$x^3-6x^2+11x+m=0$ roots in arithmetic progression

Question

Given equation: $$x^3-6x^2+11x+m=0.$$ For which values of $m$ roots of the equation are roots in arithmetic progression?

I've found a version of this exercise in which the equation could be converted to a quadratic by letting $y$ be $x^4$, but I couldn't really come out with a method that works here as well from that.

I've applied Vieta's and got $x_1^3+x_1^23r+2x_1r^2+m=0$(from $x_1*x_2*x_3=-m$, r being the ratio of the progression) and also $x_1= \frac{11-2r^2}{6+3r}$(from working with Vieta's) but I don't think replacing $x_1$ in the previous one would be the way to go.

Can I have some hints on how to approach this? Thank you.

Answer 1

Let $x_1$, $x_2$ and $x_3$ be our roots.

Thus, $$x_1+x_2+x_3=6$$ or $$3x_2=6,$$ which gives $x_2=2$ and from here we can get the value of $m$: $$8-24+22+m=0$$ or $$m=-6$$ and easy to see that this value is valid: $$x^3-6x^2+11x-6=(x-1)(x-2)(x-3).$$ Done!

Answer 2

you will have $$x_1+x_2+x_3=6$$ and $$x_1x_2+x_1x_3+x_2x_3=11$$ and $$x_1x_2x_3=-m$$ where $$x_2=x_1+d,x_2=x_1+2d$$

Answer 3

Let the roots be $a,a+d,a+2d$ then we have \begin{eqnarray*} a+a+d+a+2d=3(a+d)=6 \\ \color{red}{a+d=2} \\ a(a+d)+a(a+2d)+(a+d)(a+2d)=\color{blue}{3a^2+6ad+2d^2 =11} \end{eqnarray*} Now square $\color{red}{a+d=2}$ , multiply it by $3$ and subtract the last equation to get $d^2=1$. It is easy form here ?

Answer 4

Let $P(x)=x^3-6x^2+11x+m$. Change $x=y+2$ to get $$Q(y)=P(x+2)=(y+2)^3-6(y+2)^2+11(y+2)+m=y^3-y+6+m$$

The roots of $P$ are in arithmetic progression if and only if the roots of $Q$ either are.

Since the sum of the roots of $Q$ are $0$, the roots of $Q$ are $-u,0,u$ for some $u\ge 0$. Since $0$ is a root of $Q$, we get $6+m=0$.

Answer 5

Using the roots $a-d,a,a+d$, the conditions are

$$3a=6,\\ 3a^2-d^2=11,\\ a^3-ad^2=-m.$$

Then

$$a=2,d=\pm1,\\m=-6$$ and this is the only solution (though there are two progressions, $1,2,3$ and $3,2,1$).

Answer 6

If we deflate the polynomial using $y=x-2$, the roots remain in an arithmetic progression

$$x^3-6x^2+11x+m\to y^3-y+m+6.$$ The deflated polynomial needs to be odd (of the form $y(y^2-d^2)$), and this works with

$$m+6=0.$$

Answer 7

HINT:

The roots of $P$ ($\deg 3$) are in arithmetic progression if and only if their arithmetic mean $\frac{s}{3}$ is a root, that is $$P(\frac{s}{3})=0$$