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Question Number 51691 by peter frank last updated on 29/Dec/18

Answered by afachri last updated on 29/Dec/18

suppose γ = U_1 ; β = U_2 ; α = U_3 𝛄 , 𝛃 = ((−a ± (√(a^2 − 4b )))/2) 𝛃 − 𝛄 = (((−a − (√(a^2 − 4b ))) − (−a+ (√(a^2 − 4b))))/2) 𝛃 − 𝛄 = −(√(a^2 − 4b )) or β − γ = +(√(a^2 − 4b )) in A.P : β − γ = α − β ±(√(a^2 − 4b )) = α − ((−a ± (√(a^2 − 4b )))/2) α = ±(√(a^2 − 4b )) + ((−a ± (√(a^2 − 4b )))/2) α = ((±2(√(a^2 − 4b )) −a ± (√(a^2 − 4b)))/2) α = ((−a ± 3(√(a^2 − 4b)))/2) the roots of y^2 + ay + (9b − 2a^2 ) = 0 y_(1 , 2) = ((−a ± (√( a^2 − 4(9b − 2a^2 ) )))/2) = ((−a ± (√(a^2 + 8a^2 − 36b )))/2) = ((−a ± (√(9a^2 − 36b )))/2) = ((−a ± (√(9(a^2 − 4b) )))/2) = ((−a ± 3(√((a^2 − 4b) )))/2) we get same product for 𝛂 and y_(1,2)

$$\mathrm{suppose}\:\gamma\:=\:{U}_{\mathrm{1}} \:;\:\beta\:=\:{U}_{\mathrm{2}} \:;\:\alpha\:=\:{U}_{\mathrm{3}} \\ $$$$\boldsymbol{\gamma}\:,\:\boldsymbol{\beta}\:\:=\:\:\frac{−{a}\:\pm\:\sqrt{{a}^{\mathrm{2}} −\:\mathrm{4}{b}\:}}{\mathrm{2}} \\ $$$$\boldsymbol{\beta}\:−\:\boldsymbol{\gamma}\:\:=\:\:\frac{\left(−{a}\:−\:\sqrt{{a}^{\mathrm{2}} −\:\mathrm{4}{b}\:}\right)\:−\:\left(−{a}+\:\sqrt{{a}^{\mathrm{2}} −\:\mathrm{4}{b}}\right)}{\mathrm{2}} \\ $$$$\boldsymbol{\beta}\:−\:\boldsymbol{\gamma}\:\:=\:\:−\sqrt{{a}^{\mathrm{2}} −\:\mathrm{4}{b}\:\:}\:\:\boldsymbol{\mathrm{or}}\:\:\:\:\beta\:−\:\gamma\:\:=\:+\sqrt{{a}^{\mathrm{2}} −\:\mathrm{4}{b}\:\:} \\ $$$$\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{A}}.\boldsymbol{\mathrm{P}}\:\::\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\beta\:−\:\gamma\:\:\:=\:\:\alpha\:−\:\beta \\ $$$$\pm\sqrt{{a}^{\mathrm{2}} −\:\mathrm{4}{b}\:}\:\:=\:\:\alpha\:−\:\frac{−{a}\:\pm\:\sqrt{{a}^{\mathrm{2}} −\:\mathrm{4}{b}\:}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\alpha\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\pm\sqrt{{a}^{\mathrm{2}} −\:\mathrm{4}{b}\:}\:\:+\:\:\frac{−{a}\:\pm\:\sqrt{{a}^{\mathrm{2}} −\:\mathrm{4}{b}\:}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\alpha\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\pm\mathrm{2}\sqrt{{a}^{\mathrm{2}} −\:\mathrm{4}{b}\:}\:\:−{a}\:\pm\:\sqrt{{a}^{\mathrm{2}} −\:\mathrm{4}{b}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\alpha\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{−{a}\:\pm\:\mathrm{3}\sqrt{{a}^{\mathrm{2}} −\:\mathrm{4}{b}}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{roots}}\:\boldsymbol{\mathrm{of}}\:{y}^{\mathrm{2}} +\:{ay}\:+\:\left(\mathrm{9}{b}\:−\:\mathrm{2}{a}^{\mathrm{2}} \right)\:=\:\mathrm{0} \\ $$$${y}_{\mathrm{1}\:,\:\mathrm{2}} \:\:=\:\:\frac{−{a}\:\pm\:\sqrt{\:{a}^{\mathrm{2}} −\:\mathrm{4}\left(\mathrm{9}{b}\:−\:\mathrm{2}{a}^{\mathrm{2}} \right)\:}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\frac{−{a}\:\pm\:\sqrt{{a}^{\mathrm{2}} +\:\mathrm{8}{a}^{\mathrm{2}} \:−\:\mathrm{36}{b}\:}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\frac{−{a}\:\pm\:\sqrt{\mathrm{9}{a}^{\mathrm{2}} \:−\:\mathrm{36}{b}\:}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\frac{−{a}\:\pm\:\sqrt{\mathrm{9}\left({a}^{\mathrm{2}} \:−\:\mathrm{4}{b}\right)\:}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\frac{−{a}\:\pm\:\mathrm{3}\sqrt{\left({a}^{\mathrm{2}} \:−\:\mathrm{4}{b}\right)\:}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{get}}\:\boldsymbol{\mathrm{same}}\:\boldsymbol{\mathrm{product}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\alpha}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{{y}}_{\mathrm{1},\mathrm{2}} \\ $$$$ \\ $$

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