Question Number 30220 by abdo imad last updated on 18/Feb/18
$${let}\:{p}\left({x}\right)=\:{x}^{\mathrm{3}} \:{px}\:+{q} \\ $$ $$\left.\mathrm{1}\right)\:{prove}\:{that}\:{p}\:{have}\:{double}\:{roots}\Leftrightarrow\:\mathrm{4}{p}^{\mathrm{3}} \:+\mathrm{27}{q}^{\mathrm{2}} =\mathrm{0} \\ $$ $$\left.\mathrm{3}\right)\:{let}\:{suppose}\:{p}\:{have}\:\mathrm{3}\:{real}\:{roots}\:{differnts}\:{prove}\:{that} \\ $$ $$\mathrm{4}{p}^{\mathrm{3}} \:+\mathrm{27}{q}^{\mathrm{2}} \:<\mathrm{0}. \\ $$
Commented byabdo imad last updated on 18/Feb/18
$${p}\left({x}\right)={x}^{\mathrm{3}} \:+{px}\:+{q}\: \\ $$