1My solution -
1 . Observe that " P ( 0 ) " is also a root of " P " .
2 . If we let the other root be " c " , then we must have -
c P ( 0 ) = P ( 0 )
Implying that , " c = 0 , 1 " .
3 . Done .
P(x) is a quadratic such that
(1) its leading coefficient is 1
(2) P(x) and P(P(P(x))) share a root.
Then prove that P(0)*P(1)=0
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3 Answers
1
21My own solution:
Given P(x) and P(P(P(x))) share a root.
So must there be some t such that P(t)=0 and P(P(P(t)))=0 implying we must have P(P(0))=0 that is P(0) is a root of P(x)
Let P(x)= x^{2}+bx+P(0) we have
P(P(0))= P(0)^{2}+b(P(0))+P(0)=0 (*)
Case 1) P(0)=0, its obvious that P(0)P(1)=0 and we are done.
Case 2)P(0)\neq 0
From (*) we get 1+ b+ P(0)= 0= P(1)
Hence P(0)P(1)=0 and we have proved.
