A quadratic trinomial `P(x)=a x^2+b x+c` is such that the equation `P(x)=x` has o real roots. Prove that in this case equation `P(P(x))=x` has no real roots either.

Explanation

Since the equation `ax^(2) =bx + c = x ` has no real roots, the expression `P(x) - x = ax^(2) + (b - 1) x+ c` assumes values of one sing `AA x in` R, say `P(x) = x gt` 0 . Then
`P(P(x_(0))) - P (x_(0)) gt 0`
for any `x = x_(0), i.e., P (x_(0)) gt x_(0)` and hence `P(P(x_(0))) gt x_(0).` Therefore,
`x_(0)` cannot be a root of the fourth degree equation `P(P(x)) = x` .