# What is the square root of -50 times the square root of -10?

Sep 9, 2015

$\sqrt{- 50} \cdot \sqrt{- 10} = - 10 \sqrt{5}$

#### Explanation:

This is slightly tricky, since $\sqrt{a} \sqrt{b} = \sqrt{a b}$ is only generally true for $a , b \ge 0$.

If you thought it held for negative numbers too then you would have spurious 'proofs' like:

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

Instead, use the definition of the principal square root of a negative number:

$\sqrt{- n} = i \sqrt{n}$ for $n \ge 0$, where $i$ is 'the' square root of $- 1$.

I feel slightly uncomfortable even as I write that: There are two square roots of $- 1$. If you call one of them $i$ then the other is $- i$. They are not distinguishable as positive or negative. When we introduce Complex numbers, we basically pick one and call it $i$.

Anyway - back to our problem:

$\sqrt{- 50} \cdot \sqrt{- 10} = i \sqrt{50} \cdot i \sqrt{10} = {i}^{2} \cdot \sqrt{50} \sqrt{10}$

$= - 1 \cdot \sqrt{50 \cdot 10} = - \sqrt{{10}^{2} \cdot 5} = - \sqrt{{10}^{2}} \sqrt{5}$

$= - 10 \sqrt{5}$