# How can we attempt to simplify expressions of the form sqrt(p+qsqrt(r)) where p, q, r are rational?

May 19, 2017

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

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

#### Explanation:

Consider the quartic:

$\left(x - \sqrt{p + q \sqrt{r}}\right) \left(x + \sqrt{p + q \sqrt{r}}\right) \left(x - \sqrt{p - q \sqrt{r}}\right) \left(x + \sqrt{p - q \sqrt{r}}\right)$

$= \left({x}^{2} - \left(p + q \sqrt{r}\right)\right) \left({x}^{2} - \left(p - q \sqrt{r}\right)\right)$

=((x^2-p)-qsqrt(r))((x^2-p)+qsqrt(r)))

$= {\left({x}^{2} - p\right)}^{2} - {q}^{2} r$

$= {x}^{4} - 2 p {x}^{2} + \left({p}^{2} - {q}^{2} r\right)$

Suppose ${p}^{2} - {q}^{2} r = {s}^{2}$ for some positive rational number $s$.

Since this quartic has no terms of odd degree it can alternatively be factored as:

$= \left({x}^{2} - k x + s\right) \left({x}^{2} + k x + s\right)$

$= {x}^{4} + \left(2 s - {k}^{2}\right) {x}^{2} + {s}^{2}$

Equating coefficients, we have:

$- 2 p = 2 s - {k}^{2}$

and hence:

${k}^{2} = 2 p + 2 s$

$\frac{\pm k \pm \sqrt{{k}^{2} - 4 s}}{2} = \pm \frac{\sqrt{2 p + 2 s}}{2} \pm \frac{\sqrt{2 p - 2 s}}{2}$

where all combinations of signs are allowed.

In particular, if $q > 0$ then:

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

If $q < 0$ then:

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

$\textcolor{w h i t e}{}$
Example

Simplify:

$\sqrt{5 + 2 \sqrt{6}}$

$\left\{\begin{matrix}p = 5 \\ q = 2 \\ r = 6 \\ s = \sqrt{{p}^{2} - {q}^{2} r} = \sqrt{{5}^{2} - {2}^{2} \cdot 6} = 1\end{matrix}\right.$

So:

$\sqrt{5 + 2 \sqrt{6}} = \frac{\sqrt{2 p + 2 s}}{2} + \frac{\sqrt{2 p - 2 s}}{2}$

$\textcolor{w h i t e}{\sqrt{5 + 2 \sqrt{6}}} = \frac{\sqrt{12}}{2} + \frac{\sqrt{8}}{2}$

$\textcolor{w h i t e}{\sqrt{5 + 2 \sqrt{6}}} = \sqrt{3} + \sqrt{2}$