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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)) .

Ifα,βaretherootsofx2+ax−b=0andγ,δaretherootsofx2+ax+c=0thenshowthatα−γβ−γ=β−δα−δ.

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)) ✓

α−ββ−γ=β−αα−δ⇒β+α=γ+δs1=s2⇒α+β=γ+δ⇒α−γ=δ−β&β−γ=δ−α⇒α−γβ−γ=β−δα−δ✓

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

α−γβ−γ=β−δα−δ⇔(α−γ)+(β−γ)(α−γ)−(β−γ)=(β−δ)+(α−δ)(β−δ)−(α−δ)[Componendo−Dividendo]⇔α+β−2γα−β=α+β−2δ−(α−β)⇔α+β−2γ=−(α+β)+2δ⇔2(α+β)=2(γ+δ)⇔α+β=γ+δ⇔−a=−a

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