Given polynomial \(\displaystyle{P}{\left({x}\right)}={x}^{{2}}+{a}{x}+{b}\). Knowing that there only

Asher Olsen

Asher Olsen

Answered question

2022-03-23

Given polynomial P(x)=x2+ax+b. Knowing that there only exists one value of real c such that P2(c)=c, calculate the minimum value of a+b+c.
Notation: P2(x)=(PP)(x)

Answer & Explanation

Karsyn Wu

Karsyn Wu

Beginner2022-03-24Added 17 answers

Step 1
Just a less computational argument for the first part of the solution: if P(x)=x had no real solutions, then P(x)>x for all real x and hence P(P(x))>P(x)>x, contradicting the existence of c. Any solution of P(x)=x is also a solution of P(P(x))=x, hence P(x)=x must have exactly one solution, c. That implies P(x)=x+(xc)2.
Step 2
Therefore a=12c,b=c2 and a+b+c=c2c+1=(c12)2+3434. Equality holds for c=12,a=0, and b=14.

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