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Question Number 56368 by Hassen_Timol last updated on 15/Mar/19

What is the common formula to obtain the 3 solutions of a polynomial equation in the following form ? ax^3 + b^2 x + cx + d = 0

$${What}\:{is}\:{the}\:{common}\:{formula}\:{to}\:{obtain} \\ $$$${the}\:\mathrm{3}\:{solutions}\:{of}\:{a}\:{polynomial}\:{equation} \\ $$$${in}\:{the}\:{following}\:{form}\:? \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{ax}^{\mathrm{3}} \:+\:{b}^{\mathrm{2}} {x}\:+\:{cx}\:+\:{d}\:=\:\mathrm{0} \\ $$

Commented by mr W last updated on 15/Mar/19

Commented by Hassen_Timol last updated on 16/Mar/19

Thank you so much!!!!!

$${Thank}\:{you}\:{so}\:{much}!!!!! \\ $$

Answered by MJS last updated on 15/Mar/19

1. divide by a x^3 +(b/a)x^2 +(c/a)x+(d/a)=0 2. substitute x=z−(b/(3a)) z^3 +((3ac−b^2 )/(3a^2 ))z+((27a^2 d−9abc+2b^3 )/(27a^3 ))=0 3. substitute ((3ac−b^2 )/(3a^2 ))=p; ((27a^2 d−9abc+2b^3 )/(27a^3 ))=q z^3 +pz+q=0 4. check resolvability D=(p^3 /(27))+(q^2 /4) D≥0 ⇒ Cardano′s formula D<0 ⇒ trigonometric formula 5. solve Cardano: u=((−(q/2)+(√((p^3 /(27))+(q^2 /4)))))^(1/3) ; v=((−(q/2)−(√((p^3 /(27))+(q^2 /4)))))^(1/3) it′s essential to take the principal 3^(rd) root z_1 =u+v z_2 =(−(1/2)−((√3)/2)i)u+(−(1/2)+((√3)/2)i)v z_3 =(−(1/2)+((√3)/2)i)u+(−(1/2)−((√3)/2)i)v trigonometric: z_n =(2/3)(√(−3p)) sin (((2π)/3)n+(1/3)arcsin ((3(√3)q)/(2(√(−p^3 ))))) with n=1, 2, 3

$$\mathrm{1}.\:\mathrm{divide}\:\mathrm{by}\:{a} \\ $$$${x}^{\mathrm{3}} +\frac{{b}}{{a}}{x}^{\mathrm{2}} +\frac{{c}}{{a}}{x}+\frac{{d}}{{a}}=\mathrm{0} \\ $$$$\mathrm{2}.\:\mathrm{substitute}\:{x}={z}−\frac{{b}}{\mathrm{3}{a}} \\ $$$${z}^{\mathrm{3}} +\frac{\mathrm{3}{ac}−{b}^{\mathrm{2}} }{\mathrm{3}{a}^{\mathrm{2}} }{z}+\frac{\mathrm{27}{a}^{\mathrm{2}} {d}−\mathrm{9}{abc}+\mathrm{2}{b}^{\mathrm{3}} }{\mathrm{27}{a}^{\mathrm{3}} }=\mathrm{0} \\ $$$$\mathrm{3}.\:\mathrm{substitute}\:\frac{\mathrm{3}{ac}−{b}^{\mathrm{2}} }{\mathrm{3}{a}^{\mathrm{2}} }={p};\:\frac{\mathrm{27}{a}^{\mathrm{2}} {d}−\mathrm{9}{abc}+\mathrm{2}{b}^{\mathrm{3}} }{\mathrm{27}{a}^{\mathrm{3}} }={q} \\ $$$${z}^{\mathrm{3}} +{pz}+{q}=\mathrm{0} \\ $$$$\mathrm{4}.\:\mathrm{check}\:\mathrm{resolvability} \\ $$$${D}=\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${D}\geqslant\mathrm{0}\:\Rightarrow\:\mathrm{Cardano}'\mathrm{s}\:\mathrm{formula} \\ $$$${D}<\mathrm{0}\:\Rightarrow\:\mathrm{trigonometric}\:\mathrm{formula} \\ $$$$\mathrm{5}.\:\mathrm{solve} \\ $$$$\mathrm{Cardano}: \\ $$$${u}=\sqrt[{\mathrm{3}}]{−\frac{{q}}{\mathrm{2}}+\sqrt{\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}}};\:{v}=\sqrt[{\mathrm{3}}]{−\frac{{q}}{\mathrm{2}}−\sqrt{\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}}} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{essential}\:\mathrm{to}\:\mathrm{take}\:\mathrm{the}\:\mathrm{principal}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{root} \\ $$$${z}_{\mathrm{1}} ={u}+{v} \\ $$$${z}_{\mathrm{2}} =\left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){u}+\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){v} \\ $$$${z}_{\mathrm{3}} =\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){u}+\left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){v} \\ $$$$\mathrm{trigonometric}: \\ $$$${z}_{{n}} =\frac{\mathrm{2}}{\mathrm{3}}\sqrt{−\mathrm{3}{p}}\:\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}{n}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{3}\sqrt{\mathrm{3}}{q}}{\mathrm{2}\sqrt{−{p}^{\mathrm{3}} }}\right)\:\mathrm{with}\:{n}=\mathrm{1},\:\mathrm{2},\:\mathrm{3} \\ $$$$ \\ $$

Commented by Hassen_Timol last updated on 16/Mar/19

Thanks a lot...

$${Thanks}\:{a}\:{lot}... \\ $$

Commented by mr W last updated on 16/Mar/19

very nice summary! thanks sir!

$${very}\:{nice}\:{summary}!\:{thanks}\:{sir}! \\ $$

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