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Question Number 63154 by Rio Michael last updated on 29/Jun/19
Given that α and β are the roots oc the equation ax^2 +bx+c=0 . Show that λμb^2 = ac(λ + μ)^2 , where (α/β)= (λ/μ).
$${Given}\:{that}\:\alpha\:{and}\:\beta\:{are}\:{the}\:{roots}\:{oc}\:{the}\:{equation}\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$$.\:{Show}\:{that}\:\:\:\lambda\mu{b}^{\mathrm{2}} =\:{ac}\left(\lambda\:+\:\mu\right)^{\mathrm{2}} ,\:{where}\:\frac{\alpha}{\beta}=\:\frac{\lambda}{\mu}. \\ $$
Commented by Prithwish sen last updated on 30/Jun/19
α+β=−(b/a) and αβ =(c/a) Now (α+β)^2 =((αβ(α+β)^2 )/(αβ)) = αβ ((α/β) +(β/α) +2 ) (b^2 /a^2 ) = (c/a) ((λ/μ) + (μ/λ) + 2 ) = (c/a) (((λ+μ)^2 )/(λμ)) λμb^2 =ac(λ+μ)^2 proved.
$$\alpha+\beta=−\frac{\mathrm{b}}{\mathrm{a}}\:\mathrm{and}\:\alpha\beta\:=\frac{\mathrm{c}}{\mathrm{a}} \\ $$$$\mathrm{Now} \\ $$$$\left(\alpha+\beta\right)^{\mathrm{2}} =\frac{\alpha\beta\left(\alpha+\beta\right)^{\mathrm{2}} }{\alpha\beta}\:=\:\alpha\beta\:\left(\frac{\alpha}{\beta}\:+\frac{\beta}{\alpha}\:+\mathrm{2}\:\right) \\ $$$$\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\:=\:\frac{\mathrm{c}}{\mathrm{a}}\:\left(\frac{\lambda}{\mu}\:+\:\frac{\mu}{\lambda}\:+\:\mathrm{2}\:\right)\:=\:\frac{\mathrm{c}}{\mathrm{a}}\:\frac{\left(\lambda+\mu\right)^{\mathrm{2}} }{\lambda\mu} \\ $$$$\lambda\mu\mathrm{b}^{\mathrm{2}} =\mathrm{ac}\left(\lambda+\mu\right)^{\mathrm{2}} \:\:\mathrm{proved}. \\ $$$$ \\ $$$$ \\ $$
Commented by Rio Michael last updated on 02/Jul/19
cool i think thats correct.
$${cool}\:{i}\:{think}\:{thats}\:{correct}. \\ $$
Commented by Rio Michael last updated on 02/Jul/19
please check my solution. α+β = −(b/(a )) and αβ=(c/a) −b= a(α +β) and b^2 = a^2 (α+β)^2 ..........(i) and αβ= (c/a)..........(ii) taking eqn (i) × (ii) ⇒ αβb^2 = (c/a)×a^2 (α+β)^2 αβ b^2 = ac(α +β) since (α/β) =(λ/μ) then α=λ and β= μ and substituting we get. λμb^2 = ac(λ + μ)^2 proved. please make more corrections.
$${please}\:{check}\:{my}\:{solution}. \\ $$$$ \\ $$$$\alpha+\beta\:=\:−\frac{{b}}{{a}\:\:}\:\:{and}\:\alpha\beta=\frac{{c}}{{a}} \\ $$$$−{b}=\:{a}\left(\alpha\:+\beta\right) \\ $$$${and}\:{b}^{\mathrm{2}} =\:{a}^{\mathrm{2}} \left(\alpha+\beta\right)^{\mathrm{2}} ……….\left({i}\right) \\ $$$$\:{and}\:\:\alpha\beta=\:\frac{{c}}{{a}}……….\left({ii}\right) \\ $$$${taking}\:{eqn}\:\left({i}\right)\:×\:\left({ii}\right) \\ $$$$\Rightarrow\:\alpha\beta{b}^{\mathrm{2}} =\:\frac{{c}}{{a}}×{a}^{\mathrm{2}} \left(\alpha+\beta\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\alpha\beta\:{b}^{\mathrm{2}} \:=\:{ac}\left(\alpha\:+\beta\right) \\ $$$${since}\:\:\frac{\alpha}{\beta}\:=\frac{\lambda}{\mu}\:\:{then}\:\:\alpha=\lambda\:{and}\:\beta=\:\mu\: \\ $$$${and}\:{substituting}\:{we}\:{get}. \\ $$$$\:\:\lambda\mu{b}^{\mathrm{2}} =\:{ac}\left(\lambda\:+\:\mu\right)^{\mathrm{2}} \:{proved}.\:\:\: \\ $$$$ \\ $$$${please}\:{make}\:{more}\:{corrections}. \\ $$$$ \\ $$$$ \\ $$
Commented by Prithwish sen last updated on 02/Jul/19
I think (α/λ) = (β/μ) = k (let) then αβ b^2 = ac(α+β)^2 ⇒ k^2 λμb^2 =k^2 ac(λ+μ)^2
$$\mathrm{I}\:\mathrm{think} \\ $$$$\frac{\alpha}{\lambda}\:=\:\frac{\beta}{\mu}\:=\:\mathrm{k}\:\left(\mathrm{let}\right) \\ $$$$\mathrm{then}\:\alpha\beta\:\mathrm{b}^{\mathrm{2}} =\:\mathrm{ac}\left(\alpha+\beta\right)^{\mathrm{2}} \:\Rightarrow\:\mathrm{k}^{\mathrm{2}} \lambda\mu\mathrm{b}^{\mathrm{2}} =\mathrm{k}^{\mathrm{2}} \mathrm{ac}\left(\lambda+\mu\right)^{\mathrm{2}} \\ $$$$ \\ $$
Commented by Rio Michael last updated on 02/Jul/19
thats also a good idea
$${thats}\:{also}\:{a}\:{good}\:{idea} \\ $$

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