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Question Number 211815 by alcohol last updated on 21/Sep/24

ax^2 +bx+c=0 has roots α and β  and (α/β)=(λ/μ). show that λμb^2  = ac(λ+μ)^2

$${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:{has}\:{roots}\:\alpha\:{and}\:\beta \\ $$$${and}\:\frac{\alpha}{\beta}=\frac{\lambda}{\mu}.\:{show}\:{that}\:\lambda\mu{b}^{\mathrm{2}} \:=\:{ac}\left(\lambda+\mu\right)^{\mathrm{2}} \\ $$

Answered by som(math1967) last updated on 22/Sep/24

 let α=kλ , β=kμ   k(λ+μ)=−(b/a)  ⇒k^2 (λ+μ)^2 =(b^2 /a^2 )   k^2 =(b^2 /(a^2 (λ+μ)^2 ))   again k^2 λμ=(c/a)  ⇒ ((b^2 λμ)/(a^2 (λ+μ)^2 ))=(c/a)  ∴λμb^2 =ac(λ+μ)^2

$$\:{let}\:\alpha={k}\lambda\:,\:\beta={k}\mu \\ $$$$\:{k}\left(\lambda+\mu\right)=−\frac{{b}}{{a}} \\ $$$$\Rightarrow{k}^{\mathrm{2}} \left(\lambda+\mu\right)^{\mathrm{2}} =\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} } \\ $$$$\:{k}^{\mathrm{2}} =\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} \left(\lambda+\mu\right)^{\mathrm{2}} } \\ $$$$\:{again}\:{k}^{\mathrm{2}} \lambda\mu=\frac{{c}}{{a}} \\ $$$$\Rightarrow\:\frac{{b}^{\mathrm{2}} \lambda\mu}{{a}^{\mathrm{2}} \left(\lambda+\mu\right)^{\mathrm{2}} }=\frac{{c}}{{a}} \\ $$$$\therefore\lambda\mu{b}^{\mathrm{2}} ={ac}\left(\lambda+\mu\right)^{\mathrm{2}} \\ $$

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