An algebra problem by Priyanshu Mishra

Since the roots of $f(f(x))$ are 1 and 2. Then $\implies \begin{cases} f(f(1)) = f(\alpha + \beta+1) = 0 \\ f(f(2)) = f(2\alpha + \beta+4) = 0 \end{cases}$ . This means that the roots of $f(x)$ are $\alpha + \beta+1$ and $2\alpha + \beta+4$ and by Vieta's formula :

$\begin{aligned} \alpha + \beta+1 + 2\alpha + \beta+4 & = - \alpha \\ \implies 4\alpha + 2\beta & = -5 & ...(1) \end{aligned}$

$\begin{aligned} (\alpha + \beta+1)({\color{#3D99F6}2\alpha + \beta}+4) & = \beta & \small \color{#3D99F6} \text{Note }(1): \ 4\alpha + 2\beta = -5 \\ (\alpha + \beta+1)\left({\color{#3D99F6}- \frac 52}+4\right) & = \beta \\ 3\alpha + \beta & = - 3 \qquad ...(2) \end{aligned}$

From $3(1)-4(2): \quad 2\beta = - 3 \implies \beta = - \dfrac 32$ . Note that $f(0) = \beta = - \dfrac 32 = \boxed{-1.5}$ .