Prove $|q|>64$: Roots of $x^{12}-pqx+p^2=0$

[anemone]

Let $p$ and $q$ be real parameters. One of the roots of the equation $x^{12}-pqx+p^2=0$ is greater than 2. Prove that $|q|>64$.

 

[Opalg]

Spoiler

Write the equation as a quadratic in $p$: $p^2 - qxp + x^{12} = 0$. For this to have a real solution for $p$, its discriminant $(qx)^2 - 4x^{12}$ must be nonnegative. So $q^2x^2 \geqslant 4x^{12}$. If $x>2$ then $x\ne0$ so we can divide by $x^2$ to get $q^2\geqslant 4x^{10} >4*2^{10} = 2^{12}$. Now take square roots to get $|q| > 2^6 = 64$.