Functions will function

Relevant wiki: Vieta's Formula - Forming Quadratics

Given that $f(2x)=x^2+x-2$ , therefore

$\begin{aligned} f\left(\frac x2\right)&= 4 \\ f\left(\ 2\left(\frac x4\right)\right) &=4 \\ \frac {x^2}{16}+\frac x4 -2&=4 \\ \frac {x^2}{16}+\frac x4 -6&=0 \\ \implies x^2+4x -96&=0 \end{aligned}$

By Vieta's formula, we have $a=-4$ and $b=-96$ , hence $a+b =\boxed{-100}$ .

In the equation, f(2x)=x^2+X-2. I can do it reversely as: f(x)=x^2/4+x/2-2, with this I can substitute x/2 from the function f(x/2)=4. In doing so,[( x/2)^2]/4+[x/2]-2=4 is the result then I can get the equation, x^2/16+x/4-6=0 In this equation, the sum of the roots is -4 And the product is -96 therefore, the sum of a+b=-100 $LaTeX$