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?

Question

2526 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-09T09:57:12

#8/3#

Considering the polynomial equation

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

we have

#p(p(1))-1=1-1=0#
#p(p(2))-2=2-2=0#
#p(p(3))-3=3-3=0#

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

Calling #p(x) = a x^2+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