Question Number 176327 by mathlove last updated on 16/Sep/22
$${p}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} +\sqrt{\mathrm{4}{x}^{\mathrm{2}} }\:\:\: \\ $$$${pq}=\mathrm{4}\:\:\:\:{and}\:\:{p}+{q}=\mathrm{2}\:\:\:{faind} \\ $$$${p}^{{a}_{\mathrm{2}} } +{q}^{{a}_{\mathrm{1}} } =? \\ $$