find the value of $f(0)$, given $f(f(x))=0$ has roots as 1 & 2

Question

find the value of $f(0)$, given $f(f(x))=0$ has roots as 1 & 2, $f(x) = x^2 + \alpha x + \beta$

My attempt:

If $k_1$, $k_2$ are the roots of $f(x)$, $f(x) = k_1$ has roots 1 & 2 $\therefore \alpha = -3, \beta-k_1 = 2 \text{ or } \beta-k_2 = 2$ and $f(x) = x^2 -3x+\beta$. I could not proceed further with this approach

Another line of thought is that $f(f(x))$ would be a quartic equation, which means that for it to have 2 roots, it will either have 2 real, 2 imaginary roots or just 2 roots (x-axis would be tangent to this curve at 1 and 2). This information doesn't seem very helpful in solving the problem.

Answer 1

Note that $f(f(x))= (x^2+ax+b)^2 + a(x^2+ax+b) + b$

Put $x=1,x=2$ and set to 0:

$(1+a+b)^2 + a(1+a+b) + b = 0$ _(i) $(4+2a+b)^2 + a(4+2a+b) + b =0$ _(ii)

Subtracting first equation from second yields

$(a+3)(4a+2b+5)=0$

If $a=-3$, then putting value of $a$ in (i) will give a quadratic with no real solutions in $b$.

So, $4a+2b+5=0$

or $a=\frac{-5-2b}{4}$

Putting this value in (i) will lead to $b=\frac{-3}{2}$

So, $$f(0)=b=-\frac{3}{2}$$

Answer 2

$$f(x) = x^2 + \alpha x + \beta$$

Since $f(f(x))=0$ has root at $x=1$ and $x=2$, it implies $f(f(1))=0$ and $f(f(2))=0$

$$f(f(1))=0 \implies f(\alpha + \beta + 1) = 0$$ $$f(f(2))=0 \implies f(2 \alpha + \beta + 4) = 0$$

$\implies f(x)$ has at least $1$ roots $\implies Δ \geq 0$

Case $1$: $Δ = 0$

• $f(x) \text{ has only 1 root} \implies \alpha + \beta + 1 = 2 \alpha + \beta + 4 \text{ and } \implies \alpha = -3$

• Then $f(\beta - 2) = 0 \implies (\beta - 2)^2 - 3(\beta - 2) + \beta = 0$ which has 2 imaginary roots: $\beta = 3 \pm i$. Therfore:

$$f(0) = \beta = 3 \pm i$$

Case $2$: $Δ > 0$

• $f(x) \text{ has 2 distinct roots which is } \alpha + \beta + 1 \text{ and } 2 \alpha + \beta + 4$

• By Vieta's Formula:

$$(\alpha + \beta + 1) + (2 \alpha + \beta + 4) = -\alpha \text{ (1) }$$ $$(\alpha + \beta + 1) \cdot (2 \alpha + \beta + 4) = \beta \text{ (2) }$$

$\text{(1) } \implies 4\alpha + 2\beta + 5 = 0 \implies 2 \alpha + \beta + 4 = \frac{3}{2} \text{(1')}$

Substitute $(1')$ into $(2)$:

$\text{(2) } \implies \alpha + \beta + 1 = \frac{2}{3} \beta \implies \alpha + \frac{1}{3} \beta + 1 = 0 \text{ (2')}$

Solving $(1')$ and $(2')$ to get:

$$\alpha = -\frac{1}{2}, \beta = -\frac{3}{2}$$

Therefore:

$$f(0) = \beta = -\frac{3}{2}$$