Let p(x) be a degree 2 polynomial such that p(1) = 1, p(2) =3, amd p(3) = 2. The p(p(x)) =x has four real solutions. Find the only such solution that is not an integer?

1 Answer
Jun 9, 2017

8/383

Explanation:

Considering the polynomial equation

q(x)=p(p(x))-x=0q(x)=p(p(x))x=0

we have

p(p(1))-1=1-1=0p(p(1))1=11=0
p(p(2))-2=2-2=0p(p(2))2=22=0
p(p(3))-3=3-3=0p(p(3))3=33=0

so 1,2,31,2,3 are roots of q(x)=p(p(x))-xq(x)=p(p(x))x

Calling p(x) = a x^2+bx+cp(x)=ax2+bx+c we have

{(a + b + c = 1),(9 a + 3 b + c = 2),(4 a + 2 b + c = 3):}

so

a=-3/2,b=13/2,c=-4 or

p(x)=-3/2x^2+13/2x-4

and p(0) =-4

then

q(0)=p(p(0))-0=p(-4)=-54

but

q(x)=alpha (x-1)(x-2)(x-3)(x-r) then

q(0)=alpha 6 r = -54 and analogously

q(-1)=p(p(-1))+1=p(-12)+1=-297 =24alpha(r+1)

Solving now

{(6 alpha r = -54), (24 alpha (1 + r) = -297):}

we get

alpha = -27/8 and r=8/3 which is the non integer root