Question Number 222856 by OmoloyeMichael last updated on 09/Jul/25
$$\boldsymbol{{find}}\:\boldsymbol{{the}}\:\boldsymbol{{possible}}\:\boldsymbol{{root}}\:\boldsymbol{{of}}\:\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{5}\boldsymbol{{x}}+\mathrm{6}=\mathrm{0} \\ $$$$\boldsymbol{{using}}\:\boldsymbol{{the}}\:\boldsymbol{{fixed}}\:\boldsymbol{{point}}\:\boldsymbol{{iteration}}\:\boldsymbol{{method}}? \\ $$
Answered by AntonCWX8 last updated on 09/Jul/25
$${x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}=\mathrm{0} \\ $$$${x}^{\mathrm{3}} =\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{6} \\ $$$${x}=\mathrm{2}+\frac{\mathrm{5}}{{x}}−\frac{\mathrm{6}}{{x}^{\mathrm{2}} } \\ $$$$ \\ $$$${Let}\:{x}_{\mathrm{0}} =\mathrm{0}.\mathrm{5} \\ $$$${x}_{\mathrm{1}} =\mathrm{2}+\frac{\mathrm{5}}{\mathrm{0}.\mathrm{5}}−\frac{\mathrm{6}}{\mathrm{0}.\mathrm{5}^{\mathrm{2}} }=−\mathrm{12} \\ $$$${x}_{\mathrm{2}} =\mathrm{2}+\frac{\mathrm{5}}{−\mathrm{12}}−\frac{\mathrm{6}}{\left(−\mathrm{12}\right)^{\mathrm{2}} }=\mathrm{1}.\mathrm{541667} \\ $$$${x}_{\mathrm{3}} =\mathrm{2}.\mathrm{718773} \\ $$$${Continue}\:{this}\:{process}\:{and}\:{you}'{ll}\:{get}\:{x}=\mathrm{3} \\ $$