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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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