Question

How do you solve `(x-2)^2=-3`?

Answer 1

`x = 2+-sqrt(3)i`

Explanation:

To solve this we want to take the square root of both sides, but the right hand side is negative. If `a` is any Real number then `a^2 >= 0`, so we need a Complex number, in this case `sqrt(3)i`.

If `a < 0` then `sqrt(a) = sqrt(-a)i`

So given:

`(x-2)^2 = -3 = (sqrt(3)i)^2`

We must have:

`x-2 = +-sqrt(3)i`

Note the `+-`, since if `a^2 = b` then `(-a)^2 = b` too.

Then adding `2` to both sides we get:

`x = 2+-sqrt(3)i`