Question Number 224312 by behi834171 last updated on 01/Sep/25
Answered by Frix last updated on 01/Sep/25
$${x}_{\mathrm{1}} ={c}+\sqrt{{a}}+\sqrt{{b}}+\sqrt{{ab}} \\ $$$${x}_{\mathrm{2}} ={c}+\sqrt{{a}}−\sqrt{{b}}−\sqrt{{ab}} \\ $$$${x}_{\mathrm{3}} ={c}−\sqrt{{a}}+\sqrt{{b}}−\sqrt{{ab}} \\ $$$${x}_{\mathrm{4}} ={c}−\sqrt{{a}}−\sqrt{{b}}+\sqrt{{ab}} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{4}} +{Ax}^{\mathrm{3}} +{Bx}^{\mathrm{2}} +{Cx}+{D}=\mathrm{0} \\ $$$${A}=−\mathrm{4}{c} \\ $$$${B}=−\mathrm{2}\left({a}+{ab}−\mathrm{3}{c}^{\mathrm{2}} \right) \\ $$$${C}=−\mathrm{4}\left(\mathrm{2}{ab}−{abc}−{ac}−{bc}+{c}^{\mathrm{3}} \right) \\ $$$${D}=\left({a}−{b}\right)^{\mathrm{2}} −{ab}\left(\mathrm{2}\left({a}+{b}+{c}^{\mathrm{2}} −\mathrm{4}{c}\right)−{ab}\right)−\mathrm{2}{c}^{\mathrm{2}} \left({a}+{b}\right)+{c}^{\mathrm{4}} \\ $$$$\mathrm{In}\:\mathrm{the}\:\mathrm{given}\:\mathrm{case} \\ $$$${x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} −\mathrm{16}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{4}=\mathrm{0} \\ $$
Commented by behi834171 last updated on 02/Sep/25
