What you need to do is replace (4a)^2
in the second denominator with either \smash{(4a)}^2
or \smash[b]{(4a)}^2
. This yields compact-looking square root terms, and it works with both \tfrac
and \dfrac
.
Observe that if you, alternatively, replaced 16a^2
in the first denominator with 16a^2\mathstrut
, the two square root symbols would also have equal sizes. However, they would be much taller -- excessively and unnecessarily so, IMNSHO -- than with the adjustment suggested above.

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\[
\sqrt{\dfrac{1+2\cdot4a^2+(4a^2)^2}{16a^2}}
=\sqrt{\dfrac{1+2\cdot4a^2+(4a^2)^2}{\smash{(4a)}^2}}
\]
\[
\sqrt{\tfrac{1+2\cdot4a^2+(4a^2)^2}{16a^2}}
=\sqrt{\tfrac{1+2\cdot4a^2+(4a^2)^2}{\smash{(4a)}^2}}
\]
\end{document}
\smash
command in the solution suggested by Mico. The following two examples does not result in the same display:\smash{(4a)}^2
and\smash{(4a)^2}
(Sorry, did not know how to type in the comment box to get better display of codes.)\smash{(4a)}^2
and\smash{(4a)^2}
, you'll notice that the exponent is placed higher relative to the baseline if the scope of\smash
includes the exponent -- not by a huge amount, for sure, but by about 1 or 2 points. This difference results in a slight increase in the overall height of the denominator which, in turn, explains why LaTeX sees fit to employ a taller (and deeper) square root symbol when it processes\smash{(4a)^2}
.