Answer :
[tex]\hbox{From formula} \ \ \ \sqrt{xy} = \sqrt{x} \cdot \sqrt{y} \ \hbox{you've got:} \\ \sqrt{-\frac{16}{169}}=\sqrt{-1 \cdot \frac{16}{169}}= \sqrt{-1} \cdot \sqrt{\frac{16}{169}} \\ \sqrt{-1} \ \hbox{is an imaginary unit so} \ \ \sqrt{-1} = i \\ \hbox{And from the formula:} \\ \sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}} \ \hbox{you'll have:} \\ \frac{\sqrt{16}}{\sqrt{169}}i \\ \hbox{See that you can reach:} \\ \sqrt{16}=4 \qquad \hbox{because} \qquad 4^2=16 \\ [/tex]
[tex]\sqrt{16}=-4 \qquad \hbox{because} \qquad (-4)^2=16 \\ \hbox{The same to 169. Here, look that} \ \ 13^2=169 \\ \hbox{So you've got positive and negative possibility, so two roots.} \\ \hbox{First is principal root:} \\ \frac{4}{13}i \\ \hbox{And second root:} \\ -\frac{4}{13}i[/tex]
[tex]\sqrt{16}=-4 \qquad \hbox{because} \qquad (-4)^2=16 \\ \hbox{The same to 169. Here, look that} \ \ 13^2=169 \\ \hbox{So you've got positive and negative possibility, so two roots.} \\ \hbox{First is principal root:} \\ \frac{4}{13}i \\ \hbox{And second root:} \\ -\frac{4}{13}i[/tex]
