Computing $f(0)$ given other information

Question

I encountered this problem today.

$f(x)$ is a quadratic polynomial with real coefficients.

• $f(2) = 0$
• $f(f(x))=0$ has only $1$ real root at $x=5$

Compute $f(0)$.

I attempted to answer this question by writing $f(x)$ in vertex form and using bullet $1$. $$\begin{align*}f(x)&=a(x-h)^2+k\\&=a(x-2)[x-(2h-2)]+k\end{align*}$$

I don't know how to use the other bullet in the solution. I would like to know how to proceed from here. (The answer is $-\frac{32}{9}$)

Answer 1

$f(x) = a(x-2)(x-b)\\ f\circ f(x) = a(a(x-2)(x-b) - 2)(a(x-2)(x-b) - b)$

Since $f\circ f(x)$ is a quartic (degree 4) equation and it has only one real root. $5$ must be a root of multiplicity.

Which means $a(x-2)(x-b) - 2 = a(x-5)^2$ or $a(x-2)(x-b) - b = a(x-5)^2$

Suppose:

$a(x-2)(x-b) - 2 = a(x^2 -10 x + 25)\\ b = 8\\ a(x^2-10x + 16 - \frac {2}{a})\\ a = -\frac {2}{9}$

In which case, the other factor equals

$-\frac 29(x-2)(x-8) - 8\\ -\frac 29(x-5)^2 - 6$ which has no real roots.

However if we solve... $a(x-2)(x-b) - b = a(x^2 -10 x + 25)$

we would find the other factor equals:

$-\frac 89(x-5)^2 + 6$ and there would still be real roots.

$f(x) = -\frac 29 (x-2)(x-8)$ $f(0) = -\frac 29 (16)$