# How do you simplify 5i sqrt(-54) ?

Sep 27, 2017

$- 15 \sqrt{6}$

#### Explanation:

$5 i \times \sqrt{- 1} \times \sqrt{54}$

$5 i \times i \times \sqrt{54}$

$5 {i}^{2} \times \sqrt{54}$

$5 \left(- 1\right) \times \sqrt{54}$

$- 5 \times \sqrt{54}$

$- 5 \times \sqrt{9} \sqrt{6}$

$- 5 \times 3 \sqrt{6}$

$- 15 \sqrt{6}$

Sep 27, 2017

$- 15 \sqrt{6}$

#### Explanation:

$5 i \sqrt{- 54} \implies 5 i \sqrt{6 \times 9 \times - 1}$

We can take out the root of 9:

$5 i \times 3 \sqrt{6 \times - 1} \implies 15 i \sqrt{6 \times - 1} \implies 15 i \sqrt{6} \sqrt{- 1}$

$\sqrt{- 1} = i$

So we have:

$15 i \times i \sqrt{6}$

$i \times i \implies {i}^{2} = - 1$

So this gives:

$- 15 \sqrt{6}$

Sep 27, 2017

$5 i \sqrt{- 54} = - 15 \sqrt{6}$

#### Explanation:

Because you should know that it is easy to make errors when it comes to square roots of negative (and complex) numbers.

The problem is that every non-zero number has two square roots, and the choice between them is a little arbitrary.

To see that there is a potential problem, consider the common "rule":

$\sqrt{a b} = \sqrt{a} \sqrt{b}$

then note that:

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

Ouch! The "rule" breaks if $a < 0$ and $b < 0$.

Let's tread a little more carefully...

We use the symbol sqrt to denote the principal square root.

If $n > 0$ then its principal square root is the positive one.

If $n < 0$ then by convention, its principal square root is:

$\sqrt{n} = i \sqrt{- n}$

where $i$ is the imaginary unit, satisfying ${i}^{2} = - 1$

With these conventions, we can safely state:

$\sqrt{a b} = \sqrt{a} \sqrt{b} \text{ if "a >= 0" or } b \ge 0$

Then we find:

$5 i \sqrt{- 54} = 5 {i}^{2} \sqrt{54} = - 5 \sqrt{{3}^{2} \cdot 6} = - 5 \sqrt{{3}^{2}} \sqrt{6} = - 15 \sqrt{6}$