Question Number 63474 by aliesam last updated on 04/Jul/19 | ||
$${let}\:{P}\left({x}\right)={x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{x}+{b} \\ $$$$ \\ $$$${and}\:{Q}\left({x}\right)={x}^{\mathrm{2}} +{cx}+{d} \\ $$$$ \\ $$$${be}\:{to}\:{polynomials}\:{with}\:{real}\:{coefficient}\:{such}\:{that} \\ $$$$ \\ $$$${P}\left({x}\right)\:{Q}\left({x}\right)={Q}\left({P}\left({x}\right)\right) \\ $$$$ \\ $$$${find}\:{all}\:{the}\:{real}\:{roots}\:{of}\:{P}\left({Q}\left({x}\right)\right)=\mathrm{0} \\ $$ | ||
Answered by MJS last updated on 04/Jul/19 | ||
$${P}\left({x}\right){Q}\left({x}\right)−{Q}\left({P}\left({x}\right)\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{4}{c}−\mathrm{2}\right){x}^{\mathrm{3}} −\left(\mathrm{4}{b}+\mathrm{2}{c}−\mathrm{4}{d}+\mathrm{1}\right){x}^{\mathrm{2}} +\left(\mathrm{4}{bc}−\mathrm{4}{b}−\mathrm{2}{c}+\mathrm{2}{d}\right){x}−\left(\mathrm{4}{b}^{\mathrm{2}} +\mathrm{4}{bc}−\mathrm{4}{bd}+\mathrm{4}{d}\right)=\mathrm{0} \\ $$$$\begin{cases}{\mathrm{4}{c}−\mathrm{2}=\mathrm{0}}\\{\mathrm{4}{b}+\mathrm{2}{c}−\mathrm{4}{d}+\mathrm{1}=\mathrm{0}}\\{\mathrm{4}{bc}−\mathrm{4}{b}−\mathrm{2}{c}+\mathrm{2}{d}=\mathrm{0}}\\{\mathrm{4}{b}^{\mathrm{2}} +\mathrm{4}{bc}−\mathrm{4}{bd}+\mathrm{4}{d}=\mathrm{0}}\end{cases} \\ $$$$\Rightarrow\:{b}=−\frac{\mathrm{1}}{\mathrm{2}}\wedge{c}=\frac{\mathrm{1}}{\mathrm{2}}\wedge{d}=\mathrm{0} \\ $$$${P}\left({x}\right)={x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{x}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${Q}\left({x}\right)={x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{x} \\ $$$${P}\left({Q}\left({x}\right)\right)=\mathrm{0} \\ $$$${x}^{\mathrm{4}} +{x}^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{4}}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}{x}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\left({x}+\mathrm{1}\right)\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${x}_{\mathrm{1}} =−\mathrm{1} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}_{\mathrm{3},\:\mathrm{4}} =−\frac{\mathrm{1}}{\mathrm{4}}\pm\frac{\sqrt{\mathrm{15}}}{\mathrm{4}}\mathrm{i} \\ $$ | ||
Commented by aliesam last updated on 04/Jul/19 | ||
$${god}\:{bless}\:{you} \\ $$ | ||