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Solve-x-3-15x-92-0-

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Question Number 110182 by udaythool last updated on 27/Aug/20
Solve x^3 +15x−92=0
$$\mathrm{Solve}\:{x}^{\mathrm{3}} +\mathrm{15}{x}−\mathrm{92}=\mathrm{0} \\ $$
Commented by mr W last updated on 27/Aug/20
let x=u+v (u+v)^3 +15(u+v)−92=0 u^3 +v^3 +3(uv+5)(u+v)−92=0 let uv+5=0 ⇒uv=−5 ⇒u^3 v^3 =−125 ⇒u^3 +v^3 −92=0 ⇒u^3 +v^3 =92 u^3 and v^3 are roots of z^2 −92z−125=0 ⇒z=46±(√(46^2 +125))=46±3(√(249)) ⇒u=((46+3(√(249))))^(1/3) ⇒v=((46−3(√(249))))^(1/3) ⇒x=((46+3(√(249))))^(1/3) +((46−3(√(249))))^(1/3) ≈3.4339
$${let}\:{x}={u}+{v} \\ $$$$\left({u}+{v}\right)^{\mathrm{3}} +\mathrm{15}\left({u}+{v}\right)−\mathrm{92}=\mathrm{0} \\ $$$${u}^{\mathrm{3}} +{v}^{\mathrm{3}} +\mathrm{3}\left({uv}+\mathrm{5}\right)\left({u}+{v}\right)−\mathrm{92}=\mathrm{0} \\ $$$${let}\:{uv}+\mathrm{5}=\mathrm{0}\:\Rightarrow{uv}=−\mathrm{5}\:\Rightarrow{u}^{\mathrm{3}} {v}^{\mathrm{3}} =−\mathrm{125} \\ $$$$\Rightarrow{u}^{\mathrm{3}} +{v}^{\mathrm{3}} −\mathrm{92}=\mathrm{0}\:\Rightarrow{u}^{\mathrm{3}} +{v}^{\mathrm{3}} =\mathrm{92} \\ $$$${u}^{\mathrm{3}} \:{and}\:{v}^{\mathrm{3}} \:{are}\:{roots}\:{of}\:{z}^{\mathrm{2}} −\mathrm{92}{z}−\mathrm{125}=\mathrm{0} \\ $$$$\Rightarrow{z}=\mathrm{46}\pm\sqrt{\mathrm{46}^{\mathrm{2}} +\mathrm{125}}=\mathrm{46}\pm\mathrm{3}\sqrt{\mathrm{249}} \\ $$$$\Rightarrow{u}=\sqrt[{\mathrm{3}}]{\mathrm{46}+\mathrm{3}\sqrt{\mathrm{249}}} \\ $$$$\Rightarrow{v}=\sqrt[{\mathrm{3}}]{\mathrm{46}−\mathrm{3}\sqrt{\mathrm{249}}} \\ $$$$\Rightarrow{x}=\sqrt[{\mathrm{3}}]{\mathrm{46}+\mathrm{3}\sqrt{\mathrm{249}}}+\sqrt[{\mathrm{3}}]{\mathrm{46}−\mathrm{3}\sqrt{\mathrm{249}}}\approx\mathrm{3}.\mathrm{4339} \\ $$
Commented by udaythool last updated on 28/Aug/20
��THANKS��
Answered by 1549442205PVT last updated on 28/Aug/20
Put x=(√5)y⇒5(√5)y^3 +15(√5)y−92=0 ⇔y^3 +3y−((92)/(5(√5)))=0 (1) Put y=z−(1/z)⇒y^3 =z^3 −(1/z^3 )−3(z−(1/z)) Replace into (1)we get z^3 −(1/z^3 )−((92(√5))/(25))=0⇔z^6 −((92(√5))/(25))z^3 −1=0 This is quadratic equation w.r.t “z^3 ” Δ=(((92(√5))/(25)))^2 +4=((8464)/(125))+4=((8964)/(125)) ⇒z^3 =((((92(√5))/(25))±((6(√(1245)))/(25)))/2)=((92(√5)±6(√(1245 )))/(50)) i)For z^3 =((92(√5)+6(√(1245 )))/(50)). Since (a−b)(a+b)=a^2 −b^2 ,we have ((92(√5)+6(√(1245 )))/(50))×((92(√5)−6(√(1245 )))/(50))=−1(∗) Hence,.By Vieta′s theorem we get z=^3 (√((92(√5)+6(√(1245 )))/(50))),−(1/z)=^3 (√((92(√5)−6(√(1245 )))/(50))) ⇒x=(√5)y=(√5)(z−(1/z))= (√5)(^3 (√((92(√5)+6(√(1245 )))/(50)))+^3 (√((92(√5)−6(√(1245 )))/(50)))) =^3 (√((2300+150(√(249 )))/(50)))+^3 (√((2300−150(√(249 )))/(50))) ==^3 (√(46+3(√(249))))+^3 (√(46−3(√(249)))) ii)For z^3 =^3 (√((92(√5)−6(√(1245 )))/(50))).From(∗) we infer z−(1/z) is the same value Thus,given equation x^3 +15x−92=0 has unique root x=^3 (√(46+3(√(249))))+^3 (√(46−3(√(249))))
$$\mathrm{Put}\:\mathrm{x}=\sqrt{\mathrm{5}}\mathrm{y}\Rightarrow\mathrm{5}\sqrt{\mathrm{5}}\mathrm{y}^{\mathrm{3}} +\mathrm{15}\sqrt{\mathrm{5}}\mathrm{y}−\mathrm{92}=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{y}^{\mathrm{3}} +\mathrm{3y}−\frac{\mathrm{92}}{\mathrm{5}\sqrt{\mathrm{5}}}=\mathrm{0}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Put}\:\mathrm{y}=\mathrm{z}−\frac{\mathrm{1}}{\mathrm{z}}\Rightarrow\mathrm{y}^{\mathrm{3}} =\mathrm{z}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{3}} }−\mathrm{3}\left(\mathrm{z}−\frac{\mathrm{1}}{\mathrm{z}}\right) \\ $$$$\mathrm{Replace}\:\mathrm{into}\:\left(\mathrm{1}\right)\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{z}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{3}} }−\frac{\mathrm{92}\sqrt{\mathrm{5}}}{\mathrm{25}}=\mathrm{0}\Leftrightarrow\mathrm{z}^{\mathrm{6}} −\frac{\mathrm{92}\sqrt{\mathrm{5}}}{\mathrm{25}}\mathrm{z}^{\mathrm{3}} −\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{This}\:\mathrm{is}\:\mathrm{quadratic}\:\mathrm{equation}\:\mathrm{w}.\mathrm{r}.\mathrm{t}\:“\mathrm{z}^{\mathrm{3}} '' \\ $$$$\Delta=\left(\frac{\mathrm{92}\sqrt{\mathrm{5}}}{\mathrm{25}}\right)^{\mathrm{2}} +\mathrm{4}=\frac{\mathrm{8464}}{\mathrm{125}}+\mathrm{4}=\frac{\mathrm{8964}}{\mathrm{125}} \\ $$$$\Rightarrow\mathrm{z}^{\mathrm{3}} =\frac{\frac{\mathrm{92}\sqrt{\mathrm{5}}}{\mathrm{25}}\pm\frac{\mathrm{6}\sqrt{\mathrm{1245}}}{\mathrm{25}}}{\mathrm{2}}=\frac{\mathrm{92}\sqrt{\mathrm{5}}\pm\mathrm{6}\sqrt{\mathrm{1245}\:}}{\mathrm{50}} \\ $$$$\left.