Correct use of smash with math and root signs

Equalizing radicals is something of a black art.

The difference is due to the right hand side having parentheses. We can cope with this by adding \mathstrut in the left hand side denominator. But this makes TeX choose the next size for the radical. Using \smash[b]{...} for the denominator doesn't help.

The problem is that \tfrac imposes \textstyle, which has raised denominators. One could use \cramped from mathtools, but there's a slicker solution:

\documentclass{article}
\usepackage{amsmath}

\begin{document}

\[
\textstyle
\sqrt{\frac{1+2\cdot4a^2+(4a^2)^2}{\mathstrut 16a^2}}
=\sqrt{\frac{1+2\cdot4a^2+(4a^2)^2}{(4a)^2}}
\]

\end{document}

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}

A simple \vphantom will do the trick. And, please, don't use the plain TeX construct $$ ... $$ for unnumbered displayed equations. Use [ ... \] instead.

\documentclass[11pt, a4paper]{article}
\usepackage{amsmath}

\begin{document}

\[ \sqrt{\tfrac{1+2\cdot4a^2+(4a^2)^2}{\vphantom{)}16a^2}} =\sqrt{\tfrac{1+2\cdot4a^2+(4a^2)^2}{(4a)^2}} \]

\end{document}