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

$L e t P (x) = x^{2} + \frac{x}{2} + b a n d$ $Q (x) = x^{2} + c x + d b e t w o$ $p o l y n o m i a l w i t h r e a l c o e f f i c i e n t s$ $s u c h t h a t P (x) Q (x) = Q (P (x))$ $f o r a l l r e a l x .$ $F i n d a l l t h e r e a l r o o t s o f$ $P (Q (x)) = 0$
	Answered by Rasheed.Sindhi last updated on 21/Jul/23
	$P(x)= x^2 +(x/2)+b , Q(x)=x^2 +cx+d P(x)Q(x)= Q(P(x)) Real roots of P(Q(x))=0 ? P(x)Q(x)= Q(P(x)) lhs: P(x)Q(x)=(x^2 +(x/2)+b)(x^2 +cx+d) =x^4 +cx^3 +dx^2 +(1/2)x^3 +(c/2)x^2 +(d/2)x +bx^2 +bcx+bd =x^4 +(c+(1/2))x^3 +(b+d+(c/2))x^2 +(bc+(d/2))x+bd rhs: Q(P(x)) Q(P(x))=(x^2 +(x/2)+b)^2 +c(x^2 +(x/2)+b)+d =x^4 +(1/4)x^2 +b^2 +x^3 +bx+2bx^2 +cx^2 +(c/2)x+bc+d =x^4 +x^3 +(1/4)x^2 +2bx^2 +cx^2 +bx+(c/2)x+b^2 +bc+d =x^4 +x^3 +((1/4)+2b+c)x^2 +(b+(c/2))x+b^2 +bc+d • c+(1/2)=1 c=(1/2) •b+d+(c/2)=(1/4)+2b+c b+d+(1/4)=(1/4)+2b+(1/2) d=((2b+1)/2) •bc+(d/2)=b+(c/2) (b/2)+((2b+1)/4)=b+(1/4) 2b+2b+1=4b+1 No result •bd=b^2 +bc+d b(((2b+1)/2))=b^2 +(b/2)+((2b+1)/2) 2b^2 +b=2b^2 +b+2b+1 b=−(1/2)⇒d=((2b+1)/2)⇒d=0 P(x)= x^2 +(x/2)+b=x^2 +(x/2)−(1/2)=((2x^2 +x−1)/2) Q(x)=x^2 +cx+d=x^2 +(1/2)x=((2x^2 +x)/2) P(Q(x))=0 ((2(((2x^2 +x)/2))^2 +(((2x^2 +x)/2))−1)/2)=0 2(((2x^2 +x)/2))^2 +(((2x^2 +x)/2))−1=0 ((4x^4 +4x^3 +x^2 )/2)+((2x^2 +x)/2)−1=0 4x^4 +4x^3 +3x^2 +x−2=0 4x^4 +4x^3 +3x^2 +3x−2x−2=0 4x^3 (x+1)+3x(x+1)−2(x+1)=0 (x+1)(4x^3 +3x−2)=0 (x+1)(2x−1)(2x^2 +x+2)=0 { ((x+1=0⇒x=−1)),((2x−1=0⇒x=(1/2))),((2x^2 +x+2=0 (no real roots))) :} x=−1,(1/2)$
	$P (x) = x^{2} + \frac{x}{2} + b, Q (x) = x^{2} + c x + d$ $P (x) Q (x) = Q (P (x))$ $R e a l r o o t s o f P (Q (x)) = 0 ?$ $P (x) Q (x) = Q (P (x))$ $l h s :$ $P (x) Q (x) = (x^{2} + \frac{x}{2} + b) (x^{2} + c x + d)$ $= x^{4} + {c x}^{3} + {d x}^{2}$ $+ \frac{1}{2} x^{3} + \frac{c}{2} x^{2} + \frac{d}{2} x$ $+ {b x}^{2} + b c x + b d$ $= x^{4} + (c + \frac{1}{2}) x^{3} + (b + d + \frac{c}{2}) x^{2} + (b c + \frac{d}{2}) x + b d$ $r h s : Q (P (x))$ $Q (P (x)) = {(x^{2} + \frac{x}{2} + b)}^{2} + c (x^{2} + \frac{x}{2} + b) + d$ $= x^{4} + \frac{1}{4} x^{2} + b^{2} + x^{3} + b x + 2 {b x}^{2} + {c x}^{2} + \frac{c}{2} x + b c + d$ $= x^{4} + x^{3} + \frac{1}{4} x^{2} + 2 {b x}^{2} + {c x}^{2} + b x + \frac{c}{2} x + b^{2} + b c + d$ $= x^{4} + x^{3} + (\frac{1}{4} + 2 b + c) x^{2} + (b + \frac{c}{2}) x + b^{2} + b c + d$ $∙ c + \frac{1}{2} = 1$ $c = \frac{1}{2}$ $∙ b + d + \frac{c}{2} = \frac{1}{4} + 2 b + c$ $b + d + \frac{1}{4} = \frac{1}{4} + 2 b + \frac{1}{2}$ $d = \frac{2 b + 1}{2}$ $∙ b c + \frac{d}{2} = b + \frac{c}{2}$ $\frac{b}{2} + \frac{2 b + 1}{4} = b + \frac{1}{4}$ $2 b + 2 b + 1 = 4 b + 1 N o r e s u l t$ $∙ b d = b^{2} + b c + d$ $b (\frac{2 b + 1}{2}) = b^{2} + \frac{b}{2} + \frac{2 b + 1}{2}$ $2 b^{2} + b = 2 b^{2} + b + 2 b + 1$ $b = - \frac{1}{2} \Rightarrow d = \frac{2 b + 1}{2} \Rightarrow d = 0$ $P (x) = x^{2} + \frac{x}{2} + b = x^{2} + \frac{x}{2} - \frac{1}{2} = \frac{2 x^{2} + x - 1}{2}$ $Q (x) = x^{2} + c x + d = x^{2} + \frac{1}{2} x = \frac{2 x^{2} + x}{2}$ $P (Q (x)) = 0$ $\frac{2 {(\frac{2 x^{2} + x}{2})}^{2} + (\frac{2 x^{2} + x}{2}) - 1}{2} = 0$ $2 {(\frac{2 x^{2} + x}{2})}^{2} + (\frac{2 x^{2} + x}{2}) - 1 = 0$ $\frac{4 x^{4} + 4 x^{3} + x^{2}}{2} + \frac{2 x^{2} + x}{2} - 1 = 0$ $4 x^{4} + 4 x^{3} + 3 x^{2} + x - 2 = 0$ $4 x^{4} + 4 x^{3} + 3 x^{2} + 3 x - 2 x - 2 = 0$ $4 x^{3} (x + 1) + 3 x (x + 1) - 2 (x + 1) = 0$ $(x + 1) (4 x^{3} + 3 x - 2) = 0$ $(x + 1) (2 x - 1) (2 x^{2} + x + 2) = 0$ ${\begin{cases} x + 1 = 0 \Rightarrow x = - 1 \\ 2 x - 1 = 0 \Rightarrow x = \frac{1}{2} \\ 2 x^{2} + x + 2 = 0 (n o r e a l r o o t s) \end{cases}$ $x = - 1, \frac{1}{2}$
	Commented by dimentri last updated on 21/Jul/23

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