Question Number 49118 by Rio Michael last updated on 03/Dec/18
$${Given}\:{that}\:{the}\:{equation}\:\:\mathrm{3}{x}^{\mathrm{2}} +{mx}+{n}=\mathrm{0}\:{has}\:{roots}\:\alpha\:+\:\frac{\mathrm{1}}{\beta}\:{and} \\ $$$$\beta\:+\:\frac{\mathrm{1}}{\alpha\:}\:{find}\:{the}\:{value}\:{of}\:\:{m}\:{and}\:{n} \\ $$$$ \\ $$
Answered by Kunal12588 last updated on 03/Dec/18
$$\mathrm{3}{x}^{\mathrm{2}} +{mx}+{n}=\mathrm{0} \\ $$$$\alpha+\frac{\mathrm{1}}{\beta}+\beta+\frac{\mathrm{1}}{\alpha}=−\frac{{m}}{\mathrm{3}} \\ $$$$\frac{\alpha^{\mathrm{2}} \beta+\alpha+\alpha\beta^{\mathrm{2}} +\beta}{\alpha\beta}=−\frac{{m}}{\mathrm{3}} \\ $$$$\boldsymbol{{m}}=−\mathrm{3}\frac{\left(\boldsymbol{\alpha}+\boldsymbol{\beta}\right)\left(\mathrm{1}+\boldsymbol{\alpha\beta}\right)}{\boldsymbol{\alpha\beta}} \\ $$$$\left(\alpha+\frac{\mathrm{1}}{\beta}\right)\left(\beta+\frac{\mathrm{1}}{\alpha}\right)=\frac{{n}}{\mathrm{3}} \\ $$$$\alpha\beta+\mathrm{1}+\mathrm{1}+\frac{\mathrm{1}}{\alpha\beta}=\frac{{n}}{\mathrm{3}} \\ $$$${n}=\mathrm{3}\frac{\alpha^{\mathrm{2}} \beta^{\mathrm{2}} +\mathrm{2}\alpha\beta+\mathrm{1}}{\alpha\beta} \\ $$$$\boldsymbol{{n}}=\frac{\mathrm{3}\left(\boldsymbol{\alpha\beta}+\mathrm{1}\right)^{\mathrm{2}} }{\boldsymbol{\alpha\beta}} \\ $$