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

##### 3 Answers

#### Explanation:

#### Explanation:

We can take out the root of 9:

So we have:

So this gives:

#### Explanation:

Why another answer?

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(ab) = sqrt(a)sqrt(b)#

then note that:

#1 = sqrt(1) = sqrt(-1 * -1) != sqrt(-1) * sqrt(-1) = -1#

Ouch! The "rule" breaks if

Let's tread a little more carefully...

We use the symbol *principal* square root.

If

If

#sqrt(n) = i sqrt(-n)#

where *imaginary unit*, satisfying

With these conventions, we can safely state:

#sqrt(ab) = sqrt(a)sqrt(b)" if "a >= 0" or "b >= 0#

Then we find:

#5isqrt(-54) = 5i^2sqrt(54) = -5sqrt(3^2*6) = -5sqrt(3^2)sqrt(6) = -15sqrt(6)#