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[SOLVED] Show that |q|>64

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MHB POTW Director
Staff member
Feb 14, 2012
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 $|q|>64$.


MHB Oldtimer
Staff member
Feb 7, 2012
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 $|q| > 2^6 = 64$.