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Question Number 200460 by faysal last updated on 19/Nov/23
Find the cardano′s solution of the equation 28x^3 −9x^2 +1=0
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{cardano}'\mathrm{s}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{equation}\:\mathrm{28x}^{\mathrm{3}} −\mathrm{9x}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$
Answered by Frix last updated on 19/Nov/23
28x^3 −9x^2 +1=0 x^3 −((9x^2 )/(28))+(1/(28))=0 Before using any other method we try all factors of ±(1/(28)) ⇒ x_1 =−(1/4). We now have (x+(1/4))(x^2 −((4x)/7)+(1/7))=0 ⇒ x_(2, 3) =(2/7)±((√3)/7)i Cardano means more work: 28x^3 −9x^2 +1=0 x^3 −((9x^2 )/(28))+(1/(28))=0 x=t+(3/(28)) t^3 −((27t)/(784))+((365)/(10976))=0 ⇒ p=−((27)/(784))∧q=((365)/(10976)) u=((−(q/2)+(√((q^2 /4)+(p^3 /(27))))))^(1/3) ∧v=((−(q/2)−(√((q^2 /4)+(p^3 /(27))))))^(1/3) t_1 =u+v t_2 =ωu+ω^2 v t_3 =ω^2 u+ωv ...
$$\mathrm{28}{x}^{\mathrm{3}} −\mathrm{9}{x}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$${x}^{\mathrm{3}} −\frac{\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{28}}+\frac{\mathrm{1}}{\mathrm{28}}=\mathrm{0} \\ $$$$\mathrm{Before}\:\mathrm{using}\:\mathrm{any}\:\mathrm{other}\:\mathrm{method}\:\mathrm{we}\:\mathrm{try}\:\mathrm{all} \\ $$$$\mathrm{factors}\:\mathrm{of}\:\pm\frac{\mathrm{1}}{\mathrm{28}}\:\Rightarrow\:{x}_{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{4}}.\:\mathrm{We}\:\mathrm{now}\:\mathrm{have} \\ $$$$\left({x}+\frac{\mathrm{1}}{\mathrm{4}}\right)\left({x}^{\mathrm{2}} −\frac{\mathrm{4}{x}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{7}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{x}_{\mathrm{2},\:\mathrm{3}} =\frac{\mathrm{2}}{\mathrm{7}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{7}}\mathrm{i} \\ $$$$\mathrm{Cardano}\:\mathrm{means}\:\mathrm{more}\:\mathrm{work}: \\ $$$$\mathrm{28}{x}^{\mathrm{3}} −\mathrm{9}{x}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$${x}^{\mathrm{3}} −\frac{\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{28}}+\frac{\mathrm{1}}{\mathrm{28}}=\mathrm{0} \\ $$$${x}={t}+\frac{\mathrm{3}}{\mathrm{28}} \\ $$$${t}^{\mathrm{3}} −\frac{\mathrm{27}{t}}{\mathrm{784}}+\frac{\mathrm{365}}{\mathrm{10976}}=\mathrm{0}\:\Rightarrow\:{p}=−\frac{\mathrm{27}}{\mathrm{784}}\wedge{q}=\frac{\mathrm{365}}{\mathrm{10976}} \\ $$$${u}=\sqrt[{\mathrm{3}}]{−\frac{{q}}{\mathrm{2}}+\sqrt{\frac{{q}^{\mathrm{2}} }{\mathrm{4}}+\frac{{p}^{\mathrm{3}} }{\mathrm{27}}}}\wedge{v}=\sqrt[{\mathrm{3}}]{−\frac{{q}}{\mathrm{2}}−\sqrt{\frac{{q}^{\mathrm{2}} }{\mathrm{4}}+\frac{{p}^{\mathrm{3}} }{\mathrm{27}}}} \\ $$$${t}_{\mathrm{1}} ={u}+{v} \\ $$$${t}_{\mathrm{2}} =\omega{u}+\omega^{\mathrm{2}} {v} \\ $$$${t}_{\mathrm{3}} =\omega^{\mathrm{2}} {u}+\omega{v} \\ $$$$… \\ $$

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