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Question Number 39636 by math khazana by abdo last updated on 09/Jul/18

let P_α (x) =x^3 +2α x −3 1) determine the roots of P_α 2) determine the roots of P_(−1)

$${let}\:{P}_{\alpha} \left({x}\right)\:={x}^{\mathrm{3}} \:\:+\mathrm{2}\alpha\:{x}\:−\mathrm{3} \\ $$$$\left.\mathrm{1}\right)\:{determine}\:{the}\:{roots}\:{of}\:{P}_{\alpha} \\ $$$$\left.\mathrm{2}\right)\:{determine}\:{the}\:{roots}\:{of}\:\:{P}_{−\mathrm{1}} \\ $$

Answered by ajfour last updated on 09/Jul/18

let x= u+v ⇒ (u+v)^3 +2α(u+v)−3=0 ⇒ u^3 +v^3 +(u+v)(3uv+2α)−3=0 Further let (we can..) that 3uv+2α = 0 ⇒ u^3 v^3 = −((8α^3 )/(27)) Then u^3 +v^3 = 3 u^3 , v^3 are then roots of eq. z^2 −3z−((8α^3 )/(27)) = 0 ⇒ u^3 , v^3 = ((3±(√(9+((32α^3 )/(27)))))/2) As x = u+v ⇒ x(α) = (((3/2)+(√((9/4)+((8α^3 )/(27))))))^(1/3) +(((3/2)−(√((9/4)+((8α^3 )/(27))))))^(1/3) x∣_(α=−1) = (((3/2)+(√((211)/(108)))))^(1/3) +(((3/2)−(√((211)/(108)))))^(1/3) .

$${let}\:\:{x}=\:{u}+{v} \\ $$$$\Rightarrow\:\:\left({u}+{v}\right)^{\mathrm{3}} +\mathrm{2}\alpha\left({u}+{v}\right)−\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow\:{u}^{\mathrm{3}} +{v}^{\mathrm{3}} +\left({u}+{v}\right)\left(\mathrm{3}{uv}+\mathrm{2}\alpha\right)−\mathrm{3}=\mathrm{0} \\ $$$$\:\:{Further}\:{let}\:\left({we}\:{can}..\right)\:{that} \\ $$$$\:\:\:\:\:\:\:\mathrm{3}{uv}+\mathrm{2}\alpha\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\:\:{u}^{\mathrm{3}} {v}^{\mathrm{3}} \:=\:−\frac{\mathrm{8}\alpha^{\mathrm{3}} }{\mathrm{27}}\:\:\:\: \\ $$$${Then}\:\:\:\:\:{u}^{\mathrm{3}} +{v}^{\mathrm{3}} \:=\:\mathrm{3} \\ $$$$\:\:\:\:{u}^{\mathrm{3}} ,\:{v}^{\mathrm{3}} \:{are}\:{then}\:{roots}\:{of}\:{eq}. \\ $$$$\:\:\:\:\:\:\:\:{z}^{\mathrm{2}} −\mathrm{3}{z}−\frac{\mathrm{8}\alpha^{\mathrm{3}} }{\mathrm{27}}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\:\:{u}^{\mathrm{3}} ,\:{v}^{\mathrm{3}} \:=\:\frac{\mathrm{3}\pm\sqrt{\mathrm{9}+\frac{\mathrm{32}\alpha^{\mathrm{3}} }{\mathrm{27}}}}{\mathrm{2}} \\ $$$${As}\:\:\:\:\:\:\:\:\:{x}\:=\:{u}+{v} \\ $$$$\Rightarrow\:\:{x}\left(\alpha\right)\:=\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{3}}{\mathrm{2}}+\sqrt{\frac{\mathrm{9}}{\mathrm{4}}+\frac{\mathrm{8}\alpha^{\mathrm{3}} }{\mathrm{27}}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\sqrt[{\mathrm{3}}]{\frac{\mathrm{3}}{\mathrm{2}}−\sqrt{\frac{\mathrm{9}}{\mathrm{4}}+\frac{\mathrm{8}\alpha^{\mathrm{3}} }{\mathrm{27}}}}\: \\ $$$$\:\:\:\:{x}\mid_{\alpha=−\mathrm{1}} \:=\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{3}}{\mathrm{2}}+\sqrt{\frac{\mathrm{211}}{\mathrm{108}}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\sqrt[{\mathrm{3}}]{\frac{\mathrm{3}}{\mathrm{2}}−\sqrt{\frac{\mathrm{211}}{\mathrm{108}}}}\:\:. \\ $$

Commented by MrW3 last updated on 09/Jul/18

Ajfour sir is now also a specalist for cubic equations, on side of MJS sir.

$${Ajfour}\:{sir}\:{is}\:{now}\:{also}\:{a}\:{specalist}\:{for} \\ $$$${cubic}\:{equations},\:{on}\:{side}\:{of}\:{MJS}\:{sir}. \\ $$

Commented by ajfour last updated on 09/Jul/18

You taught me how, sometime back..

$${You}\:{taught}\:{me}\:{how},\:{sometime}\:{back}.. \\ $$

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