Question Number 211815 by alcohol last updated on 21/Sep/24
$${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:{has}\:{roots}\:\alpha\:{and}\:\beta \\ $$$${and}\:\frac{\alpha}{\beta}=\frac{\lambda}{\mu}.\:{show}\:{that}\:\lambda\mu{b}^{\mathrm{2}} \:=\:{ac}\left(\lambda+\mu\right)^{\mathrm{2}} \\ $$
Answered by som(math1967) last updated on 22/Sep/24
$$\:{let}\:\alpha={k}\lambda\:,\:\beta={k}\mu \\ $$$$\:{k}\left(\lambda+\mu\right)=−\frac{{b}}{{a}} \\ $$$$\Rightarrow{k}^{\mathrm{2}} \left(\lambda+\mu\right)^{\mathrm{2}} =\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} } \\ $$$$\:{k}^{\mathrm{2}} =\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} \left(\lambda+\mu\right)^{\mathrm{2}} } \\ $$$$\:{again}\:{k}^{\mathrm{2}} \lambda\mu=\frac{{c}}{{a}} \\ $$$$\Rightarrow\:\frac{{b}^{\mathrm{2}} \lambda\mu}{{a}^{\mathrm{2}} \left(\lambda+\mu\right)^{\mathrm{2}} }=\frac{{c}}{{a}} \\ $$$$\therefore\lambda\mu{b}^{\mathrm{2}} ={ac}\left(\lambda+\mu\right)^{\mathrm{2}} \\ $$