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

May 1, 2016

$x = 2 \pm \sqrt{3} i$

#### 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 ${a}^{2} \ge 0$, so we need a Complex number, in this case $\sqrt{3} i$.

If $a < 0$ then $\sqrt{a} = \sqrt{- a} i$

So given:

${\left(x - 2\right)}^{2} = - 3 = {\left(\sqrt{3} i\right)}^{2}$

We must have:

$x - 2 = \pm \sqrt{3} i$

Note the $\pm$, since if ${a}^{2} = b$ then ${\left(- a\right)}^{2} = b$ too.

Then adding $2$ to both sides we get:

$x = 2 \pm \sqrt{3} i$