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Let-P-x-x-2-x-2-b-and-Q-x-x-2-cx-d-be-two-polynomial-with-real-coefficients-such-that-P-x-Q-x-Q-P-x-for-all-real-x-Find-all-the-real-roots-of-P-Q-x-0-

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Question Number 194953 by dimentri last updated on 20/Jul/23
Let P(x)= x^2 +(x/2)+b and Q(x)=x^2 +cx+d be two polynomial with real coefficients such that P(x)Q(x)= Q(P(x)) for all real x . Find all the real roots of P(Q(x))=0
$$\:{Let}\:{P}\left({x}\right)=\:{x}^{\mathrm{2}} +\frac{{x}}{\mathrm{2}}+{b}\:{and} \\ $$$$\:\:{Q}\left({x}\right)={x}^{\mathrm{2}} +{cx}+{d}\:{be}\:{two}\: \\ $$$$\:\:{polynomial}\:{with}\:{real}\:{coefficients} \\ $$$$\:\:{such}\:{that}\:{P}\left({x}\right){Q}\left({x}\right)=\:{Q}\left({P}\left({x}\right)\right) \\ $$$$\:{for}\:{all}\:{real}\:{x}\:. \\ $$$$\:\:{Find}\:{all}\:{the}\:{real}\:{roots}\:{of}\: \\ $$$$\:\:{P}\left({Q}\left({x}\right)\right)=\mathrm{0}\: \\ $$
Answered by Rasheed.Sindhi last updated on 21/Jul/23
P(x)= x^2 +(x/2)+b , Q(x)=x^2 +cx+d P(x)Q(x)= Q(P(x)) Real roots of P(Q(x))=0 ? P(x)Q(x)= Q(P(x)) lhs: P(x)Q(x)=(x^2 +(x/2)+b)(x^2 +cx+d) =x^4 +cx^3 +dx^2 +(1/2)x^3 +(c/2)x^2 +(d/2)x +bx^2 +bcx+bd =x^4 +(c+(1/2))x^3 +(b+d+(c/2))x^2 +(bc+(d/2))x+bd rhs: Q(P(x)) Q(P(x))=(x^2 +(x/2)+b)^2 +c(x^2 +(x/2)+b)+d =x^4 +(1/4)x^2 +b^2 +x^3 +bx+2bx^2 +cx^2 +(c/2)x+bc+d =x^4 +x^3 +(1/4)x^2 +2bx^2 +cx^2 +bx+(c/2)x+b^2 +bc+d =x^4 +x^3 +((1/4)+2b+c)x^2 +(b+(c/2))x+b^2 +bc+d • c+(1/2)=1 c=(1/2) •b+d+(c/2)=(1/4)+2b+c b+d+(1/4)=(1/4)+2b+(1/2) d=((2b+1)/2) •bc+(d/2)=b+(c/2) (b/2)+((2b+1)/4)=b+(1/4) 2b+2b+1=4b+1 No result •bd=b^2 +bc+d b(((2b+1)/2))=b^2 +(b/2)+((2b+1)/2) 2b^2 +b=2b^2 +b+2b+1 b=−(1/2)⇒d=((2b+1)/2)⇒d=0 P(x)= x^2 +(x/2)+b=x^2 +(x/2)−(1/2)=((2x^2 +x−1)/2) Q(x)=x^2 +cx+d=x^2 +(1/2)x=((2x^2 +x)/2) P(Q(x))=0 ((2(((2x^2 +x)/2))^2 +(((2x^2 +x)/2))−1)/2)=0 2(((2x^2 +x)/2))^2 +(((2x^2 +x)/2))−1=0 ((4x^4 +4x^3 +x^2 )/2)+((2x^2 +x)/2)−1=0 4x^4 +4x^3 +3x^2 +x−2=0 4x^4 +4x^3 +3x^2 +3x−2x−2=0 4x^3 (x+1)+3x(x+1)−2(x+1)=0 (x+1)(4x^3 +3x−2)=0 (x+1)(2x−1)(2x^2 +x+2)=0 { ((x+1=0⇒x=−1)),((2x−1=0⇒x=(1/2))),((2x^2 +x+2=0 (no real roots))) :} x=−1,(1/2)
$${P}\left({x}\right)=\:{x}^{\mathrm{2}} +\frac{{x}}{\mathrm{2}}+{b}\:,\:{Q}\left({x}\right)={x}^{\mathrm{2}} +{cx}+{d} \\ $$$${P}\left({x}\right){Q}\left({x}\right)=\:{Q}\left({P}\left({x}\right)\right) \\ $$$${Real}\:{roots}\:{of}\:{P}\left({Q}\left({x}\right)\right)=\mathrm{0}\:? \\ $$$$ \\ $$$${P}\left({x}\right){Q}\left({x}\right)=\:{Q}\left({P}\left({x}\right)\right) \\ $$$${lhs}: \\ $$$$\:\:\:\:{P}\left({x}\right){Q}\left({x}\right)=\left({x}^{\mathrm{2}} +\frac{{x}}{\mathrm{2}}+{b}\right)\left({x}^{\mathrm{2}} +{cx}+{d}\right) \\ $$$$={x}^{\mathrm{4}} +{cx}^{\mathrm{3}} +{dx}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{3}} +\frac{{c}}{\mathrm{2}}{x}^{\mathrm{2}} +\frac{{d}}{\mathrm{2}}{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{bx}^{\mathrm{2}} +{bcx}+{bd} \\ $$$$={x}^{\mathrm{4}} +\left({c}+\frac{\mathrm{1}}{\mathrm{2}}\right){x}^{\mathrm{3}} +\left({b}+{d}+\frac{{c}}{\mathrm{2}}\right){x}^{\mathrm{2}} +\left({bc}+\frac{{d}}{\mathrm{2}}\right){x}+{bd} \\ $$$$ \\ $$$${rhs}:\:{Q}\left({P}\left({x}\right)\right) \\ $$$${Q}\left({P}\left({x}\right)\right)=\left({x}^{\mathrm{2}} +\frac{{x}}{\mathrm{2}}+{b}\right)^{\mathrm{2}} +{c}\left({x}^{\mathrm{2}} +\frac{{x}}{\mathrm{2}}+{b}\right)+{d} \\ $$$$={x}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{2}} +{b}^{\mathrm{2}} +{x}^{\mathrm{3}} +{bx}+\mathrm{2}{bx}^{\mathrm{2}} +{cx}^{\mathrm{2}} +\frac{{c}}{\mathrm{2}}{x}+{bc}+{d} \\ $$$$={x}^{\mathrm{4}} +{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{2}} +\mathrm{2}{bx}^{\mathrm{2}} +{cx}^{\mathrm{2}} +{bx}+\frac{{c}}{\mathrm{2}}{x}+{b}^{\mathrm{2}} +{bc}+{d} \\ $$$$={x}^{\mathrm{4}} +{x}^{\mathrm{3}} +\left(\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{2}{b}+{c}\right){x}^{\mathrm{2}} +\left({b}+\frac{{c}}{\mathrm{2}}\right){x}+{b}^{\mathrm{2}} +{bc}+{d} \\ $$$$ \\ $$$$\bullet\:{c}+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{1}\:\:\:\:\: \\ $$$$\:\:\:{c}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\bullet{b}+{d}+\frac{{c}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{2}{b}+{c}\:\:\: \\ $$$$\:\:{b}+{d}+\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{2}{b}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:{d}=\frac{\mathrm{2}{b}+\mathrm{1}}{\mathrm{2}} \\ $$$$\bullet{bc}+\frac{{d}}{\mathrm{2}}={b}+\frac{{c}}{\mathrm{2}}\:\:\:\:\: \\ $$$$\:\:\frac{{b}}{\mathrm{2}}+\frac{\mathrm{2}{b}+\mathrm{1}}{\mathrm{4}}={b}+\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\:\:\mathrm{2}{b}+\mathrm{2}{b}+\mathrm{1}=\mathrm{4}{b}+\mathrm{1}\:{No}\:{result} \\ $$$$\bullet{bd}={b}^{\mathrm{2}} +{bc}+{d} \\ $$$$\:\:{b}\left(\frac{\mathrm{2}{b}+\mathrm{1}}{\mathrm{2}}\right)={b}^{\mathrm{2}} +\frac{{b}}{\mathrm{2}}+\frac{\mathrm{2}{b}+\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{2}{b}^{\mathrm{2}} +{b}=\mathrm{2}{b}^{\mathrm{2}} +{b}+\mathrm{2}{b}+\mathrm{1} \\ $$$${b}=−\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{d}=\frac{\mathrm{2}{b}+\mathrm{1}}{\mathrm{2}}\Rightarrow{d}=\mathrm{0} \\ $$$${P}\left({x}\right)=\:{x}^{\mathrm{2}} +\frac{{x}}{\mathrm{2}}+{b}={x}^{\mathrm{2}} +\frac{{x}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{2}{x}^{\mathrm{2}} +{x}−\mathrm{1}}{\mathrm{2}} \\ $$$$\:{Q}\left({x}\right)={x}^{\mathrm{2}} +{cx}+{d}={x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{x}=\frac{\mathrm{2}{x}^{\mathrm{2}} +{x}}{\mathrm{2}} \\ $$$$ \\ $$$${P}\left({Q}\left({x}\right)\right)=\mathrm{0} \\ $$$$\frac{\mathrm{2}\left(\frac{\mathrm{2}{x}^{\mathrm{2}} +{x}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}{x}^{\mathrm{2}} +{x}}{\mathrm{2}}\right)−\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\mathrm{2}\left(\frac{\mathrm{2}{x}^{\mathrm{2}} +{x}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}{x}^{\mathrm{2}} +{x}}{\mathrm{2}}\right)−\mathrm{1}=\mathrm{0} \\ $$$$\frac{\mathrm{4}{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{3}} +{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{2}{x}^{\mathrm{2}} +{x}}{\mathrm{2}}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{4}{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} +{x}−\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{4}{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{2}{x}−\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{4}{x}^{\mathrm{3}} \left({x}+\mathrm{1}\right)+\mathrm{3}{x}\left({x}+\mathrm{1}\right)−\mathrm{2}\left({x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({x}+\mathrm{1}\right)\left(\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\left({x}+\mathrm{1}\right)\left(\mathrm{2}{x}−\mathrm{1}\right)\left(\mathrm{2}{x}^{\mathrm{2}} +{x}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\begin{cases}{{x}+\mathrm{1}=\mathrm{0}\Rightarrow{x}=−\mathrm{1}}\\{\mathrm{2}{x}−\mathrm{1}=\mathrm{0}\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{2}{x}^{\mathrm{2}} +{x}+\mathrm{2}=\mathrm{0}\:\left({no}\:{real}\:{roots}\right)}\end{cases} \\ $$$${x}=−\mathrm{1},\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by dimentri last updated on 21/Jul/23
$$\:\:\:\:\underbrace{\:} \\ $$

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