<s>[INST] Find sum of all possible value of $(a+b)^2$ 

Let $P(x)$ be a quadratic with Complex coefficient whose coefficient of $x^2$ is $1$. Suppose the equation $P(P(x))=0$ has four distinct solution, $x=3,4,a,b$. Find sum of all possible value of $(a+b)^2$.


Solution: Either $P(3) = P(4)$ or not. We first see that if $P(3) = P(4)$ it's easy to obtain by Vieta's that $(a+b)^2 = 49$. Now, take $P(3) \neq P(4)$ and WLOG $P(3) = P(a), P(4) = P(b)$. Now, consider the parabola formed by the graph of $P$. It has vertex $\frac{3+a}{2}$. Now, say that $P(x) = x^2 - (3+a)x + c$. We note $P(3)P(4) = c = P(3)\left(4 - 4a + \frac{8a - 1}{2}\right) \implies a = \frac{7P(3) + 1}{8}$. Now, we note $P(4) = \frac{7}{2}$ by plugging in again. Now, it's easy to find that $a = -2.5, b = -3.5$, yielding a value of $36$. Finally, we add $49 + 36 = \boxed{085}$. ~awang11, charmander3333
Remark: We know that $c=\frac{8a-1}{2}$ from $P(3)+P(4)=3+a$.
My doubt how can we say that $P(3)(4)=c$ and $P(3)+P(4)=3+a$?.
 [/INST] Since $3,4,a,b$ are roots of $P(P(x))$ and $P(3)$ and $P(4)$ are different.
$P(3)$ and $P(4)$ are distinct roots of $P(x)$.
As a result, from Vieta's formula, we know that  we have $c=P(3)P(4)$ and $P(3)+P(4)=3+a$.
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