Question Number 217191 by dabiswa1986 last updated on 05/Mar/25
$$\:\frac{{a}}{{b}}+\frac{{b}}{{a}}=\mathrm{1} \\ $$
Answered by ArshadS last updated on 06/Mar/25
$$\:\frac{{a}}{{b}}+\frac{{b}}{{a}}=\mathrm{1} \\ $$$$\Rightarrow{a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} =\mathrm{0} \\ $$$${a}=\frac{{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{b}^{\mathrm{2}} }}{\mathrm{2}}=\frac{{b}\pm\boldsymbol{{i}}{b}\sqrt{\mathrm{3}}\:}{\mathrm{2}}=\frac{{b}}{\mathrm{2}}\left(\mathrm{1}\pm\boldsymbol{{i}}\sqrt{\mathrm{3}}\:\right) \\ $$$$ \\ $$
Commented by profcedricjunior last updated on 05/Mar/25
$$\boldsymbol{{comment}}!!!\sqrt{\boldsymbol{{b}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{{b}}^{\mathrm{2}} \:}=\boldsymbol{{i}}\mid\boldsymbol{{b}}\mid\sqrt{\mathrm{3}} \\ $$
Commented by ArshadS last updated on 05/Mar/25
$${You}'{re}\:{right}\:{sir}! \\ $$
Commented by mr W last updated on 06/Mar/25
$${since}\:{it}\:{is}\:\pm{i}\sqrt{\mathrm{3}}{b},\:{you}\:{don}'{t}\:{need}\:{to} \\ $$$${change}\:{to}\:\pm{i}\sqrt{\mathrm{3}}\mid{b}\mid.\:\pm{b}\:{and}\:\pm\mid{b}\mid\:{are}\: \\ $$$${the}\:{same}.\:{so} \\ $$$${a}=\frac{{b}\pm{i}\sqrt{\mathrm{3}}\:{b}}{\mathrm{2}}\:{was}\:{absolutely}\:{correct}. \\ $$
Commented by ArshadS last updated on 06/Mar/25
$${Nice}\:{sir}!\:{I}'{m}\:{going}\:{to}\:{edit}\:{my}\:{answer}. \\ $$