\mathrm{i}\right)\mathrm{For}\:\mathrm{z}^{\mathrm{3}} =\frac{\mathrm{92}\sqrt{\mathrm{5}}+\mathrm{6}\sqrt{\mathrm{1245}\:}}{\mathrm{50}}. \\ $$$$\mathrm{Since}\:\left(\mathrm{a}−\mathrm{b}\right)\left(\mathrm{a}+\mathrm{b}\right)=\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} ,\mathrm{we}\:\mathrm{have} \\ $$$$\frac{\mathrm{92}\sqrt{\mathrm{5}}+\mathrm{6}\sqrt{\mathrm{1245}\:}}{\mathrm{50}}×\frac{\mathrm{92}\sqrt{\mathrm{5}}−\mathrm{6}\sqrt{\mathrm{1245}\:}}{\mathrm{50}}=−\mathrm{1}\left(\ast\right) \\ $$$$\mathrm{Hence},.\mathrm{By}\:\mathrm{Vieta}'\mathrm{s}\:\mathrm{theorem}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{z}=\:^{\mathrm{3}} \sqrt{\frac{\mathrm{92}\sqrt{\mathrm{5}}+\mathrm{6}\sqrt{\mathrm{1245}\:}}{\mathrm{50}}},−\frac{\mathrm{1}}{\mathrm{z}}=\:^{\mathrm{3}} \sqrt{\frac{\mathrm{92}\sqrt{\mathrm{5}}−\mathrm{6}\sqrt{\mathrm{1245}\:}}{\mathrm{50}}} \\ $$$$\Rightarrow\mathrm{x}=\sqrt{\mathrm{5}}\mathrm{y}=\sqrt{\mathrm{5}}\left(\mathrm{z}−\frac{\mathrm{1}}{\mathrm{z}}\right)= \\ $$$$\sqrt{\mathrm{5}}\left(\:^{\mathrm{3}} \sqrt{\frac{\mathrm{92}\sqrt{\mathrm{5}}+\mathrm{6}\sqrt{\mathrm{1245}\:}}{\mathrm{50}}}+\:^{\mathrm{3}} \sqrt{\frac{\mathrm{92}\sqrt{\mathrm{5}}−\mathrm{6}\sqrt{\mathrm{1245}\:}}{\mathrm{50}}}\right) \\ $$$$=\:^{\mathrm{3}} \sqrt{\frac{\mathrm{2300}+\mathrm{150}\sqrt{\mathrm{249}\:}}{\mathrm{50}}}+\:^{\mathrm{3}} \sqrt{\frac{\mathrm{2300}−\mathrm{150}\sqrt{\mathrm{249}\:}}{\mathrm{50}}} \\ $$$$==\:^{\mathrm{3}} \sqrt{\mathrm{46}+\mathrm{3}\sqrt{\mathrm{249}}}+\:^{\mathrm{3}} \sqrt{\mathrm{46}−\mathrm{3}\sqrt{\mathrm{249}}} \\ $$$$\left.\mathrm{ii}\right)\mathrm{For}\:\mathrm{z}^{\mathrm{3}} =\:^{\mathrm{3}} \sqrt{\frac{\mathrm{92}\sqrt{\mathrm{5}}−\mathrm{6}\sqrt{\mathrm{1245}\:}}{\mathrm{50}}}.\mathrm{From}\left(\ast\right) \\ $$$$\mathrm{we}\:\mathrm{infer}\:\mathrm{z}−\frac{\mathrm{1}}{\mathrm{z}}\:\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{value} \\ $$$$\mathrm{Thus},\mathrm{given}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{3}} +\mathrm{15x}−\mathrm{92}=\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{unique}\:\mathrm{root} \\ $$$$\boldsymbol{\mathrm{x}}=\:^{\mathrm{3}} \sqrt{\mathrm{46}+\mathrm{3}\sqrt{\mathrm{249}}}+\:^{\mathrm{3}} \sqrt{\mathrm{46}−\mathrm{3}\sqrt{\mathrm{249}}} \\ $$
Commented by udaythool last updated on 28/Aug/20
Really I enjoyed this approach. Thanks �� ����

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