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

Consider the cubic equation

${x}^{3}-p{x}^{2}+qx-r=0\phantom{\rule{2.77695pt}{0ex}},$

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 ${q}^{3}-r{p}^{3}=0\phantom{\rule{2.77695pt}{0ex}}.$
(ii)
If $r={q}^{3}∕{p}^{3}\phantom{\rule{2.77695pt}{0ex}}$, 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.

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

 ${x}^{3}-p{x}^{2}+qx-r=\left(x-\alpha \right)\left(x-\beta \right)\left(x-\gamma \right)\phantom{\rule{0.3em}{0ex}},$ ($\ast$)

where $\alpha \phantom{\rule{0.3em}{0ex}}$, $\beta$ and $\gamma$ are the roots of the equation ${x}^{3}-p{x}^{2}+qx-r=0\phantom{\rule{0.3em}{0ex}}$. 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

$\left(x-a{k}^{-1}\right)\left(x-a\right)\left(x-ak\right)\equiv {x}^{3}-p{x}^{2}+qx-r$

i.e.

${x}^{3}-a\left({k}^{-1}+1+k\right){x}^{2}+{a}^{2}\left({k}^{-1}+1+k\right)x-{a}^{3}={x}^{3}-p{x}^{2}+qx-r\phantom{\rule{2.77695pt}{0ex}}.$

Thus $p=a\left({k}^{-1}+1+k\right)\phantom{\rule{0.3em}{0ex}}$, $q={a}^{2}\left({k}^{-1}+1+k\right)\phantom{\rule{0.3em}{0ex}}$, and $r={a}^{3}\phantom{\rule{0.3em}{0ex}}$. Dividing gives $q∕p=a\phantom{\rule{0.3em}{0ex}}$, which is a root, and ${q}^{3}∕{p}^{3}={a}^{3}=r$ as required.

(ii) Set $r={q}^{3}∕{p}^{3}\phantom{\rule{0.3em}{0ex}}$. Substituting $x=q∕p$ into the cubic shows that it is a root:

${\left(q∕p\right)}^{3}-p{\left(q∕p\right)}^{2}+q\left(q∕p\right)-{\left(q∕p\right)}^{3}=0\phantom{\rule{2.77695pt}{0ex}}.$

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 $k\phantom{\rule{0.3em}{0ex}}$, 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 $\left(a-d\right)\phantom{\rule{0.3em}{0ex}}$, $a$ and $\left(a+d\right)\phantom{\rule{0.3em}{0ex}}$. (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

$\left(\rightx-\left(a-d\right)\left)\right\left(\rightx-a\left)\right\left(\rightx-\left(a+d\right)\left)\right\equiv {x}^{3}-p{x}^{2}+qx-r$

i.e.

${x}^{3}-3a{x}^{2}+\left(3{a}^{2}-{d}^{2}\right)x-a\left({a}^{2}-{d}^{2}\right)\equiv {x}^{3}-p{x}^{2}+qx-r\phantom{\rule{2.77695pt}{0ex}}.$

Thus $p=3a\phantom{\rule{0.3em}{0ex}}$, $q=3{a}^{2}-{d}^{2}\phantom{\rule{0.3em}{0ex}}$, $r=a\left({a}^{2}-{d}^{2}\right)\phantom{\rule{0.3em}{0ex}}$. A necessary condition is therefore $r=\left(p∕3\right)\left(q-2{p}^{2}∕9\right)$. Note that one root is $p∕3$.

Conversely, if $r=\left(p∕3\right)\left(q-2{p}^{2}∕9\right)\phantom{\rule{0.3em}{0ex}}$, the equation is

${x}^{3}-p{x}^{2}+qx-\left(p∕3\right)\left(q-2{p}^{2}∕9\right)=0\phantom{\rule{2.77695pt}{0ex}}.$

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 $r=\left(p∕3\right)\left(q-2{p}^{2}∕9\right)$.

Post-mortem

As usual, it is a good idea to give a bit of thought to the conditions given, namely $p\ne 0$ and $r\ne 0\phantom{\rule{0.3em}{0ex}}$.

Clearly, if the roots are in geometric progression, we cannot have a zero root. We therefore need $r\ne 0\phantom{\rule{0.3em}{0ex}}$. However, we don’t need the condition $p\ne 0\phantom{\rule{0.3em}{0ex}}$. If the roots are in geometric progression with $p=0$, then $q=0\phantom{\rule{0.3em}{0ex}}$, but there is no contradiction: the roots are $b{e}^{-2\pi i∕3}\phantom{\rule{0.3em}{0ex}}$, $b$ and $b{e}^{2\pi i∕3}\phantom{\rule{0.3em}{0ex}}$ where ${b}^{3}=r\phantom{\rule{0.3em}{0ex}}$, and these are certainly in geometric progression. So the necessary and sufficient conditions are $r\ne 0$ and ${p}^{3}r={q}^{3}\phantom{\rule{0.3em}{0ex}}$.

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.