F(x)= 2x-3, G(x)= 9-x², solve gf(x)=16

Question

Is this equation possible?

Answer 1

If you mean g(f(x))=16 then:

g(f(x))=9-(2x-3)²=9-4x²+12x-9=16, or we can write:

g(f(x))=(3-2x+3)(3+2x-3)=2x(6-2x)=12x-4x²=16. Either way:

4x²-12x+16=0=4(x²-3x+4). This has complex roots only:

x=(3±√(9-16))/2=(3±i√7)/2. So there are no real roots.

If you mean g×f(x)=(9-x²)(2x-3) then -2x³+3x²+18x-27=16,

2x³-3x²-18x+43=0, which has 2 complex roots and one real root. So this can be solved for a real solution using various methods.

In each case we get an equation and there are solutions, but not necessarily real solutions.