Question Number 50630 by peter frank last updated on 18/Dec/18
Answered by ajfour last updated on 18/Dec/18
$$\alpha+\beta\:=\:\mathrm{0}\:\:,\:\gamma+\delta\:=\:−{b}\:\:\:\:…\left({i}\right) \\ $$$$\alpha\beta+\left(\alpha+\beta\right)\gamma+\gamma\delta+\left(\alpha+\beta\right)\delta\:=\:{c} \\ $$$$\Rightarrow\:\:\alpha\beta+\gamma\delta\:=\:{c}\:\:\:\:\:\:\:\:\:\:\:….\left({ii}\right) \\ $$$$\alpha\beta.\gamma\delta\:=\:{e}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\left({iii}\right) \\ $$$${e}\left(\frac{\mathrm{1}}{\alpha}+\frac{\mathrm{1}}{\beta}+\frac{\mathrm{1}}{\gamma}+\frac{\mathrm{1}}{\delta}\right)\:=\:−{d} \\ $$$$\Rightarrow\:\:\frac{\gamma+\delta}{\gamma\delta}\:=\:\frac{−{d}}{{e}}\:\:\:\:\:\:\:\:\:\:\:\:….\left({iv}\right) \\ $$$$\Rightarrow\:\:\:\boldsymbol{\gamma\delta}\:=\:\frac{{be}}{{d}}\:\:\:,\:\:\:\:\boldsymbol{\alpha\beta}\:=\:\frac{{e}}{\gamma\delta}\:=\:\frac{{d}}{{b}} \\ $$$${using}\:{in}\:\left({ii}\right) \\ $$$$\:\:\:\:\:\:\frac{{d}}{{b}}+\frac{{be}}{{d}}\:=\:{c} \\ $$$$\Rightarrow\:\:\:\boldsymbol{{b}}^{\mathrm{2}} \boldsymbol{{e}}+\boldsymbol{{d}}^{\mathrm{2}} \:=\:\boldsymbol{{bcd}}\:\:. \\ $$
Commented by peter frank last updated on 20/Dec/18
$${thank}\:{you}\:{sir}…{please}\:{help} \\ $$$$\mathrm{50754} \\ $$