Given that f(x) = 5x^2-3x+7 and f(g(x)) = (5x^4)/9 + (17x^2)/3 + 21, find all possible values for the sum of the coefficients in the quadratic function g(x).

Answer 1

well, the brute force method goes like this:
f(g(x)) = 5g^2-3g+7
If g(x) = ax^2+bx+c, that means
f(g(x)) = 5(ax^2+bx+c)^2 + 3(ax^2+bx+c) + 7
= (5a^2)x^4 + (10ab)x^3 + (10ac+3a+5b^2)x^2 + (10bc+3b)x + (5c^2+3c+7)
For that to be the same as the given f(g(x)) the coefficients must match, so

5a^2 = 5/9
10ab=0
10ac+3a+5b^2 = 17/3
10bc+3b = 0
5c^2+3c+7 = 21
Clearly, a = ±1/3 and b=0
And it is easy to see that
(a,b,c) = (1/3, 0, -2) or (-1/3, 0, 7/5)

I'll have to ponder on whether there is a more direct method.