Given 2 positive reals a and b. There exists 2 polynomials F(x)=x^{2}+ax+b and G(x)=x^{2}+bx+a such that all roots of polynomials F(G(x)) and G(F(x)) are real. Show that a and b are greater than 6.
suppose a=b
then F(x)=G(x)=x^2+ax+a
the roots of F are
x=(a(+/)sqrt(a^24a))/2
so we need F(x) to equal these values
x^2+ax+a=(a(+/)sqrt(a^24a))/2
or
x^2+ax+3a/2 (+/)sqrt(a^2/4a)=0
These two quadratics must each have positive discriminant (the on with a "+" is guaranteed to be, so we just need the one with a "")
a^2+6a4sqrt(a^2/4a) > 0
which simplifies to the cubic
4a^312a^2+32a+16>0
the relevant root is approximately 7.35026
I'm not sure if this is a deadend. To complete a proof by this method, you'd need to show that increasing a wouldn't allow b to decrease below 6.
Edited on November 22, 2020, 12:16 pm

Posted by Jer
on 20201122 11:36:02 