $k,\,r\ne 0$.

And so $g\left(\dfrac{rx}{2}+a-\dfrac{3r}{2}\right)=\dfrac{kr^4}{16}(x-3)(x-1)(x+1)(x+3)=\dfrac{kr^4}{16}(x^4-10x^2+9)$

$\begin{align*}\dfrac{r}{2}g'\left(\dfrac{rx}{2}+a-\dfrac{3r}{2}\right)&=\dfrac{kr^4}{16}(4x^3-20x)\\&=\dfrac{kr^4}{4}x(x^2-5)\\&=\dfrac{kr^4}{4}x(x-\sqrt{5})(x+\sqrt{5})\end{align*}$

Since $g'\left(\dfrac{rx}{2}+a-\dfrac{3r}{2}\right)$ has three distinct roots in AP, it is immediate to get that the roots of $g'(x)$ must also form an AP.

 

[CellCoree]

I find this content to be a very interesting and important topic in the field of mathematics. The concept of roots of a polynomial forming an arithmetic progression has been studied extensively and has various applications in different areas of mathematics and science.

To prove that the roots of $g'(x)=0$ must also form an AP, we first need to understand the definition of an arithmetic progression. An arithmetic progression is a sequence of numbers where the difference between any two consecutive terms is constant. In other words, if we have an arithmetic progression with terms $a, a+d, a+2d, a+3d, ...$, then the difference between any two consecutive terms is always equal to $d$.

Now, let us consider the polynomial $g(x)$ with roots $r_1, r_2, r_3, r_4$. Since the roots of $g(x)$ are in an AP, we can express them as $r_1=a, r_2=a+d, r_3=a+2d, r_4=a+3d$, where $a$ is the first term and $d$ is the common difference.

To prove that the roots of $g'(x)=0$ also form an AP, we need to show that the difference between any two consecutive roots of $g'(x)=0$ is constant. Let us take the derivative of $g(x)$, which is $g'(x)=4ax^3+3bx^2+2cx+d$. Now, to find the roots of $g'(x)$, we set it equal to zero and solve for $x$. This gives us $x=0$ and $x=\frac{-b\pm \sqrt{b^2-3ac}}{2a}$.

Since $x=0$ is a repeated root, we can ignore it and focus on the other two roots, which we can express as $x=\frac{-b+\sqrt{b^2-3ac}}{2a}$ and $x=\frac{-b-\sqrt{b^2-3ac}}{2a}$. Now, let us simplify these two roots by taking out a common factor of $\frac{1}{2a}$. This gives us $x=\frac{-b+\sqrt{b^2-3ac}}{2a}=\frac{-b}{2a}+\frac{\sqrt{b