As a somewhat inferior alternative to Peter's algebraic approach,
you could work out a square root in long-division style as shown here:

\begin{array}{r}
6 \; \phantom{0}6 \;\phantom{0}6 \;\phantom{0}6 \\[-3pt]
\sqrt{44 \; 44 \; 44 \; 44} \\[-3pt]
\underline{36}\; \phantom{66 \; 00 \; 00} \\[-3pt]
12\underline6|\phantom{0}8 \; 44 \phantom{\; 00 \; 00}\\[-3pt]
\underline{7 \; 56} \phantom{\; 00 \; 00}\\[-3pt]
132\underline{6} |\phantom{0} \; 88 \; 44 \phantom{\; 00}\\[-3pt]
\underline{79 \; 56} \phantom{\; 00}\\[-3pt]
1332\underline{6} | \phantom{0}8 \; 88 \;44\\[-3pt]
\underline{7 \; 99 \; 56}
\end{array}

From here you may be able to extrapolate some provable patterns that
occur as you append more pairs of $4$s to the digits under the square root sign.