What is the square root of 16
\(\sqrt{16} = \sqrt{4\times 4} = \sqrt{4^2} =4^{\frac {2}{2}} =4^1=4\)
\(The\ square\ root\ of\ 16:\\\\\sqrt{16}=4\ because\ 4^2=4\cdot4=16\\--------------------\\de finition\ of\ the\ square\ root:\ \sqrt{a}=b\ if\ b^2=a\ where\ a\geq0\ and\ b\geq0\)
