Review question

# Can we show $x^4 - px^3 - 6x^2 + px +1 =0$ always has four real roots? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9189

## Solution

If one root of the equation $x^4 - px^3 - 6x^2 + px +1 =0$ is $\alpha$, prove that the others are $-\frac{1}{\alpha},\frac{\alpha-1}{\alpha+1},\frac{1+\alpha}{1-\alpha}.$

Let’s say $x^4 - px^3 - 6x^2 + px +1 =f(x).$ If $\alpha$ is a root of the equation $f(x) = 0$, then $\alpha^4 - p\alpha^3 - 6\alpha^2 + p\alpha +1 =0 \implies p = \dfrac{\alpha^4 - 6\alpha^2 + 1}{\alpha(\alpha^2 - 1)}.$

If the four stated terms are roots of the equation, we should be able to factorise it accordingly. Hence the given equation has to be equivalent to

\begin{align*} \left(x - \alpha \right)\left(x + \frac{1}{\alpha}\right)\left(x + \frac{1-\alpha}{1+\alpha} \right)\left(x + \frac{\alpha+1}{\alpha-1} \right) = 0. \end{align*} Expanding the expression leads to the equation \begin{align*} x^4 - \dfrac{\alpha^4 - 6\alpha^2 + 1}{\alpha(\alpha^2 - 1)}x^3 -6x^2 + \dfrac{\alpha^4 - 6\alpha^2 + 1}{\alpha(\alpha^2 - 1)}x + 1 = 0, \end{align*}

which is the same equation that we were given at the start, with $p = \dfrac{\alpha^4 - 6\alpha^2 + 1}{\alpha(\alpha^2 - 1)}$, and hence the stated terms are indeed the four roots.

Hence or otherwise show that for all real values of $p$ the equation has four real roots.

All we have to show is that $f(x)=0$ has a real solution $\alpha$ for any given $p$ that is not equal to zero or $\pm 1$. The other three roots are then $-\frac{1}{\alpha},\frac{\alpha-1}{\alpha+1},\frac{1+\alpha}{1-\alpha}.$

We have $f(0) = 1,$ and $f(1)=-4.$ Since $f(x)$ is continuous between $0$ and $1$, it must have a real root, say $\alpha$, between these values, and we are done.

There is a profounder way to look at this question. Define a recurrence relation by $u_1 = x, u_n = \dfrac{u_n+1}{1-u_n}.$

Thus $u_2 = \dfrac{x+1}{1-x}, u_3 = \dfrac{\dfrac{x+1}{1-x}+1}{1-\dfrac{x+1}{1-x}} = -\dfrac{1}{x}, u_4 = \dfrac{-\dfrac{1}{x}+1}{1+\dfrac{1}{x}} = \dfrac{x-1}{x+1}, u_5 = \dfrac{\dfrac{x-1}{1+x}+1}{1-\dfrac{x-1}{1+x}}=x.$

So the sequence is $x, \dfrac{x+1}{1-x}, -\dfrac{1}{x},\dfrac{x-1}{x+1}, x, \ldots$ and the recurrence relation is periodic with period $4$.

Recurrences like this are called Lyness cycles, after Robert Cranston Lyness.

Consider the equation $x+ \dfrac{x+1}{1-x} -\dfrac{1}{x}+\dfrac{x-1}{x+1}=p.$

If we replace $x$ with $\dfrac{x+1}{1-x},$ every term cycles on on, and we get $\dfrac{x+1}{1-x} -\dfrac{1}{x}+\dfrac{x-1}{x+1}+x=p;$ in other words, the equation remains invariant.

What is this equation? If we put everything over a common denominator, we get $\dfrac{x^4 - px^3 - 6x^2 + px + 1}{x(x + 1)(x - 1)}=0;$

in other words, we arrive back at the equation we were given to solve at the start.

Thus if $\alpha$ is one solution for the equation that is not $0$ or $\pm1,$ we have the other three immediately as $\dfrac{\alpha+1}{1-\alpha}, -\dfrac{1}{\alpha}, \dfrac{\alpha-1}{\alpha+1}.$