Show that p² + p - 6 is equal to q or r.

[anemone]

Let $p$, $q$ and $r$ be roots of polynomial \(\displaystyle x^3-9x+9=0\). Show that \(\displaystyle p^2+p-6\) is equal to $q$ or $r$.

 

[zzephod]

Let $p$, $q$ and $r$ be roots of polynomial \(\displaystyle x^3-9x+9=0\). Show that \(\displaystyle p^2+p-6\) is equal to $q$ or $r$.

It is sufficient to show that \(\displaystyle p^2+p-6\) is a root of \(\displaystyle x^3-9x+9\) different from $p$. We may as well also note that \(\displaystyle x^3-9x+9\) has three distinct real roots.

Well substitute the first into the second and we get:

$$(p^2+p-6)^3-9(p^2+p-6)+9=(p^3-9p+9)(p^3+3p^2-6p-17)$$

but as $p$ is a root of \(\displaystyle x^3-9x+9\) we have:

$$(p^2+p-6)^3-9(p^2+p-6)+9=(p^3-9p+9)(p^3+3p^2-6p-17)=0$$

That is $p^2+p-6$ is a root of the cubic, and as neither $\pm \sqrt{6}$ is a root of the cubic $p^2+p-6 \ne p$ hence $p^2+p-6$ must be equal to $q$ or $r$.

.