MHB 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:

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

Hence:

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

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:

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

Hence:

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

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.