Problem:
A quadratic polynomial p(x) with real coefficients and leading coefficient 1 is called disrespectful if the equation p(p(x))=0 is satisfied by exactly three real numbers. Among all the disrespectful quadratic polynomials, there is a unique such polynomial p~β(x) for which the sum of the roots is maximized. What is p~β(1) ?
Answer Choices:
A. 165β
B. 21β
C. 85β
D. 1
E. $$\frac{9}{8}$$
Solution:
Suppose p(x)=(xβr)(xβs). Observe that p(x) must have (two) real roots in order for p(p(x)) to have any roots at all. More specifically, if y is a root of p(p(x)), then p(y)=r or p(y)=s. That is, the equations
(xβr)(xβs)βr=0 and (xβr)(xβs)βs=0
together must have exactly three real roots among them. It follows that one of these two quadratics, say (xβr)(xβs)βr, must have discriminant zero.\
Expansion yields x2β(r+s)x+r(sβ1)=0, so the discriminant Ξ of this quadratic must satisfy
0=Ξ=(r+s)2β4r(sβ1)=(rβs)2+4r.
This implies that r is negative, say r=βr0β, and that s=rΒ±β4rβ=βr0βΒ±2r0ββ. It follows that
where the second inequality follows from the fact that aβa2β€41β for all real numbers a. Thus r=β41β and s=43β, which works. In turn, p~β(x)=(x+41β)(xβ43β) and p~β(1)=45ββ 41β=(A)165ββ.