Question Number 67686 by Rio Michael last updated on 30/Aug/19
$${given}\:{that}\:{the}\:{roots}\:{of}\:{the}\:{equation}\:\:\mathrm{4}{x}^{\mathrm{2}} \:+\:\mathrm{6}{x}\:+\:\mathrm{9}\:=\mathrm{0}\:{are}\:\:\lambda\:{and}\:\delta\:\:{where}\: \\ $$$$\:\lambda\:=\:\left(\mathrm{1}\:+\:\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} \right)\:\:{and}\:\:\delta\:=\:\alpha^{\mathrm{3}} \:+\:\beta^{\mathrm{3}} \\ $$$${find}\:{an}\:{equation}\:{whose}\:{roots}\:{are}\: \\ $$$$\:\:\frac{\mathrm{1}}{\alpha\lambda}\:{and}\:\:\frac{\mathrm{1}}{\beta\delta} \\ $$
Commented by MJS last updated on 30/Aug/19
$$\mathrm{please}\:\mathrm{check}\:\mathrm{the}\:\mathrm{question} \\ $$$$\mathrm{I}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{only}\:\mathrm{approximately}… \\ $$$$\mathrm{what}'\mathrm{s}\:\mathrm{the}\:\mathrm{source}\:\mathrm{of}\:\mathrm{it}? \\ $$
Commented by Rio Michael last updated on 14/Sep/19
$${well},\:{from}\:{one}\:{of}\:{our}\:{Country}'{s}\:{General}\:{exam}\:{question} \\ $$$${i}'{ll}\:{try}\:{it}\:{again}\:{bro} \\ $$