Problem 35:  Roots of a cubic equation ($✓$ $✓$ $✓$) 1999 Paper III

Consider the cubic equation

where $p\ne 0$ and $r\ne 0$.

(i)

If the three roots can be written in the form $a{k}^{-1}$, $a$ and $ak$ for some constants $a$ and $k$, show that one root is $q∕p$ and that $q^3-r{p}^3=0.$

(ii)

If $r=q^3∕p^3$, show that $q∕p$ is a root and that the product of the other two roots is ${\left(q∕p\right)}^2$. Deduce that the roots are in geometric progression.

(iii)

Find a necessary and sufficient condition involving $p$, $q$ and $r$ for the roots to be in arithmetic progression.

Comments

The Fundamental Theorem of Algebra says that a polynomial of degree $n$ can be written as the product of $n$ linear factors, so we can write

$$x^3-p{x}^2+qx-r=\left(x-\alpha \right)\left(x-\beta \right)\left(x-\gamma \right),$$

($\ast$)

where $\alpha$, $\beta$ and $\gamma$ are the roots of the equation $x^3-p{x}^2+qx-r=0$. The basis of this question is the comparison between the left hand side of $\left(\ast \right)$ and the right hand side, multiplied out, of $\left(\ast \right)$. Some of the roots may not be real, but you don’t have to worry about that here.

The ‘necessary and sufficient’ in part (iii) looks a bit forbidding, but if you just repeat the steps of parts (i) and (ii), it is straightforward.

Solution to problem 35

(i) We have

i.e.

Thus $p=a\left(k^{-1}+1+k\right)$, $q=a^2\left(k^{-1}+1+k\right)$, and $r=a^3$. Dividing gives $q∕p=a$, which is a root, and $q^3∕p^3=a^3=r$ as required.

(ii) Set $r=q^3∕p^3$. Substituting $x=q∕p$ into the cubic shows that it is a root:

Since $q∕p$ is one root, and the product of the three roots is $q^3∕p^3$ ($=r$ in the original equation), the product of the other two roots must be $q^2∕p^2$. The two roots can therefore be written in the form $k^{-1}\left(q∕p\right)$ and $k\left(q∕p\right)$ for some number $k$, which shows that the three roots are in geometric progression.

(iii) The three roots are in arithmetic progression if and only if they are of the form $\left(a-d\right)$, $a$ and $\left(a+d\right)$. (I have followed the lead of part (i) in using this form rather than $a$, $a+d$, $a+2d$.)

If the roots are in this form, then

i.e.

Thus $p=3a$, $q=3a^2-d^2$, $r=a\left(a^2-d^2\right)$. A necessary condition is therefore $r=\left(p∕3\right)\left(q-2p^2∕9\right)$. Note that one root is $p∕3$.

Conversely, if $r=\left(p∕3\right)\left(q-2p^2∕9\right)$, the equation is

We can verify that $p∕3$ is a root by substitution. Since the roots sum to $p$ and one of them is $p∕3$, the others must be of the form $p∕3-d$ and $p∕3+d$ for some $d$. They are therefore in arithmetic progression.

A necessary and sufficient condition for the roots to be in arithmetic progression is therefore $r=\left(p∕3\right)\left(q-2p^2∕9\right)$.

Post-mortem

As usual, it is a good idea to give a bit of thought to the conditions given, namely $p\ne 0$ and $r\ne 0$.

Clearly, if the roots are in geometric progression, we cannot have a zero root. We therefore need $r\ne 0$. However, we don’t need the condition $p\ne 0$. If the roots are in geometric progression with $p=0$, then $q=0$, but there is no contradiction: the roots are $b{e}^{-2\pi i∕3}$, $b$ and $b{e}^{2\pi i∕3}$ where $b^3=r$, and these are certainly in geometric progression. So the necessary and sufficient conditions are $r\ne 0$ and $p^{3}r=q^3$.

Neither of these conditions is required if the roots are in arithmetic progression.

Therefore, the condition $p\ne 0$ is only there as a convenience — one thing less for you to worry about. The condition $r\ne 0$ should really have been given only for the first part. It should be said that this sort of question is incredibly difficult to word, which is why the examiner was a bit heavy handed with the conditions.