Question Number 91439 by Cynosure last updated on 30/Apr/20
$${solve}\:\mathrm{2}{x}^{\mathrm{99}} +\mathrm{3}{x}^{\mathrm{98}} +\mathrm{2}{x}^{\mathrm{97}} +\mathrm{3}{x}^{\mathrm{96}} +…..\mathrm{2}{x}+\mathrm{3}=\mathrm{0}\:{in}\:\mathbb{R} \\ $$
Commented by Cynosure last updated on 30/Apr/20
$${pls}\:{i}\:{need}\:{answer} \\ $$
Commented by Prithwish Sen 1 last updated on 30/Apr/20
$$\mathrm{The}\:\mathrm{only}\:\mathrm{real}\:\mathrm{soln}.\:\mathrm{for}\:\mathrm{this}\:\mathrm{equation}\:\mathrm{is}\:−\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Answered by Rasheed.Sindhi last updated on 30/Apr/20
$${x}^{\mathrm{98}} \left(\mathrm{2}{x}+\mathrm{3}\right)+{x}^{\mathrm{96}} \left(\mathrm{2}{x}+\mathrm{3}\right)+…+\left(\mathrm{2}{x}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\left(\mathrm{2}{x}+\mathrm{3}\right)\left({x}^{\mathrm{98}} +{x}^{\mathrm{96}} +…+\mathrm{1}\right)=\mathrm{0} \\ $$$$\bullet\:\mathrm{2}{x}+\mathrm{3}=\mathrm{0}\Rightarrow{x}=−\mathrm{3}/\mathrm{2} \\ $$$$\bullet{x}^{\mathrm{98}} +{x}^{\mathrm{96}} +…+\mathrm{1}=\mathrm{0} \\ $$$$……. \\ $$$$…. \\ $$
Commented by Cynosure last updated on 30/Apr/20
$${thanks} \\ $$