# What is the general rule to simplify a radical expression in a square root?

## For example, $\sqrt{2 \sqrt{3} - 4}$

Jul 23, 2018

$\sqrt{2 \sqrt{3} - 4} = \left(\sqrt{3} - 1\right) i$

#### Explanation:

I explored this in https://socratic.org/s/aSS7FqaZ finding:

If $p , q , r > 0$ and ${p}^{2} - {q}^{2} r$ is a perfect square ${s}^{2}$ then:

$\sqrt{p + q \sqrt{r}} = \frac{\sqrt{2 p + 2 s}}{2} + \frac{\sqrt{2 p - 2 s}}{2}$

If these conditions break down then we might expect something like:

$\sqrt{p + q \sqrt{r}} = \frac{\sqrt{2 p + 2 s}}{2} - \frac{\sqrt{2 p - 2 s}}{2}$

or:

$\sqrt{p + q \sqrt{r}} = - \frac{\sqrt{2 p + 2 s}}{2} + \frac{\sqrt{2 p - 2 s}}{2}$

In particular, with $p = - 4$, $q = 2$ and $r = 3$ we have:

$s = \sqrt{{p}^{2} - {q}^{2} r} = \sqrt{16 - 12} = 2$

$\frac{\sqrt{2 p + 2 s}}{2} = \frac{\sqrt{- 8 + 4}}{2} = \frac{\sqrt{- 4}}{2} = i$

$\frac{\sqrt{2 p - 2 s}}{2} = \frac{\sqrt{- 8 - 4}}{2} = \frac{\sqrt{- 12}}{2} = \sqrt{3} i$

In your example the radicand is negative, but we can simplify by splitting the radicand into a perfect square:

$\sqrt{2 \sqrt{3} - 4} = \sqrt{- \left(3 - 2 \sqrt{3} + 1\right)}$

$\textcolor{w h i t e}{\sqrt{2 \sqrt{3} - 4}} = \sqrt{- {\left(\sqrt{3} - 1\right)}^{2}}$

$\textcolor{w h i t e}{\sqrt{2 \sqrt{3} - 4}} = \left(\sqrt{3} - 1\right) i$