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Find-the-value-of-x-by-using-Cardon-s-Method-1-x-2-x-1-3-x-2-1-

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Question Number 156579 by Mr.D.N. last updated on 12/Oct/21
Find the value of x by using Cardon′s Method. (1/x) + (2/(x+1))+ (3/(x+2)) = 1
$$\:\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{using}}\:\boldsymbol{\mathrm{Cardon}}'\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{Method}}. \\ $$$$\:\:\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\:+\:\frac{\mathrm{2}}{\boldsymbol{\mathrm{x}}+\mathrm{1}}+\:\frac{\mathrm{3}}{\boldsymbol{\mathrm{x}}+\mathrm{2}}\:=\:\mathrm{1} \\ $$
Answered by mr W last updated on 13/Oct/21
(((x+1)(x+2)+2x(x+2)+3x(x+1))/(x(x+1)(x+2)))=1 x^2 +3x+2+2x^2 +4x+3x^2 +3x=x^3 +3x^2 +2x x^3 −3x^2 −8x−2=0 x=t+1 t^3 +3t^2 +3t+1−3t^2 −6t−3−8t−8−2=0 t^3 −11t−12=0 t=((2(√(33)))/3) sin (−(1/3)sin^(−1) ((18(√(33)))/( 121))+((2kπ)/3)) ⇒x=1+((2(√(33)))/3) sin (−(1/3)sin^(−1) ((18(√(33)))/( 121))+((2kπ)/3)),k=0,1,2 cardano method can′t be applied!
$$\frac{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)+\mathrm{2}{x}\left({x}+\mathrm{2}\right)+\mathrm{3}{x}\left({x}+\mathrm{1}\right)}{{x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}+\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}={x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x} \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} −\mathrm{8}{x}−\mathrm{2}=\mathrm{0} \\ $$$${x}={t}+\mathrm{1} \\ $$$${t}^{\mathrm{3}} +\mathrm{3}{t}^{\mathrm{2}} +\mathrm{3}{t}+\mathrm{1}−\mathrm{3}{t}^{\mathrm{2}} −\mathrm{6}{t}−\mathrm{3}−\mathrm{8}{t}−\mathrm{8}−\mathrm{2}=\mathrm{0} \\ $$$${t}^{\mathrm{3}} −\mathrm{11}{t}−\mathrm{12}=\mathrm{0} \\ $$$${t}=\frac{\mathrm{2}\sqrt{\mathrm{33}}}{\mathrm{3}}\:\mathrm{sin}\:\left(−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{18}\sqrt{\mathrm{33}}}{\:\mathrm{121}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right) \\ $$$$\Rightarrow{x}=\mathrm{1}+\frac{\mathrm{2}\sqrt{\mathrm{33}}}{\mathrm{3}}\:\mathrm{sin}\:\left(−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{18}\sqrt{\mathrm{33}}}{\:\mathrm{121}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right),{k}=\mathrm{0},\mathrm{1},\mathrm{2} \\ $$$$ \\ $$$${cardano}\:{method}\:{can}'{t}\:{be}\:{applied}! \\ $$
Commented by mr W last updated on 14/Oct/21
how to solve t^3 −11t−12=0 ? see Q155945
$${how}\:{to}\:{solve}\:{t}^{\mathrm{3}} −\mathrm{11}{t}−\mathrm{12}=\mathrm{0}\:? \\ $$$${see}\:{Q}\mathrm{155945} \\ $$
Commented by Mr.D.N. last updated on 21/Oct/21
thank you mr.W
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{mr}.\mathrm{W} \\ $$

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