If the coefficients of the middle terms in the binomial expansions...

The expansion of $(1 + \alpha x)^{26}$ has $27$ terms. The middle term is $T_{14}$.

The coefficient of the middle term is $^{26}C_{13} \alpha^{13}$.

The expansion of $(1 - \alpha x)^{28}$ has $29$ terms. The middle term is $T_{15}$.

The coefficient of the middle term is $^{28}C_{14} (-\alpha)^{14} = ^{28}C_{14} \alpha^{14}$.

Equating the two coefficients:

$^{26}C_{13} \alpha^{13} = ^{28}C_{14} \alpha^{14}$

Since $\alpha \neq 0$, dividing by $\alpha^{13}$ gives:

$\alpha = \dfrac{^{26}C_{13}}{^{28}C_{14}}$

Expanding the combinations using factorials:

$^{28}C_{14} = \dfrac{28!}{14! 14!} = \dfrac{28 \times 27 \times 26!}{14 \times 13! \times 14 \times 13!} = \dfrac{28 \times 27}{14 \times 14} \times \dfrac{26!}{13! 13!}$

$^{28}C_{14} = \dfrac{27}{7} \times ^{26}C_{13}$

Substituting this back:

$\alpha = \dfrac{^{26}C_{13}}{\dfrac{27}{7} \times ^{26}C_{13}} = \dfrac{7}{27}$

Answer: $\dfrac{7}{27}$