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

Aug 11, 2016

$\left(1\right)$ : In $\mathbb{R}$, the soln. set is $\phi$.

$\left(2\right)$ : $I n \mathbb{C}$, the soln set is $\left\{2 \pm i \sqrt{3}\right\} .$

#### Explanation:

In $\mathbb{R}$, the soln. set is $\phi$; as, in $\mathbb{R} , {\left(x - 3\right)}^{2} \ge 0$

In $\mathbb{C}$ since, $i = \sqrt{- 1}$, we have, ${\left(x - 2\right)}^{2} = - 3 = {\left(i \sqrt{3}\right)}^{2}$

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