Question Number 154980 by mathdanisur last updated on 23/Sep/21
$$\mathrm{if}\:\:\boldsymbol{\mathrm{a}}\:\:\mathrm{and}\:\:\boldsymbol{\mathrm{b}}\:\:\mathrm{are}\:\mathrm{real}\:\mathrm{numbers} \\ $$$$\mathrm{determine}\:\mathrm{a}\:\mathrm{necessary}\:\mathrm{and}\:\mathrm{sufficient} \\ $$$$\mathrm{condition}\:\mathrm{so}\:\mathrm{as}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{x}^{\mathrm{2}} \:+\:\frac{\mathrm{a}\sqrt{\mathrm{b}}}{\mathrm{x}}\:=\:\mathrm{a}\:+\:\mathrm{b} \\ $$$$\mathrm{to}\:\mathrm{have}\:\mathrm{three}\:\mathrm{real}\:\mathrm{distinct}\:\mathrm{roots}. \\ $$
Answered by mr W last updated on 24/Sep/21
$${x}\neq\mathrm{0},\:{b}>\mathrm{0} \\ $$$${x}^{\mathrm{3}} −\left({a}+{b}\right){x}+{a}\sqrt{{b}}=\mathrm{0} \\ $$$${for}\:{three}\:{distinct}\:{roots}: \\ $$$$\frac{\left({a}\sqrt{{b}}\right)^{\mathrm{2}} }{\mathrm{4}}−\frac{\left({a}+{b}\right)^{\mathrm{3}} }{\mathrm{27}}<\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}\left({a}+{b}\right)^{\mathrm{3}} −\mathrm{27}{a}^{\mathrm{2}} {b}>\mathrm{0} \\ $$
Commented by mathdanisur last updated on 24/Sep/21
$$\mathrm{Thank}\:\mathrm{you}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:\mathrm{4}\:\mathrm{and}\:\mathrm{27} \\ $$$$\mathrm{please} \\ $$
Commented by mathdanisur last updated on 24/Sep/21
$$\mathrm{Thank}\:\mathrm{you}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$
Commented by mr W last updated on 24/Sep/21
$${i}\:{just}\:{applied}\:{the}\:{cardano}'{s}\:{formula} \\ $$$${for}\:{cubic}\:{equation}. \\ $$