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Question Number 202198 by MATHEMATICSAM last updated on 22/Dec/23
If α, β are the roots of x^2 + ax − b = 0 and γ, δ are the roots of x^2 + ax + c = 0 then show that ((α − γ)/(β − γ)) = ((β − δ)/(α − δ)) .
$$\mathrm{If}\:\alpha,\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{x}^{\mathrm{2}} \:+\:{ax}\:−\:{b}\:=\:\mathrm{0}\: \\ $$$$\mathrm{and}\:\gamma,\:\delta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{x}^{\mathrm{2}} \:+\:{ax}\:+\:{c}\:=\:\mathrm{0}\: \\ $$$$\mathrm{then}\:\mathrm{show}\:\mathrm{that}\:\frac{\alpha\:−\:\gamma}{\beta\:−\:\gamma}\:=\:\frac{\beta\:−\:\delta}{\alpha\:−\:\delta}\:\:. \\ $$
Answered by MM42 last updated on 22/Dec/23
((α−β)/(β−γ))=((β−α)/(α−δ))⇒β+α=γ+δ s_1 =s_2 ⇒α+β=γ+δ ⇒α−γ=δ−β & β−γ=δ−α ⇒((α−γ)/(β−γ))=((β−δ)/(α−δ)) ✓
$$\frac{\alpha−\beta}{\beta−\gamma}=\frac{\beta−\alpha}{\alpha−\delta}\Rightarrow\beta+\alpha=\gamma+\delta \\ $$$${s}_{\mathrm{1}} ={s}_{\mathrm{2}} \Rightarrow\alpha+\beta=\gamma+\delta \\ $$$$\Rightarrow\alpha−\gamma=\delta−\beta\:\:\&\:\:\beta−\gamma=\delta−\alpha \\ $$$$\Rightarrow\frac{\alpha−\gamma}{\beta−\gamma}=\frac{\beta−\delta}{\alpha−\delta}\:\:\:\checkmark \\ $$$$\: \\ $$
Answered by Rasheed.Sindhi last updated on 23/Dec/23
((α − γ)/(β − γ)) = ((β − δ)/(α − δ)) ⇔(((α − γ)+(β − γ))/((α − γ)−(β − γ)))=(((β − δ)+(α − δ))/((β − δ)−(α − δ))) [Componendo-Dividendo] ⇔((α+β−2γ)/(α−β))=((α+β−2δ)/(−(α−β))) ⇔α+β−2γ=−(α+β)+2δ ⇔2(α+β)=2(γ+δ) ⇔α+β=γ+δ ⇔−a=−a
$$\:\frac{\alpha\:−\:\gamma}{\beta\:−\:\gamma}\:=\:\frac{\beta\:−\:\delta}{\alpha\:−\:\delta} \\ $$$$\Leftrightarrow\frac{\left(\alpha\:−\:\gamma\right)+\left(\beta\:−\:\gamma\right)}{\left(\alpha\:−\:\gamma\right)−\left(\beta\:−\:\gamma\right)}=\frac{\left(\beta\:−\:\delta\right)+\left(\alpha\:−\:\delta\right)}{\left(\beta\:−\:\delta\right)−\left(\alpha\:−\:\delta\right)}\:\left[{Componendo}-{Dividendo}\right] \\ $$$$\Leftrightarrow\frac{\alpha+\beta−\mathrm{2}\gamma}{\alpha−\beta}=\frac{\alpha+\beta−\mathrm{2}\delta}{−\left(\alpha−\beta\right)} \\ $$$$\Leftrightarrow\alpha+\beta−\mathrm{2}\gamma=−\left(\alpha+\beta\right)+\mathrm{2}\delta \\ $$$$\Leftrightarrow\mathrm{2}\left(\alpha+\beta\right)=\mathrm{2}\left(\gamma+\delta\right) \\ $$$$\Leftrightarrow\alpha+\beta=\gamma+\delta \\ $$$$\Leftrightarrow−{a}=−{a} \\ $$

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