Explained: "Given polynomial P(x)=x2+ax+b. Knowing that there only exists..."

Asher Olsen

Answered question

2022-03-23

Given polynomial

P (x) = x^{2} + a x + b

. Knowing that there only exists one value of real c such that

P^{2} (c) = c

, calculate the minimum value of

a + b + c

.
Notation:

P^{2} (x) = (P \circ P) (x)

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 + {(x - c)}^{2}

.
Step 2
Therefore

a = 1 - 2 c, b = c^{2}

and

a + b + c = c^{2} - c + 1 = {(c - \frac{1}{2})}^{2} + \frac{3}{4} \geq \frac{3}{4}

. Equality holds for

c = \frac{1}{2}, a = 0

, and

b = \frac{1}{4}

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