$x^4-mx^3-2mx^2+2m^2x=0$ roots in arithmetic progression

Question

For which real values of $m$ roots of equation $$x^4-mx^3-2mx^2+2m^2x=0$$ are in arithmetic progression?

I managed to find a solution using a lot of casework, i.e. factoring the equation into $$x(x-m)(x^2-2m)=0$$ and then finding $m$ for which it has one root, three distinct roots and four distinct roots but I think there may be an easier solution and I'm just overcomplicating it. In addition, I'm not quite sure if I should include the one root case.

Could someone guide me to a better solution? Thanks.

Answer 1

I assume that by "roots in arithmetic progression" you mean that the four roots are of the form $$ \alpha,\quad\alpha+k,\quad\alpha+2k,\quad\alpha+3k $$ for some $\alpha$ and some $k\neq0$.

Now we know that $0$ and $m$ are roots. Since the coefficient of $x^3$ is $-m$ the sum of the roots must be $m$ which gives $\alpha=-m/2$ and $k=m/2$, i.e the four roots are $$ -\frac m2,\quad0,\quad\frac m2,\quad m. $$ Since one of the roots is $0$ the coefficient of $x$ is the negative of the product of the three non-zero roots, i.e. $$ 2m^2=\frac{m^3}4 $$ whose only non-zero solution is $m=8$.

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On the other hand, if you do not assume that the four roots are distinct, since anyway the four roots are $\{-\beta,0,\beta,m\}$ for some $\beta$ one gets two extra-cases, namely $\beta=0$ and $\beta=m$.

Working again with the linear coefficient we soon get that

• $\beta=0$ occurs only for $m=0$ (the polynomal is just $x^4$);
• $\beta=m$ occurs only for $m=2$.

Answer 2

WLOG the roots are $a-3d, a-d,a+d, a+3d$

As $(a-3d)(a-d)(a+d)(a+3d)=0,$

Case$\#1:$

If $a-3d=0,$ the rest of the roots become $2d,4d,6d$

If $2d=m,4d\cdot6d=2m\implies\dfrac m{12}=\left(\dfrac m2\right)^2\iff m(3m-1)=0$

If $4d=m,2d\cdot6d=2m\implies\left(\dfrac m4\right)^2=\dfrac m6\implies m=?$

If $6d=m,2d\cdot8d=2m\implies\left(\dfrac m6\right)^2=\dfrac m8\implies m=?$

Case$\#2:$ $a+3d=0$ Replace $d$ with $-d$ in Case$\#1$

Case$\#3:$ If $a-d=0, a=d$ the rest of the roots become $-2d,2d,4d$

$$x^4-mx^3-2mx^2+2mx=x(x-4d)(x-2d)(x+2d)$$

$$\iff x^3-mx^2-2mx+2m=(x-4d)(x^2-4d^2)=x^3-4dx^2-4d^2x+16d^3$$

$$\implies m=4d\ \ \ \ (1),2m=16d^3\ \ \ \ (2),-2m=-4d^2\ \ \ \ (3)$$

If $d\ne0,$

$(2)/(3)\implies-1=\dfrac{16d^3}{-4d^2}=-4d, d=?\ \ \ \ (4)$

$(2)/(1)\implies2=\dfrac{16d^3}{4d}=4d^2\implies d^2=?$ but this does not satisfy $(4)$

Case$\#4:$ $a+d=0$ Replace $d$ with $-d$ in Case$\#3$

Answer 3

From your factorized equation, the roots are $0$, $m$, $-\surd(2m)$, and $\surd(2m)$. If $m<0$, then $m$ lies off the imaginary axis, on which the three roots $0$ and $\pm\,\mathrm i\surd(-2m)$ lie, violating the condition that the roots are in arithmetic progression. Hence $m\geqslant0$. It follows that the arithmetic progression is $-\surd(2m)$, $0$, $\surd(2m)$, $m$. Therefore $m=2\surd(2m)$, giving $$m=0\quad\text{or}\quad m=8.$$