Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 156579 by Mr.D.N. last updated on 12/Oct/21

$$\mathbf{Find}\mathbf{the}\mathbf{value}\mathbf{of}\mathbf{x}\mathbf{by}\mathbf{using}{\mathbf{Cardon}}^{\prime }\mathbf{s}\mathbf{Method}.$$$$\frac{1}{\mathbf{x}}+\frac{2}{\mathbf{x}+1}+\frac{3}{\mathbf{x}+2}=1$$

Answered by mr W last updated on 13/Oct/21

$$\frac{(x+1)(x+2)+2x(x+2)+3x(x+1)}{x(x+1)(x+2)}=1$$$$x^2+3x+2+2x^2+4x+3x^2+3x=x^3+3x^2+2x$$$$x^3-3x^2-8x-2=0$$$$x=t+1$$$$t^3+3t^2+3t+1-3t^2-6t-3-8t-8-2=0$$$$t^3-11t-12=0$$$$t=\frac{2\sqrt{33}}{3}\sin (-\frac{1}{3}{\sin }^{-1}\frac{18\sqrt{33}}{121}+\frac{2k\pi }{3})$$$$\Rightarrow x=1+\frac{2\sqrt{33}}{3}\sin (-\frac{1}{3}{\sin }^{-1}\frac{18\sqrt{33}}{121}+\frac{2k\pi }{3}),k=0,1,2$$$$$$$$cardanomethod{can}^{\prime }tbeapplied!$$

Commented by mr W last updated on 14/Oct/21

$$howtosolve{t}^3-11t-12=0?$$$$seeQ155945$$

Commented by Mr.D.N. last updated on 21/Oct/21

$$\mathrm{thank}\mathrm{you}\mathrm{mr}.\mathrm{W}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com