# Can you find the square root of #-1/2# ?

##### 2 Answers

You can't technically, but it's

#### Explanation:

You can't square root a negative. Because we come up with a problem here it is. Remeber a when you **square root** a number you get a number when **multiplied** by **itself** gives us the number that we just **square rooted**

E. g.

**Quick Example**

When we try to square root a negative we realise this isn't possible. Because when you square root a number the root has to be the **same** number. So to complete the problem below we'll have to split the

E. g.

Now we know that 16 is a **square number**, so when its square root is... 4 of course! However, we **cannot** square the negative one so we rewrite it as **imaginary number** as there is no square root of any negative.

**Now onto the question**

Let's split it into

This gives us

#### Explanation:

Assuming you are asking about the square root of

In common with all non-zero numbers,

The *imaginary unit*

Hence we find:

#(sqrt(2)/2i)^2 = (sqrt(2)/2)^2 * i^2 = 1/2 * -1 = -1/2#

So one square root of

We also have:

#(-sqrt(2)/2i)^2 = (-sqrt(2)/2)^2 * i^2 = 1/2 * -1 = -1/2#

So the other square root is

By convention, when we write *principal* square root of

**Footnote**

Why not simply write:

#sqrt(-1/2) = sqrt(-1 * 1/2) = sqrt(-1) * sqrt(1/2) = isqrt(1/2)#

I tend to avoid being too hasty with the "rule"

For example:

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

On the other hand, the convention that if

#sqrt(x) = isqrt(-x)#

is safe.