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Question-50630




Question Number 50630 by peter frank last updated on 18/Dec/18
Answered by ajfour last updated on 18/Dec/18
α+β = 0  , γ+δ = −b    ...(i)  αβ+(α+β)γ+γδ+(α+β)δ = c  ⇒  αβ+γδ = c           ....(ii)  αβ.γδ = e                     ....(iii)  e((1/α)+(1/β)+(1/γ)+(1/δ)) = −d  ⇒  ((γ+δ)/(γδ)) = ((−d)/e)            ....(iv)  ⇒   𝛄𝛅 = ((be)/d)   ,    𝛂𝛃 = (e/(γδ)) = (d/b)  using in (ii)        (d/b)+((be)/d) = c  ⇒   b^2 e+d^2  = bcd  .
$$\alpha+\beta\:=\:\mathrm{0}\:\:,\:\gamma+\delta\:=\:−{b}\:\:\:\:…\left({i}\right) \\ $$$$\alpha\beta+\left(\alpha+\beta\right)\gamma+\gamma\delta+\left(\alpha+\beta\right)\delta\:=\:{c} \\ $$$$\Rightarrow\:\:\alpha\beta+\gamma\delta\:=\:{c}\:\:\:\:\:\:\:\:\:\:\:….\left({ii}\right) \\ $$$$\alpha\beta.\gamma\delta\:=\:{e}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\left({iii}\right) \\ $$$${e}\left(\frac{\mathrm{1}}{\alpha}+\frac{\mathrm{1}}{\beta}+\frac{\mathrm{1}}{\gamma}+\frac{\mathrm{1}}{\delta}\right)\:=\:−{d} \\ $$$$\Rightarrow\:\:\frac{\gamma+\delta}{\gamma\delta}\:=\:\frac{−{d}}{{e}}\:\:\:\:\:\:\:\:\:\:\:\:….\left({iv}\right) \\ $$$$\Rightarrow\:\:\:\boldsymbol{\gamma\delta}\:=\:\frac{{be}}{{d}}\:\:\:,\:\:\:\:\boldsymbol{\alpha\beta}\:=\:\frac{{e}}{\gamma\delta}\:=\:\frac{{d}}{{b}} \\ $$$${using}\:{in}\:\left({ii}\right) \\ $$$$\:\:\:\:\:\:\frac{{d}}{{b}}+\frac{{be}}{{d}}\:=\:{c} \\ $$$$\Rightarrow\:\:\:\boldsymbol{{b}}^{\mathrm{2}} \boldsymbol{{e}}+\boldsymbol{{d}}^{\mathrm{2}} \:=\:\boldsymbol{{bcd}}\:\:. \\ $$
Commented by peter frank last updated on 20/Dec/18
thank you sir...please help  50754
$${thank}\:{you}\:{sir}…{please}\:{help} \\ $$$$\mathrm{50754} \\ $$

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