# What is the square root of -26 times the square root of -13?

Oct 8, 2015

$\sqrt{- 26} \cdot \sqrt{- 13} = - 13 \sqrt{2}$

#### Explanation:

If $a , b \ge 0$ then $\sqrt{a} \sqrt{b} = \sqrt{a b}$

If $a < 0$, then $\sqrt{a} = i \sqrt{- a}$, where $i$ is the imaginary unit.

So:

$\sqrt{- 26} \cdot \sqrt{- 13} = i \sqrt{26} \cdot i \sqrt{13}$

$= {i}^{2} \cdot \sqrt{26} \sqrt{13}$

$= - 1 \cdot \sqrt{26 \cdot 13}$

$= - \sqrt{{13}^{2} \cdot 2}$

$= - \sqrt{{13}^{2}} \sqrt{2}$

$= - 13 \sqrt{2}$

Note that you have to be careful with square roots of negative numbers. For example:

$1 = \sqrt{1} = \sqrt{- 1 \cdot - 1} \ne \sqrt{- 1} \cdot \sqrt{- 1} = {i}^{2} = - 1$