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
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
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.