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 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

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 Further thoughts | Comment 2 of 3 |
I can improve a bit on my earlier result with two ideas:

#1
The larger root of G(x)=0 is given by R = ( sqrt( b^2 - 4a) - b )/2 . Since b >= 4, this is negative

The minimal value of F(x) is given by M = F(-a/2) = b - a^2/4.

In order for there to be a solution to G(F(x)) = 0 we must have: M <= R, or

4b - a^2 <= 2 sqrt( b^2 - 4a ) - 2b that is:

6b <= a^2 + 2 sqrt( b^2 - 4a ) and symmetrically:
6a <= b^2 + 2 sqrt( a^2 - 4b)

#2
We like to show that for any given value of b, there is a solution to the problem if and only if a is in the range [ a_1, a_2].
We are part way there when we showed that
4b <= a^2 and 4a <= b^2.

Next, we'd like to show that a_1(b) is an increasing function of b.  Once this is demonstrated, we can solve the problem by showing that given a=b then 6 =< a_1.

Edited on November 21, 2020, 4:38 pm
 Posted by FrankM on 2020-11-21 16:37:10

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