Menu Close

Solve-the-equation-x-3-12x-2-6x-10-0-by-cardon-s-method-




Question Number 200351 by faysal last updated on 17/Nov/23
Solve the equation x^3 −12x^2 −6x−10=0   by cardon′s method
$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{3}} −\mathrm{12x}^{\mathrm{2}} −\mathrm{6x}−\mathrm{10}=\mathrm{0}\: \\ $$$$\mathrm{by}\:\mathrm{cardon}'\mathrm{s}\:\mathrm{method} \\ $$
Answered by Frix last updated on 17/Nov/23
The name is Cardano  x^3 −12x^2 −6x−10=0  x=t+4  t^3 −54t−162=0  t=3(2)^(1/3) (1+(2)^(1/3) )  x=4+3(2)^(1/3) (1+(2)^(1/3) )
$$\mathrm{The}\:\mathrm{name}\:\mathrm{is}\:{Cardano} \\ $$$${x}^{\mathrm{3}} −\mathrm{12}{x}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{10}=\mathrm{0} \\ $$$${x}={t}+\mathrm{4} \\ $$$${t}^{\mathrm{3}} −\mathrm{54}{t}−\mathrm{162}=\mathrm{0} \\ $$$${t}=\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{2}}\left(\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{2}}\right) \\ $$$${x}=\mathrm{4}+\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{2}}\left(\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{2}}\right) \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *