Question Number 75623 by liki last updated on 14/Dec/19
Commented by liki last updated on 14/Dec/19
$$…\boldsymbol{{Please}}\:\boldsymbol{{anyone}}\:\boldsymbol{{to}}\:\boldsymbol{{help}}\:\boldsymbol{{me}}\:\boldsymbol{{this}}\:\boldsymbol{{Qn}}\:! \\ $$
Answered by $@ty@m123 last updated on 14/Dec/19
$${Equation}\:{whose}\:{roots}\:{are}\:\:\alpha,\beta\:\&\:\gamma\:{is} \\ $$$${x}^{\mathrm{3}} =\mathrm{19}{x}+\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{3}} +\mathrm{0}.{x}^{\mathrm{2}} +\left(−\mathrm{19}\right){x}+\left(−\mathrm{1}\right)=\mathrm{0}\:\:….\left(\mathrm{1}\right) \\ $$$$\therefore\:{Equation}\:{whose}\:{roots}\:{are}\:\:\frac{\mathrm{1}}{\alpha},\:\frac{\mathrm{1}}{\beta}\&\:\frac{\mathrm{1}}{\gamma}\:{is} \\ $$$$\:\left(−\mathrm{1}\right){x}^{\mathrm{3}} +\left(−\mathrm{19}\right){x}^{\mathrm{2}} +\mathrm{0}.{x}+\mathrm{1}=\mathrm{0}\:\left({Write}\:{the}\:{coefficients}\:{of}\:\left(\mathrm{1}\right)\:{in}\:{reverse}\:{order}\right) \\ $$$$\Rightarrow{x}^{\mathrm{3}} +\mathrm{19}{x}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$
Commented by liki last updated on 14/Dec/19
$$..\boldsymbol{{thank}}\:\boldsymbol{{you}}\:\boldsymbol{{so}}\:\boldsymbol{{much}}\:\boldsymbol{{sir}}. \\ $$