Problem:
Let P(x) be a quadratic polynomial with complex coefficients whose x2 coefficient is 1. Suppose the equation P(P(x))=0 has four distinct solutions, x=3,4,a,b. Find the sum of all possible values of (a+b)2.
We are given that P(P(3))=P(P(4))=0, or that P(3) and P(4) are roots to the polynomial P(x).
We know that P(x) is a quadratic, and in the case that P(3)ξ =P(4), notice we'd have found the two unique roots to P(x).
However, we also have the case wehere P(3)=P(4), in which case we don't exactly know the roots.
Notice that if P(3)=P(4), that this polynomial is symmetric around x=27β, as 3=27ββ21β and 4=27β+21β.
Thus, this quadratic P in this case could be modelled by the polynomial
P(x)=P(7βx)βP(x)=(xβ27β)2+c
and plugging this in for c given that P(3) is a root gives us that
P(P(3))=(c+41ββ27β)2+c=0
which yields that
c=411βΒ±i43ββ
However, all we needed was possible values of (a+b)2 which would have to sum to 7 because this polynomial is symmetric around 27β so the sum of the possible (a+b)2 is 72=49 for this case.
Now for the case that P(3)ξ =P(4), we notice that we have the two distinct roots of this polynomial so we can write
P(x)=(xβP(3))(xβP(4))
Notice again that the values a,b are roots to the equation P(P(x))=0 and thus the two roots are when P(x)=P(3)βP(x)βP(3)=0 or P(x)=P(4)βP(x)βP(4)=0.
NOW, notice that P(3) is a constant, and so that the sum of roots to the quadratic P(x)βP(3) is the same as the sum of roots of P(x) and the same applies to P(x)βP(4), so the sum of roots of each of these polynomials is P(3)+P(4) as those were the original two roots of the polynomial P(x) in this case.
From here, the sum of roots would be the sum of roots of each of these terms minus 3+4=7, as otherwise we'd overcount these roots and calculate a+b+3+4 instead of a+b
Thus the desired sum that we want is
2(P(3)+P(4))β7
which can be calculated as long as we can find P(3)+P(4)
Now, we can solve this pretty directly.
Notice that P(x)=(xβP(3))(xβP(4)) so substituting P(3) yields
P(3)=(3βP(3))(3βP(4))
and very similarly
P(4)=(4βP(3))(4βP(4))
Solving this system of equations yields P(3)=β3 and P(4) = \frac{7}
Our desired expression again is
2(P(3)+P(4))β7=2(21β)β7=β6
and adding up its square gives us the answer
72+(β6)2=85β
Solution:
Note that because P(P(3))=P(P(4))=0,P(3) and P(4) are roots of P(x). There are two cases.
CASE 1:P(3)=P(4). Then P(x) is symmetric about x=27β; that is to say, P(r)=P(7βr) for all r.
Thus the remaining two roots must sum to 7. Indeed, the polynomials P(x)=(xβ27β)2+411βΒ±i3β satisfy the conditions.
CASE 2:P(3)ξ =P(4). Then P(3) and P(4) are the two distinct roots of P(x), so
P(x)=(xβP(3))(xβP(4))
for all x. Note that any solution to P(P(x))=0 must satisfy either P(x)=P(3) or P(x)=P(4). Because P(x) is quadratic, the polynomials P(x)βP(3) and P(x)βP(4) each have the same sum of roots as the polynomial P(x), which is P(3)+P(4). Thus the answer in this case is 2(P(3)+P(4))β7, and so it suffices to compute the value of P(3)+P(4).
Let P(3)=u and P(4)=v. Substituting x=3 and x=4 into the above quadratic polynomial yields the system of equations
Subtracting the first equation from the second gives vβu=7βuβv, yielding v=27β. Substituting this value into the second equation gives
27β=(4βu)(4β27β)
yielding u=β3. The sum of the two solutions is 2(27ββ3)β7=β6. In this case, P(x)=(x+3)(xβ27β). The requested sum of squares is 72+(β6)2=85β.