Let $$a$$ be the sum of all coefficients in the expansion of $$(1 - 2x + 2x^2)^{2023}(3 - 4x^2 + 2x^3)^{2024}$$ and $$b = \lim_{x \to 0} \frac{\int_0^x \frac{\log(1+t)}{t^{2024}+1}dt}{x^2}$$. If the equations $$cx^2 + dx + e = 0$$ and $$2bx^2 + ax + 4 = 0$$ have a common root, where $$c, d, e \in \mathbb{R}$$, then $$d : c : e$$ equals

To find a, observe that it equals the sum of all coefficients in $$(1 - 2x + 2x^2)^{2023}(3 - 4x^2 + 2x^3)^{2024}$$. Since the sum of coefficients of a polynomial $$f(x)$$ is $$f(1)$$, substituting $$x = 1$$ yields $$a = (1 - 2(1) + 2(1)^2)^{2023}\times(3 - 4(1)^2 + 2(1)^3)^{2024} = (1 - 2 + 2)^{2023}\times(3 - 4 + 2)^{2024} = 1^{2023}\times1^{2024} = 1.$$

The value of $$b$$ can be computed from the limit $$b = \lim_{x\to0} \frac{\int_0^x \frac{\log(1+t)}{t^{2024}+1}\,dt}{x^2}.$$ As $$x\to0$$, both numerator and denominator approach zero, so applying L’Hôpital’s rule gives $$b = \lim_{x\to0} \frac{\frac{d}{dx}\bigl(\int_0^x\frac{\log(1+t)}{t^{2024}+1}\,dt\bigr)}{\frac{d}{dx}(x^2)} = \lim_{x\to0}\frac{\frac{\log(1+x)}{x^{2024}+1}}{2x} = \lim_{x\to0}\frac{\log(1+x)}{2x(x^{2024}+1)}.$$ Using the approximations $$\log(1+x)\approx x$$ and $$x^{2024}+1\to1$$ as $$x\to0$$ yields $$b = \frac{1}{2}.$$

Substituting $$a=1$$ and $$b=\frac{1}{2}$$ into the quadratic equation $$2bx^2 + ax + 4 = 0$$ produces $$2\cdot\frac{1}{2}x^2 + 1\cdot x + 4 = x^2 + x + 4 = 0.$$ Its discriminant is $$1 - 16 = -15$$, which is negative, so the roots are non-real complex numbers.

Since the quadratic equation $$cx^2 + dx + e = 0$$ with real coefficients shares a common root with $$x^2 + x + 4 = 0$$ and complex roots occur in conjugate pairs, both roots must coincide. Therefore the two quadratics are proportional, giving $$\frac{c}{1} = \frac{d}{1} = \frac{e}{4} = k.$$ Hence $$c=k$$, $$d=k$$, $$e=4k$$, and the ratio of their coefficients follows as $$d : c : e = k : k : 4k = 1 : 1 : 4.$

The final answer is that the ratio $$d : c : e$$ equals 1 : 1 : 4.