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Question Number 63474 by aliesam last updated on 04/Jul/19

let P(x)=x^2 +(1/2)x+b and Q(x)=x^2 +cx+d be to polynomials with real coefficient such that P(x) Q(x)=Q(P(x)) find all the real roots of P(Q(x))=0

$${let}\:{P}\left({x}\right)={x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{x}+{b} \\ $$$$ \\ $$$${and}\:{Q}\left({x}\right)={x}^{\mathrm{2}} +{cx}+{d} \\ $$$$ \\ $$$${be}\:{to}\:{polynomials}\:{with}\:{real}\:{coefficient}\:{such}\:{that} \\ $$$$ \\ $$$${P}\left({x}\right)\:{Q}\left({x}\right)={Q}\left({P}\left({x}\right)\right) \\ $$$$ \\ $$$${find}\:{all}\:{the}\:{real}\:{roots}\:{of}\:{P}\left({Q}\left({x}\right)\right)=\mathrm{0} \\ $$

Answered by MJS last updated on 04/Jul/19

P(x)Q(x)−Q(P(x))=0 ⇒ (4c−2)x^3 −(4b+2c−4d+1)x^2 +(4bc−4b−2c+2d)x−(4b^2 +4bc−4bd+4d)=0 { ((4c−2=0)),((4b+2c−4d+1=0)),((4bc−4b−2c+2d=0)),((4b^2 +4bc−4bd+4d=0)) :} ⇒ b=−(1/2)∧c=(1/2)∧d=0 P(x)=x^2 +(1/2)x−(1/2) Q(x)=x^2 +(1/2)x P(Q(x))=0 x^4 +x^3 +(3/4)x^2 +(1/4)x−(1/2)=0 (x+1)(x−(1/2))(x^2 +(1/2)x+1)=0 ⇒ x_1 =−1 x_2 =(1/2) x_(3, 4) =−(1/4)±((√(15))/4)i

$${P}\left({x}\right){Q}\left({x}\right)−{Q}\left({P}\left({x}\right)\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{4}{c}−\mathrm{2}\right){x}^{\mathrm{3}} −\left(\mathrm{4}{b}+\mathrm{2}{c}−\mathrm{4}{d}+\mathrm{1}\right){x}^{\mathrm{2}} +\left(\mathrm{4}{bc}−\mathrm{4}{b}−\mathrm{2}{c}+\mathrm{2}{d}\right){x}−\left(\mathrm{4}{b}^{\mathrm{2}} +\mathrm{4}{bc}−\mathrm{4}{bd}+\mathrm{4}{d}\right)=\mathrm{0} \\ $$$$\begin{cases}{\mathrm{4}{c}−\mathrm{2}=\mathrm{0}}\\{\mathrm{4}{b}+\mathrm{2}{c}−\mathrm{4}{d}+\mathrm{1}=\mathrm{0}}\\{\mathrm{4}{bc}−\mathrm{4}{b}−\mathrm{2}{c}+\mathrm{2}{d}=\mathrm{0}}\\{\mathrm{4}{b}^{\mathrm{2}} +\mathrm{4}{bc}−\mathrm{4}{bd}+\mathrm{4}{d}=\mathrm{0}}\end{cases} \\ $$$$\Rightarrow\:{b}=−\frac{\mathrm{1}}{\mathrm{2}}\wedge{c}=\frac{\mathrm{1}}{\mathrm{2}}\wedge{d}=\mathrm{0} \\ $$$${P}\left({x}\right)={x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{x}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${Q}\left({x}\right)={x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{x} \\ $$$${P}\left({Q}\left({x}\right)\right)=\mathrm{0} \\ $$$${x}^{\mathrm{4}} +{x}^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{4}}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}{x}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\left({x}+\mathrm{1}\right)\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${x}_{\mathrm{1}} =−\mathrm{1} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}_{\mathrm{3},\:\mathrm{4}} =−\frac{\mathrm{1}}{\mathrm{4}}\pm\frac{\sqrt{\mathrm{15}}}{\mathrm{4}}\mathrm{i} \\ $$

Commented by aliesam last updated on 04/Jul/19

god bless you

$${god}\:{bless}\:{you} \\ $$

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