Review question

# What can we say if the roots of a cubic are in arithmetic progression? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6276

## Solution

1. Given that the roots of the equation $x^3+px^2+qx+r=0$ are three consecutive terms of an arithmetic progression, show that $2p^3 + 27r = 9pq.$

Suppose the cubic equation $ax^3+bx^2+cx+d=0$ has the three roots $x_1$, $x_2$ and $x_3$. We can now write $ax^3+bx^2+cx+d=a(x-x_1)(x-x_2)(x-x_3).$

If we multiply out the right hand side, we have

$ax^3+bx^2+cx+d=ax^3-a(x_1+x_2+x_3)x^2+ a(x_1x_2+x_3x_1+x_2x_3)x - ax_1x_2x_3.$

This holds for all values of $x$, so corresponding coefficients must be equal. This means \begin{align*} x_1+x_2+x_3 = -\dfrac{b}{a},\\ x_1x_2+x_3x_1+x_2x_3 = \dfrac{c}{a},\\ x_1x_2x_3 = -\dfrac{d}{a}. \end{align*}

These equations connecting the coefficients of an equation with its roots are extremely useful.

We are given that $x^3+px^2+qx+r=0$ has three roots in arithmetic progression, so let’s give them the values $n-m$, $n$ and $n+m$. From our work above, we have the three equations \begin{align*} n-m+n+n+m = -p & \Longleftrightarrow 3n = -p,\\ (n-m)n+n(n+m)+(n+m)(n-m) = q & \Longleftrightarrow 3n^2-m^2 = q,\\ (n-m)n(n+m) = -r & \Longleftrightarrow n^3-nm^2 = -r. \end{align*}

So $m^2 = 3n^2-q$, and so $n^3-n(3n^2-q)=-r,$ or $2n^3=r+nq$.

Substituting in for $n$, we reach $2\left(\dfrac{-p}{3}\right)^3=r+q\dfrac{-p}{3}$. Now multiplying out gives $2p^3 + 27r = 9pq$ as required.

1. Given that the roots of the equation $x^3+px^2+qx+r=0$ are three consecutive terms of a geometric progression, find a condition that $p$, $q$ and $r$ must satisfy.

Let’s say this time that our roots take the values $n$, $nm$ and $nm^2$.

Using our work above on roots and coefficients for a second time, we have the three equations \begin{align*} n+nm+nm^2 & = -p, \\ n^2m+n^2m^2+n^2m^3 & = q, \\ n^3m^3 & = -r. \end{align*}

The first equation tells us that $1+m+m^2 = \dfrac{-p}{n}$.

The second says $n^2m(1+m+m^2) = q$, so on substituting in for $1+m+m^2$, we have $n^2m\dfrac{-p}{n} = q,$ and so $nm = \dfrac{-q}{p}$, and the third equation then gives us $\left(\dfrac{-q}{p}\right)^3=-r,$ which is $rp^3=q^3.$

This is the required condition.