Let (P(x)=x^2+\dfrac x 2 +b) and (Q(x)=x^2+cx+d) be two polynomials with real coefficients such that (P(x)Q(x)=Q(P(x))) for all real (x). Find all real roots of (P(Q(x))=0).

Sol)

set $Q(x)=x$ , you can take this for a finitely amount of x

then using the second relation , we get either $P(x) =0$ or$ x=1$ , only the former is possible , we shall see why

but taking the former

we get $Q(0)=0$

so $d=0$

expanding both sides of the second given relation i.e $P(x)*Q(x)= P(Q(x))$

we get $d- 1/2 =b => b= -1/2$

and $c= 1/2$

now we use the coefficients in $P(x)=0$  , therefore we get $ x=-1 $and$ 1/2$ , which are the answers

note that if $Q(x) = -1 $, we get imaginary roots but if we take $Q(x) = 1/2$ we take $x= -1$ and $1/2$

now if we take $x=1 $we again get imaginary roots