Roots of $x^3+ax^2+bx+c=0$ are in arithmetic progression. Find $2a^3-9ab$

Question

If the roots of the equation

$$x^3+ax^2+bx+c=0$$

are in an arithmetic, then what is the value of

$$2a^3-9ab$$

Please explain.

Answer 1

Without loss of generality, assume that the roots are $x_1<x_2<x_3$. Note that the roots are different because they are members of an arithmetic progression. Hence, $$2x_3=x_1+x_2$$ Now see if you can derive any other equations involving the roots $x_1,x_2,x_3$ and the coefficients $a,b,c$ of your polynomial, and try to manipulate them to get your result.(Hint: Vieta's formulas.)

Answer 2

HINT:

Let $\displaystyle p-d,p, p+d$ be the three roots

Apply Vieta's formula to express $a,b$ in terms of $p,d$ like

$\displaystyle p-d+p+p+d=-\frac a1\iff p=-\frac a3\iff a=-3p$

and $\displaystyle (p-d)p+p(p+d)+(p+d)(p-d)=b\iff 3p^2-d^2=b$

Answer 3

The only way to get three roots in an arithmetic progression is if the middle one is the point of inflection of the curve.

So, this occurs for $x^3 - Nx$ when $N > 0.$ Let us switch letters to $t^3 - Nt.$ What happens if you take a translate, that is, $t = x + W?$

Oh, well, the cable went out in the middle of Chelsea-Atletico Madrid. For a cubic of type $x^3 - N x + K,$ there is a local minimum and a local maximum. If we look at a horizontal line going through the local minimum and gradually raise it, the double root at the local minimum separates until farthest apart when we reach the height of the local maximum. Same thing going the other way, beginning at the local maximum. So, byt what we call a continuity argument, they are equal exactly once. Since they are equal at the height of the inflection point, that is all we need to know. In sum, a cubic with three real roots in arithmetic progression is of the form $$ f(x) = (x + W)^3 - N (x+W) $$ with some $N> 0$ and $W$ real but otherwise not restricted.

Chelsea scored while the cable was out. http://www.espnfc.com/us/en/gamecast/391813/gamecast.html?soccernet=true&cc=5901

Answer 4

Let $r$, $r\pm d$ be the three roots and, per the Vieta’s formulas

\begin{align} &(r-d)+r+(r+d)= -a\\ &(r-d)r+r(r+d)+(r+d)(r-d)=b\\ &(r-d)r(r+d)=-c \end{align} or, $$r=-\frac a3,\>\>\>\>\>d^2=3r^2-b,\>\>\>\>\>r^3-d^2r=-c $$ Substitute the 1st and 2nd equations into the 3rd to obtain $$2a^3-9ab=-27c$$