 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Go Discriminant! (Posted on 2020-11-21) Given 2 positive reals a and b. There exists 2 polynomials F(x)=x2+ax+b and G(x)=x2+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.

 No Solution Yet Submitted by Danish Ahmed Khan No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Simplified problem solution Comment 3 of 3 | suppose a=b

then F(x)=G(x)=x^2+ax+a
the roots of F are
x=(-a(+/-)sqrt(a^2-4a))/2
so we need F(x) to equal these values
x^2+ax+a=(-a(+/-)sqrt(a^2-4a))/2
or
x^2+ax+3a/2 (+/-)sqrt(a^2/4-a)=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+6a-4sqrt(a^2/4-a) > 0
which simplifies to the cubic
4a^3-12a^2+32a+16>0
the relevant root is approximately 7.35026

I'm not sure if this is a dead-end.  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 2020-11-22 11:36:02 Please log in:

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