Question

Find the square root of the following numbers by factorisation:

$\left(i\right)196;\left(ii\right)256;\left(iii\right)10404;\left(iv\right)1156;\left(v\right)13225.$

Answer 1

(i) $196=2\times 2\times 7\times 7$

$\begin{array}{cc}2& 196\\ 2& 98\\ 7& 49\\ 7& 7\\ & 1\end{array}$

$\therefore \sqrt{196}=\sqrt{\underset{–}{2\times 2}\times \underset{–}{7\times 7}}=2\times 7=14$

(ii) $256=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2$

$\begin{array}{cc}2& 256\\ 2& 128\\ 2& 64\\ 2& 32\\ 2& 16\\ 2& 8\\ 2& 4\\ 2& 2\\ & 1\end{array}$

$\therefore$ $\sqrt{256}=\sqrt{\underset{–}{2\times 2}\times \underset{–}{2\times 2}\times \underset{–}{2\times 2}\times \underset{–}{2\times 2}}=2\times 2\times 2\times 2=16$

(iii) $10404=2\times 2\times 3\times 3\times 17\times 17$

$\begin{array}{cc}2& 10404\\ 2& 5202\\ 3& 2601\\ 3& 867\\ 17& 289\\ 17& 17\\ & 1\end{array}$

$\therefore \sqrt{10404}=\sqrt{\underset{–}{2\times 2}\times \underset{–}{3\times 3}\times \underset{–}{17\times 17}}=2\times 3\times 17=102$

(iv) $1156=2\times 2\times 17\times 17$

$\begin{array}{cc}2& 1156\\ 2& 578\\ 17& 289\\ 17& 17\\ & 1\end{array}$

$\therefore \sqrt{1156}=\sqrt{\underset{–}{2\times 2}\times \underset{–}{17\times 17}}=2\times 17=34$

(v) $13225=5\times 5\times 23\times 23$

$\begin{array}{cc}2& 13225\\ 5& 2645\\ 23& 529\\ 23& 23\\ & 1\end{array}$

$\therefore \sqrt{13225}=\sqrt{\underset{–}{5\times 5}\times \underset{–}{23\times 23}}=5\times 23=115$