Question Number 145113 by imjagoll last updated on 02/Jul/21
Answered by liberty last updated on 02/Jul/21
$$\Rightarrow\mathrm{5}\left({x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}\right)=\left(\mathrm{9}{x}^{\mathrm{2}} +\mathrm{9}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{5}{x}^{\mathrm{4}} −\mathrm{10}{x}^{\mathrm{3}} +\mathrm{15}{x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{5}= \\ $$$$\:\:\:\:\:\mathrm{9}{x}^{\mathrm{4}} −\mathrm{18}{x}^{\mathrm{3}} +\mathrm{18}{x}^{\mathrm{2}} −\mathrm{18}{x}+\mathrm{9} \\ $$$$\Rightarrow\mathrm{4}{x}^{\mathrm{4}} −\mathrm{8}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{4}=\:\mathrm{0} \\ $$$$\Rightarrow\left({x}−\mathrm{2}\right)\left(\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{2}\left({solution}\right) \\ $$$$\Rightarrow\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{2}{x}−\mathrm{1}\right)\left(\mathrm{2}{x}^{\mathrm{2}} +{x}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}}\left({solution}\right) \\ $$$$\Rightarrow\mathrm{2}{x}^{\mathrm{2}} +{x}+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow{x}\:=\:\frac{−\mathrm{1}\:\pm\:\sqrt{\mathrm{1}−\mathrm{16}}}{\mathrm{4}}\:=\:\frac{−\mathrm{1}\pm{i}\sqrt{\mathrm{15}}}{\mathrm{4}}\left({not}\:{solution}\right) \\ $$
Commented by MJS_new last updated on 02/Jul/21
![something went wrong. the answer is x=(1/2)∨x=2 [∨x=−(1/4)±((√(15))/4)i]](https://www.tinkutara.com/question/Q145124.png)
$$\mathrm{something}\:\mathrm{went}\:\mathrm{wrong}.\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}\vee{x}=\mathrm{2}\:\:\:\:\:\left[\vee{x}=−\frac{\mathrm{1}}{\mathrm{4}}\pm\frac{\sqrt{\mathrm{15}}}{\mathrm{4}}\mathrm{i}\right] \\ $$
Commented by liberty last updated on 02/Jul/21
$${yes}.\: \\ $$